5STEPSTOA5™

500APChemistryQuestionstoKnowbyTestDay

5STEPSTOA5™500APChemistryQuestionstoKnowbyTest

Day

MinaLebitz

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CONTENTS

AbouttheAuthorIntroductionNotefromtheAuthor

Chapter1AtomicTheoryandStructure

Questions1–50

Chapter2ChemicalBonding

Questions51–90

Chapter3StatesofMatter

Questions91–150

Chapter4Solutions

Questions151–180

Chapter5ChemicalReactions

Questions181–230

Chapter6Thermodynamics

Questions231–269

Chapter7Kinetics

Questions270–300

Chapter8Equilibrium

Questions301–330

Chapter9Acid–BaseChemistry

Questions331–361

Chapter10Electrochemistry

Questions362–387

Chapter11NuclearChemistry

Questions388–408

Chapter12Descriptive

Questions409–428

Chapter13LaboratoryProcedure

Questions429–444

Chapter14DataInterpretation

Questions445–500

Answers

ABOUTTHEAUTHOR

MinaLebitzhasaBSinbiologyfromtheStateUniversityofNewYorkatAlbanyandanMSinnutritionalbiochemistryfromRutgersUniversity.Shehasmorethan16yearsofteachingexperienceatboththehighschoolandcollegelevel.Ms.LebitzreceivedtheNewYorkTimes’TeachersWhoMakeaDifferenceawardin2003duringhertenureatBrooklynTechnicalHighSchoolandwastheseniorsciencetutoratoneofthemostprestigioustutoringandtestprepagenciesintheUnitedStates.Currently,sheisdoingresearch,writing,andassistingstudentsinreachingtheiracademicgoals,whilecontinuingtolearneverythingshecanaboutscience.Herwebsiteiswww.idigdarwin.com.

INTRODUCTION

Congratulations!You’vetakenabigsteptowardAPsuccessbypurchasing5Stepstoa5:500APChemistryQuestionstoKnowbyTestDay.WeareheretohelpyoutakethenextstepandearnahighscoreonyourAPExamsoyoucanearncollegecreditsandgetintothecollegeoruniversityofyourchoice.

Thisisbookgivesyou500AP-stylemultiple-choicequestionsthatcoverallthemostessentialcoursematerial.Eachquestionhasadetailedanswerexplanation.ThesequestionswillgiveyouvaluableindependentpracticetosupplementbothyourregulartextbookandthegroundworkyouarealreadycoveringinyourAPclassroom.ThisandtheotherbooksinthisserieswerewrittenbyexpertAPteacherswhoknowyourexaminsideandoutandcanidentifycrucialexaminformationandquestionsthataremostlikelytoappearonthetest.

YoumightbethekindofstudentwhotakesseveralAPcoursesandneedstostudyextraquestionsafewweeksbeforetheexamforafinalreview.Oryoumightbethekindofstudentwhoputsoffpreparinguntilthelastweeksbeforetheexam.Nomatterwhatyourpreparationstyleis,youwillsurelybenefitfromreviewingthese500questionsthatcloselyparallelthecontent,format,anddegreeofdifficultyofthequestionsontheactualAPexam.Thesequestionsandtheiranswerexplanationsaretheideallast-minutestudytoolforthosefinalfewweeksbeforethetest.

Remembertheoldsaying“Practicemakesperfect.”Ifyoupracticewithallthequestionsandanswersinthisbook,wearecertainyouwillbuildtheskillsandconfidenceneededtodogreatontheexam.Goodluck!

—EditorsofMcGraw-HillEducation

NOTEFROMTHEAUTHOR

TheAPChemistryexamhasamultiple-choicesectionduringwhichyouarenotallowedtouseacalculator.Inthissection,youcanroundfairlygenerouslytosolveproblemsfaster.However,donotroundgenerouslyinthefree-responseproblems.Calculatorsareallowedformostofthefree-responsesectionsoyou’reexpectedtobepreciseinyourcalculations(andobeytherulesforsignificantfigures).

Thequestionsinthisbookcoverthematerialforboththefree-responseandmultiple-choicesectionsoftheAPChemistryexam.Somecalculationsarebestdonewithacalculator,butallofthemcanbedonewithoutone.Inthemultiple-choicesectionoftheAPChemistryexam,therewillneverbeaproblemthatrequiresthecorrectanswerfromadifferentquestiontosolve.Tocoverallthematerialrelatedtotheexaminthisbook,topicsfromfree-responsequestionshavebeenadaptedintomultiple-choicestylequestions.Thisrequiredthatsomequestionsoccuringroupsreferringtoacommonexperimentordataset.Thesequestionsmayrelyonanswersfromotherquestions.

5STEPSTOA5™

500APChemistryQuestionstoknowbytestday

CHAPTER1AtomicTheoryandStructure

1.Whichofthefollowingshowsthecorrectnumberofprotons,neutrons,andelectronsinaneutralcadmium-112atom?

Questions2–7refertothefollowingdiagramoftheperiodictable.

2.Reactsviolentlywithwaterat298K

3.Highestfirstionizationenergy4.Highestelectronegativity5.Highestelectronaffinity6.Largestatomicradius7.Mostmetalliccharacter8.Theatomicmassofbromineis79.904.Giventhattheonlytwonaturallyoccurringisotopesare79Brand81Br,theabundanceof79Brisotopeisapproximately:(A)20percent(B)40percent(C)50percent(D)80percent(E)99percent

9.TheatomicmassofSris87.62.Giventhatthereareonlythreenaturallyoccurringisotopesofstrontium,86Sr,87Sr,and88Sr,whichofthefollowingmustbetrue?(A)86Sristhemostabundantisotope.(B)87Sristhemostabundantisotope.(C)88Sristhemostabundantisotope.(D)86Sristheleastabundantisotope.(E)Theisotopes87Srand88Sroccurinapproximatelyequalamounts.

10.Whichofthefollowingpropertiesgenerallydecreasesfromlefttorightacrossaperiod(frompotassiumtobromine)?(A)Electronegativity(B)Electronaffinity(C)Atomicnumber(D)Atomicradius(E)Maximumvalueofoxidationnumber

11.Allofthefollowingstatementsdescribetheelementsofthegroup1alkalimetals(notincludinghydrogen)except:(A)Theirreactivityincreaseswithincreasingperiodnumber.(B)Theyhavelowfirstionizationenergies.(C)Theyreactviolentlywithwatertoformstrongacids.(D)Theyhavestrongmetalliccharacter.(E)Theyareallsilversolidsat1atmand298K.

12.Whichofthefollowingelementswouldbeexpectedtohavechemicalpropertiesmostsimilartothoseofphosphorus?(A)S(B)Se(C)O(D)As(E)Si

13.Whichofthefollowingpairsareisoelectronic(havethesamenumberofelectrons)?(A)Kr−,Br+

(B)F−,Na+

(C)Sc,Ti−

(D)Be2+,Ne(E)Cs,Ba2+

14.WhichofthefollowingionshasthesamenumberofelectronsasI−?(A)Sr2+

(B)Rb+

(C)Cs+

(D)Ba2+

(E)Br−

15.WhichofthefollowingbestexplainswhytheF−ionissmallerthantheO2–ion?(A)F−hasamoremassivenucleusthanO2–.(B)F−hasahigherelectronegativitythanO2–.(C)F−hasagreaternuclearchargethanO2–.(D)F−hasagreaternumberofelectronsthanO2–.(E)F−hasmorenucleonsandelectronsthanO2–.

16.Allofthefollowingaretruestatementsabouttheperiodictableexcept:(A)Thereactivityofthegroup1alkalimetalsincreaseswithincreasingperiod.(B)Thereactivityofthegroup17halogensdecreaseswithincreasing

period.(C)Thegroup1and2metalsreactwithwatertoformbasicsolutions.(D)Thegroup18noblegasescanexistonlyasinert,monatomicgases.(E)Allelementswithanatomicnumberequaltoorgreaterthan84are

radioactive.

17.Whichofthefollowinglistscontainsallthediatomic,elementalgasesatstandardtemperaturesandpressures?(A)H,N,O(B)H,N,O,F,Cl(C)H,N,O,F,Cl,Br,I(D)H,N,O,Cl,Br,I,Hg,Rn(E)H,N,O,Cl,He,Ne,Ar,Kr,Xe,Rn

18.Asatomicnumberincreasesfrom11to17intheperiodictable,whathappenstoatomicradius?(A)Itremainsconstant.(B)Itincreasesonly.(C)Itdecreasesonly.(D)Itincreases,thendecreases.(E)Itdecreases,thenincreases.

19.TheeffectivenuclearchargeexperiencedbyavalenceKrisdifferentthantheeffectivenuclearchargeexperiencedbyavalenceelectronofK.Whichofthefollowingaccuratelyillustratesthisdifference?(A)KisasolidwhileKrisagas.(B)ThevalenceelectronsofKrhavealowerfirstionizationenergythan

K.(C)Theproton-to-electronratioishigherforKrthanforK.(D)KrhasahigherfirstionizationenergythanK.(E)ThevalenceelectronsofKrexperiencelessshieldingbytheinner

electronsthanthevalenceelectronsofK.

20.BasedontheionizationenergiesforelementXlistedinthetableabove,

whichofthefollowingelementsisXmostlikelytobe?(A)Li(B)Be(C)Al(D)Si(E)As

Questions21and22refertothefollowinggraphoffirstionizationenergies.

21.Correctexplanationsforthelargedropsinionizationenergiesbetweenelementsofatomicnumbers2and3,10and11,and18and19occursbecause,comparedtoelements3,11,and19,elements2,10,and18haveI.smalleratomicradii.II.agreaterelectronaffinity.III.agreatereffectivenuclearcharge.(A)Ionly(B)IIonly(C)IIIonly(D)IandIIIonly(E)I,II,andIII

22.Correctexplanationsfortheincreasesanddecreasesinionizationenergiesbetweenelementsbetweenatomicnumbers2and10(and11and18)include:I.Thereisrepulsionofpairedelectronsinthep4configuration.II.Theelectronsinafilledsorbitalaremoreeffectiveatshieldingthe

electronsintheporbitalsofthesamenthaneachother.III.Filledorbitalsandsubshellsaremorestablethanunfilledorbitalsand

subshells.(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)I,II,andIII

23.Whichofthefollowingchemicalspeciesiscorrectlyorderedfromsmallesttolargestradius?(A)P<S<Cl(B)Ne<Ar<Kr(C)F<O<O2–

(D)K<K+<Rb(E)Na+<Mg2+<Na

24.Whichofthefollowingelectronconfigurationsrepresentsanatominanexcitedstate?(A)1s22s22p5

(B)1s22s22p53s2

(C)1s22s22p63s1

(D)1s22s22p63s23p5

(E)1s22s22p63s23p64s1

Questions25–28refertothegroundstateatomsofthefollowingelements:(A)Ga

(B)Tc

(C)C(D)S(E)N

25.Thisatomcontainsexactlyoneunpairedelectron.

26.Thisatomcontainsexactlytwounpairedelectrons.

27.Thisatomcontainsexactlytwoelectronsinthehighestoccupiedenergysublevel.

28.Thiselementisradioactive.

Questions29–35refertothefollowing:

29.Ahighlyreactive,groundstatemetal30.Highestfirstionizationenergy31.Anatomintheexcitedstate32.Anatomthatformsatrigonalplanarmoleculewhensaturatedwithhydrogen33.Hasexactlyfivevalenceelectrons34.ThemostabundantelementinEarth’satmosphere35.AchemicallyunreactiveatomQuestions36–42refertothefollowing:

36.Anatomintheexcitedstate37.Anatomwhoseaqueouscationiscolored38.Achemicallyunreactiveatom39.Anatomthatformsanalkaline

solutionandhydrogengaswhencombinedwithwater40.Anatomwiththehighestsecondionizationenergy41.Anatomthatformscoloredcompounds42.Ahighlyreactivemetal43.WhichofthefollowingisthemostaccurateinterpretationofRutherford’sexperimentinwhichhebombardedgoldfoilwithalphaparticles?(A)Electronsarearrangedinshellsofincreasingenergyaroundthe

nucleusofanatom.(B)Thevolumeofanatomismostlyemptyspacewiththepositive

chargesconcentratedinadensenucleus.(C)Protonsandneutronsaremoremassivethanelectronsbuttakeupless

space.(D)Atomsaremadeofsubatomicparticlesofdifferentchargesand

masses.(E)Discreteemissionsspectrumlinesareproducedbecauseonlycertain

energystatesofelectronsareallowed.

44.Allofthehalogensintheirelementformat25°Cand1atmare:(A)Gases(B)Colorless(C)Odorless(D)Negativelycharged(E)Diatomicmolecules

Questions45–49refertothefollowingchoices:(A)Alkalimetals(B)Noblegases(C)Halogens(D)Transitionelements(E)Actinides

45.Themostlikelytoformanions46.Theirmonovalentcationsformclearsolutions47.Havethehighestionizationenergiesinagivenperiod48.Allareradioactive49.Themostdifficulttooxidizeinagivenperiod

50.Twounknown,solidsubstancesareanalyzedinalab.Theresultsareshownabove.Truestatementsaboutthecompositionofthesetwosubstancesinclude:I.Substance1containsanalkalimetal.II.Substance2containsanalkaliearthmetal.III.Substance2containsatransitionmetal.(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)IandIIIonly

CHAPTER2ChemicalBonding

51.Whichofthefollowingcompoundshasthegreatestioniccharacter?(A)SiO2

(B)ClO2

(C)CH4

(D)AlF3(E)SO2

52.Whichofthefollowingcompoundshasthegreatestlatticeenergy?(A)KCl(B)NaCl(C)CaCl2(D)MgCl2(E)FeCl3

53.WhichofthefollowingisthecorrectnameforthecompoundwiththechemicalformulaMg3N2?

(A)Trimagnesiumdinitrogen(B)Trimagnesiumdinitride(C)Magnesiumnitrogen(D)Magnesiumnitrate(E)Magnesiumnitride

Questions54and55refertothefollowingchoices:(A)N2

(B)F2(C)O3

(D)NH3

(E)CO2

54.Hasoneormorebondswithabondorderof1.5

55.Hasoneormorebondswithabondorderof3

56.IfmetalXformsanionicchloridewiththeformulaXCl2,whichofthefollowingismostlikelytheformulaforthestablephosphideofX?(A)XP2(B)X2P3(C)X3P2(D)X2(PO4)3(E)X3(PO4)2

57.Inwhichofthefollowingprocessesarecovalentbondsbroken?(A)C10H8(s)→C10H8(g)

(B)C(diamond)→C(graphite)(C)NaCl(s)→NaCl(molten)(D)KCl(s)→KCl(aq)(E)NH4NO3(s)→NH4

+(aq)+NO3

−(aq)

58.Inwhichofthefollowingprocessesarecovalentbondsbroken?(A)Solidsodiumchloridemelts.(B)Bronze(analloyofcopperandtin)melts.(C)Tablesugar(sucrose)dissolvesinwater.(D)Solidcarbon(graphite)sublimes.(E)Solidcarbondioxide(dryice)sublimes.

59.Diamondisanextremelyhardsubstance.Thisqualityisbestexplainedby

thefactthatadiamondcrystal(A)isionicwithahighlatticeenergy.(B)ismadecompletelyofcarbon,averyhardatom.(C)isformedonlyunderextremelyhighheatandpressure.(D)hasmanydelocalizedelectronsthatcontributetogreatervanderWaal

forces.(E)isonegiantmoleculeinwhicheachatomformsstrongbondswith

eachofitsneighbors.

Questions60–65refertothefollowinganswerchoices:(A)PCl5(B)BH3

(C)NH3

(D)CO2

(E)SO2

60.Themoleculewithtrigonalpyramidalmoleculargeometry61.Themoleculewithtrigonalbipyramidalmoleculargeometry62.Themoleculewithtrigonalplanarmoleculargeometry63.Themoleculewithbentmoleculargeometry64.Themoleculewithlinearmoleculargeometry65.ThemoleculewithtetrahedralelectronpairgeometryQuestions66–70refertothefollowinganswerchoices:(A)C3H8

(B)C6H6

(C)H2O

(D)CO2

(E)CH2O

66.Themoleculewiththelargestdipolemoment67.Themoleculewiththegreatestnumberofπ(pi)bonds68.Themoleculewiththegreatestnumberofσ(sigma)bonds69.Themoleculethatcontainsacentralatomwithsphybridization70.ThemoleculewithexactlyonedoublebondQuestions71–76refertothefollowinganswerchoices:(A)CO(B)C2H4

(C)PH3

(D)HF(E)O2

71.Hasthelargestdipolemoment72.Containstwoπ(pi)bonds73.Anecessaryreactantforcombustionreactions74.Hastrigonalpyramidalelectronpairgeometry75.Containsthemostσ(sigma)bonds76.OneoftwoallotropesofanelementfoundinEarth’satmosphereQuestions77–80refertothefollowinganswerchoices:(A)CCl4(B)CO2

(C)H2O

(D)BH3

(E)NH3

77.Thismoleculehasexactly2doublebonds.

78.Thismoleculehasthelargestdipolemoment.

79.Thismoleculehasatrigonalplanarmoleculargeometry.

80.Thismoleculehasatrigonalpyramidalmoleculargeometry.

81.TypesofhybridizationexhibitedbyCatomsinethaneincludewhichofthefollowing?I.spII.sp2

III.sp3

(A)Ionly(B)IIonly(C)IIIonly(D)IandIIIonly(E)I,II,andIII

82.TypesofhybridizationexhibitedbyCatomsinhexeneincludewhichofthefollowing?(A)sponly(B)sp2only(C)sp3only(D)spandsp2only(E)sp2andsp3only83.TypesofhybridizationexhibitedbyCatomsin

butyneincludewhichofthefollowing?I.spII.sp2

III.sp3

(A)Ionly(B)IIonly(C)IIIonly(D)IandIIIonly(E)I,II,andIII

84.ThereisaprogressivedecreaseinthebondangleintheseriesofmoleculesCCl4,PCl3,andH2O.AccordingtotheVSEPRmodel,thisisbestexplainedby:(A)Increasingpolarityofbonds(B)Increasingelectronegativityofthecentralatom(C)Increasingnumber

ofunbondedelectrons(D)Decreasingsizeofthecentralatom(E)Decreasingbondstrength

85.Whichofthefollowingisanonpolarmoleculethatcontainspolarbonds?(A)H2

(B)O2

(C)CO2

(D)CH4

(E)CH2F2

86.Whichofthefollowingmoleculescontainsonlysinglebonds?(A)C3H6

(B)C6H6

(C)C6H14

(D)CH3CHO

(E)CH3CH2COOH

87.WhichofthefollowingdescribesthehybridizationofthephosphorusatominthecompoundPCl5?

(A)sp(B)sp2

(C)sp3

(D)sp3d(E)sp3d2

88.Whichofthefollowingsinglebondsistheleastpolar?(A)H–N(B)H–O(C)F–O(D)I–F(E)H–F

89.Whichofthefollowingmoleculeshasanangular(bent)geometryandismostcommonlyrepresentedasaresonancehybridoftwoormoreLewis-dotstructures?(A)O3

(B)H2O

(C)CO2

(D)BeCl2(E)OF2

90.Whichofthefollowingmoleculeshasthelargestdipolemoment?(A)CO(B)HCN(C)HCl(D)HF(E)NH3

CHAPTER3StatesofMatter

Questions91–95refertothefollowingdescriptionsofbondingindifferenttypesofsolids.

(A)Alatticeofcloselypackedcationswithdelocalizedelectronsthroughout(B)Alatticeofcationsandanionsheldtogetherbyelectrostaticforces(C)Strong,singlecovalentbondsconnecteveryatom

(D)Strong,covalentbondsconnectatomswithinasheet,whileindividualsheetsareheldtogetherbyweakintermolecularforces(E)Strong,multiplecovalentbondsincludingσ(sigma)andπ(pi)bondsconnecttheatoms91.Gold(Au)92.Magnesiumchloride(MgCl2)93.Carbondioxide(CO2)94.Carbon(Cgraphite)95.Carbon(Cdiamond)Questions96–98refertothefollowingphasediagramofapuresubstance.

96.Thepointonthediagramthatcorrespondstothenormalboilingpointofthesubstance97.Thelineonthegraphthatcorrespondstotheequilibriumbetweenthesolidandgasphasesofthesubstance98.Allofthefollowingarecorrectstatementsregardingthenegativeslopeofthelineindicatedby

theletterBexcept:(A)Aspressureincreases,thetemperaturemustdecreaseforthesolidtoform.(B)Aspressureincreases,moreheatmustberemovedfromthecompound

inordertosolidify.(C)Thefreezingpointofthecompoundisactuallylowerthanthenormal

freezingpointatpressuresabove1atm.(D)Thesolidformofthiscompoundmayhaveagreaterdensitythanthe

liquidformofthiscompound.(E)Atlowtemperatures,ahighpressureisrequiredforasolidtoform.

Questions99and100refertothephasediagramforcarbondioxide.

99.Thetemperatureofasampleofpuresolidisslowlyraisedfrom–100°Cto20°Cataconstantpressureof1atm,whatistheexpectedbehaviorofthesubstance?(A)Itmeltstoaliquidandthenboilsat–80°C.(B)Itmeltstoaliquidbutdoesnotboiluntilatemperaturehigherthan

100°Cisreached.(C)Itmeltstoaliquidandboilsatabout30°C.(D)Itevaporates.

(E)Itsublimes.

100.Whatistheexpectedbehaviorofthesubstanceasthetemperatureisslowlyraisedfrom–100°Cto–40°Cataconstantpressureof1atm?(A)Itmeltstoaliquid.(B)Itsublimestoavapor.(C)Itevaporatestoavapor.(D)Itfirstmeltsatapproximately80°Candthenquicklyevaporates.(E)Itfirstmeltsatapproximately80°Candthenquicklysublimes.

101.Whichofthefollowingpuresubstanceshasthehighestmeltingpoint?(A)H2S

(B)C5H12

(C)I2(D)SiO2

(E)S8

Questions102–109refertothefollowingdiagramshowingthetemperaturechangesof0.5kgofwater,startingasasolid.Itisheatedataconstantrateof1atmofpressureinanopencontainer.Assumenomassislostduringtheexperiment.

102.Thesampleofwaterrequiresthegreatestinputofenergyduring:(A)Theheatingoficefrom–50°Cto0°C(B)Themeltingoficeat0°C(C)Theheatingofwaterfrom0°Cto100°C(D)Thevaporizationofwater100°C(E)Theheatingofsteamfrom100°Cto120°C

103.Whichofthefollowingbestdescribeswhatishappeningat0°C?(A)Theaveragekineticenergyoftheparticlesisincreasingasheatis

beingabsorbed.(B)Theaveragedistancebetweenthemoleculesisdecreasing.(C)Thenumberofhydrogenbondsbetweenthemoleculesareincreasing.(D)Thepotentialenergyofthesubstanceisdecreasing.(E)Thesubstanceissublimating.

104.Theheatoffusionisclosestto:(A)75kJkg−1

(B)150kJkg−1

(C)300kJkg−1

(D)600kJkg−1

(E)750kJkg−1

105.Theheatofvaporizationisclosestto:(A)750kJkg−1

(B)1,500kJkg−1

(C)1,800kJkg−1

(D)2,300kJkg−1

(E)3,000kJkg−1

106.Thespecificheatoficeisclosestto:(A)2.0kJkg−1°C−1

(B)4.2kJkg−1°C−1

(C)6.0kJkg−1°C−1

(D)8.4kJkg−1°C−1

(E)10.0kJkg−1°C−1

107.Howisthethedisparitybetweentheheatoffusionandtheheatofvaporizationbestexplained?(A)Ittakesmorehydrogenbondsforwatertofusethanitdoesto

vaporize.(B)Watermoleculesaremovingfartherapartduringfusionthanduring

vaporization.(C)Watermoleculesaremovingclosertogetherduringfusionandfarther

apartduringvaporization.(D)Vaporizationoccursatahigherkineticenergythanfusion.(E)Morehydrogenbondsarebrokenduringvaporization.

108.Thedataintheheatingcurvegraphcanbeusedtocalculate:(A)Theenthalpyofformationofwater(B)Theenthalpyofhydrogenbondformation(C)Thespecificheatofsuperheatedsteam(D)Theamountoftimeittakesforwatertomeltat0°C(E)Thedensityofwaterat50°C

109.Allofthefollowingaretrueregardingenergyandentropychangesinthewaterduringtheexperimentexcept:(A)Theenergyofthewatercontinuouslyincreases.(B)Thekineticenergyofthewaterdoesnotincreasecontinuously.(C)Therearetwopointsonthecurvewhereonlythepotentialenergyand

entropyofthewaterareincreasing.(D)Theentropyofthewateronlyincreasesduringphasechanges.(E)Therearrangementofthewatermoleculesduringphasechanges

increasestheirpotentialenergy.

Questions110–112refertothechoicesinthefollowingtable.

110.Thecompoundwiththeleastorweakestintermolecularforces111.ThecompoundthatcanformhydrogenbondswiththestrongestLondondispersionforces112.Nonpolarmoleculeoflowestvolatility113.ThemeltingpointofBeOis2,507°CwhilethemeltingpointofNaClis801°C.Explanationsforthisdifferenceincludewhichofthefollowing?I.Be2+ismorepositivelychargedthanNa+.II.O2−ismorenegativelychargedthanCl−.III.TheCl−ionislargerthantheO2−ion.(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)I,II,andIII

114.Apureliquidheatedinanopencontainerwillboilwhenatthetemperatureatwhichthe(A)averagekineticenergyoftheliquidisequaltotheaveragekineticenergyofthegas.(B)averagekineticenergyoftheliquidequalsthemolarentropyofthe

gas.(C)entropyoftheliquidequalstheentropyofthegas.(D)entropyofthevaporabovetheliquidequalstheentropyofthe

atmosphere.(E)vaporpressureoftheliquidequalstheatmosphericpressureabovethe

liquid.

115.Atthetopofahighmountain,waterboilsat90°C(insteadof100°C,theboilingpointofwateratsealevel).Whichofthefollowingbestexplains

thisphenomenon?(A)Waterathighaltitudescontainsagreaterconcentrationofdissolved

gases.(B)Watermoleculesathighaltitudeshavehigherkineticenergiesdueto

thelowerpressureonthem.(C)Equilibriumbecausewatervaporpressureequalsatmosphericpressure

atalowertemperature.(D)Thevaporpressureofwaterincreaseswithincreasingaltitude.(E)Waterfoundathighaltitudeshasfewersolutesandimpuritiesthat

allowsboilingtooccuratlowertemperatures.

116.WhichofthefollowingbestdescribesthechangesthatoccurintheforcesofattractionbetweenCO2moleculesastheychangephasefromagastoasolid?(A)C–Obondsareformed.(B)HydrogenbondsbetweenCO2moleculesareformed.

(C)IonicbondsbetweenCO2moleculesareformed.

(D)London(dispersion)forcesoperatetoformthesolid.(E)CO2moleculesformacrystalaroundanucleationpoint.

117.AllofthefollowingchangesoccurasH2Ofreezesexcept:(A)Ionicbondsformbetweenthewatermolecules.(B)Thewatertakesonacrystallinestructure.(C)Thedensityofthewaterdecreases.(D)Themassofthewaterdoesnotchange.(E)Thenumberofhydrogenbondsbetweenthewatermolecules

increases.

118.Whichofthefollowingstatementsaccountsfortheincreaseinboilingpointsoftheelementsgoingdowngroup18(thenoblegases)?(A)TheLondon(dispersion)forcesincrease.(B)Atomswithalargeradiusareclosertogether.

(C)Atomsofhighermassmovemoreslowlyonaveragethanatomsoflowermass.

(D)Dipole–dipoleinteractionsincrease.(E)Thekineticenergyoftheatomsdecreaseswithincreasingmass.

119.Whichofthefollowingisexpectedtohavethehighestboilingpointbasedonthestrengthofintermolecularforces?(A)Xe(B)Br2(C)Cl2(D)N2

(E)O2

120.Whichofthefollowingmustbetrueofapure,covalentsolidheatedslowlyatitsmeltingpointuntilabouthalfthecompoundhasturnedintoliquid?(A)Thesumoftheintermolecularforcesholdingthesolidtogether

decreasetozeroasthesolidcontinuestomelt.(B)Covalentbondsarebrokenasthesolidmelts.(C)Thetemperatureincreasesandtheaveragekineticenergyofthe

moleculesintheliquidphaseincreases.(D)Thevolumeincreasesasthesubstancebecomesaliquid.(E)Theaveragekineticenergyofthesubstanceremainsthesame.

Questions121–127refertothefollowinggasesat0°Cand1atm.(A)He(molarmass4)(B)Xe(131)(C)O2(32)

(D)CO2(44)

(E)CO(28)

121.Hasthegreatestdensity122.Theparticles(atomsormolecules)ofthis

gashaveanaveragespeedclosesttothatofN2moleculesatSTP

123.Hasthegreatestrateofeffusion124.Hasthelowestrateofeffusion125.Requiresthelowesttemperatureandhighestpressuretoliquefy126.ThegasthathasthegreatestLondondispersionforces127.Thegasthatcondensesatthehighesttemperatureat10atm128.Underwhichofthefollowingconditionsdogasesbehavemostideally?(A)Lowpressureandtemperature(B)Highpressureandtemperature(C)Highpressure,lowtemperature(D)Lowpressure,hightemperature(E)Anytemperatureifthepressureislessthan0.821atm

129.At298Kand1atm,a0.5-molsampleofO2(g)andaseparate0.75-molsampleofCO2(g)havethesame:(A)Mass

(B)Density(C)Averagemolecularspeed(D)Averagemolecularkineticenergy(E)Numberofatoms

130.AtSTP,a0.2-molsampleofCO2(g)andaseparate0.4-molsampleofN2O(g)havethesame:(A)Mass

(B)Volume(C)Averagemolecularspeed(D)Numberofatoms(E)Chemicalproperties

131.Thetemperatureatwhich32.0gofO2gaswilloccupy22.4Lat4.0atmisclosestto:(A)90K(B)273K(C)550K(D)950K

(E)1,900K

132.Thepressureexertedby2.5molofanidealgasplacedina4.00-Lcontainerat55°Cisgivenbywhichofthefollowingexpressions?

133.GasesN2(g)andH2(g)areaddedtoapreviouslyevacuatedcontainerandreactataconstanttemperatureaccordingtothefollowingchemicalequation:

IftheinitialpressureofN2(g)was1.2atm,andthatofH2(g)was3.8atm,whatisthepartialpressureofNH3(g)whenthepartialpressureofN2(g)hasdecreasedto0.9atm?(A)0.30atm(B)0.60atm(C)0.9atm(D)1.8atm(E)3.8atm

134.Whichofthefollowinggasesbehavesleastideally?(A)Ne(B)CH4

(C)CO2

(D)H2

(E)SO2

135.Whichofthefollowinggaseswillbehavemostideally?

136.EqualmassesofNeandArareplacedinarigid,sealedcontainer.Ifthetotalpressureinthecontaineris1.2atm,whatisthepartialpressureofAr?(A)0.20atm(B)0.40atm(C)0.60atm(D)0.80atm(E)2.40atm

137.Aflaskcontains0.5molofSO2(g),1molofCO2(g),and1molofO2(g).Ifthetotalpressureintheflaskis750mmHg,whatisthepartialpressureofSO2(g)?

(A)750mmHg(B)375mmHg(C)350mmHg(D)300mmHg(E)150mmHg

138.A2-Lcontainerwillholdapproximately3gramsofwhichofthefollowinggasesat0°Cand1atm?(A)CO2

(B)H2O

(C)Cl2(D)O2

(E)NH3

139.A2-Lflaskcontains0.50moleofSO2(g),0.75moleofO2(g),0.75moleofCH4(g),and1.00moleCO2(g).Thetotalpressureintheflaskis800mmHg.WhatisthepartialpressureofO2(g)intheflask?

(A)125mmHg(B)188mmHg(C)200mmHg(D)250mmHg(E)375mmHg

140.HClandNH3gasesarereleasedintooppositeendsofa1-meter(100-cm),verticalglasstubeat25°C.Theirreactionquicklyproducesawhitefogofammoniumchloride.Ifthetwogasesarereleasedatexactlythesametime,whichofthefollowingmostcloselyapproximateswheretheammoniumchloridefogwouldform?(A)20cmfromthesidewhereNH3wasreleased

(B)40cmfromthesidewhereNH3wasreleased

(C)Inthemiddle(50cmfromeitherside)(D)65cmfromthesidewhereNH3wasreleased

(E)80cmfromthesidewhereNH3wasreleased

141.A2-Lcontainerwillholdabout7gofwhichofthefollowinggasesat0°Cand1atm?(A)SO2

(B)CO2

(C)N2

(D)Cl2

(E)C4H8

142.Whichofthefollowinggases,whencollectedoverwater,wouldproducethegreatestyield(thehighestpercentcollected)?(A)CH4

(B)HCN(C)SO2

(D)HCl(E)NH3

143.Whichofthefollowingbestexplainswhyahot-airballoonrises?(A)Therateofdiffusionofthehotairinsidetheballoonisgreaterthanthe

rateofdiffusionofthecolderairsurroundingtheballoon.(B)Thepressureonthewallsoftheballoonisgreaterthanthe

atmosphericpressure.(C)Thedifferenceintemperatureandpressurebetweentheairinsideand

outsidetheballooncreatesanupwardactingcurrent.(D)Theaveragedensityoftheballoonislessthanthatofthesurrounding

air.(E)Thehigherpressureofthesurroundingairpushesonthesidesofthe

balloon,squeezingituptohigheraltitudes.

144.ArigidmetalcontainercontainsNegas.WhichofthefollowingistrueofthegasinthetankwhenadditionalNeisaddedataconstanttemperature?(A)Thepressureofthegasdecreases.(B)Thevolumeofthegasincreases.(C)Thetotalnumberofgasmoleculesremainsthesame.(D)Theaveragespeedofthegasmoleculesremainsthesame.(E)Theaveragedistancebetweenthegasmoleculesincreases.

145.EqualnumbersofmolesofAr(g),Kr(g),andXe(g)areplacedinarigidglassvesselatroomtemperature.Ifthecontainerhasapinhole-sizedleak,whichofthefollowingwillbetrueregardingtherelativevaluesofthe

partialpressuresoftheremaininggasesaftersomeeffusionhasoccurred?(A)PAr<PKr<PXe(B)PXe<PKr<PAr(C)PKr<PAr<PXe(D)PAr<PXe<PKr(E)PAr=PKr=PXe

146.Whichofthefollowinggaseshasthegreatestaveragemolecularspeedat298K?(A)He(B)H2

(C)N2

(D)O2

(E)Ne

Questions147–149refertothefollowingsituation.

Inalaboratoryexperiment,astudentreactsNa2CO3(106gmol−1)withHCl.WaterdisplacementisusedtomeasuretheamountofCO2produced(thegasoverwateriscollectinaeudiometer).

147.Ifthestudentreacts10.6gNa2CO3in250mlof2.50MHCl,howmanymolesofCO2gaswouldoneexpecttocollect?

(A)0.10molCO2

(B)0.25molCO2

(C)0.325molCO2

(D)0.63molCO2

(E)1.625molCO2

148.Thevolumeofgasthestudentcollectsissignificantlylessthanexpected

becausetheCO2gas(A)canreactwithwater.

(B)isdenserthanwatervapor.(C)hasamolarmasslargerthanN2andO2andthereforecannotdisplace

theairabovethewaterintheeudiometer.(D)hasamolarmasslargerthanN2andO2,andthereforehasalower

averagespeedatthesametemperature.(E)isnotthegasthatisactuallyproducedbythereaction.

149.Thetotalatmosphericpressureofthelaboratory(760mmHg),aswellasthetemperatureofthewater(22°C)andthevolumeofgas(502mL)intheeudiometer,areknown.Whichadditionaldata,ifany,isneededtocalculatethenumberofmolesofCO2gascollectedduringtheexperiment?(A)Thetemperatureofthegascollected(B)Themassofthegasintheeudiometer(C)ThevolumeofH2O(l)intheeudiometer(D)Thevaporpressureof

wateratthetemperatureofthewaterintheeudiometer(E)Nootherinformationisneeded

150.Threegases,1.6gHe(4gmol−1),4gAr(40gmol−1),and26gXe(131gmol−1),areaddedtoapreviouslyevacuatedrigidcontainer.Ifthetotalpressureinthetankis2.1atm,thepartialpressureofXe(g)isclosestto:(A)0.2atm(B)0.3atm(C)0.4atm(D)0.6atm(E)0.8atm

CHAPTER4Solutions

151.Asolutionispreparedbydissolvinganonvolatilesoluteinapuresolvent.Comparedtothepuresolvent,thesolution(A)hasahighernormalboilingpoint.(B)hasahigherfreezingpoint.(C)hasahighervaporpressure.(D)haslessosmoticpressure.(E)hasthesamevaporpressure,boilingpoint,andfreezingpointbecause

thesoluteisnonvolatile.

152.AsolutionofNaClisheatedfrom25°Cto75°C.Truestatementsregardingthissolutionincludewhichofthefollowing?I.Themolalityofthesolutiondidnotchange.II.Themolarityofthesolutiondidnotchange.III.Thedensityofthesolutiondidnotchange.(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)IIandIIIonly

153.ApproximatelywhatmassofCuSO4·5H2O(250gmol−1)isneededtoprepare125mLofa0.20-Mcopper(II)sulfatesolution?(A)2.0g(B)2.5g(C)6.2g(D)12.5g(E)25.0g

154.Whatvolumeofdistilledwatershouldbeaddedto20mLof5MHCl(aq)topreparea0.8-Msolution?(A)100mL(B)105mL(C)125mL(D)140mL(E)200mL

155.WhatisthefinalconcentrationofPb2+ionswhena100mL0.20MPb(NO3)2solutionismixedwitha100mL0.30MNaClsolution?

(A)0.005M(B)0.010M(C)0.015M(D)0.020M(E)0.025M

156.A0.2-MsolutionofK2CO3isabetterconductorofelectricitythana0.2-MsolutionofKBr.Whichofthefollowingbestexplainsthisobservation?(A)K2CO3ismoresolublethanKBr.

(B)K2CO3hasmoreatomsthanKBr.

(C)K2CO3containsthecarbonateion,apolyatomicion.

(D)KBrhasahighermolarmassthanK2CO3.

(E)KBrdissociatesintofewerionsthanK2CO3.

157.Anaqueoussolutionthatis66percentC2H4O(44gmol−1)bymasshasamolefractionofethanolclosestto:(A)0.29(B)0.44(C)0.50(D)0.66(E)1

158.Asolutioncontains144gH2Oand92gofethanol(CH3CH2OH,molarmass46gmol−1).Themolefractionofethanolisclosestto:(A)20percent(B)25percent(C)40percent(D)64percent(E)80percent

159.Whatisthemolalityofasolutionthathas29gNaCldissolvedin200gofwater?(A)0.0025m(B)0.025m(C)0.15m(D)2.5m(E)2.9m

160.Saltscontainingwhichofthefollowingionsareinsolubleincoldwater?(A)Nitrate(B)Ammonium(C)Sodium(D)Phosphate(E)Acetate

161.BaF2issparinglysolubleinwater.TheadditionofdiluteHFtoasaturatedBaF2solutionatequilibriumisexpectedto(A)raisethepH.

(B)reactwithBaF2toproduceH2gas.

(C)increasethesolubilityofBaF2.

(D)precipitateoutmoreBaF2.

(E)producenochangeinthesolution.

Questions162–165refertothefollowingsolution.

Ethanol,CH3CH2OH(l),andwater,H2O(l),aremixedinequalvolumesat25°Cand1atm.

162.Whichofthefollowingincludeendothermicprocessesregardingthepreparationofthesolution?I.Ethanolmoleculesmoveawayfromotherethanolmoleculesasthey

moveintosolution.II.Watermoleculesmoveawayfromotherwatermoleculesastheymove

intosolution.III.Ethanolmoleculesformhydrogenbondswithwatermoleculesasthey

moveintosolution.(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)I,II,andIII

163.Whatisthemolefractionofethanolinthesolution?(Thedensityofethanolandwaterat25°Care0.79gmL−1and1.0gmL−1,respectively.)(A)0.24(B)0.33(C)0.40(D)0.50(E)0.72

164.Mixingdifferentproportionsofethanolandwaterproducedifferententhalpyvalues.Atlowconcentrationsofwaterorethanol,solvationisexothermic,butformixingequalamounts,itisendothermic.Whichofthefollowingisalogicalinterpretationofthisobservation?(A)Theratioofhydrogenbondbreakages(betweenmoleculesofthepure

liquids),andtheformationofhydrogenbonds(betweenthetwodifferentmoleculeswhencombinedinsolution)varieswiththeratiosinwhichthetwoliquidsarecombined.

(B)Atlowconcentrationsofethanolorwater,fewerhydrogenbondsare

formedthanwhenmixingtheminequalamounts.(C)Mixingliquidsthatformthesametypeofintermolecularforces

undergonoenthalpychangeswhencombinedinequimolaramounts.(D)Ethanoliscapableofformingmorehydrogenbondsthanwater.(E)Wateriscapableofformingmorehydrogenbondsthanethanol.

165.Theintermolecularforcesbetweenethanolandwaterinclude:I.HydrogenbondingII.Dipole–dipoleattractionIII.Londondispersionforces(A)Ionly(B)IIonly(C)IIIonly(D)IandIIIonly(E)I,II,andIII

166.A1.0-Lsolutioncontains0.1molKCl,0.1molCaCl2,and0.1molAlCl3.WhatistheminimumnumberofmolesofPb(NO3)2thatmustbeaddedtoprecipitatealloftheCl−ionsasPbCl2?

(A)0.1mol(B)0.2mol(C)0.3mol(D)0.4mol(E)0.6mol

167.UnderwhichofthefollowingsetsofconditionswouldthemostN2(g)bedissolvedinH2O(l)?

168.Sodiumchlorideisleastsolubleinwhichofthefollowingliquids?(A)CH3COOH

(B)CH3OH

(C)CCl4(D)H2O

(E)HBr

169.Thelargestpercentageofwhichofthefollowingcompoundscanbecollectedbycoolingasaturatedsolutionofthatcompoundfrom90°Cto20°C?

Questions170and171refertothefollowingdata.Solutionsofthefivecompoundsinthetableweremixedwithequimolarsolutionsofoneoftwocompounds,XorY,alsointhelist.Compoundsofthesameidentitywerenotcombined.Assumeallconcentrationsare1.0M.

170.TheidentityofsubstanceX

171.TheidentityofsubstanceY

172.Asampleof60mLof0.4MNaOHisaddedto40mLof0.6MBa(OH)2.Whatisthehydroxideconcentration[OH–]ofthefinalsolution?(A)0.24M(B)0.40M(C)0.48M(D)0.50M(E)0.72M

173.Astudentmixesequalvolumesof1.0-Msolutionsofcopper(II)chlorideandmagnesiumsulfate,andnoprecipitateisobserved.Whenthestudentmixesequalvolumesof1.0-Msolutionsofaluminumsulfateandcopper(II)fluoride,aprecipitateisobserved.Whichofthefollowingistheformulaoftheprecipitate?(A)CuF2(B)CuSO4

(C)AlF3(D)AlCl3(E)AlSO4

174.Whichofthefollowingpairsofliquidsformsthemostidealsolutionwhenmixedinequalvolumesat25°C?(A)HClandH2O

(B)CH3CH2OHandH2O

(C)CH3CH2OHandC6H14

(D)C6H14andC8H18

(E)C8H18andH2O

175.Supposeasampleofahomogenoussolutioncontains10percenthexane(molarmass86gmol−1)bymass.Whichofthefollowingstatementsistrueregardingtheminimuminformationneededtocalculatethemolarityofhexaneinthissolution?(A)Thetemperatureofthesolution(B)Thetotalmassofthesolutionfromwhichthesampleistaken(C)The

massandvolumeofasampleofthesolution(D)Thevolumeofthesample

(E)Themassofthesample

176.A360-mgsampleofglucose,C6H12O6(molarmass180gmol−1),isdissolvedinenoughwatertoproducea200-mLsolution.Whatisthemolarityofa10-mLsampleofthissolution?(A)0.01M(B)0.10M(C)1.0M(D)2.0M(E)10.0M

177.Whichofthefollowingaqueoussolutionshasthehighestboilingpointat1.0atm?(A)0.2mNaCl(B)0.3mCaCl2(C)0.4mK3PO4

(D)0.5mNaNO3

(E)0.6mC12H22O11

178.Whatisthevaporpressureofasolutioninwhich2.00-molpropyleneglycol,anonvolatilecompound,ismixedwith8.00-molwater?Assumethesolutionbehavesideallyandisatthetemperaturewherethevaporpressureofwateris20.0mmHg.(A)4.00mmHg(B)15.00mmHg(C)16.00mmHg(D)18.00mmHg(E)20.00mmHg

179.Adilutehydrochloricacidsolutionwasaddedtoasampleofanunknownsolutioninalab.Awhiteprecipitatewasformed,filteredfromthesolution,washedwithhotwater,andthendissolvedinasolutionofNH3.AfewdropsofK2SO4wereaddedtothefiltrateandanotherwhiteprecipitateformed.Whattwoionswereprecipitatedoutofsolution?(A)Mg2+andPb2+

(B)Mg2+andAg+

(C)Ag+andBa2+

(D)Ag+andPb2+

(E)NH4+andPb2+

180.Whichofthefollowingcompoundsistheleastsolubleinwater?(A)(NH4)2CO3

(B)BaCO3

(C)Fe(NO3)3(D)Na3(PO4)

(E)LiO

CHAPTER5ChemicalReactions

181.Howmanymoleculesarecontainedin180gofwater(H2O)?

(A)6.02×1022

(B)1.20×1023

(C)6.02×1023

(D)1.20×1024

(E)6.02×1024

182.Howmanyoxygenatomsarein4.4gofCO2?

(A)6.02×1022

(B)1.20×1023

(C)1.20×1024

(D)6.02×1023

(E)6.02×1024

183.Howmanyatomsofhydrogenarein1.5gofribose(C5H10O5,150gmol−1)?(A)6.02×1022

(B)6.02×1023

(C)6.02×1024

(D)6.02×1025

(E)6.02×1026

184.Acompoundcontains22.2percentTi,33.3percentC,and44.4percentO.Whatistheempiricalformulaforthiscompound?(A)TiCO(B)Ti2C3O4

(C)Ti(CO)2(D)Ti(CO)3(E)Ti(CO)6

185.Acompoundis92percentCand8percentH.Whatistheempiricalformulaforthiscompound?(A)CH(B)CH2

(C)CH4

(D)C6H6

(E)C6H8

186.Acompoundcontainingonlycarbon,hydrogen,andoxygenhasamolecularmassof150gmol−1.Whichofthefollowingmaybetheempiricalformulaofthecompound?(A)CHO(B)C2H3O

(C)CH2O

(D)CH2O2

(E)C5H10O5

187.Acompoundcontains0.2molPd,0.8molC,1.2molH,and0.8molO.Whichofthefollowingisthesimplestformulaofthiscompound?(A)Pd2C8H12O8

(B)Pd2(C4H6O4)2(C)Pd2(C2H3O2)3(D)Pd(C2H3O2)3(E)Pd(C2H3O2)2

188.Acompoundcontains38percentFand62percentXe.Theempirical

formulaofthecompoundis:(A)XeF(B)Xe2F

(C)Xe4F

(D)XeF4(E)Xe2F3

189.Whatistheempiricalformulaforahydrocarbonthatis75percentcarbonbymass?(A)CH2

(B)CH4

(C)CH6

(D)C4H

(E)C4H5

190.WhatmassofCu(s)isproducedwhen0.050molCu2O(143gmol−1)isreducedwithexcessH2(g)?

(A)3.18g(B)6.35g(C)12.7g(D)31.8g(E)63.5g

Questions191–196refertothefollowinganswerchoices:(A)HCl(aq)+NH3(aq)→NH4Cl(aq)+H2O(l)

(B)Ag+(aq)+Cl−(aq)→AgCl(s)(C)Mg(s)+O2(g)→MgO2(s)

(D)PtCl4(s)+2Cl−(aq)→PtCl62–(aq)(E)3Cl2(aq)+6OH−

(aq)→5Cl−(aq)+ClO3−(aq)+3H2O(l)

191.Areactionthatproducesacoordinationcomplex192.Areactioninwhichthesamereactantundergoesanoxidationandareduction193.Aneutralizationreaction194.Aprecipitationreaction195.Acombustionreaction196.Areactionthatproducesanacidicsalt197.Allofthefollowingchemicalequationsrepresentthecorrectnetionicequationforthereactionthatoccurswhenaqueoussodiumhydroxideisaddedtoasaturatedsolutionofaluminumhydroxideexcept:(A)Al(OH)3+OH−→[Al(OH)4]−

(B)Al(OH)3+3OH−→[Al(OH)6]3–

(C)Al3++4OH−→[Al(OH)4]−

(D)Al3++6OH−→[Al(OH)6]3–

(E)Al+3OH−→Al(OH)3

198.Whichofthefollowingchemicalequationsrepresentstheintenseheatingofsolidhydrogencarbonate(sodiumbicarbonate)?(A)NaHCO3→Na+H2O+CO2

(B)NaHCO3→Na2CO3+H2O+CO2

(C)NaHCO3→Na+H2+O2+CO2

(D)Na(CO3)2→Na+Na2CO3+H2O+CO2

(E)Na(CO3)2→Na2CO3+H2O+CO2

199.Whichofthefollowingchemicalequationsrepresentsthereactionthatoccurswhenpure,solidwhitephosphorusburnsinair?(A)4P+3O2→2P4O3

(B)4P+3O2→P4O6

(C)4P+5O2→P4O10

(D)P4+3O2→P4O6

(E)P4+O2→P4O10

200.Whichofthefollowingchemicalequationsrepresentsthenetionicequationforthereactionthatoccurswhensodiumiodidesolutionisadded

toasolutionoflead(II)acetate?(A)2I−+Pb2+→PbI2(B)Na++CH3COO−→NaCH3COO

(C)2NaI+Pb(CH3COO)2→2NaCH3COO+PbI2(D)2NaI+Pb(CH3COO)2+H2O→2NaOH+Pb(OH)2+I2(E)3I−+Pb2++2H2O→PbI2+Pb(OH)2+HI201.Allofthefollowing

pairsofsubstancesgivevisibleortactile(throughtheproductionorabsorptionofasignificantamountofheat)evidenceofachemicalreactionuponmixingexcept:

(A)HCl(aq)andKOH(aq)(B)CaCO3(aq)andHF(aq)(C)Mg(s)andHI(aq)(D)Pb(NO3)2(aq)andNaCl(aq)(E)NH4NO3(aq)andHCl(aq)

Questions202–205refertothefollowinganswerchoices:(A)MgO+CO2→MgCO3

(B)I−+Cd2+→CdI42–

(C)SiH4+O2→SiO2+H2O

(D)CaO+SO2→CaSO3

(E)Mg(s)→Mg2++2e−

202.Acombustionreaction203.Areactionthatproducesacomplexion204.Anoxidationthatisnotacombustion205.Areactioninwhichtheproductformsagasinacidicsolutions206.Whentheequationforthereactionofhexaneinairiscorrectlybalancedandallcoefficientsarereducedtotheirlowestwhole-numberterms,thecoefficientforO2is:(A)6

(B)12(C)14(D)19

(E)25

207.Whentheequationaboveisbalancedandallcoefficientsreducedtotheirlowestwholenumberterms,thecoefficientforH2O(l)is:(A)7

(B)6(C)5(D)3(E)2

208.Whentheequationaboveisbalancedandallcoefficientsreducedtotheirlowestwhole-numberterms,thecoefficientforH2O(l)is:(A)1

(B)2(C)3(D)4(E)6

209.Whentheequationaboveisbalancedandallcoefficientsreducedtotheirlowestwhole-numberterms,thecoefficientforCl2(g)is:(A)2

(B)3(C)4(D)5(E)6

210.Whentheequationaboveisbalancedandallcoefficientsreducedtotheirlowestwhole-numberterms,thecoefficientforH3PO4(l)is:(A)1

(B)2

(C)3(D)4(E)5

211.Theoxidationofethanolinanacidicsolutionisrepresentedabove.Whentheequationisbalancedandallcoefficientsreducedtotheirlowestwhole-numberterms,thecoefficientforH+is:(A)6(B)12(C)14(D)16(E)28

212.Thereactionrepresentedaboveoccursinabasicsolution.Whentheequationisbalancedandallcoefficientsreducedtotheirlowestwhole-numberterms,thecoefficientforOH−is:(A)0(B)1(C)2(D)4(E)5

Questions213–216refertothefollowingbalancedchemicalreaction:

213.Accordingtothereactionabove,when0.400molofCS2(l)isreactedascompletelyaspossiblewith1.20molO2(g),thetotalnumberofmolesofproductsis:(A)0.40(B)0.80(C)1.20(D)1.60(E)4.80

214.If6.30molesofgasareformedbythereactionindicatedabove,howmanymolesofO2(g)areneededtoreact?

(A)1.05(B)2.10(C)4.20(D)6.30(E)9.45

215.If33.6LofproductareformedbytheabovereactionatSTP,howmanymolesofCS2(l)reacted?

(A)0.50(B)1.00(C)1.50(D)3.00(E)3.36

216.AnexcessofZn(s)isaddedto100mLof0.6MHClat0°Cand1atm.Whatvolumeofgaswillbeproduced?(A)67.2mL(B)672mL(C)1.3L(D)2.24L(E)6.7L

Questions217–219refertothefollowinganswerchoicesandthereactionrepresentedbelow.

217.Thereactantsthatwouldproduce3molSO3withonemolO2inexcess218.Productsofthereactiontocompletionif4molSO2combinewith1molO2

219.Productsif5molSO2reactswith6.5molO2

Questions220–224refertothefollowingchemicalreaction:

220.If1.7gNH3(17gmol−1)aremixedwith3.2gO2(32gmol−1),whatisthemaximummassofNO(30gmol−1)thatcanbeproduced?(A)0.40g(B)2.4g(C)3.2g(D)4.0g(E)4.8g

221.SupposeNH3andO2weremixedina4:5molarratio.Whenthereactionwascompleted,228gramsofthereactiongaseswererecovered.Assumingthereactionmixturereactedcompletelyandthegasescollectedcontain100percentproducts,whatmassesofNOandH2Oareexpectedtobepresent?

222.If100mLofdistilledwaterisaddedto400mLof0.375MNaCl,whichofthefollowingisclosesttotheNa+concentrationinthefinalsolution?(A)0.15M(B)0.19M(C)0.30M(D)0.33M(E)0.69M

223.Howmanymolesofoxygengaswouldberequiredtoproduce1.0LofNOatSTP?

224.If2.5molNH3reactedwith2.5molO2ascompletelyaspossible,how

manymolesofreactantwouldremainunreacted(inexcess)?(A)0.50molO2

(B)0.50molNH3

(C)0.63NH3

(D)2.0molO2

(E)2.0molNH3

Questions225–227refertothefollowingreactionofpotassiumsuperoxide(KO2).KO2isusedforitsabilitytoabsorbcarbondioxidegasandreleaseoxygen.

225.Accordingtotheequationabove,howmanymolesofO2(g)canbereleasedif6.00molKO2and9.00molofCO2(g)areavailable?

(A)4.50(B)6.00(C)9.00(D)12.0(E)13.5

226.WhatvolumeofCO2isrequiredtoreactwithexcessKO2toproduce6LofO2(g)atSTP?

(A)4.00L(B)11.2L(C)14.9L(D)22.4L(E)89.6L

227.When2.9gCO2(g)(molarmass44gmol−1)reactswithexcessKO2accordingtotheequationabove,thevolumeofO2(g)producedatSTPis

closestto:(A)2.2L(B)4.4L(C)11.2L(D)22.4L(E)33.6L

228.Whichofthefollowingisthecorrectnetionicequationforthereactionofsodiumhydroxideandnitricacid?(A)H++OH−→H2O

(B)Na++NO3−→NaNO3

(C)NaOH+HNO3→NaNO3+H2O

(D)Na++OH−+H++NO3−→NaNO3+H2O

(E)Na++OH−+2H+→NaOH+H2

229.Whichofthefollowingisanadditionreaction?(A)CH4+Cl2→CH3Cl+HCl

(B)CH2=CH2+Cl2→CH2ClCH2Cl(C)C4H10+Cl2→C4H10Cl+HCl(D)CH3(CH2)6CH=CHCOOH+NaOH→CH3(CH2)6CH=CHCOONa+H2O

(E)6CO2+6H2O→C6H12O6+6O2

230.Whichofthefollowingisaproductofthereactionbetweencarbondioxideandwater,CO2(g)+H2O(l)?

I.H+

II.CO32–

III.HCO3−

(A)Ionly(B)IandIIonly(C)IandIIIonly(D)IIandIIIonly

(E)I,II,andIII

CHAPTER6Thermodynamics

Questions231–237refertothefollowinganswerchoices:(A)Activationenergy

(B)Latticeenergy(C)Freeenergy(D)Kineticenergy(E)Potentialenergy

231.Theenergyneededtoconvertanionicsolidintowell-separatedgaseousions232.Theenergyliberatedfromaphysicalorchemicalprocessthatisavailabletodowork233.Theenergyneededfortheformationofthetransitionstateofachemicalreaction234.Theenergyneededtoovercometheactivationbarrierofachemicalreaction235.Thisquantityisdeterminedbymeasuringtherateofaparticularreactionattwoormoredifferenttemperatures236.Thisquantitymustchangeinanysubstanceundergoingaphasechange237.TheformulaforthisquantityisusedtoderivetheeffusionratioofO2andCO2atthegiventemperatureQuestions238–241refertothefollowinganswerchoices:(A)Enthalpyofformation(B)Entropy(C)Energyofcrystallization(D)Activationenergy(E)Gibbsfreeenergy

238.Aquantitythatiszeroforaperfect,purecrystallinesolidat0K

239.Aquantitythatiszerowhenareactionisatequilibriumatconstanttemperatureandpressure240.Aquantitythatiszeroforapureelementinitsstandardstate241.AquantitythatdescribesafeatureofachemicalreactionandisalwayspositiveforreactionsinwhichanincreaseintemperatureresultsinanincreasedreactionrateQuestions242–247refer

tothefollowinganswerchoices:(A)ΔH>0,ΔS>0(B)ΔH<0,ΔS<0(C)ΔH>0,ΔS<0(D)ΔH<0,ΔS>0(E)ΔH=0,ΔS>0

242.Mustbetrueforareactionthatisspontaneousatalltemperatures243.Trueformeltingofwaterat25°Cand1atm244.TrueforthedepositionofCO2(g)intoCO2(s)at–100°Cand1atm245.TrueforthecombustionofwoodintoCO2(g)andH2O(g)inacampfire246.Mustbetrueforareactionthatisneverspontaneousatanytemperature247.Trueforthemixingoftwoidealgases(nointermolecularforcesofattractionorrepulsionoccurbetweenanyofthegasparticles)ataconstanttemperatureandpressure.Assumenoreactionoccurs.

248.ThedecompositionofCaCO3(s)isshownintheequationabove.Usingthedatainthetableabovethereaction,whichofthefollowingvaluesisclosesttotheΔHrxnofthedecompositionofCaCO3(s)?

(A)−2,240kJmol−1

(B)−180kJmol−1

(C)180kJmol−1

(D)1,207kJmol−1

(E)2,240kJmol−1

249.Thereactionsforthecombustionofdiamondandgraphiteareshownabove.WhichofthefollowingvaluesisclosesttotheΔHrxnfortheconversionofC(graphite)toC(diamond)?

(A)−789kJ(B)−1.9kJ(C)1.9kJ(D)394.45kJ(E)798kJ

250.Theentropychangeforthedissolutionofcalciumchlorideinwatershownabovemightbeexpectedtobepositive,buttheactualΔSisnegative.Whichofthefollowingisthemostplausibleexplanationforthenetlossofentropyduringthisprocess?(A)CaCl2(s)isanamorphoussolid.

(B)Theparticlesinthesolutionaremoreorderedthantheparticlesinthesolid.

(C)Theionsinsolutioncanmovemorefreelythantheparticlesinthesolid.

(D)Thedecreasedentropyofthewatermoleculesinthesolutionisgreaterthantheincreasedentropyoftheions.

(E)Insolution,thedistancebetweenionsismuchgreaterthanthedistancebetweentheionsinthesolid.

251.Whichofthefollowingbestdescribestheroleofasparkinabutanelighter?(A)Thesparkdecreasestheactivationenergyofthecombustionreaction.

(B)Thesparkincreasestheconcentrationofbutaneinthereactionchamber.

(C)Thesparkprovidestheheatofvaporizationfortheliquidfuel.(D)Thesparksuppliessomeoftheenergytoformtheactivatedcomplex

forthecombustionreaction.(E)Thesparkprovidesanalternativestoichiometryforthereaction,

decreasingtheamountofoxygenrequiredforcompletecombustion.

252.Whichofthefollowingprocessesdemonstratesadecreaseinentropy(ΔS<0)?(A)Br(s)→Br(l)(B)I2(s)→I2(g)(C)CombiningequalvolumesofC2H6O2(l)andH2O(l)(D)TheprecipitationofPbI2fromsolution

(E)Thethermalexpansionofaheliumballoon

253.Whichofthefollowingreactionsinvolvesthelargestdecreaseinentropy?(A)MgCO3(s)→MgO(s)+CO2(g)

(B)3H2(g)+N2(g)→2NH3(g)

(C)4La(s)+3O2(g)→2La2O3(s)

(D)2NaI(aq)+Pb(CH3COO)2(aq)→2NaCH3COO(aq)+PbI2(s)(E)C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)

254.Thecombustionofmethylhydrazine,acommonrocketfuel,isrepresentedabove.TheΔHofthisreactionis–1,303kJmol−1CH6N2(l).WhatwouldbetheΔHpermolCH6N2(l)ifthereactionproducedH2O(l)insteadofH2O(g)?(TheΔHforthecondensationofH2O(g)toH2O(l)is–44kJmol−1.)(A)–1,171kJ(B)–1,259kJ(C)–1,347kJ

(D)–1,567kJ(E)–1,435kJ

255.Astudentaddssolidammoniumchloridetoabeakercontainingwaterat25°C.Asitdissolves,thebeakerfeelscolder.WhichofthefollowingaretrueregardingtheΔHandΔSofthedissolutionprocessofNH4Cl(s)?

256.Supposeareactionisspontaneousattemperaturesonlybelow300K.IftheΔHa°forthisreactionis–18.0kJmol−1,thevalueofΔS°forthisreactionisclosesttowhichofthefollowing?AssumeΔS°andΔH°donotchangesignificantlywithtemperature.(A)–60Jmol−1K−1

(B)–18Jmol−1K−1

(C)–0.0010Jmol−1K−1

(D)18Jmol−1K−1

(E)18,000Jmol−1K−1

257.ThenormalmeltingpointofAg(s)is962°C.Whichofthefollowingistruefortheprocessrepresentedaboveat962°C?(A)ΔH=0(B)ΔS=0(C)TΔS=0(D)ΔH=TΔS(E)ΔH>TΔS

258.ThestandardGibbsfreeenergychange,ΔG°298,fortheconversionofCdiamondintoCgraphitehasanapproximatevalueof–3kJmol−1.However,graphitedoesnotformfromdiamondunderstandardconditions(298Kand1atm).Whichofthefollowingbestexplainsthisobservation?(A)Diamondismoreorderedthangraphite(lowerentropy).(B)TheΔHfortheconversionofdiamondtographiteishighly

endothermic.(C)Theactivationenergyfortheconversionofdiamondtographiteis

verylarge.(D)TheC–CbondsindiamondaremuchstrongerthantheC–Cbondsin

graphite.(E)Diamondissignificantlydenserthangraphite.

259.Whichofthefollowingistrueregardingtheadiabaticandreversiblecompressionofanidealgas?(A)Thetemperatureofthegasremainsconstant.(B)Thevolumeofthegasremainsconstant.(C)Thepressureofthegasremainsconstant.(D)Noworkcanbedonebythegas.(E)Thenetentropychangeofthegasiszero.

260.Giventhebondenergiesinthetableabove,whichofthefollowingstatementsbestdescribestheformationof1moleofH2O(l)fromH2(g)andO2(g)?

(A)Theprocessisendothermicwithanenthalpychangeofapproximately480kJ.

(B)Theprocessisendothermicwithanenthalpychangeofapproximately240kJ.

(C)Theprocessisexothermicwithanenthalpychangeofapproximately

480kJ.(D)Theprocessisexothermicwithanenthalpychangeofapproximately

240kJ.(E)Theprocessisexothermicwithanenthalpychangeofapproximately

1,800kJ.

261.Theenthalpyofcombustionofethanol(C2H5OH(l))isshownabove.Usingthisinformationandthedatainthetableabove,thestandardheatofformationofCO2(g)isclosestto:(A)–1,080kJmol−1

(B)–540kJmol−1

(C)–510kJmol−1

(D)–390kJmol−1

(E)–250kJmol−1

262.Whatisthestandardenthalpychange, ,ofthereactionrepresentedabove?(A)–24.3kJ(B)–58.3kJ(C)24.3kJ(D)58.3kJ(E)77.7kJ

263.WhenCO3(s)isaddedtoabeakerofdiluteHCl(aq)at298K,thebeakergetscoldandagasisproduced.WhichofthefollowingindicatesthecorrectsignsforΔG,ΔH,andΔSforthereactionrepresentedabove?

264.Consideringthedatainthetableabove,whichofthefollowingmustbetrueofthereaction?(A)Acatalystispresent.(B)Thereactionorderiszero.(C)Thereactionisatequilibrium.(D)Theenthalpychangeofthereactioniszero.(E)Thisreactionisoccurringatatemperatureabove298K.

Questions265–269refertothedatainthefollowingtable.

265.WhatisthevalueofΔH°298forreactionZ?

(A)90kJmol−1

(B)–90kJmol−1

(C)172kJmol−1

(D)–172kJmol−1

(E)213kJmol−1

266.ReactionsforwhichthevalueofKpwillincreaseundergreaterpressureincludewhichofthefollowing?(A)Xonly(B)Zonly(C)XandZ(D)ThevaluesofKpdecreasewithincreasedpressure(E)NoneoftheKp

valueswillincrease

267.ReactionsforwhichthevalueKpwillincreaseifthetemperatureisraisedabove298Kincludewhichofthefollowing?(A)Xonly(B)Yonly(C)Zonly(D)XandYonly(E)YandZonly

268.WhichofthefollowingmostaccuratelydescribestheΔSofreactionZ?(A)Positive,becausetherearemoreproductsthanreactants.(B)Positive,becausetherearemorestatesofmatterintheproducts.(C)Positive,becausethereisonlyonespeciesofreactantbuttherearetwo

speciesofproduct.(D)Negative,becausetwomolesofgasareconvertedtoasolidandone

moleofgas.(E)Negative,becauseapureelementwasformed.

269.WhichofthefollowingstatementsmostaccuratelydescribestheratesofreactionsXandY?(A)XwilloccurmorerapidlythanYbecausetheΔHismorepositive.(B)YwilloccurmorerapidlythanXbecausetheΔHislesspositive.(C)XwilloccurmorerapidlythanYbecausetheΔSandΔGaremore

positive.(D)YwilloccurmorerapidlythanXbecausetheΔSandΔGareless

positive.

(E)Thermodynamicdataforoverallreactionsdonotindicateanythingabouttherateofachemicalreaction.

CHAPTER7Kinetics

Questions270–273refertothefollowingchoices:(A)Rate=k[M](B)Rate=k[M][N](C)Rate=k[M][N]2

(D)Rate=k[M]2[N]2

(E)Rate=k[N]2

270.DoublingtheconcentrationofMhasnoeffectonthereactionrate.

271.DoublingtheconcentrationofMandNincreasesreactionrateby2.

272.DoublingtheconcentrationofMonlyquadruplesthereactionrate.

273.DoublingtheconcentrationofMandNincreasesthereactionratebyeightfold.HalvingtheconcentrationofNdecreasesthereactionratefourfold.

274.Theproductionofiron(II)sulfideoccursatasignificantlyhigherratewhenironfilingsareusedinsteadofblocks(volume=0.1mL).Whichofthefollowingbestexplainsthisobservation?(A)Theironfilingsarepartiallyoxidizedduetotheirgreaterexposureto

oxygen.(B)TheironintheblockisFe(s)andtheironinthefilingsisFe2+.

(C)Theironblockistooconcentratedtochemicallyreact.(D)Theironfilingshaveamuchgreaterareaincontactwithsulfur.(E)Thereactantorderofironintheratelawis1forFe(s)and2forFe2+.

275.Achemistrystudentsittingaroundacampfireobservesthatthelargepiecesofwoodburnslowly,butamixtureofsmallscrapsofwoodandsawdustaddedtotheflamecombustsexplosively.Thecorrectexplanation

forthedifferenceinthecombustionbetweenthesetwoformsofwoodisthat,comparedwiththewoodscrapsandsawdust,thelargepiecesofwood(A)haveagreatersurfacearea-to-volumeratio.(B)haveasmallersurfaceareaperkilogram.(C)haveahigherpercentcarbon.(D)containcompoundswithalowerheatofcombustion.(E)containmorecarbondioxideandwater.

276.Allofthefollowingresultinanincreasedrateofreactioninanaqueoussolutionexcept:(A)Increasingthetemperatureofanendothermicreaction.(B)Increasingthetemperatureofanexothermicreaction.(C)Increasingthesurfaceareaofasolidreactant.(D)Increasingthepressureonthesolution.(E)Mixingorstirringthesolution.

277.Factorsthataffecttherateatwhichachemicalreactionproceedsincludewhichofthefollowing?I.TheorientationofthereactantsattimeofcollisionII.Thekinetic

energyofthecollisionsbetweenreactantsIII.Thefrequencyofcollisionsbetweenreactants

(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)I,II,andIII

278.Whichofthefollowingcorrectlyexplain(s)theeffectofincreasedtemperatureontherateofachemicalreaction?I.IncreasesthereactionrateofendothermicreactionsII.Increasesthe

reactionrateofexothermicreactionsIII.Decreasesthereactionrateofreactionswitha–ΔH

(A)Ionly

(B)IIonly(C)IandIIonly(D)IIandIIIonly(E)IandIIIonly

279.Allofthefollowingstatementsregardingthekineticsofradioactivedecayaretrueexcept:(A)Thelengthoftimeofahalf-lifeisspecifictoaparticularelement.(B)Allradioactivedecaydisplaysfirst-orderkinetics.(C)Inasampleofapure,radioactiveisotope,one-halfthenumberof

radioactiveatomsandone-halfthemassoftheradioactivesubstanceremainsafteronehalf-life.

(D)Thehalf-lifeofanatomdoesnotchangewhentheatomisincorporatedintoacompound.

(E)Thehalf-lifeofaparticularsubstancedoesnotchangewithtimeortemperature.

Questions280–285refertothereactionofnitrogenmonoxide(nitricoxide)andoxygenandthefollowingdata.

280.Theratelawforthisreactionis:(A)Rate=k[NO][O2]

(B)Rate=k[NO][O2]2

(C)Rate=k[NO]2[O2]

(D)Rate=k[NO]2[O2]2

(E)Rate=k[NO][O2]1.25

281.Thenumericalvaluefortherateconstant(k)isclosestto:(A)5.9×10−2

(B)170(C)3×103

(D)7×103

(E)1.2×105

282.Theunitoftherateconstant(k)is:(A)sec−1

(B)Lmol−1sec−1

(C)L2mol−2sec−1

(D)L3mol−3sec−1

(E)L4mol−4sec−1

283.IncreasingtheinitialconcentrationofNOfivefoldwouldincreasethereactionrateby:(A)5X(B)10X(C)25X(D)3,125X(E)Nosubstantialmargin

284.WhatwouldbethereactionrateiftheinitialconcentrationofNOwas2×10−2MandtheinitialconcentrationofO2was4×10−2M?

(A)3.2×10−5

(B)8×10−4

(C)2.3×10−2

(D)1.1×10−1

(E)5.7

285.Whichofthefollowingisacorrectstatementaboutreactionorder?(A)Reactionordermustbeawholenumber.(B)Reactionordercanbedeterminedmathematicallyusingonly

coefficientsofthebalancedreactionequation.

(C)Reactionordercanchangewithincreasingtemperature.(D)Asecond-orderreactionmustinvolveatleasttworeactants.(E)Reactionordercanonlybedeterminedexperimentally.

286.Propertiesofacatalystincludeallofthefollowingexcept:(A)Acatalystthatworksforonechemicalreactionmaynotworkfora

differentreaction.(B)Acatalystisnotconsumedbythereactionitcatalyzes.(C)Catalystscanbesolidsorgases.(D)Acatalystwillonlyspeedupachemicalreactionineithertheforward

orreversedirection.(E)Catalystsspeedupchemicalreactionsbyprovidinganalternate

pathwayforreactioninwhichtheactivatedcomplexisoflowerenergy.

287.Whichofthefollowingcanbeusedtocalculateormeasuretherateofachemicalreaction?I.TheappearanceofproductovertimeII.ThedisappearanceofoneormoresubstratesovertimeIII.Therate

lawIV.TheKeq(equilibriumconstant)andQ(reactionquotient)ofthe

reaction(A)IandIIonly(B)IandIIIonly(C)I,II,andIIIonly(D)I,II,andIVonly(E)I,II,III,andIV

288.Thereactionofnitricoxidewithhydrogengasat25°Cand1atmisrepresentedbelow.Theratelawforthisreactionis:rate=k[H2][NO]2.

Accordingtotheratelaw,whichofthefollowingisthebestpredictionof

therateofthisreaction?(A)TherateofdisappearanceofNOisalwaystwiceasgreatasthe

disappearanceofH2.

(B)TherateofdisappearanceofNOisalwaysfourtimesasgreatasthedisappearanceofH2.

(C)TherateofdisappearanceofNOistwiceasgreatasthedisappearanceofH2iftheconcentrationofNOisinitiallytwicethatofH2.

(D)TherateofdisappearanceofNOisfourtimesasfastasthatofH2,butonlyiftheinitialconcentrationofNOisinitiallytwicethatofH2.

(E)TherelativedisappearancesofNOandH2cannotbededucedwithoutthevalueoftherateconstant(k).

289.Thereactionbetweennitrogendioxideandcarbonmonoxideisrepresentedabove.Theproposedreactionmechanismisasfollows:

Whichofthefollowingreactionmechanismsisconsistentwiththeproposedmechanism?(A)Rate=k[NO2]

(B)Rate=k[NO2]2

(C)Rate=k[NO][CO](D)Rate=k[NO]2[CO](E)Rate=k[CO]

290.Thereactionbetweennitrogenmonoxide(commonlycallednitricoxide)andbromineisrepresentedabove.Theproposedreactionmechanismisas

follows:

Whichofthefollowingreactionmechanismsisconsistentwiththeproposedmechanism?(A)Rate=k[NO]2

(B)Rate=k[NO][Br2]

(C)Rate=k[NO][Br2]2

(D)Rate=k[NO]2[Br2]

(E)Rate=k[NO]2[Br2]2

291.Therateconstantforacertainchemicalreactionat25°Cis9.0×105L2mol−2sec−1.Whichofthefollowingmustbetrueregardingthisreaction?(A)Thisreactionisslowerthanareactionthathasarateconstantof9.0

L2mol−2sec−1.(B)Thisreactionisexothermic.(C)Therateofthisreactionwilldecreasewithincreasingtemperature.(D)Thereactionorderis3.(E)Doublingtheconcentrationofreactantswillincreasethereactionrate

byafactorof8.1×1010.

292.Theconversionofozone,2O3(g)→3O2(g),obeystheratelaw,rate=k[O3]2[O2]−1.Whichofthefollowingstatementsistrueregardingtherateofthebreakdownofozone(O3)intomolecularoxygen(O2)?

(A)Thecatalystthatconvertsozonetooxygenisinhibitedbyoxygen.(B)Therateatwhichozoneisconvertedtooxygenincreasesasthe

concentrationofoxygendecreases.(C)Thereisaninversesquarerelationshipbetweenozoneandoxygen

concentrations.(D)Theconversionofoxygentoozoneisfasterthantheconversionof

ozonetooxygen.(E)Thenegativereactantorderofoxygenindicatesthatthereactionisat

equilibriumandtheforwardreactionismorefavorable.

293.Allofthefollowingstatementsregardingtheactivatedcomplexaretrueexcept:(A)Theenergyoftheactivatedcomplexdeterminestheactivationenergy

ofthereaction.(B)Theactivatedcomplexrepresentsthehighestenergystatealongthe

transitionpathofachemicalreaction.(C)Theactivatedcomplexofachemicalreactionisspecifictothat

reaction.(D)Theconfigurationofatomsintheactivatedcomplexofanuncatalyzed

reactionisthesameasthatofacatalyzedreaction,exceptforthepresenceofthecatalyst.

(E)Areactioninwhichtheenergyoftheactivatedcomplexisverylargeindicatesthatthereactantsareverythermodynamicallystable(asopposedtochemicallystable).

Questions294and295refertothefollowinggraphofachemicalreactionovertime.

294.Whichofthefollowingistrueconcerningthereactionat25°C?I.Thereactionisendothermic.II.Theactivationenergy(Ea)isapproximately510kJmol−1.

III.Themagnitudeofthedifferencebetweenenergyofthereactantsand

productsisapproximately175kJmol−1.(A)Ionly(B)IIonly(C)IIIonly(D)IIandIIIonly(E)IandIIIonly

295.Whichofthefollowingistrueconcerningtheeffectofaddingacatalyst?(A)Theactivatedcomplexwouldforminlessthan40seconds.(B)Itlowerstheenergyoftheproducts.(C)Itincreasestheenergyofthereactants.(D)Morereactionswouldoccurpersecond.(E)Theequilibriumwouldshifttofavortheproducts.

296.Areactionwasobservedfor30minutes.Every5minutes,thepercentofreactantremainingwasmeasured.Accordingtothedatainthetableabove,whichofthefollowingmostaccuratelydescribesthereactionorderandhalf-lifeofthisreaction?

297.TheratelawsforthereactionbetweenO2andNOisk=[O2][NO]2.IfthereactionrateisfirstmeasuredwithO2andNOof2.5×10−4Meach,bywhatfactorwilltherateincreaseiftheconcentrationofO2andNOarebothincreasedto5.0×10−4each?(A)2

(B)3(C)4(D)6(E)8

Questions298and299refertothedatabelow.

298.Considerthedatashownabove.Ifthedatawereobtainedfromareactionwhoseratelawisk=[X][Y]2,whatwouldbetheexpectedrateofreactionforTrial2?(A)R(B)2R(C)4R

299.Supposethedatawereobtainedforareactionwhoseratelawisk=[X]2[Y].WhatwouldbetheexpectedrateofreactionforTrial2?(A)R(B)2R(C)4R

300.Theunitsoftherateconstant(k)forareactionthatoccursbetweentwosecond-orderreactantsis:(A)sec−1

(B)Lmol−1sec−1

(C)L2mol−2sec−1

(D)L3mol−3sec−1

(E)L4mol−4sec−1

CHAPTER8Equilibrium

301.WhichofthefollowingstresseswhenappliedtothereactionabovewillresultinanincreasedamountofMgO(s)?

I.RemoveMg(s)II.Increasethepressure(decreasethevolume)III.AddO2(g)

(A)Ionly(B)IIonly(C)IIIonly(D)IIandIIIonly(E)I,II,andIII

302.Supposethereactionrepresentedaboveisatequilibrium.WhichofthefollowingchangeswillresultinanincreasedamountofO2(g)?

(A)Increasingthepressure(B)Decreasingthevolume(C)AddingmoreN2(g)

(D)Decreasingthetemperature(E)RemovingNOCl(g)

303.Whichofthefollowingstatementsistrueregardingtheequilibriumreactionrepresentedabove?(A)Increasingthepressurewillshifttheequilibriumtotheright.

(B)Increasingthetemperaturewillshifttheequilibriumtotheright.(C)AddingmoreC(s)willshifttheequilibriumtotheright.

(D)DecreasingthetemperaturewillresultintheformationofmoreO2(g).

(E)AddingmoreO2willresultintheformationofmoreheat.

304.Inwhichofthefollowingreactionsatequilibriumwilltherebenochangeinresponsetoachangeinthevolumeofthereactionvessel?(Assumeconstanttemperature.)(A)SO2Cl2(g)⇔SO2(g)+Cl2(g)(B)H2(g)+Cl2(g)⇔2HCl(g)(C)2SO2(g)+O2(g)⇔2SO3(g)

(D)4NH3(g)+5O2(g)⇔4NO(g)+6H2O(g)(E)2N2O(g)+O2(g)⇔2NO(g)

305.AclosedrigidcontainercontainsdistilledwaterandH2(g)atequilibrium.WhichofthefollowingactionswouldincreasetheconcentrationofH2(g)inthewater?I.DecreasingthetemperatureofthewaterII.VigorousshakingIII.InjectingmoreH2(g)intothecontainer(A)Ionly

(B)IIonly(C)IIIonly(D)IandIIIonly(E)I,II,andIII

Questions306–308refertothefollowingchemicalreactionatequilibrium.

306.Supposethereactionaboveoccursinaclosed,rigidtank.Afterithasreachedequilibrium,pureN2(g)isinjectedintothetankataconstanttemperature.Afterithasre-establishedequilibrium,whichofthefollowinghasalowervaluecomparedtoitsvalueattheoriginal

equilibrium?(A)TheamountofNH3(g)inthetank(B)TheamountofH2(g)inthetank

(C)TheamountofN2(g)inthetank(D)Keqforthereaction

(E)Thetotalpressureinthetank

307.Whichofthefollowingchanges,ifoccurredalone,wouldcauseadecreaseinthevalueofKeqforthereactionrepresentedabove?

(A)Addingacatalyst(B)Loweringthetemperature(C)Raisingthetemperature(D)Decreasingthevolume(E)Increasingthevolume

308.Whenthereactionrepresentedaboveisatequilibrium,theratio

canbeincreasedbywhichofthefollowing?I.IncreasingthepressureII.DecreasingthepressureIII.IncreasingthetemperatureIV.Decreasingthetemperature(A)Ionly(B)IandIIIonly(C)IandIVonly(D)IIandIIIonly(E)IIandIVonly

Questions309–311refertothefollowingreaction.

309.Themolarequilibriumconcentrationsforthereactionmixturerepresented

aboveat298Kare[X]=4.0M,[Y]=5.0M,and[Z]=2.0M.Whatisthevalueoftheequilibriumconstant,Keq,forthereactionat298K?

(A)0.06(B)2.50(C)16.0(D)62.5(E)105.5

310.WhichofthefollowingchangestotheequilibriumsystemshownabovewillresultinanincreasedquantityofZ?I.IncreasingthepressureII.DecreasingthetemperatureIII.AddingmoreX(g)(A)Ionly(B)IIonly(C)IIIonly(D)IIandIIIonly(E)I,II,andIII

311.Changestotheequilibriumsystemabovethatwillincreasetheratioof

includewhichofthefollowing?I.DecreasingthepressureII.IncreasingthetemperatureIII.AddingmoreX(g)(A)Ionly(B)IIonly(C)IandIIIonly(D)IIandIIIonly(E)I,II,andIII

312.ThediagramaboverepresentsamixtureofF2andFgasesina1.0-LcontaineratequilibriumaccordingtotheequationF2⇔2Fatatemperatureabove1,200K.Whichofthefollowingistrueoftheequilibriumconstantforthisreactionatthistemperature?(A)K<0(B)K=0(C)0<K<1(D)K=1(E)K>1

313.At500K,theKeqforthereactionaboveis2.5×1010.WhatisthevalueofKeqforthereversereactionat500K?

(A)–2.5×1010

(B)2.5×10−10

(C)2.5×10−11

(D)4.0×1010

(E)4.0×10−11

314.AsolublecompoundXY3dissociatesinwateraccordingtotheequationabove.If,ina0.06-msolutionofthecompound,30.0percentofXY3dissociates,whichofthefollowingisclosesttothenumberofmolesofparticlesofsoluteper1.0kgofwater?(A)0.018

(B)0.060(C)0.072(D)0.114(E)0.240

315.WhatisthemolarsolubilityofBaSO4(s)inwater?(TheKspforBaSO4(s)is1.1×10−10.)(A)5.5×10−11M(B)1.10×10−10M(C)1.05×10−5M(D)5.5×10−5

(E)5.5×10−10M

316.WhatisthemolarsolubilityofCaF2(s)inwater?(TheKspforCaF2(s)is4.0×10−11.)(A)(4.0×10−11)1/2M(B)(1.0×10−11)1/2M(C)4.0×10−11M(D)(4.0×10−11)1/3M(E)(1.0×10−11)1/3M

317.AsaturatedsolutionofmetalhydroxideM(OH)2,hasapHof10.0at25°C.WhatistheKspvalueforM(OH)2?

(A)5×10−31

(B)1×10−20

(C)5×10−13

(D)2.5×10−9

(E)1×10−8

318.Iron(II)hydroxideissparinglysoluble.Itssolvationinwaterisrepresentedby .Truestatementsaboutthesolubilityofiron(II)hydroxideinclude:I.IncreasingthepHreducesitssolubility

II.DecreasingthepHreducesitssolubilityIII.ThevalueofKspforthereactionisspecifictothepHofthesolutionin

whichitisdissolved.(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)I,II,andIII

319.Asolutioncontains0.2MBa2+and0.2MCa2+.WhichofthefollowingCrO4

2–concentrationswillprecipitateasmuchBa2+aspossiblewithoutprecipitatinganyCaCrO4?(TheKspofBaCrO4=~1×10-10andtheKspofCaCrO4=~7×10−4.)(A)3.5×10−3M

(B)7×10−4M(C)1.5×10−7M(D)5×10−10M(E)7×10−14M

320.Thefigureaboveshowstwocompletelysealedcontainers,a10-mLtube,anda1,000-mLbowl,eachfilledtoafractionofitsvolumewithmethanol.Inalaboratoryat1atmand25°C,thevaporpressureofthemethanolis(A)lowerinContainer1becausethevolumeofmethanolis

lower.(B)higherinContainer1becausethevolumeofthecontainerabovethe

methanolissmaller.(C)higherinContainer2becausethesurfaceareaofthemethanolis

larger.(D)higherinContainer2becausethevolumeofthecontainerabovethe

methanolisgreater.(E)equalinContainers1and2becausetheyareatthesametemperature.

321.Thevaporizationofaliquidisrepresentedintheequationabove.Whichofthefollowingmostaccuratelyaccountsforthefactthatthevaporpressureofliquidsincreaseswithincreasingtemperature?(A)Vaporizationisexothermic.(B)Aliquidataparticulartemperaturecontinuestovaporizeuntilithas

completelyevaporated.(C)Ifachemicalsystematequilibriumexperiencesachangein

temperature,theequilibriumshiftstocounteractthechange.(D)Thecondensationofgasestakeslongerthanthevaporizationof

liquids.(E)Thecondensationofgasesreleasesmoreenergythanisabsorbedby

vaporizingliquids.

322.Whichofthefollowingbestaccountsforthefactthat,atthesamealtitude,thepartialpressureofwatervaporintheatmospherecanincreasesignificantlywithincreasingtemperature,butthepartialpressuresofN2,O2,Ar,andCO2stayrelativelyconstant?

(A)ThevaporpressuresofN2,O2,Ar,andCO2aremuchgreaterthanthatofH2O.

(B)Thewatercycle(evaporation,condensation,precipitation)causeswatertoconstantlychangebetweenliquidandgasphases.

(C)TheconditionsrequiredforatmosphericN2,O2,Ar,andCO2tobeinequilibriumwiththeirliquidphasesdonotexistontheearth’ssurface.

(D)WaterintheatmosphereconstantlyformswaterdropletsaroundcondensationnucleiwhereasN2,O2,Ar,andCO2onlycondenseinorabovethestratosphere.

(E)Watervaporisconstantlybeingaddedtotheatmospherebycellularrespirationandcombustion.

323.Consideraclosed,adiabaticsystemconsistingofamixtureofliquidandsolidsubstanceZatequilibriumatitsmeltingpoint.Whichofthefollowingstatementsistrueregardingthesystem?(A)Theentropyofthesystemisatamaximum.(B)Theentropyofthesystemisataminimum.(C)Theentropyofthesystemwillincreaseovertime.(D)Theentropyofthesystemiszero.(E)Theentropyofpuresubstancesdoesnotchangeifataconstant

temperature.

Questions324–328refertotheinformationinthetablebelow.

324.WhichofthefollowingistrueregardingthestrengthsofHOClandHOBr?(A)HOBrisastrongeracidbecauseBrisalargeratomthanCl,andloses

itsprotonmoreeasilythanHOCl.(B)HOClisastrongeracidbecauseClismoreelectronegativethan

HOBr.(C)ThestrengthsofHOClandHOBraredependentonthepHoftheir

respectivesolutions.(D)ThestrengthsofHOClandHOBraredependentontheirrespective

concentrations.

(E)HOBristhestrongeracidbecausethenegativelogofitsKa,thepKa,islargerthanthatofHOCl.

325.WhichofthefollowingstatementsisthemostaccuratepredictionaboutthestrengthofacidHOI?(A)HOIisthestrongestacidbecauseIisalargeratomthanBrandCl,

andthereforetheprotondissociatesmorereadilyinsolution.(B)HOIisthestrongestacidbecausetheO−HbondinHOIisstronger

thantheOHbondinHOClorHOBr.(C)HOIistheweakestacidbecauseitsconjugatebase,OI−,islessstable

thantheconjugatebasesofHOClandHOBr.(D)ThestrengthofHOIisdependentontheconcentrationsandpHofits

solution.(E)HOIisthestrongeracidthanHOBrandHOClbecausethenegative

logofitsKa,thepKa,islargerthanthepKaofHOBrandHOCl.

326.WhichofthefollowingisthecorrectnameforHOCl?(A)Hydrochloricacid(B)Hypochloricacid(C)Hypochorousacid(D)Perchlorousacid(E)Hydrogenchloride

327.ThehydrolysisofKOClisrepresentedabove.Whichofthefollowingisthecorrectsetupfordeterminingtheequilibriumconstantofthereaction?(A)2.9×10−8=[OH−][HOCl]/[OCl−](B)2.9×10−8=[OH−][HOCl]/[OCl−][H2O]

(C)1×10−14=[OH−][2.9×10−8]/[OCl−](D)Kb=1.0×10−14/2.9×10−8

(E)Kb=2.9×10−8/1.0×10−14

328.Whichofthefollowingshowsthecorrectwaytocalculatethe[OH−]concentrationina1.5-MsolutionofKOClat298K?AssumetheKhydrolysis=3.4×10−7.

Questions329and330refertothereactionA⇔B.TheconcentrationsofAandBthroughoutthereactionareshowninthefigurebelow.

329.Whichofthefollowingstatementsistrueregardingtheprogressofthereaction?(A)AtpointX,therateofA→BequalstherateB→A.(B)From0.0to0.8sec,theconcentrationofReactantAisincreasing.(C)From0.0to0.8sec,therateofA→BexceedstherateofB→A.(D)After1sec,thereactionhasgonetocompletion.(E)Ifthereactionwasobservedafter10minutes,theconcentrationsofA

andBwouldbeequal.

330.Supposeattime1.5sec,moreBwasaddedtothesystem(enoughtomomentarilyraisetheconcentrationto0.25M).Inwhichofthefollowingwaysisthesystemexpectedtorespond?(A)Theforwardreactionwouldbepushedtocompletion,consumingall

ofReactantA.(B)TheequilibriumstatewouldbeoverwhelmedandtheexcessBwould

precipitate.(C)Theequilibriumstatewouldbeoverwhelmedforcingthereverse

reactiontogotocompletion.(D)TheconcentrationofAwouldincrease.(E)TheequilibriumwouldadjusttoconsumetheexcessBandproducea

newratioof[A]/[B].

CHAPTER9Acid–BaseChemistry

Questions331–334refertothefollowinganswerchoices:(A)Protonacceptor(B)Protondonor(C)Electronpairacceptor(D)Electronpairdonor(E)Hydroxidegenerator

331.Bronsted–Lowryacid332.Lewisbase333.Arrheniusbase334.FormscoordinatecovalentbondswithacidsQuestions335–339refertoaqueoussolutionscontaining1:1moleratiosofthefollowingpairsofsubstances.Assumeallconcentrationsare1M.(A)NH3andH3CCOOH(aceticacid)(B)KOHandNH3

(C)HClandKCl(D)H3PO4andKH2PO4

(E)NH3andNH4Cl

335.ThesolutionwiththehighestpH

336.ThesolutionwiththelowestpH

337.ThesolutionwiththepHclosesttoneutral338.AbufferatanalkalinepH

339.AbufferatanacidicpH

340.WhatistheH+concentrationofa0.02-Mnitrousacid(HNO2)solution?(TheKaforHNO2is4.5×10−4.)(A)2.25×10−2

(B)3.0×10−3

(C)5.1×10−4

(D)9.0×10−6

(E)2.6×10−7

341.Whichofthefollowingstatementsistrueofthereactionrepresentedabove?(A)H2OistheconjugateacidofF−.

(B)OH−istheconjugateacidofH2O.

(C)HFistheconjugatebaseofF−.(D)HFandH2Oareconjugateacid–basepairs.

(E)HFandH2OarebothBronsted–Lowryacids.

342.Inthereactionabove,H2Oactsas:(A)Anacid

(B)Abase(C)Aconjugateacid(D)Anoxidizingagent(E)Areducingagent

343.Allofthefollowingincreasethestrengthofanoxyacidexcept:(A)Astronglyelectronegativecentralatom(B)Electronegativeatomsbondedtothecentralatom(C)Electronegativeatomsbondedtoatomsotherthanthecentralatom(D)Anincreasednumberofoxygenatomsbondedtothecentralatom(E)Anincreasednumberofhydrogens

344.Itispossibletocreateallofthefollowingsolutionsbymixing0.25Mand0.35MHClexcept:(A)0.34MHCl(B)0.31MHCl(C)0.29MHCl(D)0.26MHCl(E)0.24MHCl

345.TheaciddissociationconstantforaweakmonoproticacidHAis5.0×10−9.ThepHofa0.5MHAsolutionisclosestto:(A)2(B)3(C)4(D)5(E)6

346.ThereactionaboverepresentsageneralequilibriumbetweenmonoproticacidsHXandHY.Iftheequilibriumconstantforthisreactionwas2.5×103,whichofthefollowingcouldbecorrectlyconcludedaboutthechemicalspeciesinvolved?(A)HYisastrongeracidthanHX.(B)Y−isastrongerbasethanX−.(C)HYistheconjugatebaseofY−.(D)X−istheconjugateacidofHX.(E)ThepHofasolutioncontaininga1:1moleratioofHXandY−is7.

Questions347–349refertothefollowingtitration.

Asolutionofaweakmonoproticacidistitratedwitha0.1-Mstrongbase,NaOH.Thetitrationcurveisshownbelow.

347.Fromthecurve,whatisthevalueofthepHwherethenumberofmolesofstrongbaseaddedisequaltothenumberofmolesofweakacidintheinitialsolution?(A)4.75(B)7.00(C)7.45(D)8.73(E)11.3

348.AtwhichpHaretheconcentrationsoftheweakacidanditsconjugatebaseapproximatelyequal?(A)2.88(B)4.75(C)7.00(D)8.73(E)13.0

349.ThebufferregionofthistitrationcurvecanbefoundbetweenwhichofthefollowingpHvalues?(A)2.88and6.13

(B)4.75and8.73(C)6.13and11.29(D)8.73and11.29(E)11.29and12.25

350.Whichofthefollowingsolutionsataconcentrationof0.1MhasapH>7?(A)NaCl(B)KI(C)HC2H3O2

(D)LiF(E)NaBr

351.WhichofthefollowingistheconjugateacidofNH3?

(A)H+

(B)N2

(C)NH2−

(D)NH3+

(E)NH4+

352.At25°C,aqueoussolutionswithapHof6haveahydroxideionconcentration,[OH−],of:(A)1×10−6M(B)1×10−8M(C)0.006M(D)6M(E)8M

353.Whichofthefollowingstepswouldconvert100mLofKOHsolutionwithapHof13toaKOHsolutionwithapHof12?(A)Dilutethesolutionbyadding10mLofdistilledwater.(B)Dilutethesolutionbyadding900mLofdistilledwater.

(C)DilutethesolutionwithanequalamountofKOHsolutionwithapHof1.

(D)Add10mLof1MHClsolution.(E)Add100mLof1MHClsolution.

354.Citricacid(H3C6H5O7)isatriproticacidwithK1=8.4×10−4,K2=1.8×10−5,andK3=4.0×10−6.Ina0.01-Maqueoussolutionofcitricacid,whichofthefollowingspeciesispresentinthelowestconcentration?(A)H3O+

(l)

(B)H3C6H5O7(aq)

(C)H2C6H5O7−(aq)

(D)H1C6H5O72–(aq)

(E)C6H5O73–(aq)

355.Mixturesthatwouldbeusefulasbuffersincludewhichofthefollowing?I.NH3andNH4Cl

II.HClandKClIII.HFandKF(A)Ionly(B)IandIIonly(C)IandIIIonly(D)IIandIIIonly(E)I,II,andIII

356.Asampleof200mLof0.20MSr(OH)2isaddedto800mLof0.80MBa(OH)2.WhichofthefollowingbestapproximatesthepHofthefinalsolution?(A)0(B)1(C)12(D)13

(E)14

357.ThepHofasolutionpreparedbytheadditionof100mL0.002MHClto100-mLdistilledwaterisclosestto:(A)1.0(B)1.5(C)2.0(D)3.0(E)4.0

358.ThepHofasolutionpreparedbytheadditionof5.0mL0.02MBa(OH)2to15-mLdistilledwaterisclosestto:(A)2.0(B)4.0(C)11.0(D)12.0(E)13.0

Questions359–361refertothefiguresbelow.Thefiguresshowaburetteusedinthetitrationofanacidwitha0.5-MsolutionofSr(OH)2.Figure1showsthelevelofSr(OH)2atthestartofthetitrationandFigure2showsthelevelofSr(OH)2attheendofthetitration.Theindicatorusedinthisexperimentwasphenophthalein.

359.Whatevidenceindicatesthattheendpointofthetitrationhasbeenreached?(A)Thecolorofthesolutionintheburettechangesfromcolorlesstopink.(B)Thecolorofthesolutionintheburettechangesfrompinktocolorless.(C)Thecolorofthesolutionintheflaskbelowchangesfromcolorlessto

pink.(D)Thecolorofthesolutionintheflaskbelowchangesfrompinkto

colorless.(E)ThepHofthesolutiondoesnotchange.

360.ThevolumeofSr(OH)2usedtoneutralizetheacidwasclosestto:(A)3.5mL(B)10.60mL(C)31.80mL(D)42.04mL(E)42.40mL

361.Inadifferenttitration,25mLofthesame0.5MSr(OH)2solutionwasusedtotitrate35mLHClsolutiontotheequivalencepoint.WhatisthemolarityoftheHClsolution?(A)0.5M(B)0.7M(C)1.4M(D)2.1M

(E)2.8M

CHAPTER10Electrochemistry

362.TheoxidationstatemostcommontoionsofFe,Mn,andZninsolutionis:(A)+1(B)+2(C)+3(D)+4(E)+5

363.TheoxidationstateofchlorineinHClO4is:(A)–1

(B)+1(C)+3(D)+5(E)+7

364.InwhichofthefollowingspeciesdoeschromiumhavethesameoxidationnumberasinCr2O4

2–?

(A)Cr2(O2CCH3)4(B)Cr(CO)6(C)Cr2O7

2–

(D)K3[Cr(O2)4]

(E)Cr2O3

365.A0.1-MsolutionofH3PO4isabetterconductorofelectricitythana0.1-MsolutionofNaCl.Whichofthefollowingbestexplainsthisobservation?(A)NaClislesssolublethanH3PO4.

(B)NaClhasalowermolarmassthanH3PO4.

(C)H3PO4dissociatestoproduceparticlesofalargersizethanNaCl.

(D)H3PO4isweaklyacidic,whichincreasestheflowofchargethroughasolution.

(E)FewermolesofionsarepresentintheNaClsolutionthaninthesamevolumeofanNa3PO4solution.

366.Whichofthefollowingchoicesbestexplainswhythereactionrepresentedabovegenerateselectricityinagalvaniccell?(A)Cl2isastrongeroxidizingagentthanBr2.

(B)Cl2loseselectronsmoreeasilythanBr2.

(C)BratomshavemoreelectronsthanClatoms.(D)Br2ismorestablethanCl2.

(E)Cl−ismorestablethanBr−.

367.Whichofthefollowingstatementsis/aretrueregardingthereactionabove?I.Cu(s)actsasanoxidizingagent.

II.H+(aq)getsreduced.

III.Theoxidationstateofnitrogenchangesfrom+5to+2.(A)Ionly(B)IIonly(C)IIIonly(D)IIandIIIonly(E)I,II,andIII

Questions368and369refertoagalvaniccellconstructedoftwohalf-cellsandthetwohalf-reactionsrepresentedbelow.

368.Asthecelloperates,whichofthefollowingspeciesiscontainedinthehalf-cellcontainingthecathode?I.Zn2+

II.Zn+

III.Ag+

(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)IandIIIonly

369.Whatisthestandardcellpotentialforthisgalvaniccell?(A)–0.04V(B)0.04V(C)0.84V(D)1.56V(E)2.36V

370.Thephotodecompositionofpermanganateisshownabove.Asthisreactionproceedsintheforwarddirectionandthereactionspeciesareconsideredfromlefttoright,theoxidationnumberofMnchangesfrom:(A)+1to+2and+2(B)+5to+2and+2(C)+5to+6and+4(D)+7to+6and+4(E)+7to+2and+2

Questions371and372refertoanelectrolyticcellinvolvingthefollowing

reaction.

371.Whichofthefollowingprocessesoccursinthisreaction?(A)O3actsasanoxidizingagent.

(B)O36–actsasanoxidizingagent.

(C)Al3+isreducedatthecathode.(D)Alisoxidizedattheanode.(E)Aluminumisconvertedfromanegativetoaneutraloxidationstate.

372.Asteadycurrentof15amperesispassedthroughthecellfor20minutes.Whichofthefollowingcorrectlyexpresseshowtocalculatethenumberofgramsofaluminumproducedbythiscell(1faraday=96,500coulombs)?

373.Accordingtotheinformationabove,whatisthestandardreductionpotentialforthehalf-reaction ?(A)+0.67V(B)–0.67V(C)+1.47V

(D)–1.47V(E)+3.07V

374.Anelectriccurrentof1.00ampereispassedthroughanaqueoussolutionofFeCl3.Assuming100percentefficiency,howlongwillittaketoelectroplateexactly1.00moleofironmetal?

(1faraday=96,500coulombs=6.02×1023electrons)(A)32,200sec

(B)96,500sec(C)193,000sec(D)289,500sec(E)386,000sec

Questions375–381refertoagalvaniccellmadeoftwohalf-cells.Inonehalf-cell,aZnelectrodeisbathedina1.0MZnSO4solution.Theothercellcontainsa1MHClsolutionandahydrogenelectrode.

375.WhichofthefollowingisthecorrectelectronconfigurationforZn2+?(A)1s22s22p63s23p64s2

(B)1s22s22p63s23p63d10

(C)1s22s22p63s23p63d104s2

(D)1s22s22p63s23p63d84s2

(E)1s22s22p63s23p63d94s1

376.Whichofthefollowingcorrectlyidentifiesandjustifieswhichspecies,ZnorZn2+,hasthehighestionizationenergy?(A)Zn,electronshaveagreateraffinityforpuremetals.(B)Zn,ithasagreaterelectrontoprotonratio.

(C)Zn2+,ithasalargerradius.(D)Zn2+,ithasagreatereffectivenuclearcharge.(E)Zn2+,ithasagreaterelectron-to-protonratio.

377.Processesthatoccurattheanodeincludewhichofthefollowing?I.Zn(s)isoxidizedtoZn2+(aq).

II.Zn2+(aq)isreducedtoZn(s).

III.H2(g)isoxidizedtoH+(aq).

IV.2H+(aq)isreducedtoH2(g).

(A)Ionly(B)IIonly(C)IIIonly(D)IandIIIonly(E)IIandIVonly

378.ThesaltbridgeisfilledwithasaturatedsolutionofKNO3.Processesthatoccuratthesaltbridgeincludewhichofthefollowing?I.K+movesintotheanodehalf-cell.II.K+movesintothecathodehalf-cell.III.NO3

−movesintotheanodehalf-cell.

IV.NO3−movesintothecathodehalf-cell.

(A)Ionly(B)IIIonly(C)IandIVonly(D)IIandIIIonly(E)I,II,III,andIV

379.WhichofthefollowingistrueregardingtheHClhalf-cell?(A)ThereductionofH+toH2isneitherspontaneousnornon-

spontaneous.

(B)ThereductionofH+andoxidationofH2areatequilibrium.

(C)Allelectrochemicalcellsuseahydrogenhalf-cell.(D)ItisusedtomeasurethereductionpotentialofZn2+becausethe

reductionpotentialhasameasuredvalueofzerovolts.(E)TheE°valueof0Vforthehydrogenhalf-cellisarbitrary.

380.Whichofthefollowingistrueregardingthestandardfreeenergychange(ΔG°)ofthecell?(A)ItisnegativebecausethestandardreductionpotentialofZn2+is

negative.(B)Itisnegativebecausethesumofthestandardreductionpotentialsof

Zn2+andH+is–0.76V.(C)ItisnegativebecausetheoxidationofZnisspontaneous.(D)Itispositivebecausethereductionofhydrogenionsisspontaneous.(E)Ithasavalueofzerobecausethereactionisatequilibrium.

381.Whichofthefollowingchangestothevoltage,ifany,wouldresultfromincreasingtheconcentrationofZnSO4?

(A)Nochange,thevoltageofthecellisdeterminedonlybythestandardreductionpotentialsoftheelectrodes.

(B)Increasedvoltage,thevalueofQwillfallbelow1andthelogtermintheNernstequationwillbecomemorenegative.

(C)Increasedvoltage,thereactionbecomesmorefavorableastheconcentrationofZn+increases.

(D)Decreasedvoltage,thevalueofQwillfallbelow1andthelogtermintheNernstequationwillbecomemorenegative.

(E)Decreasedvoltage,theincreasedconcentrationofZn2+willinhibittheoxidationofZn(s).

382.Ironnailsareoftenelectroplated(galvanized)withzinc.Whichofthefollowingistrueregardingthisprocess(E°ofZn2+→Zn(s)=–0.76V,E°ofFe2+→Fe(s)=–0.44V)?

(A)ZnservesasasacrificialanodeforFe.

(B)Znservesasanairtightsealpreventingtheelectronsfromironfromreactingwithoxygen.

(C)ThegalvanizationprocessproducesanalloyofZnandFethatisstrongerandmoreresistanttooxidationthanFealone.

(D)Thegalvanizationprocessisspontaneous,therebyreducingtheenergycostofproducingnails.

(E)TheoxidationofZnallowsthereductionofFe2+toproceedspontaneously.

383.Whichofthefollowingcorrectlyrelatesthequantitiesneededfortheconversionofchemicalenergyintoelectricalenergyinagalvaniccell?(A)ThesumoftheE°cellofthetwohalf-cells.(B)Theproductofthecellvoltageandthetotalchargepassedthroughthe

cell.(C)Theproductofcellvoltageandthenaturallog(ln)ofK,the

equilibriumconstantfortheredoxreaction.(D)Theproductofthegasconstantandthevoltagedividedbythetotal

chargepassedthroughthecell.(E)Thedifferencebetweenthevoltagesofthetwohalf-cells.

384.Allofthefollowingstatementsareconsistentwiththeoperationofagalvaniccellexcept:(A)Duringcelloperation,reactantconcentrationandproductformationdecreases.(B)Theelectronsthatflowfromanodetocathodearegeneratedatthe

anode.(C)E°valuesarespecificforaparticularreactionanddonotvarywith

reactantconcentration.(D)Atequilibrium,thereisnonettransferofelectronsandthecellcannot

dowork.(E)Astheproduct-to-reactantratioincreases,theworkthatcanbedone

bythecelldecreases.

Questions385–387refertothefollowingredoxreactionthatoccursinalithiumionbattery.

385.Whichofthefollowingreactionsoccursattheanode?

386.Allofthefollowingareadvantagesofalithium-ionbatteryexcept:(A)Li+hasthemostnegativestandardreductionpotential.(B)Fewerthan7-gLiareneededtoprovide1-molelectrontothebattery.(C)Lithiumisapowerfulreducingagent.(D)Lithium-ionbatteriescanbechargedhundredsoftimeswithout

deterioration.(E)Lithiumisahighlyreactivemetalthatrequiresanonaqueous

electrolytesolution.

387.Featurescommontobothgalvanicandelectrolyticcellsincludewhichofthefollowing?I.OxidationattheanodeII.CanperformelectrolysisIII.Spontaneous(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)I,II,andIII

CHAPTER11NuclearChemistry

388.Thereactionaboverepresentsthebombardmentofuraniumwithaneutron.WhichofthefollowingnuclidesisrepresentedbyX?(A)14137Rb

(B)9537Rb

(C)9437Rb

(D)9337Rb

(E)9237Rb

389.ThereactionaboveshowsthesynthesisofelementX.WhichofthefollowingnuclidesisrepresentedbyX?(A)3115P

(B)3015P

(C)2815P

(D)2813Al

(E)3113Al

390.Thealphadecayrepresentedaboveproduces178Oandaproton.WhichofthefollowingnuclidesisrepresentedbyX?(A)187N

(B)147N

(C)137N

(D)189F

(E)149F

Questions391–395refertothefollowingtypesofradioactivedecay.

(D)gammaradiation(γ)

391.Hasnomass392.Leastpenetrative393.Mostpenetrative394.Carriesthegreatestpositivecharge395.IncreasestheatomicnumberwithoutchangingthemassnumberQuestions396and397refertothefollowinggraphofbindingenergypernucleon.

396.Accordingtothegraph,whichofthefollowingatomshasthemoststablenucleus?(A)H

(B)He(C)Fe(D)Au(E)U

397.Allofthefollowingaretruestatementsregardingthebindingenergyofanuclideexcept:(A)Thenucleuswiththehighestbindingenergymeansitisthenucleuswiththemostattractiveforces.(B)Energypernucleoncomparesnuclidesonacommonbasis.(C)Thebindingenergyisafunctionofattractiveversusrepulsiveforces

inthenucleus.(D)Themagnitudeoftheattractiveforceswithinthenucleusisgreater

thanthemagnitudeoflike-chargerepulsioninthenucleus.(E)Thetotalsystemhaslesspotentialenergythanthesumofitsparts.

398.Thedifferencebetweenthemassofanatomandthesumofthemassesofitsprotons,neutrons,andelectronsiscalled:(A)Nucleardecay(B)Nuclear-bindingenergy(C)Massdefect(D)Mass-energyequivalence(E)Transmutation

399.Whichofthefollowingcorrectlydescribeswhathappenstothemassthatislostwhenanucleusisformed?(A)Itisemittedfromthenucleusasanα(alpha)particle.(B)Itisemittedfromthenucleusintheformofaβ(beta)particle.(C)Itisemittedasaneutrino.(D)Itisemittedasenergy.(E)Whathappenstothelostmassisnotcurrentlyunderstood.

400.Truestatementsregardingthealkaliandnaturallyoccurringlanthanidemetalsinclude:I.Thechemicalreactivityofthealkalimetalsisduetotheirlowfirstionizationenergies.

II.Radioactivenucleicanonlybecomemorestablebyformingcompounds.

III.Thealkaliandlanthanidemetalsincreasetheirchemicalstabilitybyformingcompounds.

(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)IandIIIonly

401.Coulomb’slawstatesthatlikechargesrepel.Whichofthefollowingmostaccuratelyexplainshowtheprotonsinthenucleuscanformastablenucleus?(A)Thepositivechargesattracteachotherwhentheyarewithinavery

shortdistanceofeachother.(B)Therearenegativechargesinthenucleusthatneutralizethepositive

charges.(C)Theneutronscanceloutthechargesoftheprotons.(D)Therepulsiveforceisverysmallbecauseprotonsarespacedfarfrom

otherprotons.(E)Theneutronsformatightcagearoundtheprotonsthatkeepthemfrom

leavingthenucleus.

402.Theaveragedensityofanatomicnucleusisontheorderofmagnitudeclosesttowhichofthefollowing?(Thedensityofosmium,Os,theheaviestknownelement,is23gcm−3.ThemassofanaverageZnnucleusisapproximately1×10−22gandtheradiusofanucleusofthatmassisapproximately5×10−13cm.)(A)10−14gcm−3

(B)103gcm−3

(C)1014gcm−3

(D)1028gcm−3

(E)1063gcm−3

403.In325days,a30-gramsampleof95Zrdecayeduntilapproximately1gramof95Zrremainedinthesample.Thehalf-lifeof95Zrisclosestto:(A)11days(B)29days(C)65days(D)81days(E)162days

404.If93.75percentofasampleofpureradioisotopeXdecaysin24days,whatisthehalf-lifeofX?(A)4days(B)6days(C)12days(D)18days(E)24days

405.If12.5percentofasampleofpureradioisotopeZremainsafter30days,whatisthehalf-lifeofZ?(A)2.4days(B)8days(C)10days(D)12.5days(E)30days

406.Thehalf-lifeofradioisotopeJis2years.IftheinitialamountofJpresentis60grams,approximatelyhowmuchisexpectedtoremainafter12years?(A)30grams(B)12grams(C)5grams(D)2grams(E)1gram

407.Whichofthefollowingparticlesisemittedbyanatomof40Kwhenitdecaysintoanatomof40Ar?(A)Electron(β−)(B)Positron(β+)

(D)Gammaphoton(γ)

408.Whichofthefollowinggraphsoftheconcentrationofradioisotope(RI)remainingversustimeisconsistentwithradioactivedecaykinetics?

CHAPTER12Descriptive

409.Whichofthefollowingelementsformsacovalentnetworksolidbycombiningwithoxygen?(A)N(B)O(C)S(D)Si(E)P

Questions410–415refertothefollowingsolidcompounds.

410.Thewhite,solublesaltresponsibleforthesalinityoftheocean

411.Producesabluesolutionwhenmixedwithwater

412.Soluble,yellowsolid

413.Apurplesolidthatproducesapurplesolutionwhenmixedwithwater

414.Adeliquescentcompoundthatturnstobluewhenhydrated

415.Asolublewhitesolidusedasafertilizer

Questions416–420refertothefollowingcompounds:

416.Agaseousfuelusedforheatingandingas-burningbarbequegrills

417.Arefrigerantlinkedtothethinningoftheozonelayerinthestratosphere

418.Insolution,thiscompoundhasbeenusedasdisinfectantforminorskinwounds

419.Acolorlessgaswithafoulodor

420.Aweakacidthatcannotbestoredinglasscontainers

421.Whichofthefollowingnonmetalsisagoodconductorofelectricityinthesolidform?(A)I2(B)S6(C)C(graphite)(D)C(diamond)(E)P4(white)

422.Naturallyoccurringaminoacidscontainallofthefollowingelementsexcept:(A)Carbon(B)Nitrogen(C)Oxygen(D)Hydrogen(E)Chlorine

423.Allofthefollowingoxidesexistasagasat25°Cand1atmexcept:

Questions424–428refertothefollowingorganiccompounds.

424.Aketone

425.Verywatersoluble

426.Anesterwiththescentofbananas

427.Benzoicacid

428.Anether

CHAPTER13LaboratoryProcedure

429.Whichofthefollowingisanacceptablelaboratorypractice?(A)Placinghotobjectsonabalance(B)Dilutingasolutioninavolumetricflaskwithhotwater(C)Using5

mLofphenolphthaleintotitrate20mLofanacidicsolution(D)AddingwaterslowlytoapreweighedsolidacidinadryErlenmeyerflask(E)Rinsingaburettewithastandardizedsolutionbeforefillingitwiththestandardizedsolution430.TheproperprocedureforthedilutionofconcentratedNaOHistoslowlyaddthebasetoabeakerofwaterratherthanslowlyaddingwatertothebeakerofbase.Thisprecautionistoensurethat(A)thereissufficienttimeforthefullionizationofNaOH.

(B)thewaterdoesnotfloatontopofthedenserbaseandremainunmixed.(C)thebasedoesnotreactwiththedrybeakerorimpuritiesinthebeaker.(D)thereisaenoughwatertoabsorbtheheatreleased.(E)thereisenoughwaterfortheOH−ionstobecompletelydissolved.

431.Astudentaccidentallysplashesaconcentrated,strongacidontohisbareskin.Whichofthefollowingisthesafestandmosteffectivecourseofaction?(A)Drytheskinwithpapertowels.(B)Flushtheareawithwaterandadilutesolutionofaweakacid.(C)FlushtheareawithadiluteNaOHsolution,thenwater.(D)FlushtheareawithwaterandthenadilutesolutionofNaHCO3.

(E)SprinkletheareawithpowderedNa2SO4,thencarefullyshakeofftheexcessandrinsewithwater.

432.Astudentdissolvesa20.0molsampleofaceticacid,aweakacid,into1kgofwatertomakea20.0molalsolution.Assumingnootherinformationisavailabletothestudent,whichofthefollowingisthebestwayto

determinethemolarityofthesolution?(A)Titratewithastandardacid(B)Measurethetotalvolumeofthesolution(C)Determinethefreezing

pointofthesolution(D)MeasurethepHwithacalibratedpHmeter(E)Measuretheelectricalconductivityofthesolution433.A98-gramsampleofphosphoricacid,H3PO4(aweakacid),isdissolvedin1kgofwater.Assumingnootherinformationisavailable,whichofthefollowingproceduresisthebestwaytodeterminethemolarityofthesolution?

(A)Measurethetotalmassofthesolution(B)Measurethetotalvolumeofthesolution(C)MeasurethepHofthe

solutionwithacalibratedpHmeter(D)Determinetheboilingpointofthesolution(E)TitratethesolutionwithastandardacidQuestions434–438refertothefollowinganswerchoices:(A)Visible-lightspectrophotometry(colorimetry)(B)Paperchromatography

(C)Titration(D)Gravimetricdeterminationordifferentialprecipitation(E)Electrodes

434.Canbeusedtomeasuretheconductivityofasolution435.Usedtoseparatepigmentsinamixture436.UsedtodeterminetheconcentrationofasolutionofKMnO4

437.UsedtodetermineifNa+,Mg2+,and/orPb2+ionsarepresentinasolution438.Usedtodeterminetheunknownconcentrationofaknownreactant439.Whichofthefollowingprocedureswouldallowastudenttoboilwateratatemperaturesignificantlyabove100°Cinalaboratoryinwhichthetemperatureis22°Candthepressureis1atm?(A)Addsalttothewater(B)Stiroragitatethewaterasitisheating(C)Insulateandcoverthe

containerinwhichthewaterisheated(D)Heatthewaterinasealedcontainerintowhichairhasbeenpumpedtoincreasethepressureinthecontainer(E)Heatthewaterinasealedcontainerfromwhichalltheatmosphericairhasbeenremoved440.Whichofthefollowingmeasuresofconcentrationchangeswithtemperature?

(A)Molefraction

(B)Massfraction(C)Masspercentage(D)Molality(E)Molarity

441.Whichofthefollowingpiecesoflaboratoryglasswarewouldbeusedtomostaccuratelytransfera10.00-mLsampleofsolution?(A)5-mLpipet(B)10-mLpipet(C)10-mLgraduatedcylinder(D)15-mLgraduatedcylinder(E)15-mLErlenmeyerflask

442.Astudentdeterminedthedensityofapureliquidat25°Cbymeasuringitsmasswithanelectronicbalanceandmeasuringitsvolumeinaclean,dry50.00-mLvolumetricflask.Onthebasisofthisinformationandthemeasurementsshowninthetablebelow,tohowmanysignificantdigitsshouldthisdensitybereported?

(A)2(B)3(C)4(D)5(E)6

443.Astudentweighsout0.20molofglucosetopreparea2-Mglucosesolution.Whichofthefollowingpiecesoflaboratoryequipmentismostappropriateforpreparingthissolution?(A)100-mLgraduatedcylinder(B)1-Lgraduatedcylinder

(C)1-LErlenmyerflask(D)100-mLvolumetricflask(E)1-Lvolumetricflask

444.WhichofthefollowingtechniquesismostappropriateforrecoveringsolidNH4NO3fromanaqueoussolutionofNH4NO3?

(A)Thinlayerchromatography(B)Evaporation(C)Distillation(D)Filtration(E)Titration

CHAPTER14DataInterpretation

Question445referstothefollowingsituation.

Asmallsampleofastatinewaspurifiedfromanastatinecontainingoreinthelab.Thepureastatinewasstoredat25°Covernight.Thenextday,thesamplewasreanalyzedandfoundtohaveasignificantquantityofbismuth.Thesamplewasshippedtoanotherlabforindependentanalysis,wherethesample,sixdaysaftershipping,wasfoundtohavealmostnoastatine,alargequantityofbismuth,andasmallquantityoflead.

445.Whichofthefollowingbestaccountsforthedifferencesinanalysesofthesample?(A)Thesamplewasnotproperlypurified.(B)Theoriginalsamplewascontaminatedandsomeastatinewaslostwith

eachanalysis.(C)Theastatinetransmutedtobismuth,whichtransmutedtoleadoverthe

timecourseofanalyses.(D)Theastatinesamplewascontaminatedwithbismuthduringthefirst

analysisandthenthesamplewasfurthercontaminatedwithbismuthandleadduringthesecondanalysis.

(E)Thepureastatinesamplewascontaminatedduringthefirstanalysis,buttheanalysesbytheindependentlabweremixedupwiththoseofanunrelatedsamplefromacompletelydifferentsource.

Questions446and447refertothefollowingchoicesandthedatatablebelow.

Astudentforgottolabeltheflasksof1.0-Mconcentrationsofthefollowingsolutions:(A)Fe3+(aq)

(B)Silvernitrate(C)Bariumchloride

(D)Mercury(I)nitrate(E)Copper(II)nitrate

Toidentifyeachcompound,thestudentmixedequalvolumesofthe1.0-Msolutionsabove,listedA–E,withaconcentratedsolutionofNH3(aq).Eachofthecompoundswasnumbered1–5.

446.Compound3

447.Compound5

448.AstudentaddsaqueousNH3toasolutionofNi2+ionsandaprecipitateforms.WhenthestudentaddsexcessNH3,theprecipitatedissolvesandproducesadeepbluesolution.WhichofthefollowingbestexplainswhytheprecipitatedissolvedinexcessNH3?

(A)NH3isastrongbaseathighconcentrations.

(B)NH3onlyactsasabaseatlowconcentrations.

(C)Ni2+formsasoluble,complexionwithNH3.

(D)Ni2+solubilityincreaseswithincreasedpH.(E)Ni2+getsreducedbytheexcessNH3,formingsolubleNiatoms.

449.Inthelaboratory,whichofthefollowingcanproduceagaswhenaddedto1MHCl?I.Zn(s)II.NaHCO3(s)

III.1MNH3(aq)

(A)Ionly(B)IIIonly(C)IandIIonly(D)IandIIIonly(E)I,II,andIII

450.Asolid,whitecrystallinesubstanceisaddedtowatertoproduceabasicsolution.Whenastrongacidisaddedtothesolution,agasisliberated.Basedonthisinformation,thesolidcouldbe:(A)NaCl(B)NaOH(C)NaNO3

(D)Na2CO3

(E)Na2SO4

Questions451–453refertothefollowingexperiment.

Anexperimentanalyzedtheratiosinwhichironandoxygencombinetoformdifferentcompounds.Thefollowingdatawereobtained.

451.Thisexperimentbestdemonstrateswhichofthefollowingchemicalprinciples?(A)Conservationofmass(B)Conservationofenergy(C)Thelawofdefiniteproportions(D)Thelawofmultipleproportions(E)Thelawofstoichiometry

452.Ifnootherinformationwasavailable,whichofthefollowingcouldbe

determinedfromthedatacollected?(A)Empiricalformula(B)Equilibriumconstant(C)Reactionorder(D)Enthalpyofformation(E)Density

453.ThecompoundsthatwereformedduringtheexperimentwereI.FeOII.FeO2

III.Fe2O

IV.Fe2O3

V.Fe3O4

(A)I,II,andIV(B)I,III,andIV(C)I,III,andV(D)II,III,andV(E)III,IV,andV

Questions454–456refertothefollowingexperiment.

Astudentweighed13gramsofblueCoCl2andplaceditonawatchglassatroomtemperature.Withinafewminutes,sheobservedthatthecompoundturnedpurple.Whenshereweighedthesample,itweighed17grams.Severalminuteslater,itturnedred.Whensheweighedthesampleathirdtime,itweighed24grams.

454.TheCoCl2compoundcanbedescribedasallofthefollowingexcept:(A)Deliquescent(B)Hygroscopic(C)Desiccant(D)Efflorescent(E)Hydrophilic

455.Thecorrectformulasofthepurpleandredhydratesthestudentobservedare:(A)CoCl2.1H2OandCoCl2.3H2O

(B)CoCl2.2H2OandCoCl2.6H2O

(C)CoCl2.4H2OandCoCl2.7H2O

(D)CoCl2.4H2OandCoCl2.11H2O

(E)CoCl2.17H2OandCoCl2.24H2O

456.Ioniccompoundsthatundergoasignificantcolorchangewhenhydratedhavewhichofthefollowingproperties?(A)Theycontainametalandanonmetal.(B)Theycontainatransitionmetal.(C)Theyformonlyonekindofhydrate.(D)Theyaremoreefflorescentthandeliquescent.(E)Theycontainpolyatomicions.

Questions457and458refertothefollowingexperiment.

Astudentperformedanexperimentat760torr.Heobtainedasampleofanunknown,volatileliquid.Heplaceditina2.00-LErlenmeyerflaskandcovereditwithalidcontainingatinypinhole.Thestudentplacedtheflaskinboilingwaterof100°Cuntilalltheairintheflaskescapedthroughthepinholeandalloftheliquidwasvaporized.Hethenimmersedtheflaskincoldwatertocondensethegas.Hedriedtheflaskanddeterminedthemassofthecondensedvapor.

457.Themassofthecondensedvaporwas3.0grams.Whatisthemolarmassoftheliquid?(A)30gmol−1

(B)46gmol−1

(C)62gmol−1

(D)128gmol−1

(E)190gmol−1

458.Allofthefollowingcouldresultinthestateddeviationfromtheactualmolarmassexcept:(A)Thestudentleftsomeoftheunknownsubstanceintheliquidphasebeforeimmersinginthecoldwater,increasingtheobservedmolarmass.(B)Thecontainer’slidwasnotsecurelyattachedtotheflask,decreasing

theobservedmolarmass.(C)Thesubstancehadnotactuallyheatedupto100°Cinthewaterbathas

thestudenthadthought,decreasingtheobservedmolarmass.(D)Thestudentdidnotproperlydrytheflaskbeforeaddingtheunknown,

increasingtheobservedmolarmass.(E)Thestudentallowedsomeofthevaportoescape,decreasingthe

observedmolarmass.

459.Astudentadded10gramsofanunknown,nonvolatilesoluteto50gramsofwater.At760torr,thesolutionboiledat102°C.Ifthesolutewasknowntobeanioniccompoundwithavan’tHoffdissociationfactorof2,themolarmassofthesoluteisclosestto:(A)72gmol−1

(B)101gmol−1

(C)204gmol−1

(D)372gmol−1

(E)500gmol−1

Questions460–464refertothefollowinggraph.

Thedatabelowwereobtainedinthedeterminationofthefreezingpointofasolutionofnaphthaleneinpara-dichlorobenzene.(Assumethecompoundsdonotdissociateinthesolution.)

460.Whatisthefreezingpointofpurenaphthalene(°C)?(A)90(B)85(C)80(D)70(E)65

461.Thefreezingpoint(°C)ofthenaphthalene-para-dichlorobenzenesolutionisclosestto:(A)80(B)75(C)70(D)65(E)<65

462.Themolalityofnaphthaleneisclosesttowhichofthefollowing?(Thefreezingpointdepressionconstantforpara-dichlorobenzeneis7.1°Cm−1.)(A)0.71(B)1.4(C)2.8(D)3.6

(E)7.1

463.Asolutionispreparedbydissolving20.0gofanondissociatingsolutein100gofapuresolventwithaKf=5.0°Cm−1andafreezingpointof−25°C.Thefreezingpointofthesolutionis−37.5°C.Themolarmassofthesoluteisclosestto:(A)63gmol−1

(B)80gmol−1

(C)108gmol−1

(D)120gmol−1

(E)180gmol−1

464.ElectromageneticradiationintheformofX-rayscanbepassedthroughacrystalofapuresubstanceproducingadiffractionpatternthatcanbecapturedonphotographicpaperandanalyzedtodeterminemolecularstructure.RadiationofX-raywavelengthisusedinthisprocedurebecause(A)X-raywavelengthsaresmallenoughtopassthroughthecrystal.(B)X-rayswavelengthsaresignificantlygreaterthanthesizeoftheatoms

sotheycaneasilybereflectedoffthesurface.(C)diffractionpatternsemergefromcrystalsonlywhenthewavelengthof

theradiationiscomparableinsizetothedistancebetweentheatoms.(D)theenergyofX-raysishighenoughtobreakthecrystalapartand

scattertheatomsintoapatternonthephotographicpaper.(E)theenergyofX-raysislargeenoughtoionizethecrystalbutnotto

breakapartthebondsbetweenatoms.

465.Astudentdeterminedthepercentageofwaterinahydratetobe26percentbyweighingthesample,heatingituntildry,andthenreweighingit.Theacceptedvalueforthepercentageofwaterinthehydrateis45percent.Whichofthefollowingisthebestexplanationforthedifferencebetweenmeasuredandacceptedvalues?(A)Thestudenthadahigh-percenterror.(B)Thestudentstartedwithamassofsamplethatwastoolowto

accuratelyweigh.(C)Overheatingofthesamplecausedsomeofthesolidtospatteroutof

thecrucible.(D)Overheatingofthesamplecausedthedrysampletodecomposeintoa

gas.(E)Thedehydratedsampleabsorbedmoisturefromtheatmosphere

betweendryingandreweighing.

466.Inalaboratory,H2gascanbeproducedbyaddingwhichofthefollowingtoa1-MHClsolution?I.Mg(s)II.Zn(NO3)2(s)III.Na2CO3(s)

(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)IIandIIIonly

Questions467and468refertotheanswerchoicelistedinthetablebelowandthefollowingtitrationcurve.

467.Whichoftheaboveindicatorsisthemostappropriatechoiceforthistitration?

468.Whichoftheaboveindicatorswouldtransitioninthebufferregionofthecurve?

Question469referstothefollowinganalysis.

Acolorlesssolutionwasaliquotedinto3testtubes.Thefollowingtestswereperformed:

469.Whichofthefollowingionscouldbepresentinthesolutionata0.15-Mconcentration?(A)CO3

2–

(B)Na+

(C)Ca2+

(D)Ni2+

(E)Ba2+

Questions470–472torefertothefollowingexperimentalprocedure.

Inalaboratoryexperiment,H2gasisproducedbythefollowingreaction:

TheH2gasiscollectedoverwaterinagas-collectiontube(eudiometer).Theatmosphericpressureinthelaboratoryis770torrandthetemperatureofthelabandthewaterusedintheexperimentis22°C.Thevaporpressureofwateris19.8torrat22°C.

470.Beforemeasuringthevolumeofgascollectedinthetube,whichofthefollowingisanecessarystepincorrectlydeterminingthetotalgaspressureinsidethetube?(A)Increasethetemperatureofthewaterto25°C.(B)Waitfortheatmosphericpressureinthelabtoreach760torr.(C)Letairintothetubetobreakthevacuum.(D)Adjustthetubesothatthewaterlevelinsidethetubeisthesameas

thewaterleveloutsidethetube.(E)Liftthetubeuntilthelipisjustbarelyimmersedinthewater.

471.Whichofthefollowinggasesshouldnotbecollectedusingthistechnique?I.HClII.NH3

III.CO2

(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)IIandIIIonly

472.Thepartialpressureofhydrogengasinthetubeisclosestto:(A)730torr(B)750torr(C)760torr(D)770torr(E)Thepartialpressurecannotbedeterminedwithoutknowingthe

volumeofgascollectedQuestions473–475refertothefollowingexperiment.

Astudentdesignedaproceduretodeterminetheheatoffusionofice.Sheconstructedacalorimeterusingapolystyrenecupandathermometer.Sheweighedthecup,filleditwith150mLofwarmwater,thenweighedthecupandwatertogether.Thetemperatureofthewaterwasmeasured,andthenicetakenfromanicebathtemporarilystoredina–20°Cfreezerwasaddedtothecup.Thecupwithwaterandicewasweighed.Thestudentthencoveredthecupwithapolystyrenelidwithtwosmallopenings.Inoneopening,sheinsertedathermometerandintheother,sheinsertedastirringrodtogentlystirthecontentsofthecupuntilalltheicewasmelted.Thelowesttemperaturereachedbythewaterinthecupwasrecorded.

473.Thepurposeofweighingthecupanditscontentsthethird(last)timewasto(A)determinethemassofwaterthatwasadded.(B)determinethemassoficethatwasadded.(C)determinethemassoficeandwaterthatwereadded.(D)determinethemassofthecalorimeterandthermometer.(E)determinethemassofwaterthatevaporatedwhilethelidwasoffthe

cup.

474.Supposeasignificantamountofwaterfromtheicebathadheredtotheicecubesthatthestudentaddedtothecalorimeter.Howwouldthisaffectthevalueoftheheatoffusionoficeiscalculated?(A)Therewouldbenoeffectbecausethewaterfromthebathwouldbeat

thesametemperatureastheicecubes.(B)Thecalculatedvaluewouldbetoolargebecauselesswarmwater

neededtobecooled.(C)Thecalculatedvaluewouldbetoolargebecausemorecoldwater

neededtobeheated.(D)Thecalculatedvaluewouldbetoosmallbecausemorecoldwater

neededtobeheated.(E)Thecalculatedvaluewouldbetoosmallbecauselessicemeltedthan

wasweighed.

Thestudent’sfinaldataisrecordedbelow.

475.Whichofthefollowingpiecesofinformationisnecessarytocalculatetheheatoffusionoficefromthisdata?I.ThespecificheatofwaterII.ThespecificheatoficeIII.ThethermalconductivityofwaterIV.Thethermalexpansioncoefficientofice(A)Ionly(B)IIonly(C)IandIIonly(D)IandIIIonly(E)I,II,andIVonly

476.WhatmassofKOH(molarmass56gmol−1)isrequiredtomake250mLofa0.400MKOHsolution?(A)1.00g(B)5.60g(C)8.96g(D)14.0g(E)22.4g

477.AsteadyelectriccurrentispassedthroughmoltenNaClforexactly2hoursproducing230gofNametal.ThesamecurrentispassedthroughmoltenFeCl3forexactly2hours.ThemassofFemetalexpectedtobeproducedisclosestto:(A)56g(B)112g(C)168g(D)186g(E)224g

478.When100mLof2MPb(NO3)2ismixedwith100mLof3MNaClinabeaker,awhiteprecipitateforms.Whichofthefollowingistrueoftheconcentrationofionsremaininginsolution?

479.WhichofthefollowingwouldbeaqualitativetestforasolutionofBa2+,Fe3+,andZn2+ionsthatwouldseparateBa2+fromtheotherionsatroomtemperature?I.AddingdiluteHClII.AddingdiluteNaOHIII.AddingdiluteLi2SO4

(A)Ionly(B)IIonly(C)IIIonly(D)IIandIIIonly(E)I,II,andIII

Adischargetubefilledwithonlyhydrogengaswaselectrified.Thegasgaveoffbluelight,whichwaspolarizedandthenpassedthroughaprism.Fournarrow,coloredbandswereobservedonascreenbehindtheprism.TheenergyofaphotonisgivenbytheequationE=hν,whereh=6.63×10−34J.sandv=

thefrequency.Thefollowingdatawerecollectedduringtheexperiment.

480.Whichofthefollowingbestexplainswhyhydrogengasemittedlightwhenelectrified?(A)Forenergytobeconservedinanatom,photonsareemittedwhenan

electrondropstogroundstateafterbeingexcited.(B)Electronsabsorbedphotonsofelectricitythatprovidedtheenergy

neededforthemtobeejected.(C)Theionizedgasesproducedbytheelectriccurrentemitphotons.(D)Theelectricitycausedthegasparticlestocollidewithgreatkinetic

energy,producingphotons.(E)Theelectronsturnedintophotonswhensubjectedtoanelectricfield.

481.Wave-likepropertiesoflightincludewhichofthefollowing?I.InterferenceII.PolarizationIII.Thephotoelectriceffect(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)IIandIIIonly

482.Inanotherfamousexperiment,ametalplatewasbombardedwithphotonsofdifferentfrequencies.Atfrequenciesabove4.4×1014sec−1,electronswereejectedfromthemetal,henceionizingit.Whichofthefollowingis

closesttotheionizationenergyofthemetalinJoules?(A)1.5×10−48

(B)6.6×10−34

(C)4.4×10−20

(D)2.9×10−19

(E)3.0×10−14

483.Aphotonofredlightisproducedbyanatom.Whichofthefollowingexpressionsaccuratelycalculatesitsenergy?(A)(4.6×1014)(656)(B)(4.6×1014)(6.63×10−34)(C)(4.6×1014)(3×108)(D)(656)(6.63×10−34)(E)(656)(6.63×10−34)

484.Allofthefollowingaretruestatementsregardingatomicspectraexcept:(A)Linespectraaretypicalofelectrifiedgasesandcontinuousspectraare

producedfromtheglowofhotobjects.(B)Theelectronconfigurationoftheatomdeterminesthetypeofspectra

thatwillbeemitted.(C)Thenumberoflinesinthespectraisdirectlyproportionaltothe

numberofelectronsintheatom.(D)Photonswithlowerwavelengthsthanthoseofvisiblelightcanbecan

beemittedbyatoms.(E)Thelinesproducedinatomicspectrasupportthequantummechanical

modelofthatatomthatsaysthereareachievableenergystates.

485.Whichofthefollowingistrueregardingthepolarizationofthebluelightintheexperiment?(A)Monochromaticlight(onecolor)mustbepolarizedinordertopassit

throughaprism.(B)Thepolarizerfocusesthelightsoithitstheprisminamoreintense

band.(C)Polarizationfilterswavesoflightsothatonlywavesorientedinthe

sameplaneemergefromthepolarizer.(D)Thepolarizerisaweakprismthatpartiallyseparatesthedifferent

wavelengthsoflightsothattheprismcanseparatethemmoreeffectively.

(E)Thepolarizerabsorbswavelengthsoflightthatarenotinthevisiblerangetopreventthemfromenteringtheprism.

Questions486–488refertothefollowingexperiment.

Abeamofgaseoushydrogenatomsisemittedfromahotfurnaceandpassedthroughamagneticfieldontoadetectorscreen.Theinteractionoftheelectronofthehydrogenatomandthemagneticfieldcausesthehydrogenatomtobedeflectedfromastraightlinepath.Atiny,permanentspotdevelopswhereanatomstrikesthescreen.

486.Ifthespinoftheelectroninahydrogenatom(only1electron)wascompletelyrandom,whichofthefollowingpatternswouldbeobservedonthescreen?(A)Oneverysmall,focusedspotinthemiddleofthescreen(B)Averydarkspotinthecenterofthescreenthatbecomesmore

diffusedastheradiusincreases(C)Nodiscretespotswouldbeobserved

(D)Twospotsofequalintensity(E)Alineintheshapeofawave

487.Whichofthefollowingobservationswastheexperimentdesignedtofurtherstudy?(A)Theenergyofatomsisquantized.(B)Electronshavebothwave-andparticle-likeproperties.(C)Itisimpossibletoknowboththepositionandmomentumofan

electronwithcertainty.(D)Electronshaveachargeequalinmagnitudebutoppositeinchargeto

protons,buttheirmassis1/1,800thatofaproton.

(E)Emissionspectralinesofhydrogenandsodiumcanbesplitwhenamagneticfieldisapplied.

488.Thedataproducedinthisexperimentverifywhichofthefollowingconceptsoftheatom?(A)Electronenergiesarequantized.(B)Electronspiniseither+½or−½.(C)Electronsoccupyorbitsaroundthenucleus.(D)Electronshavepropertiesofbothwavesandparticles.(E)Themorethatisknownaboutanelectron’sposition,thelesscanbe

knownofitsmomentum(orvelocity).

489.Particle-likepropertiesoflightincludewhichofthefollowing?I.InterferenceII.PolarizationIII.Thephotoelectriceffect(A)Ionly(B)IIonly(C)IIIonly(D)IandIIonly(E)IIandIIIonly

Questions490–493refertothefollowingexperiment.

Acathoderaystrikesadetectorinastraightline,butwhenamagneticorelectricfieldisapplied,thepathoftherayisdeflected.

490.Accurateinterpretationsofthisobservationinclude:I.Cathoderayparticlesarecharged.II.Cathoderayshavebothwave-andparticle-likeproperties.III.Cathoderaysarecomposedofelectrons.(A)Ionly(B)IIonly

(C)IIIonly(D)IandIIonly(E)IandIIIonly

491.Theexperimentrevealedacharge-to-massratioof−1.76×108Cg−1.Thechargeofanindividualelectronis−1.602×10−19C.Whichofthefollowingcorrectlyexpressesthemassofanindividualelectron?(A)(−1.602×10−19)(−1.76×108)(B)(−1.602×10−19)(−1.76×108)−1

(C)(−1.76×108)(−1.602×10−19)−1

(D)(−1.602×10−19)(−1.76×108)−1(6.02×1023)(E)(−1.602×10−19)(−1.76×108)(6.02×1023)492.Whichofthefollowingparticleswouldnotbedeflectedwhenpassedthroughanelectricfield?

(A)α(alpha)particle(B)β(beta)particle(C)Proton(D)Neutron(E)Positron

Questions493–498refertothefollowingdatacollectedat20°C.

493.Theliquidwiththesteepestmeniscusinaglasstubeofsmalldiameter494.Theliquidwiththegreatestresistancetoflow495.Theliquidwiththegreatestintermolecularforcesofattraction496.Whichpairofliquidslistedbelowismiscible?(A)Benzeneandwater(B)Waterandoliveoil

(C)Castoroilandglycerol(D)Benzeneandglycerol(E)Waterandglycerol

497.Theexpectedsurfacetension(inN/m)ofcastoroilisapproximately:(A)<0.02(B)Between0.020and0.032(C)Between0.033and0.064(D)Between0.065and0.073(E)>0.074

498.Viscosityandsurfacetensiondecreasewithincreasingtemperature.Thisismostaccuratelyexplainedbythefactthat(A)liquidsexpandwhenheated,increasingthedistancebetweenthemolecules.(B)bothviscosityandsurfacetensionareprimarilydeterminedbythe

intermolecularforcesofattractionbetweenthemoleculesintheliquid.(C)densitydecreaseswithincreasingtemperature,allowingthemolecules

toflowmorefreelypasteachother.(D)increasingtemperatureincreasesreactionrates.(E)increasingtemperatureincreasestheKeqoftheequilibriumstateof

boththeviscosityandsurfacetension.

Questions499and500refertothefollowinganswerchoices:

499.Aweakacid500.Thebestelectrolyte

ANSWERS

Chapter1:AtomicTheoryandStructure

1.(B)Themassnumberofcadmiumis112,nottheatomicmass(theweightedaverageofthenaturallyoccurringisotopes).Themassnumberwillalwaysbeawholenumberbecauseitisthesumofthenumberofprotonsandneutrons(collectivelycalledthenucleons,referringtotheirlocationinthenucleus)inanatom.Thenumberofelectronsandprotonswillalwaysbethesameinaneutralatombecausetheyaretheonlynegativelyandpositivelycharged(respectively)particlesintheatom.Theatomicnumberisthenumberofprotons.Itdeterminestheidentityoftheatom,sofindingcadmiumontheperiodictablewilltellusitsatomicnumber.Ifwesubtracttheatomicnumberfromthemassnumber,wegetthenumberofneutronsinthatparticularisotope.

2.(E)LocationEontheperiodictableisinthevicinityoffrancium.Franciumhasahalf-lifeofjust22minutesandisthesecondrarestnaturallyoccurringelement(astatineistherarest).Itisdoubtfulanyonehasactuallyreactedfranciumwithwater,butifitbehavesasexpected,itwouldcertainlybeaspectacle.Ingeneral,butmoresoforthealkalimetalsatthebottomofthegroup(withthelargestatomicradiiandthelowestfirstionizationenergies),thereactionishighlyexothermic,partlybecausethereactionproducesastrongbasewhosetotaldissociationinwaterishighlyexothermic.ThereactionalsoproducesenergyintheformoflightandH2(g).Whenaddedtowater,theinteriorofapieceofsodiummetal,forexample,willmeltbeforeitisconsumedduetothehightemperatureproducedbythereaction(themeltingpointofNais~800°C).Inaddition,thehighheatignitestheflammableH2(gg)thatisproduced.

3.(B)ThelocationofBonthetableisinthehalogens,specificallyfluorine.Fluorinedoesnothavethehighestionizationenergyofalltheelements,justtheelementswe’vebeengiventochoosefrominthisquestion.Heliumistheelementwiththehighestfirstionizationenergy.Firstionizationenergyistheminimumamountofenergyrequiredtoionizeagroundstate,gaseousatombyremovinganelectron.Theresultisacation.Ingeneral,thefirstionizationenergyislowformetalsandhighfornonmetals.(SeethefigureinQuestions21

and22tocomparethefirstionizationenergiesofelements1–20.)

4.(B)ThelocationofBonthetableisinthevicinityofthehalogens,specificallyfluorine.Fluorineistheelementwiththehighestelectronegativity.Electronegativityisameasureofanatom’sabilitytoattractelectronstoitselfwhileinabond(withinamolecule).Becausethemeasurementofelectronegativityofanatomreliesontheatombeinginabond,thenoblegasesHe,Ne,andArdonothavemeasuredvaluesforelectronegativity.TheionizationenergiesofthevalenceelectronsinKr,Xe,andRnaresufficientlylow,allowingthesenoblegasestoformcovalentbondswithotheratoms(mostlythosewithahighelectronegativity,likeF,whoseelectron-attractingabilitiesarestrongenoughtoforcetheKr,Xe,orRnatomstosharetheirelectrons)andthereforehavetheirelectronegativitiesassessed.

5.(C)Electronaffinityisthemeasureoftheenergychangethatoccurswhenanelectronisaddedtoagroundstate,gaseousatom.Atomsthathaveahighaffinity,orattraction,forelectronshaveverynegativeelectronaffinities.Theperiodictrendforelectronaffinityisgenerallycorrelatedwithelectronegativity,soregionsBandCarethetopcontendersforanswerchoices.However,therearesomeimportantdifferences.Allthenoblegaseshavemeasuredelectronaffinity(thelightestthreehavenovaluesforelectronegativity,seeAnswer4)andchlorine,notfluorine,hasthehighestelectronaffinity.Electronaffinitydoesn’tchangesignificantlywithinagroup,anditcanbethoughtofasthereverseionizationenergyofanatom’s–1anion.Forexample,theenergychangetoremoveanelectronfromCl−(toproduceCl)is349kJmol−1(endothermic),andtheenergychangetoaddanelectrontoCltoproduceCl−is–349kJmol−1(exothermic).

6.(E)LocationEontheperiodictableisinthevicinityoffrancium,whichmighthavethelargestatomicradiusifitwereknown(withahalf-lifeofjust22minutes,itdoesn’texistlongenoughtodothenecessarymeasurements).Atomsdon’thaveasharp,well-definededge,sotheirbondingatomicradiusisusedtoinfertheradiusofanindividualatom.Theatomicradiusismeasuredwhiletheatomisinabondwithanotheratomofthesamekind.Thedistancebetweenthetwonucleiismeasuredandthendividedinhalf.Whatwedoknowisthatcesiumhasthelargestatomicradiusoftheelementsmeasuredsofar.(SeeAnswer10foranexplanationoftheperiodictrendregardingatomicradius.)

7.(E)Metalliccharacterisnotsomethingthatisspecificallymeasured.Itisaset

ofpropertiesgiventometals,butthepropertiesareduetooneofthemostbasicpropertiesofmetals—theirreadinesstoloseelectrons.Thisisduetometallicbonding,whichcanbethoughtofasthemost“sharing”formofbonding.Metalsarelatticesofpositivelychargedionsthatarebathinginaseaofelectrons.Theseelectronsarehighlymobileandaccountfornearlyallthepropertiesofmetals,especiallytheirelectricalconductivity.Theweakpullonthevalenceelectronsbythenucleusallowsthemtobepulledoffeasily,resultinginthelowionizationenergiesandtherelativelystrongtendencyofmetalstoloseelectronsandtakeon(almost)exclusivelypositiveoxidationstates.Metalswiththehighestmetalliccharactercanbeconsideredasthosehavingthelowestionizationenergies,thoughthisisabitofasimplification(butitwillworkfortheAPChemistryexam).

8.(C)Theatomicmassofbromineisalmost80.Thetwoisotopesareofmasses79and81.Theaverageofthesetwonumbersis80,andthatimpliesthemasseswereequallyweightedinthecalculation.Remember,different(naturallyoccurring)isotopesofatomsexistindifferentquantities.Theseareaccountedforintheatomicmasscalculationaccordingtotheirpercentnaturaloccurrence.Remember,theatomicmassistheweightedaverageofallthenaturallyoccurringisotopesofanelement.(Questions9issimilar.)

9.(C)ThethreeisotopesofSrare86,87,and88.Iftheyoccurredinequalnumbers,theiratomicmasswouldbetheaverageoftheirmassnumbers,87.Sincethemassnumberisgreaterthantheaverage,theisotopesofhighermassmustbepresentingreaterquantities.Additionally,becausetheatomicmassiscloserto88than87,wecanpredicttheoccurrenceofthe88isotopeasthehighest.Although(E)ismostlytrue—thenaturaloccurrenceofisotopes86and87are9.9percentand7.0percent,respectively,andthatcannotbeestablishedfromtheinformationgiveninthequestion.(Questions8issimilar.)

10.(D)Atomicradiusdecreasesfromlefttorightalongaperiod.Thisisduetotheshieldingeffectofthecoreelectrons(thisdoesn’tapplytoatomswithoneelectron,suchashydrogen,oraHe+ion).Alltheelementsinaparticularperiodhavethesameconfigurationofcoreelectrons.Theseelectronsshieldthevalenceelectronsfromthepullofthenucleus.TheeffectivenuclearchargeiscalculatedZeff=Z–S(Zeffistheeffectivenuclearcharge;Zistheatomicnumber,a.k.a.thenumberofpositivechargesinthenucleus;andSisthenumberofnonvalence,orcore,electrons).TheZeffforalltheatomsinaperiodgetslargerasthenumberofpositivechargesinthenucleusincreases,butthenumberof

nonvalence(core)electronsdoesn’t.Anincreasedeffectivenuclearchargemeansthevalenceelectronsfeelagreaterpullfromthenucleus,andcanthusbefoundclosertothenucleusthantheelectronsinatomswithlowervaluesofZeff.(SeeAnswer6foranexplanationofhowatomicradiusismeasured.)

11.(C)Thealkalimetalsformstrongbaseswhentheyreactwithwater,notstrongacids.(SeeAnswer2foradescriptionofthereactionofalkalimetalswithwater.)

Ageneralstrategyforexceptquestions:Theexceptquestionsaretricky,evenifwhattheyareaskingisnot.Ourbraindoesn’tthinkinthenegative,soagoodhabittogetintoistocirclethewordexceptinthequestiontoremindusthatwearelookingforafalsestatement,thentreateachanswerchoiceaseithertrueorfalse,markingeachchoiceaswego.Attheendofchoice(E),wechoosethefalseoneasouranswer.

12.(D)Thisquestionisaskingifweknowthatgroundstateelementsinthesamegrouphavesimilarproperties.Phosphorusandastatinearebothingroup15(5A)sotheirvalenceshellelectronconfigurationsareboths2p3,conferringonthemsimilarchemicalreactivities.Sulfur,selenium,andoxygenareingroup16(s2p4),whilesiliconisagroup14semimetal(metalloid)withvalenceshellelectronconfigurationofs2p2.

13.(B)Thishasasimplemathematicalsolution—taketheatomicnumber(whichwilltellusthenumberofelectronsinaneutralatom)oftheelementandaddtheabsolutevalueofthenegativeoxidationstates(moreelectrons)andsubtracttheabsolutevalueofthepositiveoxidationstates.F−(9+1)andNa+(11–1)bothhave10electronsandarethereforeisoelectronic.

Wecanalsoarriveatthisanswerbyfindingoneoftheelementsineachpairontheperiodictableandmovingoneelementtotherightforeachnegativechargeandonetotheleftforeachpositivecharge.Ifthetwoelementswearecomparingleadustothesameelementoncewe’veaccountedfortheiroxidationstate,thenthey’vegotthesamenumberofelectrons.ForNa+andF−,thiselementwouldbeneon.Thesamenumberandconfigurationofelectronsdoesnotcorrelatewithsimilarchemicalreactivityinions.F−andNa+arelikeneoninthattheyhaveafullvalenceshellandbotharemorestableandlessreactivethanintheirgroundstate,buttheyarechargedandthereforebehavelikeions.Theirionicradiusalsodiffersduetotheirdifferentnuclearcharge(seesecond

paragraphofAnswer15).

14.(D)SeeAnswer13,keepinginmindthattheiodideionhas54electrons.

15.(C)Ionicradiusisnotthesameasatomicradius(describedinAnswer6).Inanion,thenumberofelectronsdoesnotequalthenumberofprotons.Atomsbecomeionsbecausetheygainorloseelectrons(notprotons),soionsthatarepositivelychargedwillbesmallerthanwhattheyarewhenintheirgroundstate(sameeffectivenuclearchargepullingonfewerelectrons),whereasnegativelychargedionswillbelargerwhentheyareintheirgroundstate(sameeffectivenuclearchargepullingonmoreelectrons).

Forisoelectronicions(ionswiththesamenumberofelectrons),theionicradiusdecreaseswithincreasingnuclearcharge.Forexample,O2–>F−>Na+>Mg2+>Al3+.Thisisbecausethesamenumberofelectronsarebeingpulledbyanincreasingnumberofprotons.(SeeAnswer10foranexplanationofhowsizeisaffectedbyshielding.)Foratomsofthesamecharge(andinthesamegroup),thesizeofionsincreasesasyougodownthegroup.Keepinmindthationicsizeisanimportantdeterminantoflatticeenergy(seeAnswer52foradescriptionofthefactorsthataffectlatticeenergy).

16.(D)ThefirstionizationenergiesofKr,Xe,andRnaresufficientlylowtoallowthesenoblegasestoformcovalentbondswithotheratoms.Kryptondifluoride,KrF2,wasthefirstcompoundofkryptondiscovered.Xenoncanformcompoundswithoxygen(XeO3andXeO4)andfluorine(XeF4andXeF6).Radonappearstoformcompoundswithfluorine(RnF2).Noticethatoxygenandfluorinearehighlyelectronegativeatoms.ItistheirstrongelectronattractingabilitiesthatforceKr,Xe,andRnatomstosharetheirelectronsandformcovalentbonds.(SeeAnswer11foranexceptquestionstrategy.)

17.(B)Theoperativewordinthisquestionisdiatomic.Thenoblegasesaremonatomicandsoanyanswerchoicethatcontainsagroup18gasisincorrect.Withtheexceptionofastatine,whichismoremetallicthantherestoftheelementsinthegroup,allofthehalogens(group17)arediatomicintheirstandardstates,butonlyF2andCl2aregases.Br2isaliquidandI2isasolid.

18.(C)Theactualpatternofatomicsizeisnotasregularasourgeneraltrenddescribingit.Thetransitionelementspresentsomeexceptions.Forexample,theatomicradiusofthemanganesegroup(7)andthecoppergroup(11)havelarger

atomicradiithanthosetotherightandleftofthoseelements.Butforelementswithonlysandpouterelections,theatomicsizedecreasesonlyfromlefttoright.

19.(D)Ionizationenergyisanindicatorofeffectivenuclearcharge.(SeeAnswer10foranexplanationofeffectivenuclearchargeandthefigureaccompanyingQuestions21and22foragraphofionizationenergies.)Theotherchoicesareincorrectbecause(A)hasnothingtodowitheffectivenuclearcharge,(B)and(C)arefalse,and(E)statestheoppositeeffectofshieldingonionizationenergy.Lessshielding=higherionizationenergy.(Becausetheeffectivenuclearchargeislargerwithlessshielding,thenucleuspullsmorestronglyontheelectrons.Thisisevidentbecausemoreenergyisrequiredtoremovetheelectron.)

20.(D)Thetablelistssuccessiveionizationenergies,theminimumenergyrequirementsforthefurtherionizationofanelement(byremovalofsuccessiveelectrons).Fromthetable,weseethat786kJofenergypermolofsiliconisrequiredtoremovethefirstelectron(ap2electron),leavingSi+.Removinganotherelectron(thep1)requiresanadditional1,577kJforpermolSi+.Thetricktoansweringthiskindofquestionistofindaverylarge“jump”inionizationenergies.Forsilicon,it’sbetweenthefourthandfifthionizations.Thisindicatesthatthefifthionizationenergyis“digginginto”thecoreelectronsbecauseallthevalenceelectronshavebeenremoved.Nowweknowwe’rebasicallylookingforanelementwithfourvalenceelectrons.Theotherelementsingroup14willshowasimilartrendinsuccessiveionizationenergies,buttheabsolutenumberswill,ofcourse,varyfromlowerthanthoseforSi(Ge,Sn,Pb)andhigherforC.

21.(D)Elementsofatomicnumbers2,10,and18areHe,Ne,andAr,respectively.Thesenoblegaseshavethehighestfirstionizationenergies.Thelargedropinionizationenergyismainlybecausetheelementsofatomicnumber3(Li),11(Na),and19(K)haves1electronsthatarefarfromthenucleus.Itisboththislargeradiusandthelowereffectivenuclearcharge(whichisoneofthereasonsforthelargeradius)thatmaketheenergyrequirementsforremovalofthiselectronsolow.Generally,thesizeoftheatomindicatesthestrengthofthenucleus’pullontheelectrons.

Weneedtobecareful,however,asthereisanotherwaytothinkaboutthis:Theforceofanelectricfieldproducedbyachargedparticleisinverselyrelatedtothe

squareofthedistance.Inotherwords,doublethedistance,andtheforcedecreasesbyone-fourth.Theeffectivenuclearchargeisnottheonlydeterminantofanatom’ssize;forlargeratoms,wemustalsoconsiderthatthedistancebetweenanelectronandthenucleuswillsignificantlyaffecttheforceofthepullexperiencedbytheelectron.

Electronaffinityisameasureoftheenergychangethatoccurswhenanelectronisaddedtoagroundstate,gaseousatom.Atoms2,10,and18don’thaveahighelectronaffinitybecausetheyhavefullvalenceshellsandmuchenergymustbeaddedtoovercometherepulsionoftheelectronsalreadyintheatom(likechargesrepel).Atomsthathaveahighaffinity,orattraction,forelectronshaveverynegative(exothermic)electronaffinities.Theperiodictrendforelectronaffinityingenerallycorrelatedwithelectronegativity,butwithsomeimportantdifferences:(1)Allthenoblegaseshavemeasuredelectronaffinity(thoughthelightestthreehavenovaluesforelectronegativity)andchlorine,notfluorine,hasthehighestelectronaffinity.(2)Electronaffinitydoesn’tchangemuchwithinagroup.(3)Electronaffinitycanbeimaginedasthereverseionizationenergyofanatom’s–1anion.Forexample,theenergychangetoremoveanelectronfromCl−(toproduceCl)is349kJmol−1,andtheenergychangetoaddanelectrontoCltoproduceCl−is–349kJmol−1.(SeeAnswer3formoreonfirstionizationenergy.)

22.(D)StatementsIandIIarecorrect.StatementIIIisnotentirelycorrectbecausefilledorbitalsarenotnecessarilymorestable.Forexample,thedropbetweenelements7and8isduetotherepulsionofthesecondelectroninthepxorbital.Havingoneelectronineachpxorbitalismorestablethanhavingtwoelectronsinoneorbitalandoneintheother2(thepyandpzorbitals).Theinformationweneedtoanswerthequestionisallinthegraph,evenifwedon’tknowwhy.Allweneedtodoiscomparetheionizationenergywiththeelectronconfigurationusingourperiodictable.

23.(C)SeeAnswers243andforstrategiestocompareatomicandionicradii.

24.(B)Weneedtolookforelectronconfigurationsthathaveelectronsmissingfromlowerenergysubshells.Weneedtobecareful,however,becausethepeoplewhowritetheAPChemistryexamoftenlist3dbefore4s,andsoitlookslikethe4sisthehighestenergysubshellifwe’renotpayingattention.Inchoice(B),the2psubshellhasonlyfiveelectronsinsteadofsix,andyetthe3ssubshellisfilled.

Thisindicatesthata2pelectronjumpedintothe3sorbital(whichwasalreadyoccupiedbyoneelectron)byabsorbingenergy.

25.(A)Galliumisagroup13element,meaningitsonevalenceelectronisinthepxorbital.

26.(C)Carbonhastwopelectrons,eachofwhichisunpairedintheirrespectivepxandpyorbitals.

27.(C)Weneedtomakesurewedon’tconfuseenergylevels(n=1,2,3,etc.)withthes,p,d,andfsublevels(subshells).Carbonhastwopelectrons,eachofwhichisunpairedintheirrespective2pxand2pyorbitals.

28.(B)Technetiumhasnostableisotopesandistheatomoflowestatomicnumberforwhichthatistrue.NearlyallTcisproducedsynthetically.NaturallyoccurringTcisproducedbyfissioninuraniumorbyneutroncapturebymolybdenum.

29.(C)Sodium,andallthegroup1alkalimetals,arehighlyreactiveintheirgroundstate.Inparticular,theyreactwithwater(evenintheatmosphere,whichiswhyalkalimetalsarestoredinoil)toformH2gasandastrongbase(NaOHinthiscase).(SeeAnswer2foramoredetaileddescriptionaboutthereactionofthealkalimetalswithwater.)

30.(A)Helium.Becauseofthesmallradius,helium’scompleteandstablevalenceshellofelectrons,anditshigheffectivenuclearcharge,Hehasthehighestfirstionizationenergyofalltheelements.

31.(D)Wearelookingforanatomwithanincompletelowersubshellororbital.AtomD,sodium,hastwoelectronsinits3ssubshell,butonlyoneelectroninthe2ssubshell.

32.(B)Boronisanexceptiontotheoctetrule.Themnemonic“Bishappywith3”remindsusthatboroncanformcompoundswithjustthreebondsjoiningittotheotheratoms.Inthesecompounds(likeBH3),theboronatomhasnoloneelectrons,soboththeelectronandmoleculargeometryaretrigonalplanar.(SeeAnswer62formoreonboron.)

33.(E)Nitrogen.Onlycounttheelectronsinthesandporbitalsofthehighest

(butsame)energylevel(n)asvalenceelectrons.

34.(E)Nitrogen.Thestandardstateformofnitrogen,N2,makesupabout78percent(molfraction)oftheearth’satmosphere.

35.(A)Helium.Atomsformcompoundstocompletetheirvalenceshell.Allthenoblegaseshaveacompletevalenceshell,whichiswhytheyaremonatomicgasesunderstandardconditions.

36.(A)Anexcitedhydrogenatom.Wearelookingforanatomwithanincompletelowersubshellororbital.AtomAhasanelectroninits2ssubshell,butnotthe1sorbital.

37.(E)Cobalt.Thecationsoftransitionelementsproducecoloredcompoundsandsolutions,sowearelookingforelementsinwhichtheelectronsofhighestenergylevel(whichmaynotbethehighestinvalue)areindorbitals.Cobaltproducesmagentaorbluesolutionsdependingonitsoxidationstate(IIorIII,respectively).

38.(C)Neon.Anunreactiveatomisoneinwhichthevalenceshelloftheatomisfilled,inotherwords,anoblegas.

39.(D)Sodium.Anyofthegroup1metalsreactviolentlywithwateraccordingtothefollowingreaction:

Na(s)+2H2O(l)→NaOH(aq)+H2(g)

ThereactionproducesastrongbasewhosetotaldissociationinwaterisextremelyexothermicandthehightemperatureitproducesignitestheflammableH2(g)thatisproduced.(SeeAnswer2formoreonthereactionofalkalimetalswithwater.)

40.(D)Sodium.Theatomwiththehighestsecondionizationenergyistheoneinwhichremovingthefirstelectronleavesafullvalenceshellbehind.Choice(D),sodium,hasone3selectron.Removingthatelectronleavestheneonelectronconfigurationbehind(butwithahighereffectivenuclearcharge,sothesecondionizationenergyofNaishigherthanthefirstionizationenergyofneon).

41.(E)Cobalt.(SeeAnswer37.)

42.(D)Sodium.Thealkalimetalsarehighlyreactiveduetotheirlowfirstionizationenergies.

43.(B)Rutherfordbombardedathinsheetofgoldfoilwithalphaparticles(He2+,heliumnuclei).MostoftheHe2+passedstraightthroughthefoil,indicatingthattheatomsmakingupthefoilweremostlyemptyspace.SomeoftheHe2+particlesweredeflectedfromtheirpaths,butafewactuallybackscattered.ThesedeflectedandbackscatteredHe2+particlessuggestedthatthepositivechargesofanatomwereconcentratedintoasmallvolume,hencethegreatrepulsionwhentheHe2+approached.Bohrconceivedoftheenergylevelsofelectrons.Choice(C)isnottrue,andseveralscientistscontributedtothefactstatedinchoice(D).

44.(E)Withtheexceptionofastatine,whichismoremetallicthantherestoftheelementsinthegroup,allofthehalogens(group17)arediatomicintheirstandardstates.OnlyF2andCl2aregases.Br2isaliquidandI2isasolid.Astatineistherarestofthenaturallyoccurringelements,andisthereforenotusuallyconsideredwiththerestofthehalogens,sowedon’tuseastatinetomakeexceptionsforthehalogengroup.

45.(C)Metalspracticallynevertakeonnegativeoxidationstates.Nonmetalscantakeonnegativeorpositiveoxidationstates.Comparedtotheotherelements,thehalogenshavehighelectronaffinitiesandmostoftentakeona–1oxidationstate(fluorinealwaystakesona–1oxidationstate).

46.(A)Amonovalentcationisa+1cation.Thealkalimetalshave1valenceelectronandlowfirstionizationenergies.Theyexclusivelyform+1cations.

47.(B)Firstionizationenergyistheminimumamountofenergyrequiredtoionizeagroundstate,gaseousatombyremovinganelectron.Theresultisacation.Ingeneral,thefirstionizationenergyislowformetalsandhighfornonmetals.Heliumistheelementwiththehighestfirstionizationenergy.(SeethefigureaccompanyingQuestions21and22tocomparethefirstionizationenergiesofelements1–20).

48.(E)Allnaturallyoccurringatomswithatomicnumber84orabove(poloniumandhigher)areradioactive.Thenaturallyoccurringactinideshave

atomicnumbers89–92(actinium,thorium,protactinium,anduranium,respectively).

49.(B)Oxidationislossofelectrons(themnemonicOILRIGishelpful:OxidationIsLossofelectrons,ReductionIsGainofelectrons).Generally,thefirstionizationenergyofanatomindicateshowdifficultanatomistooxidize.Thenoblegaseshavethehighestfirstionizationenergiesandarethemostdifficulttooxidize,mostlybecausethevalenceelectronsinthenoblegasesexperiencethehighesteffectivenuclearchargeintheirrespectiveperiods.

50.(E)Substance1containsCa2+andSubstance2containsCu2+.Althoughweshouldbefamiliarwiththeflametest,therealcluestothequestionareinthesolutions.Thealkaliandalkaliearthmetalsproducecolorlesssolutionsandtransitionmetalsproducecoloredsolutions.It’sworthmemorizingthatCu2+producesabluesolution(usethemnemonic“Copper2inwaterblue”).

Chapter2:ChemicalBonding

51.(D)Ioniccharacterreferstoaplaceonthecontinuumofbondcharacter.Ononeendofthiscontinuumiscovalentcharacter.Thisistheperfectlyequalsharingofelectronsbetweenatoms.Thisoccursonlywhentwoofthesameatomsarebondedtogether,F2,forexample.Covalentcharacterbeginstodecreaseasthesharingoftheelectronsbecomeslessequal.Thiscanbepredictedbycomparingtheelectronegativitiesoftheatomsinvolvedinthebond.Thegreaterthedifferenceinelectronegativitybetweenthetwoatoms,thelessequalthesharingofelectronsisuntilthedifferencegetssolarge,thatsharingisnolongeranoption.Ontheoppositesideofthebondcharactercontinuumisioniccharacter.Anatomwithaverylowelectronegativity,likeCs,doesn’texertastrongpullonitsvalenceelectron,theoneinvolvedinthebond.Fluorinehasthehighestelectronegativityandpullsverystronglyontheelectronsinthebonditisapartof.Ifcesiumisnotstronglyattachedtoitselectron,fluorineishappytotakeit.SoCslosesanelectrontoF,becomingCs+andF−,respectively.Now,theoppositechargesonthesetwoionswillcausethemtoattracteachotherverystronglyinabondofhighioniccharacter.

Thebottomlineisthatthegreaterthedifferenceintheelectronegativitybetweenthetwoatoms,themoreioniccharacterthebond,orthecompound,has.Thecompoundsinchoices(B),(C),and(E)areallcovalentcompounds.Ionic

compoundsaretypically(thoughnotalways)ametal(lowelectronegativity)andanonmetal(highelectronegativity).Siliconisasemimetalandaluminumisametal.ThedifferenceinelectronegativitybetweenAlandFisabout2.4,whereasthedifferenceinelectronegativitybetweenSiandOisabout1.5.(Thisstillindicatessomeioniccharacter.Somepeopledrawthelineforanionicbondasanelectronegativitydifferenceofabout2,thoughthatnumbervariesandcanbeaslowas1.7.)However,weshouldrecognizeSiO2(silicondioxide,alsoknownasquartz)asanetworksolid,asolidinwhichalloftheatomsarecovalentlybondedinacontinuousnetwork.

52.(E)Latticeenergyistheminimumenergyrequiredtocompletelyseparatetheions(togaseousform)inanionicsolid.Ionicsthatcontainionswiththesmallestradiiandlargestchargewillhavethelargestlatticeenergy(theconfigurationoftheionsisalsoaconsideration,butwewon’tbeaskedaboutlatticeenergyatthatlevelofdetailontheAPChemistryexam).Largerchargescreateagreaterforceofattraction.Forsmallerionsrememberthattheforceoftheelectricfieldproducedbyachargedparticleisinverselyrelatedtothesquareofthedistancefromthatparticle.Ionswithsmallerradiiwillbeclosertotheionstheyareattractingandwillthusexertagreaterforceonthem,sothelatticeenergiesoftheanswerchoicesarearrangedfromhighestlatticeenergytolowest.Sincechlorineisthecommonaniontoallofthem,onlythesizeandchargeofthecationisneededtoanswerthequestion.

53.(E)Onlycovalentcompoundsuseprefixessuchasdi-andtri-.Sinceanybinarycompoundcontainingametalandanonmetalisionic,wecaneliminatechoices(A),(B),and(C).Thekeyinnamingthisionicistocorrectlynametheanion,N3–.Becauseitisananionwecaneliminatechoice(C)(again).Sincetherearenooxygenatomsinvolved,itsnamecan’tbesomethingthatendsin-ate(or-ite).The-idesuffixisageneralsuffixforamonatomicanionofanymagnitudeofcharge.

54.(C)Bondorder(BO)isthenumberofchemicalbondsbetweenapairofatoms.Itispartofmolecularorbital(MO)theory,anothermodelofbonding.(Lewis-dotstructuresandVSEPRareothersystemsusedtomodelbondingandcompoundstructures.)

Thebondorderdoesnothavetobeawholenumber.Wedon’thavetoworkoutmolecularorbitalstoansweraquestionlikethis.IfyoudonotremembertheformulaforbondorderBO=½(numberofbondingelectrons–numberofantibondingelectrons),there’samuchsimplerway.First,wedrawtheLewisstructure.Ifthere’sresonance,wewilllikelyhaveafractionalbondorder.InamoleculeofO3,thereare18valenceelectronstoaccountfor.Eitherofthestructuresbelowcouldrepresentozone.

Buttherealozonemoleculehasequalbondlengthsandstrengthsbetweentheoxygenatoms.

Wecanthinkofeachoxygenasbeingbondedtoanotheroxygenby1½bonds.Thepielectrons(fromtheunhybridizedporbitals)aresharedbetweenalltheoxygenatomsinsteadoflocalizingbetweenthetwoatomsdirectlyinvolvedwiththebond.

TheshortcutforcalculatingfractionalbondordersistotakethetotalnumberofbondsinoneoftheLewisstructures(inthiscase,three)anddivideitbytheminimumnumberofsinglebondsthatwouldconnecttheatomsinquestion(inthiscase,two).Forozone,3÷2=1.5.

55.(A)N2hasatriplebond.(SeeAnswer54foranexplanationofbondorder.)

56.(C)Theending-idetellsusthatwearenotdealingwithanoxygencontaininganions(whichendin-ateor-ite,likephosphate, ),justalonePwithwhatevernegativeoxidationstateittypicallycarries(–3).SinceCl−hasa–1charge,weknowthatXcarriesa+2oxidationstate.Thenweassignthenumericalvalueoftheoxidationstateofthecationasthesubscriptoftheanionandviceversa:

57.(B)Fordiamondtobecomegraphite,covalentbondsmustbebroken.Graphiteisanetworksolidcomposedofsheetsofcarbonstackedontopofoneanother.Withineachsheet(calledgraphene),thecarbonatomsarecovalentlybondedtothreeothercarbonatoms.Thesheetsarenotcovalentlybondedtoeachother,however.Londondispersionforceskeepthesheetstogetherinstacks.

Diamondisanetworksolid,too,butoneinwhichallthecarbonatomsarebondedtofourothercarbonatoms,soeverycarbonatomiscovalentlylinkedtoitspartners.(SeeAnswer94foracomparisonofthestructuresofdiamondandgraphite.)

Choices(A)and(C)arephasechanges,while(D)and(E)areioniccompoundsdissolvinginsolution.

58.(D)SeeAnswer57.ThesublimationofcarbondioxidediffersfromthatofgraphitebecausethereareonlyvanderWaalsforcesholdingtheindividualcarbondioxidemoleculestogetherinthesolid,andonlythoseforcesarebrokenwhenCO2sublimes.

59.(E)SeeAnswers243andfordescriptionsofdiamondstructureandbonding.

60.(C)Thenitrogeninammoniahasatetrahedralelectrongeometry.(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)Sincethereareonlythreehydrogenatomsbondedtoit,asingleunbondedpairofelectronsremainsonthenitrogenatom,makingthemoleculargeometryofammoniatrigonalpyramidal.Thisasymmetrygreatlyincreasesitspolarity.

61.(A)Compoundswithtrigonalbipyramidalandoctahedralelectrongeometrieshaveorbitalhybridsthatcontainoneortwodorbitals,respectively.Moleculeswithtrigonalbipyramidalmoleculargeometryhavefiveatomsboundtotheircentralatom.(Thoughnotallmoleculeswithfiveatomsboundtoacentralatomaretrigonalbipyramidal.BrF5,forexample,issquarepyramidal.)Whereascompoundswithoctahedralmoleculargeometryhavesixatomsboundtotheircentralatom.(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)

Phosphorusandsulfurareimportantexceptionstotheoctetrule(andanythingafterchlorineontheperiodictablemay,butdoesn’thaveto,obeytheoctetrule).Phosphoruscanhaveuptofiveelectrondomains(easytorememberbecausethe“ph”inphosphoroushasthefsound,likefive)andsulfurcanformuptosix.Boron(seeAnswer62)andhydrogenaretwootherexceptions.

62.(B)Boronisanimportantexceptiontotheoctetrule.Wecanrememberthemnemonicdevice“Bishappywith3”becauseboroncanformthreebondswithotheratomswithoutanunbondedelectronpairontheboron,somoleculeslikeBH3andBF3aretrigonalplanarinbothelectronandmoleculargeometry.(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)Boroncanformfourbonds,butitrequiresoneoftheatomstocontributebothelectronstothebondwithoutanelectroncontributionfromboron(formingacoordinatecovalentbond).(SeeAnswer334foradescriptionofcoordinatecovalentbonds.)

63.(E)TheS–Obondsinsulfurdioxidehaveabondorderof1.5.Thesulfuratomissp2hybridized,butisbondedtoonlytwootheratoms,sothethirdhybridorbitalhasalonepairofunbondedelectrons,repellingtheoxygenatoms.Rememberthatelectrondomainsfornonbondingelectronpairsexertagreaterforceonneighboringelectrondomainsthanbondedelectrondomains.(SeetheorbitalhybridizationcheatsheetbelowAnswer59,andseeAnswer54foranexplanationofbondorderandresonance.Answer84explainstheeffectofunbondedelectronpairsonbondangles.)

64.(D)Thestructureofcarbondioxideislinear:O=C=O.Eachoxygenatomhastwounbondedelectronpairs.TheC=Obondsarepolar,butthesymmetryofthemoleculemakeitnonpolar.(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)

65.(C)SeeAnswer60.

66.(C)Dipolesoccurinmoleculesduetoanonuniformdistributionofchargesinthemolecule.Typically,thisoccursbecausetheelectrondensityisnotequallysharedbetweenatoms.TheO—Hbondsinwaterareverypolar.Theelectrondensityisgreateraroundtheoxygenatomcomparedwiththehydrogenatom.Theoxygenatomalsohastwounbondedpairsofelectrons,whichmakesthewatermoleculebent.Thismakesthedipolemomentofwater1.85debye.

67.(B)Pi(π)bondsarecovalentbondsinvolvingtheoverlapofthetwolobesofanunhybridizedporbital.Theelectronoverlapoccursaboveandbelowtheplaneofthenucleiofthetwoatomsinvolved,butdoesnotoccurbetweenthetwonuclei(asinasigma,σ,bond).Onlydoubleandtriplebondsinvolveπbonds.Allbondorderscontainaσbond.Asinglebondissimplyaσbond,adoublebondconsistsofoneσbondandoneπbond,andatriplebondconsistsofoneσandtwoπbonds.Themoleculewiththegreatestnumberofπbondsistheonewiththemostdoubleandtriplebonds.C6H6,benzene,hasthreedoublebonds(althoughthebondorderoftheC—Cbondinbenzeneisreally1.5duetoresonanceanddelocalizedπelectrons,seeAnswer67foranexplanationofthebondorderandresonanceinbenzene).

68.(B)Asigma(σ)bondisoneinwhichtheregionofelectronoverlapbetweentheatomsinthebondisbetween,andinthesameplaneas,thetwonuclei.Allbondscontainoneσbond,butdoubleandtriplebondsalsocontainanadditiononeortwopi(π)bonds,respectively.(SeeAnswer67foradescriptionofπbonds.)

69.(D)Anatomthathassphybridizationwillhavetwosphybridorbitalsandtwounhybridizedporbitals.Amoleculewithansphybridizedcentralatomwillmostlikelybelinear,atleastwithrespecttothatpartofthemolecule(seetheorbitalhybridizationcheatsheetbelowAnswer59).Theunhybridizedporbitalsdon’thavetocontainelectrons,butiftheydo,theyarelikelytobeinvolvedwithpibonds.Sincewehavetwounhybridizedporbitals,wecanformtwopibonds,eithertwodoublebonds,oronetriplebond.ThecarboninCO2issphybridized,whichallowsittoformtwodoublebondswitheachoxygenatom.(Answers67and68describepiandsigmabonds,respectively.)

70.(E)CH2Oistheempiricalformulaforamonosaccharide,butitalsothemolecularformulaforformaldehyde(itssystematicnameismethanal).Becausethereisexactlyonedoublebond(thecarbonylcarbonandtheoxygen),thereisexactlyonepibond.(SeeAnswer67foradescriptionofpibonds.)

71.(D)Hydrogenfluoride(HF)hasthelargestdipolemomentofthemoleculeslistedbecausetheelectronegativitydifferencebetweentwoatoms(HandF)isthelargest.PH3isapolarcompound,butitsdipolemomentisless(0.58μ,ordebye,acoulombmeter)whereasthedipolemomentofHFisawhopping1.91μ.Thedipolemomentiscalculatedastheproductofthemagnitudeofthecharges(orpartialcharges)andthedistancebetweenthem.Valuesrangefromabout0–11μ.

72.(A)Carbonmonoxide,CO,hasatriplebond.Doubleandtriplebondscontainπbonds.Adoublebondconsistsofoneσ(sigma)andoneπbond,whereasatriplebondcontainsoneσandtwoπbonds.CO2,withitstwodoublebonds,wouldhavealsobeencorrecthaditbeenananswerchoice.(Answers67and68describepiandsigmabonds,respectively.)

73.(E)Acombustionreactionisaself-propagatingexothermicreactionthatcombinesoxygenwithasubstanceandproducesanoxideoftheelement.

74.(C)PH3(phosphane,alsocalledphosphine)hasatetrahedralelectrongeometry,butsinceonlythreehydrogenatomsarebound,thereisasingleunbondedpairofelectronsonthephosphorousatom,makingittrigonalpyramidalinmoleculargeometry.(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)

75.(B)Allsinglebondsaresigmabonds,aswellasoneofthebondsinadoublebondandoneofthebondsinatriplebond.C2H4hassixatoms,soit’sagoodplacetostartlookingforthemostsigmabonds.Infact,ithasfive:oneofeachoftheC—Hbonds(fourinall)plusoneofthetwobondsinthedoublebondbetweenthecarbons.(Answers67and68describepiandsigmabonds,respectively.)

76.(E)Allotropesarepureformsofthesameelementwithdifferentstructures.Atmosphericoxygen(O2)andozone(O3)arewellknownallotropesofoxygen,asgraphiteanddiamondarefamouscarbonallotropes.

77.(B)ThecarboninCO2isdoublebondedtoeachoftheoxygenatoms.

78.(C)Whenlookingforthecompoundwiththegreatestdipolemoment,lookforhighlypolarbonds.TheH–Obondinwateristhemostpolarbondofthecompoundslisted.TheelectronegativitydifferencebetweenH–Oislargerthan

H–N,soitsdipolemomentisgreater.(SeeAnswer71foranexplanationofhowdipolemomentsarecalculated).

79.(D)SeeAnswer60.

80.(E)TheelectrongeometryaroundthenitrogenatominNH3istetrahedral(seetheorbitalhybridizationcheatsheetbelowAnswer59),butbecausethereareonlythreeatomsboundtothenitrogen,thereremainsanunbondedpairofelectrons.Therefore,themoleculargeometryofNH3istrigonalpyramidal.(SeeAnswer84foranexplanationoftheeffectofunbondedelectronpairsonmoleculargeometry.)

81.(C)Ethaneisanalkane(asaturatedhydrocarbon)withtheformulaC2H6.Bothcarbonsaresp3hybridized.(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)

82.(E)HexeneisanalkenewiththeformulaC6H12.IthasoneC=Cdoublebond.Thecarbonsnotinvolvedinthedoublebondaresp3hybridized.Thetwocarbonsinvolvedinthedoublebondaresp2hybridized.(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)

83.(D)ButyneisanalkynewiththeformulaC4H6.IthasoneC≡Ctriplebond.Thecarbonsnotinvolvedinthetriplebondaresp3hybridized.Thetwocarbonsinvolvedinthetriplebondaresphybridized.(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)

84.(C)Electrondomainsfornonbondingelectronpairsexertagreaterforceonneighboringelectrondomainsthanbondedelectrondomains.CCl4hasbothtetrahedralelectrongeometryandmoleculargeometry(seetheorbitalhybridizationcheatsheetbelowAnswer59).Whenthemolecularandelectrongeometriesarethesame,eachelectrondomainhasanatomboundtoit.The109.5°bondangleagreesperfectlywiththeanglesrepresentingthefourcornersofatetrahedron.

ThePCl3moleculehasatetrahedralelectrongeometrybutatrigonalpyramidalmoleculargeometrybecausethereisonepairofunbondedelectronsonthecentralphosphorousatom.Thebondanglesarelessthantheyareforatetrahedral—approximately107°(althoughsomehavereportedbondanglesof

around100°,probablyduetothefurtherrepulsionbetweentheunbondedpairandthechlorineatoms).

Waterhastetrahedralelectrongeometrybutisabentmoleculebecausetherearetwounbondedelectronpairs.Thebondanglesinthebentwatermoleculeare104.5°.

85.(C)Thedifferenceinelectronegativitybetweencarbonandoxygenisabout0.9sothetwoC=ObondsinCO2arepolar.However,CO2islinearand,therefore,thedipolemomentsofeachofthebondsare180°relativetoeachother,cancellingeachotheroutandmakingCO2anonpolarmoleculeeventhoughithaspolarbonds.

Choices(A)and(B)canbeeliminatedrightawaybecausetheyaremoleculesconsistingofonlyonekindofatom,sothebondhastobenonpolarsincethereisnoelectronegativitydifferencebetweenthem.C—Hbondsarenotverypolar.Theelectronegativitydifferencebetweencarbonandhydrogenisabout0.35.Themoleculeinchoice(E),difluoromethane,hastwopolarbonds.Theelectronegativitydifferencebetweencarbonandfluorineisabout1.4,buttheelectronsarenotdistributedevenlyaroundthemolecule(ofcoursetheyaremovingbuttheystillformdipolesduetothepresenceofatomswithgreaterelectron-drawingpowerinthemolecules).ThismakesCH2F2apolarmoleculewithpolarbonds.

86.(C)Thisquestionismosteasilyansweredbyknowingtheformulaforthealkanesandthealdehydesandcarboxylicacidfunctionalgroups.Thefirstthreecompoundsareallhydrocarbons.Theformulaforthealkanes(theseriesofhydrocarbonswithonlyC—Csinglebondsinwhicheverycarboniscompletelysaturatedwithhydrogen)isCnH2n+2.Choice(C)istheonlyhydrocarbonthatfitsintotheformula.Choice(D)isacetyladehyde.Analdehydegroupisacarbondoublebondedtoanoxygen(C=O)thatoccursonthefirst(orlast)carbonofacompound.WecanrecognizeanaldehydesgroupinachemicalformulabytheappearanceofCHO.Choice(E)ispropanicacid.TheCOOHinthechemicalformulaindicatesthepresenceofacarboxylgroup(whichcontainsanoxygendoublebondedtoacarbon,andahydroxylgroupbondedtothesamecarbon).

87.(D)ThePCl5moleculeconsistsofonephosphorusatombondedtofivechlorineatomswithnounbondedelectronpairsaroundthephosphorus

(rememberthemnemonic:“PhosphoruscanformFive”bonds).Therearefivesp3dhybridorbitals(the“averaging”ofans,threeporbitals,andonedorbitalforatotaloffivehybrids).(SeetheorbitalhybridizationcheatsheetbelowAnswer59.)

88.(C)Theleastpolarbondwillcontainthetwoatomswiththemostsimilarelectronegativities,fluorineandoxygen.Thesmallertheelectronegativitydifferencebetweenthetwoatomsinthebond,themoreuniformlydistributedtheelectroncloudissharedbetweenthem.

89.(A)Ozonehasabondorderof1.5,whichmeansthatinsteadofonesingleandonedoublebond,thecentraloxygenisbondedtothetwoouteroxygenatomsbyabondofintermediatestrengthandlength.(SeeAnswer54foranexplanationofbondorderandresonanceinozone.)

90.(D)Dipolesoccurinmoleculesduetononuniformdistributionofchargesinthemolecule.Typically,thisoccursbecausetheelectrondensityisnotequallysharedbetweenatoms.TheN—Hbondsinammoniaareverypolar,astheelectrondensityisgreateraroundthenitrogenatomcomparedwiththehydrogenatom.Thenitrogenatomalsohasanunbondedpairofelectrons,whichmakestheammoniamoleculetrigonalpyramidalinshape.Thedipolemomentofammoniais1.47debye.

Chapter3:StatesofMatter

91.(A)Gold(Au)isametal,alatticeofcationsbathinginaseaofelectrons.(SeeAnswer92tocontrastmetalandioniclattices.)

92.(B)MgCl2isanioniclatticethatcontainsbothcationsandanions.Anioniclatticehasverydifferentpropertiescomparedtothelatticeinmetals.Alatticedescribesastructure.Thelatticesinioniccompoundshaveroughlythesamecharacteristics—alternatingpositiveandnegativeions—thoughtheactualpatternsofthelatticesvary.Ioniclatticesareverystrongandrigid.Theyarenotmalleableorductile,andtheyarepoorconductorsofheatandelectricitybecause,unlikemetals,thechargesinioniclatticesarenotfreetoroam.Inotherwords,theirchargesarenotmobile.Thepositiveandnegativeionsareperfectlypositionedtohavemaximumstability.Exceptfortheever-presentvibrationofatoms(attemperaturesabove0K),there’snothingmovinginthelattice.Thelatticestructureofametal,however,hasonlypositivecharges.Likeanionic

lattice,thecationsinametalarepositionedveryregularly,buttheyarenotasrigid.Metalsaremalleableandductilepartlybecausetheirelectronsarefreetoroam,particularlyinawireinwhichacurrentisapplied,butmostlybecausethemetal’scationsareabletotakenewpositionsrelativetoeachotherinthelattice(whenastressisapplied)withoutbreakingthemetallicbonds.

93.(E)Carbondioxideexistsasindividualmolecules.Thecarbonisdoublebonded(adoublebondconsistsofoneσ[sigma]andoneπ[pi]bond;seeAnswers243andfordescriptionsofpiandsigmabonds,respectively)tothecarbonatom(sphybridizationontheC;seetheorbitalhybridizationcheatsheetbelowAnswer59).CarbondioxideformsasolidonlyunderhighpressureandverylowtemperaturesmainlyduetoslightdipolesbetweenCandOatoms(butthesymmetryofthemoleculenegatesthepolarityofthebonds)andLondondispersionforces.Thesetwotypesofintermolecularforcesofattractionholdcarbondioxidemoleculestogetherinasolid.(SeethephasediagramforcarbondioxideaboveQuestions99.)

94.(D)Graphiteisanetworksolid(likequartzanddiamond)thatiscomposedofsheetsofcarbon.Withinasheet(calledgraphene),thecarbonatomsarecovalentlybondedtothreeothercarbonatoms,sothehybridizationofcarboningraphiteissp2(seetheorbitalhybridizationcheatsheetbelowAnswer59).Theelectronsintheunhybridizedporbitalsaredelocalized;theyspreadoutoverseveralcarbonatoms,creatingastructurethatcanexertfairlystrongLondondispersionforces.Thisiswhatallowsalltheindividualsheetstosticktogether.Graphiteisbrittlebecauseeventhoughthedispersionforcescreatedbythepielectronsarestrong,theyareweakrelativetocovalentandionicbonding,whichholdmostsolidstogetherunderstandardconditions.Graphiteisremarkableinthatitisanonmetalsolidthatconductselectricity(duetothedelocalizedelectrons,whicharemobile).Likeametal,graphitehasluster,butlikeanonmetal,it’snotmalleable.Itissoftandflexible,butinelastic(itdoesn’treformafterbeingdeformed).Graphiteanddiamondareallotropes,pureformsofthesameelementwithdifferentstructures.(SeeAnswer95tocomparewithdiamond.)

95.(C)Diamondisanetworksolid(likequartzandgraphite).Indiamond,allthecarbonatomsarebondedtofourothercarbonatoms,soeverycarbonatomiscovalentlylinkedtoitspartner.Thecarbonatomsaresp3hybridized(seetheorbitalhybridizationcheatsheetbelowAnswer59),sotherearenodelocalizedpielectrons.Diamondandgraphiteareallotropes,pureformsofthesame

elementwithdifferentstructures.(SeeAnswer94tocomparewithgraphite.)

96.Thenormalboilingpointisthepointatwhichaliquidturnstoagasatanatmosphericpressureof1atm.Onthisgraph,itisthepointonthelinebetweengasandliquidphasesthatwouldmeetwithalinedrawnperpendiculartothey-axisat1atm.Aliquid(inanopencontainer)boilswhenthevaporpressureabovetheliquidreachesthepressureatmosphere.Sincetheboilingpointisdeterminedinpartbytheatmosphericpressure,thenormalboilingpoint(at1atm),istheboilingpointat1atm.

97.Thesolidareaofthegraphmeetsdirectlywiththegasareaofthegraphatlowtemperaturesandpressures,totheleftofthetriplepoint.

98.(E)Typically,highpressurefavorstheformationofasolid.Thenegativeslopeofthelineinthephasediagramindicatesthatforthiscompound,adecreasedtemperatureisneededforthesolidtoformathigherpressures.Thissuggeststhatthesolidformofthiscompoundislessdensethantheliquidform.Waterisacompoundwhosesolidisalmostalwayslessdensethantheliquid,thoughitactuallydependsonthewaythecrystalsform(therearehighandveryhighdensityformsofamorphousice,butit’shighlyunlikelywe’llbeaskedaboutthemfortheAPChemistryexam).

Thegraphdoesnotsupportchoice(E)becausetheareaforsolidsextendswellbelow1atm.Thelineindicatesequilibriumbetweenthetwophases,buttheareaforsolidsshowsthatthelowerthetemperature,thelesspressureneeded.Alowertemperatureisneededforahighpressuresolidification,butalowtemperaturesolidification(say,–100°C)requiresverylittlepressure.(SeeAnswer11foranexceptquestionstrategy.)

99.(E)Ataconstantpressureof1atm,thesolidCO2sublimesdirectlyintoagaswithoutgoingthroughtheliquidphase.Toliquefy,CO2requiresapressuregreaterthan5atm.Attemperaturesbelow–56°Capproximately,CO2doesnotliquefyatall.

100.(B)SeeAnswer99.

101.(D)WeshouldimmediatelyrecognizeSiO2asanetworksolid(silicondioxide,alsoknownasquartz).Networksolids,likediamondandgraphite,havehighmeltingpoints.SiO2isnotamolecularformula,itisanempiricalformula

thatrepresentstheratioofSitoOatomsinthecompound.ThemeltingpointofSiO2is~1,600–1,725°C.H2SandC5H12arebothgasesunderstandardconditions.I2andS8aresolidsbuttheirmeltingpointsarelow,relativetoSiO2,at~114°Cand115°C,respectively.

102.(D)Thisinformationisobtaineddirectlyfromthegraph.Onekilogramofwaterabsorbedapproximately2,300kJofenergytovaporize.

103.(B)Waterisdifferentinthatthedensityoftheliquidisgreaterthanthatofthesolid(whyicefloats).Eachwatermoleculeinsolidwater(ice)isconnectedtofourotherwatermoleculesbyhydrogenbonds.Thisspreadsthemoleculesfurtherapartthanwhentheyareinaliquid,whereeachwatermoleculeishydrogenbondedtotwoorthreeotherwatermoleculesatatime.

104.(C)Duringfusion(melting),thetemperatureofwaterdoesn’tchange.Usetheformulaq=mHfus,solvingforHfus.Thereareapproximately350kJofheatabsorbedby1kgofwaterduringtheintervalwherethewaterisat0°C(thetemperature,oraveragekineticenergy,ofasubstancedoesnotchangewhileitisundergoingaphasechange,seeAnswer109foranexplanation).

105.(D)The1.0kgwaterabsorbedapproximately2,300kJtovaporize.

106.(A)Thereisatemperaturechange,soweusetheformulaq=mcΔTandsolveforc,thespecificheat∴c=q/mΔT=100kJ/(1.0kg)(50°C)=~2.0kJkg−1°C−1.

107.(E)Theheatofvaporizationismuchhigherthantheheatoffusion.Thetablebelowshowsthatanaverageof1.5molofhydrogenbondspermolofwaterarebrokenduringfusion(4molHbondspermolice–averageof2.5molHbondspermolwater=1.5molHbondsbrokenduringfusion),whereasanaverageof2.5molofhydrogenbondsarebroken(permolofwater)duringvaporization.Watermoleculesdomoveclosertogetherduringfusionanddomovefartherapartduringvaporization,butthisfactalonedoesnotexplainthedifferenceinenergyrequirementsbetweenfusionandvaporization.

108.(B)Withtheinformationinthetableabove,theenthalpyofhydrogenbondformationcanbecalculated(itwillhavethesamemagnitudebutoppositesignoftheenthalpyofbreakinghydrogenbonds).

WaterisnotformingfromH2andO2gas,itissimplychangingstate,sowecannotcalculatetheenthalpyofformationfromthedata.Superheatedsteamisnotrepresentedinthegraphsowehavenodatatouseinacalculation.Thereisnotimecomponent,sowedonotknowtherateatwhichheatisbeingadded,andthereisnovolumedatawithwhichtocalculatedensity(density=mass/volume).

109.(D)Theabsolutetemperatureofasubstanceisproportionaltotheaveragekineticenergy(KE)oftheparticlesinthesubstance.Itishelpfultoreplacethewordtemperaturewith“averagekineticenergy”whensolvingchemistryproblems.Ifthetemperatureisincreasing,theaverageKEoftheparticlesisincreasing.Duringphasechanges,thetemperatureremainsconstant.Therefore,thereisnochangeinKE.Theheataddedisusedtochange(increase)thepotentialenergy(PE)oftheparticles,whicharechangingpositionsrelativetoeachotherduringphasechanges.Thedefinitionofzeroentropyisaperfect,purecrystallinesolidat0K(it’salsothethirdlawofthermodynamics).Anydeviationfrom0Kand/orapure,perfectcrystallinesolidindicatesthatentropyisincreasing.Anincreaseintemperaturetypicallyincreasestheentropy(theunitsofwhichareJmol−1K−1).(SeeAnswer11foranexceptquestionstrategy.)

110.(A)Diethylether.Ahighvaporpressureindicatestheparticlesoftheliquidarenotstronglyattractedtoeachothersincealowtemperature(averagekineticenergy)allowsthemtoescapefromsolution.

111.(D)OnlysubstanceDandEcanformhydrogenbonds(becauseoftheirO–Hgroups)butE(methanol)hasahighervaporpressuresoitsintermolecularforcesmustbeweaker.AllmoleculesexhibitLondondispersionforces,buttheseareveryweakrelativetohydrogenbondsanddipole–dipoleattractions,sotheyarenotconsideredsignificantinpolarmolecules.ThestrengthofLondondispersionforcesincreaseswiththenumberofelectrons,whichistypicallyproportionaltomolarmass.Particleswithahighmolarmass(andthereforelotsofelectrons)willexhibitstrongerLondondispersionforces.BecausesubstanceD,ethanol,islargerthanmethanol,itexhibitsgreaterdispersionforcesandthereforehasalowervaporpressure.

112.(C)ThenonpolarcompoundsinthetableareA(diethylether),B(carbondisulfide)andC(carbontetrachloride).Avolatilecompoundreadilyevaporates,willhaveahighvaporpressure,andhaveweakintermolecularforcesofattraction.

113.(D)Ioniccompoundsthatcontainionswiththesmallestradiiandlargestchargewillhavethehighestmeltingpoints(andlargestlatticeenergies,seeAnswer52foranexplanationoflatticeenergy).Theconfigurationoftheionswithinthelatticeisalsoaconsideration,butwedon’tneedtobeconcernedwiththatlevelofdetailfortheAPChemistryexam.Largerchargesontheioncreateagreaterforceofattraction,increasingthemeltingpoint.Theforceoftheelectricfieldproducedbyachargedparticleisinverselyrelatedtothesquareofthedistancefromthatparticle,soionswithsmallerradiiwillbeclosertotheionstheyareattractingandwillthusexertagreaterforceonthem,therebyincreasingthemeltingpoint.

114.(E)Atmosphericpressureisthecolumnofairaboveaparticulararea.Thehigherthealtitude,theshorterthecolumnofair,sothelowerthepressure.Thepressuredropwithincreasingaltitudeisfairlylinearuntilabout10kmabovesealevel,afterwhichitdropsprecipitously.Therefore,thecolumnofairisdensestclosesttosealevel(sincetheweightofthecolumnofairabovetheairclosesttothegroundisgreatest,pressingalltheparticlesclosertogether).Thecolumnofairaboveanopencontainerpushesdownontheparticlesinthecontainerthataretryingtoescape.Thegreaterthepressure,thegreatertheescapevelocityrequiredbytheparticles,therefore,thehigherthetemperature(averagekineticenergy,KE)requiredforescape(KE=½mv2,wherem=massandv=velocity).Choice(A)isnotcorrectbecauseatthesametemperature,theparticlesofanytwosubstanceshavethesameaveragekineticenergy.

115.(C)SeeAnswer114.

116.(D)Carbondioxideexistsasindividualmolecules.Thecarbonisdoublebondedtoeachofthecarbonatomsinalinearmolecule.CarbondioxideformsasolidonlyunderhighpressureandverylowtemperaturesmainlyduetoLondondispersionforcesandtheslightpolarityoftheCandObond(butthesymmetryofthemoleculemostlynegatesthepolarityofthebonds).Thesetwotypesofintermolecularforcesofattractionholdcarbondioxidemoleculestogetherinasolid.Rememberthatphasechangesarephysical,notchemical,meaningthatonlyintermolecularforcesofattraction(IMFs,alsocalledvanderWaalsforces)

arebeingformedorbroken.

Althoughdeposition(andsolidification)typicallyrequiresnucleationsites,thatchoiceisnotthebestanswerbecauseitdoesn’taddressthechangesintheattractiveforces.(SeethephasediagramforcarbondioxideaboveQuestions99.)

117.(A)Phasechangesarephysicalchanges,notchemicalchanges,socovalentandionicbondsarenotbrokenorformed.OnlyvanderWaalsforces(intermolecularforcesofattraction,orIMFs)arebeingformedorbroken.Thedensityofliquidwaterisgreaterthanthatofsolidwater.Water,unlikemostsubstances,islessdenseasasolidbecausetherearefourhydrogenbondsperwatermoleculeinice(comparedtothetwotothreehydrogenbondsperwatermoleculeinliquidwater)thatcausethewatermoleculestospreadfartherapartfromeachother,formingawell-organizedcrystal.Becausefewermoleculesofwaterarepresentpervolumeofwaterinice,thedensityislower(sothesolidformfloatsinitsliquidform).(SeethephasediagramforwateraboveQuestions96,andseeAnswer11foranexceptquestionstrategy.)

118.(A)ThestrengthofLondondispersionforcescorrelateswithmolarmass,butonlybecausethenumberofelectronsiscorrelatedwithmolarmass.Heliumatomshaveonlytwoelectrons,sotheycannotformastrongtemporarydipole.Xenonatomshave54electrons,sotheycanformsignificantdipoles(atlowtemperatures).TheimportanceofthetrendinelectronnumberandLondondispersionstrengthisreflectedinthedifferentboilingpointsofHeandXe.Heliumhasthelowestboilingpointoftheelements,–269°C(amere4K),whiletheboilingpointofXeis–108°C(165K).

119.(B)Alloftheelementsarenonpolar,sotheonlyIMFtoconsiderisLondondispersion.Br2hasthehighestmolarmass(andthegreatestnumberofelectrons)ofthechoiceslisted,thereforeithasthestrongestdispersionforces.Br2isalsotheonlyliquidamonggases(understandardconditions),sothatfactalonewouldindicatethehighestboilingpoint(sincetherestboiledattemperaturesbelow25°Ciftheyaregasesatroomtemperature).(SeeAnswers243andformoreontherelationshipbetweenLondondispersionforces,molarmass,andnumberofelectrons.)

120.(E)Thequestionbasicallyisaskinguswhathappensduringmelting,whenasubstanceisatitsmeltingpointandhasalreadystartedbuthasnotyetmelted

completely.TheheatingcurveofwaterisshownaboveQuestions102.Meltingoccursat0°Candevaporationoccursat100°C.Thetemperaturedoesnotchangeduringthephasechangeseventhoughthewaterisstillabsorbingheat.Thesetemperatureplateausareduetoanincreasedpotentialenergyofthesubstancechangingphases.Temperatureistheaveragekineticenergyoftheparticlesinasubstance,soifthetemperatureisnotchanging,neitheristheaveragekineticenergy.Covalentbondsaretypicallynotbrokenduringmelting,whichisaphysical,notachemical,change.Thevolumeofasubstanceoftenincreaseswithmelting,sincethesolidformofmostsubstancesisdenserthantheliquidform.Water,however,isanimportantexception.(SeeAnswer109foracomparisonofkineticandpotentialenergychangesthatoccurduringheatingandphasechangesandAnswer117foracomparisonofthedensitiesofliquidandsolidwater.)

121.(B)Thegaswiththegreatestmasswillhavethegreatestdensitysinceatthesametemperatureandpressure,equalvolumesofgascontainthesamenumberofgasparticles.OnemolXeatSTPoccupies22.4Landhasamassof131g(131g/22.4L=5.85g/L).Helium,ontheotherhand,hasadensityof0.18g/L(4gpermole/22.4L)atSTP.Aslongaswecomparethesamevolumesatthesametemperatureandpressure,weonlyhavetocomparemolarmassestocomparedensities.

122.(E)Theabsolute(Kelvin)temperatureofasubstanceisdirectlyproportionaltotheaveragekineticenergy(KE)ofitsparticles.TheequationforKEis½mv2.Sincethemolarmassofcompoundisanintrinsicproperty,itremainsconstant.Onlythevelocityoftheparticleschangeswhenthekineticenergychanges.MassandvelocityarebothproportionaltoKE,butinverselyrelatedtoeachother.Amoremassivegaswillmoveslowerthanalighteroneatthesametemperature.ThemolarmassofN2is28gmol−1,soanygasofsimilarmolarmass(CO)willhaveasimilarvelocityunderthesameconditions.(SeetheAnswer123fortheformulatocalculatearatioofgasspeedsatthesametemperature.)

123.(A)Effusionisthediffusionofagasthroughatinyholeoropening.Thefasteragasparticlecanmove,themorequicklyiteffuses(anddiffuses).Thespeedofaparticleofgasisrelatedtoitskineticenergyandmolarmass.

Underthesameconditions,theleastmassivegas(He,inthiscase)willeffuse

thefastestandthemostmassivegas(Xe,inthiscase)willeffusetheslowest.

Absolutetemperatureisdirectlyproportionaltotheaveragekineticenergy(KE)oftheparticlesinasubstance.KE=½mv2.ForthesameKE, ,or,wecansaythatthevelocityisinverselyproportionaltothesquarerootofthemolarmass.Graham’slawofeffusionallowsustocalculatetherelativespeedsoftwodifferentgasesatthesametemperature.TocompareXeandHe:

Thatmeansthattherateofeffusionofheliumis5.7timesgreaterthanthatofxenon.(SeeAnswer237foranotherderivationofGraham’slawofeffusion.)

124.(B)SeeAnswer123.

125.(A)Helium,whichhasthefewestIMFs,willrequirehighpressuresandlowtemperaturestoliquefy.Ithasthelowestboilingpointofalltheelements,4K.(SeeAnswer118foracomparisonoftheboilingpointsofthenoblegases.)

126.(B)Sinceallthegaseslistedarenonpolar,theonlysignificantIMFtheycanformisLondondispersion.Thegaswiththegreatestnumberofelectrons(themostmassivegas)willgeneratethegreatestLondondispersionforcesunderthesameconditions.Helium,then,willhavethelowestboilingpointandrequirethemostpressuretocondenseintoaliquid.(SeeAnswer118formoreontherelationshipbetweenLondondispersionforces,molarmass,andnumberofelectrons.)Heliumhasthelowestboilingpointoftheelements,–269°Cat1atm(amere4K).

127.(B)Thehighestcondensationtemperatureisalsothehighestboilingpoint,sowearelookingforthegascapableofformingthestrongestIMFs.Apressureof10atmisveryhigh(forcomparison,theairpressureincartiresisabout2atm),buthighpressuremakesIMFsmorelikelytoform,andthereforehelpsthegastocondense.TheboilingpointofXeis–108°Cat1atm(165K).(SeeAnswer118formoreontherelationshipbetweenLondondispersionforces,molarmassandnumberofelectrons.)

128.(D)Anidealgasisanimaginarygaswhosebehaviorwithregardstotemperature,pressure,andvolumeiscompletelydescribedbytheidealgasequation,PV=nRT(itmaybehelpfultorememberthename“pivnert”).Actual

measurementsofT,P,andVofrealgasesvary(very)slightlyfromthepredictionsmadebytheidealgaslaw.Theserealgasesdeviatemainlybecauseoftheforcesofattractionandrepulsionbetweentheparticles.Lowpressuresmaintainalowdensityofthegas,soparticlesarefarenoughapartthatattractiveandrepulsiveforcesaren’tfelt,andhightemperaturesovercomeforcesofattractionandrepulsion.Withtheseforcesofattractionandrepulsionminimized,realgasesbehavequiteideally.

129.(D)Thetwogasesareatthesametemperatureandpressure,soequalvolumesofthetwogaseswillcontainthesamenumberofparticles,buttherearemoreCO2molecules(andthereforealargervolume)thanO2molecules.CO2ismoremassivethanO2,andthereareagreaternumberofparticles,sothemassesofthetwogasesarecertainlynotthesame.Thedensityoftwogasesatthesametemperatureandpressurecanbecomparedsimplybycomparingtheirmolarmasses.CO2ismoremassivethanO2,at298Kand1atm,theCO2gasisdenser.Iftheyareatthesametemperature,theiraverageKEisthesame,butCO2ismoremassiveandsotheaveragevelocityoftheparticlesislessthanthatoftheO2particles.Twogasesatthesametemperaturewillhavethesameaveragekineticenergy,butthemoremassivegaswillhave,onaverage,slowermovingparticles.(SeeAnswer123foranexplanationofKEandaveragemolecular,aswellasGraham’slaw,asimpleformulatocalculatetherelativespeedsoftwogases.)

130.(C)Twogasesatthesametemperaturewillhavethesameaveragekineticenergy,butthelargergaswillhave,onaverage,slowermovingparticles.Sincethesetwogasesareequallymassive(44gmol−1),theirparticleswillhavethesameaveragespeed.TheN2O(g)

samplehasmoremolessoithasmoreparticles,anditwilloccupyagreatervolumethanCO2atthesametemperatureandpressure.Noticedensitywasnotgivenasachoice.Gasesofthesamemolarmassatthesametemperatureandpressurehavethesamedensities.(SeerelatedQuestionandAnswer129.)

131.(D)Thisisasimpleconversion.Thirty-twogramsofO2isonemoleofO2.At1atmand298K,thisnumberofoxygenmoleculeswouldoccupy22.4L,butthepressureisfourtimes.However,temperatureandpressurearedirectlyproportional,soafourfoldincreaseinpressurewouldbeaccompaniedbyafourfoldincreaseintemperature(298×4=950).

Wewillarriveatthesameanswerusingtheidealgaslaw(PV=nRT,knownaspivnert).Or,sinceweknowthereis1.0moleofgas(andat298Kand1atm,itwilloccupy22.4L),wecanuseGay–Lussac’slaw:P1/T1=P2/T2,1atm/298K=4atm/T2∴T2=950K.RemembertoonlyusetheKelvin(absolute)temperaturescalewhendealingwithgasesbecausetherearenonegativenumbers.

132.(B)RearrangingPV=nRTforPgivesP=nRT/V.Choice(E)isincorrectbecausethenumericalvalueofthegasconstantdependsontheunits.Thenumber0.0821isusedwhenPismeasuredinatm.Thevalue8.314isusedwhenPisgiveninkPa(kilopascals).RemembertoonlyusetheKelvin(absolute)temperaturescalewhendealingwithgases(becausetherearenonegativenumbers).

133.(B)Thetotalinitialpressureinthecontaineristhesumofthepartialpressuresofallthegasespresent(Dalton’slawofpartialpressures),sothetotalPinitial=1.2+3.8=5atm.Sincethepressureisproportionalonlytonumberofparticlesatagivenvolumeandtemperature,weknowthattherearethreetimesmoreH2moleculesthanN2molecules(1.2×3=3.8).ThisratioofH2toN2isexactlythesameastheratioofcoefficientsinthebalancedequation,sothisisnotcomplicated.WhenthepartialpressureofN2fallsto0.9atm,thatmeansone-fourthoftheN2hasbeenconsumedbythereaction(0.3is25percentof1.2).IftwoNH3areformedforeveryN2moleculethatisconsumed,then0.6atm(0.3×2)ofNH3willbeformed.Theimportantthingtorememberisthatthepartialpressuresofthegasesinamixturetellsustherelativenumberofparticlesofeachgasinthemixture.

134.(E)Sulfurdioxide,SO2,hasbondsofhighpolarityandhasabentmoleculargeometry,makingitverypolarandthereforesubjecttodipole–dipoleIMFsandofcourse,Londondispersionforces.Idealgasesaretheimaginarygasesinwhichtheparticlesexperiencenoforcesofattractionorrepulsion.(SeeAnswer128foradefinitionofidealgasesandtheconditionsunderwhichrealgasesbehavemostlikeidealgases.)

135.(D)SeeAnswer128.

136.(B)Dalton’slawofpartialpressurestatesthatthepressureexertedbyaspecificgaswithinamixtureisproportionaltothemolefractionofthatgas.The

tricktoansweringthisquestionisrealizingthatiftheequalmassesofneonandargonareinthecontainer,thentwicethenumberofneonatomsarepresent(sincethemolarmassofargonistwicethatofneon).Sinceweonlyneedtobeconcernedaboutmolfractionandnootherinformationisgiven,let’ssimplyassumewehave1moleAr(40g)and2molesNe(40g).Thatmeansthere’satotalof3molofgas,ofwhichone-thirdareAratomsandtwo-thirdsareNeatoms.Ifthetotalpressureis1.2atm,one-thirdofthatpressure,or0.4atm,isduetoArand0.8atmisduetoNe.Itdoesn’tmatterwhatactualnumberofmoleweuse,onlytheratiobetweenArandNe.AslongastherearetwiceasmanyNeatomsasAratomsinourcalculations,ouranswerwillbethesame(andcorrect).

137.(E)Dalton’slawofpartialpressurestatesthatthepressureexertedbyaspecificgaswithinamixtureisproportionaltothemolefractionofthatgas.Thetotalnumberofmolesofgasis0.5+1+1=2.5mol.Thepressureexertedbyeachmoleofgas=750mmHg/2.5mol=300mmHgpermolegas.IfahalfmoleofSO2(g)ispresent,thenhalfof300mmHg,or150mmHg,ofpressureisexerted.WewillarriveatthesameanswerbycalculatingthemolefractionofSO2(g)(0.5mol/2.5mol=0.2)andmultiplyingitbythetotalpressure(0.2×750mmHg=150mmHg).

138.(D)A2-Lcontainerwillholdapproximately0.1moleofgasatSTP.ThemolarmassofO2is32gmol−1∴0.1mol=~3g.

139.(C)The2-Lflaskholdsatotalofthreemolegasesatapressureof800mmHg.Dalton’slawofpartialpressurestatesthatthepressureexertedbyaspecificgaswithinamixtureisproportionaltothemolefractionofthatgas.Thepressureofeachgascanbecalculatedasfollows.Wecaneasilycheckourworkbecausethesumofthepartialpressuresshouldequalthetotalpressure.

140.(D)ThisisaGraham’slawofeffusionproblem.Thetwogasesarereleasedatoppositesidesofatubeandiftheydiffusedatthesamerate,theywouldmeet

inthemiddle.Thespeedatwhichagasdiffusesatagiventemperatureisinverselyproportionaltothesquarerootofitsmolarmass(seeAnswer123foraderivationofGraham’slaw).ThelargermassofHClmakesitdiffusemore

slowlythanNH3bythefollowingequationrateNH3/rateHCl= diffuses1.5timesasquickly,thereforeinthesameperiodoftime,NH3willtravel1.5timesasmuchdistance.

IfdistanceHCltravels=x,thedistanceNH3travels=1.5x.

Totaldistance=1x+1.5x=2.5x=100cm∴x=40cmand1.5x=60cm.

141.(D)A2-Lcontainerwillholdapproximately0.1molegasatSTP.ThemolarmassofCl2is71gmol−1∴0.1mole=~7g(Questions138issimilar).

142.(A)Thisquestionisaskingustoidentifythegasthatisleastsolubleinwater,andthereforeifcollectedabovewater,willproduceahighyieldbecausetheleastamountwillbedissolvedintothewater(andthereforenotcollected).

143.(D)Itissometimesusefultoimaginetheatmosphereasaverythin,lightfluid.Therubberduckinabathtubfloatsbecausetheaveragedensityoftheduckislessthanthedensityofthewaterasistheaveragedensityofaluxuryoceanliner.Asubmarine,ontheotherhand,canchangeitsaveragedensitytosinkorfloat.

144.(D)Thetermconstanttemperatureisourcluethattheaveragekineticenergyoftheparticleswillbethesame.Althoughthespeedoftheparticlesisrelatedtotheirkineticenergy,soistheirmass.Atthesametemperature,moremassivegaseswillmovewithslowerspeed.Sincealltheseparticlesarethesame,theirspeedremainsthesameatthesametemperature.

145.(A)Thespeedatwhichagasdiffuses(oreffusesthroughatinyhole)isinverselyproportionaltothesquarerootofitsmolarmass(seeAnswer123foraderivationofGraham’slaw).Therefore,Arwilleffuseoutofthecontainerthefastest,leavingtheleastnumberofArparticlesbehind,andthereforeexertingtheleastpartialpressure.KrisofintermediatemassbetweenArandXeandsowilleffuseatanintermediatespeed.Krwilleffusetheslowestleavingthegreatestnumberofparticlesinthecontainerandthereforehavingthehighestpartialpressure.

146.(B)Thespeedatwhichagasdiffuses(oreffusesthroughatinyhole)isinverselyproportionaltothesquarerootofitsmolarmass(seeAnswer123foraderivationofGraham’slaw).ThelightestgasinthelistisH2,soatthesametemperature(averagekineticenergy),H2willmovethefastest.

147.(A)Thesodiumcarbonatereactswithhydrochloricacidaccordingtotheequation:

Na2CO3(aq)+2HCl(aq)→2NaCl(aq)+CO2(g)+H2O(l)

0.250LHCl×2.5mol/L=0.63moleHCl

10.6gNa2CO3×1mol/106g=0.10moleNa2CO3

Remembertocheckforlimitingreagentsiftheamountsoftworeactantsaregiven.Asimplecheckistotakethenumberofmolesofeachreactantanddivideitbyitsstoichiometriccoefficientinthebalancedequation.

0.63moleHCl÷2÷0.325

0.10moleNa2CO3÷1=0.10(limitingreactant)

Sincethereisa1:1ratiobetweenNa2CO3consumptionandCO2formation,0.11moleCO2isthetheoreticalyield.

148.(A)Carbondioxide,CO2(g),isrelativelynonpolar,sonotagreatdealofitdissolves,butasmallfractiondoes(fromabout3.5gCO2perkgwaterat0°Cto0.5gCO2perkgwaterat60°Catsealevel).Importantly,CO2reactswithwatertoformcarbonicacid,aweakacid,accordingtotheequation.

(Ourlivesdependonthisreaction.CO2istransportedinthebloodmainlyasHCO3

−,andtheregulationofourbreathingreliesonit.Thereactionoccursinredbloodcellswiththehelpoftheenzymecarbonicanhydrase.)

149.(D)Anygascollectedoverwaterwillbeamixtureofatleasttwogases,thegas(orgases)producedinthereactionandwatervaporfromtheeudiometer.

JustrememberDalton’slawofpartialpressures,thetotalpressureofamixtureofgasesisthesumofthepartialpressuresofitsconstituentgases.Thepartialpressureofeachgasisdeterminedsolelybyitsmolefractioninthemixture.

Becausewedon’tknowthemolefractionofeithergas,wedetermineitbythepressure.Westartwiththepartialpressureofthewaterandweuseahandyfact:Thevaporpressureofwaterisdeterminedsolelybythetemperature.Ifthevaporpressureofwaterat22°Cisabout20mmHg(thisinformationwouldbeprovided),wecanthendeducethepressureofthegasinourmixture.

Thetotalpressureofthegasintheeudiometeristheatmosphericpressureinthelab,whichwasgivenas:

760–20=740mmHgofCO2(g).Thenweusetheidealgaslaw,PV=nRT,tocalculatethenumberofmolesofCO2.

150.(D)Firststartbycalculatingthenumberofmolesofeachgas:

1.6gHe=0.4mole,4gAr=0.1mole,26gXe=0.2mole

Totalnumberofmoles=0.7

P=2.1atm∴0.3atmper0.1moleofgas

0.2moleXe∴0.6atmofpressure

WecanalsofindthemolefractionofXeandmultiplyingitbythetotalpressure:

0.2molXe/0.7totalmol=0.29=~0.3

0.3×2.1atm=0.63=~0.6atm

Chapter4:Solutions

151.(A)Raoult’slawstatesthatthevaporpressureofasolutionofanonvolatilesoluteisequaltoproductofthevaporpressureofthepuresolventanditsmolefraction.Asolutionwillhavealowervaporpressurethanthepuresolventandthereforeahigherboilingpointbecausealiquidorsolutionboilswhenthevaporpressureabovetheliquidreachesthepressureoftheatmosphere.Soluteswill

alsodepressthefreezingpointandincreasetheosmoticpressureindirectproportiontotheirconcentration.(SeeAnswer459fortheformulatocalculatetheDTofboilingpointsorfreezingpointsofsolutions.)

152.(A)Mostsubstancesundergoachangeindensity(andvolume)withachangeintemperature.Mass,however,doesn’tchangewithtemperature.Molality(m)istheconcentrationexpressedinmoleperkilogramsolvent,aunitofmass.Molarity(M)isaunitofconcentrationexpressedinmoleperliter,avolume.

153.(C)WeneedtoconvertmLtoLtosolvefornumberofmolefrommolarity,mol/L.

0.125×(0.2mol/L)=0.025molCuSO4.5H2O.

0.025mol×(250g/mol)=6.25g

154.(B)Thisisadilutionproblem.UseM1V1=M2V2

MHClVHCl=M0.8MsolV0.8Msol

(20)(5)=(0.8)(V0.8Msol)

V0.8Msol=125mL

Butthisisthefinalvolumeofthesolution.Thequestionaskedhowmuchdistilledwatermustbeadded.Wemustsubtractthe20mLof5MHClfromthefinalvolumetocalculatethevolumeofwateradded:

125mlFinalSol–20mLHCl=105mLwater

155.(E)Theleadnitrateandsodiumchloridereactaccordingtotheequation:

Pb(NO3)2+2NaCl→PbCl2(s)+2NaNO3(aq)

0.100LPb(NO3)2×(0.2mol/L)=0.02molPb2

0.100LNaCl(0.3mol/L)=0.03molmolCl−

SincetwoCl−areneededforeachPb2+,Cl−isourlimitingreactant(rememberthesimplecheck:dividethenumberofmoleofeachreactantbyitsstoichiometriccoefficientinthebalancedequation,thesmallestquotientisthelimitingreactant)andwe’llhaveexcessPb2+ionsinsolution.

0.03moleCl−×(1molPb2+/2molCl)=0.015moleofPb2+,ionsprecipitatedoutwiththechlorideions

0.02molePb+2–0.015molePb+2precipitated=0.005molePb2+

0.005molePb2+inafinalvolume(don’tforgettoaddthevolumesofbothsolutions)of200mL(or0.2L)=0.025M.

156.(E)Thetwosolutionswerepreparedwiththesamenumberofmolesoftheirrespectivecompounds,sothedeterminantofelectricalconductivityistheconcentrationofionsoftheirsolutions.Aparticlethatdoesnotdissociateintoionswhendissolveddoesnotconductelectricity.Thehighertheionconcentrationofasolution,thegreateritselectricalconductivity.Ions(orcharges)mustbemobileinordertoconductelectricity,sosolidsaltsdonotconductelectricity(butmoltensaltsdo).Solublesaltsandstrongacidsandbasesarethebestelectrolytesbecausetheycompletelydissociateinwater,producingatleast2moleionspermolecompound.

157.(C)Sincewe’regivenapercentandnotanabsolutemass,assumea100gsampleofsolution∴66%C2H4O66gC2H4O=1.5moles.

100gofsample–66gC2H4O34gofwater,orabout1.9(roundto2)molewater.Totalmoles=3∴1.5molethanol/3moltotal50%.

158.(A)Thereare2molesofethanolinatotalof10molsolution(20percent).

144gH2O@(18gmol−1)=8moles

92gethanol@(46gmol−1)=2moles

159.(D)29gNaCl@(58gmol−1)=0.5moleNaCl

0.5molNaCl/0.2kgsolvent=2.5msolution

160.(D)Phosphatesarenotparticularlysoluble.Allnitrateandammoniumsaltsaresoluble,aswellasallgroup1salts.Acetateisalsoverysoluble.

161.(C)TheadditionofHF,aweakacid,willlowerthepHbyincreasingtheH+

concentrationinthesolution.TheincreasedH+concentrationwillshifttheequilibriumoftheBaF2tofavoritsdissociation(andthereforeitssolubility)byprovidingtheF−alreadyinthesolutionwithmoreH+ionstobindtoformHF.F−isagoodconjugatebase,soitwilltakeupmanyoftheH+ions.TheH+andF−ionsintheacidarealreadyatequilibrium,soaddingthemtotheBaF2solutionwillnotincreasetheF−concentrationwhenitisadded,asmuchasitwillincreasetheH+concentration.ThisisbecausetheF−thatisbeingaddedwillbondtotheH+ionstoformHF,furtherloweringtheF−concentration,andfurtherpullingthesolubilityequilibriumtotheright,favoringdissociationandincreasedsolubility.Bythetimethetwosolutionshavecompletelymixedandanewequilibriumisachieved,thefinalF−concentrationislower,allowingmoreBa2+ionstobedissolved.

162.(D)Theformationofhydrogenbondsisexothermic.

163.(A)Assume100mLofeachliquid.

100mLethano×0.79g/mL=79gethanol@46gmol−1=1.7moles

100mLwater×1.0g/mL=100gH2O@18gmol−1=5.5moles

Totalmolsolution=1.7+5.5=7.2moles

1.7molethanol/7.2moltotal=0.24

164.(A)Differentproportionsofsoluteandsolventcanproducedifferententhalpychanges,butthesolvationofethanolandwaterisunusualinthatitstartsoutexothermicatlowconcentrationsofwater,changestoendothermicinthemid-range,andthenrevertsbacktoexothermicathighconcentrationsofwater.Solutionsarecomplex,butwecanstillarriveatafairlysimplebutlogicalinterpretationofthissituation.Ifaparticularsolvationprocessisexothermic,thenmore(or“stronger”)intermolecularforcesofattraction(IMFs)areformedthanbroken(IMFformationisexothermic).Ifitisendothermic,thenmore(orstronger)IMFsarebrokenthanformed(IMF“breakage”isendothermic).

165.(D)AnyparticlewithelectronswillexhibitsomeamountofLondondispersionforces,soouranswermustcontainchoiceIII.Thenameethanolletsusknowthecompoundisanalcohol,butthe—OHgroupisobviousfromthechemicalformulagivenatthebeginningofthequestionset.Water,ofcourse,hastwoO–Hbonds.MoleculeswithO—H,N—H,andF—Hbondscanformhydrogenbonds.Hydrogenbondingisthestrongestoftheintermolecularforcesofattraction(technically,hydrogenbondingisaspecialcaseofdipole–dipoleattraction).TheC—Hbondsinethanolarefairlynonpolar,sotherearenodipole–dipoleforcesholdingthesetwomoleculestogether.

166.(C)First,weneedtodeterminethenumberofmolesofCl−ionsinthesolution.Thereare0.1molefromKCl,0.2molefromCaCl2,and0.3molefromAlCl3∴0.6moleCl−in1-Lsolution.BecauseeachPb2+ioncanprecipitate2Cl−ions,0.6÷2=0.3,theminimumnumberofmolesofPb2+ionsneeded.

167.(B)Thesolubilityofgasesinwaterisgreatestathighpressuresandlowtemperatures.

168.(C)Watersolublesaltsarenotsolubleinnonpolarsolvents,likeCCl4.Saltsdissolvebydissociating.Theirionsforminteractionswithsolventmoleculesthathavepolargroupsonthem(ion-moleculeattractions).ANa+,forexample,willbeattractedtothepartiallynegativelychargedoxygenatominanO—Hbond(ofwater,forexample),whereastheCl−ionwillbeattractedtothepartiallypositivelychargedhydrogenatomoftheO—Hbond.

169.(C)Wearelookingforthecompoundwhosesolubilityisdrasticallyreducedbyareductionintemperature.Ifthisdatawerepresentedinagraph,we’dbelookingforthesolubilitycurvewiththesteepestslopewithinthetemperaturesgiveninthequestion.

170.(B)TheAg+isnotsolubleasachlorideorasulfate,butitissolubleasanitrate.

171.(A)ThechloridefromNaClformsaninsolubleprecipitatewithAg+(AgCl).

172.(E)First,wecalculatethenumberofmolesofNaOH(andOH−):

60mL×0.4M=24,buttheunitMismolL−1,soweneedtomovethedecimaloverthreeplacestoaccountfor1,000mLperliter∴0.024molNaOH(or,converttoLbeforedoingthecalculation)∴0.024moleOH−(onlyoneOH−perNaOH).

NextwecalculatetheBa(OH)2(andOH−):40×0.6M=0.024molBaOH2∴0.048moleOH−(2molesOH−permoleBaOH2).

Thereareatotalof0.072molesOH−(0.024+0.048)in100mL(0.1L,addthevolumes)sothemolarityis0.72M.

WecanuseavariationonM1V1=M2V2tosolvethisproblem,butweneedtorememberthatwhenmixingsolutions,wemustaddtheirvolumes.

M1V1+M2V2=M3V3

(0.06)(0.4)+2(0.04)(0.6)=M3(0.1)

173.(C)Thefirstsolution,amixtureofCuCl2andMgSO4,istotallysoluble.ThatnotonlytellsusthatCuCl2andMgSO4aresoluble,buttheproductsofthedoublereplacementreactionbetweenthemarealsosoluble.ThatwouldbeCuSO4andMgCl2.

WhenthestudentthenmixesAl2(SO4)3andCuF2,however,aprecipitateforms.WeassumeAl2(SO4)3andCuF2aresolublebecausetheywerecombinedassolutions,butoneoftheproductsofthedoublereplacementisobviouslynotsoluble.AlF3andCuSO4aretheproducts.Wemayhaveoursolubilityrulesmemorized,butforthisparticularquestion,wedon’tneedthem.ThefirstsolutionmadebythestudentalreadytellsusthatCuSO4issoluble,sowemustconcludeit’stheAlF3.

Theimportantthingtorememberforthisquestionisaboutmakingsurethechargesonourionsarecorrectsowecanpredicttheformulasofthenewcompounds.Theoxidationstateofcopperisobvious,allthealkaliearthmetalstakeona+2oxidationstate,andallthehalogenstakeona–1oxidationstate,butwe’dhavetohavememorizedthatAlalwaystakesona+3oxidationstate,andthatthesulfateion,SO4

2–,hasa–2charge.

AtrickforrememberingtheoxidationstatesofAg,Zn,andAl:theyareconnecteddiagonallyontheperiodictable,andasweprogressfromAgtoZntoAl,theoxidationstatesare+1,+2,and+3.Manytransitionmetalstakeonmorethanoneoxidationstate,butAgandZnaretwoimportantexceptions.SomuchsothattheparenthesesthatindicatetheoxidationstatesoftransitionmetalsinioniccompoundsarenotusedforcompoundswithAgandZn.

174.(D)Idealsolutionsarelikeidealgases,theydon’texist.However,theyareusefulimaginarymodelsforpredictingthepropertiesofrealsolutions(andgases).Anidealsolutionisoneinwhicheachoftheparticlesinsolutionissubjecttothesameforcesitwouldbeinitspurestate.Inotherwords,thedifferentmoleculespresentinanidealsolutionhavenogreaterorlesserattractionsfortheothermoleculesinthesolutionastheydofortheirownkind.Therearesomeassumptionsaboutidealsolutionsweshouldbefamiliarwith:(1)Thevolumeofthesolutionispurelythesumofthevolumescombined.Thereisnoexpansionorcontractionuponmixing.(2)Theheatofsolutioniszero.It’sneitherexothermicnorendothermic.(3)AllthecomponentsofthesolutionobeyRaoult’slaw(thevaporpressureofeachcomponentofthesolutionisproportionaltothemolefractionofthatcomponent).

Thevaporpressureaboveanidealsolutionisideal,too—thetotalpressureisthesumofthepartialpressuresoftheconstituentgases(theyobeyDalton’slawofpartialpressures).

Thepairsofliquidsinchoices(C)and(E)arenotmisciblesowecaneliminatethemimmediately.Choice(A)isincorrectbecauseHClhasafairlyexothermicdissolution(violatesassumptionNo.2)anditalsodissociatesintoions,whichformion-moleculeattractionswithwaterinsteadofhydrogenbonds(violatesassumptionNo.1).Finally,HClisagasunderstandardconditionsandthisisfairlyvolatilewheninasolution.Choice(B)istemptingbecauseitisanalcoholthatcanhydrogenbondwithwater,howevertheCH3CH2–groupisnonpolarandwouldhaveagreaterattractionforotherCH3CH2OHmoleculesthanitwouldforwater.It’satoughchoice,butoverall,thepairsofliquidsinchoice(D)aremoresimilarthanthepairsofliquidsinchoice(B).

175.(D)Themostdirectandefficientmethodtodeterminethemolarityofthesolutionistomeasurethemassandvolumeofasampleofthesolution(or,thewholesolution).Weneedtwopiecesofinformationtogettheunitofmolarity(M),molandL.Ifasolutioncontains10percenthexanebymass,let’sassume

wehavea100gsample.That’s10g,or0.12mole,ofhexane.Thevolumeofthesamplewillthenallowustocalculatemolarity.Tomakesureweansweraquestionlikethiscorrectly,wecantryimaginingthesituation.We’llfindwiththisproblemthatwemustknowthemassofthesampletocalculatethenumberofmolesofsolute.Oncewe’vebroughtintheassumptionofa100-gramsample,we’veadmittedthatamassisneeded.

176.(A)Thetrickinthisquestionisintheunitsofglucose—milligrams.Theobviousmistakeisassumingwehave2molesofglucose:360/180,butwereallyhave2/1,000moleglucose,or0.002mole,dissolvedin200mLofwater,or0.2L.Thefinalmolarityofthesolutionis0.002/0.2=0.01Msolution.Becauseglucoseisverysolubleandthesolutionishomogenous,anysizedsamplewillhavethesameconcentration.

177.(C)Thesolutionwiththehighestboilingpointwillbetheonewiththegreatestnumberofparticles.Asimplecalculationistomultiply(m)×(i)(i=thevanHofffactor,theratioofthenumberofparticlesacompoundproduceswhendissolvedversusthenumberofparticlesofcompoundadded).

178.(C)Tosolvethisproblem,wemustrememberRaoult’slaw:Thevaporpressureofanidealsolution(seeAnswer174foranexplanationofidealsolution)isdependentonthevaporpressureofeachcomponentandthemolefractionoftheeachcomponent.Having2molespropyleneglycoland8molesofwatermeansthatoutofthe10molesoftotalsolution,20percentispropyleneglycoland80percentiswater.Theeffectofthesolute(propyleneglycol)istolowerthevaporpressureofthewaterbecausethesoluteisnonvolatile,thatis,itdoesn’treadilyvaporize.Ifthesolutionis20percentnonvolatilesolute,thenthevaporpressureofthesolutionwillbe20percentlowerthanthevaporpressureofpurewater.20percentof20mmHgis4∴20–4=16mmHg.

179.(C)TheprecipitatethatformedaftertheadditionofHClindicatesthatthe

solutioncontainedanionthatwasinsolubleasachloride.Ag+,Pb2+,andHg2+immediatelyspringtomind.Outofthose,onlyAg+formsasolublecomplexionwithNH3,Ag(NH3)2+.Theionthatremainedinsolutionmusthavebeensolubleinchloridebutinsolubleasasulfate.Outoftheionsthatarenotsolubleasasulfate,Ba2+,Ca2+(notananswerchoice),Pb2+,andAg+,Ag+,andPb2+arenotsolubleaschloridesandwouldhavealreadyprecipitated(asAg+did).

180.(B)Allammoniumandnitratesaltsaresoluble,asareallthegroup1salts.ThatleavesBaCO3astheinsolublecompound.

Chapter5:ChemicalReactions

181.(E)H2Ois18gmol−1,so180gwateris10molesofwater,oneorderofmagnitudegreaterthan1mole(6.02×1023molecules)∴6.02×1024molecules.

182.(B)Carbondioxide(CO2)hasamolarmassof44gmol−1,so4.4g=0.1mole(6.02×1022CO2molecules).EachCO2moleculecontainstwooxygenatoms,sothereare0.2moleoxygenatomsor(0.2)×(6.02×1023)=1.2×1023atomsofoxygen.

183.(A)Ribosehasamolarmassof150gmol−1,so1.5g=0.01moleribose(6.02×1023ribosemolecules).Eachribosehas10hydrogenatoms,sothereare0.01×10=0.1molehydrogenatoms(6.02×1022hydrogenatoms),oneorderofmagnitudelessthan1mole(6.02×1023molecules).

184.(E)ThecompoundisTi(CO)6,titaniumhexacarbonyl.

Assumea100-gsampletoconvertpercenttonumberofgrams.Thenconvertgramstomolesandfindthesimplest,wholenumbermoleratiobetweenthem.

22.2gTi@48gmol−1=~0.5molesTi÷0.5=1

33.3gC@12gmol−1=~3moles÷0.5=6

44.4gO@16gmol−1=~3moles÷0.5=6

Remembertousethemassofatomicoxygen(16gmol−1)whencalculatingits

percentinacompound.

185.(A)Assumea100gsample,converttomolesandfindthesimplest,wholenumbermoleratiobetweenthem.

92gC×(12gmol−1)=~8moles

8gH×(1gmol−1)=~8moles

Thereisa1:1ratioofcarbon-to-hydrogenatoms,soCHistheempiricalformula.

186.(C)Themolecularformulaofacompoundisawholenumbermultipleofitsempiricalformula.Ifwecalculatethemolarmassofeachoftheanswerchoices,CHO=29gmol−1,C2H3O=43gmol−1,CH2O=30gmol−1,andCH2O2=46gmol−1.CH2O,at30gmol−1,istheonlychoicethat150gmol−1

candivideintowithoutaremainder(5x,itisalsotheempiricalformulaforthemonosaccharides,aclassofcarbohydrates).Choice(E)isnotanempiricalformula,itisthemolecularformulaofribose(CH2O×5=C5H10O5).

187.(E)We’realreadygiventhenumberofmoleofeachelement,soweonlyneedtofindthesimplest,wholenumberratiobetweenthem.

0.2molePd÷0.2=1

0.8mole÷0.2=4

1.2molesH÷0.2=6

0.8moleO÷0.2=4∴PdC4H6O4,butwecanfactor2fromthesubscriptsontheC,H,andO(thenonmetalswhichareformingacomplexaroundthePdcation)toPd(C2H3O2)2,palladiumacetate.TheformulaPd(C2H3O2)2ismuchmorestructurallyinformativethanPdC4H6O4becauseithighlightsthefactthattherearetwoacetateionsattachedtoeachPd2+ion.

188.(D)Assumea100-gsample,converttomoleandfindthesimplest,wholenumbermoleratiobetweenthem.

38gF@19gmol−1=2moles÷0.5=4

62gXe@131gmol−1=0.5mole÷0.5=1∴XeF4

189.(B)Ifahydrocarbonis75percentcarbon,itmustbe25percenthydrogen(100%–75%).Assumea100-gsample,converttomoleandfindthesimplest,wholenumbermolarratiobetweenthem.

75gC@12gmol−1=6.25moles÷6.25=1

25gH@1gmol−1=25moles÷6.25=4∴CH4

190.(B)TheexcessH2tellsusthatthisisnotalimitingreactantproblem.ThemolarmassofCu2Ois143gmol−1butthatpieceofinformationisnotneededtosolvetheproblem.Itisaredherring.Whatwewanttoknowisthat0.05molCu2Ohas0.1molCu,sincethereare2molesCuatomspermoleofCu2O.ThemolarmassofCuis63.5gmol−1,so0.1moleweighs6.35g.

191.(D)Acoordinationcomplexconsistsofanatomorionsurroundedbyanarrayofionsormolecules(calledligands).Thecentralatomorionistypicallyametal(mostoftenatransitionmetal)andthecomplexitcreatesisoftencharged,whichisaclueforidentifyingthecomplex.Inchoice(D),Ptisthecentralmetalandthechlorideionsareitsligands.

192.(E)Chlorinehasanoxidationstateof0inCl2,–1inCl−and+5inClO−.

193.(A)Anacidandabasearethereactantsofaneutralizationreaction.Asaltandwateraretheproducts.

194.(B)Aprecipitationreactionproducesasolid(ifthestateofmatteroftheproductsisnotspecified,lookforaninsolublecompound).

195.(C)CombustionreactionsconsumeO2asareactantandproduceoxides.

196.(A)Anacidicsaltisformedwhenastrongacidreactswithaweakbase.Weakbasesformgoodconjugateacids,whereasstrongacidsformweakconjugatebases.Forexample,ifCl−wasabletopickupH+insolution,thenHClwouldn’tbeastrongacidbecauseitwouldn’tfullydissociateinsolution.TheCl−wouldpickupH+ionsandanequilibriumbetweenHClandH++Cl−wouldbereached.Theequilibriumofastrongacid/baseionizationisn’tatrue

equilibrium.Thereactiongoestocompletion.Theconcentrationsofproductsandreactantsdon’tchange,butonlybecausethereversereactiondoesn’toccur,notbecausetheforwardandreversereactionsoccuratequalrates.Infact,theconcentrationofreactantinstrongacidorbasedissociationsispracticallyzero.

197.(E)Solidaluminumisnotoxidizedduringthisreaction.(SeeAnswer11foranexceptquestionstrategy.)

ImportantNote:Questionslikethiswilltypicallynotappearinthemultiplechoicesectionoftheexam.Thefreeresponsesection,however,hasamandatory“Reactions”sectionthatrequirestesttakerstowriteoutabalancedequationandansweraquestionaboutit.

198.(B)Sodiumcarbonateisnotproducedbytheheatingofsodiumbicarbonate.(SeeimportantnotefollowingAnswer197.)

199.(A)ThecompoundP4O3doesnotformundertheconditionsinquestion.Itisnotanaturallyoccurringoxideofphosphorus.(SeeimportantnotefollowingAnswer197.)

200.(A)Choices(B)through(D)containthespectatorionsofthetotalionicreaction.Choice(E)involvestheformationofHI,astrongacid,whichisnotproducedinthisreaction.(SeeimportantnotefollowingAnswer197.)

201.(E)HCl(aq)andKOH(aq)=Theexothermicneutralizationwouldincreasethetemperatureofthereactionvessel.

CaCO3(aq)andHF(aq)=Bubblingfromthecarbondioxidegaswouldbeobserved.

Mg(s)andHI(aq)=Bubblingofthehydrogengaswouldbeobserved.

Pb(NO3)2(aq)andNaCl(aq)formaninsolubleprecipitate(PbCl2).

ThemixingofNH4NO3(aq)+HCl(aq)solutionsisnotveryexciting.Theyarebothverysolubleinwaterandadoublereplacementswitchproducesnoinsolublecompounds.Nitrate(NO3

−)andchloride(Cl−)arepoorconjugatebasesandwillnotpickupprotonsinsolutiontoformHNO3orHCl,twostrong

acids(seeAnswer196foranexplanationofwhystrongacidsformtheweakestconjugatebases).Ammoniumchlorideisacompletelysolublesalt.(SeeAnswer11foranexceptquestionstrategy.)

202.(C)Acombustionreactionisavigorous,exothermic,self-sustainingreactioninwhichthesubstancescombinetogiveoffheatandlight.Thecombustionreactionswearemostfamiliarwithuseanoxidizingagent(likeO2).Choice(E)isanoxidation,buttheoxidizingagentinnotrepresented,itisanelectrochemicalhalf-reaction.

203.(B)It’stheonlyreactionthatproducesanion,

204.(E)RememberOILRIG:OxidationIsLossofelectrons,ReductionIsGainofelectrons.

205.(A)CarbonatesformCO2gaswhenacidifiedaccordingtotheequilibriumequation:

206.(D)Whenbalancingthecombustionofcarbon-containingcompounds,balancecarbonatomsfirst,hydrogenatomssecond,andoxygenatomslast(rememberMrs.Cho,she’llhelpyoubalancecombustionreactions).

207.(C)Whenbalancingthecombustionofcarboncontainingcompounds,balancecarbonatomsfirst,hydrogenatomssecond,andoxygenatomslast(rememberMrs.Cho,she’llhelpyoubalancecombustionreactions).

208.(C)ThisreactioncanbequicklybalancedbynoticingthattheH2OontheleftsideofthearrowhasthehydrogenandoxygenatomsthatwillbedistributedofNH3andOH−.Ontherightside,wecanseethateachwatermoleculebreaksupintoahydrogenionandahydroxideion,soweneed3watermoleculesbecausetheNfromLi3Nwillneed3hydrogenionstoformammonia.

209.(B)Tobalanceredoxreactions,itistypicallyeasiesttouseoxidationstates.WeknowthatCr3+(aq)needs3electronstobereducedtoCr(s).Theconfusingpartmaybewiththechlorineions.Each needstolose1electronbutweneedthemtocombineinpairstoproduceCl2(g),soweshouldimmediatelyputa2near toreminduslaterthatweaccountedforchlorine’selectronsinpairs.Weknowweneed3electronstoreducechromiumand2electronstooxidizethechlorines.Theleastcommonmultipleof3and2is6electrons,andweseethatifwelost6electronsfrom6chlorineions(or3pairsofchlorines),we’dbeabletoreduce2chromiumions.

210.(D)Whenbalancinganequationthatcontainsapolyatomicion,keeptheiontogetherwhenpossible.Forexample,thefirstthingwemightbalanceisthemagnesium.Oncethe3isinfrontofMg(H2PO4)2(s),weareleftwith6H2PO4

−

ions.It’seasytoseethatthephosphatefrommagnesiumphosphatepickedupahydrogenionfromphosphoricacid(H3PO4),sowecanworkbackward.Thereisa2:1ratioofphosphatesbetweenMg3(PO4)2(s)andH3PO4(l),sothe6wehaveintheproductsshouldbeallocatedaccordingly.Sixphosphatescanbeallocatedto4phosphoricacidmoleculesandtheothertwocomefromonemagnesiumphosphate.

211.(D)Importantnote:AtypeofbalancingthatappearsontheAPChemistryexamisaredoxreactionoccurringinanacidicorbasicsolution.Ifaquestionlikethisappearsinthemultiplechoicesection,justbalanceasyounormallywould.Ifthenumberofeachatomonthereactantsideoftheequationisidenticaltothoseintheproducts,nothingelseisneeded.However,ifitdoesn’twork,orthequestionappearsinthefree-responsesection(probablyassociatedwithanelectrochemistryproblem)andasksourworkbeshown,weshouldbefamiliarwiththeprocedure.

Tobalanceredoxreactionsthatoccurinacidicandneutralconditions:(1)Balancetheatomsotherthanoxygenandhydrogen.(2)Balanceoxygenatomsbyaddingwatermolecules.(3)Balancehydrogensbyaddinghydrogenions.(4)Balancethechargesbyaddingelectrons.

212.(C)SeeimportantnoteinAnswer211.Tobalanceredoxreactionsthatoccurinalkalineconditions:(1)Balanceatomsotherthanoxygenandhydrogen.(2)Balanceoxygenandhydrogenatomsatthesametime.Tobalanceoxygen,wecanaddhydroxideionsorwater,buttheybothcontainoxygenandhydrogen.However,thereisa1:1hydrogen-to-oxygenratioinOH−andthe2:1hydrogen-to-oxygenratioinH2O.Toaddahydrogenatom,addawatermolecule,theonlywaytobe“oneup”onhydrogeninthehydrogen-to-oxygenratio.

213.(C)Whengiventheamountsofbothreactantsinaproblem,checkforlimitingreactants.Asimplewaytodothisistotakethenumberofmoleofeachreactantanddivideitbyitsstoichiometriccoefficientinthebalancedequation.Inthiscase,0.4moleCS2÷1=0.4,and1.20molesO2÷3=0.4.Becausetheirquotientsareequal,thereisnolimitingreactant.Thetworeactantsareavailableinthecorrectratio.Ifthereisnolimitingreactant,usethereactantthathastheeasiestnumberstocrunch.Inthiscase,itis0.4moleCS2.OnemoleCS2wouldproduce3molesofproducts(1moleofCO2and2molesofSO2)so0.4moleCS2willproduce1.2molesofproducts.

214.(D)Boththeproductsaregasesandthetotalnumberofthemproducedis3,whichalsohappenstobethenumberofmolesofO2requiredtoproducethem.SothenumberofmolesofgasproducedwillbethesameasthenumberofmolesofO2thatreacted.

215.(A)Thereare3molesofgasproducedforeverymoleCS2reacted.If33.6LofgasformedatSTP,then1.5molesofgasisproduced(33.6L×(1molgas/22.4L)).Themathiseasy:1.5ishalfof3,so0.5moleofCS2reacted(or,1.5molesproducts×(1molCS2/3molproducts)).

216.(B)OnehundredmL(0.1L)of0.6MHClcontains0.06moleH−ions(0.1L×(0.6molHCl/1L))whichwillgetreducedto0.03moleH2gas(ZnisabovehydrogenintheactivityseriesandwillloseelectronstoH+ionsinanacidicsolution).BecausethereactionoccursatSTP,wedon’tneedtouseagaslaw,

onlytheconversionfactorof1molegas=22.4L∴0.03mol×22.4L/mol=0.672L,or672mL.

ForQuestions217–219:

(A)4SO3only(B)2SO2and2SO3(C)3SO2,1O2and2SO3(D)3SO2and2O2(E)4O2and5SO3

217.(D)Thereactionof3molesofSO2withexcessoxygenwouldproduce3molesofSO3.Since1moleofO2isneededforthereaction,andthequestionasksfor1moleO2inexcess,weneedtostartwith2molesofO2.Choice(D)istheonlyonethatcontainsonlyreactants(seelistaboveforthecontentsofeachbox),sowecouldhaveeasilychosenthecorrectanswerwithoutevendoingthestoichiometry.

218.(B)Asimplewaytocheckforalimitingreactantistotakethenumberofmolesofeachreactantprovidedinthequestionanddivideitbyitsstoichiometriccoefficientinthebalancedequation:4molSO2÷2=2and1molO2÷1=1∴O2isthelimitingreactant,sowewon’tget4molesofSO3,we’llgetasmuchSO3aswhat1moleofO2willproducewithexcessSO2(2moles).Westartedwith4molesofSO2and2moleswereconsumedtoproduce2molesofSO3,leaving2molesinexcess.(Seelistaboveforthecontentsofeachbox.)

219.(E)Asimplewaytocheckforalimitingreactantistotakethenumberofmolesofeachreactantprovidedinthequestionanddivideitbyitsstoichiometriccoefficientinthebalancedequation:5molSO2÷2=2.5and6.5molO2÷1=6.5,soSO2isourlimitingreactant.With5molesofSO2,wecanproduce5molesofSO3andwillconsume2.5molesofO2intheprocess.Thatleavesuswith4molesofO2inexcess.(Seelistaboveforthecontentsofeachbox).

220.(B)Whenwearepresentedwiththequantitiesoftwoormorereactantsinaproblem,wecheckforalimitingreactant.Asimplewaytocheckforalimiting

reactantistotakethenumberofmolesofeachreactantprovidedinthequestionanddivideitbyitsstoichiometriccoefficientinthebalancedequation:0.1molNH3÷4=0.025,0.1molO2÷5=0.02∴O2isourlimitingreactant.ThemaximumamountofNOthatcanbeproducedwith0.1moleofO2is0.08moleofNO(0.1molO2×4molNO/5molO2

),or2.4gNO.

221.(E)ThereactionofNH3andO2proceedsaccordingtotheequation:

The4:5molarratiogiveninthequestionaretheexactcoefficientsofthereactantsinthebalancedequation,whichmakestheproblemfairlysimple.Becausetheproblemstatesthatthereactantsreacted100percent,onlyaratiothatreflectsthesamerelationshipasthestoichiometryinthebalancedequationcouldbegiven.

Theproducts,NOandH2O,willbepresentina4:6ratioasgiveninthebalancedequation.Remember,however,thatthesearemoleratios,notmassratios.ThemolarmassofNO=30gmol−1andthemolarmassofH2O=18gmol−1.4molesNO=120gand6molesH2O=108g.120g+108g=228g,theexactyieldofthereaction.Ifthenumbersgiveninaproblemdon’talignperfectlywiththebalancedequation,usethemassstoichiometryinstead.Forexample:

Nowwecandeterminethepercentmassofeachproduct:

120gNO/228gtotal=~53%NO∴47%H2O

Youmustusethe228gfromtheadditionof120gNO+108gH2fromtheequationtogetthecorrectpercentages.Ifweweregivenanothermassofproduct,let’ssay72g,wewouldtakethe53percentwearrivedatbythemethodaboveandthenapplyittothemassgivenintheproblem:53percentof72g=

~38gNO∴34gH2O.

222.(C)WeuseM1V1=M2V2∴(0.375)(400)=(x)(500)∴x=0.3M.Addvolumeswhencombiningsolutions.

223.(E)Somequestionsaskustosetupaproblemasopposedtoactuallycalculatingananswer.Inthisexample,startbysettinguptheexpressiontoconvert1.0LofNOintomolesofNO(1.0L×1mol/22.4L).Sinceweneed22.4inthedenominator,wecanimmediatelyeliminatechoices(A)and(B).Next,arrangeanexpressionthatconvertsmolofNOproducedintomolO2consumedusingthestoichiometriccoefficientsfromthebalancedequation(molesNOproduced×5molO2consumed/4molNOproduced=molesO2consumed).Weneedthefraction5/4,whichleaves(E)asouronlychoice.

224.(B)Whenpresentedwiththequantitiesoftwoormorereactantsinaproblem,checkforalimitingreactant.Asimplewaytocheckforalimitingreactantistotakethenumberofmolesofeachreactantprovidedinthequestionanddivideitbyitsstoichiometriccoefficientinthebalancedequation.2.5molNH3÷4>2.5molO2÷5,thereforeO2isourlimitingreactantandNH3wouldremaininexcess.

2.5molesO2available×4molNH3consumed/5molO2consumed=2molesNH3

consumed

Westartedwith2.5molesNH3∴0.5moleremains.

225.(A)Whenpresentedwiththequantitiesoftwoormorereactantsinaproblem,checkforalimitingreactant.Asimplewaytocheckforalimitingreactantistotakethenumberofmolesofeachreactantprovidedinthequestionanddivideitbyitsstoichiometriccoefficientinthebalancedequation:6molesKO2÷4<9molesCO2÷2∴KO2isourlimitingreactant.

6molKO2×3molO2/4molKO2=4.5molesO2produced

226.(A)Whencarryingoutstoichiometrywithgasesunderthesameconditions(temperatureandpressure),wecantreatvolumeslikemoles(atthesametemperatureandpressure,equalvolumesofanygaseshavethesamenumberof

particles).

6LO2×2CO2/3O2=4LCO2

227.(A)BecausewearegivenanswersinL,itiseasiestforustoconvertCO2toLfirst,andthenusethegasstoichiometryshortcutdescribedinAnswer226(treatvolumeslikemoles).

2.9gCO2×1mol/44g×22.4L/1mol=~1.5LCO2

1.5LCO2×3O2/2CO2=2.25LO2

228.(A)ThenetionicequationforjustabouteveryneutralizationreactionisH+

+OH−→H2O.Na+andNO3−ionsare(always)spectatorions.Theyremainin

solution(inotherwords,theydon’treact)fortheentirereaction.

229.(B)Anadditionreactionismainlylimitedtoalkenesandalkynes.AtypicaladditionreactionontheAPChemistryexamwilladdahalogentoanalkene.Reaction(A)isasubstitutionreaction.Asubstitutionreactionoccurswhenafunctionalgroup(orahydrogenatom,asinthisreaction)isreplacedbyanothergroup(orelement,oftenahalogen).Thesereactionsmainlyinvolvealkanes.Reaction(D)issaponification,theprocessthatproducessoapfromfatandastrongbase(typicallyNaOH,alsocalledlye).Theresultisasoapofthecarboxylate(inthiscase,itisafattyacid,alonghydrocarbonchainwithacarboxylgroupattheend).Reaction(E)demonstratesphotosynthesis.

230.(E)Thereactionbetweencarbondioxideandwaterisworthmemorizing.

Chapter6:Thermodynamics

231.(B)Thedefinitionoflatticeenergy(kJmol−1).(SeeAnswer52foradescriptionofthefactorsthatdeterminelatticeenergy.)

232.(C)Thedefinitionof(Gibb’s)freeenergy(ΔG,kJmol−1).(SeeAnswer239foranexplanationofGibb’sfreeenergy.)

233.(A)Adefinitionofactivationenergy,Ea.(SeeQuestions234foranotherdefinitionofEaandAnswers235,241,and251fordescriptionsofthedifferentaspectsofactivationenergy.)

234.(A)Anotherdefinitionofactivationenergy,Ea.(SeeQuestions233foranotherdefinitionofEaandAnswers235,241,and251fordescriptionsofthedifferentaspectsofactivationenergy.)

235.(A)Yetathirdwaytoconsidertheactivationenergy,Ea,ofareaction(seeQuestion233and234)isthatitreflectsthesensitivityofthereactionratetotemperaturechanges.TheArrheniusequationquantifiestherelationshipbetweenactivationenergyandtherateatwhichareactionproceeds:Ea=RTln(k/A),whereRistheuniversalgasconstant,Tisthetemperature,kistherateconstant,andAisthefrequencyfactorforthereaction,aconstantforagivenreactionthatrepresentsanempiricalrelationshipbetweentemperatureandtherateconstantandhastheunitss−1.(SeeAnswer241foranexplanationofhowtheArrheniusequationilluminatestherelationshipbetweenEaandKeq.)

236.(E)Substancesundergoingphasechangesdon’texperienceachangeintemperature,anindicationoftheaveragekineticenergyoftheparticles.Instead,thepotentialenergyoftheparticleschangesastheychangepositionsrelativetooneanother.

237.(D)Theformulaforkineticenergyis½mv2.Graham’slawofeffusionallowsustocalculatetherelativespeedsoftwodifferentgasesatthesametemperatureandisderivedfromtheformulaforkineticenergy,whichcanberearrangedtosolvefortheaveragevelocity,v,oftheparticles.Becausetheyareatthesametemperature,thevalueforkineticenergy(whichisproportionaltobutnotequaltotheabsolutetemperature)isthesameforgases1and2:

Thevelocityisinverselyproportionaltothesquarerootofthemolarmass.(SeeAnswer123foranin-depthexplanationofeffusion.)

238.(B)Thedefinitionofzeroentropyisaperfect,purecrystallinesolidat0K

(andthethirdlawofthermodynamics).

239.(E)Gibb’sfreeenergyisathermodynamicstatefunctiongivenbytheformulaΔG=ΔH–TΔS.Itisameasureofhowmuchuseful(butnonmechanical)workcanbedonebyasystem(ataconstantpressureandtemperature).Systemsindisequilibrium,thatis,notatequilibrium,havetheabilitytodoworkastheymovetowardequilibrium.Onceatequilibrium,nomoreworkcanbedone.AusefulformoftheGibb’sfreeenergyequation,ΔG°=–RTlnKeq,allowsthestandard-statefreeenergychangeofareactiontobecalculatediftheKeqisknown.Moreimportantly,itshowsthattheequilibriumestablishedforareactionisafunctionofthefreeenergychange(forreactionsinsolution).

240.(A)Thestandardenthalpyofformation istheenthalpychangethataccompaniestheformationofonemoleofacompound(oranelementinitsnonstandardstate)fromitsstandard-stateelements.Forexample,

mol−1butthe ,whichtellsusthattheformationsofonemoleofcarbondioxidegasfrom1moleofC(g)and1moleofO2(g)isexothermicandproduces393.5kJofheat.

241.(D)TheArrheniusequationquantifiestherelationshipbetweentheactivationenergy,Ea,andtherateatwhichthereactionproceeds:Ea=–RTln(k/A)whereRistheuniversalgasconstant,Tisthetemperature(inK),kistherateconstant(fromtheratelaw)forthereaction,andAisthefrequencyfactorforaparticularreaction.Itisaquantityrelatedtothefrequencyofcollisionsbetweenparticlesthatarecorrectlyorientedtoproduceaneffectivecollision.Anyreactionwithapositiveactivationenergy(thevastmajority)willexperienceanincreasedrateofreactionwithincreasingtemperature.(SeeAnswer235foranexplanationofhowEaismeasured,Answer251forapracticaldescriptionofactivationenergy,andAnswers277and278forfurtherexplanationsoftemperature,activationenergy,andreactionrate.)

242.(D)Anexothermicreactionthatincreasesentropywillbespontaneous(exergonic)atalltemperatures.Ingeneral,natureprefersmovingtowardalowerenergyandmoreentropicstate.(Seetablebelow.)

243.(A)Meltingisanendothermicreaction(heatmustbeadded)thatincreasesentropy.

244.(B)Bothdepositionandcondensationareexothermicprocessesthatdecreaseentropy.

245.(D)Woodismostlymadeofligninandcellulose.Ligninisacomplexchemicalcompound(itsformulaisC9H10O2,C10H12O3,C11H14O4)andcelluloseisalarge,linearpolymerofglucose(C6H12O6)n.TheircombustionwithO2producesCO2andH2Oasfollows(foroneglucose):

Thecombustionofwood(orpetroleumproducts)ishighlyexothermic(whichiswhyweuseitforheat,light,andwork)andproducessixmoremolesofgasthanitconsumes,whichresultsinincreasedentropy.Inaddition,thehighheatproducedisalsohighlyentropic.Remember,thedefinitionofzeroentropyisaperfect,purecrystallinesolidat0K(it’salsothethirdlawofthermodynamics).Anydeviationfrom0Kand/orapure,perfectcrystallinesolidindicatesthatentropyisincreasing.Anincreaseintemperaturetypicallyincreasestheentropyofthesystem.(SeetablebelowAnswer242.)

246.(C)Areactionthatisbothendothermicandresultsindecreasedentropywillneverbespontaneousatanytemperature(liketheformationofwoodfromCO2(g)andH2O(g).Plantsdoitallthetime,butittakesalotofenergy(fromthesun)todoit.Ingeneral,naturemovestowardastatethatisoflowerenergyandhigherentropy.Whenanentropyreductionoccursinoneprocess(organizingthe

chemistrylab),itdoessobycreatingmoreentropyintheuniverse.(Theheatcreatedbyourenergymetabolism,forexample.Ourmitochondriaareonly~40percentefficientatconvertingfoodenergyintocellularenergy,therestislostasheat,themostentropicformofenergy.)(SeetablebelowAnswer242.)

247.(E)Themixingoftwogasesataconstanttemperature(anadiabaticprocessisonethatoccurswithoutheattransfer)musthaveaΔH=0.Ifnoreactionoccursandthegasesexperiencenoforcesofattractionorrepulsion,thentheentropyofthesystemmustincrease(ΔS>0).Theentropyofasystemisproportionaltothenumberofstatesasystemcanassumeversusthenumberofstatesthatare“right.”Forexample,imagineastrangedeckofcardsinwhicheachcardhasanumberfrom1through52and,all52areinnumericalorder.Forsimplicity,wewillassumethisistheonlyorderedstateofthedeck.Ifthecardsareshuffledinanyway,theyareoutoforder,ordisordered.Itiseasytoseethattherearethereare(52–1)!waysthecardsareconsidereddisorderedandonlyonewaytheyareconsideredordered.Iftherewereonlytwocardsinthedeck,therewouldbe(2–1)!,oronewaythecardscouldbedisorderedandonewaytheycouldbeinorder,sothemore(different)componentsthatmakeupasystem,themorepotentialfordisorderandentropyinthesystem.

248.(C)TocalculatetheΔHrxnfrom values,rememberasimpleformula:

AddtheΔH°foftheproductsandreactantsseparately,thensubtractthevalueofthesumofthereactantsfromthesumoftheproducts.BecarefulwithsignsandmakesuretomultiplytheΔH°fofeachsubstancebyitsstoichiometriccoefficient.

249.(C)ThisisaHesslawproblem.Sinceenthalpyisastatefunction(itonlydependsonfinalandinitialstates,aDsignisusuallyaclue),itdoesn’tmatterhowwegetthere,onlywherewestartandwhereweendup.(H,G,andSaswellasT,V,andPareallstatefunctions,too.)

Ifwekeeptheconversionofgraphiteandoxygentocarbondioxide,but“flip”theconversionofdiamondandoxygentocarbondioxide(andremembertoreversethesignofΔH),thetwoequationscanceloutoxygen(withadoubleslash)andcarbondioxide(withasingleslash)andtheirsumleavesuswiththeconversionofgraphitetodiamondthatwewanted:

250.(D)Onewaytothinkofentropyisasameasureofdisorderinasystem.Thethirdlawofthermodynamicsstatesthattheentropyofapure,perfectcrystalat0Kiszero.Anychangethattakesasystemfurtherfromthatstateistypicallythoughtofasincreasingentropy.Thisismostlytrue,andisagoodguidetopredictingwhetheraparticularchangewillincreaseordecreaseentropy.Forexample,amixturetypicallyhasmoreentropythanapuresubstance.

Undercertaincircumstances,however,theentropychangesexpectedarenotwhatisobserved.Inthissituation,theadditionofCa2+andCl−ionsdecreasesentropybecausethewatermoleculesendupbecomingmoreorderedastheyformhydrationshellsaroundtheions.Theentropychangeofasystemisthesumoftheentropychangesofitscomponents,soeventhoughtheCaCl2becamelessorderedasitdissolved,themagnitudeoftheentropydecreaseofthewatermoleculeswasgreaterthanmagnitudeoftheentropyincreaseofCaCl2.(SeeAnswer247foranexplanationofwhyamixtureisexpectedtohavegreaterentropythanapuresubstanceandforadifferentconsiderationofentropy.)

251.(D)Wecanthinkoftheactivationenergyofareactionastheenergyneededfortheformationofthetransitionstateofthereaction.Itisconsideredanenergybarriertothereactionsprogression.Thesparkofthelighterproducestheheatneededtogetthecombustiongoingbysupplyingtheenergyneededbythereactantstoformthetransitionstateneededtoformtheproducts.Thecombustionofthebutaneishighlyexergonic,sooncethereactionstarts,theenergyitliberatessuppliestheactivationenergyfortheotherbutanemoleculesinthelighter(whicharesprayedintotheatmospherewheretheyreactwithoxygen).

Weprobablydon’tgotobedatnightworryingthatourchemistrytextbookwill

turnintocarbondioxideandwaterwhilewesleep.Theconditionsatourdeskdon’tprovidetheactivationenergyneededtostarttheprocess.Butifweappliedtheheatofthelighterflamestoit(or,ifwecouldgetthesparktodirectlycatchoneofthepages),wecouldeasilyturnourbookintocarbondioxideandwater(notrecommendeduntilaftertheAPChemistryexam).(SeeAnswer235foranexplanationofhowEaismeasured,Answer241fortherelationshipbetweenEaandKeq,andAnswers243andforfurtherexplanationsoftemperature,activationenergy,andreactionrate.)

252.(D)(SeeAnswer250foranexplanationofentropyandAnswers243andfortablesoftheentropychangesassociatedwithphasechanges.)

253.(C)Reaction(C)formedonekindofsolidcompoundfromasolidandagas.(SeeAnswer250foranexplanationofentropy,andAnswers243andfortablesoftheentropychangesassociatedwithphasechanges.)

254.(E)Thisisastoichiometryproblem.ThetrickypartisthattheΔHrxnisgiveninkJmol−1CH6N2,butthestoichiometriccoefficientofCH6N2inthereactionis2.Thatmeanswedon’tconvert6molesofH2O(g)toH2O(l),weonlyneedtoconvert3(becausewe’reaskedpermoleofCH6N2).

Thenegativesignindicatesthecondensationofwaterisexothermic(whichweprobablyalreadyknew),butthatmeanswewillactuallygetmoreenergyoutofthecombustionofCH6N2,sowe’llwantaΔHvaluethatismorenegative

∴–1,303kJmol−1+(–132kJ)=–1,435kJmol−1.

255.(A)Ifareactionlowersthetemperatureofthecontainerinwhichitoccurs,thereactionisendothermic.Itisabsorbingheatfromitsenvironment;thisiswhyitdecreasesintemperature(itislosingheattothereaction).Mostdissolutionshavea+ΔS,theyincreaseentropy.(SeeAnswer250foranexception.)

256.(A)TheunitsofΔSareJmol−1K−1,notkJmol−1,likeΔGandΔH.Aspontaneousreactionhasa–ΔG.Ifthereactionisspontaneousonlyatlowtemperatures,thenweknowthatthereactionisexothermicandtheentropy

changeispositive(seetablebelowAnswer242).

ΔH–TΔS<0

(–18,000Jmol−1)–(300)(ΔS)<0

300ΔS<18,000∴ΔS<60Jmol−1

257.(D)Thedoublearrowtellsusthatthereactionisatequilibrium.Atequilibrium,ΔG=0.UsingGibbsfreeenergyequationΔG=ΔH–TΔS,ifΔG=0thenΔH=TΔS.Foraphasechange,thechangeinentropyisequaltotheheatoffusiondividedbythemeltingpointtemperature(inKelvin).

258.(C)Weliveinafairlystableenvironmentthankstoactivationenergies.Weprobablydon’tgotobedatnightworryingthatourbedwillturnintocarbondioxideandwaterwhilewesleep.Theconditionsinourhousedon’tprovidetheactivationenergyneededtostartthecombustionprocess.Butabedinthemiddleofaforestfirewouldquicklybeconsumedandbecomeagreatmass(andvolume)ofgases.ThehighertheEaofagivenreaction,themorethermodynamicallystablethereactantsandthemoreenergytheyneedtoformtheactivatedcomplexintheirtransitiontoproduct(s).Weliterallycansleepatnightbecauseofthestabilityactivationenergiesprovide.

259.(E)Theshortexplanation(probablyworthmemorizing)isthatalladiabatic,reversibleprocessareisentropic(ΔS=0,noentropychange).Inanadiabaticprocessthereisnoheatflowbetweenthesystemanditssurroundings∴Δheat=0.IfΔS=Δheat/ΔTandΔheat=0∴S=0.Theentropyofthegasinacylinderisincreasedwhenitstemperatureisincreasedandtheentropyisdecreasedwhenitsvolumeisreduced.Whenwecompressthecylinderofgaswithoutheatexchange,itsvolumeisreducedwhileitstemperaturerises,soitsentropyisunchanged.

260.(D)Bondenergyisameasureofbondstrength.Itistheamountofheatrequiredtobreakonemoleofaparticularbond.Allbondenergiesarepositivenumbers,becausebreakingbondsisalwaysendothermic.WeusetheformulaHrxn=(energyofbondsbroken)–(energyofbondsformed)todeterminetheΔHrxnusingbondenergydata.Asimplemnemonicdevicetoremembertheformulais“B-FOR”(before):bondsbrokenareaccountedforB-FORbondsFORmed.

Forthereaction2H2(g)+O2(g)→2H2O(l)

Thebondsbrokenare:

2molesH—Hbonds@432kJmol−1=864kJ1moleO=Obonds@494kJmol−1=439kJTotalenergyneededtobreakbonds=1,358kJ

Thebondsformedare:

4molesO—Hbonds@459kJmol−1=1,836kJTotalenergyreleasedbybondformation=1,836kJ

Theformulaalreadyaccountsfortheexothermicityofbondformationwiththe“–”sign,sodon’tchangethesigns;leavethempositive.

ΔHrxn=(energyofbondsbroken)–(energyofbondsformed)ΔHrxn=(1,358kJ)–(1,836)=–478kJper2molesH2O

∴TheformationofonemoleofH2Ois–239kJ.Alternatively,wecouldhaveperformedthecalculationsfor1moleH–Hbondsand½moleO2bondstoform1moleH2O.

Breakallbondsinthereactantsandformallthebondsintheproductsunlessthereactionmechanismisknownwithcompletecertainty.Ifabondisn’tactuallybrokenduringthereaction,itsformationwillbeincludedinthecalculation,sothevalueofthebondenergyoftheformedbondwillbesubtractedfromthebondenergiesofthebrokenbonds,cancelingitselfout(themagnitudeoftheenergychangeforbreakingandformingaparticularbondisthesame,onlythesignchanges).Forexample,tobreakanH–Hbondrequiresaninputof432kJmol−1andformingandH–Hbondresultsin432kJofenergygivenoff.Thereisnonetenergychangeforbreakingandformingthesamebond.

261.(D)TheΔHrxncanbecalculatedbytheformula:

RememberthattheΔH°fofelementsintheirstandardstatesiszero.The

stoichiometriccoefficientofeachspeciesmustbemultipliedtotheΔH°fofthatspeciestocorrectlycalculatethetotalenthalpychangeofthereaction.

262.(B)Weusetheformula

RemembertomultiplytheΔH°fofthecompoundbythestoichiometriccoefficientinthebalancedequationtogetthetotalenthalpychange.

ΔHrxn=(9.7–2(34))∴ΔHrxn=–58.3kJ

(Questions261issimilar.)

263.(D)Thetemperaturedecreaseinthebeakerindicatesanendothermicreaction(thereactionabsorbedheatfromtheenvironment),thereforeΔH>0.Thegasliberatedindicatesthatentropywasincreased,thereforeΔS>0.

264.(D)Theactivationenergies(Ea)forforwardandreversereactionsaretypicallydifferent.Activationenergyistheenergydifferencebetweenthereactantsandtheactivatedcomplex.Iftheforwardreactionisexothermic,theEa=(energyofactivatedcomplex)–(energyofreactants).Forthereverse,endothermicreaction,theEaincludestheenthalpychangeofthereaction.Inotherwords,theEaofthereversereaction=(Eaexothermicrxn+ΔHrxn).IftheEaoftheforwardandreversereactionsarethesame,theΔHofthereactionmustbezero.

265.(D)ThisisaHesslawproblem.

Thereactionwewantis .

Ifwecombinetheothertworeactionstoproducethisreaction,wecandeterminetheΔHrxn.Becauseenthalpyisastatefunction(itonlydependsonfinalandinitialstates),itdoesn’tmatterhowwegetthere,onlywherewestartandwhereweendup.

Wehadtoreversebothreactionstocanceloutintermediatesandendupwiththecorrectforwardreaction,sowemustreversethesignsofΔH°298foreachreactionandaddthem∴ΔH°298=–172kJ.

266.(E)TheKpofagaseoussystematequilibriumwillnotchangewithpressurechanges.Whenthepressureonagaseoussysteminequilibriumincreases,theequilibriumshiftstofavorthesideofthereactionwiththeleastnumberofmolesofgases.Forapressureincreasetoshifttheequilibrium,theremustbeadifferentnumberofmolesofgasintheproductsandthereactants.InreactionX,there’s1moleofgasinthereactantsand2molesofgasintheproducts,thereforeincreasingthepressurewillshiftthereactiontofavorthereactants.InreactionY,thereisanequalnumberofmolesofgasesintheproductsandreactants,sonoshiftwilloccurwithapressurechange.InreactionZ,thereare2molesofgasinthereactantsand1intheproducts,soproductformationwillbefavoredunderincreasedpressure.However,whenthenewequilibriumisestablished,thevalueofKpwillbethesame.Eventhoughpartialpressuresoftheindividualgaseswillchange,theirratio(asdefinedbytheequilibriumexpression)willhavethesamevalueforKp.ThenumericalvaluesofKp,Keq,andKspforaparticularreactiononlychangewithtemperature.

267.(D)TheKpincreaseswithincreasingtemperatureforendothermicreactionsanddecreaseswithincreasingtemperatureforexothermicreactions.Ifweimagineheatasaproductofanexothermicreaction,addingmoreheatbyincreasingtemperaturepushestheequilibriumtotheleft,favoringthereactants.Themajordifferencebetweenaddingmoreheatandaddingmoreofachemicalproductisthatincreasingthetemperatureisaccomplishedbycontinuallyaddingmoreheat,sothereactionremains“pushed”totheleft;whereasaddingmoreofachemicalproducttransientlypushestheequilibriumtotheleft,butthenthepriorequilibriumreestablishesitself.

268.(D)Gasesarethemostentropicstateofmatterandsolidsaretheleastentropicstate.Convertingagastoasoliddecreasesentropy,thereforeSfinal–Sinitial–0.

269.(E)Thermodynamicdataonlyquantify(ordescribe)energyandentropychangesinchemicalreactions.

Chapter7:Kinetics

ForQuestions270–273:

(A)First-orderforM,first-orderreaction(B)FirstorderforMandN,second-orderreaction(C)First-orderforMandsecond-orderforN,third-orderreaction(D)Second-orderforMandfirst-orderforN,third-orderreaction(E)Second-orderforN,second-orderreaction

270.(E)IfchangingtheconcentrationofMhasnoeffectonreactionrate,wearelookingforaratelawthatdoesnotincludeM.

271.(A)Ifthetotalconcentrationchangeof2xdoublesthereactionrate,wearelookingforafirst-orderreaction([2x]?=2,∴?=1).OnlyAisafirst-orderreaction.Inthiscase,theconcentrationchangeofNmadenodifferencetothereactionrate,onlytheconcentrationchangeofM.

272.(D)IfdoublingtheconcentrationofMquadruplesthereactionrate,thereactionmustbesecond-orderwithrespecttoM([2x]?=4,∴?=2).TheeffectofNwasnotconsidered,soweareonlyconcernedwitharatelawthatcontains[M]2.

273.(D)Ifdoublingtheconcentrationofreactantsincreasesthereactionratebyafactorof8,thenthetotalreactionordermustbe3([2x]?=8,∴?=3).Ifnootherinformationweregiven,both(C)and(D)arepossibleratelaws.However,ifhalvingtheconcentrationofNreducesreactionratebyfourfold,thenthereactantorderforNmustbe2([½x]?=¼,∴?=2).Alternatively,wecaninvertthesituationtoread“doublingtheconcentrationofNincreasesthereactionratebyfourfold,”whichisalittleeasiertoconsider.Wecaneliminate(B)becauseweneedaratelawthatcontains[N]2.

274.(D)Thisquestionisaskingusabouttheeffectofsurfaceareaonreactionrate.Asageneralrule,increasingthesurfaceareaofasolidwillincreasereactionrate.Breakingupasoliddoesn’taffectitsconcentration[choice(C),althoughconcentrationisn’tatermtypicallyappliedtosolids].Choice(A)may

seemalluringbecauseitaddresses(indirectly)thesurfaceareaissue(exposure),buttheironneedstoreactwithsulfur,notoxygen,andtheoxidationstateofironshouldbe0toreactwithsulfursinceitwilllosingelectronstoreducesulfur.Iftheironwerepartiallyoxidized,itwouldbealesseffectingreducingagent.(SeeAnswer275foradetailedexplanationofeffectofsurfaceareaonreactionrate.)

275.(B)Thisquestion,likeQuestions274,isaskingusabouttheeffectofsurfaceareaonreactionrate.However,twoanswerchoicesaddresssurfacearea.Ifwehadtwocubesofwood,onewithdimensions2×2×2andotherwithdimensions4×4×4,theirsurfaceareaswouldbe24and96,respectively(theunitsareirrelevant).Thesurfaceareaofthelargerpieceofwoodisfourtimesthatofthesmallerpieceofwoodwhichmakessense,thecubeisbasicallytwotimesaslargeandsinceareahasaunitofisalength2,(2×length)2=4×thesurfacearea.Thevolumeofthe2×2×2cubeis8,whereasthevolumeofthe4×4×4cubeis64.Asexpected,thevolumeofthelargercubeiseighttimesthatofthesmallercube.Theunitofvolumeislength3,so(2×length)3=8.

Increasingsurfaceareaincreasestherateofareactionduetothegreaterexposureperunitofvolumeormass.

Whenwerevisitourcubes,the2×2×2cubehasasurfaceareatovolumeratioof ,butthesurfaceareatovolumeratioofthe4×4×4cubeis .Noticedoublingthesizedecreasesthesurfaceareatovolumeratiobyone-half.

Ifthetwosolidsarethesame(wood,twigs,sawdust)thenthecomposition(andconcentration,whichisn’tatermtypicallyappliedtosolids)doesnotchangewhenwebreakitintosmallerpieces.

276.(D)Increasingthepressureonlyshiftstheequilibrium(andtherebymomentarilychangestherateofeithertheforwardorreversereaction)ofareactionthatinvolvesgases,andeventhen,thenumberofgaseousmolesofreactantandproductmustdifferorthepressurechangehasnoeffect.Increasingthetemperatureatwhichareactionoccurswillalmostalwaysincreasethereactionrate,whetherthereactionisexothermicorendothermic.(Forexplanationsregardingtheeffectofsurfaceareaonreactionrate,seeAnswers274and275.)(SeeAnswer11foranexceptquestionstrategy.)

277.(D)Averycommonmisconceptionregardingtheeffectoftemperatureonreactionrateisthatincreasedtemperaturessignificantlyincreasesthefrequency

ofcollisions.Thesmallincreaseincollisionfrequencydoesn’tactuallyhavemucheffectonreactionrate.

Inorderforachemicalreactiontooccur,thereactantsmustcollide,butthecollisionitselfwillnotformaproductunlessthereactantscollideintheproperorientationtooneanotherandwithenoughkineticenergy.Whenreactantscollideintherightorientationandwithenoughenergytoovercometheactivationenergyofthereaction,theyproduceaproduct(orareactionintermediate,atleast).Thisiscalledaneffectivecollision.

ImagineBatmantryingtofightoffthebadguysunderthemisconceptionthatmorecollisions(betweenhimandthebadguy)willresultinahigherreactionrate.Hecalculatesthatifheuseshisfingertipsinsteadofhisfists,hecangetinfivetimesmorecollisionsthanwithjusthistwofistsalone.Accordingtohislogic,afightthatwouldhavenormallytakenanhourshouldtakeonly12minutes.Hegoesoffintothenight,findsthebadguys,andapplieshisnewlogictotheirfight.Hesoonfindsthatpokingisnosubstituteforpunching.Andjustlikecollisionsbetweenreactantshavetooccurnotonlywithenoughenergybutinanorientationthatallowsthereactantstocollideinjusttherightpositionrelativetoeachother,apunchdeliveredtothefaceorbellyisgoingtobemuchmoreeffectiveatstoppingthebadguysthanapunchinthearmorleg.

278.(C)Temperatureincreasesthereactionrateofalmostallchemicalreactions,whethertheyareexothermicofendothermic(therearesomecasesforwhichthisisn’ttrue,andthesereactionshavenegativeactivationenergies).Highertemperaturesresultinmoreeffectivecollisions(collisionswithenoughenergyandwiththeproperorientationorreactantsrelativetoeachother)byincreasingtheaveragekineticenergyoftheparticlesinthereactionmixture,assuringthatagreaternumberofcollisionswilloccurwithanenergythatmeetsorexceedstheactivationenergyofthereaction,theenergybarrierthatmustbeovercomeinorderforthereactiontoproceed.(SeeAnswer277foramorein-depthexplanationofeffectivecollisions.)

279.(A)Thisquestionisverytricky.Choice(A)isalmostcorrect—eachradioactivespeciesdoeshaveaparticularhalf-life,butaparticularelementcanhaveseveraldifferentisotopes,eachwithitsownhalf-life.(SeeAnswer11foranexceptquestionstrategy.)

280.(C)IncomparingTrials1and2,thetwotrialsinwhichtheO2

concentrationdoesnotchange,weseethattheconcentrationofNOisincreasedby1.6(notquite2).Therateincreasesbyabout2.6-fold.Ifwewereallowedtouseacalculator(insomesectionsofthefreeresponsesection),wecouldeasilyplugin([1.6x]?=2.6foldrateincrease).Ifweencounteraproblemlikethisinthemultiplechoicesection,wedoanestimate:IfthereactantorderforNOwere1,doublingtheconcentrationwoulddoubletherate,soa1.6-foldincreaseinconcentrationshouldincreasetherateby1.6-fold.Butthechangeweseeis2.6-fold.Wecanprobablyseethatthereactionorderisn’t3,butmaybewe’reunsureof2.IfthereactantorderofNOis2,andtheconcentrationofNOwasdoubled,we’dexpecttheratewouldincreasebyfourfold.Iftheconcentrationwasincreasedby1.6,1.62=2.56,or2.6.

WecanapplythesamemethodologytothereactantorderofO2.ThetwotrialsinwhichonlytheconcentrationofO2changesareTrials1and3.TheO2concentrationincreasesfrom0.035to0.045,anincreaseofabout1.3times.Therateincreasesfrom0.143to0.184.Ifwerecognizethat0.143isabouttwo-thirdsof0.184,wecandeterminethatthereactionrateincreasedbyone-third.Therefore,withregardtoO2,increasingtheconcentrationbyone-thirdincreasesthereactionratebyone-third.(Pleaseseetheimportantnotefromtheauthoratthebeginningofthebook.)

281.(D)Allweneedtodohereissubstitutethedatafromoneofthetrialsintotheratelaw.Usethetrialthathastheeasiestnumberstocrunch.UsingthedatafromTrial1:

Rate=k[NO]2[O2]

k=rate/[NO]2[O2]

k=0.143/[0.024]2[0.035]

k=7,093or7.0×103

Remember,thenumericalvalueoftherateconstant,k,alwaysincreaseswithincreasingtemperature.Sothevalueofkcalculatedforaparticularreactionoccurringatonetemperaturewillnotbethesameinthesamereactionperformedatadifferenttemperature.(Pleaseseetheimportantnotefromtheauthoratthebeginningofthebook.)

282.(C)Wecanmemorizetheunitsoftherateconstantorcalculatethem.NoticethattheabsolutevalueoftheexponentforLandmolarealwaysonelessthanthereactionorder.

Tocalculatek,justsubstitutetheunitsintotheratelaw.

Rate=k[NO]2[O2]

Ms−11=k(molL−1)2(molL−1)

**RememberthattheunitsofMaremolL−1**

molL−1s−1=kmol3L3

k=L2mol−2sec−1

(Pleaseseetheimportantnotefromtheauthoratthebeginningofthebook.)

283.(C)ThereactionorderforNOis2,soincreasingitsconcentrationwouldincreasethereactionrateby25×([5x]2=25-foldincreaseinrate).(Pleaseseetheimportantnotefromtheauthoratthebeginningofthebook.)

284.(D)Allweneedtodoisplugthenumbersintotheratelaw,usingthevalueofkdeterminedinQuestions281.

Rate=k[NO]2[O2]

Rate=7.3×103[2×10−2]2[4×10−2]

Rate=1.1×10−1

(Pleaseseetheimportantnotefromtheauthoratthebeginningofthebook.)

285.(E)Theeffectthattheconcentrationofaparticularreactanthasonareactioncannotbedetermineduntiltheactualexperimentofchangingtheconcentrationofthatreactionandmeasuringtheratechangethatresultsisperformed.Inwritingtheequilibriumexpression,theexponentsthataccompanyreactionsaretakenfromthebalancedequation,buttheequilibriumconcentrationsmustbemeasuredtoactuallydeterminetheKeqforthereaction.(Pleaseseetheimportantnotefromtheauthoratthebeginningofthebook.)

286.(D)Weshouldcertainlyrememberthestatementsinchoices(A),(B),(C),and(E)asimportantfactsaboutcatalysts.Choice(D)istheoppositeofwhatistrue,andthisisalsoimportanttoremember:Acatalystdoesnotdifferentiatebetweentheforwardandreversereactions.Theactivatedcomplexisthesamewhetherareactionoccursintheforwardorreversereaction,sodecreasingtheenergyoftheactivatedcomplexwilldecreasetheactivationenergyinbothdirectionsbyanequalamount.Becauseacatalystincreasestheratesofbothforwardandreversereactionsequally,thepresenceofacatalystdoesnotchangetheKeqforareaction.Itdoesn’tchangetheΔHrxn,either(yetanotherimportantthingtorememberaboutcatalysts).(SeeAnswer11foranexceptquestionstrategy.)

287.(E)StatementsIandIIarestandardwaysofmeasuringreactionrates.Theratelaw(III)canbeusedtopredicttherateofachemicalreaction.Thereactionquotient(Q)canbeusedinconjunctionwiththeequilibriumconstant(Keq)tocalculatehowfarareactionisfromequilibrium.

288.(A)Reactionratesaretypicallymeasuredinoneoftwoways,theappearanceofproductovertime,and/orthedisappearanceofoneormoresubstratesovertime.Whenusingreactiondata,itisvitaltoconsiderthestoichiometriccoefficientswhentocomparerelativeratesofappearanceordisappearance.Forexample,foreachmoleNOconsumedbythisreaction,onemoleofH2willbeconsumedaswellandonemoleofH2Owillform,aswell.However,onlyhalfofamoleofN2willform.Relatedtotime,N2willbeformedathalftherateasH2Oformation,andhalftherateatwhichNOandH2areconsumed.

289.(B)Theslowest(ratedetermining)stepinamultistepreactiongovernstheratelawofthereaction.COisnotinvolvedintheratedeterminingstep,soitisnotincludedintheratelaw.BecausetwoNO2moleculesarenecessaryfor

collision(itisbimolecular),thereactantorderforNO2istwo.

290.(D)Itismoredifficulttodeterminetheratelawfromareactionmechanisminwhichanintermediate(NOBr,inthiscase)isinvolvedintheslow(ratedetermining)step.Noticethatthefirst,faststepisatequilibrium,sothattheforwardandreversereactionsareoccurringatthesamerate.Wecanwriteandequatearatelawforthesereactions,thensolvefor[NOBr],whichallowsustousetheintermediatethatwenormallydon’tincludeintheratelaw,butneedwhenitispartoftheratedeterminingstep.

kforward[NO][Br2]=kreverse[NOBr]

[NOBr]=kforward/kreverse[NO][Br2]

Theratelawfortheslowstep,ifNOBrwasnotanintermediate,wouldlooklikethis:

Rate=k[NOBr][NO]

butnowwesubstitutefor[NOBr]

Rate=k2(kf/kr)[NO][Br2][NO]

andsimplifytogetrate=krxn[NO]2[Br2].

Thevalueofkrxn=k2kf/kr.

291.(D)Thenumericalvalueoftherateconstant,k,isirrelevanttoansweringthisquestion.Itistheunitoftherateconstantthatgivesusthemostinformation(atfirst).Fromtheunit,wecandeducethatthisisathird-orderreaction.Remember,theabsolutevalueoftheexponentforLandmolis1lessthanthereactionorder.(Forreactionordern,theunitsoftheratelawwillbeL(n–1)mol−(n–1)sec−1.)(SeeAnswer282foratableoftherateconstantunitsandhowtodeterminethem.)

Therateconstantrepresentsan“adaptor”betweentheconcentrationofreactantsandthereactionrate(ataparticulartemperature),itdoesn’ttellusanythingabouttheenthalpychangesthatoccurinthereaction,norcanitbeusedforcomparingreactionrateswithotherreactionsforwhichnothingbuttherate

constantisknown.Theunitsoftherateconstantdoprovideaquantitativerelationshipbetweenconcentrationchangesandreactionrates.Inthisthird-orderreaction,doublingtheconcentrationsofallthereactionswillresultinaneightfoldincreaseinreactionrate([2x]3=8×rate),notan8.1×1010-foldrateincrease.

292.(B)Reactantorderscanbepositive,negative,orfractional.Anegativereactantorderindicatesthatthereactionrateincreasesasthereactantwiththedecreasedconcentrationofthatreactant.

293.(D)Theactivatedcomplexformedduringachemicalreactionwithoutacatalystmustbedifferentfromtheoneformedbythesamereactionwithacatalyst.First,thepotentialenergiesaredifferent.Second,thepresenceofacatalystprovidesanalternativepathwayforreaction,whichmeansadifferentsetofintermediatesareformed.(SeeAnswer11foranexceptquestionstrategy.)

294.(E)Thereactionisendothermicbecausetheproductshavemorepotentialenergythanthereactants.TheEaiscalculatedbysubtractingtheenergyofthereactantsfromtheenergyoftheactivatedcomplex∴510–75=435kJmol−1.Thepotentialenergyoftheproductsisapproximately250kJmol−1andtheenergyofthereactantsisabout75kJmol−1∴ΔH=250–75=175kJmol−1.

295.(D)Reactionrateismeasuredbyproductformation(orsubstrateconsumption)overtimeandisafunctionofthenumberofeffectivecollisionsthatoccurinagivenperiodoftime.

296.(B)Ittakesbetween10and15minutestodecreasetheamountofreactantby50percent.Ittakesanother10to15minutestodecreasetheamountofreactantbyanother50percent,thereforethereactionorderis1,andthehalf-lifeisbetween10and15minutes∴12minutes.

297.(E)Thetotalreactionorderis3,sodoublingbothreactantswillincreasethereactionrateeightfold([2x]3)=8-foldincreaseinrate).

298.(B)InTrial2,theconcentrationofXiscutbyhalf,soweexpecttheratetodecreasebyhalfsinceXisafirst-orderreactant(∴R/2).TheconcentrationofYisdoubled,however,andYisasecond-orderreactant,sothechangeinreactionrateduetodoubling[Y]=4R.(R/2)×(4R)=2R.

299.(D)InTrial2,theconcentrationofXiscutbyhalfsoweexpecttheratetodecreasebyone-fourthsinceXisasecond-orderreactantinthissituation(∴R/4).TheconcentrationofYisdoubled,however,andsinceYisafirst-orderreactantinthiscase,thechangeinreactionrateduetodoublingis[Y]=2R.(R/4)×(2R)=R/2.

300.(D)Theunitsoftherateconstantforafourth-orderreactionisL3mol−3sec−1.RememberthattheabsolutevalueoftheexponentsofLandmolis1lessthanthereactionorder.(SeethetableunderAnswer282forasummaryoftheunitsofk.)

Chapter8:Equilibrium

301.(D)LeChatelier’sprinciple:Ifachemicalsystematequilibriumexperiencesachangeinconcentration,volume,temperature,orpartialpressure,theequilibriumshiftstocounteractthechangeandanewequilibriumwillbeestablished.ToincreasetheamountofMgOproduced,adisturbancetotheequilibriummustbeappliedthatwillbecounteractedbytheproductionofmoreMgO.RemovingMg(s)(oraddingit)willnotcauseanychangeintheequilibriumbecauseitisasolid.Solidsdon’thaveaconcentrationandtheyarenotrepresentedintheequilibriumexpression.Increasingthepressureonagaseoussystematequilibriumwillcausetheequilibriumtoshifttofavorthesideofthereactionwiththefewestmolesofgas.Decreasingthepressureshiftstheequilibriumtofavorthesideofthereactionwiththemostmolesofgas.ChangingthepressureofagaseoussystemdoesnotchangethevalueofKp,however.TheratioofpartialpressuresafterthepressurechangewillequalthesamevalueofKp.AddingmoreO2(g)willincreaseMgOformationbecausethesystemwillconsumetheO2inanefforttocounteracttheincreaseinO2pressure.(SeetableinAnswer307.)

302.(D)Thereactionisendothermicbecauseheat(energy)isonthereactantsideoftheequation.Toshifttheequilibriumtofavorthereactants,thepressurecoulddecrease(thereactantshavegreaternumberofmolesofgas)orthetemperaturecanbelowered(removingheatwouldpulltheequilibriumtotheleft).

303.(E)Thisreactionisexothermic,soincreasingthetemperaturefavorsthereactantsbypushingtheequilibriumtotheleft.Thenumberofmolesofgases

oneachsideoftheequationisthesame,sopressurechangeswillnotaffecttheequilibriumpartialpressures.

304.(B)Increasedpressureduetodecreasedvolumeshiftstheequilibriumofagaseoussystemtofavorthesideofthereactionwiththefewestmolesofgases(withoutchangingtheKp).Equation(B)hasanequalnumberofmolesofgasesintheproductsandreactants,sopressurechangeswillnotchangeequilibriumpartialpressures.

305.(D)Decreasedwatertemperatureandincreasedgaspressureabovethewaterincreasethesolubilityofagasinwater.Shakingthecontainerwillnotresultinasustainedincreaseinsolubility(sincethesystemisatequilibrium).

306.(B)IfN2isinjectedintothetank,itwillreactwithH2byconsumingit,andthereforedecreasingitspartialpressureinthetank.

307.(C)Thereactionisendothermic.ThevalueforKponlychangeswithtemperaturechanges.

308.(C)Theratioistheequilibriumexpressionforthereverse,exothermicreaction.OnlytemperaturechangescanchangethevalueofKeq.Pressurechangesinagaseoussystemwillshifttheequilibriumuntilthenewequilibriumisreached.Althoughtheactualpartialpressuresofeachgas(atequilibrium)maybedifferent,theirratioascalculatedbytheequilibriumexpressionwillresultinthesamevalueofKp.

309.(C)Allweneedtodoissubstitutetheconcentrationsintotheequilibriumexpression.

310.(C)Thereactantsideoftheequationhasfewermolesofgasthantheproducts,soincreasingthepressurewillshifttheequilibriumtofavorthereactants,nottheproducts.ThevalueforΔHispositive,sotheforwardreactionisendothermic,whichmeansthatincreasingthetemperaturewillshiftthe

equilibriumtofavortheforwardreaction,anddecreasingthetemperaturefavorsthereversereaction.AddingmoreXwillincreasetheformationofZasthesystemconsumesitinanefforttocounteracttheincreaseinXpressure.

311.(B)Thisquestionisnotjustaskingwhatwillshifttheequilibrium,it’saskingwhatwillchangetheequilibriumconstantforthereaction.TheKeqofareactiondoesn’tchangeifweaddmorereactantorproduct,itonlychangeswithtemperature.Theratiointhequestioniswrittenfortheforward,endothermicreaction(thereversereactionwouldbetheinverse).(SeetableinAnswer308.)

312.(E)TheequationtellsusthatforeveryoneF2thatreacts,twoFareproduced.Ifweputthenumberofparticlesintheboxintotheequilibriumexpressionweget[F]2/[F2]=[42/9]=1.8.Theequilibriumconstantwillneverbeanegativenumberbecausetherearenonegativeconcentrations.

313.(E)Theequilibriumexpressionforthereversereactioninvertstheproductsandreactants,sotheKeqforthereversereactionissimplytheinverse,1/Keq.Thevalueof1/2.5×1010iseasytocalculatebecauseallweneedtodoisdivide1by2.5(=0.4)andreversethesignontheexponent(10−10).However,0.4×10−10ismorecorrectlywrittenas4.0×10−11.

314.(D)If30percentofXY3dissociatesin1kgofa0.06msolution,thenoutofthe0.06moleofXY3present,0.018moledissociatesand0.042moleremainsintheformofXY3.Immediately,choice(A)canbeeliminated.EachXY3thatdissociatedproduced4particles,oneX3+andthreeY−,sowhenthe0.018moleofXY3dissociated,itcreated0.018×4–0.072moleofparticles.Thetotalparticlecount,then,is0.042moleundissociatedparticlesplus0.072moleions=0.114moleparticles.

315.(C)Kspisthesolubilityproductconstant,theequilibriumconstantforasolidinequilibriumwithitsions.Itismeasuredinasaturatedsolutionatequilibrium.Themolarsolubilityisthenumberofmolesthatcanbedissolvedtoproduce1Lofasaturatedsolutionofthatsubstance.WhatsimplifiesKspcalculationsisthatthereisalwaysasolidreactantintheseproblemsbutsolidsarenotincludedinequilibriumexpressions,sotheywillneverhaveadenominator.

Letx=[Ba2+]=[SO42–](Theyareproducedina1:1ratio.)

Ksp=x2=1.1×10−10

x=~1×10−5

(SeeAnswer316foramorecomplicatedvariationofthisproblem.)

316.

Letx=[Ca2+]

F−andCa2+areproducedina2:1ratio,so[F−]is2x.BecauseKspisanequilibriumconstant,the[F−]mustberaisedtotheexponentthatisitsstoichiometriccoefficient,2.

Ksp=[Ca2+][2F−1]2

Ksp=4x3

4.0×10−11=4x3∴x3=1.0×10−11∴x=(1.0×10−11)1/3M

Rememberthatthe[Ca2+]=x,butthe[F−]=2x.

(SeeAnswer315forasimplervariationofthisproblem.)

317.(C)ThisisthereverseprocessofQuestions315and316,butwe’llusethesameprinciples.Thereisalsoanextrastepinvolvedbecauseinsteadofbeinggivenavalueforthemolarsolubilityofoneofthedissociationproducts(givenwhichwecandeducetheconcentrationsoftheother),wearegiventhepH.ThepH=10,soweknowthesolutionisbasic.

pH+pOH=14,ifpOH=4then[OH−]is1×10−4M

M(OH)2⇔M2++2OH−

Letx=[M2+]

[OH−]=2x=1×10−4∴x=0.5×10−4,or5×10−5M

Ksp=[M+2][2OH−]2∴Ksp=[x][2x]2=4x3

Sincex=5×10−5,4x3=5×10−13.

318.(A)ThesolubilityofFe(OH)2decreaseswithincreasedpHbecausethehigherconcentrationsofOH−ionspushesthereactiontowardtheformationofsolidFe(OH)2.DecreasingthepHincreasesitssolubilitybecausetheextraH+

ionsinsolutionwillcombinewiththehydroxidestoproduceH2O,pushingthereactiontotheleft.ThevalueofKspdoesnotdependonthepHdirectly,butsinceoneoftheaqueousproductsofthereactionisOH−,theKspalreadytakesthepHintoaccount.

319.(A)TheKspforBaCrO4ismuchsmallerthanthatofCaCrO4,soweknowthatitismuchlesssoluble.Thedifferenceofsixordersofmagnitude(10−10versus10−4)tellsusthatitispossibletoprecipitatealmostalloftheBa2+ionswithoutprecipitatinganyCaCrO4.ThetricktosolvingthisproblemistounderstandhowtoproduceasaturatedCaCrO4solution.ThiswillleaveCa2+

ionsinsolution,butwilloverwhelmtheBa2+ions.

Ksp=[Ca2+][CrO42–]

7×10−4=[0.2][CrO42–]∴[CrO4

2–]=~3.5×10−3M.

320.(E)Theonlydeterminantofthevaporpressureofaparticularsubstanceatequilibriumwithitsliquidisthetemperature.Theamountofsurfaceareawillaffecttheratesofevaporation(andcondensation)andthevolumeofthecontainerwillaffecthowmanymolesofvaporarepresent,butnotthevaporpressureatequilibrium.

321.(C)Thevaporpressureofaliquidisdeterminedatequilibriumataparticulartemperature,sowecanconsiderthisaswewouldanequilibriumquestion.JustastheKeqofanendothermicreactionincreaseswithincreasing

temperature(andtheKeqofexothermicreactionsdecreaseswithincreasingtemperature),highertemperaturesfavorvaporizationovercondensationuntilanewequilibriumisreached,anditwillincludealargerfractionofvaporthantheequilibriumconditionatalowertemperature.Choice(A)isnottrueandthesysteminchoice(B)isnotatequilibrium.Forasystematequilibrium(andatconstanttemperature),theratesofforwardandreverseprocessesarethesame,so(D)isnottrue.Finally,theΔHofaforwardandreverseprocess(orreaction)areequalinmagnitude,butoppositeinsign.

322.(C)Atatmosphericpressures,N2,O2,andArcondenseat–195°C,–183°C,and–186°Crespectively.CO2doesnotcondenseintoaliquidatatmosphericpressures,butbecomesasolidatabout–80°C(seethephasediagramsforH2OandCO2aboveQuestions96and99,respectively).

323.(A)Anadiabaticsystemisoneinwhichheatdoesnotenterorleavethesystem.Entropyisameasureofthedegreeofdisorderordispersionofenergyinasystem,andittypicallyincreaseswiththeadditionofheat.Theliquidstateofapuresubstancehasgreaterentropythanitssolidstatebecausetheparticlesinaliquidaremoredisorderedthantheparticlesinasolid.

Duringthemeltingprocess,thechangeinentropyisequaltotheheatoffusiondividedbythemeltingpointinK.Thetermmeltingimpliesadisplacementfromequilibriumbycontinuedheattransferovertimetodrivethemeltingprocess.However,inthisquestion,thesystemisadiabatic,thatis,noheatisenteringorleavingthesystem,itisataconstanttemperature.

Thephaseequilibriumofapuresubstanceisindependentoftheratioofthesolidphasetotheliquidphase.Equilibriumexistsevenifonlythetiniestbitofsolidispresentinthesolid–liquidmixture.Forthatportionofthesolidthatmelted,thereisanentropyincrease.Theremainingmassofliquidandsolidisatequilibriumandhasnoentropychange.

324.(B)Thestrengthofoxyacidsincreaseswith(1)increasingelectronegativityofthecentralatom(theatomtheO—Hattachesto),(2)otheratomsofhighelectronegativityinthecompound,and(3)anincreasingnumberofoxygenatoms.ThepHofasolutiondependsontheconcentrationonH+insolutionandthereforetheconcentrationoftheacidinsolution,butthestrengthofanacidisintrinsictotheacidanddoesnotdependontheconcentration.

325.(C)TheKaforhypoiodousacidis2×10−11.ThestabilityoftheconjugatebasealsotellsussomethingaboutthestrengthoftheacidOI−,whichpicksupprotonsinsolutionrathereasily,meaningthatmoreHOIispresentatequilibrium.IflessHOIdissociates,theacidisweaker.(SeeAnswer324foralistofdescriptionoffactorsthataffectthestrengthofoxyacids.)

326.(C)Thenamesofoxyacidsarederivedfromthepolyatomicanionsthattheprotonsareattachedto.

InthecaseofClO−series:

TheClO−ionisnamedhypochloritebecauseithastheleastnumberofoxygenatomsintheseries.Thecompoundsarenamedrelativetoeachother,notbytheirabsolutenumberofoxygenatoms.Oncewe’veidentifiedthepolyatomicoxyanion,wenametheacidaccordingly:

Oxyanionsendingin–itebecome–ousacid.

Oxyanionsendingin–atebecome–icacid.

327.(D)TheequationshowsthereactionoftheconjugatebaseofHOCl.TheformationofanOH−ionmeanswecanuseKb,thebaseionizationconstant,tofigureouttheunknownsinthereaction.

328.(D)TheAPChemistryexamisunlikelytogiveusaquestionlikethisinthemultiplechoicesection,howeverthereisamandatoryequilibriumprobleminthefreeresponsesection,soweneedtoknowhowtosetupanICE-box.Seechoice(D).(1)Theinitialconcentrationofreactantsistakenfromthequestion.Waterisapureliquidandisthereforenotconsideredintheequilibriumexpression.(2)Theinitialconcentrationofbothproductsiszero.(3)Thechangeinconcentrationswillalwaysbe(–)forthereactantsand(+)fortheproducts,

buttheirstoichiometriccoefficientsfromthebalancedequationMUSTbeaccountedfor.

Forexample,if2A→3B,thenthechangein[A]=–2xandthechangein[B]=+3x.

329.(C)Therateoftheforwardreactioncontinuestodecreaseuntilabout1second.Theinstantaneousreactionrateisdeterminedbycalculatingtheslopeofthetangenttothepointofinterestontheconcentrationversustimecurve.AtpointX,therateoftheforwardreactionisstilldecreasingandtherateofthereversereactionisstillincreasing.AtpointXonthisgraph,theslopeofcurveAistheinverseoftheslopeofcurveB.

330.(D)LeChatlier’sprinciplesaysthatwhenachemicalsystematequilibriumexperiencesachangeinconcentration,theequilibriumshiftstocounteractthechangeimposedonit.Oncethesystemregainsitsequilibriumstate,theratioofproductstoreactantsstaysthesame(onlytemperaturechangesaffecttheKeqofthereaction).

Chapter9:Acid–BaseChemistry

331.(B)DefinitionofaBronsted-Lowryacid.

332.(D)DefinitionofaLewisbase.

333.(E)DefinitionofanArrheniusbase.

334.(D)Coordinatecovalentbondsarecovalentbondsinwhichoneatomdonatesbothelectronstothebond,insteadofeachatomdonatingoneelectron.Thereisnodifferenceinthecharacterofthebond,onlyhowitforms.LewisbasesalwaysformcoordinatecovalentbondswithacidssinceLewisbasesareelectronpairdonors.

335.(B)Choice(B)isamixtureofastrongbaseandaweakbase.

336.(C)Choice(C)isamixtureofastrongacidandaneutralsalt.TheCl−ionisaverypoorconjugatebaseandwillnotpickupanyprotonsfromthesolution,sothepHis0.(Agoodfacttomemorize:ThepHofa1Mconcentrationofastrongacid=0.)

337.(A)Aweakbaseandaweakacidwillneutralizeeachother.

338.(E)Analkalinebufferconsistsofaweakbaseandthesaltofitsconjugateacid.AbufferresistschangesinpHbyactinglikea“protonsponge,”givingthemtothesolutionwhenthe[H+]decreases,andabsorbingthemfromthesolutionwhenthe[H+]increases.

339.(D)Anacidicbufferconsistsofaweakacidandthesaltofitsconjugatebase.AbufferresistschangesinpHbyactinglikea“protonsponge,”givingthemtothesolutionwhenthe[H+]decreases,andabsorbingthemfromthesolutionwhenthe[H+]increases.

340.(B)WecanusetheexpressionforKa.

Let[HNO2]=0.02–xbutbecausethevalueofKaissolowandwewillnotbecalculatingthevalueofthe[H+]tomorethantwoorthreesignificantfigures,thevalueof(0.02–x)isindistinguishablefrom0.02,sowecandropthexinthedenominatorandgreatlysimplifyourmath.

341.(E)Inthereaction,F−istheprotonacceptor,thebase,andH2Oistheprotondonor,oracid.WhendealingwithBronsted–Lowryacidsandbases,theacidandbasearealwaysthereactants,andtheirconjugatesarealwaystheproducts.Thebase,onceitacceptsanH+,becomestheconjugateacidandtheacid,onceitdonatesanH+,becomestheconjugatebase.

342.(A)WhendealingwithBronsted–Lowryacidsandbases,theacidandbasearealwaysthereactantsandtheirconjugatesarealwaystheproducts.SinceH2Oisareactant,itcanonlybeanacidorbase,notaconjugate.Becarefulwithquestionsthataskaboutthereversereaction.Forexample,wateristheconjugateacidoftheOH−ioninthereverseofthisreaction.Butintheforwardreaction,H2OisgivingtheH−toNH3,soitisanacid.

343.(E)Anincreasednumberofhydrogensdoesnotmakeanoxyacidastronger

acid.Ifanacidhasmorethanonedissociableproton,itisapolyproticacid.TheKaofthefirstH+isthehighest,andtheKadecreasesforeachsuccessiveH+.(SeeAnswer324foralistoffactorsthataffectthestrengthofoxyacids.)(SeeAnswer11foranexceptquestionstrategy.)

344.(E)Whenmixingtwosolutions,onlyconcentrationsintermediatetothesolutionsbeingmixedcanbeachieved.Inthiscase,wecan’tmakeasolutionthatislessconcentratedthan0.25Mandwecan’tgetmoreconcentratedthan0.35M,soallconcentrationsachievablebymixingthesetwosolutions,nomatterwhattheratios,willbebetweenthesetwovalues.

345.(C)TheKafortheacidHAis5.0×10−9.AKaofsuchsmallmagnitudemeanstheacidisveryweak.Comparedto0.5Mconcentration,theamountthatdissociatesisalmostnegligibleincalculationsinwhichtheanswerwillcontaintwotothreesignificantfigures.Thismakesthesecalculationsmucheasierbecauseyoucanignorethevalueofxinthedenominator,asfollows.StartbysettingupthechemicalequationandtheequilibriumexpressionforthevalueofKa:

346.(B)Anequilibriumconstantof2.5×103islarge,andsotheproductsaregreatlyfavoredinthisreaction.Y−istheprotonacceptorintheforwardreactionandX−playsthatroleinthereversereaction.Butsincetheforwardreactionisfavoredsomuchmorethanthereverse,wecaninferthatY−isastrongerbasethanX−.WecanalsoinferthatHXisastrongeracidthanHY,butthatisnotananswerchoice.Asforconjugateacidsandbases,X−istheconjugatebaseofHX,andHYistheconjugateacidofY−.AconjugateacidwilldifferfromitsbasebyonemoreH+.AconjugatebasewilldifferfromitsacidbyonelessH+.

347.(D)Theequivalencepointwherethenumberofmolesofbase(oracid)addedequalsthenumberofmolesofacidinitiallypresent(orbase).Itismidwayalongthesteeppartofthecurve,theinflectionpoint.ThepHattheequivalencepointwillbe7forastrongacid/strongbasetitration,butwillbegreaterthan7whentitratingaweakacidwithastrongbase.Itwillbebelow7whentitratingaweakbasewithastrongacid.ThisisbecausetheequivalencepointisnotwhentheactualnumberofOH−ionsisequaltotheactualnumberofH+ions;itis,inthiscase,whenthenumberofOH−ionsaddedequalsthenumberofH+ionsinitiallypresent,thatis,neutralization.

348.(B)Theregionofthecurvebetween0andabout25mLNaOHaddedisthebufferregion.Atthelowerendofthebufferregion,theamountofweakacidexceedsthatoftheconjugatebaseandtowardtheend,theconjugatebaseispresentinmuchgreaterconcentrations.Midway,theweakacid/conjugatebaseconcentrationsareaboutequal.

349.(A)AbufferresistschangesinpHbyactinglikea“protonsponge,”givingthemtothesolutionwhenthe[H+]decreases,andabsorbingthemfromthesolutionwhenthe[H+]increases.Abufferconsistsofanacidorbaseandthesaltofitsconjugatebaseoracid.Theregionofthecurvebetween0andabout25mLNaOHaddedisthebufferregion.Theweakacidanditsconjugatebasearebothpresentandverylittleneutralizationhasoccurred.

350.(D)Thequestionisaskingustoidentifythecompoundthatwouldproduceabasicsolution.LiFisasolublesalt(anyionicwithanalkalimetalissoluble)thatproducesF−upondissociation.F−isagoodconjugatebaseandreadilypicksupprotonsinsolution,increasingthepH.

351.(E)Conjugateacid/basepairsalwaysdifferbyoneproton(H+).Ifwearelookingforaconjugateacid,itmeansthatNH3isactingasabase,acceptingaproton.TheconjugateacidwouldbeNH4

+,onemoreH+.

352.(B)TwohandyrulesregardingpHandpOHare:

Thisproblemcanbesolvedintwoways.(1)IfthepH=6,thepOH=8∴the

[OH−]=1×10−8.(2)IfthepH=6,the[H+]=1×10−6the[OH−]<1×10−8.Thismakessensebecauseweexpectthe[OH−]<1×10−7inanacidicsolution(pH6).

353.(B)ToanswerthisquestioncorrectlywemustrememberthatthepHscaleisalog10scale,soapHof6is10timesmoreacidicthanapHof7and100timesmoreacidicthanapHof8.IfwewanttoacidifythepHby1pointonthepHscale,wemusteitherreducethe[OH−]by10andincreasethe[H+]by10(anincreaseinonewillalwaysresultinadecreaseoftheother,seethetworulesinAnswer352).SowemustdilutethebasicsolutiontenfoldtodecreasethepHfrom13to12.

354.(E)ThevaluesforK1,K2,andK3decreasesuccessively,soweexpecttheanionofthethirddeprotonationtobepresentinthelowestconcentration.WecanalsoconsiderthatonlyasmallfractionofH3C6H5O7willloseaproton,leavingH2C6H5O7

−behind.Let’ssay1percenttomakeitsimple,althoughit’sactuallymuchlowerthanthat.OutofthesmallnumberofH2C6H5O7

−ions,onlyasmallfractionofthosewilllosethesecondproton,leavingHC6H5O7

2–behind.Again,let’ssay1percentforsimplicity.Thatmeansthat1percentof1percent(0.01percent)oftheoriginalH3C6H5O7losestwoprotons.Finally,if1percentofHC6H5O7

2–losesthethirdproton,only1percentof0.01percentoftheoriginalH3C6H5O7wouldhavelostallthreeprotons(0.0001percent).

355.(C)AbufferresistschangesinpHbyactinglikea“protonsponge,”givingthemtothesolutionwhenthe[H+]decreasesandabsorbingthemfromthesolutionwhenthe[H+]increases.Abufferconsistsofanacidorbase,andthesaltofitsconjugatebaseoracid.ChoiceIrepresentsaweakbaseanditsconjugateacid,andchoiceIIIrepresentsaweakacidanditsconjugatebase.ChoiceII,however,representsastrongacidandthesaltofitsconjugatebase.Theconjugatebasesofstrongacidsareveryweakandarenotabletopickupprotonsinsolution,sotheyarenotgoodbuffers.

356.(E)0.2L×0.2mol/L=0.04moleSr(OH)2but×twoOH−perSr(OH)2,so0.08moleOH–

0.8×.8×2=1.28

1.36MOH=pHof14.

357.(D)Thefinalvolumeis200mL.WecanuseM1V1=M2V2tosolvethisproblem:(100)(0.002)=(M2)(200)=0.001M,orjustrealizethattheconcentrationhasbeenhalvedbyaddinganequalvolumeofwatertothesolution.HClisastrongacidandsotheconcentrationofHClisequaltothe[H+].pH=–log[H+]∴–log[1×10−3]=3.

358.(D)Thefinalvolumewillbe20mL.WecanuseM1V1=M2V2.(5)(0.02)=(M2)(20)=0.005MbutthatisthemolarityoftheBaOH2solution,notthe[OH−]concentration.TherearetwoOH−perBaOH2sotheconcentrationofOH−istwotimesthatofthe[BaOH2]∴0.01M.ThepOH=–log[OH−]=–log[1×10−2]=2∴pH=12.

359.(C)Phenolphthaleinchangesfromclearinacidtopinkinbase.Weaddthephenolphthaleintotheacidandtitratewiththebase.Whentheacidwearetitratingturnspink,weknowtheendpointhasbeenreached.

360.(C)Thenumbersonthegraduationsdifferonthetwoburettes,soweneedtobecarefultoreadbothburettesandnotjusttakethedifferencebetweentheheightsofthem.Sinceit’sthesamesolution,wecanreadfromthemeniscusorwecanreadfromtheedge.Itdoesn’tmatterinaquestionlikethisbecausewe’llgetthesameanswereitherway.Wewillreadfromtheedgebecauseit’sclosesttoawholenumber,10,onthefirstburette,and41.5onthesecondburette.

Thechangeinvolume=Vfinal–Vinitial=41.5–10=31.5mL.

361.(C)WeuseMaVa=MbVb,butweneedtorememberthatSr(OH)2hastwomolesofOH−ionspermoleSr(OH)2.[(Ma)(35)=(0.5×2)(25)]×Ma=~0.7M.

Chapter10:Electrochemistry

362.(B)Theanswertothisquestionisveryclearifweknowasimplefact:Zntakesononlyoneoxidationstate,+2.(AtrickforrememberingtheoxidationstatesofAg,Zn,andAl:theyareconnecteddiagonallyontheperiodictable,andasweprogressfromAgtoZntoAl,theoxidationstatesare+1,+2,and+3.)

363.(E)Thesumofalltheoxidationstatesofalltheatomsmustsumtotheoverallchargeofthecompound.Inthiscase,theoxidationstateofperchloricacidiszero:

H=+1andO=–2(×4)=–8∴Cl=+7

364.(E)Thesumoftheoxidationnumbersontheatomsinthechromiteion,mustequalthechargeonthecompound,–2.O=–2(×4)––8∴2Cr=

+6∴1Cr=+3.Theanswerchoicesprovidesomecomplicatedcompoundsforassigningoxidationnumbers.Trytheeasiestonesfirst.Cr2O3lookspromisingbecauseitonlyhasthreeoxygenatomsandnocharge.The–2chargeonnegatedtheeffectoftheextraoxygen,sotheCratominCr2O3shouldbethesame(anditis).Chromiumtakesonoxidationstatesrangingfrom–2to+6.Negativeoxidationstatesarenotcommonformetals.Cr2(O2CCH3)4ischromium(II)acetate,andfeaturesaquadruplebond(whichoccursextremelyrarely).Cr(CO)6ischromiumhexacarbonyl,acompoundinwhichCrtakesonazerooxidationstateinthiscompound.TheCrindichromate, ,hasanoxidationstateof+6,andpotassiumperoxochromate,K3[Cr(O2)4],featuresa+5oxidationstateforCr.

365.(E)Theelectricalconductivityofasolutionisdeterminedmainlybytheconcentrationofitsionsanditstemperature.Theconcentrationofionsinsolutionsequimolarwithrespecttocompoundsvariesaccordingtosolubilityandthevan’tHofffactor(i).Strongacidsandbases,andsolublesaltsaregoodelectrolytesbecausetheydissociatecompletely.H3PO4isaweakacidandthereforedoesnotdissociateinto3H+anda aswemightexpect;however,ifdissolvedinabasicsolution,itcan.TheKavaluesfortheprotonsare7.5×10−3,6.2×10−8,and4.8×10−13.Foragivenacidinsolution,ifthepH<pKa“theprotonison”andifthepH<pKa,“theprotonisoff.”

366.(A)Someoxidationandreductionreactionsarespontaneous(exergonic).OurtableofStandardReductionPotentialsisanincompletelistofthem.ReductionsthathaveapositiveE°valueareexergonic,whichmeansthereversereaction,oroxidation,isendergonic.Thebottomofthetablewouldhaveendergonicreductionsandexergonicoxidations.NoticereducingF2isveryexergonicandreducingLi+isveryendergonic.Thereductionof(1M)Cl2produces1.36V,whiletheoxidationof(1M)Br−requires1.07V.The

differencebetweenthem,0.29V,isstillfavorable(+ΔE=spontaneous)sothisredoxreactionisspontaneousandcangenerateelectricity.WecanthinkofthisintermsofhowmuchCl2“wants”tobereducedversushowmuchBr−“doesn’twant”tobeoxidized.ThedifferenceintheΔE°valuesquantifieshowgreatlytheir“desires”differandwhichspecieswillgetwhatitwants.ComparingtheE°valuesforthetwoelementstellsusthattheabilityofCl2totakeelectronsfromBr−isgreaterthantheabilityofBr−topreventCl2fromtakingthem.

367.(C)StatementIisnottruebecauseCustartsoutinthesolid,reducedformandendsupasa2+ion,soitiscertainlynottheoxidizingagent.(Agentsgettheoppositeofwhattheydo.Iftheagentoxidizes,itgetsreduced.)StatementIIisalsonottrue.Theoxidationstateofhydrogendoesnotchangeinthereaction.Itis+1asanionand+1inwater.

368.(C)Wealmostcertainlyhaveheardthemnemonicdevicethatappliestoallelectrochemicalcells(galvanicandelectrolytic),AnOx,RedCat,whichtranslatestoAnode=Oxidation,ReductionattheCathode.Thetworeactionsarewrittenasreductions,andsooneofthemwillneedtobewritteninreverse.Sincetheyareoppositesigns,weshouldflipthenonspontaneousreaction,theonewiththe–ΔE°.Silverisreducedatthecathodeproducing0.80V,andZnisoxidizedattheanode,producing0.76V.Totalcellpotential=1.56V.TheZn+iondoesnotparticipateinthisreaction.

369.(D)SeeexplanationforAnswer368.

370.(D)SeeAnswers243andforexamplesofhowtoassignoxidationnumbers.

371.(C)Aluminumgetsreduced(itsoxidationstatechangesfrom+3to0)andoxygengetsoxidized(itsoxidationstatechangesfrom–2to0).AnOx,RedCat,whichisAnode–Oxidation,ReductionattheCathode.

372.(D)Anamperequantifiestherateatwhichcharge(measuredinCoulombs,C)flowsinC/sec.Thechargeononeprotonorelectronis+/–1.6×10−19C(the+1and–1weuseinotherchemistrychaptersrefertoanelementarycharge,thenumericalvalueofwhichis+/–1.6×10−19C).Afaradayisthemagnitudeoftheelectricalchargeononemoleofelectrons(orprotons).Acurrentof15amperesmeans15C,or15moleelectrons,areavailablepersecond(passinto

thecell),for20minutes(1,200seconds).Thenumbers20and60inthenumeratorconvertminutestoseconds.Thenumber27convertsgramsofAlintomoles.

Thenumber15inthenumeratorconvertstheamperageandtimeintoCoulombs:amps(C/s)×time(sec,fromtheproductof20×60)=Coulombs.1molee−=96,500C,soif18,000Care0.187molee−enterthecell.EachmoleofAlrequires3molesofe−togetreduced,so0.187÷3=0.062moleofAlwillgetreduced.At27gmol−1,that’s1.7gAl.Aninterestingfactaboutaluminumisthatit’sthethirdmostabundantelementonearth,butitselectrolyticproductionisveryenergyintensive.Approximately4percentoftheelectricityconsumptionintheUnitedStatesisusedfortheproductionofaluminum.

373.(D)ThereductionofX2+toXisthereverseofReaction1,sotheDE°ofthereactionis–2.27V,buttheoxidationofAgalsooccurs.Reaction2givesusthereductionofAg+,whichproduces0.80V,soitsoxidationis–0.8V.SubtractingtheoxidationofAg(–0.80V)fromReaction1(–2.27V)leavesuswiththehalf-reactionforthereductionofXandastandardreductionpotentialof–1.47V.

374.(D)Anamperequantifiestherateatwhichcharge(measuredinCoulombs,C)flowsinonesecond,sotheunitofcurrentisC/sec.Thechargeononeprotonorelectronis+/–1.6×10−19C(the+1and–1weuseinotherchemistrychaptersreferstoanelementarycharge,thenumericalvalueofwhichis+/–1.6×10−19C).Afaradayisthemagnitudeoftheelectricalchargeononemoleofelectrons(orprotons).1.00ampere=1C/sec.TheoxidationstateonFeinFeCl3is+3,soitwillrequire3molesofe–toreduce1moleFe3+.

3molesofe−=96,500C×3=289,500Crequired.Deliveredatarateof1.00C/sec,thatwouldrequire289,500seconds(morethan80hours).

375.(B)Asimplewaytoeliminatesomeanswerchoicesistocountelectrons.WeneedtorememberwearedealingwiththeZn2+ion,whichhastwofewerelectronsthanZn∴30–2=28electrons.Choices(B),(D),and(E)have28electronssowe’vealreadyeliminatedtwochoices.Theelectronsofhighestenergy(n)arelostfirst,sotheZnatomlosesits4selectronsleavingafulldsubshell.Thatleavesonlychoice(B).

376.(D)ThenumberofpositivechargesinthenucleusofZn2+isstill30,butnowthose30protonsareonlypullingon28electrons.

377.(A)UsethemnemonicAnOx,RedCat,whichcomestoAnode=Oxidation,ReductionattheCathode.Thequestionis,whichspecies—ZnorH2—getsoxidized.Wemayrememberthatinthehydrogenelectrode,2H+arereducedtoH2,butifwedon’t,that’sfine.WeknowthecellisspontaneousbecauseitisagalvaniccellandthereforetheE°ofthecellmusthaveapositivenumber.SincethereductionofZn2+isnegative,theoxidationispositive,andourcellwillworkspontaneously.Thehydrogenelectrodeisthecathode,sochoicesIIIandIVdonotapply.

378.(D)Thesaltbridgehastwofunctionsinthegalvaniccell.(1)Itcompletesthecircuitbyprovidingabridgeofmobileionsbetweenthetwohalfcellsand(2)itprovidesionstothehalfcellsextendingthelifeofthebattery.

Attheanode,theoxidationofsolidZnproducesZn2+ionsinsolution.ThesaltbridgeneutralizestheexcesspositivechargesbyprovidingNO3

−ions.Atthecathode,H+ionsarebeingreducedtoH2gas,sothesaltbridgeprovidesK+ionstoreplacethelostcations,neutralizingtheCl−ions.

379.(E)It’snotpossibletomeasuretheE°ofahalfreaction(orhalfcell)sothereductionof2H+toH2wasgiventhearbitraryvalueofzeroandalltheotherstandardreductionpotentialsweremeasuredrelativetoit.

380.(C)Weknowthecellisspontaneousbecauseitisagalvaniccell,notanelectrolyticcell.SpontaneouscellshaveanegativeΔGandthesumoftheoxidationandreductionpotentialsmustbepositive.

381.(E)TheNernstequationrelateshalf-cellconcentrationstotheEofthecellbyE=E°=(0.0257V/n)lnQ,whereQisthereactionquotient.Asthecelloperates,theflowofelectronsfromtheanodetothecathoderesultsintheformationofproductandtheconsumptionofreactant.Thisincreasesproductconcentrationanddecreasesreactantconcentration.Asthecellapproachesequilibrium,E=0(andQ=Keq).Zn2+istheproductoftheoxidationthatoccursattheanode.

382.(A)Asacrificialanodeisalsocalledagalvanicanode,andisametalwithamorenegativeelectrochemicalpotential.Thisallowsthesacrificialanode,theZn,tobeoxidizedpreferentiallytothe“cathode,”inthiscase,theiron.

383.(B)Electricalenergy=electricalpotential×thebaseunitofcharge=Voltage×Coulombs.Avolt=1JC−1∴electricalenergycanbeexpressedinjoules.Thesumofthetwohalf-cellsonlydeterminesthemaximumvoltageofthecell,nothowmuchenergythecellcangenerate.

384.(C)Inanelectrochemicalwetcell,concentrationdoesmatter.TheE°valuesonthetableofStandardReductionPotentialsareforstandardconditionsthatinclude1.0Mconcentrations.TheNernstequationisusedtocalculatetheelectromotiveforce(voltage)ofacellundernonstandardconditions.(SeeAnswer11foranexceptquestionstrategy.)

385.(B)FromthebalancedredoxequationwecanseethatLi(s)isoxidized,andthenweremember:AnOx,RedCat:Anode=Oxidation,ReductionattheCathode.

386.(E)Choices(A)through(D)makelithium-ionbatteriessuperiortomanyothertypesofdrycells,buttheyareveryreactive.Theycanexplodeifnothandledwithcare.(SeeAnswer11foranexceptquestionstrategy.)

387.(A)AllelectrochemicalcellsabidebyAnOx,RedCat:

Anode=Oxidation,ReductionattheCathode.However,galvanic(voltaic)cellsarespontaneousandconvertchemicalenergyintoelectricalenergywhereaselectrolyticcellsrequireenergytocausenonspontaneousredoxreactions(electrolysis).

Chapter11:NuclearChemistry

388.(E)Whenbalancingnuclearreactions,remember:(1)Conservationofmassnumber(thenumberofprotonsandneutronsareequalintheproductsandreactants)and(2)Conservationofatomicnumber(thenumberofnuclearchargesintheproductsandreactantsarethesame).

Totalmassnumberinreactants=236

Totalatomicnumberinreactants=92

Massnumberinproductsmust=236∴236–141–(3×1)=92

Atomicnumberinproductsmust=92∴92–55=37,rubidium

389.(B)SeeAnswer388.

Totalmassnumberinreactants=31

Totalatomicnumberinreactants=15

Massnumberinproductsmust=31∴31–1=30

Atomicnumberinproductsmust=15∴15–0=15,phosphorus

390.(B)SeeAnswer388.

Totalmassnumberinproducts=18

Totalatomicnumberinproducts=9

Massnumberinreactantsmust=18∴18–4=14

Atomicnumberinreactantsmust=9∴9–2=7,nitrogen

391.(D)Choice(D)=gammaradiation(γ),ahigh-frequencyformofelectromagneticradiation(EMR).VisiblelightisaverylowfrequencyofEMRrelativetogamma,butthebasicpropertiesofEMRareitswave-andparticle-likeproperties,anditslackofmass.

392.(A)Choice(A)isalphadecay.Analphaparticleisaheliumnucleus,withamassnumberof4.Itmakesitthemostmassiveparticleandthereforeleastpenetrativeformofradioactivedecay.(Foragivenvalueofkineticenergy,moremassiveparticleshavelowerspeed.)

393.(D)Choice(D)=gammaradiation(γ),thehighestfrequencyofelectromagneticradiation(EMR)known.ThefrequencyofEMRisproportionaltoitsenergy,sogammaradiationisthehighestenergyEMRknown.Thefrequencyofgammaradiationisontheorderofmagnitudeof1019Hz,andthewavelengthsarelessthanthediameterofanatom.Gammaradiationisionizing

becauseithasenoughenergytodislodgeelectronsfromatomsandmoleculesandproducefreeradicalsinthebody.

394.(A)Choice(A)isanalphaparticle,aheliumnucleuswitha2=charge.Otherchargeddecayparticleshavea+1or–1charge.

395.(B)Choice(B)isbetadecay,aformofdecayinwhichaneutronturnsintoaprotonandelectron.Theelectronisejectedfromthenucleusasabeta-particle(0−1eor1−1β),andtheatomicnumberincreasesby1,leavingthemassnumberintact(sothenumberofneutronsdecreasesby1).Beta-decaydecreasestheneutrontoprotonratioofanunstablenucleus.

396.(C)Nuclearbindingenergyisanindicationofthestabilityofanucleus.Thebindingenergyistheenergyrequiredtobreakupanucleusintoprotonsandneutrons.Thenucleuswiththehighestbindingenergypernucleonisthemoststable.

397.(A)Choice(A)isfalsebecausethecurverepresentsbindingenergypernucleon,notpernuclide.Moreimportantly,thestabilityofanucleusisthesumoftwoforces,attractiveandrepulsive.Thenetattractiveforceiswhatultimatelydeterminesthestabilityofthenucleus,notthetotalattractiveforce(withoutconsideringtherepulsiveforces).(SeeAnswer11foranexceptquestionstrategy.)

398.(C)Thedefinitionofmassdefect.Thebindingenergyistheenergyrequiredtobreakupanucleusintoprotonsandneutrons.

399.(D)ThelostmassisaccountedforbyEinstein’sfamousmass–energyequivalenceequation,E=mc2.Sincethevalueofc2=9×1016,atinypieceofmasslostfromthenucleusistransformedintoalotofenergy.

400.(E)StatementIcorrectlyaccountsforthechemicalreactivityofthealkalimetals.Chemicalreactivity(orstability)isafunctionofelectronconfiguration,whereasradioactivityisafunctionofnuclearstability(neutron-to-protonratio).StatementIIisincorrect;radioactivenucleiincreasetheirstabilitybydecaying,oftentransmutingintoanotherspeciesofatom.StatementIIIiscorrectaboutallthenon-noblegaselements;theyallformmoleculesthataremorestablethantheloneatomsoftheelement.The“motivation”ofatomsistofilltheirvalenceshellbygiving,taking,orsharingelectronswithotheratoms.

401.(A)Thisisthestrongnuclearforce,anditonlyworkswhentheprotonsveryclosetoeachotherareontheorderofabout10−15m.Thenucleusonlycarriespositivecharges(protons)althoughitcancreateandemitelectrons(β-particles,00e−)throughβ-decay.Thenucleusisdense,sotheprotonscan’tbespreadoutenoughtoescapethelike-chargerepulsion.Theneutronsdon’tformacagearoundtheprotons.Therewouldneedtobeaforcethatholdstheneutronstogetherincageformation,andthatforcewouldneedtoexceedtherepulsiveforceofthelike-chargerepulsiontobeaneffectivecage.Besidesthefactthatitisn’ttrue,itdoesn’texplainanythingabouttheforces;itonlyimpliesanewforceexertedbyadifferentparticle.

402.(C)Theaveragemass(1×10−22g)÷thevolumeofthenucleus(4/3π(5×10−13cm)3)=2×1014gcm−3(πinthisequation=3.14,theratioofthediameterofacircletoitscircumference.Itdoesnotrefertothebond).Wedidn’thavetodoallthatmath,however.Toestimateanorderofmagnitudeweuse10−22÷(10−13)3=10−22÷10−39=1017.Thatisthreeordersofmagnitudeoff,whichis1,000timestoobig(abigerror!),buttheAPChemistryexamwon’tgiveusacalculationonthemultiplechoicesectionoftheexamthatrequiresacalculatortocomputeinareasonableamountoftime.Aquestionwithgiganticorinfinitesimalnumberswilloffersignificantlydifferentquantitiesasanswerchoicessoyoucanroundgenerously.

403.(C)Onehalf-life=→,so30g→15g→7.5g→3.75g→1.875g→0.94g.Thedecaytolessthan1gramtookfivehalf-lives(fivearrows).325days÷5half-lives=65daysperhalf-life.

404.(B)Infourhalf-lives,93.75percentofasampledecays(seetablebelow).24days÷4half-lives=6daysperhalf-life.

405.(C)Afterthreehalf-lives,12.5percentofasampleremainsundecayed(seetableinAnswer404).Ifthreehalf-livestake30days,eachhalf-life=10days.

406.(E)Therearesix,two-yearhalf-livesin12years.ThetableinAnswer404showsthatafterfivehalf-lives,about3percentoftheradioisotoperemains

undecayed(about1.9goutoftheoriginal60).Onemorehalf-lifewouldreducethatbyhalf,leavinglessthan,butcloseto,1gram.

407.(B)Whenanatom’smassnumberstaysthesamebuttheatomicnumberincreases,theatomhasundergoneaβ−(beta)decay.Whenthemassnumberofanatomstaysthesamebuttheatomicnumberdecreases,theatomhasundergoneaβ−(positron)decay.About0.01percentofKatomsare40K.β–decaydecreasestheproton-to-neutronratiobyconvertinganeutronintoaprotonandanelectron(β-particle,00e−),theprotonstaysinthenucleus(sotheatomicnumberincreasesbyone)andtheβ–particle,theelectron,isejected.Becausethetotalnumberofprotonsandneutronsstayedthesame,themassnumberdoesn’tchange.

408.(E)Radioactivedecaydisplaysfirst-orderkinetics.Choice(E)isaplotofafirst-orderreaction(withrespecttosubstratedisappearance).Choice(B)isaplotofasecond-orderreaction(withrespecttosubstratedisappearance).

Chapter12:Descriptive

409.(D)Siliconcombineswithoxygentoformavarietyofsilicates.Themostcommonsilicateissilicondioxide,whichhasanumberofdifferentcrystallineandamorphousforms.Silicondioxideisacovalentnetworksolid,andtheformulaSiO2istheempiricalformulaofthesolid.

410.(C)NaClisthesaltmostresponsibleforthesalinityofseawaterandthemainingredientintablesalt(whichmayalsocontainminuteamountsofiodinesalts,addedtopreventiodinedeficiency).Anyioniccompoundwithalkalimetalissoluble,andcompoundsformedwithalkaliandalkaliearthmetals(groups1and2)astheonlymetalsaregenerallynotcolored.

411.(B)CuSO4isadeliquescentcompound.Itiswhitewhenanhydrous(withoutwater)andturnsblueuponhydration.(SeeAnswer454forasummaryofpropertiesrelatedtodeliquescence.)

412.(D)Mostchloridessaltsaresoluble,withthenotableexceptionofAgCl,HgCl2,andPbCl2.TheprocessofeliminationmayhaveallowedustoidentifythiscompoundsinceNH4NO3andNaClarebothwhite,KMNO4ispurple,andCuSO4iswhiteorbluedependingonitsstateofhydration.

413.(A)Potassiumpermanganate,KMnO4containsmanganese.Thepresenceoftransitionmetalsincompoundsoftenresultsinbrightcolors,eventhoughthepuremetalsaremostlysilver(withthenotableexceptionsofcopperandgold).KMnO4isastrongoxidizingagent.

414.(B)CuSO4isadeliquescentcompound,whichmeansithasahighaffinityforwatermoleculesandcanabsorbthemfromtheair.CuSO4changescolorwhenithydrates(itturnsblue),makingCuSO4anexcellentdesiccant(becauseweknowwhenit’sfullofwaterandneedstobedried).Adesiccantisasubstanceusedtocreateorsustainadryenvironmentbyabsorbingwaterfromit.

415.(E)Ammoniumnitrate,NH4NO3,isverysoluble.(Allioniccompoundscontainingthe ionortheNO3

−ionaresoluble,andthisonehasboth.)Itisusedforfertilizersnotonlybecauseitishighinnitrogen,butbecausethetwoformsofnitrogen(ammoniumandnitrate)arethemostreadilyabsorbedandusedbyplants.ThemostabundantgasintheatmosphereisN2(78percent).TheleastabundantisCO2(lessthan1percent,butithasconsiderableeffects).Plantscanonlyusethecarbondioxideintheatmosphereasasubstrateinorganicsyntheses,theycannotuseN2asasourceofnitrogen.OnlyafewkindsofbacteriacanmetabolizeN2intoaformthatisusablebyplants.

416.(A)Propaneisagaseousalkanethatisoftencompressedintoaliquidfortransport.Itisaby-productofnaturalgasprocessingandthefractionaldistillationofpetroleum.Hydrocarbonsareusefulfuelsbecausetheircombustionhasanegativefreeenergychangeofgreatmagnitudeandyieldsagreatnumberofmolesofgasthatexpandrapidlyatthehightemperaturesproducedbythereaction.(Theyarehighlyreduced,sothere’salottooxidize.)

417.(E)Trichlorofluoromethaneisachlorofluorocarbon.Itiscolorless,practicallyodorless,andboilsatroomtemperature.Clorofluorocarbons,orCFCs,wereusedasrefrigerantsuntilitbecameclearthatCFCscausethebreakdownofozoneinthepresenceofultravioletlight,thefrequenciesofelectromagneticradiation(1015to1016Hz,sec−1)fromwhichtheozonelayerprotectsEarth’ssurface.

418.(C)Hydrogenperoxideisastrongoxidizingagent,whichmakesitusefulforkillingbacteria(byoxidativestress).Butitalsomayharmthecellsofthe

body(byoxidativestress),whichmaypreventhealing.Hydrogenperoxideisanaturalby-productofallaerobicorganisms,butcellscontainenzymesthatdecomposeH2O2toO2andH2Oatthe(low)concentrationsproducedincells(ascomparedtothe3-percentsolutioninthepharmacy).

419.(D)Hydrogensulfidehastheodorofrotteneggs.Itisproducedbythedecompositionoforganicmatterunderanaerobicconditionsandisalsofoundinvolcanicgases.

420.(B)RememberthatthehydrogenhalidesotherthanHF(HCl,HBr,andHI)arestrongacids.HFisconsideredweak,butitisstillhighlycorrosive,henceitsuseinetchingglass.Becauseofitsreactionwithglass(andmetal),HFmustbestoredinplasticcontainers.

421.(C)(SeeAnswer94foranexplanationofhowgraphiteconductselectricity.)

422.(E)Allaminoacidscontainthecarboxylandaminofunctionalgroups.Thecarboxylgroup=COOH(physiologicalpH,7.2–7.4,isabovethepKaofthecarboxylgroup,soitisinitsdeprotonatedform,COO−,inbodyfluids)andtheaminogroup=NH2(physiologicalpHisbelowthepKaoftheaminogroupsoitisinitsprotonatedform,NH3

+,inbodyfluids).ThefourelementsC,H,O,andNmakeup96percentoflivingmatter(easilyrememberedasCHON).

423.(D)ThisquestionisreallyaskingustorecognizeSiO2asacovalentnetworksolid(oratleastnotasagas)asopposedtoaskingustorecognizetherestoftheanswerchoicesasgases(althoughit’sagoodideatobeableto).(SeeAnswer11foranexceptquestionstrategy.)

ForQuestions424–428:

(A)Isoamylacetate(bananaoil)(B)Ethylmethylether(C)Benzoicacid(D)Benzone(E)Glucose

424.(D)TheC=Oandtheending“one”makethistheketone.

425.(E)Glucosehasfivehydroxylgroupspermoleculemakingitverymuchwatersoluble.

426.(A)Isoamylacetateisanesterandisresponsibleforthefruityodorfoundinbananasandmanyothersweet-smellingitems.

427.(C)Thecarboxylicacidgroupisrepresentedby–COOH.

428.(B)Anetherhastwoalkylchainsattachedtoasinglybondedoxygen.

Chapter13:Laboratoryprocedure

429.(E)Itisnotonlyacceptablebutitisstandardpracticetorinseaburettewiththesolutionthatwillbeaddedtoitbeforeitisfilledwiththesolution.Thisensuresthatanywaterorimpuritiesintroducedintotheburetteduringcleaningwillberemoved.Anywaterorimpuritiesleftintheburettewilldiluteorcontaminatethesolution,decreasingprecisionandaccuracy.Foroursafetyandthewell-beingofthebalance,neverplacehotobjectsonabalance.Forsomesubstances,thiswillintroduceerrorintotheweighing,aswell(aircurrentsareproducedaroundhotobjects,andsomesubstanceswillpickupmassfromorlosemasstotheenvironmentdependingonitstemperature).

Addinghotwatertoavolumetricflaskisalsoconsideredunsafe,mainlybecausethemouthoftheflaskhasasmalldiameter.Using5mLofphenolphthaleintotitrate20mLofacidmeansthat5outof25mLtotalvolumeofoursolutionisthephenolphthaleinindicator.That’s20percent!Finally,rememberthestrainedbutusefulrhyme“Slowlyyouought–taaddacidstowat-tah.”Thedissociationofanacid(orbase)inwateristypicallyexothermic,butforthestrongacidandbases,it’sstronglyexothermic.Theheatproducedbythedissociationcanrapidlyincreasethetemperatureofthewater.Slowlyaddingacidtowaterallowsthewatertoabsorbexcessheatfromthedissociationasyoupour,sotheheatproducedcanberegulatedbythepersonpouring.Italsopreventsaharmfulsplashback.Evenifsomeoftheliquiddoessplashback,it’slikelygoingtobethewaterthatwasdisplacedbytheacidthatgetssplashedinsteadoftheundilutedacid.(SeeAnswer430foradetailedexplanationofwhyweaddbasetowaterslowly.)

430.(D)Remembertherhyme,“Slowlyyouought–taaddbasestowat-tah.”Thedissociationofstrongbaseinwaterishighlyexothermic.Addingthebase

slowlytowaterallowsthewatertoabsorbexcessheatfromthedissociationandallowsthepourertoadjustthepouraccordingly.Ifthebaseisaddedtooquickly,oraninsufficientamountofwaterisaddedtoabase,thecontainercangettoohotandcanevenboil.Itcanalsosplatteroutofthecontainerduetothevigorousreaction.Thecompositionofthesplatterwillbeahot(possiblyboiling),concentratedsolutionofastrongbase.(SeeAnswer429foradetailedexplanationofwhyweaddacidstowaterslowly.)

431.(D)Thebestwaytodealwithanacidspillistorinsetheareatoremoveexcessacid,thenneutralizeitwithaweakbase.Neverusestrongacidsorbasestoneutralizeabaseoracidspill,especiallyonskin.Acidsandbasesneutralizeeachotherinahighlyexothermicreactionthatcanthermallyburnskin(inadditiontochemicallyburningit).

432.(B)Themostdirectandefficientmethodtodeterminethemolarityofthissolutionistomeasureitsvolume.Twopiecesofinformationareneededtocalculatemolarity(M=mol/L),molandL.Thestudentalreadyknowshowmanymolesofaceticacidarepresent,soallthatneedstobedoneistomeasurethetotalvolumeandplugitintotheexpressionformolarity.

433.(B)Themostdirectandefficientmethodtodeterminethemolalityistomeasurethemassofthesolution.Thestudentneedstwopiecesofinformationtocalculatemolality(m=mol/kgsolvent),molandkgsolvent.Thestudentknowsthemassofthephosphoricacid,sothenumberofmolescanbecalculated.Also,themassofthephosphoricacidhastobesubtractedfromthetotalmasstodeterminethemassofthesolvent,whichisneededfortheexpression.

434.(E)Theconductivityofasolutionisameasureofitsabilitytoconductelectricity.Itcanbequantifiedbymeasuringtheresistancetotheflowofelectricitybetweentwoelectrodesinthesolution.

435.(B)Chromatographyisageneraltermforalaboratoryprocedureusedtoseparatethecomponentsofamixture.Paperchromatographyusesatwo-phasesystem,asolventandpaper,toseparatethecomponentsofasolutionaccordingtotheirdifferentialsolubilitiesin(oraffinitiesfor)thesolventandthepaper.

436.(A)Theconcentrationofacoloredsolutioncanbedeterminedbycolorimetry,orvisible-lightspectrophotometry.Themoreconcentratedthesolution,thegreatertheabsorbance(orlesstransmittance)ataparticular

wavelength.Usingstandardsolutionsofknownconcentrations,agraphofabsorbance(ortransmittance)versusconcentrationisconstructedsothatanyabsorbance(ortransmittance)valuecanbelinkedtoaconcentrationbyinterpolation(obtaining“implied”datafromareasonagraphwithindiscrete,measureddatapoints).

437.(D)Gravimetricanalysisisamethodforanalyzingtheamountofasubstancebyweighingtheproductsofitsreactionwithsomethingelsethatisknown.Forexample,wecoulddeterminetheconcentrationofsulfateionsinasolutiontakingasampleofknownvolume,addingBa2+ionsuntilaprecipitatestopsforming,andthenweighing(afterfilteringandwashing)theprecipitate.Becausewecanmeasurethemassoftheprecipitate,weknowthemolarmassesofBaSO4andthestoichiometryofthereaction(Ba2+andSO4

2–reactina1:1ratio),andwecanfigureouthowmuchsulfatewasintheoriginalsolution.Differentialprecipitationexploitsthedifferentsolubilitiesofionsinthepresenceofotherionsorunderdifferentconditionstoselectivelyremovethemfromasolution.(SeeAnswer438foradescriptionofasimilarmethodofanalysis,titration.)

438.(C)Titrationisalaboratorymethodthatisusedtodetermineanunknownconcentrationofaknownreactantbystudyingitsreactionwithaknownconcentrationofadifferentknownreactant.Anindicatorthatthereactionhasoccurred,orhasfinishedoccurring,isrequired.Thetwovolumes,theconcentrationoftheknownsolutionandthestoichiometryofthereaction,arethenusedtocalculatetheunknownconcentration.Forexample,inAnswer437,wecouldhaveusedaBa2+solutionofknownconcentrationtocalculatethenumberofmolesof ifwehadasensitivewaytodeterminewhentheverylast ionsprecipitated(anindicatorthatletsusknowwematchedupeachevery ionpresentwithaBa2+ion).ThenwecoulduseM1V1=M2V2becauseweknowthemolarityoftheBa2+solution(Solution1),wemeasuredthevolumeneededto“capture”each (weknowtheyreactina1:1ratio),andweknowthevolumeofthe solution(Solution2)wetitrated.(SeeAnswer437foradescriptionofasimilarmethodofanalysis,gravimetricanalysis.)

439.(E)Aliquidinanopencontainerboilswhenthevaporpressureabovetheliquidreachesthepressureatmosphere.Pressurecookersareusedtocookfasterandathighertemperaturesthan100°C.Thetemperatureofwaterinanopen

containeratsealevelcannotexceed100°C,nomatterhowmuchheatisadded.Increasingthepressureinasealedcontaineristheequivalentofincreasingtheatmosphericpressurearoundanopencontainer.Inordertoevaporate(andboil),thewatermoleculesmusthaveenoughkineticenergy(KE)toreach“escapevelocity.”Thegreaterthepressurepushingdownonthesurfaceoftheliquid,thegreatervelocity(andthereforeKE)requiredtoescape.Therefore,ahighertemperature(averageKE)isrequiredtoboilit.Answerchoice(A)isnotcorrectbecausethestudentwouldhavetoaddalotofsalttomakethewaterboilsignificantlyhigher(theKbofwaterisonly0.52K/m).

440.(E)Thevolumeofmostsubstanceschangeswithtemperature,butthemassandtheamountofparticlesdon’tchangewithtemperature(aslongastheyareclosedsystems).Measurementsthatdealwithvolumedirectly(likemolarityanddensity)canbeunreliableifthetemperatureischangedsignificantly(whatconstitutesasignificanttemperaturechangedependsonthesubstance).

441.(B)Apipetwillmostaccuratelytransferaparticularvolumeofsolution.Aflaskisnotusedfortheaccuratemeasurementofvolumes.Agraduatedcylinderisappropriateformeasuringvolumes,butisnotthebestchoiceforthetransferofaspecificvolume(toomuchpouring).

442.(C)Thequestioncontainsthelimitingmeasurementforsignificantdigits,50.00mL.

443.(D)Avolumetricflaskofthepropervolume(eachflaskmeasuresonlyonevolume)isthebestpieceofequipmentforpreparingsolutionsormeasuringspecificvolumesthatrequireahighdegreeofaccuracy.Thebottomoftheflaskiswideandfitsmostofthesample,buttheneckofthevolumetricflaskisverynarrowsotheareaatthesurfaceofthesolutionissmall.Theerrorincurredbyaddinganextramm(inheight)ofwatertoacylinderwithanareaof1cm2isinsignificant(surfacearea×length=volume,0.1mLinthiscase)whencomparedtoaddinganextrammofwatertoacylinderwithanareaof25cm2

(2.5mL)fortwovesselsofthesamevolume.

444.(B) and arebothwatersolubleionsnomatterwhatotherionsarepresentsotheonlywaytoretrievethemfromanaqueoussolutionistoevaporatethewaterandcollectthedrysolid.

Chapter14:DataInterpretation

445.(C)Astatineistheheaviestknownhalogenandisproducedbytheradioactivedecayofotherelements.Itsshorthalf-life(7–8hours,dependingontheisotope)preventsitfrombeingpurifiedinsignificantquantities.Theastatinepurifiedinthequestiondecayedintobismuth.Thebismuththendecayedintolead.Wedidn’thavetoknowthedecayseriesofastatinetoanswerthisquestioncorrectly.Becauseithasanatomicnumberof85,weknowitisradioactiveandwilllikelytransmutateintootherelements.

446.(E)Eachcompoundisonlyusedonce,sowecaneliminateanswerchoicesevenifwe’renotsurehowtheparticularcompoundinthequestionreactswithammonia.

1=Silvernitrate2=Bariumchloride3=Copper(II)nitrate4=Mercury(I)nitrate5=Fe3+ions

447.(A)SeeAnswer446.

448.(C)Atlowconcentrations,NH3isaweakbaseandproduceshydroxideions,OH−,thatcombinewithNi2+toformNi(OH)2,whichisnormallyinsoluble.TheexcessNH3allowsittoformastable,soluble,complexionwithNi2+insteadofprecipitatingwiththeOH−.

449.(C)SolidZnreducesH+ionsinsolutiontoproduceH2(ZnisaboveHintheactivityseries,andcanthereforebeoxidizedbyit).Whenthe ionispresentinacidicsolutions,CO2gasisproducedaccordingtothis(worthmemorizing)reaction:

NH3,however,willnotproduceagaswhencombinedwithHCl.Instead,theweakbasewillpartiallyneutralizethestrongacid,producingasolublesalt(NH4Cl)thatwillremaininsolutionwiththeexcessH+ions.

450.(D)Na2CO3isaverysolublesalt(duetotheNa+)andtheCO32–it

producesinsolutionisaweakbase.Itwillpickupprotonstoform andevenH2CO3.WhentheCO3

2–ionispresentinacidicsolutions,CO2gasisproduced(seetheequationinAnswer449).

451.(D)TheLawofMultipleProportionsstatesthatiftwoelementscancombinetoformmorethanonecompound,thenthereisasmall,whole-numberratiocomparingthemassesinwhichoneelementcombineswiththefixedmassoftheother.ThesimplestwaytoillustratethisiswithH2OandH2O2.

Ifweholdthemassofhydrogenconstant,wecanseethattheratioofthemassofoxygeninH2O2comparedtowateris16:8or2:1.Thereisno“lawofstoichiometry.”Iftherewere,itwouldbeaderivedfromthelawsofdefiniteproportions,multipleproportions,andconservationofmass.

452.(A)Findingtheempiricalformulaonlyrequiresknowingthemoleratioinwhichtheelementsinacompoundcombine.

453.(C)Eachofthecompoundscontained28g,or½mol,Fe.Thetrickypartisthatthemassesofoxygenarethemassesofmolecularoxygen,notatomicoxygen.Tofindthenumberofmolesofoxygenatoms,weneedtofindthenumberofmolesofmolecularoxygenanddoubleit(therearetwooxygenatomsineachmolecule).Inthefirstcompound,½moleFecombinedwith½moleoxygenatoms(¼moleO2molecules)ina1:1moleratio(FeO).Inthesecondcompound,½moleFecombinedwith¼moleoxygenatoms(1/8moleO2molecules)ina2:1moleratio(Fe2O).Inthethirdcompound,½moleFecombinedwith2/3moleoxygenatoms(1/3moleO2molecules)ina3:4moleratio(Fe3O4).

454.(D)Efflorescenceisthelossofwaterfromasaltcrystaluponexposuretoair.Deliquescenceisthepropertyofhavingastrongaffinitytoabsorbwaterfromtheair.Itisapropertyofagooddessicant,asubstanceusedtocreateand/ormaintainastateofdryness.Hydrophilicissynonymouswiththetermwater-soluble.(SeeAnswer11foranexceptquestionstrategy.)

455.(B)Blue,ordry,CoCl2hasamolarmassof130gmol−1,so13g=0.1mole.Thesampleabsorbed4gofwaterfromtheatmosphereduringhandling,whichisequalto~0.2moleofwater(twicethenumberofmolesofCoCl2).TheformulaforthepurplehydrateisthusCoCl2·2H2O.Thesampleabsorbed7additionalgofwater,foratotalof11g,or0.6moleofwater,makingtheformulafortheredhydrateCoCl2·6H2O.

456.(B)Withtheexceptionofgoldandcopper,thecolorofallofthepuretransitionmetals(asreducedmetals)issilver.Thecompoundsoftransitionmetalsarecolored,andthesecolorsofareduetoelectrontransitionsthatoccurwhenthemetalbondswithotherelements.Theoxidationstateofthetransitionmetaldeterminesthecolorofthecomplex.Thetwomostcommontypesofelectrontransitionsareelectrontransfersandd-dtransitions.

Electrontransfer:Inacomplexion,ligands(ionsormoleculesthatbondwiththecentralmetal)surroundthemetalandformacoordinationcomplex.Often,theligandtransfersoneormoreofitselectronstothecentralmetalatom(theyarecommonlythoughtofasLewisbases).Incomplexesinwhichthecentralatomisatransitionmetal,theelectronsinvolvedinthetransferareeasilyexcitedbywavelengthsinthevisiblelightspectrum,causingsomewavelengthsoflighttobeabsorbedandotherstobereflected.Thereflectedwavelengthsarethoseweseeascolors.

d–dtransitions:Intransitionmetalcomplexes,thedorbitalsvaryinenergylevel.Thesedifferencesinenergycorrespondtothewavelengthsoflightthatcanbeabsorbed(andreflected).

457.(B)Themolarmassofacompoundcanbefoundbyitsvapordensity.Theformulaisderivedfromtheidealgaslaw:PV=nRTbutsubstitute(mass,m,ing/MM(molarmass))forn(thenumberofmoles)andsolveformolarmass(MM).

458.(C)Ifthetemperaturewasnotactuallybroughtupto100°Cbutthecalculationwasstilldoneusingthevalue373K,themolarmasscalculatedwouldhavebeenlarger(TisinthenumeratorintheequationderivedinAnswer457).(SeeAnswer11foranexceptquestionstrategy.)

459.(B)Themolarmass(MM)ofacompoundcanbefoundbymeasuringitseffectonthefreezingpointorboilingpointofaliquid.Inthiscase,weknowthat10gofsoluteaddedto50gramsofwaterraisedtheboilingpointby2°C.

TheformulaweusetodeterminefreezingpointdepressionorboilingpointelevationisΔT=Kbmi,whereKbistheebullioscopicconstant,aconstantthatallowsustorelatethemolalityofasolutiontotheboilingorfreezingpointofthesolvent(subscriptbisforboilingpoint).TheKbandKfarespecifictoaparticularsolvent.Thevan’tHofffactor,i,istheratioofthenumberofmolesofparticlesacompoundproducesinsolutionrelativetothenumberofmolesofparticlesofcompoundadded.

TheformulaΔT=Kbmicanbeusedtodeterminethenumericalvalueofm(molality=molsolute/kgsolvent).Themassofthesolutedividedbyitsmolarmass(MM)givesusthenumberofmolesofsolute;whilethenumberofmolesofsolutedividedbythemassofthesolventgivesusthemolality.Therefore,wecanrewritethemexpressionasm=(massofsolute/(MM)(kgsolvent)).WeknowthemassofthesoluteandwecalculatedmwiththeequationΔT=Kbmi,sonowwerearrangeourmolalityexpressiontosolveforMM:MM=massofsolute/(kgsolvent)(m).

Alternatively,wecouldsubstitutetheexpressionform,(massofsoluteing,grams/(MM)(kgsolvent)),directlyintotheequationΔT=KbmiandsolveforMM:

460.(C)Thepointinthecurvewherethetemperature(averagekineticenergy)isstablemeansthepotentialenergyofthesubstanceischanging(decreasinginthiscase).Theabsenceofatemperaturechangeinasubstancethatisgainingorlosingenergyisanindicationthatthesubstanceisundergoingaphasechange.

461.(C)Thetemperatureofthesolutionincreasesabout1°Candthendecreasesagainto70°Coverthecourseofaboutthreeminutes.Theaverageslopeofthelineduringthattimeperiodisclosetozero,andindicatestheapproximatefreezingpointofthesolution.

462.(B)UseΔT=Kfmi,i=1(nodissociation).

463.(B)ΔT=Kfmi,i=1(nodissociation)

ΔT=12.5°C,Kf=5∴m=ΔT/Kf=12.5/5=2.5m

molality=molsolute/kgsolvent∴molessolute=(m)×(kgsolvent)

molessolute=(2.5)(0.10)=0.25molsolute

molessolute=masssolute/molarmass,solvingformolarmass:

molarmass=masssolute/molsolute=20g/0.25mol=80gmol−1

464.(C)Diffractionisthespreadingoutofwaveswhenthey’repassedthroughsmallopenings.Diffractionisapropertyofwaves,andtooccur,thewavelengthsofthewavesmustbecomparableinlengthtothesizeoftheopeningthroughwhichtheyarebeingpassed.Thepatternthatemergesissimilar(andprobablyindistinguishablefrom)aninterferencepattern,withalternatingdarkandlightbands.X-rayshavewavelengthscomparabletothespacesbetweentheionsinanioniccrystallatticeandwillproduceadiffractionpatternwhichisusedtoanalyzethelatticestructureofthecrystal.

465.(E)Thecalculationfordeterminingthepercentmassofahydrate=(massofdrysample/massofhydratedsample)×100.Ifthesamplewascalculatedtohaveonly26percentwater,it’sbecausethemassofthedrysamplewasgreaterthanitsactualmass,makingthedrycompoundalargerpercentofthehydratethanitactuallyis.Deliquescentcompoundsreadilyabsorbwaterfromtheatmosphere,sothemostlikelycauseoftheincreasedmassofthedrycompoundisthatitwasn’treallydry;itstillhadwatermoleculesattachedtoitwhenitwasweighed.

466.(A)Mgisabovehydrogenintheactivityseries,soweknowthatitcanbeoxidizedbytheH+ionsintheHClsolution.TheMgwillreduce2H+toH2gas.Znisalsoabovehydrogenintheactivityseries,buttheZninZn(NO3)2hasalreadybeenoxidized(it’sintheformofZn2+),andsoithasnoelectronswithwhichtoreduceH+.

Na+initselementalformcouldcertainlyreduceH+,butnotinitsoxidizedform.

TheCO3−2willproduceCO2gasinanacidicsolution,notH2gas,accordingto

theequation

467.(C)Wearetitratingaweakbasewithastrongacid.HClisastrongacid,butsincethepHattheequivalencepointis<7,weknowwearetitratingaweakbase(thepHoftheequivalencepointofastrongacid-basetitration=7).MethylredisthebestindicatorlistedforthistitrationbecauseitchangescolorwiththepHrangeoftheequivalencepointofthetitration(theequivalencepointpHisitabout5andmethylredundergoesacolorchangebetweenpH4.4and6.2).

468.(A)AbuffersolutionresistschangesinpHbyactinglikea“protonsponge,”givingthemtothesolutionwhenthe[H+]decreasesandabsorbingthemfromthesolutionwhenthe[H+]increases.Thebufferregioninthistitrationconsistsoftheweakbaseanditsconjugateacid.Thebufferregionisalwaysatthebeginningofthetitration,whentheweakbaseanditsconjugateacidarepresentinfairlyhighconcentrations,beforetoomuchneutralizationhasoccurred.Inthistitration,thebufferregionliesbetweenpH11and8.Phenolphthaleinwouldremainpinkinthebufferregionandtransitiontoclearattheendofit.

469.(B)TheCO32–ionwouldhaveproducedCO2gasinanacidicsolution

(Sample1)accordingtotheequation

CaSO4andBaSO4bothformaninsolublewhiteprecipitate,whileaqueousNH3precipitatesgreenNi(OH)2andformsacoloredsolutionwhenitreactswithwaterandammoniaaccordingtotheequation

470.(D)Thepurposeofcollectingagasoverwateristokeepthegasataconstantpressureasmoregasparticlesarecollected.Gasesarehighlycompressible,sotheirvolumechangeswithchangesinpressure.Measuringtheamountofgasinaclosed,rigidcontainerrequiresthatbothpressureandvolumebemeasured.

Aneudiometer,adeviceusedtocollectgasoverwater,worksbydisplacingwaterwithgas,butthepressureofthegasalwaysequilibratestotheatmosphericpressurebecausethegaspushingdownonthesurfaceofthewaterintheeudiometeristhesame(atequilibrium)asthepressureoftheatmospherepushingthewaterupintheeudiometer(bypushingdownonthesurfaceofthewaterinthecontainerinwhichtheeudiometerisplaced).

471.(D)HClandNH3areverywatersolublegases.Alargenumberofthegaseswilldissolveinthewater.Theymustpassthroughtobecollected,greatlyreducingtheyield.CO2isslightlysoluble,soit’snotthebestgastocollectoverwater,butarelativelysmallfractionwillbelosttoitsdissolvingin(andreactingwith)watercomparedtoHClandNH3.

472.(B)Thegascollectedintheeudiometerisactuallyamixtureoftwogases—thegasintentionallycollectedandwatervaporthatevaporatedintheeudiometer.Thepartialpressureofthewatervaporisdeterminedsolelybythetemperature.At22°C,thetemperatureatwhichtheexperimentwasperformed,thevaporpressureofwateris19.8torr.Thetotalpressureofthegasesis770torr(theatmosphericpressure).Dalton’slawofpartialpressuresstatesthatthetotalpressureofamixtureofgasesisthesumofthepartialpressuresofeachofthegasesinthemixture∴770–19.8=750.2torr.(SeeAnswer470foranexplanationofwhythetotalpressureofthegasesintheeudiometerisequaltotheatmosphericpressure.)

473.(B)masscup+water–masscup=masswater

masscup+water+ice–masscup+water=massice

474.(E)Themassoftheicemeasuredwouldhavebeengreaterthantheactualmassoftheice,sotheheatoffusioncalculatedwouldbetoolow.TheunitsofHfusion=kJ/gramorkJ/mol.Eitherway,alargermasswouldhaveincreasedthevalueofthedenominatoranddecreasedthevalueofHfusion.

475.(C)Tocalculatetheheatoffusionoficeusingthismethod,thespecificheatoftheicemustbeknownsincethetemperatureoftheicemustberaisedfrom–20to0°Cbeforeitmelts.Inaddition,theheatusedtoraisethetemperatureoftheiceandtomeltitcomesfromthelossofheatoftheliquidwater,sothespecificheatofliquidwatermustalsobeknown.Hfus=(totalheat

lostbyliquidwater)–(heatgainedbyicetoreach0°C).

476.(B)0.4M=0.4moleperliter,or0.4mol/Lsolution×0.250L=0.1moleKOH(sincetheLcancels).RemembertoconverttoL;donotusemL.ThemolarmassofKOH=56gmol−1∴0.1mole=5.6g.

477.(D)Thisquestionreferstoanelectrolyticcell.Twohoursreduced230grams(10mol)ofNa+intoNa.Since1moleofelectronsisneededtoreduce1moleofNa+ions,10molesofelectronsmusthavepassedintothecellintwohours.Fe3+requires3molesofelectronspermole,soweintuitivelyknowwe’llgetalittlemorethanone-thirdthenumberofmolesofFemetal.10moleselectron×(1molFe3+/3molelectrons)=3.3molesFe3+ionscanbereduced.ThemolarmassofFeis56gmol−1∴3.3molesFeweighsabout185g.

478.(D)Wedon’treallyneedtocalculatethefinalconcentrationsofionsintheresultingsolutionbecausetheyareallpresentinthesamevolume,sothetotalnumberofmolesofeachspeciesisenoughtocomparetheirrelativeconcentrations.

0.1Lof2mol/LPb(NO3)2=0.2molePb(NO3)2,whichyields0.2molePb2+

ionsand

2×0.2=0.4moleNO3−ions.

0.1Lof3mol/LNaCl=0.3moleNaCl,whichyields0.3moleNa+and0.3moleCl−ions.

PbCl2willprecipitateoutofsolutionina1:2ratioofPb2+toCl−.0.2molPb2+

willprecipitate0.4moleCl−ions,butthereareonly0.3mole,sojustaboutalltheCl-ionswillbetakenoutofsolutionbythe0.15molePb2+ions(0.3moleCl−willcombinewith0.15molePb2+leaving0.2–0.15=0.05molePb2+behind).

479.(D)Differentialprecipitationcanseparateionsmixedinasolution.

480.(A)Anelectrongetsexcited(raisedtoanorbitalofhigherenergy)whenitsatomormoleculeabsorbsenergy.Theexcitedelectronsarenotstableatthehigherenergylevels,andwhentheydropbackdowntogroundstate,theyemitenergy.Thatenergyisoftenlostasaphoton(butthereareotherwaysforenergytobelost).Theenergyofthephotonisproportionaltotheenergychangeoftheelectron.

481.(D)Thephotoelectriceffectdemonstratedtheparticle-likepropertiesoflight(ormoregenerally,electromagneticradiation).

482.(D)E=hv∴E=(6.63×10−34)(4.4×1014)=2.9×10−19J.

483.(B)TheenergyofaphotonisgivenbytheequationE=hv,whereh=6.63×10−34J.sandv=thefrequency.Makesuretousethefrequencyandnotthewavelength.

484.(C)Eachlineinlinespectracorrespondstoanenergychangeofanelectron.Anyoneelectroncanproduceseverallinesaccordingtotheenergytransitionsitexperiences.(SeeAnswer11foranexceptquestionstrategy.)

485.(C)Wavescanbeimaginedasvibrationsproducedbyanoscillator.Whenthewavesfromanoscillatorarepolarized,allthevibrationsoccurinoneplane.

486.(D)Thespinofanelectroniseither+½or–½,soonlytwospotswouldbeexpectedtoappearonthescreen.Theelectronsspinningwith+½spinwouldbedeflectedinonedirectionbythefield,andtheelectronswiththe–½spinwouldbedeflectedintheoppositedirection.

487.(E)Emissionlinesrepresenttheenergytransitionsoftheelectronsinanatomwhenenergized.Amagneticfieldappliedtothegaswillnotchangethemagnitudeoftheenergytransitionsofelectronsoccupyingthesameorbitals,butapplyingthefieldwillcausetheelectronstobehaveasmagnetswithinthefield.Electronsofoppositespinsarelikeopposingpolesofamagnet.Thefieldwillaffecttheminoppositedirections,splittingtheemissionlinestheyproduce.

488.(B)SeeAnswer486.

489.(C)Thephotoelectriceffectisevidencethatlight(ormoregenerally,electromagneticradiation)hasparticle-likeproperties.Whenphotonsofahighenoughfrequencystrikeametalplate,electronsaredislodgedfromtheplate.Wavesdonothavethisproperty.Allphotonsaremassless,theytravelatthesamespeedinthesamemedium,andtheirenergyisdeterminedonlybytheirfrequency.

490.(E)WhenJ.J.Thomsonperformedhiscathoderayexperimenttodeterminethecharge-to-massratioofelectrons,heknewthecathoderaywascomposedofelectrons.Theirdeflectioninmagneticandelectricfieldsindicatedtheywerecharged.AvariationofYoung’sdouble-slitexperimentusingelectronsinsteadofphotonsshowedthatelectronshavethewave-likepropertyofinterference.

491.(B)Cancellingoutunitsistheeasiestwaytosolvethisproblem.WeneedtocombineCandC/gtogetg,sowecanflipC/gtog/CandtheCunitcancel:

–1.602×10−19C×(1gram/−1.76×108C)=(1.602×10−19C)×(–1.76×108)−1.

492.(D)Neutronsaretheonlyparticlesinthelistthathavenocharge.Anychargedparticlewillbedeflectedwhenpassedthroughanelectricormagneticfield.

493.(B)Liquidswithstrongintermolecularforceshaveahighsurfacetensionduetothestrongattractionofsurfacemoleculestoeachother.Strongforcesofattractionbetweenthemoleculesinaliquidcausethesurfacemoleculestosticktoeachtightly,creatingtheappearanceofafilm.Thegreaterthetensionofthefilm,thesteeperthemeniscusitcreates.

494.(E)Viscosityisdefinedasafluid’sresistancetoflow.Liquidswithstrongintermolecularforceswillhaveahighviscositybecausetheparticlesintheliquidresistmovingpastoneanother.

495.(E)SeeAnswer494.

496.(E)Miscibletypicallyreferstotwoormoreliquidsthataresolubleineachother.Misciblesoundslikemixable,andtheycanbethoughtofassynonyms.Waterandglycerolcanbothformhydrogenbonds(The–olendinginglycerol

tellsusit’sanalcoholand∴hasO—Hbonds.Glycerolisactuallyathree-carboncompoundwithahydroxylgrouponeachcarbon,foratotalofthree.).Thehighsurface-tensionsofwaterandglycerolandthehighviscosityofglycerolindicatethattheirparticlesexperiencestrongintermolecularforcesofattraction.

497.(C)ANewton(N)isaunitofforce.Surfacetensionismeasuredbytheforcerequiredtobreakthesurfaceofaliquid.ThesurfacetensionofcastoroilisexpectedtobelowerthanwaterandglycerolsincetheonlyintermolecularforcesofattractioninoilsareweakLondondispersionforces.Wewouldexpectthesurfacetensiontobesimilartothatofoliveoilbutlargerthanbenzene(castoroilandoliveoilaretriglycerides,ortriacylglycerols),whichhaveveryhighmolarmasses(inseveralhundredgmol−1).

498.(B)ThebreakingofvanderWaalsforces(intermolecularforcesofattraction,IMFs)isendothermic.HighertemperaturesdisruptIMFs,decreasingsurfacetensionandviscosity.

499.(B)Averysmallfractionofthemoleculesofaweakaciddissociateinsolution,soweexpectslightlymorethan1Mofparticles(butmuchlessthantwo).

500.(E).Agoodelectrolyteproducesahighconcentrationofionsinsolution.CompoundEproducesthreemolesofionspermolcompoundwhendissolved.Solublesaltsandstrongacidsandbasesareexcellentelectrolytesbecausetheycompletelydissociateinsolution,producingatleasttwomolesofionspermoleofcompound.