hc 20 answer key

Section Review 1.1

Part A Completion1. mass 7. carbon

2. space 8. organisms

3. composition 9. Analytical Chemistry

4. changes 10. physical chemistry

5. five 11. energy transfer

6. carbon 12. more than one

Part B True-False13. NT 15. AT 17. AT

14. ST 16. ST

Part C Matching18. f 21. d 24. c

19. i 22. h 25. a

20. b 23. g 26. e

27. j

Part D Questions and Problems28. a. physical chemistry

b. analytical chemistry

c. biochemistry

d. organic chemistry

Section Review 1.2

Part A Completion1. specific 6. productivity

2. microscopic 7. crops

3. energy 8. specific

4. conserve 9. space

5. batteries 10. chemical composition

Part B True-False9. NT 11. AT

10. NT 12. AT

Part C Matching13. d 15. c 17. f

14. a 16. b 18. e

Part D Questions and Problems19. production of chemicals such as insulin;

replacement of a gene that is not workingproperly (gene therapy)

20. Factors include poor soil quality, lack ofwater, weeds, plant diseases, and pests thateat crops.

21. Data collected by the robotic vehicleOpportunity indicated that the landing sitewas once drenched in water.

Section Review 1.3

Part A Completion1. achemists 6. scientific method

2. tools 7. hypothesis

3. techniques 8. experiment

4. measurement 9. theory

5. systematic 10. scientific law

Part B True-False11. NT 13. ST

12. NT

Part C Matching14. c 16. a 18. d

15. b 17. e 19. f

Part D Questions and Problems20. a. observation d. observation

b. hypothesis e. scientific law

c. experiment

21. collaboration and communication

Section Review 1.4

Part A Completion1. plan 7. calculate

2. implementing 8. evaluate

3. three 9. sense

4. analyze 10. unit

5. unknown 11. significant figures

6. plan

ANSWER KEY

Answer Key 759

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05_Chem_CTRAK_Ch01-06 7/12/04 8:22 AM Page 759

Part B True-False12. ST 14. NT 16. AT

13. AT 15. NT 17. AT

Part C Matching18. b 20. d 22. c

19. e 21. a

Part D Questions and Problems23. Step 1: Knowns: length � 10.0 inches.

1 inch � 2.54 cm

Step 2: 10.0 in� � �2.5

i4n�

cm� � 25.4 cm

Step 3: There are about two and a halfcentimeters per inch, so the answer25.4 makes sense.

24. Step 1: Knowns: distance � 5.0 km 1 km � 0.62 mi

Step 2: 5.0 km� � �0.6

k2m�

mi� � 3.1 mi

Step 3: There is a little more than half a mileper kilometer. 3.1 is a little more thanhalf of 5.0.

Practice Problems 1

Section 1.11. a. analytical d. organic

b. biochemistry e. inorganic

c. physical

2. a. applied d. applied

b. pure e. pure

c. pure

Section 1.21. Answers may include development of new

methods of energy conservation, such as newtypes of insulation; development of sourcesof energy other than fossil fuels, such asbiodiesel; and the development of newmethods of energy storage, such as improvedbatteries.

2. a. T f. T

b. T g. T

c. F h. T

d. T i. T

e. F

Section 1.31. Examples include: The battery could be dead,

the car could be out of gas, the spark plugscould be fouled, the wires could be loose.

2. Several experiments were performed to testhypotheses. Some of the experimentsdisproved some hypotheses, one experimentresulted in the car starting. Based on theexperiment, you could hypothesize that awire was loose.

3. Check students’ answers.

4. Theories are only as reliable as the knowledgeon which they are based. Throughout thehistory of science, theories have beendiscarded or modified as scientific knowledgehas increased.

Section 1.41. a. cost of apples � $1.50 a pound

weight of an apple � 0.50 pound

dollars available � $16

b. cost per apple � $0.75

number of apples purchased � 8

c. Two apples weigh a pound, and a poundcosts $1.50. $6.00 is four times $1.50.

2. Figure out how many pounds of apples canbe purchased with $6.00. Then figure outhow many apples this represents.

Interpreting Graphics 11. five 4. Component A

2. Component D 5. Component E

3. 3.5 minutes 6. Component C

Vocabulary Review 11. c 6. b 11. j

2. l 7. f 12. o

3. a 8. i 13. h

4. d 9. n 14. e

5. m 10. g 15. k

760 Core Teaching Resources

© Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.


Quiz for Chapter 11. T 8. observations

2. F 9. hypothesis

3. F 10. communicate

4. T 11. a. The volume of the

5. T test tube is needed.

6. scientific method b.There is enough

7. hypothesis information.

Chapter 1 Test A

A. Multiple Choice1. d 4. b 7. a

2. c 5. b 8. b

3. b 6. c

B. Questions9. Chemistry is the study of the composition of

matter and the changes matter undergoes.

10. Making Observations: Use your senses toobtain information directly.

Testing Hypotheses: A hypothesis is aproposed explanation for what you observed.Experiments are done to test a hypothesis.

Developing Theories: A theory is a well-testedexplanation for a broad set of observations. Ahypothesis may become a theory afterrepeated experimentation.

C. Essay11. 1. Analyze. List the knowns and the

unknown. A known may be ameasurement of an equation that shows arelationship between measurements.Determine what unit, if any, the answershould have. Make a plan for getting fromthe knowns to the unknown. You mightdraw a diagram to help visualize therelationship between the knowns and theunknown; or use data from a table orgraph; or select an equation.

2. Calculate. This step can involveconverting a measurement from one unitto another or rearranging an equation tosolve for an unknown.

3. Evaluate. Decide if the answer makessense. Check your work. Make a quickestimate to see whether your answer isreasonable. Make sure the answer is givenwith the correct number of significant

figures. Express the answer in scientificnotation, if appropriate.

Chapter 1 Test B

A. Multiple Choice1. c 5. d 8. c

2. c 6. b 9. c

3. c 7. b 10. a

4. b

B. Problems11. Organic chemistry is the study of essentially

all chemicals containing carbon.

Inorganic chemistry is the study of essentiallyall chemicals that do not contain carbon.

Analytical chemistry is concerned with thecomposition of chemistry.

Physical chemistry is concerned withmechanisms, rates, and energy transfer whenmatter undergoes a change.

Biochemistry is the study of processes thattake place in organisms.

12. Check students’ answers.

13. a. Hypothesis: The lawn needs water.

Experiment: Water the lawn every day forone week.

b. Hypothesis: The lawn needs fertilizer.

Experiment: Fertilize the lawn asprescribed.

Chapter 1 Small-Scale LabSafety goggles should be worn at all timeswhen working in the laboratory. If glasswarebreaks, tell your teacher and nearbyclassmates. Dispose of the glass as instructedby the teacher. If you spill water nearelectrical equipment, stand back, notify yourteacher, and warn other students in the area.When working near an open flame, tie backhair and loose clothing. Never reach across alit burner. Keep flammable materials awayfrom the flame. After cleaning up the workarea, wash your hands thoroughly with soapand water. It is not always appropriate todispose of chemicals by flushing them downthe sink. You should follow your teacher’sinstructions for disposal.

Answer Key 761

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Section Review 2.1

Part A Completion1. extensive 7. substance

2. intensive 8. shape or volume

3. mass 9. shape or volume

4. volume 10. shape

5. amount 11. gas

6. type

Part B True-False11. AT 13. ST

12. NT 14. AT

Part C Matching15. d 19. h 23. f

16. i 20. e 24. k

17. c 21. b 25. j

18. a 22. g

Part D Questions and Problems26. a. vapor d. solid

b. liquid e. gas

c. liquid

27. a. yes c. no

b. yes d. yes

Section Review 2.2

Part A Completion1. mixture

2. heterogeneous or homogeneous

3. homogeneous or heterogeneous

4. solutions

5. phase

6. physical

7. distillation

Part B True-False8. ST 10. AT

9. AT 11. ST

Part C Matching12. f 15. e 17. d

13. c 16. b 18. g

14. a

Part D Questions and Problems19. a. homogeneous d. heterogeneous

b. heterogeneous e. heterogeneous

c. homogeneous

20. a. substance c. substance

b. mixture d. mixture

Section Review 2.3

Part A Completion1. element or compound

2. compound or element

3. elements

4. ratio or proportions

5. chemical

6. substance

7. mixture

8. symbol

9. C

10. K


10. NT 12. AT

Part C Matching13. d 15. b 17. e

14. a 16. c

Part D Questions and Problems18. a. compound d. compound

b. element e. element

c. compound

19. a. K d. Cl

b. Pb e. S

c. Na

20. a. copper d. iron

b. hydrogen e. nitrogen

c. silver




Section Review 2.4

Part A Completion1. chemical

2. physical

3. chemical

4. reactants

5. chemical

6. composition

7. conservation of mass

8. mass

Part B True-False9. ST 11. ST 13. NT

10. AT 12. NT

Part C Matching14. e 16. d 18. a

15. b 17. c

Part D Questions and Problems19. The products of the reaction are gases, which

mix with the air.

20. a chemical change

21. 36 grams

Practice Problems 2

Section 2.11. b

2. a

3. b

4. d

5. gas no no yes

liquid no yes no

solid yes yes no

6. chlorine

7. bromine, ethanol

8. a. intensive c. extensive

b. extensive d. intensive

Section 2.21. distillation, evaporation

2. a mixture with a uniform compositionthroughout

3. a. homogeneous c. heterogeneous

b. heterogeneous

4. a. substance c. mixture

b. mixture d. substance

Section 2.31. nitrogen, hydrogen

2. a. lead c. gold

b. potassium d. iron

3. a. compound c. element

b. compound d. mixture

4. a. Sn c. Ag

b. Na d. C

5. No. It could be a liquid element, a liquidcompound, or a mixture of liquids.

6. c

Section 2.41. c

2. a. physical change

b. chemical change

c. chemical change

d. physical change

3. a. carbonic acid b. carbon dioxide,water

4. 62 grams

5. 32.4 grams

6. the law of conservation of mass

Interpreting Graphics 21. Homogeneous mixture. Since motor oil

comes in grades, it has a variablecomposition.

2. Chemical process. Iron oxide is a compound,while pure iron is an element.

3. a. physical separation

b. chemical separation

c. physical separation

4. a. substance c. mixture

b. mixture d. substance

Vocabulary Review 21. phase 5. mass

2. chemical symbol 6. solid

3. solution 7. mixture

4. elements 8. gas

Solution: substance

Answer Key 763

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Quiz for Chapter 21. a 8. physical blend

2. d 9. element

3. d 10. vary

4. b 11. a. potassium

5. a b. oxygen

6. d c. hydrogen

7. d

Chapter 2 Test A

A. Matching1. i 5. b 8. d

2. f 6. e 9. j

3. g 7. h 10. c

4. a

B. Multiple Choice11. d 15. b 18. a

12. b 16. c 19. d

13. b 17. d 20. d

14. b

C. True-False21. AT 23. AT 25. NT

22. NT 24. ST

D. Completion26. energy 31. compounds

27. distillation 32. iron

28. vary 33. homogeneous

29. chemical 34. conserved

30. physical 35. products

E. Essay36. A physical change alters a substance without

changing its composition. Melting or boilingare physical changes. In a chemical change,one or more substances change into one ormore new substances. Examples of chemicalchange include the rusting of iron, thereaction between iron and sulfur to produceiron sulfide, and the burning of wood. Somepossible clues to chemical change are atransfer of energy, color change, theproduction of gas, and the formation of aprecipitate.

Chapter 2 Test B

A. Matching1. i 5. j 8. e

2. d 6. b 9. c

3. h 7. f 10. g

4. a

B. Multiple Choice11. a 16. c 21. b

12. c 17. d 22. b

13. d 18. d 23. a

14. c 19. d

15. d 20. a

C. True-False24. AT 26. ST 28. AT

25. NT 27. NT

D. Completion29. liquid 33. reactants

30. vapor 34. chemical

31. heterogeneous 35. physical

32. N 36. substance

E. Essay37. Salt dissolves in water; iron filings are

attracted to a magnet; sawdust floats in water.To separate the mixture, skim the sawdustdirectly off the surface of the water, thenexpose the resulting mixture to a magnet toremove the iron filings. Boil off the water,recapturing the water through distillation.Solid salt remains at the bottom of the flask.




Chapter 2 Small-Scale Lab

Section 2.4 1 � 2 � 3 � BLACK!,page 56

Analysis

1. yellow

2. the mixture turns a blue-black color.

3. They all turn a mixture of Kl and starchblack.

4. Starch; both turn blue-black, which suggeststhe presence of starch.

5. The results may be the same in reactionsthat are simnilar to the one with Kl andstarch, but different in other reactions.

You’re The Chemist1. Add Kl + NaClO to various foods. A black

color indicates the presence of starch.

2. Most table salt contains 0.01% Kl. Wet only aportion of a small pile of salt with starch.Add CuSO4 or H2O2. A black color indicatesthe presence of Kl.

3. If an antacid tablet contains starch, it willturn black whenn treated with Kl + NaClO.

4. The color in the ink becomes bleached. Apicture can be drawn with colored ink.Certain areas can be treated with NaClO tobleach parts of the picture.

Section Review 3.1

Part A Completion1. accuracy 6. absolute value

2. precision 7. 100%

3. experimental value 8. scientific notation

4. error 9. known

5. accepted value 10. estimated


12. AT

Part C Matching14. b 16. a 18. d

15. e 17. f 19. c

Part D Questions and Problems20. a. 3 c. 2

b. 4

21. a. 4.35 � 102 mL c. 2.1 � 102

b. 8.4 cm3

Section Review 3.2

Part A Completion1. metric 6. gram

2. seven 7. Celsius

3. meter 8. kelvin

4. liter 9. joule/calorie

5. weight 10. calorie/joule


12. NT 14. NT

Part C Matching15. k 20. l 25. j

16. e 21. d 26. f

17. m 22. c 27. b

18. g 23. h

19. a 24. i

Figure A

NaClO

yellow yellow brownppt

black black black

black black black

black black black

H2O2 CuSO4

KI

KI

KI

Paper

Cereal

KIStarch

Answer Key 765

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05_Chem_CTRAK_Ch01-06 7/12/04 8:22 AM 65

Part D Questions and Problems28. Volume � 1.8 cm � 8.8 cm � 30.5 cm

� 4.8 � 102 cm3

29. °C � K � 273 � 20 � 273 � �253°C

30. a. p; 10�12 c. �; 10�6

b. k; 103 d. c; 10�2

Section Review 3.3

Part A Completion1. one (unity) 6. steps

2. conversion factor 7. conversion factor

3. remains the same 8. denominator

4. dimensional 9. units

analysis 10. unknown

5. known


12. ST 14. NT

Part C Questions and Problems

16. a. 125 g� � �10

10k0g

g�� 0.125 kg

� 1.25 � 10�1 kg

b. 0.12 L� � �100

10L�mL� � 1.2 � 102 mL

17. �15

10.00

cin�ells

� � �2.5

14in�cm��

1010

m�cm��

101

6

m��m�

� 17 �m/cell

18. �186 0

100

s�miles��

1016

s��s� � �

11.6

m1

iklme�

�

��2.99 �

1�

10s

�1 km�

Section Review 3.4

Part A Completion1. density 3. composition

2. intensive

Part B True-False4. ST 5. ST

Part C Questions and Problems

6.

Density � �vo

mlua

mss

e� � �

321.21

7cm

g3� � 3.96 g/cm3

7.

1.58 kg� � �10

k0g�0 g� � 1580 g

Density � �vo

mlua

mss

e� � �

10105080

cmg

3� � 1.58 g/cm3

Practice Problems 3

Section 3.11. 20 cm 5. Bruce’s

2. 21.0 cm 6. 1 cm

3. 4.2 � 102 cm2 7. 5%

4. Pete’s 8. 6.0 m

Section 3.21. 20 cm 5. 50 lb

2. 400 mm 6. 6 � 1010

3. 24 liters 7. 293 K

4. 12,000 g 8. 4°C

Section 3.3

1. pop. density ��750,0

4090mp

ie2�ople

�� 64

10

mac

ir

2�es

�

� 24 people/acre

2. cost � �510c.2an�

g��

4154

lbg�

� � 2.99 �do

clalan�

rs�

� 27.0 �do

llblars�

3. 60 �mh

i��

1.60m9

i�km

� � 96.54 km/h

4. The chicken costs $2.14/lb, the beef steakcosts $1.82/lb. The beef steak is lessexpensive per pound.

5. 1 day�� 124

dah�y�

� � �60

1mh�

in��

16m0

isn�

� � 86,400 s

6. 72 �kh�m��

101

0k0m�

m��

1010

m�cm� � �

601mh�

in��

16m0

isn�

�

� 2.0 � 103 cm/s

7. 1 m3� � 19.3 �cm

g�3�

� � ��1010

m�cm��

3� �

101

0k0g

g��

� 1.93 � 104 kg

8. 40.0 �gm

ai�l�

� � �0.2

164

Lgal�

� � �1.

161

mki�m

� � 17.0 km/L

9. 1.00 dollar� � �375

1dooz�llars�

� � �116

lob�

z��

� �4154

lb�g�

� � �100

10g�mg� � 75.7 mg

1.00 L � 1000 cm3 ��1 L

1000 cm3

32.1 mL � 3.21 cm3 ��1 mL1 cm3




Section 3.41. 6.00 grams

2. The balloon will float. Air is less dense thancarbon dioxide.

3. 1.09 liters

4. 27 g

Interpreting Graphics 31. Two significant figures

2. Three significant figures

3. cylinder B

4. either cylinder A or B

5. the Celsius scale

6. the Kelvin scale

7. No, 20 �C is known. Another significant figurecould be estimated.

8. liquid

Vocabulary Review 31. g 6. j 11. c

2. i 7. l 12. d

3. h 8. b 13. e

4. f 9. a

5. k 10. m

Quiz for Chapter 31. no

2. 1 � 10�3 kg/ 1 g

3. 4

4. a. 0.083 m

b. 20°C

c. 6900 km

5. a. 8.3 � 10�2 m

b. 2.0 � 101 �C

c. 6.9 � 103 km

6. b

7. c

8. K � �55�C � 273 � 218

9. Percent Error

� � 100 � 5.88%

Chapter 3 Test A

A. Matching1. i 5. f 8. a

2. b 6. c 9. d

3. j 7. e 10. h

4. g

B. Multiple Choice11. d 16. b 21. d

12. d 17. a 22. a

13. b 18. c 23. d

14. a 19. b 24. d

15. c 20. a 25. b

C. True-False26. ST 29. AT 32. ST

27. ST 30. ST 33. AT

28. ST 31. NT 34. NT

D. Problems

35. Density � �vo

mlua

mss

e� � �

6948c0mg

3� � 15 g/cm3; no

36. a. 4.15 cm � 1.8 cm � 7.5 cm2

b. 13.00 m � 0.54 m � 12.46 m

c. (1.7 � 10�5 m) � (3.72 � 10�4 m)

� 6.3 � 10�9 m2

37. Density � �vo

mlua

mss

e� � �

181.40.0cm

g3�

� 0.778 g/cm3

E. Essay38. Density is the ratio of the mass of an object to

its volume. Density is an intensive propertythat depends only on the composition of asubstance, not on the size of the sample.Volume is a measure of the space occupied byan object. Volume is an extensive propertythat depends on the amount of matter in asample. The density of water at a giventemperature does not vary with the size of thesample; the volume of the water, however,does vary with the size of the sample.

5.10 g/cm3 � 4.80 g/cm3

��5.10 g/cm3

Answer Key 767

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Chapter 3 Test B

A. Matching1. e 5. c 8. a

2. i 6. b 9. h

3. g 7. j 10. d

4. f

B. Multiple Choice11. c 18. d 25. a

12. d 19. d 26. a

13. a 20. c 27. b

14. c 21. d 28. d

15. a 22. b 29. c

16. d 23. a 30. a

17. c 24. c 31. d


33. ST 36. AT 39. NT

34. ST 37. NT

D. Problems40. a. 36.47 cm � 2.721 cm � 15.1 cm

� 54.291 cm � 54.3 cm

b. 148.576 g � 35.41 g � 113.166 g � 113.17 g

c. (5.6 � 107 m) � (3.60 � 10�2 m) � 20.16 � 105 m2 � 2.016 � 106 m2

� 2.0 � 106 m2

d. �8.

47.42

�

�

11

00

9

�6

m�� 2.08 � 1015 m

� 2.1 � 1015 m

41. a.

� 3.0 � 103 cm3

3.0 � 103 cm3�� 100

10

Lcm3�� 3.0 L

b. 3.0 L� � �1 kg

L�H2O� � 3.0 kg H2O

42. D � �MV

� � �6755.00.

c0mg

3� � 11.5 g/cm3

The metal is not pure silver.

E. Essay43. To find the volume of the box in liters, you

must first express the volume of the box incubic units, based on the linear dimensionsgiven, and then select a conversion factorthat will allow you to convert the cubic unitsinto liters.

Thus, since V � L � W � H,

V � 25 cm � 10 cm � 8 cm � 2000 cm3

To convert the cm3 into liters, you must selecta conversion factor so that relates cm3 toliters, that the cm3 units cancel and the literunits remain.

Thus, since 1 L � 1000 cm3

2000 cm3�� 100

10

Lcm3�� 2 L


Section 3.4 Now What Do I Do?,page 94

AnalysisAnswers are based on the following sample data:

average mass of water drop � 0.019 gmass of pre-1982 penny � 3.11 gmass of post-1982-penny � 2.50 g

1. x mg � 0.019 g� � 1000 mg/g� � 19 mg

2. V � �md

� � �1.

00

.001

g�9/c

g�m3� � 0.019 cm3

x mL � 0.019 cm3�� 11

cmm

L3�

� � 0.019 mL

x �L � 0.019 mL�� 10

10m0 �

L�L

� � 19 �L

3. x mg/cm3 � 1.00 g�/cm3 � 1000 mg/g�x mg/cm3 � 1000 mg/cm3;

1000 mg/cm3�� 1 cm3�/1 mL � 1000 mg/mL

4.

5. x g Cu � 2.50 g penny � �10

20.4g

gpe

Cnuny

�

x g Cu � 0.060 g Cu

x g Zn � 2.50 g penny � �10

907.

g6

pgeZnnny

�

x g Zn � 2.44 g Zn

g Cu � 3.11 g penny � �10

905.

g0

pg

eCn

uny

�

g Cu � 2.95 g Cu

g Zn � 3.11 g penny � �10

50.0g

gpe

Znnny

�

g Zn � 0.16 g Zn

V � 0.60 m � � 10.0 cm �

� 3000 cm3

�m

100 cm�

�10 mm

1cm(

( ((

50.0 mm �




6. The new penny is lighter because it is mostlyzinc which has a lower density than copper.

You’re The Chemist1. at 90°, mass of 1 drop: 0.019 g

at 45°, mass of 1 drop: 0.0218 g

at 0°, mass of 1 drop: 0.0242 g

Pipets give different results.

2. The best angle is 90° because the pipet iseasiest to control. Expel the air bubble sothat the first drop will be the same size asthe rest.

3. Find mass of can and divide by density ofaluminum. Sample

answer: mass of one can: 14.77 g;

dA: 2.70 g/cm3; V = 5.47 cm3

4. (1) Measure the mass before and after youfill the can with water. Use the mass anddensity of water to find the volume.

(2) Measure the height and radius andcalculate volume.

V = πr2h

(Can is not a perfect cylinder.)

(3) Read label: 12 oz = 355 mL

5. Sample answer:

V = 16.5 m � 3.0 m � 12.8 m

= 630 m3 � 1000 L/m3

= 630 000 L/

Assume 30 people with an average weight of130 lb (1 kg = 2.2 lb) and a density of about1.0 kg/L.

Volume of 30 people

V = 30 � 130 lb � 1 kg/2.2 lb � 1 L/1.0 kg

= 1800 L

The volume of 30 chairs, 15 tables, and 2desks is about that of 30 people or 1800 L.The volume of people and furniture is 3600 L.

% error = (3600 L/630,000 L)(100%)

= 0.57%.

6. If die measures 1.55 cm on a side:

V = (1.55 cm)3 = 3.72 cm3

A die has 21 holes that are hemispheres witha radius of 0.20 cm.

Volume of hemisphere: 2/3πr3

V = 2/3πr3 = 0.017 cm3

Volume of 21 hemispheres: 0.36 cm3

Volume of die:

V = 3.72 cm3 � 0.36 cm3

= 3.36 cm3

Error = 0.36 cm3

Percent error

� (0.36 cm3/3.36 cm3)(100%) = 11%

Note: The holes in some dice are cones. Thevolume of a cone is 1/3 πr2h.

7. Find weight in pounds and convert to kg.

Assume the density is about 1.00 kg/L.

V = weight � 1 kg/2.2 lb � 1L/1.00 kg

Section Review 4.1

Part A Completion1. atoms 3. compounds

2. different 4. separated, joined, or rearranged

Part B True-False5. NT 6. ST 7. AT

Part C Matching8. c 10. d

9. a 11. b

Part D Questions and Problems12. simple whole-number ratios

13. � � 1010 atoms/m

Section Review 4.2

Part A Completion1. subatomic 5. nucleus

2. negative 6. Ernest Rutherford

3. protons 7. positive

4. neutrons 8. electrons

Part B True-False9. NT 11. NT

10. AT 12. AT

Part C Matching13. c 15. e 17. b

14. a 16. d

Part D Questions and Problems18. protons, neutrons

100 cm��

m1 � 108 atoms��

cm�

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19. protons, electrons

20. Atoms are mostly empty space. The relativelymassive protons and neutrons areconcentrated in a small region called thenucleus; electrons are found outside thenucleus. Most alpha particles pass throughthe empty space without deflection. The fewthat come near the nucleus are deflected orbounce straight back.

Section Review 4.3

Part A Completion1. protons 6. isotopes

2. number 7. atomic mass

3. electrons 8. 1

4. protons 9. no

5. neutrons 10. 1


12. AT 14. AT

Part C Matching16. e 19. g 22. b

17. f 20. a 23. h

18. c 21. d

Part D Questions and Problems24. oxygen-16:16 amu � 0.9976 � 15.96 amu

oxygen-17:17 amu � 0.00037 � 0.0063 amu

oxygen-18:18 amu � 0.00204 � 0.0367 amu

16.00 amu � 16 amu

Practice Problems 4

Section 4.1

1. (1 � 107 atoms� ) � �

� 1 mm

Section 4.21. mass of e� is negligible;

16(1.67 � 10�24 g) + 16(1.67 � 10�24) � 5.33 � 10�23 g

2. approximately 6 � 1023 neutrons

3. b

Section 4.31. a. 5 c. 10

b. 16 d. 3

2. Number of protons: 11; 39; 33; 89

Number of electrons: 25; 35; 39; 89

Number of neutrons: 50; 42; 138

Atomic number: 25; 11; 35; 33; 89

Mass number: 55; 23; 80

3. a. 12 c. 46

b. 146 d. 10

4. 12.011 amu

5. 75% X�100, 25% X�104

Vocabulary Review 41. f 5. j 8. d

2. h 6. c 9. a

3. g 7. e 10. i

4. b

Quiz for Chapter 41. simple whole-number

2. chemical reaction

3. Electrons

4. neutrons

5. proton

6. atomic

7. 8

8. mass

9. 156

10. 30

11. periods

12. atomic number

Chapter 4 Test A

A. Matching1. j 5. f 8. b

2. g 6. e 9. h

3. a 7. d 10. i

4. c

10 mm�

1 cm�1 cm�

��1 � 108 atoms�




B. Multiple Choice11. c 16. a 21. c

12. d 17. b 22. d

13. c 18. a 23. c

14. c 19. b

15. d 20. b

C. Problems24. 63.929 amu � 0.4889 � 31.25 amu

65.926 amu � 0.2781 � 18.33 amu

66.927 amu � 0.0411 � 12.75 amu

67.925 amu � 0.1857 � 12.61 amu

69.925 amu � 0.0062 �

atomic mass � 65.37 amu

25. Atomic number: 7; 20; 26

Mass number: 19; 41

Number of protons: 9; 7; 20; 13

Number of neutrons: 14; 30


26. 136C: 6; 7; 6

104Be: 4; 6; 4

2010Ne: 10; 10; 1011

5B: 5; 6; 53316S: 16; 17; 16

D. Essay27. Atoms of different elements have a unique

number of protons and electrons. Isotopes ofthe same element differ in the number ofneutrons in the nuclei.

Chapter 4 Test B

A. Matching1. c 5. h 8. j

2. i 6. a 9. e

3. b 7. f 10. d

4. g


12. c 17. c 22. b

13. d 18. b 23. a

14. d 19. a 24. d

15. a 20. c

C. Problems25. Ar-36 35.978 amu � 0.00337 � 0.121 amu

Ar-38 37.963 amu � 0.00063 � 0.024 amu

Ar-40 39.962 amu � 0.99600 �

atomic mass � 39.947 amu

26. 919F: 9; 10; 92713Al: 13; 14; 134018Ar: 18; 22; 186530Zn: 30; 35; 30108

47Ag: 47; 61; 47

27. Atomic number: 19; 26; 35; 79

Mass number 24; 56; 80; 197

Number of protons: 12; 19; 26; 35

Number of neutrons: 12; 45


Symbol: 3919K; 56

26Fe; 19779Au

D. Essay28. Isotopes of the same element are alike in that

they have the same atomic number, and thus,the same number of protons and electrons.Isotopes of the same element are different inthat they have different numbers of neutrons,and thus, different mass numbers.

39.802 amu

0.43 amu

Answer Key 771

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Section 4.3 The Atomic Mass ofCandium, page 120

AnalysisSample data provided.

1. See row 3 in Figure A.




5. 0.2742 g � 0.2883 g � 0.3208 g � 0.8833 g

6. Percent abundance is parts per hundred.Relative abundance is parts per one or thedecimal form of percent. The individualpercent abundances add up to 100. Theindividual relative abundances add up to 1.

7. Relative abundance tells you the decimalfraction of particles.

8. The total in row 3 is an average that ignoresthe relative abundances of particles. Thetotal in row 6 is a weighted average becauseit considers differences in mass andabundance among the particles.

9. Another student might not have had thesame relative abundance of each candy. Alarger sample would provide a greatersampling of all isotopes.

You’re The Chemist1. Any differences are probably due to small

variations in the numbers of each kind ofcandy in the samples, which affect therelative abundances.

2. The larger the samples, the better the resultswith any of the methods. Mass is likely toprovide better results than volume.

Section Review 5.1

Part A Completion1. electrons

2. John Dalton

3. J.J. Thomson

4. plum-pudding

5. nucleus

6. circular

7. quantum mechanical

8. probability

Part B True/False9. AT 11. NT 13. AT

10. ST 12. AT 14. AT

Part C Matching15. c 17. a

16. b 18. d

Part D Questions and Problems19. Dalton proposed that matter was made of

indestructible particles called atoms.Thomson proposed an atomic model inwhich negatively charged electrons wereembedded in a positively charged mass.Rutherford discovered that atoms are mainlyempty space. He proposed that electronssurround a dense nucleus. Bohr proposedthat electrons are arranged in concentriccircular paths around the nucleus. Accordingto Bohr, the electrons in a particular orbithave a fixed energy, which prevents themfrom falling into the nucleus. In the modernatomic theory, the locations of electrons arenot fixed; they are described in terms ofprobability.

20. a. 3 orbitals c. 7 orbitals

b. 5 orbitals d. 1 orbital

Figure A

Total mass(grams) 13.16 g 13.83 g 15.40 g 42.39 g

15 13 20 48

0.8773 g 1.064 g 0.7700 g 0.8831 g

0.3125 0.2708 0.4167 100.0

31.25 27.08 41.67 1.000

0.2742 g 0.2883 g 0.3208 g 0.8833 g

A B C

TotalNumber

Average mass (grams)

Relativeabundance

Totals

Percentabundance

Relativemass




Section 5.2

Part A Completion1. electron configurations

2. Aufbau principle

3. equal

4. Pauli exclusion

5. two

6. opposite

7. a single electron

8. superscripts

9. electrons

10. Chromium

Part B True/False11. ST 13. NT 15. AT

12. NT 14. AT 16. NT

Part C Matching17. e 19. b 21. c

18. d 20. a

Part D Questions and Problems22. a. 1s22s22p2

b. 1s22s22p63s23p4

c. 1s22s22p63s23p64s1

d. 1s22s22p63s23p6

23. a. Ar b. B

Section 5.3

Part A Completion1. waves

2. inversely

3. light

4. atomic emission spectrum

5. light radiation

6. photoelectric

7. frequency

Part B True/False8. NT 10. AT 12. NT

9. NT 11. NT


14. a 16. d

Part D Questions and Problems

18. � ��32.0.4

00�

�

1100

1

�

0

5

cm�cm�

/s�

� � 1.25 � 1015 s�1

19. The photoelectric effect will not occur unlessthe frequency of the light striking a metal ishigh enough to cause an electron to be ejectedfrom the metal. The frequency of the lightmust be above the threshold frequency thatwill provide the necessary quanta of energy.

Practice Problems 5

Section 5.11. a. 1 d. 4

b. 2 e. 5

c. 3 f. 6

2. a. 1 e. 1

b. 1 f. 3

c. 5 g. 16

d. 7 h. 5

3. a. 1s, 1 orbital

b. 2s,1 orbital; 2p, 3 orbitals

c. 3s, 1 orbital; 3p, 3 orbitals; 3d, 5 orbitals

d. 4s, 1 orbital; 4p, 3 orbitals; 4d, 5 orbitals; 4f, 7 orbitals

e. 5s, 1 orbital; 5p, 3 orbitals; 5d, 5 orbitals; 5f, 7 orbitals

Section 5.21. a. 1s1

b. 1s22s22p63s23p63d34s2

c. 1s22s22p63s2

d. 1s22s22p63s23p63d104s24p64d105s25p66s2

e. 1s22s22p63s23p63d104s24p5

f. 1s22s22p63s23p4

g. 1s22s22p63s23p63d104s24p6

h. 1s22s22p63s23p63d104s24p3

i. 1s22s22p63s23p63d104s24p64d104f145s25p6

5d106s26p6

Section 5.3

1. � � ��

c� ��

35.0

.00

0�

�

110

0

10

15

csm

�

/1�s�

�

� � 6.00 � 10�6 cm; ultraviolet

2. � � ��

c� ��

36..0700

�

�

1100

8

�

m�7 m�

/s�

� � 4.48 � 1014 s�1

Answer Key 773

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The laser emits red light.

3. E � h�� 6.6262�10�34 J s� � 2.22 � 1014 s��1

E � 1.47 � 10�19 J

4. � � �hE

� �

� � 9.05 � 1018 s�1

5. radio waves, microwaves, infrared, visiblelight, ultraviolet, cosmic rays

6. � � ��

c� ��

31.60000�

�

1100

8

3

ms��

/1

s��

� � 1.88 � 102 m

Interpreting Graphics 51. � (s�1): 4.01 � 1013; 1.14 � 1014; 2.75 � 1014;

7.30 � 1014; 3.20 � 1015; 7.41 � 1013; 2.34 � 1014; 6.88 � 1014; 3.15 � 1015; 1.60 � 1014; 6.17 � 1014; 3.08 � 1015; 4.57 � 1014; 2.93 � 1015; 2.48 � 1015

2. �: 7.48 � 10�6 m; 2.63 � 10�6 m; 1.09 � 10�6 m; 4.11 � 10�7 m; 9.38 � 10�8 m;4.05 � 10�6 m; 1.28 � 10�6 m; 4.36 � 10�7 m;9.52 � 10�8 m; 1.88 � 10�6 m; 4.86 � 10�7 m; 9.74 � 10�8 m; 6.56 � 10�7 m; 1.02 � 10�7 m; 1.21 � 10�7 m

3. type of radiation: infrared; infrared;infrared; visible light; ultraviolet light;infrared; infrared; visible light; ultraviolet light; infrared; visible light;ultraviolet light; visible light; ultraviolet light;ultraviolet light

4. 6 → 25 → 24 → 23 → 2

5. 6 → 2 blue5 → 2 blue4 → 2 green3 → 2 red

6. All of the transitions end at the n � 2 energylevel.

7. The Bohr model is adequate for explainingthe emission spectra of atoms with a singleelectron. Li2� and He� each have a singleelectron surrounding the nucleus andshould, thus, behave according to the Bohrmodel.

Vocabulary Review 51. Hund’s rule

2. photons

3. hertz

4. Pauli exclusion principle

5. quantum

6. quantum mechanical model

7. aufbau principle

8. wavelength

9. atomic emission spectrum

10. photoelectrons

Quiz for Chapter 51. ST 7. Planck’s constant

2. NT 8. ground state

3. AT 9. photons

4. ST 10. photoelectric

5. AT 11. wavelike

6. NT

Chapter 5 Test A

A. Matching1. b 5. h 8. d

2. g 6. j 9. f

3. e 7. i 10. a

4. c

B. Multiple Choice11. b 15. a 18. d

12. c 16. c 19. d

13. a 17. b 20. a

14. d

6.00 � 10�15 J��6.6262 � 10�34 J� s




C. Problems21. a.

1s22s22p63s23p4

b.

1s22s22p63s1

22. � � �32.0.4

00�

�

1100

1

�

0c5

m�cm�

/s�� 1.25 � 1015 s�1

23. a. aluminum c. cobalt

b. krypton

24. a. phosphorus c. bromine

b. neon

25. Energy level 1 � 2Energy level 2 � 8Energy level 3 � 18Energy level 4 � 32Energy level 5 � 50

D. Essay26. Electrons occupy orbitals in a definite

sequence, filling orbitals with lower energiesfirst. Generally, orbitals in a lower principalenergy level have lower energies than those ina higher principal energy level, but in thefourth level the energy ranges of the principalenergy levels begin to overlap. As a result, the5s sublevel is lower in energy than the 4dsublevel.

E. Additional Matching27. b 29. f 31. c

28. a 30. e 32. d

F. True/False33. AT 36. ST 39. AT

34. NT 37. NT 40. ST

35. AT 38. AT 41. AT

Chapter 5 Test B

A. Matching1. h 5. d 8. j

2. c 6. g 9. a

3. f 7. b 10. i

4. e


12. c 17. c 22. d

13. c 18. b 23. c

14. b 19. d 24. b

15. b 20. b

C. Problems25. a. 1s22s22p63s2

b. 1s22s22p63s23p3

c. 1s22s22p63s23p64s23d104p5

d. 1s22s22p63s23p64s23d104p65s24d105p6

26. a. S e. Sr

b. Ne f. P

c. Sc g. Mn

d. Br h. Fr

27. c � ��

� � c/�

� �

� �4.80 � 1014 s�1

28. E � h � �

E �(6.6262 � 10�34 Js�) � (5.2 � 1015 s�1�)E � 3.4 � 10�18 J

D. Essay29. According to the Bohr model, electrons travel

around the nucleus along fixed paths muchas planets orbit the sun. The quantummechanical model, explains the positions ofelectrons in terms of probability cloudswithin which the electrons are most likely to

3.00 � 1010 cm��s��6.25 � 10�5 cm�

3s

2p

2s

1s

xw xw xw

xw

xw

x

3s

2p

2s

1s

3p

xw xw xw

xw

xw

xw

xw

x x

Answer Key 775

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be found. The Bohr model places electrons atspecific distances from the nucleus. Thequantum mechanical model allows electronsto be at virtually any distance from thenucleus, but describes the locations ofgreatest probability in terms of specifiedorbital shapes.

E. Additional Matching30. c 32. a 34. e

31. d 33. f 35. b

F. True/False36. AT 39. NT 42. ST

37. NT 40. AT 43. AT

38. AT 41. AT 44. AT

Section Review 6.1

Part A Completion1. properties

2. groups

3. periods or rows

4. atomic number

5. group

6. Metals

7. gases

8. metalloids

9. less

10. more


11. NT 13. NT

Part C Matching14. b 16. a 18. d

15. e 17. c

Part D Questions and Problems19. nitrogen and phosphorus are nonmetals;

arsenic and antimony are metalloids;bismuth is a metal.

20. good conductors of heat and electriccurrent; high luster; ductile; malleable;solids at room temperature

21. fluorine, bromine, iodine

Section Review 6.2

Part A Completion1. names

2. atoms

3. alkali metals

4. alkaline earth metals

5. representative elements

6. halogens

7. noble gases

8. transition metals

9. inner transition metals

10. p

11. not filled

Part B True-False12. ST 14. NT

13. NT 15. AT

Part C Matching16. f 19. d 21. c

17. e 20. b 22. a

18. g

Part D Questions and Problems23. Na, 3s1; Mg, 3s2; Al, 3s23p1; Si, 3s23p2;

P, 3s23p3; S, 3s23p4; Cl, 3s23p5; Ar, 3s23p6

24. Oxygen: nonmetal, gas; Sulfur: nonmetal, solid;Selenium: nonmetal, solid;Tellurium: metalloid, solid;Polonium: metal, solid

Section Review 6.3

Part A Completion1. decrease 6. increases

2. increases 7. electrons

3. energy levels 8. smaller

4. charge 9. electronegativity

5. ionization 10. increases


12. AT 14. NT

Part C Matching15. d 17. f 19. e

16. c 18. a 20. b




Part D Questions and Problems21. a. Al d. Na

b. S e. O

c. Br

22. a. gallium c. chlorine

b. oxygen d. bromine

Practice Problems

Section 6.11. c

2. a. nonmetal

b. metalloid

c. metal

d. nonmetal

e. metal

3. a

4. Li, Na, Rb, Cs, Fr

5. The three classes are as follows.1) The metals: good conductors of heat and

electric current; high luster when clean;malleable; ductile.

2) The nonmetals: poor conductors of heatand electric current; and are nonlustrous.

3) The metalloids: elements that haveproperties similar to those of metals andnonmetals depending on the conditions.

Section 6.21. Silicon is in the third period. Its first and

second energy levels are full (1s22s22p6). It isthe fourth element in the period; so itselectron configuration must end in 3s23p2.The complete configuration is 1s22s22p63s23p2.

2. Iodine is located in period 5. Its first fourenergy levels are full. It is Group 4A; so itselectron configuration must end in 5s25p5.The complete configuration is1s22s22p63s23p63d104s24p64d105s25p5.

3. The configuration s2p3 indicates 5 electronsin the highest occupied energy level, which isa feature of Group 5A.

4. a. Elements in Group 5A have 5 electrons intheir highest energy level. The thirdperiod element in Group 5A isphosphorus

b. 4s24p5 represents the Group 7A element inperiod 4; this element is bromine.

c. selenium.

5. a. The period 2 element with six electrons isoxygen.

b. The period 4 element with 2 electrons iscalcium.

c. The element in period 4 with 2 electronsin the 4s sublevel and 10 electrons in the3d sublevel is zinc.

6. Both Ne and Ar have a completely filledhighest occupied energy level. They are inGroup 8A, which is also known as the noblegases.Ne: 1s22s22p6

Ar: 1s22s22p63s23p6

7. The chemical and physical properties arelargely determined by their electronconfigurations. Lithium in Group 1A has only1 electron in its highest occupied energylevel. Sulfur in Group 6A has 6 electrons in itshighest occupied energy level.

8. Transition metals are elements whose highestoccupied s sublevel and a nearby d sublevelcontain electrons. The electron configura-tions for Ag and Fe are:Ag 1s22s22p63s23p64s24p64d105s1

Fe 1s22s22p63s23p63d64s2

Section 6.31. A magnesium atom is smaller than a sodium

atom because the shielding effect is constantfor elements in the same period, but thenuclear charge is greater in magnesium. Sothe electrons are drawn closer to the nucleus.Magnesium and calcium have the samenumber of electrons in their highestoccupied energy level. A magnesium atom issmaller than a calcium atom because thereare fewer occupied energy levels.

2. Astatine is in period 6. Tellurium is in period 5.Astatine is in Group 7A; tellurium is in Group6A. Although atomic size decreases across aperiod, the additional occupied energy levelin astatine significantly increases the size ofthe astatine atom as compared to thetellurium atom. The prediction is that atomsof astatine are larger than atoms of tellurium.

3. A chlorine atom is smaller than a magnesiumatom because atomic size decreases from leftto right across a period. When a magnesiumatom reacts, it loses electrons from its highestoccupied energy level. A magnesium ion hasfilled first and second levels. When chlorinereacts, it gains an electron in its highestoccupied energy level. An ion with three

Answer Key 777

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occupied energy levels is larger than an ionwith two occupied energy levels.

4. Across a period from left to right the principalenergy level remains the same, but thenuclear charge increases. The increasingnuclear charge pulls the electrons closer tothe nucleus, resulting in a smaller atomicradius. The trend is less pronounced as thenumber of electrons increases because theinner electrons shield the electrons in thehighest occupied energy level. Atomic sizeincreases as you move down a period becausethe electrons are added to higher principalenergy levels. This enlarging effect is greaterthan the shrinking effect caused byincreasing nuclear charge.

5. When a sulfur atom reacts to form an ion itadds two electrons while chlorine adds oneelectron. Sulfide and chloride ions have thesame number of electrons. Because thechloride ion has the greater nuclear charge, itwill be smaller than the sulfide ion.

6. Sodium’s first ionization energy is higher thanthat of potassium because ionization energytends to decrease from top to bottom withina group.

7. Beryllium’s first ionization energy is greaterbecause first ionization energy tends toincrease from left to right across a period.

8. Barium is less electronegative than struntiumbecause electronegativity values tend todecrease from top to bottom within a group.

9. Because magnesium has a relatively low firstand second ionization energy, the removal oftwo electrons from magnesium is likely. Therelatively high third ionization energyindicates the difficulty of removing a thirdelectron from the filled second energy level.Magnesium normally forms an ion with a 2�

charge.

10. Because electronegativity decreases from topto bottom within a group, sulfur is lesselectronegative than oxygen. Becauseelectronegativity increases from left to rightacross a period, fluorine is moreelectronegative than oxygen. The correctorder for increasing electronegativity is then sulfur oxygen fluorine.

Interpreting Graphics 61. 42 5. 2617 �C

2. table A 6. table B

3. atomic weight 7. 4

4. 0.53 g/cm3

8. physical state at room temperature; generalclass, e.g. transition metal; whether anelement is not found in nature

9. In the periodic table elements with similarchemical and physical properties aregrouped together in vertical columns. Thisorganization helps scientists predict andexplain similarities and differences in theproperties of elements based on theirunderlying atomic structure. Listing theelements, in alphabetical order, makes itpossible to quickly find information aboutthe properties of a particular element withouthaving to know the location of the element inthe periodic table.

10. a. Li: Group 1A (or Group 1), period 2Mo: Group 6B (or Group 6), period 5

b. No, because they are not located in thesame group or family.

c. Lithium, Li, is an alkali metal.Molybdenum, Mo, is a transition metal.

d. Answers may include sodium, potassium,rubidium, cesium, and francium.

11. Check students’ work. Their keys need toinclude the color, mp, and bp, of the element(and the state if the square is not colorcoded for style).

Vocabulary Review 61. n 7. l 12. f

2. a 8. g 13. h

3. m 9. o 14. d

4. c 10. i 15. p

5. k 11. e 16. j

6. b




Quiz for Chapter 61. number

2. seven

3. group (column or family)

4. six

5. two less

6. ST 8. NT 10. NT

7. AT 9. AT

Chapter 6 Test A

A. Matching1. d 5. c 8. g

2. f 6. i 9. e

3. j 7. a 10. h

4. b

B. Multiple Choice11. d 16. c 21. a

12. c 17. c 22. c

13. d 18. d 23. d

14. a 19. a 24. b

15. a 20. d 25. b

C. Questions26. a. period 5, Group 2A (or Group 2)

b. period 4, Group 6A (or Group 16)

c. period 6, Group 4A (or Group 14)

27. a. 1s2

b. 4s1

c. 3s2 3p1

d. 4s2 4p6

e. 2s2 2p4

28. a. Li, K, Cs

b. Ar, Cl P, Si

c. Be, Ca, Sr, Ba

29. a. alkaline earth metal, period 4, Group 2A(or Group 2)

b. alkali metal, period 6, Group 1A (orGroup 1)

c. halogen, period 2, Group 7a (or Group 17)

d. transition metal, period 4, Group 6B (orGroup 6)

e. noble gas, period 2, Group 8A (or Group 18)

f. transition metal, period 5, Group 1B (orGroup 11)

30. a. Mg b. I c. Cl

D. Essay31. Two factors influence the size of an atom as

the atomic number increases within a group.There is an increase in nuclear charge, whichdraws the electrons closer to the nucleus.There is an increase in the number ofoccupied energy levels, which shields theelectrons in the highest occupied energy levelfrom the attraction of protons in the nucleus.The net effect is a decrease in the attractionof the nucleus on the electrons in the highestoccupied energy level and an increase inatomic radius. From left to right across aperiod, electrons are being added to the sameenergy level. The increasing charge on thenucleus tends to pull the electrons closer andatomic radius decreases.

Chapter 6 Test B

A. Matching1. c 5. h 8. d

2. i 6. a 9. f

3. b 7. j 10. e

4. g

B. Multiple Choice11. b 18. c 25. c

12. d 19. d 26. c

13. d 20. a 27. a

14. a 21. d 28. b

15. d 22. d 29. b

16. c 23. b 30. c

17. d 24. a

C. Questions31. a. 2; 2A(2); alkaline earth metal; Be

b. 3; 5A(15); P

c. 3; 8A(18); noble gas; Ar

d. 4; 1A(1); alkali metal; K

e. 4; 4B(4); transition metal; Sc

f. 4; 7A(17); halogen; Br

32. a. Cs, K, Li, C, F

b. Cs, K, Li, C, F

c. F, C, Li, K, Cs

Answer Key 779

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33. a. K; Li; K

b. C; F; C

c. Ca; Mg; Ca

d. S; O; S

34. a. B d. Bb. C e. Bc. D

D. Essay35. Ionization energy and electronegativity are

properties that reflect an atom’s ability toattract and retain electrons. A high ionizationenergy indicates that an atom has a tight holdon its electrons. A high electronegativityindicates an ability to attract additionalelectrons.

Chapter 6 Small-Scale LabAnalyze and Conclude

1. Fluorine

2. Electronegativity generally increases fromleft to right along a period.

3. Metals, which are on the left side of thetable, have lower electronegativity valuesthan nonmetals, which are on the right.

4. Electronegativity generally increases frombottom to top within a group. Except forboron, the rest of Group 3A shows a reversein this trend.

5. Although hydrogen is placed in Group 1Abased on its electron configuration, hydrogenis classified as a nonmetal.

You’re The Chemist1. Students divide the values of first ionization

energies by 300 and measure the appropriatelength of straws.

2. Students must determine their own scalebefore they begin. Students often use twowells to represent both ionic and atomicradii. Other students cut a straw to a lengththat represents the larger radius of an atomand mark the straw to show the smallerradius of the corresponding cation.

3. The value for xenon is similar to iodine,which is consistent with the general trend.Based on this value, xenon appears to havethe ability to attract electrons and formcompounds.

Section Review 7.1Part A Completion

1. valence electrons

2. group

3. electron dot structures

4. octet rule

5. cations

6. anions

7. 1�

8. Halide ions

9. gain10. charges

Part B True/False11. NT 13. ST 15. AT

12. AT 14. NT 16. NT

Part C Matching17. b 20. g 22. a

18. d 21. f 23. c

19. e


24. a. c.

b.

25 a. 2 electrons lost; magnesium ion; cation

b. 2 electrons lost; calcium ion; cation

c. 1 electron gained; bromide ion, anion

d. 1 electron lost; silver ion; cation

26. Nonmetals attain stable noble gasconfigurations by gaining electrons andforming anions with 8 outer electrons in theexisting energy level. Metals attain noble gasconfigurations by losing electrons andforming cations with a complete octet in thenext-lowest energy level.

Rb=

=Ba=Si




Answer Key 781

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Section 7.2

Part A Completion1. electrostatic forces2. oppositely3. ionic bonds4. neutral5. formula unit6. crystals7. high8. large9. stable

10. molten

Part B True-False11. AT 13. AT 15. NT

12. ST 14. ST

Part C Matching16. b 18. c 20. a

17. e 19. d

Part D Questions and Problems21. Ionic bonds are the electrostatic forces of

attraction that bind oppositely charged ionstogether. In an ionic compound, the positivecharges of the cations equal the negativecharges of the ions.

22. When ionic compounds are melted, theorderly crystal structure breaks down. Eachion is then free to move throughout themolten mass. If a voltage is applied, cationswill migrate to one electrode, and anions willmigrate to the other. This movement of ionsmeans that there is a flow of electricitybetween the two electrodes. When ioniccompounds dissolve in water, their ions arefree to move. Thus, aqueous solutions of ioniccompounds also conduct electricity.

Section 7.3

Part A Completion1. cations

2. electrons

3. metallic

4. electrical

5. malleable/ductile

6. ductile/malleable

7. body-centered/face-centered

8. face-centered/body-centered

9. hexagonal close-packed

10. alloy

Part B True-False11. NT 13. NT 15. AT

12. ST 14. AT

Part C Matching16. d 18. b 20. a

17. e 19. c

Part D Questions and Problems21. Solid metals consist of closely packed cations

surrounded by free-moving valenceelectrons, which make metals goodconductors of electric current. As electronsenter one end of a bar of metal, an equalnumber leave the other end. Metal cationsare insulated from one another by electrons.When a metal is subjected to pressure, themetal cations easily slide past one another.This behavior makes the metal malleable andductile.

22. The superior properties of alloys result fromthe cumulative properties of all theconstituents of the alloy. For example, analloy can be more durable than oneconstituent but more malleable than another.

Practice Problems 7

Section 7.11. a. (i) 2 (ii) Ba 6 (iii) Ba2�

b. (i) 7 (ii) (iii) I�

c. (i) 1 (ii) K = (iii) K�

2. a. 3 c. 6

b. 7

3. a. 1s22s22p63s23p64s2

b. 1s22s22p63s23p5

c. 1s22s22p6

d. 1s22s22p63s23p6

e. 1s22s22p6

4. The number of valence electrons in an atomof a representative element is the same as thegroup number of the element.

I


5. a. loses 2 electrons; cation

b. loses 3 electrons; cation

c. gains 2 electrons; anion

d. loses 1 electron; cation

e. gains 1 electron; anion

f. gains 3 electrons; anion

6. a. chloride ion, Cl�

b. potassium ion, K�

c. oxide ion, O2�

d. barium ion, Ba2�

7. a. 2 lostb. 1 gainedc. 1 lostd. 3 lost

8. a. cation d. anionb. cation e. cationc. anion f. cation

Section 7.21. a. NaBr d. Al2O3

b. Na2S e. BaCl2

c. CaI2

2. Ionic compounds are formed when metalsreact with nonmetals. The combinations in band c will form ionic compounds

3. The coordination number is the number ofions of the opposite charge that surround anion in a crystal.

4. The coordination number is determined byusing x-ray diffraction crystallography.Patterns are used to calculate the positions ofions in the crystal and to define the structureof the crystal.

Section 7.31. A metallic bond is made up of cations that are

surrounded by mobile valence electrons.

2. The metallic crystal is thought to consist of anarray of metal cations in a “sea” of electrons.Although the electrons are attracted to themetal cations, no individual electron isconfined to any specific cation; rather, theelectrons are free to move about thecrystalline structure. When electrical currentis applied to a metal, these mobile electronscan carry charge from one end of the metal tothe other.

3. Metals are crystalline. The metal cations arearranged in a very compact and orderlystructure or pattern.

4. • Body-centered cubic: every atom (exceptthose at the surface) has 8 neighbors.

• Face-centered cubic: every atom has 12 neighbors.

• Hexagonal close-packed: every atom has 12neighbors, but in a different arrangementthan face-centered cubic.

5. An alloy is a mixture of two or more elements,at least one of which is a metal. Alloys haveproperties of metals.

6. a. Brass: copper and zinc

b. Bronze: copper and tin

c. Stainless steel: iron, chromium, carbon,and nickel

d. Sterling silver: silver and copper

e. Cast iron: iron and carbon

f. Spring steel: iron, chromium, and carbon

Interpreting Graphics 71. sodium 1s22s22p63s1

Sodium has 1 valence electron.chlorine 1s22s22p63s23p5

Chlorine has 7 valence electrons.

2. In Step 1, each sodium atom gives up onevalence electron to a chlorine atom. In thisprocess, sodium becomes positively chargedand chlorine becomes negatively charged.Each ion attains the electron configuration ofthe nearest noble gas.

3. In Step 2, ionic bonds form between sodiumcations and chlorine anions. The ions arrangethemselves in an orderly, three-dimensionalarray characteristic of a crystalline solid. InNaCl, each ion is surrounded by six other ionsof opposite charge, which results in a verystable ionic compound.

4. NaCl is typical of many ionic compounds.The large amount of energy released when anionic lattice is formed (Step 2) compensatesfor the endothermic nature of the electrontransfer (Step 1). To reverse the latticeformation through melting would requireenough energy to overcome the multipleatttractions within the crystal lattice.

Vocabulary Review 71. j 6. l 11. b

2. i 7. a 12. k

3. f 8. c 13. m

4. g 9. e




5. h 10. d

Quiz for Chapter 71. ST 8. gaining

2. NT 9. eight

3. AT 10. pseudo-noble gas

4. AT 11. metals

5. AT 12. anion

6. valence 13. formula unit

7. octet

Chapter 7 Test A

A. Matching1. e 5. j 9. k

2. f 6. g 10. b

3. h 7. a 11. d

4. c 8. i

B. Multiple Choice12. c 16. c 20. a

13. b 17. b 21. a

14. c 18. d 22. a

15. b 19. b 23. c

C. True-False24. AT 28. AT 31. AT

25. NT 29. NT 32. ST

26. NT 30. AT 33. AT

27. ST

D. Questions34. a. =Ca = c.

b.

35. a. Cl� anion c. Na� cation

b. Be2� cation d. O2� anion

36. a. 1s22s22p63s23p6 c. 1s22s22p6

b. 1s2 d. 1s22s22p6

37.

The electron dot formulas show that oneatom of Al can give 3 electrons, so 3 atoms ofCl are needed to form the compound AlCl3.

38. Na F1s2 2s2 2p6 3s1 � 1s2 2s2 2p5 →

Na� F�

1s2 2s2 2p6 � 1s2 2s2 2p6

Both ions have the configuration of neon.

E. Essay39. Metallic bonds are the result of the attraction

of free-floating valence electrons forpositively charged metal ions. An electriccurrent is a flow of electrons. As electronsenter one end of a piece of metal, some of thefree-floating electrons leave the other end.Thus metals are good conductors ofelectricity. The cations in a piece of metal areinsulated from each other by the freeelectrons. Thus, when metal is struck, thecations slide past each other easily. Thismakes the metal malleable.

Chapter 7 Test B

A. Matching1. f 5. b 9. h

2. g 6. k 10. c

3. d 7. a 11. i

4. e 8. j

B. Multiple Choice12. c 18. d 24. c

13. b 19. c 25. d

14. c 20. d 26. a

15. a 21. c 27. c

16. a 22. d 28. d

17. c 23. c

Cl

Al � Cl y Al3�� 3 Cl �

Cl

ClAl

Br �Br

Al 3�AlCa 2�

Answer Key 783

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30. AT 34. ST 38. NT

31. AT 35. ST 39. AT

32. NT 36. AT

D. Questions40. a. 2s1; 1;

b. 2s22p3; 5;

c. 3s23p2; 4;

d. 4s24p5; 7;

41. a. Na� ; 1s22s22p6

b. F�; 1s22s22p6

c. K�; 1s2 2s2 2p6 3s2 3p6

d. Sr2�; 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

42. a. ; Na�

b. ;

c. ;

d. ; Ca2�

43. a. NaF

b. MgCl2

c. CaS

d. Al2O3

E. Essay44. The group number for Ca is 2A, which means

that two valence electrons will be lost; thecation Ca2� is produced. The group numberfor F is 7A, which means that F has sevenvalence electrons and reacts by gaining oneelectron to attain the noble gas configuration.The formula for the anion produced is F�.Thus, when Ca and F react, two atoms of F arerequired to react with one atom of Ca. Theformula of the compound formed is CaF2.

Chapter 7 Small-Scale LabSection 7.2 Analysis of Anions and

Cations, page 200

AnalyzeSample data provided.

NVR: No Visible Reaction

WP: White Precipitate

1. An intermediate compound, FeCl2, forms,which reacts with the nitrate ion. An orange-brown color forms.

2. Each of the following pairs of ions producesa visible product that can be used to identifythe ion in question: PO4

3� and Ag�, NO3�

and HCl � Fe, SO42� and Pb2�, Ca2� and

OH�, Fe3� and SCN�.

3. No; neither of the solutions produced avisible product.

You’re the ChemistAll designs should include tests that produceunique, observable products.

NaOH NVR WP Rust ppt.

NVR NVR Blood-red soln

KSCN

KI(K�)

CaCl2

(Ca2�)FeCl3(Fe3�)

Figure BCation Analysis

Ag NVR NVRLight

yellow ppt.

Bubbles BubblesBubblesw/yellowsolution

WP WPNVR

NO3

Na2SO4

(SO42�)

HNO3

(NO3�)

Na3PO4

(PO43�)

HCl plus 1 piece of Fe(s)

Pb(NO3)2

Figure AAnion Analysis

Ca

P 3�

P

Cl �

Cl

Na

Br

Si

N

Li




Section Review 8.1

Part A Completion1. compound 6. low

2. molecular 7. high

3. nonmetals 8. atoms

4. diatomic 9. structure

5. molecular formula

Part B True-False10. ST 12. AT 14. ST

11. NT 13. NT

Part C Matching 15. e 17. d 19. c

16. a 18. b

Part D Questions and Problems20. molecular compound

21. a. 4 carbon atoms, 10 hydrogen atoms

b. 6 carbon atoms, 5 hydrogen atoms, 1 fluorine atom

22. a. molecule d. atom

b. atom e. molecule

c. molecule

Section Review 8.2

Part A Completion1. stable electron

2. covalent

3. shared

4. single

5. unshared pairs

6. double/triple

7. coordinate covalent bond

8. Energy

9. bond dissociation energy

10. resonance structure

Part B True-False11. NT 13. AT 15. ST

12. NT 14. AT 16. NT

Part C Matching17. e 19. b 21. d

18. a 20. c

Part D Questions and Problems22. a.

b.

c.

Section 8.3

Part A Completion1. molecular orbitals

2. bonding orbital

3. lower

4. sigma or �

5. pi or �

6. three-dimensional

7. VSEPR theory

8. orbital hybridization

Part B True-False9. AT 11. NT 13. ST

10. ST 12. AT 14. AT

Part C Matching15. d 17. a 19. c

16. e 18. b

Part D Questions and Problems20. Reading from left to right:

sp3 sp2 sp2 sp sp sp3

Section 8.4

Part A Completion1. equally

2. nonpolar

3. unequally

4. polar

5. electronegativities

6. dipole interactions

7. hydrogen bond

8. electronegative

9. oxygen, nitrogen, or fluorine

H H 6-9N

�6 H

H

H 6 C 6 6 6 N 6

6-9Br 6-9Br 6

Answer Key 785

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11. AT 13. ST 15. AT

Part C Matching16. b 18. e 20. a

17. d 19. c

Part D Questions and Problems21. dispersion forces, dipole interactions,

hydrogen bonds

22. a. ionic

b. polar covalent bonds

c. polar covalent bonds

d. nonpolar covalent bonds

Practice Problems 8

Section 8.11. a. atom d. molecule

b. molecule e. atomc. molecule

2. a. not diatomic

b. diatomic

c. diatomic

d. not diatomic

e. diatomic

3. Molecular compounds are usually composedfrom two or more nonmetallic elements.

4. A molecular structure gives informationabout the kinds and numbers of atomspresent in a molecule.

5. Molecular compounds tend to have lowermelting and boiling points that that of ioniccompounds

Section 8.21. The two atoms share a pair of electrons in

order to form a single covalent bond.

2. Phosphorous needs 3 more electrons to fillthe 3p orbitals. Fluorine needs one moreelectron to fill its second energy level. Sinceeach fluorine atom only needs one electronand phosphorus needs 3 electrons, threefluorine atoms are required to bond withphosphorus.

3. Nitrogen needs 3 more electrons to fill itssecond energy level. Chlorine needs onemore electron to achieve a noble gasconfiguration. Because each chlorine atomneeds only one electron and nitrogen needs 3electrons, three chlorine atoms are requiredto bond with nitrogen.

4. Because carbon can form four single covalentbonds, there is an apparent shortage of atomswith which to bond. This is a clue that acarbon-carbon multiple bond exists in thiscompound. Each carbon atom shares oneelectron with one of the two hydrogen atoms.The remaining three electrons for eachcarbon atom form a triple covalent bond. Theelectron dot structure is:

H�C��C�H5. Carbon has 4 valence electrons and each of

the oxygens has 6 valence electrons. Twoadditional electrons are added to account forthe ion having a 2� charge. The carbon andoxygen can satisfy the octet rule by having theoxygens bonded to a central carbon. There isone double covalent bond between a carbonand oxygen, which can shift to any one of thecarbon-oxygen bonds giving rise to threeresonance structures.

Section 8.31. The four fluorine atoms are covalently

bonded to the central carbon atom. The fourshared pairs of electrons repel each other tothe corners of a tetrahedron. All four bondangles are 109.5°.

2. The four valence electron pairs repel eachother, but the unshared pair is held closer tothe phosphorus than the three bonding pairs.The unshared pair repels the shared pairsmore strongly. Thus, the angle betweenbonds is expected to be slightly smaller thanthe tetrahedral bond angle of 109.5�. Theactual bond angle for NH3, a similarmolecule, is 107�.

3. Boron forms three sp2 orbitals by mixing one2s orbital and two 2p orbitals. The three sp2

orbitals lie in the same plane, 120� apart fromone another. Each sp2 orbital overlaps with an

O

CO O

2 O

CO O

2 O

CO O

2

NCl

Cl Cl

F PF

F

H F




atomic orbital of chlorine to form threeequivalent sigma bonds.

4. To form four equivalent bonds, silicon mixesone s orbital and all three of the p orbitals.The hybridization in SiF4 is sp3.

5. Oxygen is the central atom in this molecule. Ithas 6 valence electrons, two of which arebonding electrons. The other 4 electrons areunshared pairs. These 2 unshared pairs repelthe two bonding pairs and prevent F2O frombeing linear. The molecule is a bent triatomicmolecule, with a bond angle of approximately104.5°. This angle is slightly smaller than thetetrahedral bond angle because the twounshared pairs repel each other morestrongly than the two shared pairs.

6. Because each carbon can form singlecovalent bonds with four other atoms, thereexists in this compound an apparentshortage of atoms with which to bond. This isa clue that CH2CF2 contains a carbon-carbonmultiple bond. The two hydrogen atomsbond with one carbon atom while the twofluorine atoms bond with the other carbon.Each carbon atom has two electrons left over.These electrons form a carbon-carbondouble covalent bond. The molecule looksvery much like the ethene molecule (C2H4).H2C2H and F2C2F bond angles of 120°, thehybridization involved in the carbon-carbonbond is sp2.

7. Carbon 1 mixes one s orbital and three porbitals to form four sp3 hybrid orbitals,which form 4 sigma bonds. Carbon 2 mixesone s and two p orbitals to form three sp2

hybrid orbitals, which overlap with thehybrid orbitals of the carbon and oxygenatoms to form three equivalent sigma bonds.The non-hybridized p carbon orbital overlapswith an oxygen p orbital to form one pibonding orbital.

Section 8.41. a. The difference in electronegativity

between Na and O is about 2.4 and thebond is ionic.

b. With like atoms, the difference is zero andthe bond is nonpolar covalent.

c. The electronegativity difference betweenP and O is about 1.4 and the bond is polarcovalent.

2. For a bond to be classified as nonpolarcovalent, like atoms must bond, as indiatomic molecules. Most bonds are betweenunlike atoms; therefore, they must be ionic orpolar covalent.

3. Both carbon dioxide and carbon monoxidecontain polar bonds. However, the effect ofthe polar bond on the polarity of the entiremolecule depends on the shape of themolecule. In carbon monoxide, there is apartial positive pole and a partial negativepole. Therefore, the molecule is a dipole. Incarbon dioxide, the carbon and oxygens liealong the same axis. The bond polaritiescancel, producing a nonpolar molecule.

4. The more electronegative atom in a covalentbond will have the �� symbol and the lesselectronegative atom the �� symbol.

a.

b.

5. CaO is an ionic compound and CS2 is a polarcovalent compound. Generally, ioniccompounds have much higher meltingpoints than molecular compounds.

Interpreting Graphics 81. ; linear; 180°; none

2. ; tetrahedral; 109.5°; none

3. ; trigonal planar; 120°;

4. ; linear; 180°; noneBeF F

O

SO O

O

SO O

O

SO O

O

O

S

O

H

C

H

H H

O 3 C 3 O

C

F

F F

F

δ+

δ–

δ–

δ–δ–

N

H

H Hδ+

δ+

δ+ δ–

C CH

H

F

F

Answer Key 787

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5. ; pyramidal; 96°; none

6. ; trigonal bipyramidal;

90° and 120°; none

7. ; bent triatomic; 105�; none

8. Check students’ work.

Vocabulary Review 81. coordinate covalent bond

2. bond dissociation energy

3. bonding molecular orbital

4. sigma bond

5. VSEPR theory

6. hybridization

7. polyatomic ion

8. van der Waals forces

9. hydrogen bond

10. molecule

Quiz for Chapter 81. a 5. b 9. ST

2. c 6. c 10. ST

3. b 7. ST 11. ST

4. d 8. NT

Chapter 8 Test A

A. Matching1. i 5. d 8. c

2. e 6. a 9. g

3. j 7. b 10. h

4. f

B. Multiple Choice11. d 17. c 23. b

12. b 18. c 24. a

13. d 19. b 25. d

14. a 20. a 26. c

15. c 21. b 27. c

16. d 22. b

C. Questions28. a.

b. �N�N�

c. �C�O�

29. a. K(0.8), F(4.0), difference 3.2; ionic bond

b. S(2.5), O(3.5), difference 1.0; polarcovalent bond

c. N(3.0), O(3.5), difference 0.5; polarcovalent bond

d. H(2.1), Br(2.8), difference 0.7; polarcovalent bond

30. a.

b.

c.

D. Essay31. Network solids are substances in which all of

the atoms are covalently bonded to eachother. Samples of these solids are thought ofas single molecules. Two examples arediamond and silicon carbide.

E. Additional Questions andProblems32. C—H 5 mol��

31

9m3

ok

l�J

� 1965 kJ

C—O 1 mol�� 356 kJ

O—H 1 mol�� 41

6m4

ok

l�J

� 464 kJ

C—C 1 mol�� 31

4m7

ok

l�J

� 347 kJ

Total 3132 kJ

33.

The first 2p orbitals lie along the axisconnecting the atoms, and so form a sigmabond. The remaining 2p orbitals are found inregions above and below the axis, and form pibonds.

2px 2px

2py 2py

� sigma bond

� pi bond

356 kJ�1 mol�

PCl

Cl Cl

H HO O

H Br

Br 2 Br

O

HH

P

Cl

Cl

ClCl

Cl

F

F

P

F




34. From left to right: sp, sp, sp3, sp2, sp3

35.

Chapter 8 Test B

A. Matching1. g 5. b 8. i

2. e 6. d 9. a

3. h 7. f 10. j

4. c

B. Multiple Choice11. c 20. c 29. d

12. a 21. b 30. c

13. d 22. a 31. d

14. d 23. c 32. a

15. d 24. b 33. a

16. a 25. b 34. c

17. b 26. b 35. c

18. d 27. c 36. a

19. b 28. b 37. c

C. Questions38. a. H 6 H ; H2H

b.

c.

d.

e.

39. a.

b.

D. Essay40. An ionic bond is formed when one or more

electrons are transferred from one atom toanother. Covalent bonds are formed whenatoms share electrons. depends upon Theelectronegativity difference between twoelements is used to predict which type ofbonding will occur when specific atomscombine. Differences in excess of 2.0 result inthe formation of ionic bonds. Differences ofless than 2.0 result in covalent bonding.

E. Additional Questions andProblems41. a. polar covalent; 0.9

b. polar covalent; 0.4

c. polar covalent; 0.5

d. ionic bond; 3.1

e. nonpolar covalent; 0.2

f. polar covalent; 1.3

42. 6(C—H) 6 � (393 kJ/mol) 2358 kJ/mol1(C—C) 1 � 347 kJ/mol 347 kJ/molTotal 2705 kJ or 2.70 � 103 kJ/mol(2.70 � 103 kJ/mol�) � (0.25 mol�) 675 kJ


Section 8.4 Paper Chromatographyof Food Dyes, page 245

Analysis

1. Red, yellow, and blue are pure compounds.

2. Green food color is usually a mixture ofYellow No. 5 and Blue No. 1.

3. Red is Red No. 40, Yellow is Yellow No. 5 (orsometimes Yellow No. 6 if it is orange inappearance), and blue is Blue No. 1. (Red #3has been banned, but sometimes appears,because under the law food manufacturersare allowed to use up their current supplies.)

4. See the above figure.

BlueYellow GreenRed

Blue No. 1

Yellow No. 5

Red No. 40

P

O 3�

O

O O

6-9N 6

H �

H

H H

C O ; C ! O

6-9N 6

H H

H 2 N2 HH H @

;

H

H; O

H

O 2 H@

N N ; N 4 N

Cl

Cl

Cl

P

F

S

F

F F

F F

Cl

Cl

Answer Key 789

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5. Blue No. 1 is the most polar because it runsthe fastest and appears at the top of thechromatogram. Red No. 40 is the least polar.

You’re the Chemist1. Wet a portion of a piece of candy and blot it

with a paper towel to remove excess water.Press the wet side of the candy onto thechromatography paper so that it makes acolored spot. Repeat for other colors ofcandy. Develop in 0.1% NaCl.

2. Make a small spot of each colored markerpen on a piece of chromatography paperand develop in solvent.

3. Use a toothpick to spot a solution ofpowdered drink on chromatography paper.

4. Rubbing alcohol runs much more slowly andgives slightly better separation than 0.1%NaCl.

5. Some papers cause a reversal of thepositions of Blue No. 1 and Yellow No. 5because of the water content of the paper.Different water content changes variationsin the polarity of the stationary phase (thewater molecules hydrogen bonded to thepaper).

Section Review 9.1

Part A Completion1. monatomic 8. transition (Group B)

2. lose metals

3. 1� 9. Stock

4. 2� 10. classical

5. 3� 11. polyatomic

6. 8 12. -ite or -ate

7. 1� 13. -ite or -ate

Part B True-False14. ST 16. ST 18. AT

15. AT 17. NT


20. e 22. a

Part D Questions and Problems24. a. 1� c. 1�

b. 2� d. 2�

25. a. hydrogen carbonate

b. ammonium

c. permanganate

d. hydroxide

26. a. loses 2 c. gains 1

b. gains 2 d. loses 3

Section Review 9.2

Part A Completion1. cation 5. Roman numeral

2. anion 6. anion

3. -ide 7. oxygen

4. sodium iodide 8. zero


10. AT 12. AT

Part C Matching13. b 15. d

14. a 16. c

Part D Questions and Problems17. a. iron(III) bromide, binary ionic

b. potassium hydroxide, ionic withpolyatomic ion

c. sodium dichromate, ionic withpolyatomic ion

18. a. NaClO3

b. Pb3(PO4)2

c. Mg(HCO3)2

Section Review 9.3

Part A Completion1. nonmetallic

2. -ide

3. atoms

4. diarsenic pentasulfide


6. NT

Part C Matching8. d 10. b

9. a 11. c





12. a. phosphorus pentachloride

b. sulfur dioxide

c. tetraphosphorus decasulfide

13. a. CBr4 b. N2O4

Section Review 9.4

Part A Completion1. hydrogen 6. hydroxide ions

2. hydrogen ions 7. ionic

3. hydrobromic 8. cation

4. nitric 9. anion

5. ionic

Part B True-False10. ST 11. NT 12. NT

Part C Matching13. a 14. c 15. b

Part D Questions and Problems16. a. Mg(OH)2 c. H3PO4

b. HF d. LiOH

17. a. potassium hydroxide

b. hydroiodic acid

c. sulfuric acid

Section Review 9.5

Part A Completion1. definite proportions 9. molecular

2. proportions 10. carbon tetrachloride

3. multiple 11. acidproportions

4. small whole 12. elements

5. acid 13. anion

6. phosphoric acid 14. 4A

7. acid 15. lead(II) acetate

8. binary


17. ST 19. AT

Part C Questions and Problems20. a. lead(IV) acetate

b. hydrofluoric acid

c. diphosphorus pentoxide

d. lithium bromide

21. a. PCl5

b. FeO

c. HNO3

d. KCl

e. Ca(NO3)2

Practice Problems 9

Section 9.11. a. 2� d. 3�

b. 1� e. 2�

c. 1� f. 2�

2. a. 3 lost d. 2 lost

b. 3 gained e. 1 gained

c. 1 lost f. 2 gained

3. a. tin(II) or stannous cation

b. cobalt(III) or cobaltic cation

c. bromide anion

d. potassium cation

e. hydride anion

f. manganese(II) or manganous cation

4. a. CO32� d. OH�

b. NO22� e. CrO4

2�

c. SO42� f. NH4

�

5. a. cyanide anion

b. hydrogen carbonate anion

c. phosphate anion

d. chloride anion

e. calcium cation

f. sulfite anion

Section 9.21. a. MgO d. AlCl3

b. SnF2 e. Na2S

c. KI f. FeBr3

2. a. BaCl2 d. KBr

b. AgI e. Al2O3

c. CaS f. FeO

3. a. manganese(II) oxide or manganous oxide

b. lithium nitride

c. calcium chloride

Answer Key 791

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d. strontium bromide

e. nickel chloride

f. potassium sulfide

g. copper(II) chloride or cupric chloride

h. tin(IV) chloride or stannic chloride

4. a. Na3PO4 d. KCN

b. MgSO4 e. NH4Cl

c. NaOH f. K2Cr2O7

5. a. (NH4)2SO4 c. Ba(OH)2

b. KNO3 d. Li2CO3

6. a. sodium cyanide

b. iron(III) chloride or ferric chloride

c. sodium sulfate

d. potassium carbonate

e. copper(II) hydroxide or cupric hydroxide

f. lithium nitrate

7. a. sodium cation, Na�

b. nickel cation, Ni2�

c. calcium cation, Ca2�

d. potassium cation, K�

e. iron(III) cation, Fe3�

f. copper(I) cation, Cu�

Section 9.31. a. phosphorous pentachloride

b. carbon tetrachloride

c. nitrogen dioxide

d. dinitrogen difluoride

e. tetraphosphorous hexoxide

f. xenon difluoride

g. silicon dioxide

h. dichlorine heptoxide

2. a. NBr3 c. SO2

b. Cl2O d. N2F4

Section 9.41. a. nitrous acid c. hydrofluoric acid

b. sulfuric acid d. carbonic acid

2. a. Ca(OH)2 c. Al(OH)3

b. NH4OH d. LiOH

Section 9.51. a. K2S g. N2O5

b. SnCl4 h. Fe2(CO3)3

c. H2S i. SF6

d. CaO j. MgCl2

e. HBr k. H3PO4

f. AlF3 l. HNO3

2.

3. a. potassium phosphate

b. aluminum hydroxide

c. sodium hydrogen sulfate

d. mercury(II) oxide or mercuric oxide

e. dinitrogen pentoxide

f. nitrogen tribromide

g. phosphorous triiodide

h. ammonium sulfate

4. The law of definite proportions states thatsamples of any compound will alwayscontain the constituent elements in thesame proportions. The law of multipleproportions states that in two compoundscontaining the same two elements, themasses of one element that combines with agiven mass of the other element will be inthe ratio of small whole numbers.

Interpreting Graphics 91. 2A 8. 10

2. two 9. 10

3. 12 10. 10

4. 10 11. a. 18

5. 7A b. 18

6. one c. 18

7. nine d. 18



SO42� NO3

� OH� PO43�

Ca2� CaSO4 Ca(NO3)2 Ca(OH)2 Ca3(PO4)2

Al3� Al2(SO4)3 Al(NO3)3 Al(OH)3 AlPO4

Na� Na2SO4 NaNO3 NaOH Na3PO4

Pb4� Pb(SO4)2 Pb(NO3)4 Pb(OH)4 Pb3(PO4)4


Vocabulary Review 91. i 5. e 8. h

2. c 6. f 9. j

3. g 7. b 10. a

4. d

Quiz for Chapter 91. Metals

2. PCl5

3. anion

4. definite proportions

5. the group number

6. lose

7. hydroxide

8. H2PO4�

9. dinitrogen monoxide

10. FeCl2

11. tin(IV) sulfide (or stannic sulfide)

12. dinitrogen pentoxide

13. sodium hydrogen carbonate (or sodium bicarbonate)

14. Cu(OH)2

15. nitric acid

Chapter 9 Test A

A. Matching1. f 4. h 7. c

2. a 5. d 8. i

3. g 6. b 9. e


11. c 16. c 20. c

12. a 17. c 21. c

13. a 18. d 22. b

14. a

C. Completion23. 4� 27. 1�

24. 1� 28. oxygen

25. ions 29. nonmetallic

26. -ide 30. hydrogen

D. Problems31. a Mg(CN)2 c. SF6

b. HgBr2

32. a. copper(I) chloride

b. dinitrogen trioxide

c. potassium acetate

E. Essay33. Compounds exist in enormous numbers.

Common names do not describe thechemical composition of a compound. Theymay relate to a physical or chemical property,but usually do not reveal what elements arein the compound. The systemic method tellswhat atoms are in the compound, givesinformation on the ratio in which the atomscombined to form the compound andpromotes efficient and effective communi-cation between chemists.

Chapter 9 Test B

A. Matching1. i 4. h 7. f

2. b 5. g 8. e

3. c 6. a 9. d

B. Multiple Choice10. c 14. d 18. b

11. a 15. d 19. a

12. a 16. c 20. d

13. c 17. c 21. c

C. Completion22. gain 27. NH4

�

23. lose 28. magnesium nitrate

24. cation 29. -ide

25. definite 30. Ca3(PO4)2

proportions

26. 2� 31. more

D. Problems32. a. Ca(NO3)2, calcium nitrate

b. Na2SO4, sodium sulfate

c. Fe2O3, iron(III) oxide

d. Al2(CO3)3, aluminum carbonate

Answer Key 793

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33. a. SiO2 e. HNO3

b. CF4 f. AgNO3

c. Zn(OH)2 g. Fe2(SO4)3

d. PBr3 h. HgCl2

34. a. carbon disulfide

b. ammonium carbonate

c. diarsenic pentoxide

d. carbon monoxide

e. tin(IV) hydroxide

f. sulfuric acid

g. phosphorus pentiodide

h. potassium permanganate

E. Essay35. Ionic compounds consist of a metallic and a

nonmetallic ion, whereas molecularcompounds consist of nonmetallic elements.Ionic compounds are named from the twoions that comprise them, using a Romannumeral to distinguish between positive ionsof the same element that have more than onecharge. Molecular compounds are namedfrom the elements that comprise them, usingprefixes to denote the numbers of atoms ofeach element present. No prefix is used ifonly one atom of the first element is present.The name of the second element always endsin -ide.


Section 9.2 Names and Formulasfor Ionic Compounds,page 267

Analysis

1. Na2SO4 � AgNO3 did not form a precipitate.

2. Formula Name

a. Ag2CO3 silver carbonate

b. Ag3PO4 silver phosphate

c. AgOH silver hydroxideTeacher’s note: This is Ag2O, silver oxide

d. no visible reaction

e. PbCO3 lead(II) carbonate

f. Pb3(PO4)2 lead(II) phosphate

g. Pb(OH)2 lead(II) hydroxide

h. PbSO4 lead(II) sulfate

i. CaCO3 calcium carbonate

j. Ca3(PO4)2 calcium phosphate

k. Ca(OH)2 calcium hydroxide

l. CaSO4 calcium sulfate

a e i

b f j

c

d h l

Na2CO3

(CO32�)

AgNO3

(Ag�) Pb(NO3)2

(Pb2�)CaCl2(Ca2�)

Na3PO4

(PO43�)

NaOH(OH�)

Na2SO4

(SO42�)

milkywhiteppt

cloudywhiteppt

muddybrown

ppt

novisible

reaction

cloudy tan ppt

milkywhiteppt

milkywhiteppt

milkywhiteppt

grainywhiteppt

milkywhiteppt

cloudywhiteppt

grainywhiteppt

Figure A

g k




You’re The Chemist1.

Formula Name

a. Fe2(CO3)3 iron(III) carbonate

b. FePO4 iron(III) phosphate

c. Fe(OH)3 iron(III) hydroxide

d. no visible reaction

e. MgCO3 magnesiumcarbonate

f. Mg3(PO4)2 magnesiumphosphate

g. Mg(OH)2 magnesiumhydroxide

h. no visible reaction

i. CuCO3 copper(II) carbonate

j. Cu3(PO4)2 copper(II) phosphate

k. Cu(OH)2 copper(II) hydroxide

l. no visible reaction

2. a. 2Fe3+ � 3CO32� → Fe2(CO3)3(s)

b. Fe3+ � PO43� → FePO4(s)

c. Fe3+ � 2OH� → Fe(OH)2(s)

e. Mg2+ � CO32� → MgCO3(s)

f. 3Mg2+ � 2PO43� → Mg3(PO4)2(s)

g. Mg2+ � 2OH� → Mg(OH)2(s)

i. Cu2+ � CO32� → CuCO3(s)

j. 3Cu2+ � 2PO43� → Cu3(PO4)2(s)

k. Cu2+ � 2OH� → Cu(OH)2(s)

Section Review 10.1

Part A Completion1. mole

2. Avogadro’s number

3. atomic masses

4. molar mass

5. 6.02 � 1023


7. ST 9. AT

Part C Matching11. b 13. d

12. c 14. c

Part D Questions and Problems15. 9.3 � 1015 atoms Pb �

1.5 � 10�8 mol Pb

16.

molar mass of C2H6 30.0 g

17. 3.65 � 10�2 mol K2SO4 ��11.0704.

m3 g

olKK2

2

SSOO

4

4�

636 � 10�2 g K2SO4 6.36 g K2SO4

18.

Section Review 10.2

Part A Completion1. 22.4 4. Density

2. molar volume 5. mole

3. 22.4 L SO2/1.00 mol SO2


7. AT 9. AT

2.5 mol H2O2 �

6.02�

� 1023 representative particles1 mol H2O2

1.5 � 1024 representative particles

2 mol C � �11

2m.0

ogl C

C� 24.0 g C

6 mol H � �11.

m0

og

lHH

� 6.0 g H

1.0 mol Pb��6.02 � 1023 atoms Pb

a e i

b f j

c

d h l

Na2CO3

(CO32�)

FeCl3(Fe3�)

MgSO4

(Mg2�)CuSO4

(Cu2�)

Na3PO4

(PO43�)

NaOH(OH�)

Na2SO4

(SO42�)

orangeppt

whiteppt

blueppt

orangeppt

whiteppt

blueppt

orangeppt

whiteppt

blueppt

no visiblereaction

no visible reaction

no visiblereaction

Figure B

g k

Answer Key 795

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Part C Matching10. d 12. a 14. e

11. c 13. b

Part D Questions and Problems15. Molar mass 28.0 � 16.0 44.0 g

Density �vo

mlua

mss

e� �

24

24

.

.40

Lg

� 1.96 g/L

16. Molar mass NaCl 23.0 g � 35.5 g 58.5g/mol

58.5 g/mol � 2 mol 117 g

17. Molar mass O2 2(16.0 g/mol) 32.0 g/mol

16.0 g/32.0 g/mol 0.500 mol

18. 22.4 L/mol � 0.500 mol 11.2 L

Section Review 10.3

Part A Completion1. percent composition

2. 100

3. molar mass

4. empirical

5. whole-number

6. molecular


8. AT 10. NT


12. b


14. a. �15

1204

ggCr

C

2

rO3

� � 100 68.4% Cr

�152

48g

gC

Or2O3

� � 100 31.6% O

b. �284

11g

0M

gnM

2Pn

2O7�� 100 38.7% Mn

�284 g

62M

gn

P

2P2O7�� 100 21.8% P

�284

1g1M2 g

n2

OP2O7

�� 100 39.4% O

c. �223031

gg

HH

ggS

� � 100 86.3% Hg

�2

33

23

.g1

Hg S

gS� � 100 13.8% S

d. �164

4g0.

C1

ag(NCa

O3)2�� 100 24.5% Ca

�164 g

28C

ga(

NNO3)2�� 100 17.1% N

�164 g

96C

ga(

ONO3)2�� 100 58.5% O

15. 29.1 g Na� � �1.

2030.0m

go

Nl N

a�a

� 1.27 mol Na

1.27 mol Na/1.27 1 � 2 2

40.5 g S� � �1.

3020.0m

go

S�l S

� 1.27 mol S

1.27 mol S/1.27 1 � 2 2

30.4 g O� � �1.

1060.0m

go

O�l O

� 1.90 mol O

1.90 mol O/1.27 1.5 � 2 3

Empirical formula Na2S2O3

16. molar mass Fe2O3

2 mol Fe� � �515.

m8

og

lFFee�

� � 3 mol O� � �116

m.0

oglOO�

�

112 g Fe � 48 g O 160 g Fe2O3

%Fe �g

gFFee

2O3� � 100 �

11

16

20

gg

� � 100

70.0% Fe

Practice Problems

Section 10.11. 342.3 g/mol

2. a. 208.2 g/mol b. 352.0 g/mol

3. a. 158.0 g/mol b. 310.2 g/mol

4. 5.85 mol H2O

5 3.6 � 1023 atoms

6. 32.0 g

Section 10.21. a. 180.2 g/mol c. 96.2 g/mol

b. 84.0 g/mol d. 153.2 g/mol

2. a. 1.8 � 103 g

b. 26 g

c. 3.20 � 10�2 g

d. 0.480 g or 4.80 � 10�1 g

e. 1.43 � 102 g

3. 1.87 � 102 g

4. 204.1 g

639 kg Fe2O3 ��10

700.

k0gkFg

eF

2Oe

3� 447 kg Fe




5. a. 4.9 � 10�3 mol d. 1.98 � 10�5 mol

b. 9.10 � 10�2 mol e. 1.97 � 10�5 mol

c. 1.08 � 10�2 mol

6. 5.43 mol

7. 15.1 g

8. 59.6 L CH4

9. 6.03 mol NH3

Section 10.3

1. Percent C �52

5..8344

gg

cCpd

� � 100 10.1% C

Percent H �52

0..8442

gg

cHpd

� � 100 0.79% H

Percent Cl �5427.8.048ggcCp

ld

� � 100 89.1% Cl

2. Mass of Cl total mass of compound � mass of Sn 18.35 g of compound � 5.74 g Sn 12.61 g Cl

Percent of Sn �18

5..37

54

gg

cS

pn

d� � 100

31.3% Sn

Percent of Cl �1182.3.651ggcCp

ld

� � 100

68.7% Cl

3. Percent C �4

3.7

.98

01

7g

gcp

Cd

� � 100 81.7% C

Percent H �4

0.7

.88

71

4g

gc

Hpd

� � 100 18.3% H

4. Percent C � 100

30.4% C

Mass C 30.4% C � 65.3 g 19.8 g

5. 13.2 g Al

6. 15.11 g Fe

7. a. CCl4

b. CHCl3

Interpreting Graphics 101. 85.7% C

2. ethene

3. Some compounds have the same empiricalformula but different molecular formulas.Cyclohexane and ethene have the sameempirical formula and, therefore, the samepercent composition. They are indistinguish-able on the basis of percent compositionalone.

4. molar mass

5.

Bromine accounts for the largest percent ofthe mass of dibromoethane.

6. molar mass of H2SO4 98.1 g

%H �928.0.0

28

gg

� � 100% 2.06%

%S �39

28

.

.00

68

gg

� � 100% 32.69%

%O �69

48

.

.00

08

gg

� � 100% 65.25%

Acid X represents sulfuric acid, H2SO4. Usethe same approach to show that Acid Yrepresents sulfurous acid, H2SO3.

Vocabulary Review 101. percent composition

2. empirical formula

3. 22.4 L

4. molar mass

5. mole

6. Avogadro’s number

7. standard temperature and pressure (0�C, 1 atm)

Quiz for Chapter 101. a substance

2. atomic mass

3. molar mass

4. atom

5. �31

2m.0

ogl O

O

2

2�

6. AT

7. AT

8. ST

9. NT

10. AT

11. 3.05 g Ag�� 11

.0080

gm

Aog�l�

� � �11

.9070

gm

Aoul�

� 5.56 g Au

12.8%C

85.1%Br

2.1%H

150.0 g C2H4 � �10

104.

g3

Cg

2

HH4

� 21.4 g

48.0 g C��158.1 g Ca(C2H3O2)2

Answer Key 797

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12. 40.7 g C� � �1.

1020.0m

go

C�l C

� 3.39 mol C

54.2 g O� � �1.

1060.0m

go

O�l O

� 3.39 mol O

5.1 g H� � �1.0

10.0

mg

oH�l H

� 5.1 mol H

5.1 mol H/3.39 1.5 mol H3.39 mol C/3.39 1.00 mol C3.39 mol O/3.39 1.00 mol O1.5 mol H � 2 3 mol H1 mol C � 2 2 mol C1 mol O � 2 2 mol O

Empirical formula C2H3O2

Chapter 10 Test A

A. Matching1. j 5. b 9. d

2. a 6. c 10. h

3. g 7. i

4. e 8. f

B. Multiple Choice11. c 16. c 21. c

12. a 17. a 22. b

13. d 18. d 23. b

14. c 19. a 24. c

15. c 20. c

C. Problems25.

26. 24.0 g C� � �11

2m.0

ogl C

C�� 2 mol C

76.0 g F� � �11

9m.0

ogl F

F�� 4 mol F

The mole ratio of C to F is �24

mm

oo

ll

CF

�.

The lowest whole-number ratio of C to F is CF2.

27. 364 g Ar� � �31

9m.9

ogl A

Arr�

� 9.12 mol. Ar

D. Essay28. A mass that has molar as a modifier must be

the mass of a mole. A mole is a unit thatcounts all kinds of representative particles, somolar mass can be used for the mass ofAvogadro’s number of particles of any puresubstance.

Chapter 10 Test B

A. Matching1. f. 4. e 7. a

2. h 5. g 8. d

3. b 6. c

B. Multiple Choice9. d 15. a 21. d

10. b 16. d 22. a

11. a 17. a 23. c

12. c 18. c 24. b

13. b 19. c 25. d

14. b 20. b 26. c

C. Problems

27. 0.25 mol�� 6.02 � 1023

1.50 � 1023 atoms

28. 6.25 mol� � 98.1 �m

gol�� 613 g

29. 15.0 kg� � 1000 �kg�g��

41

4m.0

og�l�

� � �2m2.4

ol�L

� 7636 L

7640 L

30. 0.650 �L�g

� � 22.4 �m

L�ol� 14.6 g/mol

31. 3.75 � 1015 atoms� ��6.02 �

11m0

o23

l�atoms�

�

� �19

m7.

o0l�g

� 1.23 � 10�6 g

32. 24.3 g � 28.0 g � 96.0 g

148.3 g/mol Mg(NO3)2

Mg: �12448.3.3

g�g�

� � 100 16.4%

N: �12488.0.3

g�g�

� � 100 18.9%

O: �19468.0.3

g�g�

� � 100 64.7%

�mol

atoms�

5.00 � 1023 molecules F2

� � �318

m.0

og

lFF

2

2��

31.6 g F2

1 mol F2��6.02 � 1023 molecules F2




33. 27.3 g C� � �11

2m.0

ogl C

C�� 2.28 mol C

72.7 g O�� 11

6m.0

ogl O

O�� 4.54 mol O

�22

.

.22

88

� 1, and �42

.

.52

48

� 2

Thus, the empirical formula is CO2.

34. (56.38 g P�) � �13m1.

o0

lg�P

� 1.82 mol P

(43.62 g O�) � �1

1m6.

o0

lg�O

� 2.73 mol O

Empirical formula is P2O3

empirical formula mass 110.0 g, so that

molecular formula �21

11

90

.

.90

gg

� 2 � P2O3

or P4O6

D. Essay35. The mass of a single atom of an element is the

atomic mass given on the periodic table,expressed in atomic mass units. The mass ofone mole, or Avogadro’s number of atoms ofthat element has the same numerical value asthe atomic mass, but expressed in grams.


Section 10.2 Counting byMeasuring Mass,page 304

AnalysisStudent data may vary slightly.

1. x mol NaCl 5.09 g� NaCl � �51

8m.5

og�l

�

x mol NaCl 0.0870 mol NaCl

2. x mol H2O 4.30 g� H2O � �11

8m.0

og�l

�

x mol H2O 0.239 mol H2O

x mol CaCO3 9.68 g� CaCO3 � �1100

m.1ol

g��

x mol CaCO3 0.0967 mol CaCO3

3. x mol H 0.239 mol H2O � �1

2m

mo

ol H

l H

2O�

x mol H 0.478 mol H

x mol O 0.239 mol H2O � �1

1m

mo

ol H

l O

2O�

x mol O 0.239 mol O

H2O(l )

Mass(grams)

Molar Mass(g/mol)

Moles of each

compound

Moles of each

element

Atoms of each

element

NaCl(s) CaCO3(s)

4.30 5.09 9.68

18.0 58.5 100.1

0.239 0.0870 0.0967

0.478 H0.239 O

0.0870 Na0.0870 Cl

0.0967 Ca0.0967 C0.290 O

2.88 � 1023 H1.44 � 1023 O

5.24 � 1022 Na5.24 � 1022 Cl

5.82 � 1022 Ca5.82 � 1022 C1.75 � 1023 O

Figure A

Answer Key 799

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4.

5. Water has the greatest number of moles inone teaspoon.

6. Water has the greatest total number ofatoms.

You’re The Chemist1. Determine the mass of 100 drops of water

and then calculate the mass in grams of onedrop.

2. Determine the mass of a piece of chalk.Write your name with the chalk anddetermine the mass of the chalk again.Convert the mass difference to moles andatoms.

Section Review 11.1

A. Completion1. equation 7. subscripts

2. reactants 8. (l)

3. products 9. (s)

4. mass 10. (g)

5. coefficients 11. (aq)

6. element 12. catalyst

B. True-False13. NT 15. NT

14. AT 16. ST

C. Multiple Choice17. c 20. g 22. d

18. e 21. a 23. f

19. b

D. Problems24. a. 2Al(s) � 6HCl(aq) → 2AlCl3(aq) � 3H2(g)

b. 2C2H2(g) � 5O2(g) → 4CO2(g) � 2H2O(g)

Section Review 11.2

A. Completion1. predict

2. combination

3. elements

4. single

5. decomposition

6. single-replacement

7. activity series of metals

8. double-replacement

9. aqueous

10. oxygen

11. carbon dioxide or water

12. water or carbon dioxide

x atoms H 0.478 mol H

�

x atoms H 2.88 � 1023 atoms H

x atoms O 0.239 mol O

�

x atoms O 1.44 � 1023 atoms O

x atoms Na 0.0870 mol Na

�

x atoms Na 5.24 � 1022 atoms Na

x atoms Cl 0.0870 mol Cl

�

x atoms Cl 5.24 � 1022 atoms Cl

x atoms Ca 0.09697 mol Ca

�

x atoms Ca 5.83 � 1022 atoms Ca

x atoms C 0.0967 mol C

�

x atoms C 5.83 � 1022 atoms C

x atoms O 0.290 mol O

�

x atoms O 1.75 � 1023 atoms O

6.02 � 1023 atoms��

1 mol O

6.02 � 1023 atoms��

1 mol C

6.02 � 1023 atoms��

1 mol Ca

6.02 � 1023 atoms��

1 mol Cl

6.02 � 1023 atoms��

1 mol Na

6.02 � 1023 atoms��

1 mol O

6.02 � 1023 atoms��

1 mol H

x mol Na 0.0870 mol NaCl � �1

1m

mo

ol

lN

Na

aCl

�

x mol Na 0.0870 mol Na

x mol Cl 0.0870 mol NaCl � �1

1m

mol

oNl C

aCl

l�

x mol Cl 0.0870 mol Cl

x mol Ca 0.0967 mol CaCO3

� �1 m

1 mol

oClaCCaO3

� 0.0967 mol Ca

x mol C 0.0967 mol CaCO3

��1 m

1oml C

ola

CCO3

� 0.0967 mol C

x mol O 0.0967 mol CaCO3

��1 m

3oml C

olaOCO3

� 0.290 mol O




B. True-False13. AT 15. ST 17. ST

14. NT 16. ST

C. Matching18. b 20. a

19. d 21. c

D. Questions and Problems22. a. combustion

b. combination

23. 2Li3PO4(aq ) � 3Zn(NO3)2(aq ) → Zn3(PO4)2(s) � 6LiNO3(aq )

For any double-replacement reaction tooccur, one of the products must be a solid(precipitate), or water, or a gas.

Section Review 11.3

A. Completion1. water

2. aqueous

3. complete ionic equation

4. spectator ions

5. net ionic equation

6. charge

7. atoms

8. precipitate

9. solubility

B. True-False10. ST 12. ST

11. AT 13. NT

C. Matching14. d 16. a 18. c

15. f 17. b 19. e

D. Questions and Problems20. Cl2(g) � Na�(aq) � Br�(aq)

→ Br2(l) � Na�(aq) � Cl�(aq)

The spectator ion is Na�. The balanced net ionic equation is

Cl2(g) � 2Br�(aq) → Br2(l) � 2Cl�(aq)

21. a. AgCl(s)b. CaCO3(s)c. noned. PbCl2(s)

Practice Problem Solutions

Section 11.11. H2(g) � O2(g) → H2O(l)

2. Fe(s) � S(s) → FeS(s)

3. MgCO3(s) yMgO(s) � CO2(g)

4. H2(g) � Cl2(g) y 2HCl(g)

5. Hydrochloric acid and solid calciumcarbonate react to produce carbon dioxidegas, aqueous calcium chloride, and liquidwater.

6. silver � sulfur → silver sulfide

Silver metal and sulfur react to produce solidsilver sulfide. There are 2 silver atoms and 1sulfur atom on each side of the equation, andthe coefficients are in their lowest possibleratio. Thus, the equation is balancedcorrectly.

7. H2O(l) � SO3(g) → H2SO4(aq)

8. 2AgNO3(aq) � Cu(s)→2Ag(s) � Cu(NO3)2(aq)

9. 4P(s) � 5O2(g) → P4O10(s)

Section 11.21. Magnesium is a Group 2A metal and forms

cations with a 2� charge. Oxygen is in Group6A and forms anions with a 2� charge. Theycombine in a 1:1 ratio to form MgO.

Mg � O2 → MgO

The balanced chemical equation is

2Mg(s) � O2(g) → 2MgO(s)

2. 2Al(s) � 3F2(g) → 2AlF3(s)

3. First, determine the formulas for the reactantand products and write them in their properpositions to form a skeleton equation.

KClO3 → KCl � O2

Next, balance the equation.

2KClO3(s) → 2KCl(s) � 3O2(g)

4. Ca(s) � 2HCl(aq) → H2(g) � CaCl2(aq)

5. C3H8(g) � 5O2(g) → 3CΟ2(g) � 4H2O(g)

6. FeCl3(aq) � 3NaOH(aq) → Fe(OH)3(s) � 3NaCl(aq)

7. combination reactions: 1 and 2decomposition reaction: 3single-replacement reaction: 4double-replacement reaction: 6combustion reactions: 1 and 5

8. a. no reaction

b. Ca(s) � Mg(NO3)2(aq) → Ca(NO3)2(aq) � Mg(s)

Answer Key 801

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c. 2K(s) � H2SO4(aq) → K2SO4(aq) � H2(g)

d. no reaction

In a bromine is less reactive than chlorine sono reaction occurs. In b calcium replaces aless reactive magnesium and in c potassiumreplaces the less reactive hydrogen. Becausezinc is less reactive than sodium, no reactionoccurs in d.

Section 11.31. This reaction can be described as:

Ba(NO3)2(aq) � Na2SO4(aq) → BaSO4(s) � 2NaNO3(aq)

The net ionic equation is:

Ba2�(aq) � SO42�(aq) → BaSO4(s)

2. This reaction can be described as:

Mg(s) � 2HCl(aq) → Η2(g) � MgCl2(aq)

The net ionic equation is:

Mg(s) � 2H�(aq) → Η2(g) � Mg2�(aq)

3. Pb(NO3)2(aq) � 2NH4Cl(aq) y PbCl2(s) � 2NH4NO3(aq)

Pb2� � 2Cl� → PbCl2(s)

4. a. no precipitate c. CuS(s)

b. AgCl(s) d. Al(OH)3(s)

Interpreting Graphics 11nitrous oxide N2O

nitric oxide NO

oxygen O2

carbon dioxide CO2

water H2O

ammonia NH3

urea CH4N2O

benzene C6H6

nitrobenzene C6H5NO2

carbonic acid H2CO3

nitric acid HNO3

1. N2O � O2 → NO2N2O � O2 → 4NO

2. C6H6 � O2 → CO2 � H2O2C6H6 � 15O2 → 12CO2 � 6H2O

3. NH3 � CO2 → CH4N2O � H2O2NH3 � CO2 → CH4N2O � H2O

4. C6H6 � HNO3 → C6H5NO2 � H2Obalanced as written

Vocabulary Review 111. combustion 6. spectator

2. decomposition 7. single replacement

3. net ionic equation 8. balanced equation

4. catalyst 9. coefficients

5. reactants

Solution: precipitate

Quiz for Chapter 111. skeleton

2. equation

3. formulas

4. balanced

5. a. right

b. left

6. arrow

7. 2HgO(s) → 2Ηg(l) � O2(g)

8. 2Ag�(aq) � 2Na�(aq) � 2NO3�(aq) � CO3

2�(aq)→ 2Νa�(aq) � 2NO3

�(aq) � Ag2CO3(s)

net: 2Ag�(aq) � CO32�(aq) → Ag2CO3(s)

9. C4H8(g) � 6O2(g) → 4CO2(g) � 4H2O(g)

10. 2Na(s) � Br2(l) → 2NaBr(s)

Chapter 11 Test A

A. Matching1. e 5. a 8. f

2. g 6. h 9. d

3. j 7. c 10. i

4. b

B. Multiple Choice11. a 16. b 21. d

12. c 17. b 22. d

13. b 18. c 23. b

14. c 19. d 24. a

15. a 20. b 25. c

C. Problems26. a. 3Ca(s) � 2H3PO4(aq)

→ Ca3(PO4)2(s) � 3H2(g)

b. 2KBrO3(s) → 2KBr(s) � 3O2(g)

c. (NH4)2CO3(aq) � 2NaOH(aq) → Na2CO3(aq) � 2NH3(g) � 2H2O(l)




27. a. single-replacement

b. decomposition

c. double-replacement

28. a. C5H10(g) � 5O2(g) → 5CO(g) � 5H2O(g)(incomplete)

b. 2C3H7OH(l)�9O2(g)y6CO2(g) � 8H2O(g)(complete)

29. a. 2K3PO4(aq) � 3MgCl2(aq) → Mg3(PO4)2(s) � 6KCl(aq)

net: 3Mg2�(aq) � 2PO43�(aq)

→ Mg3(PO4)2(s)

b. 2Fe(NO3)3(aq) � 3Na2CO3(aq) → Fe2(CO3)3(s) � 6NaNO3(aq)

net: 2Fe3�(aq) � 3CO32�(aq)

→ Fe2(CO3)3(s)

D. Essay30. Whether one metal will replace another is

determined by the relative reactivity of thetwo metals. The activity series of metals listsmetals in order of decreasing reactivity. Ametal will replace any metal found below it inthe activity series.

Chapter 11 Test B

A. Matching1. g 5. c 8. j

2. h 6. a 9. b

3. i 7. d 10. e

4. f


12. b 17. a 22. a

13. c 18. d 23. c

14. b 19. b 24. b

15. a 20. d 25. d

C. Problems26. NaCl(aq) � AgNO3(aq)

→ NaNO3(aq) � AgCl(s)

27. a. CS2(s) � 3O2(g) → CO2(g) � 2SO2(g)

b. 2HNO3(aq) � Mg(OH)2(aq) → Mg(NO3)2(aq) � 2H2O(l )

c. Fe2O3(s) � 3CO(g) → 2Fe(s) � 3CO2(g)

28. a. Li2O(s) � H2O(l ) → 2LiOH(aq)

b. 2H2O(l ) 2H2(g ) � O2(g )

c. 2Al(s) � 3Fe(NO3)2(aq) → 2Al(NO3)3(aq) � 3Fe(s)

d. 2HNO3(aq) � Ca(OH)2(aq)→ Ca(NO3)2(aq) � 2H2O(l )

e. C3H8(g ) � 5O2(g ) → 3CO2(g ) � 4H2O(l )

29. a. KOH(aq) � HCl(aq) → KCl(aq) � H2O(l )

K�(aq) � OH�(aq) � H�(aq) � Cl�(aq)→ Κ�(aq) � Cl�(aq) � H2O(l)

Spectator ions: K�(aq) and Cl�(aq)

Net: H�(aq) � OH�(aq) → H2O(l )

b. Pb(NO3)2(aq) � KI(aq) → PbI2(s) � KNO3(aq)

Pb2�(aq) � 2NO3�(aq) � K�(aq) � I�(aq)

→ PbI2(s) � K�(aq) � NO3�(aq)

Spectator ions: K�(aq) and NO3�(aq)

Net: Pb2�(aq) � 2I�(aq) → PbI2(s)

c. ZnI2(aq) � NaOH(aq) → NaI(aq) � Zn(OH)2(s)

Zn2�(aq) � 2I�(aq) � Na�(aq) � OH�(aq)→ Na�(aq) � I�(aq) � Zn(OH)2(s)

Spectator ions: Na�(aq) and I�(aq)

Net: Zn2�(aq) � 2OH�(aq) → Zn(OH)2(s)

D. Essay30. AB � CD → AD � CB; double-replacement

reaction; the cations have exchangedpositions such that two new compounds areformed.E � FG → EG � F; single-replacementreaction; the metal E has replaced the metal Fso that a new compound and a differentelement are produced.

electricity

JJJJH

Answer Key 803

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Section 11.3 PrecipitationReactions: Formationof Solids, page 345

Analysis

1. a. Na2CO3 � 2AgNO3

→ 2NaNO3 � Ag2CO3(s)

b. 2Na3PO4 � 3Pb(NO3)2

→ 6NaNO3 � Pb3(PO4)2(s)

2. Sodium hydroxide reacts with calciumchloride to form sodium chloride and solidcalcium hydroxide.

3. Mixings d, n, and o all gave no visiblereaction so it is not necessary to write anequation.

4. b. Na3PO4 � 3AgNO3

→ 3NaNO3 � Ag3PO4(s)

c. NaOH � AgNO3 → NaNO3 � AgOH(s)

Teacher’s note:The students will write the above reactionbut this is what really happens: 2NaOH � 2AgNO3

→ 2NaNO3 � Ag2O(s) � H2O

e. NaCl � AgNO3 → NaNO3 � AgCl(s)

f. Na2CO3 � Pb(NO3)2

→ 2NaNO3 � PbCO3(s)

h. 2NaOH � Pb(NO3)2

→ 2NaNO3 � Pb(OH)2(s)

i. Na2SO4 � Pb(NO3)2

→ 2NaNO3 � PbSO4(s)

j. 2NaCl � Pb(NO3)2 → 2NaNO3 � PbCl2(s)

k. Na2CO3 � CaCl2 → 2NaCl � CaCO3(s)

l. 2Na3PO4 � 3CaCl2

→ 6NaCl � Ca3(PO4)2(s)

m. 2NaOH � CaCl2 → 2NaCl � Ca(OH)2(s)

5. a. 2Ag� � CO32� → Ag2CO3(s)

b. 3Ag� � PO43� → Ag3PO4(s)

c. Ag� � OH� → AgOH(s)

e. Ag� � Cl� → AgCl(s)

f. Pb2� � CO32� → PbCO3(s)

g. 3Pb2� � 2PO43� → Pb3(PO4)2(s)

h. Pb2� � 2OH� → Pb(OH)2(s)

i. Pb2� � SO42� → PbSO4(s)

j. Pb2� � 2Cl� → PbCl2(s)

k. Ca2� � CO32� → CaCO3(s)

l. 3Ca2� � 2PO43� → Ca3(PO4)2(s)

m. Ca2� � 2OH� → Ca(OH)2(s)

You’re the Chemist1. KI � AgNO3 → KNO3 � AgI(s)

Silver iodide is pale green.

Ag� � I� → AgI(s)

2KI � Pb(NO3)2 → 2KNO3 � PbI2(s) Lead(II) iodide is bright yellow.

Pb2� � 2I� → PbI2(s)

2. Adding one drop of lead nitrate to a fewgrains of table salt causes white crystals togrow on the salt. Silver nitrate produces asimilar result.

3. Place one drop of lead nitrate on a small pileof dry table salt. Be sure to keep part of thepile dry and look carefully for signs of yellowlead iodide.

Section Review 12.1

Part A Completion1. moles/molecules

2. balanced equation

3. mass/atoms

4. atoms/mass

5. moles

6. STP (standard temperature and pressure)

a f k

b

c

e j o

Na2CO3

(CO32�)

AgNO3

(Ag�) Pb(NO3)2

(Pb2�)CaCl2(Ca2�)

Na3PO4

(PO43�)

NaOH(OH�)

Na2SO4

(SO42�)

NaCl(Cl�)

whiteppt

tanppt

brownppt

novisible

reaction

whiteppt

whiteppt

whiteppt

whiteppt

whiteppt

whiteppt

whiteppt

whiteppt

whiteppt

novisible

reactionno

visiblereaction

g

d

l

n

h m

i




Part B True-False7. AT 9. AT 11. AT

8. ST 10. ST 12. ST

Part C Matching13. b 15. e 17. c

14. d 16. a

Part D Questions and Problems18. moles N2 2

moles O2 3moles N2O3 2molecules N2 2molecules O2 3molecules N2O3 2volume N2 2 � 22.4 L 44.8 Lvolume O2 3 � 22.4 L 67.2 Lvolume N2O3 2 � 22.4 L 44.8 L

2 mol N2 56 g3 mol O2 96 g

mass reactants 152 g

2 mol N2O3 152 gmass product 152 g

19.

Section Review 12.2

Part A Completion1. representative particles

2. volumes

3. coefficients

4. mole ratios

5. product/reactant

6. reactant/product

7. moles

Part B True-False8. AT 11. ST 13. NT

9. NT 12. NT 14. AT

10. AT

Part C Matching15. c 17. d 19. a

16. b 18. e


20.

21.

Section 12.3

Part A Completion1. limiting reagent 4. theoretical yield

2. used up 5. maximum

3. product 6. actual yield

Part B True-False7. AT 9. NT 11. AT

8. NT 10. ST 12. AT

Part C Matching13. b 15. e 17. d

14. c 16. a


18. a.

1.6 mol O2 neededSO2 is the limiting reagent.

b.

mol SO3 can be formed2.7 mol O2 � 1.6 mol O2

1.1 mol O2 in excess

Practice Problems 12

Section 12.11. 10A � 2C � Ci y A10C2Ci

2. 2KClO3(s) y 2KCl(s) � 3O2(g)

12 mol KClO3 � 18 mol O23 mol O2��

2 mol K 3ClO

25 A10C2Ci � �A1

1

0C0A

2Ci� 250 apples

3.1 mol SO2 �2 mol SO3��2 mol SO2

3.1 mol SO2 � �21

mm

oollSOO

2

2�

2.1 � 1024 molecules O2

� � �47mm

oollNOH

2

3�

� �11

7m.0

ogl N

NHH

3

3� 33.9 g NH3

1 mol O2��6.02 � 1023 molecules O2

4.8 g O2��

� 6.7 L CO22.4 L CO��1 mol CO

2 mol CO��1 mol O2

1 mol O2��32.0 g O2�

14 mol FeCl3 � 21 mol Cl23 mol Cl2��

2 mol FeCl3

Answer Key 805

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3.

4. 2H2(s) � O2(g) y 2H2O(g)

Section 12.2

1.

2.

3.

4.

5. 2Ag(s) � Cl(g) y 2AgCl(s)

6.

7.

8. Zn(s) � 2HNO3y H2(g) � Zn(NO3)2

Section 12.31. 2H2(g) � O2(g) y 2H2O(g)

2. 2H2(g) � O2(g) y 2H2O(g)

3. C(s) � O2(g) y CO2(g)

percent yield � 100% = 83.3%

4. 2HCl(g) y H2(g) � C12(g)

percent yield � 100% = 54.2%

5.

mass of Ag(s) reclaimed = 0.946 � 75.3 g Ag 71.2 g Ag

6.

actual yield 71.0 g MgO � 0.817 58.0 g MgO

Interpreting Graphics 121. a. 102.1 g/mol c. 180.1 g/mol

b. 138.1 g/mol

2. Student 1: 0.0155 moles SAStudent 2: 0.0147 moles SA

3. Student 1: 5.25 g acetic anhydride0.0514 moles

Student 2: 5.25g acetic anhydride0.0514 moles

4. salicylic acid

5. Student 1: 2.79 gStudent 2: 2.65 g

42.8 g Mg � �

� 71.0 g MgO40.3 g MgO��1 mol MgO

2 mol MgO��

2 mol Mg1 mol Mg��24.3 g Mg

100.0 g AgCl � �

� 75.3 g Ag108 g Ag��1 mol Ag

2 mol Ag��2 mol AgCl

1 mol AgCl��143.5 g AgCl

13.6 g Cl2��25.1 g Cl2

25.8 g HCl � �

� 25.1 g Cl271.0 g Cl2��1 mol Cl2

1 mol Cl2��2 mol HCl

1 mol HCl��36.5 g HCl

55.0 g CO2��66.0 g CO2

18.0 g C � �

� 66.0 g CO244.0 g CO2��1 mol CO2

1 mol CO2��1 mol C

1 mol C�12.0 g C

160.0 g O2 � �

10.0 mol H2 needed

Oxygen is the limiting reagent.

5.00 mol O2 � �

180 g H2O

18.0 g H2O��1 mol H2O

2 mol H2O��

1 mol O2

2 mol H2��1 mol O 2

1 mol O2��32.0 g O2

4 mol O2 � �21

mm

oo

ll

HO2

2� 8 mol H2

16 mol H2 � 8 mol O2

Oxygen is the limiting reagent.

4 mol O2 � 8 mol H2O formed2 mol H2O��

1 mol O2

1 mol O2��2 mol H2

7.5 L H2��

� 22 g Zn65.4 g Zn��1 mol Zn

1 mol Zn��1 mol H2

1 mol H2��22.4 L H2�

15.0 L N2O3 � 22.5 L O23 L O2�

2 L N2O3

4.80 g O2 � �

� 6.72 L CO22.4 L CO��1 mol CO

2 mol CO��1 mol O2

1 mol O2��32.0 g O2

84 g AgCl � �

� 63 g Ag108 g Ag��1 mol Ag

2 mol Ag��2 mol AgCl

1 mol AgCl��43.5 g AgCl

75.0 g KClO3 �

� � 29.4 g O232.0 g O2��1 mol O2

3 mol O2��2 mol KClO3

1 mol KClO3��122.6 g KClO3

4 mol NO � �

184 g NO2

46 g NO2��1 mol NO2

2 mol NO2��2 mol NO

14 mol FeCl3 � �2

3m

mo

ol F

l CeC

l2

l3� 21 mol Cl2

10 mol H2 � 20 mol HCl2 mol HCl��1 mol H2

2.0 � 1023 molecules O2 ��2

1m

mol

oelceuc

luelse

HO

2

2

O�

4.0 � 1023 molecules H2O

22.5 mol O2 � 45.0 mol H2O2 mol H2O��

1 mol O2

14 mol KClO3 � 21 mol O23 mol O2��

2 mol KClO3




6. Student 1: 1.745 gStudent 2: 2.509 g

7. Student 1: 62.5%Student 2: 94.7%

8. Student 2 exhibited much better labtechnique, which is reflected by a higherpercent yield than that obtained by Student 1.Student 2 should receive the higher grade.

Vocabulary Review 121. d 5. g 8. i

2. h 6. j 9. e

3. f 7. b 10. c

4. a

Quiz for Chapter 121. coefficients 6. NT

2. reactant 7. NT

3. moles 8. NT

4. atoms 9. NT

5. 44.8 10. ST

Chapter 12 Test A

A. Matching1. b 3. e 5. a

2. c 4. f 6. d

B. Multiple Choice7. b 11. b 15. c

8. c 12. c 16. a

9. b 13. b 17. c

10. d 14. a 18. c

C. Problems19. There is no limiting reagent, because the

mole ratio of the reactants is 1 mol N2 to 3mol H2.

20.

21. 15.0 g CO2 � �14m4 g

olCCOO

2

2� � �28mm

oollCC4HO2

10� �

�15m8 g

olCC

4

4

HH

1

1

0

0� 4.94 g C4H10

22. a. Theoretical yield:

b. Percent yield: �23

82

.

.00

gg

� � 100% 87.5%

D. Essay23. The coefficients of a balanced chemical

equation describe the relative number ofmoles of reactants and products. From thisinformation, the amounts of reactants andproducts can be calculated. The number ofmoles may be converted to mass, volume, ornumber of representative particles.

E. Additional Problems

24. a.

Al2(SO4)3 is the limiting reagent.

b. 450 g � 325 g 125 g excess Ca(OH)2

25.

26.

27. �45

.

.80

00

LL

� � 100% 96.0%

Chapter 12 Test B

A. Matching1. b 3. d 5. f

2. e 4. a 6. c

5.00 L O2�� 2 m

3 mol

oKlCO

l

2

O3�

��1122

m.6ol

gKKCClOlO

3

3� 18.2 g KClO3

1 mol O2��22.4 L O2�

10.0 L H2S � �23

mm

oollHO

2

2

S� 15.0 L O2

125 g Ca(OH)2 ��71

4m.1

ogl C

Caa

((OO

HH

))

2

2�

1.69 mol Ca(OH)2 remaining

5.00 � 102 g Al2(SO4)3 ��134

m2

oglAA

ll

2

2

((SSOO

4

4

))

3

3�

��1

3m

mo

ol A

l Cl2

a(S

SOO

4

4

)3��

113

m6

oglCC

aaSSOO

4

4�

596 g CaSO4

5.00 � 102 g Al2(SO4)3 ��134

m2

oglAA

ll

2

2

((SSOO

4

4

))

3

3�

��13

mm

oo

llAC

la

2((SO

OH

4

))2

3��

71

4m.1

ogl C

Caa

((OO

HH

))

2

2�

325 g Ca(OH)2

4.0 g H2 � �12.

m0

og

lHH

2

2� ��1 m

2oml C

olHH

3O

2

H�

��31

2m.0

ogl C

CHH

3

3

OO

HH

� 32.0 g CH3OH

558 g Fe � �51

5m.8

ogl F

Fee

� � �32

mm

oollCF

Oe

�

� �21

8m.0

ogl C

COO

� 4.20 � 102 g CO

Answer Key 807

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B. Multiple Choice7. d 12. b 17. c

8. d 13. c 18. d

9. b 14. b 19. a

10. d 15. a 20. b

11. d 16. c 21. c

C. Problems

22.

23.

24.

25.

26.

% yield �the

aoc

rtuet

ailcal

� � 100%

% yield �22

68

.

.45

gg

CC

aa

OO

� � 100%

% yield 92.6% yield

D. Essay27. Based on the 2:3 molar ratio between A and B,

the 1.0 mol of A requires only 1.5 mol of B inorder to react completely. The maximumamount of A2B3 that can be produced (0.50mol) is thus limited by the amount of A that isavailable, with 0.50 mol of B remaining inexcess.

E. Additional Problems28. 6CO2 � 6H2O

energyuuy C6H12O6 � 6O2

29.

30.

31. a.

Thus, CuO is the limiting reagent.

b.

NH3 excess 3.35 mol � 2.43 mol 0.92 mol

c.

32. �44

mm

oollNN

HO

3� � 10.0 kg NH3 �

� �11

7m.0

ogl N

NHH

3

3� � �

31

0m.0

og�l N

NOO

� � �10

10k0g

g��

� 0.80 14.1 kg NO

1000 g��

1 kg�

� 3.65 mol CuO �

34.1 g N2

28.0 g N2��1 mol N2

1 mol N2��3 mol CuO

�32

mm

oo

ll

CN

uH

O3� � 3.65 mol CuO

2.43 mol NH3 react

Since 57.0 g NH3 � �11

7m.0

ogl N

NHH

3

3�

3.35 mol NH3 present

� � 57.0 g NH3

5.03 mol CuO needed

290.0 g CuO � �71

9m.5

ogl C

Cuu

OO

�

3.65 mol CuO present

1 mol NH3��17.0 g NH3

3 mol CuO��2 mol NH3

� 1.25 � 105 kg H2SO4

� � ��91

8m.0

og�l H

H

3

3

PP

OO

4

4�

� 8.32 � 104 kg H3PO41 kg�1000 g�

1000 g��

1 kg�1 mol H2SO4��98.1 g H2SO4

2 mol H3PO4��3 mol H2SO4

� 5.00 � 104 g C2H2

� � �41

4m.0

ogl C

COO

2

2�

1.69 � 105 g CO2

1 mol C2H2��26.0 g C2H2

4 mol CO2��2 mol C2H2

� 4.50 mol H2O

� 135 g C6H12O6180 g C6H12O6��1 mol C6H12O6

1 mol C6H12O6��6 mol H2O

50.8 g CaCO3 ��1100

m.1ol

gCCaaCCOO

3

3�

��1

1m

mo

ol C

l CaC

aOO3

�� 51

6m.1

ogl C

Caa

OO

�

28.5 g CaO

�2

3m

mol

oAl

lO2O

2

3� � �21

2m.4

oLl O

O2

2��

10010L�mL��

� 625 mL O2� � �110

m2

oglAA

ll2

2

OO

3

3�

1.90 g Al2O3

�11

mm

oo

ll

CC

HO2

4� � 150 mol CO2 � �11

6m.0

ogl C

CHH

4

4�

2.4 � 103 g CH4

�12mm

oollSHnF

F2� � �

120

m.0

ogl H

HFF�

� � 45.0 g HF�

� �1156

m.7ol

gSSnnFF

2

2� 176 g SnF2

�23mm

oollNHH

2

3� � 12.0 mol H2 8.00 mol NH3





Section 12.2 Analysis of BakingSoda, page 367

Procedure Sample answers are given.

A. 2.83 g

B. 3.28 g

C. 10.70 g

D. 4.29 g

E. If the mixture does not turn red whenthymol blue is added, have students add justenough HCl from a third pipet to turn themixture red. Students must measure themass lost by the pipet to add to the totalmass of HCl used.

F. 10.53 g

G. 8.78 g

Analysis1. HCl � NaHCO3(s) → CO2(g) � H2O � NaCl

2. 3.28 g � 2.83 g 0.45 g baking soda

3.

4.

5. 6.41 mmol HCl total � 0.875 mmol HCl unreacted

5.53 mmol HCl neutralized(5.53 mmol NaHCO3)

6.

7. 100% � (0.46 g� � 0.45 g�)/0.45 g� 2.2 % error(assuming baking soda is 100% NaHCO3).

You’re the Chemist1. (See Steps 2–7.)

2. Repeat Steps A–G and 1–7 except use bakingpowder instead of baking soda. The % erroris the % of baking soda in baking powderassuming no other errors.

Section Review 13.1

Part A Completion1. motion 6. collisions

2. empty space 7. kinetic energy

3. far apart 8. atmospheric

4. independently 9. 0°C

5. random or rapid 10. 101.3 kPa or 1 atm


12. AT 14. AT 16. AT

Part C Matching17. b 19. d 21. a

18. c 20. e


22. 4.30 atm�� 10

11.

a3tm�

kPa� 436 kPa

4.30 atm�� 760

1mat

mm�

Hg� 3.27 � 103 mm Hg

23. According to the kinetic theory, the motion ofthe particles in a gas is constant and random.Because the particles are relatively far apart,no attractive or repulsive forces exist betweenthe particles. They move independently ofeach other and travel in straight line pathsuntil they collide with one another or otherobjects.

24. Odors travel long distances from theirsources.

Section 13.2

Part A Completion1. denser 6. surface

2. condensed 7. vapor pressure

3. vaporization 8. manometer

4. boiling 9. vapor pressure

5. cooling 10. 101.3 kPa or 1 atm

Part B True-False11. ST 13. NT 15. ST

12. ST 14. AT 16. AT

Part C Matching17. a 19. c 21. b

18. e 20. d

(0.0840 g NaHCO3/mmol NaHCO3) � 5.53 mmol NaHCO3 0.46 g NaHCO3

(10.53 � 8.78) g NaOH � 0.500 mmol NaOH/g NaOH

0.875 mmol NaOH (0.875 mmol HCl unreacted)

(10.70 � 4.29) g HCl � 1.00 mmol HCl/g HCl 6.41 mmol HCl

Answer Key 809

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Part D Questions and Problems22. At the boiling point, particles throughout the

liquid have enough kinetic energy tovaporize.

23. Liquid B would evaporate faster because ithas a higher vapor pressure, which indicatesthat it is more volatile.

24. Evaporation leads to cooling of a liquidbecause the particles with the highest kineticenergy tend to escape first. The remainingparticles have a lower average kinetic energyand a lower temperature.

Section 13.3

Part A Completion1. compress 6. high

2. fixed 7. crystalline

3. melts 8. lattice

4. melting point 9. unit cell

5. freezing point 10. amorphous


12. ST 14. ST

Part C Matching16. e 19. a 22. d

17. c 20. b

18. f 21. g

Part D Questions and Problems24. When a solid is heated, its particles vibrate

more rapidly as their kinetic energy increases.Eventually, the disruptive vibrations of theparticles are strong enough to overcome theattractions that hold them in fixed positions.The organization of the particles within thesolid breaks down and the solid becomes aliquid.

Section 13.4

Part A Completion1. sublimation 5. equilibrium

2. vapor pressure 6. triple point

3. carbon dioxide 7. 0.016°C

4. phase 8. 0.61 kPa

Part B True-False9. NT 11. NT 13. NT

10. AT 12. AT 14. NT


16. c 18. e 20. f

Part D Questions and Problems21. Solids that have a vapor pressure that exceeds

atmospheric pressure at or near roomtemperature can change directly to a vapor.This process is called sublimation.

22. The temperature of the system remainsconstant while the change of state isoccurring.


Section 13.11. Gas pressure is the result of collisions

between between rapidly moving particles ina gas and an object. Because there are noparticles of matter in a vacuum, there can beno collisions or pressure.

2. Setting aside fluctuations due to changes inthe weather, you would notice that thepressure reading on the barometer woulddecrease as you climbed in altitude.

3. 754.3 mm Hg� � � 0.9925 atm

754.3 mm Hg� � � 100.5 kPa

4. The average kinetic energy of the particles ofa substance is directly proportional to theKelvin Temperature.

�100.0 �C � 273 � 173 K

73 oC + 273 � 346 K

Because the Kelvin temperature increases bya factor of two, the average kinetic energyincreases by a factor of two.

Section 13.21. According to kinetic theory, there are no

attractions between the particles in a gas, butthere are attractions between particles of aliquid.

2. For a dynamic equilibrium to be established,the beaker must be sealed so that the rate ofcondensation can equal the rate ofevaporation.

101.3 kPa��

760 mm Hg�

1 atm��

760 mm Hg�




3. The fastest runner corresponds to theparticles in a liquid with the greatest kineticenergy. When these particles vaporize, theremaining particles have a lower averagekinetic energy.

4. Ethanol must have the greater vapor pressurebecause 75°C is very close to the boiling pointof ethanol and the vapor pressure is equal tothe external pressure at a liquid’s boilingpoint.

Section 13.31. The carbon atoms in graphite are arranged in

widely-spaced sheets. In diamond, eachcarbon atom is strongly bonded to four othercarbon atoms in a rigid three-dimensionalarray.

2. Allotropes are two or more differentmolecular forms of the same element in thesame physical state. Carbon has multipleallotropes, including diamond, graphite, andbuckminsterfullerene. The carbon atoms arearranged differently in each allotrope.

3. Peanut brittle is an amorphous solid.

4. A molecular solid. In general, ionic solidshave higher melting points because theforces that hold particles together in an ionicsolid are usually stronger than the forces thathold particles together in a molecular solid.

5. One example of a crystalline solid is sodiumchloride. In crystal atoms, ions, or moleculesare arranged in an orderly, repeating, three-dimensional pattern called a crystal lattice.

Section 13.41. The melting point of water decreases as the

pressure increases.

2. This line represents the set of alltemperature–pressure values at which thesolid and gas phases of water are inequilibrium.

3. This line represents the set of alltemperature–pressure values at which theliquid and gas phases of water are inequilibrium.

4. 101.3 kPa (1 atm)

Interpreting Graphics 131.

2. Normal melting point � �7.0°CNormal boiling point � 59°CTriple point � �8°C and 6 kPa

3. See answer to 1.

4. The melting-point curve leans slightly to theright (has a positive slope) indicating that, aspressure is increased, the melting point ofbromine increases. Higher pressures favorthe denser phase of a substance. Solidbromine is more dense than liquid bromine.

5. 50°C

6. The triple point is the temperature andpressure at which solid, liquid, and vaporphases of a substance are in equilibrium.

7. See answer to 1.

8. condenses

9. freezes

Vocabulary Review 131. kinetic energy 5. crystal

2. gas pressure 6. allotropes

3. evaporation 7. melting point

4. boiling point 8. pascal

Solution: barometer

Temperature (°C)

Pre

ssur

e (k

Pa)

100

60

80

40

20

0�30 604020�10 0

NormalMelting point

NormalBoiling point

Solid

A

M

SC

V

B

Liquid

Triplepoint

Vapor

Answer Key 811

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Quiz for Chapter 131. kinetic

2. decreases

3. temperature

4. kinetic

5. elastic

6. condensed

7. minimum kinetic energy

8. external (atmospheric) pressure

9. unit cell

10. amorphous

11. ST

12. AT

13. AT

Chapter 13 Test A

A. Matching1. i 6. d 11. k

2. f 7. b 12. h

3. g 8. n 13. l

4. j 9. e 14. a

5. c 10. m


16. d 21. a 25. b

17. d 22. b 26. c

18. c 23. a 27. c

19. b

C. True-False28. AT 31. ST

29. NT 32. AT

30. AT

D. Problems

33. 3.70 atm�� 10

11.

a3tm�

kPa� � 375 kPa

34. 3.70 atm�� 760

1mat

mm�

Hg� � 2.81 � 103 mm Hg

E. Essay35. Because the Kelvin temperature is directly

proportional to the average kinetic energy ofthe particles in a substance, it doesn’t matterhow many particles there are in the sample.

F. Additional Problems

36. 610.0 mm Hg� � �76

10

am

tmm Hg�� 0.803 atm

37. 0.803 atm�� 10

11.

a3tm�

kPa� � 81.3 kPa

G. Additional Questions38. The additional energy is being used to change

the liquid water to water vapor. Thetemperature of the water will not rise until allof the water is in the gaseous state.

39. When water boils at standard atmosphericpressure, it cannot be heated above 100°C.Only water at pressures of more than oneatmosphere will boil at higher temperatures.The pressure must be increased to the pointat which water boils at 150°C or higher to killbacteria.

Chapter 13 Test B

A. Matching1. j 6. k 11. a

2. n 7. b 12. m

3. h 8. g 13. c

4. e 9. f 14. i

5. d 10. l

B. Multiple Choice15. d 21. a 26. a

16. b 22. a 27. a

17. d 23. a 28. c

18. d 24. b 29. d

19. c 25. b 30. b

20. b

C. True-False31. AT 34. AT 36. ST

32. ST 35. NT 37. AT

33. AT

D. Essay38. The boiling point of a liquid is the

temperature at which the vapor pressure ofthe liquid is equal to atmospheric pressure.The normal boiling point is the boiling pointof the liquid when the atmospheric pressureis 101.3 kPa. If atmospheric pressure is lessthan 101.3 kPa, then the boiling point of aliquid will be lower than its normal boilingpoint. Conversely, if atmospheric pressure is




greater than 101.3 kPa, the boiling point ofthe liquid will be higher than its normalboiling point.

E. Additional Questions39. The particles of a gas are relatively far apart

and there are no attractive or repulsive forcesamong them. The volume of the particles isinsignificant and their collisions are perfectlyelastic. In a liquid, the particles are attractedto each other, which is why liquids have adefinite volume. Because the particles arecloser together, liquids are denser than gases.In solids, the particles are closely packedtogether, usually in an organized array. Theparticles in a solid vibrate around fixedpoints. Solids are dense and difficult tocompress.

40. a. As more energy is added, more particles ofthe liquid acquire enough kinetic energyto escape. As a result, all the liquid boilsaway at a constant temperature.

b. If the mixture is heated, more ice meltsbut the temperature of the mixtureremains the same as long as ice is present.If the mixture is cooled, more liquidfreezes, but the temperature of themixture remains the same as long asliquid water is present.


Section 13.3 The Behavior ofLiquids and Solids,page 400

Analyze and Conclude1. Water in the dish evaporates and condenses

into a cloud when it contacts the coldsurface under the ice.

2. The drop of water on top of the dishprovides enough cooling to cause cloudformation.

3. Water beads up and alcohol spreads out dueto stronger intermolecular attractions inwater. The water “cloud” consists of tinyindividual beads; the alcohol cloud is madeup of larger, more spread out pools.

4. Calcium chloride absorbs water from theenvironment in the dish.

5. The many pieces of calcium chlorideeffectively dry the atmosphere leaving nowater vapor in the dish.

You’re the Chemist1. The water drop increases in diameter over

time as the alcohol evaporates and iscaptured by the water drop. The attractionsin the resulting mixture are weaker overall.

2. The BTB turns from green to yellow in thepresence of vinegar.

3. Place a drop of vinegar and a drop of BTBabout 3 cm apart in a Petri dish. Cover andobserve. The BTB slowly changes from greento yellow even though there is no mixing ofthe drops. Ethanoic acid that evaporates is“captured” by the BTB.

4. Place a drop of ammonia and a drop of BTBabout 3 cm apart in a Petri dish. Cover andobserve. The BTB slowly changes from greento blue even though there is no mixing of thedrops. Ammonia thatt evaporates iscaptured by the BTB.

Section 14.1

Part A Completion1. compressed 6. kinetic

2. spare 7. doubles

3. volume 8. reducing

4. temperature 9. pressure

5. moles


11. ST 13. ST


16. d 18. a

Part D Questions and Problems20. The motion of particles in a gas is constant

and random. The particles travel in straightpaths until they collide with other particles orthe walls of their container. There are nosignificant attractive or repulsive forcesbetween particles in a gas, which is why a gascan expand to take the shape and volume ofits container. The volume of the particles in agas is small compared the overall volume of agas, which is why a gas can be compressed.

Answer Key 813

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Section 14.2

Part A Completion1. inversely 6. Charles’s

2. increases 7. Gay-Lussac’s

3. Boyle’s 8. directly

4. mass 9. combined

5. Kelvin 10. amount

Part B True-False11. NT 13. ST 15. NT

12. AT 14. NT 16. AT

Part C Matching17. c 19. e 21. d

18. b 20. a


22. P2 � �P1

T�

1

T2� ��55 kP

1a73

�

K�473 K�

�

P2 � 1.5 � 102 kPa

23. V2 � �P1

T�

1

V�

1

P�

2

T2�

V2 �

V2 � 0.061 L � 61 mL

Section 14.3

Part A Completion1. number of moles 6. ideal

2. PV � nRT 7. real

3. n 8. attractions

4. ideal gas constant 9. volume

5.


11. ST 13. NT

Part C Matching15. d 17. b

16. c 18. a


19. n � �PR

�

�

VT

� �

n � 1.29 � 102 mol O2(g)

20. n � �PR

�

�

VT

� �

n � 2.64 � 10�2 mol NO2

Section 14.4

Part A Completion1. total

2. sum

3. lower

4. uniform

5. diffusion

6. effusion

7. hole

8. inversely

9. molar mass

10. Graham’s law


12. NT 14. AT

Part C Matching15. a 17. c

16. b

Part D Questions and Problems18. The kinetic energy of a molecule is equal to

where m isthe mass and v is the

velocity of the molecule. At a giventemperature, molecules all have the sameaverage kinetic energy. If two molecules withdifferent masses have the same kineticenergy, the less massive molecule must havea higher velocity.

�mv2

12

��

mass of NO2

� 2.64 � 10�2 mol NO2 � �41

6m.0

ogl N

NOO

2

2�

� 1.21 g NO2

240.0 kPa�� 0.275 L��8.31 �K�

L��

mkP

oa�l

� � 301 K�

25,325 kPa�� 12.5 L��8.31 �K�

L��

mkP

oa�l� � 295 K�

8.31 � �(K • mol)(L • kPa)�

91 kPa�� 0.075 L � 273 K��

303 K� � 101.3 kPa�




Practice Problems

Section 14.11. On average, temperatures are higher in the

summer than in the winter. The motion of thetires causes the air in the tires to heat up. Athigher temperatures, the particles inside thetire have a greater average kinetic energy. So,the frequency and force of the collisionsbetween the particles and the walls of the tireare greater, resulting in a greater pressureinside the tire.

2. Overnight the air in the mattress cools down;the average kinetic energy of the particles inthe air decreases. Thus, they collide lessfrequently and less forcefully with the walls ofthe mattress. Consequently, the pressureinside the mattress decreases, as does thevolume.

Section 14.2

1. P2 � �P1

V�

2

V1� �

P2 � 341 kPa

2. No, the balloon will only expand untilinternal pressure is equal to the externalpressure, which in this case is about half theinitial pressure.

3. V2 � �V1

T�

1

T2� � � 15.0 L

4. P2 � �P1

T�

1

V�

1

V�

2

T2�

P2 � � 466 kPa

5. P1 � �P2

T�

2

V�

2

V�

1

T1�

P2 � � 71.9 kPa

Section 14.3

1. n � �PR

�

�

VT

� �

n � 1.3 � 102 mol argon

2. T � �nP �

�

VR

� �

T � 3.01 � 102 K

3. n � �PR

�

�

VT

� �

n � 1.03 mol He

4. V � �n � R

P� T�

V �

V �

50.4 L

5. V2 � �V1

T�

1

T2� � � 6.66 L

6. V � �n � R

P� T�

V �

V � 7.95 L

7. n � 25.0 g� � 1 mol\44.0 g� � 0.568 mol

V � �n � R

P� T�

V �

V � 12.7 L

Section 14.41. PO2

� Ptotal � (PN2� PAr)

PO2� 98.5 kPa � (22.0 kPa � 50.0 kPa)

PO2� 26.5 kPa

n � �PR

�

�

VT

� �

n � 3.75 � 10�2 mol O2(g)

2.

Oxygen effuses slightly slower than nitrogen.

Interpreting Graphics 141. a. 0.41 g b. 0.43 g

2. 0.267 L

3. a. 372 K b. 372 K

4. a. 103 kPa b. 108 kPa

�molar massO2��molar massN2

��RateO2��RateN2

� 28.0 g��32.0 g

��

��

� �0.875 � 0.935��

26.5 kPa�� 3.5 L��8.31 �K�

L��

mkP

oa�l� � 298 K�

0.568 mol�� 8.31 �K�L

��

mkP

oa�l�� 273 K�

��101.3 kPa�

0.355 mol�� 8.31 �K�L

��

mkP

oa�l�� 273 K�

��101.3 kPa�

10.5 L � 300 K��

473 K�

2.25 mol�� 8.31 �K�L

��

mkP

oa�l�� 273 K�

��101.3 kPa�

mass of He � 1.03 mol He �

� 4.12 g He

4.00 g He��1 mol He

102.0 kPa�� 25.0 L��8.31 �K�

L��

mkP

oa�l

� � 297 K�

500.0 kPa�� 35.0 L��7.00 mol�� 8.31 �K

L��

mkP

oa�l��

1.24 � 104 kPa�� 25 L��

8.31 �K�L�

��

mkP

oa�l

� � 297 K�

105.4 kPa � 55.0 L� � 323 K��

248 K� � 105 L�

501 kPa � 5.2 L� � 373 K��

298 K� � 7.00 L�

10.0 L � 373 K��

248 K�

155.0 kPa � 22.0 L��

10.0 L�

Answer Key 815

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5. a. n �

n � 8.90 � 10�3 mol

b. n �

n � 9.33 � 10�3 mol

6. a. molar mass ��8.90 �

0.4110�

g3 mol

�

molar mass � 46 g/mol

b. molar mass ��9.33 �

0.4130�

g3 mol

�

molar mass � 46 g/mol

7. 46 g/mol

8. C2H6O (ethanol)

Vocabulary Review 141. g 5. d 8. a

2. i 6. e 9. h

3. j 7. c 10. b

4. f

Quiz for Chapter 141. collisions

2. doubles

3. small

4. real

5. diffusion

6. P2� P1 � �TT

1

2� � 388 kPa � �

27

71

33

K�K�

� � 149 kPa

7. T1 � 27°C � 273 � 300 K P1 � 115 kPaT2 � �10°C � 273 � 263 K P2 � 99 kPa

V2 � V1 � �PP

1

2� � �

TT

2

1�

V2 � 3.5 � 105 m3 � �19195

kkPPa�a�

� � �23

60

30

K�K�

�

V2 � 3.6 � 105 m3

8. PO2� Ptotal � (PN2

� 2PCO2)

PO2� 145.0 kPa � (28.5 kPa � 76.0 kPa)

PO2� 40.5 kPa

Chapter 14 Test A

A. Matching1. c 4. g 7. e

2. h 5. f 8. d

3. b 6. a


10. b 15. c 20. b

11. a 16. b

12. c 17. c

13. d 18. c

C. Problems

21.

T1 � 227°C � 273 � 500 K

T2 � 27°C � 273 � 300 K

P2 �

� 393 kPa

22. P1 � V1 � P2 � V2

V2 �

� 10.9 L

23. T1 � 35.0°C � 273 � 308 KT2 � 0.0°C � 273 � 273 K

V2 � �P1

T�

1

V�

1

P�

2

T2�

V2 � � 151 mL

24. V2 � V1� �TT

2

1�

T1 � �55.0°C � 273 � 218 KT2 � 30.0°C � 273 � 303 K

V2 � 550 mL � �32

01

38

K�K�

� � 764 mL

D. Essay25. An ideal gas is one that follows the gas laws at

all conditions of pressure and temperature.The behavior of a real gas deviates from thebehavior of an ideal gas, especially at low

95.9 kPa� � 180 mL � 273 K��

308 K� � 101.3 kPa�

156 kPa�� 15.0 L��

215 kPa�

�P2

P1 � V1��

V2

300 K� � 655 kPa��

500 K�

�P2

T2 � P1��

T1

�P2 ��T2

P1 ��T1

108 kPa�� 0.267 L��8.31 �K�

L��

mkP

oa�l

� � 372 K�

103 kPa�� 0.267 L��8.31 �K�

L��

mkP

oa�l

� � 372 K�




temperatures and high pressures. Also, kinetictheory assumes that the particles of an idealgas have no volume and are not attracted toeach other. This is not true for real gases,which can be liquefied and sometimessolidified by cooling and applying pressure.

E. True-False26. NT 28. ST

27. AT 29. AT


30. n � �PR

�

�

VT

� �

n � 2.17 mol

�2.

9167.0m

gol

� � 44.2 g/mol formula mass

31.

UF6 containing U-235 diffuses 1.004 timesfaster.

Chapter 14 Test B

A. Matching1. g 5. d 8. a

2. i 6. e 9. h

3. j 7. c 10. b

4. f

B. Multiple Choice11. d 17. a 23. c

12. c 18. c 24. b

13. b 19. a 25. a

14. d 20. b 26. b

15. d 21. b

16. b 22. d

C. Problems27. P1V1 � P2V2

V2 �

V2 � 0.829 L

28. �VT1

1� � �VT2

2�

V2 ��(0.65

(029

L3)(

K�31

)3 K�)

�

V2 � 0.694 L

29. �PT1

1� � �PT2

2�

T2 �

T2 � 357 K or 84.4°C

31. �P

T1V

1

1� � �P

T2V

2

2�

V2 �

V2 � 2.29 L

31. PV � nRT

n � �PR

�

�

VT

� �

n � 0.49 mol

D. Essay

32. If all gases behaved ideally, the individualparticles that make up each gas could neverexert the attractive forces on each other thatare necessary for them to condense to liquidsand solids.

E. True-False33. ST 35. AT

34. AT 36. NT


37. �VT1

1� � �VT2

2�

T2 ��(0.92

(51.

L�25

)(2L�

5)0 K)

�

T2 � 185 K or �88°C

�T2

V2 � T1��

V1

0.49 mol CO2 � �41

4m.0

ogl C

COO

2

2� � 22 g CO2

152 kPa�� 8.0 L��8.31 �K�

L��

mkP

oa�l

� � 300 K�

(80.0 kPa�)(3.50 L)(273 K�)��

(330 K�)(101.3 kPa�)

�V2

P1 � V1 � T2��

T1 � V2

(98.0 kPa�)(310 K)��

(85.0 kPa�)

�T2

P2 � T1��

P1

�V2

V1 � T2��

T1

(425 kPa�)(1.20 L)��

615 kPa�

�V2

P1 � V1��

P2

�252.0��349.0

��Rate235��Rate238 � � �1.009 � 1.004

��

�

216 kPa�� 25.0 L��8.31 �K�

L��

mkP

oa�l

� � 300 K�

Answer Key 817

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38. PV � nRT

n � �PR

�

�

VT

� �

n � 0.304 mol


Section 14.4 Diffusion,page 437

Analyze and Conclude1. The drops near the center change

immediately. As the gas diffuses, all thedrops change color. The color change beginsat the outer edge of each drop.

2. The first picture should show one edgeturning yellow. Succeeding pictures shouldshow the yellow area gradually increasinguntil the entire dot is yellow.

3. The particles of gas produced are in motion.As the particles diffuse from the center, theycollide and react with molecules of BTB.

4. NaHSO3 + HCl → SO2 + H2O + NaCl

You’re the Chemist1. As ammonia diffuses, BTB changes from

yellow to blue.

2. Vary the size of the BTB drops from “pin-heads” to “puddles.” Tiny drops are betterable to detect small quantities of gas.

3. The Kl turned orange in the same manner asthe BTB turned yellow. 3NaNO2 + 2HCl →2NO + H2O + NaNO3 + 2NaCl

Section Review 15.1

Part A Completion1. polar 7. high

2. negative 8. surface

3. positive 9. spherical

4. polar 10. surfactant

5. hydrogen 11. Ice

6. low 12. dense

13. hydrogen bonding

Part B True-False 14. NT 16. AT

15. NT 17. AT

18. ST

Part C Matching19. a 21. b

20. c

Part D Question22. a. lower

b. higher

Section Review 15.2

Part A Completion1. solvent 6. partially

2. homogeneous 7. conduct

3. “like dissolves like” 8. nonelectrolyte

4. electrolytes 9. hydrates

5. strong 10. efflorescence


12. AT 14. AT

Part C Matching15. d 18. b 21. c

16. a 19. f 22. e

17. h 20. g

Part D Questions and Problems23. molar mass Na2SO4 � 10H2O � 322 g

Mass of 10H2O � 180.0 g

�138202.0

gg

� � 100% � 55.9%

24. b and c

Section Review 15.3

Part A Completion1. larger 6. molecules/ions

2. filtration 7. ions/molecules

3. Colloids 8. Emulsions

4. Tyndall effect 9. stability

5. Brownian 10. emulsions

50.6 kPa�� 10.0 L��8.31 �K�

L��

mkP

oa�l

� � 200 K�




Part B True-False 11. NT 13. AT 15. NT

12. ST 14. ST

Part C Matching16. e 18. d 20. f

17. b 19. a 21. c

Part D Questions and Problems22. b 23. a


Section 15.11. Hydrogen bonds are attractive forces in

which a hydrogen atom that is covalentlybonded to a very electronegative atom is alsoweakly bonded to an unshared electron pairof an electronegative atom in the samemolecule or in a nearby molecule.

2.

3. Hydrogen bonds hold the water molecules inplace in the solid phase. The structure of ice isa regular, open, framework like a honeycomb.

Section 15.21. The solute is potassium chloride (KCl). The

solvent is water.

2. NH3(g) � H2O(l) 1 NH4�(aq) � OH�(aq)

3. Possible answers include glucose (C6H12O6)and ethyl alcohol (C2H6O).

4. Hygroscopic compounds are thosecompounds that remove moisture from air.

5. a. soluble c. soluble

b. insoluble d. soluble

6. a. NH4NO3(s) → NH4�(aq) � NO3

�(aq)

b. KOH(s) → K�(aq) � OH�(aq)

7. a. CaSO4 � 2H2O

b. CoCl2 � 6H2O

8. Molar mass of NiCl2 � 6H2O � 237.7 g/molMass of 6H2O � 108.0 g

Percent H2O ��m

ma

ass

sso

of

fh

wyd

artaer

te�� 100%

� �237

1.078

g.0/m

gol

� � 100% � 45.44%

Section 15.31. Colloids and suspensions exhibit the Tyndall

effect and have larger particles thansolutions. The particles in a suspension areretained on a filter and will settle out slowlyupon standing.

2. Brownian motion refers to the chaoticmovement of colloidal particles caused bythe collisions of water molecules with thesmall, dispersed colloidal particles.

3. a. colloid e. suspension

b. colloid f. colloid

c. solution g. solution

d. colloid

Vocabulary Review 151. desiccants

2. hydrogen bonding

3. suspension

4. aqueous solution

5. solvent

6. surfactant

7. strong electrolyte

8. water of hydration

9. Brownian motion

SOLUTION: 1. WATER VAPOR

2. ICE

3. LIQUID WATER

Quiz for Chapter 151. a 3. c

2. a 4. b

5. solute

6. solvent

7. aqueous solutions

8. like dissolves like

HO

H

HO

H

HO

H

Answer Key 819

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Chapter 15 Test A

A. Completion1. nonelectrolyte

2. deliquescent

3. suspension

4. solvent

5. effloresce

6. Brownian motion

7. Emulsions

8. hydrate

9. Tyndall effect

10. Surface tension


12. d 16. a 20. a

13. a 17. a 21. b

14. d 18. c 22. c

C. True-False23. ST 25. ST

24. ST 26. NT

D. Problems27. Molar Mass of hydrate � 238 g

1 mol H2O � 18.0 g6 mol H2O � 108 g

�12

03

88

gg

� � 100% � 45.4% H2O

E. Essay28. Because polar water molecules can attract

charged particles, they cause solute ions tobreak away from the surface of the solid. Asthe solute dissolves, the ions are surroundedby molecules of solvent

Chapter 15 Test B

A. Completion1. hygroscopic 6. colloids

2. solvation 7. aqueous

3. solute 8. hydrogen

4. surfactants 9. nonelectrolytes

5. dessicants 10. effloresce

B. Multiple Choice11. a 15. c 19. c

12. a 16. d 20. d

13. b 17. c 21. a

14. c 18. c


23. ST 26. AT 29. AT

24. AT 27. NT

D. Problem

30. %H2O ��m

mas

assos

fohfyHd

2

rOate

�� 100%

� �13

82

02

.

.01

g�g�

� � 100% � 55.9% H2O

E. Essay31. Soaps and detergents are surfactants that

reduce the surface tension of water byinterfering with the hydrogen bondingbetween water molecules. With surfacetension reduced, the beads of water thatwould normally have formed collapse,allowing the water to spread out to cover andpenetrate the fabric. Soaps and detergentsalso are emulsifying agents that allow oils andgreases to form colloidal dispersions. The oiland grease particles, which are normallyinsoluble in water, are removed from thesurface of the fabric.


Section 15.2 Electrolytes, page 458

Analysis

NaCl(s)Aqueousconducts

MgSO4(s)Aqueousconducts

Na2CO3(s)Aqueousconducts

SugarAqueous doesnot conduct

NaHCO3(s)Aqueousconducts

Corn StarchAqueous doesnot conduct

KCl(s)Aqueousconducts

Kl(s)Aqueousconducts




1. These are electrolytes: NaCl, MgSO4,Na2CO3, NaHCO3, KCl, KI.

These are nonelectrolytes: sugar, cornstarch.

2. None of the electrolytes conduct electricityin the solid form because the ions are lockedin a crystal lattice and cannot move.

3. Table sugar and corn starch are covalentcompounds. NaCl, MgSO4, Na2CO3,NaHCO3, KCl, KI are ionic compounds. Ingeneral, to be an electrolyte a compoundmust dissociate into ions in solution.

You’re the Chemist1. MgSO4(s) → Mg2�(aq) � SO4

2�(aq)

NaHCO3(s) → Na�(aq) � HCO3�(aq)

KCl(s) → K�(aq) � Cl�(aq)

KI(s) → K�(aq) � I�(aq)

2. Test a drop of each solution with aconductivity device.

Strong electrolytes (bright light): HCl, H2SO4,HNO3, NaOH

Weak electrolytes (dim light) CH3COOH,NH3

Nonelectrolytes (no light): rubbing alcohol,distilled water

3. Strong electrolytes: soft drinks, pickle juiceWeak electrolytes: orange juice, coffee

Section Review 16.1

Part A Completion1. particle size 6. saturated

2. rate 7. miscible

3. pressure 8. solubility

4. Henry’s 9. increases

5. temperature 10. supersaturated

Part B True-False11. AT 13. AT 15. AT

12. ST 14. NT


17. a 20. g 22. b

18. f

Part D Problem

23. �11..06

agt/mL

� � �2.5

Sa2

tm�

S2 � � 4.0 g/L

Section Review 16.2

Part A Completion1. solvent 5. liter

2. solution 6. diluting

3. dilute 7. solute

4. moles 8. solvent


10. AT 12. NT

Part C Matching13. b 15. e 17. d

14. c 16. a

Part D Problem18. molar mass C12H22O11 � 342.3

M � V � mol solute needed

�0.5

10

Lmol� � 0.3000 L � 0.15 mol

�3142

m.3ol

g� � 0.15 mol � 51 g

Section Review 16.3

Part A Completion1. solute

2. colligative properties

3. freezing

4. lowering/depression

5. elevation

6. directly

7. solution

8. particles

9. twice

10. twice

Part B True/False11. NT 13. NT

12. ST 14. AT

1.6 g/L � 2.5 atm��

1.0 atm�

Answer Key 821

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Part C Matching15. d 17. c

16. a 18. b

Part D Questions and Problems19. a. two c. four

b. one d. two

20. a. K2CO3 c. NaCl

b. NaCl

Section Review 16.4

Part A Completion1. solute

2. solvent

3. kilogram

4. mole fraction

5. molal boiling point

6. depression

7. molal

8. elevation


10. NT 12. AT


14. e 16. a

Part D Problem

18. molality � � �10

10k0g

g�

molality � 2.5m

molality of total particles � 3 � 2.5m � 7.5m

�Tf � Kf � m � �1.8

m�6�C� � 7.5m� � 14�C

freezing point of solution � 0�C � 14�Cfreezing point of solution � �14�C


Section 16.1

1. S2 � �S1

P�

1

P2� �

S2 � 0.82 g/L

2. � 500.0 g H2O� � 1.7 � 102 g KCl

3. Solubility of AgNO3 at 20�C � 222.0 g/100 g H2O

Solubility of AgNO3 at 50�C � 455.0 g/100 g H2O

455.0 g � 222.0 g � 233.0 gAdd 233.0 g to maintain saturation at 50�C.

Section 16.2

1. a. �0.40 m

1.6ol

LNaCl

�� 0.25M

b.

2. a. moles KOH � 2500.0 mL��

� � 7.5 mol KOH

b. moles HNO3 � 2.0 L� ��2.0 mo

L�l HNO3�

moles HNO3 � 4.0 mol HNO3

3. moles NaNO3 � 212.5 g� � �81

5m.0

og�l

� � 2.50 mol

molarity � �2.5

30.0

mL

ol� � 0.83M

4. M1V1 � M2V2

0.750M � 300.0 mL � 2.00M � V2

V2 � � 113 mL

5. V2 � � � 12 L

The final volume should be 12 L. Therefore,add 11 L of H2O.

6. a. % (v/v) � � 100%

% (v/v) � 12.0% methanol (v/v)

b. % (v/v) � � 100%

% (v/v) � 12.5% C3H7OH (v/v)

25.0 mL� C3H7OH��200.0 mL� solution

60.0 mL� methanol��500.0 mL� solution

6.00M� � 1.0 L��

0.500M�M1 � V1�

M2

0.75M� � 300.0 mL��

2.0M�

mass of HNO3 � 4.0 mol HNO3

� �16m3 g

olHHNNOO

3

3�

mass of HNO3 � 2.5 � 102 g HNO3

mass of KOH � 7.5 mol KOH

� � 4.2 � 102 g KOH56.1 g KOH��1 mol KOH

3.0 mol KOH��

1 L�

1 L��1000 mL�

�20

2.520g.0

KmN

L�O3� � �

� 0.799M

1000 mL��

1 L1 mol KNO3��

101.1 g KNO3

34.0 g KCl��

100 g H2O�

0.54 g/L � 1.86 atm��

1.22 atm�

2.0 mol CaCl2��800.0 g H2O




7. a. 1.00 L� � � � 30.0 g NaCl

b. 2.00 L � � �

� 1.00 � 102 g KNO3

Section 16.31. Colligative properties of solutions are the

physical properties of solutions that dependon the concentration of solute particles insolution but not on the chemical identity ofthe solute. Three important colligativeproperties are vapor-pressure lowering,boiling-point elevation, and freezing-pointdepression.

2. Each formula unit of K2CO3 produces threeparticles in solution.

3. Three moles of Na2SO4, when dissolved inwater, produce 9 mol of particles becauseeach formula unit of Na2SO4 dissociates intothree ions.

4. The boiling point of water increases by 0.512 �C for every mole of particles that thesolute forms when dissolved in 1000 g ofwater. When 2 mol of MgCl2 dissolve in water,6 mol of particles are produced because eachformula unit of MgCl2 dissociates into threeions. Thus, the boiling point of the solutionincreases by 6 � 0.512�C � 3.07�C. Theboiling point of the solution is100 �C � 3.07�C � 103.07�C.

5. Vapor-pressure lowering is a colligativeproperty.

6. a. The solution containing calcium chloridehas a lower freezing point.

b. The solution containing calcium chloridehas a lower vapor pressure.

c. The solution containing sodium chloridehas a lower boiling point.

Section 16.4

1. a. XLiBr � �nLiB

n

r �LiB

nr

H2O� �

XLiBr � 0.33

b. XKNO3�

�

� 0.0271

2.

3.

4. a. � � 4.6m glucose

b.

�

� 0.668m Ba(NO3)2

5. a. �Tb � Kb � m�Tb � 0.512�C/m� � 2.00m� � 2�Tb � 2.05�CThe boiling point of this solution is 100 �C � 2.05�C � 102.05 �C.

b. �Tb � Kb � m�Tb � 0.512�C/m� � 1.50m� � 3�Tb � 2.30�CThe boiling point of this solution is100�C � 2.30�C � 102.30�C.

6. a. molality of solute particles

� 2 � �

� 0.78m NaCl

�Tf � Kf � m�Tf � 1.8�C/m� � 0.78m��Tf � 1.45�CThe freezing point of this solution is0 �C � 1.45�C � �1.45�C.

b.

molality of solute particles

� �

� 0.1472m C12H22O11

�Tf � Kf � m�Tf � 1.86 �C/m� � 0.1472m��Tf � 0.274 �C

103 g��1 kg

0.3861 mol C12H22O11��2500.0 g� H2O

moles C12H22O11 � 126.0 g C12H22O11

�

moles C12H22O11 � 0.3681 mol C12H22O11

1 mol C12H22O11��342.3 g C12H22O11

103 g��1 kg

0.35 mol NaCl��

900.0 g� H2O

103 g��1 kg

0.501 mol Ba(NO3)2��750.0 g� H2O

moles Ba(NO3)2 � 131 g Ba(NO3)2

�

moles Ba(NO3)2 � 0.501 mol Ba(NO3)2

1 mol Ba(NO3)2��261.3 g Ba(NO3)2

103 g��1 kg

2.3 mol glucose��

500.0 g� H2O

1600.0 g H2O �

� � 1.5 � 102 g Li2S45.9 g Li2S��1 mol Li2S

2.0 mol Li2S��1000 g H2O

750.0 g H2O �

� � 22 g NaCl58.4 g NaCl��1 mol NaCl

0.50 mol NaCl��

1000 g H2O

�125 g � �1101

m.1ol

g��

��

�125 g � �1101

m.1ol

g�� 800.0 g � �11

8m.0

ogl

��

nKNO3��nKNO3

� nH2O

3.0 mol��3.0 mol � 6.0 mol

5.00 g�100 mL�

103 mL��

1 L�

3.00 g�100 mL�

103 mL��

1 L�

Answer Key 823

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The freezing point of this solution is0�C � 0.274�C � �0.274�C.

Interpreting Graphics 161. b, c

2. Solubility KCl at 20�C � 34.0 g/100 g H2OSolubility KCl at 50�C � 42.6 g/100 g H2O

� 42.6 g � 34.0 g � 8.6 g

3. Solubility of sucrose at 100�C � 487 g/100 g H2O

Solubility of sucrose at 20�C � 230.9 g/100 g H2O

(487 g/100 g H2O � 1000.0 g H2O) � (230.9 g/100 g H2O � 1000.0 g H2O)

� 4870 g � 2309 g � 2561 g

4.

a. LiBr b. Na2SO4

Vocabulary Review 161. Molarity. All others are qualitative terms

used to describe solutions.

2. Concentration. The other terms are used todescribe mixtures of liquids.

3. Henry’s Law. All others are units ofconcentration.

4. Colligative properties. All others areassociated with the solubility of gases.

5. Saturated solution. All others are associatedwith colligative properites of solutions.

6. Molarity. All others are related to freezing-point depression

7. Mole fraction. All others are factors affectingthe rate at which a substance dissolves.

8. Boiling-point elevation. All others areassociated with the preparation of solutions.

Quiz for Chapter 161. b 4. a 7. c

2. d 5. b 8. b

3. a 6. c

Chapter 16 Test A

A. Matching1. b 5. f 8. a

2. i 6. d 9. h

3. g 7. j 10. e

4. c


12. d 16. b 20. b

13. a 17. a 21. d

14. c 18. b

C. True-False22. ST 26. NT 29. AT

23. AT 27. NT 30. ST

24. AT 28. NT 31. NT

25. AT

D. Problems32. V2 � 250 mL, M2 � 0.60M, M1 � 2.0M

M1V1 � M2V2

V1 ��0.60M�

2�

.0M�250 mL�� 75 mL

Add 75 mL of 2.0M Al2(SO4)3 to enoughdistilled water to make 250 mL of solution.

33. Molar mass KNO3:

K: 1 � 39.1 � 39.1 gN: 1 � 14.0 � 14.0 gO: 3 � 16.0 � 48.0 g

101.1 g

mol KNO3 � �101

9.15.

g�5/g�mol

� � 0.945 mol

molarity � �0.

09.47

550

mLol

� � 1.26M

34. �PS1

1� � �

PS2

2�

S2 � �S1

P�

1

P2� �

� 20.3 g/L

16.9 g/L � 606 kPa��

505 kPa�

��1V

M2 2V��

M1

50

100

150

200

250

300

350

0 10 20 30 40 50 60 70 80 90 100

Con

cent

ratio

n (g

/100

g H

2O)

Temperature °C

Na2SO4

LiBr

KCl




E. Essay35. Boiling-point elevation, and freezing-point

and vapor-pressure lowering are colligativeproperties. They depend solely on thenumber of particles in the solution.

Boiling-point elevation: Additional attractiveforces exist between solute and solvent thatmust be overcome for the solution to boil.

Freezing-point depression: The soluteparticles interfere with the formation of theorderly pattern that the solvent particlesassume as the solvent changes from liquid tosolid.

Vapor-pressure lowering: The formation ofsolvent shells around the solute particlesreduces the number of solvent particles thathave sufficient kinetic energy to vaporize.


36. �1m00

o0l K

g3

HPO

2O4�

� � �1100

k0g

gH

H

2O2O�

�

� 0.327mK3PO4 → 3K� � PO4

3� � 4 particles4 particles � 0.327m � 1.31m�Tb � Kb � m

�Tb � 0.512��

m�C� � 1.31m�

�Tb � 0.671�CThe boiling point of the solution is 100°C � 0.671°C � 100.671�C.

37. Molar mass KNO3:

K: 1 � 39.1 � 39.1 gN: 1 � 14.0 � 14.0 gO: 3 � 16.0 � 48.0 g

101.1 g

molality ��10

m00

olg

ssoolulv

teent

�

� �

� 1.38m

Chapter 16 Test B

A. Matching1. g 5. d 8. i

2. c 6. a 9. j

3. h 7. f 10. b

4. e

B. Multiple Choice11. d 17. a 22. a

12. a 18. b 23. c

13. b 19. d 24. a

14. b 20. b 25. b

15. c 21. d 26. a

16. a

C. True-False27. AT 31. ST 34. AT

28. NT 32. NT 35. NT

29. NT 33. NT 36. AT

30. NT

D. Problems

37.

38. �PS1

1� � �

PS2

2�

P2 � �S2 �

S1

P1�

P2 �

P2 � 1.91 atm

39.

40. � �10

10k0g�

g�� 125 g�

� � 5.31 g AgNO3169.9 g AgNO3��1 mol AgNO3

0.250 mol AgNO3��1 kg�

� 1000 �m

LL��

� � 1.50M1 mol Mg(NO3)2��

148.3 g Mg(NO3)2

50.0 g Mg(NO3)2��225 mL�

6.25 g/L�� 0.750 atm��

2.45 g/L�

� (725 g AgNO3) � 336 g H2O100.0 g H2O��216 g AgNO3

103 g H2O��1 kg H2O

1.73 mol KNO3��1250 g H2O�

�175 g KNO3

1 mol ��101.1 g KNO3

� 1.73 mol KNO3

0.900 mol K3PO4��2750 g H2O�

Answer Key 825

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E. Essay41. Salt is often used on bridges and sidewalks

because it dissolves in an ice/ice watermixture to produce a solution with a lowerfreezing point than that of water alone. Thiscauses any ice that was initially present tomelt, and prevents additional ice fromforming down to temperatures below0 �C.The salt causes the freezing point of water tobe depressed because it interferes with thecrystallization process.


42.

43.

freezing point � 0�C � 6.39�C � �6.39�C

44.

�Tb � Kb � m � 0.512 �C/m� � 0.958m��Tb � 0.490 �C

boiling point � 100�C � 0.490�C boiling point � 100.490�C


Section 16.4 Making a Solution,page 497

AnalysisSample answers are given.

mass of dry flask � 15.98 g

mass of flask � NaCl � 22.88 g

mass of flask � NaCl � water � 69.09 g

1. a. mass of the solute (NaCl) � 22.88 g �15.98 g � 6.90 g

b. mass of the solvent (water) � 69.09 g �22.88 g � 46.21 g

c. % by mass of NaCl

��(6.90

6�

.9406

g�.21) g�� 100 � 13.0% NaCl

2. a. moles of NaCl solute � �58.

65.9

g�0/m

g�ol

�

� 0.118 mol

b. moles of water � �18

4.06.

g�2/1mg�

ol� � 2.57 mol

c. mole fraction �

� 0.0439.

3. Molality ��00

.

.10

14

862

m1

oklgNHa

2

CO

l�� 2.55m

4. a. liters of solution � 50 mL�� 100

10LmL��

liters of solution � 0.050 L

b. Molarity ��0.118

0.m05

o0lLNaCl

�� 2.4M

5. Density ��6.90 g

50�

m4L6.21 g

�� 1.1 g/mL

Notice that because the flask measures lessaccurately than the balance, molarity and densityhave fewer significant figures than molality, masspercent and mole fraction.

You’re the ChemistSample answers are given.

1. Sample data:

dry flask � 15.98 g

flask � NaCl solution � 22.88 g

flask � NaCl solution � water � 69.09 g

mass of NaCl solution � 6.90 g

mass of water � 46.21 g

mass of NaCl � 0.897 g

0.118 mol��(0.118�2.57) mol�

molality of total particles

� �62.5

7g5B0

a.0

(Ng�

O3)2��126

m1

oglBB

aa((NN

OO

3

3

))

2

2�

� �10

10k0g

g��

31mm

oolel

sB

pa(

aNrt

Oic

3

l)e

2

s�

� 0.958m

�Tf � Kf � m

�Tf � 1.86 �C/m� �

�Tf � � �

�Tf � 6.39 �C

1000 g��

1 kg1 mol CH3OH��32.0 g CH3OH

27.5 g CH3OH��

250.0 g�

�

� �100

10L�mL�� 750 mL�

��91

8m.1

ogl H

H

2

2

SS

OO

4

4�� 11 g H2SO4

0.15 mol H2SO4��1 L�




percent mass of NaCl � 1.69%

moles NaCl � 1.54 � 10�2 mol

mass of water � 46.21 g

moles of water � 2.57 mol

mole fraction � 5.96 � 10�3

molality � 0.333m

density � 1.1 g/mL

2. mass of dry flask � 16.72 g

mass of flask � sugar � 20.85 g

mass of flask � sugar � water � 69.53 g

mass of the sugar � 20.85 g �16.72 g mass of the sugar � 4.13 g

mass of the solvent (water) � 69.53 g � 20.85 g � 48.68 g

% by mass of sugar ��(4.13

4�

.1348

g�.68) g�

�� 100

% by mass of sugar � 7.82%

moles of sugar (C12H22O11) �

� 0.0121 mol

moles of water �

� 2.70 mol

Mole fraction �

Mole fraction � 0.00446

Molality ��00.0.0142816m8 k

ogl s

Hu

2

gOar

�� 0.249m

Molarity ��0.012

01.0

m50

olLNaCl

�� 0.242M

density ��4.13 g

50�

m4L8.68 g

�� 1.1 g/mL

Section Review 17.1

Part A Completion1. heat

2. potential energy

3. thermochemistry

4. calorie

5. joule

6. specific heat or specific heat capacity

7. metals

8. water

Part B True-False9. NT 11. NT 13. AT

10. AT 12. AT

Part C Matching14. c 16. b 18. a

15. e 17. d

Part D Questions and Problems19. Chemical potential energy is energy stored

within the structural units of chemicalsubstances. Work is done when a force movesan object. Heat is energy that is transferedbecause of a temperature difference.

20. C ��18.0

1g

2�

4.215

J.0 �C

�� 0.460 J/(g • °C)

The unknown metal is iron.

Section 17.2

Part A Completion1. calorimeter 4. initial or final

2. enthalpy 5. final or initial

3. �H 6. mass

Part B True-False7. ST 9. AT 11. AT

8. AT 10. NT 12. AT

Part C Matching13. e 15. b 17. d

14. a 16. c

Part D Questions and Problems18. 2NO � O2 → 2NO2 �H � �113.04 kJ

19.

Section 17.3

Part A Completion1. molar heat of fusion

2. molar heat of solidification

3. equal

4. 3.16 kJ/mol

5. molar heat of vaporization

6. Condensation

34.8 g CH4 � �

� 1.94 � 103 kJ

890.2 kJ��1 mol CH4

1 mol CH4��16.0 g CH4

0.0121 mol��(0.0121�2.70) mol

�4.13 g 1 mol ��342 g

�48.68 g 1 mol ��18.0 g

Answer Key 827

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7. molar heat of condensation

Part B True-False8. AT 10. ST 12. NT

9. NT 11. NT

Part C Matching13. a 15. b 17. d

14. e 16. c

Part D Questions and Problems18. a. endothermic d. endothermic

b. endothermic e. exothermic

c. exothermic f. exothermic

19.

20.

Section 17.4

Part A Completion1. sum

2. enthalpy

3. indirectly

4. changed (reversed)

5. standard heat of formation

6. change

7. one

8. �Hf0

9. zero

10. subtracting

Part B True-False11. NT 13. AT 15. NT

12. NT 14. NT

Part C Matching16. b 18. a 20. e

17. c 19. d

Part D Questions and Problems21. CuO(s) y Cu(s) � �

12

�O2 �H � �155 kJH2(g) � �

12

�O2(g) y�H2O(g) �H � �242 kJ

CuO(s) � H2(g) y Cu(s) � H2O(g) �Hrxn � �87 kJ

22.

�H0 � �Hf0 (products) � �Hf

0 (reactants)

� �1181 kJ � (�1154 kJ)

� �27 kJ

Practice Problems

Section 17.1

1. 200.0 Cal�� 10

10C0

acl�al�

� � �4

1.1

c8a4l�J�

� � �110k3

JJ�

�

� 836.8 kJ

2. C ��25.0

52g

5�

.01c5a.l0�C

�� 1.40 cal/(g � �C)

3. �T � � 6.0�C

4. q � 100.0 g� � 120.0�C � 0.90 �g� �

J�C�

�

� 1.1 � 103 J

Section 17.2

1. �H � 150.0 g� � 4.18 �g� �

J�C�

� � 10�C�

� 6.3 � 103 J � 6.3 kJ

2.

3.

4. 4NH3 � 5O2(g) y 4NO(g) � 6H2O(g)

�H � � 4 mol NH3(g)

� �904 kJ

�226 kJ��1 mol NH3(g)

�H � 52.4 g CH4(g) �

� � 2.93 � 103 kJ�890.2 kJ

��1 mol CH4(g)

1 mol CH4(g)��16.0 g CH4(g)

�H � 15.0 g Ca(OH)2(s) �

��1 mo

�

l6C

5a.(2O

kHJ

)2(s)�� 13.2 kJ

1 mol Cu(OH)2(s)��74.1 g Ca(OH)2(s)

1255.0 J��100.0 g� � 2.1 J�/(g� � �C)

�Hf0 (products)

� 3 mol CO2(g) �

� �1181 kJ

�Hf0 (reactants)

� 3 mol CO(g) � � (�822.1 kJ)

� �1154 kJ

�110.5 kJ��1 mol CO(g)

�393.5 kJ��1 mol CO2(g)

�H � 5.53 mol NH4NO3(s)

� � 142 kJ25.7 kJ

��1 mol NH4NO3(s)

�H � 28.3 g H2O(s) �

��1 m

6o.0l

1H

k

2OJ

(s)�� 9.45 kJ

1 mol H2O(s)��18.0 g H2O(s)




Section 17.3

1.

2. Step 1: H2O(l) at 18 �C y H2O(l) at 100 �C

�H � 190.0 g� � 4.18 �g� �

J�C�

� � 82 �C�

�H � 6.512 � 103 J � 6.512 kJ

Step 2: H2O(l) at 100 �C y H2O(g) at 100 �C

�H � 190.0 g� � �11

8m.0

og�l�

� � �410

m.7

okl�J

�

�H � 429.6 kJ

�Htotal � 6.512 kJ � 429.6 kJ � 436.1 kJ

3. �H � 2.543 mol NaOH(s) �

�H � 1.132 � 103 kJ

4. Step 1: H2O(s) at �24 �C y H2O(s) at 0 �C

�H � 96 g� � 2.1 �g� �

J�C�

� � 24 �C � 4.8 � 103 J

Step 2: H2O(s) at 0 �C y H2O(l) at 0 �C

�H � 96 g� � �11

8m.0

og�l�

� � �61.0

m1

okl�J

� � 32 kJ

Step 3:H2O(l) at 0 �C y H2O(l) at 28 �C

�H � 96 g� � 4.18 �g� �

J�C�

� � 28 �C

�H � 1.1 � 104 J � 11 kJ

�Htotal � 4.8 kJ � 32 kJ � 11 kJ � 47.8 kJ

Section 17.41.

� �593.6 kJ � (�483.6 kJ)

� �1077kJ

�Hf0 � �1077 kJ � (�40.2 kJ) � 1.04 � 103 kJ

2. �Hf0(reactants) � �635.1 kJ � (�393.5 kJ)

�Hf0(reactants) � �1028.6 kJ

�H0 � �1207.0 kJ � (�1028.6 kJ)

�H0 � �178.4 kJ

The reaction is exothermic.

3. N2(g) � 2H2O(l) y N2H4(l) � O2(g) �H � 622.2 kJ

2[H2(g) � �12

�O2(g) y H2O(l)] �H � �571.6 kJ

2H2(g) � N2(g) y N2H4(l) �H � 50.6 kJ

Interpreting Graphics 171. a. 27.0 °C b. 26.5 °C

2. a. �73.0 °C b. �73.5 °C

3. a. 5 °C b. 5.5 °C

4. a. �H � (4.184)(39.100)(5.0) � 8.2 � 102 J

b. �H � (4.184)(39.452)(5.5) � 9.1 � 102 J

5. heat gained by H2O � heat lost by metal

a. �8.2 � 102 J

b. �9.1 � 102 J

6. a. specific heat ��(5

�

0.83

.32)(�

�

17

03

2

.0)�

specific heat � 0.22 J/(g � °C)

b. specific heat � �9.1 � 102\(50.35)(�73.5)

specific heat � 0.25 J/(g � °C)

7. b; see Table 17.1.

Vocabulary Review 171. a 5. b 8. d

2. h 6. e 9. j

3. f 7. i 10. c

4. g

Quiz for Chapter 171. b 6. a

2. c 7. is reversed

3. a 8. zero

4. d 9. minus

5. b

�Hf0(reactants)

� 2 mol H2S(g) �

� �40.2 kJ

�20.1 kJ��1 mol H2S(g)

�Hff0 (products)

� �2 mol SO2(g) � �� 2 mol H2O(g) � ��241.8 kJ

��1 mol H2O(g)

�296.8 kJ��1 mol SO2(g)

�445.1 kJ��1 mol NaOH(s)

�H � 35.0 g H2O(s) �

�

� 11.7 kJ

6.01 kJ��1 mol H2O(s)

1 mol H2O(s)��18.0 g H2O(s)

Answer Key 829

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Chapter 17 Test A

A. Matching1. e 5. j 8. b

2. i 6. h 9. a

3. g 7. c 10. d

4. f

B. Multiple Choice11. c 16. a 21. b

12. b 17. c 22. c

13. c 18. b 23. a

14. a 19. d 24. d

15. d 20. b 25. b

C. Essay26. Endothermic processes absorb heat, while

exothermic processes release heat.Endothermic examples include the meltingof ice, the evaporation of a puddle, thesublimation of mothballs, and the heat usedto cook food. Exothermic examples includethe combustion of fossil fuels such asgasoline, the cooling of skin as perspirationevaporates, and the freezing of water.

D. Problems

27.

28. �H � m � C � �T

�H � �60.0 mL� � 1.00 �m

gL�� 4.18 �

g �

J°C

� �� (35.0°C � 27.0°C)

�H ��60.0

g�g��

�

°4C�

.18 J� � 8.0°C�

�H � 2.0 � 103 J or 2.0 kJ

29.

30. 55.0 kJ� � �61.0

m1

okl�J�

� � �11

8m.0

ogl� H

H

2

2

O�O�

� � 165 g

31. 2[Mg(s) � Cl2(g) → MgCl2(s)] 2(�H � �641 kJ)SiCl4(l) → Si(s) � 2Cl2(g) �H � �687 kJ2Mg(s) � SiCl4(l) → Si(s) � 2MgCl2(s)

�H � �1282 kJ � 687 kJ � �595 kJ

32. C2H6 � �72

�O2(g) → 2CO2(g) � 3H2O(l) �H0 � ?�H0 � �H0

f (products) � �H0f (reactants)

�H0 � [2(�393.5 kJ) � 3(�285.8 kJ)] � [(�84.68 J) � (0.0 kJ)]

�H0 � �1559.7 kJ

Chapter 17 Test B

A. Matching1. g 5. j 8. i

2. f 6. d 9. b

3. h 7. e 10. a

4. c

B. Multiple Choice11. c 16. b 21. b

12. b 17. d 22. d

13. a 18. b 23. c

14. d 19. c 24. b

15. b 20. a 25. d

C. Essay26. In vaporizing, steam absorbs the heat

required for vaporization (40.7 kJ/mol). Thus,steam at 100 �C contains more energy thanboiling water at the same temperature.

D. Problems

27.

28. �H � m � C � �T

�H � �55.0 mL�� 1.00 �m

gL�� 4.18 �

g �

J°C

�� (33.0°C � 24.0°C)

�H � 55.0 g� � 4.18 �g� �

J°C�

� � 9.0°C�

�H � 2.1 � 103 J � 2.1 kJ

29.

30. 75.0 kJ� � �61.0

m1

okl�J�

� � �11

8m.0

ogl� H

H

2

2

O�O�

� � 225 g

31. 2[C(s) � O2(g) → CO2(g)] 2(�H � �393.5 kJ)2CO2(g) → 2CO(g) � O2(g) �H � �565.7 kJ2C(s) � O2(g) → 2CO(g)

�H� �787.0 kJ � 565.7 kJ � �221.3 kJ

�31m41

o1lkOJ

2��8.00 g O2� � �

31

2m.0

ogl O

O2

2�� 118 kJ

� 0.17 J/(g � °C)

Specific heat � �75 J

18 g � 25°C

�3

2m4.

o7l

kC

JO

��56.0 g CO� � �128

m.0

ogl C

COO�

�

� 16.5 kJ

� 0.40 J/(g � °C)

Specific heat � �96 J

12 g � 20°C




32. C6H6 � 15/2O2(g) → 6CO2(g) � 3H2O(l) �H� ?�H0 � �Hf

0 (products) � �Hf0 (reactants)

�H0 � [6(�393.5 kJ) � 3(�285.8 kJ)] � [(48.50 kJ) � (0.0 kJ)]

�H0 � �3266.9 kJ


Section 17.4 Heat of Combustion ofa Candle, page 533

Analysis1.

2. The wax burns but many students will saythe wick.

3. The wick draws melted wax to the flame.Those who think the wick burns may suggestthat the wax slows the rate of burning.

4. Because of gravity, hot gases rise, in this caseCO2 and H2O.

5. The candle flame might be round in zerogravity.

6. Depending on the candle and the timeburned, the wick loses a few millimeters andthe candle a few tenths of a gram. The massloss is consistent with the wax burning.

7. Heat from the combustion melts the wax,which is drawn up into the wick, evaporated,and burned. Wick, as a verb, means to drawa liquid from one place to another bycapillary action.

8. C20H42 � �621�O2 → 20CO2 � 21H2O

9. C: 20 � 12 � 240H: 42 � 1 � 42

282

0.50 g� � 1 mol/282 g� � 0.0018 mol

10. ∆H � 20(�394) � 21(�242) �(�2230) �61/2(0)

∆H comb � �10,700 kJ/mol

11. 10,700 kJ/mol�� 0.0018 mol�� 19 kJ

You’re the Chemist1. Black soot will appear on a glass Petri dish

held over the flame.

2. Liquid water will form on the underside of aglass Petri dish filled with ice held over theflame.

Section Review 18.1

Part A Completion1. Rates 6. products

2. react 7. slower

3. kinetic energy 8. temperature

4. activation 9. catalyst

5. minimum 10. increasing


12. NT 14. AT

Part C Matching15. b 17. f 19. e

16. d 18. a 20. c

Part D Questions and Problems21. 120 kg/24 h � 5.0 kg/h

22. b, d

Section Review 18.2

Part A Completion1. reversible

2. products

3. forward

4. reactants

5. reverse

6. reactants

7. products

8. equilibrium

9. equilibrium constant

10. ratio

11. Le Châtelier’s

Answer Key 831

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13. ST 15. AT

Part C Matching16. d 18. b 20. e

17. a 19. c


21. Keq � �[SO

[S2

O]2

3

[]O2

2]�

Keq �

Section Review 18.3

Part A Completion1. solubility product constant

2. common ion effect

3. addition

4. precipitate

Part B Matching5. a

6. c

7. b

Part C Problem8. [CO3

2�] � 0.00070M

[Ba2�] � 0.0015M

[CO32�] � [Ba2�]

� (7.0 � 10�4M) � (1.5 � 10�3M) � 1.1 � 10�6

Precipitation occurs because the ion product(1.1 � 10�6) is greater than the Ksp of BaCO3

(5.0 � 10�9).

Section Review 18.4

Part A Completion1. spontaneous

2. nonspontaneous

3. energy

4. work

5. free energy

6. energy

7. greater

8. entropy

9. disorder

10. law of disorder

11. maximum

Part B True-False12. ST 14. AT 16. ST

13. AT 15. AT


18. f 20. a 22. b

Part D Questions23. a. a heap of loose stamps

b. ice cubes in a bucket

c. 10 mL of steam at 100°C

d. the people watching the parade

24. b

25. d

Section 18.5

Part A Completion1. rate

2. concentration

3. rate law

4. specific rate constant

5. order

6. first-order

7. second order

8. experiment

9. elementary reaction

10. mechanism

� 7.145 � 7.1 (0.42) (0.21)2

��(0.072) 2





12. ST 14. AT

Part C Matching15. d 17. b 19. e16. f 18. a 20. c

Part D Questions21. This diagram represents a reaction that takes

place in two elementary steps. The reaction isexothermic. Points A and C represent theenergy level of the activated complexes. PointB represents the energy level of theintermediate product. Point D represents theenergy level of the final product.


Section 18.11. Rates of chemical reactions can usually be

increased by (1) increasing the temperature,(2) increasing the concentration of thereactants, (3) decreasing the reactant particlesize, and (4) using of a catalyst.

2. 2 mol/4 h � 0.5 mol/h

3. a. decrease the rate

b. increase the rate

4. increase the rate

Section 18.2

1. Keq ��[NO

[N2]

2

4

O�

5][2

O2]�

2. Keq �

Keq � 0.33

3. a. shift left

b. shift right

4.

5. a. Keq � �[NO

[N]4

2

�

O4

[]O

2

2]2�

b. Keq � �[NO

[N]2

O�

Br[]B

2

r2]�

c. Keq � �[C

[OC

]H

�3O

[HH

]

2]2�

d. Keq � �[[SSOO

2

3

]]�

�

[[NN

OO

2

]]

�

6. a. shift right

b. shift right

c. shift right

d. no shift

7. Keq � 1 � 1012

8. Keq � �[H2

[]H

2

2

�

S][2

S2]�

Keq � � 2.1 � 10�4

Section 18.31. a. Ca(OH)2(s) 1 Ca2�(aq) � 2OH�(aq)

Ksp � [Ca2�] � [OH�]2

b. Ag2CO3(s) 1 2Ag�(aq) � CO32�(aq)

Ksp � [Ag�]2 � [CO32�]

2. Ag2CO3(s) 1 2Ag�(aq) � CO32�(aq)

Ksp � [Ag�]2 � [CO32�]

Let x � [CO32�]; 2x � Ag�

Ksp � (2x)2(x) � 8.1 � 10�12

4x3 � 8.1 � 10�12

x3 � 2.0 � 10�12

x � 1.3 � 10�4M � [CO32�]

[Ag�] � 2x � 2.6 � 10�4M

3. Fe(OH)2(s) 1 Fe2�(aq) � 2OH�(aq)

Ksp � [Fe2�] � [OH�]2

[Fe2�] � 0.5[OH�] � 6.0 � 10�6

Ksp � (6.0 � 10�6)(1.2 � 10�5)2

Ksp � 8.6 � 10�16

4. SrCO3 1 Sr2�(aq) � CO32�(aq)

Ksp � [Sr2�] � [CO32�]

Let x � [Sr2�] � [CO32�]

Ksp � x2 � 9.3 � 10�10

x � [Sr2�] � 3.0 � 10�5M

5. Ksp � [Ag�]2 � [CO32�]

Let [Ag�] � 2x; [CO32�] � x � 0.20

assume x << 0.20 mol; [CO32�] � 0.20

Ksp � (2x)2(0.20) � 8.1 � 10�12

x2 � 1.0 � 10�11

x � 3.2 � 10�6M � [CO32�]

[Ag�] � 6.4 � 10�6M

(0.014)2 � (0.035)��

(0.18)2

[N2O4] ��

5.6 [NO2]2 �

0.66 ��5.6

[NO2]2 �

[N2O4] �� 5.6 [NO2]2

� 0.34��

[0.80]4 � [0.20]��

[0.50]2

Answer Key 833

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6. Ksp(PbSO4) � 6.3 � 10�7 � [Pb2�] � [SO42�]

[Pb2�] � [SO42�] � (0.0012)(0.0020)

[Pb2�] � [SO42�] � 2.4 � 10�6

Because this product exceeds the Ksp value,precipitation will occur.

7. Ksp(CaCO3) � 4.5 � 10�9 � [Ca2�] � [CO32�]

The total volume is 1000 mL, so

[Ca2�] � 0.0021 mol/L � 0.0021M

[CO332�] � 0.0013 mol/L � 0.0013M

[Ca2�] � [CO32�] � (0.0021)(0.0013)

[Ca2�] � [CO32�] � 2.7 � 10�6

Because this product exceeds the Ksp value,precipitation will occur.

8. NaCl has no ion in common with Mg(OH)2.

Section 18.41. increase

2. decrease

3. a. increasing c. increasing

b. increasing

4. a. D c. D

b. N d. A

Section 18.51. 2.4 mol/(L • s) � 8 � 0.30 mol/(L • s)2. rate � k[HgCl2][Na2C2O4]2

3. rate � k[J][K]The reaction is first order in both J and K.

4. rate � k[H2O2]; k � rate/[H2O2]k � 0.00842 mol/(L • s) � 0.500 mol/L k � 0.0168 s�1

5. three elementary reactions

6. 1.0; 0.020

7. a. two elementary reactions

b. C3H8O3

c. rate � k[C2H4O2][CH4O]

Interpreting Graphics 181. a. negative d. positive

b. negative e. positive

c. negative f. positive

2. greater

3. less

4. Example a

5. yes

Vocabulary Review 181. e 5. b 8. d

2. g 6. f 9. c

3. i 7. j 10. h

4. a

Quiz for Chapter 181. b 4. b 7. NT

2. d 5. d 8. AT

3. a 6. b 9. NT

Chapter 18 Test A

A. Matching1. i 5. g 8. d

2. a 6. e 9. f

3. j 7. h 10. b

4. c

B. Multiple Choice11. b 16. c 20. a

12. b 17. d 21. b

13. c 18. b 22. a

14. d 19. c 23. d

15. d

C. Problems

24. Keq ��[NO

[N]2

O�

Cl[]C2

l2]��

25. �[C

[OC

]H

�3O

[HH

]

2]2� � 2.2 � 102

[CH3OH] � 2.2 � 102 � [CO] � [H2]2

� 2.2 � 102 � (0.020)(0.60)2

[CH3OH] � 1.58 mol/L

� 9.6 �(1.2)2 � (0.60)

(0.30)2

[SO42�] � 0.0050M � � 0.0020M

400.0 mL ��1000.0 mL

[Pb2�] � 0.0020M � � 0.0012M 600.0 mL ��1000.0 mL




26. a. favors products

b. favors reactants

c. favors products

d. favors reactants

D. Essay27. Spontaneous reactions are reactions that,

under the conditions specified, are known tofavor the formation of products.

Nonspontaneous reactions do not favor theformation of products under the specifiedconditions. Some spontaneous reactionsapprear to be nonspontaneous because theirrates are slow.

E. Additional Problem28. Doubling A doubles the rate�first order in A.

Doubling B increases the rate 8 times (23 � 8)�third order in B. First order � third order �fourth order overall.

Chapter 18 Test B

A. Matching1. j 5. f 8. e

2. g 6. c 9. b

3. h 7. d 10. a

4. i

B. Multiple Choice11. d 16. c 21. c

12. a 17. d 22. b

13. d 18. b 23. c

14. b 19. d 24. b

15. c 20. a

C. Problems25. a. shifts left; decreases

b. shifts left; decreases

c. shifts right; increases

d. shifts right; increases

e. shifts right; increases

f. shifts left; decreases

26. Keq

Keq � 1.4 � 1010

27. PbF2

D. Essay28. a. The addition of more reactant causes an

increase in the rate of the forwardreaction, which consumes that reactant.

b. An increase in temperature causes theendothermic reaction to speed up in aneffort to consume the additional heat.

c. An increase in pressure (for a gaseoussystem with an unequal number ofmolecules) causes the reaction thatproduces the fewest number of moleculesto speed up.

E. Additional Problem

29.

[N2] ?� 1.10 � 1011 mol/L


Section 18.4 Enthalpy and Entropy,page 574

AnalyzeSample data are provided.

1.

2. NH4Cl � H2O is endothermic. ∆H is positive.CaCl2 � H2O is exothermic. ∆H is negative.

3. NaCl � H2O(l) did not change much intemperature. ∆H is close to 0.

4. heat � NH4Cl(s) → NH4�(aq) � Cl�(aq)

CaCl2(s) → Ca2+(aq) � 2Cl�(aq) � heat

5. All of the solids dissolved rapidly. Entropyusually increases in the dissolving process.∆S is positive in each case.

T1Mixture T2 �T

21°C 21°C 0°C

21°C 5°C �16°C

21°C

a. NaCl � H2O(l )

b. NH4Cl � H2O(l )

c. CaCl2 � H2O(l ) 53°C �32°C

[N2]�

�[NH3]2

[N2] � [H2]3

�[NH3]2 Keq � [H2]3

Keq�

� �(1.23 � 10�4)2

(6.59 � 10�3) � (2.75 � 10�6)3

�

�[H2O]2 � [Cl2]2

[HCl]4 � [O2]

�(5.8 � 10�2)2 � (5.8 � 10�2)2

(1.2 � 10�3)4 � (3.8 � 10�4)

�

Answer Key 835

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6. ∆G � ∆H � T∆S.�G � (0) �T(�) for NaCl(s)�G is �.�G � (�) �T(�) for NH4Cl(s)�G is � or �.�G � (�) �T(�) for CaCl2(s)∆G is �.

You’re the Chemist1. The temperature of the NaCl and ice

dropped dramatically.Sample data:

2. Melting ice is endothermic. Endothermicreactions absorb heat, cooling theenvironment. This explains the drop intemperature of the NaCl and ice mixture.

3. Both CaCl2 and NH4Cl depress the freezingpoint of ice and cause a drop intemperature.

4. Many salts such as KCl, NaHCO3, Na2CO3

and Na3PO4 dissolve endothermically orwith little or no change in temperature.

Section Review 19.1

Part A Completion1. three

2. Arrhenius

3. hydroxide ions

4. proton

5. acceptor

6. electron-pair

7. donor

8. monoprotic

9. diprotic

10. conjugate acid–base pair

11. amphoteric


13. NT 15. ST

Part C Matching17. g 20. h 23. c

18. d 21. e 24. f

19. a 22. i 25. b

Part D Problems26. Dimethyl ether is a Lewis base because it

donates an electron pair to form a bond.Boron trifluoride is a Lewis acid because itaccepts an electron pair from dimethyl ether.

Section Review 19.2

Part A Completion1. self-ionize

2. 1 � 10�7

3. 0 to 14

4. hydrogen ion

5. acidic

6. basic

7. neutral

8. 7

9. ion-product

10. hydronium/hydroxide

11. hydroxide/hydronium


13. ST 15. NT

Part C Matching17. c 20. a 22. g

18. f 21. b 23. d

19. e

Part D Problems24. Kw � [H�][OH�]

[OH�] � �[HKw

�]� � �

11

�

�

11

00

�

�

1

1

4

0� � 1 � 10�4

The solution is basic.

25. a. pH � �log[H�][H�] � 1 � 10�3M

b. pH � �log[H�][H�] � 1 � 10�6M

c. pH � �log[H�][H�] � 1 � 10�10

T1Mixture T2 �T

0°C �15°C �15°CNaCl � H2O(s )




Section Review 19.3

Part A Completion1. degree of ionization

2. Ka

3. larger

4. pH

5. completely

6. strong

7. weak

8. bases

9. water

10. acid

11. strong


13. NT 15. ST


17. e 19. b 21. f

Part D Problem22. HX 1 H� � X�

[H�] � [X�] � 4.1 � 10�2

[HX] � 0.35 � 4.1 � 10�2 � 0.35 � 0.041[HX] � 0.309

Ka � �[H

[

�

H][XX]

�]� �

Ka � 5.4 � 10�3

Section Review 19.4

Part A Completion1. acid 5. titration

2. hydroxide 6. end point

3. water 7. equivalence

4. neutralization

Part B True-False8. AT 10. AT

9. AT 11. NT


13. e 15. b

Part D Problem

17. a. H3PO4 � Al(OH)3y AlPO4 � 3H2O

b. 2HI � Ca(OH)2y CaI2 � 2H2O

Section Review 19.5

Part A Completion1. salt 6. strong

2. acidic 7. weak

3. basic 8. buffer

4. neutral 9. capacity

5. hydrolyze


11. NT 13. AT

Part C Matching14. a 16. b

15. d 17. c

Part D Questions and Problems18. a. acidic

b. neutral

c. basic

Practice Problems

Section 19.11. H2SO4 and H3O� are proton donors and H2O

and HSO4� are the proton acceptors. The

conjugate acid–base pairs are H2SO4/HSO4�

and H3O�/H2O.

H2SO4 � H2O 1 H3O�� HSO4�

2. H3PO4 is a triprotic acid able to ionize threehydrogens. All ions formed are shown in thefollowing chemical equations.

H3PO4 � H2O 1 H3O� � H2PO4�

H2PO44� � H2O 1 H3O� � HPO4

2�

HPO42� � H2O 1 H3O� � PO4

3�

3. Only hydrogens bonded to highly electro-negative atoms are ionizable.

a. monoprotic

b. monoprotic

c. diprotic

d. triprotic

4. Like other alkali metals, lithium reactsviolently with water to produce hydrogen andthe base lithium hydroxide.

(4.1 � 10�2) (4.1 � 10�2)��

0.309

Answer Key 837

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2Li(s)� 2H2O → H2(g) � 2LiOH(aq)

5. BF3 can accept a pair of electrons to form acovalent bond and is therefore a Lewis acid.Since F� donates the pair of electrons, it is aLewis base.

6. Acids have a tart or sour taste and causeindicators to change color. Acids react withcompounds containing hydroxide ions toproduce a salt and water.

7. Aqueous solutions of bases taste bitter andfeel slippery. They react with acids toproduce a salt and water. Bases causeindicators to change colors.

Section 19.21. pH � �log[H�]

pH � �log(1 � 10�6) reminder: the log(a � b) � log a � log bpH � �(0.0 � (�6))reminder: the log 1 � 0.0pH � 6.0

2. pH � �log[H�]pH � �log(7.2 � 10�9)pH � �(0.86) � (�9.00) Use log tables or your calculator to find thelog of 7.2.pH � 9.00 � 0.86pH � 8.14

3. pOH � �log[OH�]pOH � �log(3.5 �10�2)pOH � � (0.54) � (�2.00)pOH � 2.00 � 0.54pOH � 1.46

4. pOH � 14.0 � pHpOH � 14.0 � 3.4pOH � 10.6

5. a. basic d. neutral

b. acidic e. acidic

c. acidic

6. a. 5.0 c. 7.34

b. 10.36 d. 12.6

7. a. acidic c. basic

b. basic d. basic

8. Most acidic solutions of interest have ahydrogen ion concentration of less than 1M.The log of this concentration would always bea negative number. Taking the negative log(minus sign in the pH definition) ensures thatthe pH values will usually be positive.

9. pH � pOH � 14.0pH � 14.0 � pOH � 14.0 � 12.4 � 1.6

10. pH � �log[H�]pH � �log(1 � 10�3) � �(0.0 � (�3)) � 3.0

Section 19.31. strong base, weak base, weak acid,

strong acid

2. HF(aq) 1 H�(aq) � F�(aq)

Ka � �[H

[

�

H][FF]

�]�

3. N2H4(aq)�H2O(l) 1 N2H5�(aq) � OH�(aq)

Kb ��[N2H

[N5

�

2H][O

4]H�]

�

4. The weakest acid has the smallest Ka.HCO3

� � H2PO4� � HCOOH � H2C2O4

5. a. H2S(aq) 1 H�(aq) � HS�(aq)

Ka � �[H

[

�

H][

2

HS]

S�]�

b. NH4�(aq) 1 NH3(aq) � H�(aq)

Ka � �[N

[NH

H3][

4

H�]

�]�

c. C6H5COOH(aq)1

C6H5COO�(aq) � H�(aq)

Ka ��[C

[6

CH

6

5

HC

5

OC

OO

�

O][HH

]

�]�

6. a. (4) d. (5)

b. (2) e. (3)

c. (1)

7. Kb �

8. Ka �

Ka � 1.8 � 10�4M

9. At equilibrium, [H�] � x � [C6H5COO�][C6H5COOH] � 0.20M � x � 0.20M(since x �� 0.20M)

Ka � �0.2

x0

2

M� � 6.3 � 10�5M

x � 3.5 � 10�3M � [H�]

10. At equilibrium, [H�] � [CN�] � 6.3 � 10�6M

Ka � �[H

[

�

H][CCNN]

�]�

Ka ��(6.3 �

0.1100M

�6M)2

�� 4.0 � 10�10M

(4.2 � 10�3M)(4.2 � 10�3M)��

0.096M

[C6H5NH3�][OH�]

��[C6H5NH2]




Section 19.41. 2NaOH(aq) � H2SO4(aq)

y Na2SO4(aq) � 2H2O(l)

Molarity � �mlit

oelress

� ��0.

00.20

138

mLo

Nl N

aOaO

HH

�

Molarity � 0.55M NaOH

2. Ca(OH)2(aq) � 2HC2H3O2(aq) y Ca(C2H3O2)2(aq) � 2H2O(l)

� 9.94 � 10�5 mol Ca(OH)2

Molarity �

Molarity � 0.00404M

3. Ca(OH)2(aq) � H2SO4(aq) y CaSO4(aq) � 2H2O(l)

� 0.000198 mol H2SO4

Molarity � �mlit

oelress

� �

Molarity � 0.0160M H2SO4

4. Ba(OH)2(aq) � 2HCl(aq) y BaCl2(aq) � 2H2O(l)

0.0122 L HCl� �

0.0122 L HCl� �

� 0.0015 mol Ba(OH)2

liters � �m

mo

ola

lreisty

� �

liters � 0.0125 L Ba(OH)2 � 13 mL Ba(OH)2

5. Al(OH)3(aq) � 3HCl(aq)y AlCl3(aq) � 3H2O(l)

liters � �m

mo

ola

lreisty

� ��0.0

00.22

10

20M

moH

lCHlCl

�

liters � 0.0106 L HCl � 10.6 mL HCl

Section 19.51. CHO2

� � H�1 HCHO2

HCHO2 � OH�1 CHO2

� � H2O

2. a. neutral solution

b. acidic solution

c. basic solution

Interpreting Graphics 191. C6H5COOH � NaOH y C6H5COONa � H2O

One mole of sodium hydroxide willneutralize one mole of benzoic acid.

2. To determine the equivalence point, find thearea of the titration curve where the pHchanges abruptly when a small volume ofNaOH is added. Locate the point on this steep portion of the curve equidistantbetween the two plateaus. The pH at theequivalence point is approximately 8.5; thesolution is slightly basic.

3. Benzoic acid is a weak acid. Theneutralization of a weak acid with a strongbase, such as NaOH, produces a basicsolution at the equivalence point.

4. 0.025 L� � 0.10 mol/L� NaOH

� 0.0025 mol NaOH

5. The equivalence point occurs when thenumber of moles of NaOH added equals thenumber of moles of C6H5COOH originallypresent. Because NaOH is a strong base,each mole of NaOH added reacts with eachmole of C6H5COOH present. Thus, at theequivalence point, [C6H5COOH] � [NaOH] � 0M and

[C6H5COONa] � �0.0

00.0

2550mL

ol� � 0.050M.

6. Based on the answers to questions 4 and 5,0.0025 mol C6H5COOH were originallypresent in a volume of 25 mL.

Thus, [C6H5COOH] � �0.0

00.0

2255mL

ol� � 0.10M

0.0550 g Al(OH)3 �

� � 0.00212 mol HCl3 mol HCl

��1 mol Al(OH)3

1 mol Al(OH)3��78.0 g Al(OH)3

0.0015 mol Ba(OH)2��0.12M Ba(OH)2

1 mol Ba(OH)2��2 mol HC

0.25 mol HCl��

1 L HCl�

0.000198 mol H2SO4��0.0124 L H2SO4

0.0198 L Ca(OH)2 �

� � 0.000198 mol H2SO41 mol H2SO4��

1 mol Ca(OH)2

0.0100 mol Ca(OH)2��1 L Ca(OH)2

9.94 � 10�5 mol Ca(OH)2��0.0246 L Ca(OH)2

0.0142 L HC2H3O2 �

0.0142 L HC2H3O2 �

� 9.94 � 10�5 mol Ca(OH)2

1 mol Ca(OH)2��2 mol HC2H3O2

0.0140 mol HC2H3O2��1 L HC2H3O2

0.014 L H2SO4 �

� � 0.021 mol NaOH2 mol NaOH��1 mol H2SO4

0.75 mol H2SO4��1 L H2SO4

Answer Key 839

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7. Because the equivalence point occursbetween pH 6 and pH 11, phenolphthaleinwould be a good choice. A faint pink colorshould be detected at the equivalence point.Thymol blue might also be a good candidate.Students should draw a horizontal band onthe graph encompassing the pH range 8-10 toshow the region of the curve wherephenolphthalein would be an effectiveindicator of neutralization.

8. C6H5COO� � H2O 1 C6H5COOH � OH�

At the equivalence point, the benzoate ionestablishes the equilibrium shown. Theresulting solution is slightly basic because[OH�] > [H�].

9. Kb �

10. pH � pOH � 14

8.5 � pOH � 14

pOH � 5.5

[OH�] � 3.2 � 10�6M

At the equivalence point [OH�] �[C6H5COOH] � 3.2 � 10�6M

[C6H5COO�] � 0.050M

Kb ��(3.2

(0�

.051

00

)

�6)2

�� 2.0 � 10�10

Vocabulary Review 191. Hydronium ion. The other terms describe

aqueous solutions based on their pH.

2. Acidic solution. Alkaline is another name fora basic solution and basic solutions wouldhave a high hydroxide ion concentration.

3. Amphoteric. The other terms are theoriesused to classify acids and bases.

4. Lewis acid. The other terms refer to ways ofdescribing acids and bases according to theBrønsted-Lowry theory.

5. Strong acids. Weak acids and bases are onlypartially ionized in aqueous solution. Thedissociation constant reflects the fraction ofa weak base or weak acid that is in ionizedform.

6. hydrolyzing salt 8. equivalence point

7. neutral 9. buffer

Quiz for Chapter 191. b 4. c 7. c

2. b 5. c 8. b

3. a 6. d

Chapter 19 Test A

A. Matching1. i 5. h 8. e

2. a 6. g 9. c

3. j 7. d 10. b

4. f

B. Multiple Choice11. b 18. d 25. a

12. d 19. a 26. d

13. d 20. b 27. a

14. a 21. d 28. c

15. c 22. d 29. b

16. d 23. d

17. c 24. a

C. Problems30. a. pH � �log[H�] � 9, basic

b. [H�] � 1 � 10�4, pH � 4, acidic

c. pH � 7, neutral

31. Kw � [H�] � [OH�]

[H�] � �[O

KH

w�]

� � �11

�

�1100

�

�

1

1

2

4

�

[H�] � 1 � 10�2 acidic

32. a. H2SO3 1 H� � HSO3�

Ka ��[H�]

[H�

2S[H

OS

3]O3

�]�

b. H��3 1 H� � NO3�

Ka � �[H�

[]H�

N[ON

3

O]

3�]

�

33. a. 2HBr � Mg(OH)2yMgBr2 � 2H2O

b. 3H2SO4 � 2Al(OH)3y Al2(SO4)3 � 6H2O

34. a. basic c. acidic

b. neutral d. basic

[C6H5COOH][OH�]��

[C6H5COO�]




D. Essay35. Both acids and base, cause indicators to

change colors and react with each other toform water and a salt. Acids taste sour; basestaste bitter. Bases feel slippery.

Chapter 19 Test B

A. Matching1. b 5. f 9. i

2. g 6. c 10. j

3. d 7. h 11. e

4. a 8. k

B. Multiple Choice12. b 17. b 22. c

13. d 18. d 23. a

14. a 19. a 24. b

15. b 20. d 25. a

16. c 21. a 26. a

C. Problems27. Kw � [H�][OH�] � 1.0 � 10�14 (mol/L)2

[OH�] � Kw/[H�]

[OH�] �

[OH�] � 1.0 � 10�5 mol/L The solution is basic.

28. a. [OH�] � 1 � 10�11; 3; acidic

b. [H�] � 1 �10�6 ; 6; acidic

c. [H�] � 1 �10�7 ; 7; neutral

29. a. Ka � �[H

[

�

H][II]

�]�

b. Ka ��[H

[

�

H][

2

HSO

SO

4]4

�]�

30. a. HF(aq) � KOH(aq) y KF(aq) � H2O(l)

b. H2SO4(aq) � 2LiOH(aq) y Li2SO4(aq) � 2H2O(l)

31. H2SO4(aq) � 2KOH(aq) yK2SO4(aq) � 2H2O(l)1 mol 2 mol 1 mol 2 mol

� 0.18 mol H2SO4

32. CuCl(s) y Cu�(aq) � Cl�(aq)

Ksp � [Cu�][Cl�]

3.2 � 10�7 � [Cu�][Cu�]

3.2 � 10�7 � [Cu�]2

5.7 � 10�4M � [Cu�]

D. Essay33. The Brønsted-Lowry theory defines acids as

proton donors and bases as proton acceptors.According to the Lewis theory, acids areelectron-pair acceptors, whereas bases areelectron-pair donors.

E. Additional Problems34. pH � �log [H�]

pH � �log [3.4 � 10�4]pH � �(log 3.4 � log 10�4)pH � �[(0.53) � (�4)]pH � �[�3.47]pH � 3.47The solution is acidic.

35. a. HF(aq); H2O(l); H3O�(aq); F�(aq)

b. HCl(g); H2O(l); H3O�(aq); Cl�(aq)

c. HC2H3O2(aq); H2O(l); H3O�(aq);C2H3O2

�(aq)

36. a. H�; I�

b. BCl3; NH3

37. HC2H3O2(aq) 1 H�(aq) � C2H3O2�(aq)

[H�] � [C2H3O2�] � 2.25 � 10�3M

[HC2H3O2] � 0.1000M � 0.00225M[HC2H3O2] � 0.09775M

Ka � �[H

[

�

H][CC

2

2

HH

3

3

OO

2

2

]

�]�

Ka �

Ka � 5.18 � 10�5

[0.00225][0.00225]��

[0.09775]

�12

mm

oollHK

2

OSO

H4� � 0.35 mol KOH

1.0 � 10�14 (mol/L)2

��1.0 � 10�9 mol/L

Answer Key 841

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Section 19.4 Ionization Constantsof Weak Acids,page 617

Analysis

1. The pH solutions 1-3 are yellow.

2. The pH solutions 5-12 are blue.

3. The pH solution 4 is green, an intermediatebetween yellow and blue.

4. The conjugate acid, HBCG is yellow.

5. The conjugate base, BCG� is blue.

6. An equal mixture of HBCG and BCG� isgreen at pH � 4.

You’re the Chemist1. Results will vary depending on the indicator

chosen.

2. To measure the Ka of a colored weak acid,mix one drop of the weak acid with one dropof each pH 1-12 buffer solution. Look for thepH of the color change. This pH is the Ka ofthe acid.

Section Review 20.1

Part A Completion1. redox 5. oxidizing

2. away 6. reduced

3. toward 7. reducing

4. reduction 8. oxidized

Part B True-False9. AT 11. NT

10. AT 12. NT

Part C Matching13. e 15. f 17. a

14. b 16. c 18. d

Part D Questions and Problems19. Oxidation is the complete or partial loss of

electrons. Reduction is the complete orpartial gain of electrons.

20. The zinc metal, Zn, was oxidized and is thereducing agent. The copper ion, Cu2�, wasreduced and is the oxidizing agent.

21. When oxygen and water attack iron, the ironatoms lose electrons as the iron begins to beoxidized. Since aluminum and zinc are betterreducing agents than iron and are more easilyoxidized, they immediately transfer electronsto the iron ions, reducing them back toneutral iron atoms.

Section Review 20.2

Part A Completion1. zero 5. charge on the ion

2. sign 6. electron

3. charge 7. oxidation

4. zero 8. decrease


10. AT 13. NT 16. AT

11. NT 14. NT

Part C Matching17. e 20. a 23. f

18. h 21. g 24. i

19. d 22. c 25. b

1 2 3

4 5 6

97 8

1210 11

Figure A

Yellow Yellow Yellow

Green Blue Blue

Blue Blue Blue

Blue Blue Blue



05_Chem_CTRAK_Ch19-25 7/12/04 8:26 AM 2

Part D Questions and Problems26. An increase in the oxidation number of an

atom indicates oxidation. A decrease in theoxidation number indicates reduction.

27. N is reduced (�5 to �2); Br is oxidized (�1 to 0)

Mn is reduced (�7 to �2); Cl is oxidized (�1 to 0)

N is reduced (�5 to �2); Sb is oxidized (0 to �5)

S is reduced (�6 to �4); C is oxidized (0 to �4)

Section Review 20.3

Part A Completion1. oxidation number 5. two

2. half-reaction 6. added

3. balanced 7. ionic

4. ionic


9. AT 11. NT 13. NT

Part C Matching14. c 17. b 20. g

15. a 18. e

16. f 19. d

Part D Questions and Problems21. a. HNO3y NO; N changes �5 to �2, a gain

of 3e�; multiply by 2

2HI y I2; I changes from �1 to 0, a loss of2e� for I2; multiply by 3

2HNO3 � 6HI y 2NO � 3I2 � 4H2O

b. HNO3y NO2; N changes �5 to �4, again of 1e�; multiply by 10

I2y 2HIO3; I changes from 0 to �5, a lossof 10e� for I2

10HNO3 � I2y 2HIO3 � 10NO2 � 4H2O

22. a. S2� → S � 2e� and 3e� � 4H� � NO3

� → NO � 2H2O;Multiply the oxidation reaction by 3 andthe reduction reaction by 2.

3H2S � 2HNO3 → 3S � 2NO � 4H2O

b. Fe2� → Fe3� � e� and 6e� � 14H� �

Cr2O72� → 2Cr3� � 7H2O; multiply the

oxidation reaction by 6

14H� � 6Fe2� � Cr2O72�

→ 6Fe2� � 2Cr3� � 7H2O

Practice Problems

Section 20.11. Sr: oxidized (reducing agent)

O2: reduced (oxidizing agent)

2. Li: oxidized (reducing agent)

S: reduced (oxidizing agent)

3. Cs: oxidized (reducing agent)

Br2: reduced (oxidizing agent)

4. Mg: oxidized (reducing agent)

N2: reduced (oxidizing agent)

5. Fe: oxidized (reducing agent)

O2: reduced (oxidizing agent)

6. Br�: oxidized (reducing agent)

Cl2: reduced (oxidizing agent)

7. Si: oxidized (reducing agent)

F2 reduced (oxidizing agent)

8. Ca: oxidized (reducing agent)

O2 reduced (oxidizing agent)

9. Mg: oxidized (reducing agent)

H�: reduced (oxidizing agent)

10. Na: oxidized (reducing agent)

H2O: reduced (oxidizing agent)

Section 20.21. a. Sn is tin in an uncombined state. The

oxidation number is 0.

b. The ionic charge on potassium is 1�, thusthe oxidation number is �1.

c. The ionic charge on sulfur is 2�, thus theoxidation number is �2.

d. The ionic charge on iron is 3�, thus theoxidation number is �3.

e. Se is selenium in an uncombined state.The oxidation number is 0

f. The ionic charge on magnesium is 2�,thus the oxidation number is �2.

g. The ionic charge on tin is 4�, thus theoxidation number is �4.

Answer Key 843

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h. The ionic charge on bromine is 1�, thusthe oxidation number is �1.

2. a. �3 c. �2

b. �6 d. �60 �1 �6�2 �4 �2 �4 �2 �1 �2

3. a. C � H2SO4y CO2 � SO2 � H2O

Carbon is oxidized (0 y�4).Sulfur is reduced (�6 y�4).�1 �5 �2 �1 �1 �2 �2 0 �1 �2

b. HNO3 � HI y NO � I2 � H2O

Nitrogen is reduced (�5 y�2).Iodide ion is oxidized (�1 y 0).�1 �7 �2 �1 �1 �2 �1 0 �1 �2 �1 �1

c. KMnO4� HClyMnCl2� Cl2� H2O � KCl

Manganese is reduced (�7 y�2).Chloride ion is oxidized (�1 y 0).0 �1 �5 �2 �5 �2 �2 �2 �1 �2

d. Sb � HNO3y Sb2O5 � NO � H2O

Antimony is oxidized (0 y�5).Nitrogen is reduced (�5 y�2).

4. a. Oxidizing agent is sulfur; Reducing agent is carbon.

b. Oxidizing agent is nitrogen; Reducing agent is iodine.

c. Oxidizing agent is manganese; Reducing agent is chlorine.

d. Oxidizing agent is nitrogen; Reducing agent is antimony.

Section 20.31. a. Increase in oxidation number of carbon

� �4; decrease in oxidation number ofsulfur � �2.

C � 2H2SO4y CO2 � 2SO2 � 2H2O

b. Increase in oxidation number of sulfur ��2; decrease in oxidation number ofnitrogen � �3.

3H2S � 2HNO3y 3S � 2NO � 4H2O

c. Increase in oxidation number of iodineion � �1; decrease in oxidation numberof nitrogen � �3.

2HNO3 � 6HI y 2NO � 3I2 � 4H2O

d. Increase in oxidation number of antimony� �5; decrease in oxidation number ofnitrogen � �3.

6Sb � 10HNO3y 3Sb2O5 � 10NO � 5H2O

e. Increase in oxidation number of chlorineion � �1; decrease in oxidation numberof manganese � �5.2KMnO4 � 16HCl

y 2MnCl2 � 5Cl2 � 8H2O � 2KCl

f. Increase in oxidation number of iodineion � �1; decrease in oxidation numberof iodine � �7.

KIO4 � 7KI � 8HCl y 8KCl � 4I2 � 4H2O

g. Increase in oxidation number of zinc �

�2; decrease in oxidation number ofchromium � �3.3Zn � 2Cr2O7

2� � 28H�

y 3Zn2� � 4Cr3� � 14H2O

2. a. Fe2�y Fe3� � e�

5e� � 8H� � MnO4�yMn2� � 4H2O

b. Sn2�y Sn4� � 2e�

6H� � 6e� � IO3� → I� � 3H2O

c. S2� → S � 2e�

3e� � 4H� � NO3�y NO � 2H2O

d. 4OH� � Mn2� → MnO2 � 2H2O � 2e�

2e� � H2O � H2O2 → H2O � 2OH�

3. a. 2OH� � Zn � HgO y ZnO22� � Hg � H2O

b. 8H� � 5Fe2� � MnO4�

y 5Fe3� � Mn2� � 4H2O

c. 6H� � 3Sn2� � IO3�y 3Sn4� � I� � 3H2O

d. 8H� � 3S2� � 2NO3�y 3S � 2NO � 4H2O

e. 2OH� � Mn2� � H2O2 → MnO2 � 2H2O

f. 2OH� � CrO2 � ClO�

y CrO42� � Cl� � H2O

Interpreting Graphics 201. a. 3 d. 2

b. 2 e. 4

c. 3 f. 1

2. MnO44� � 8H� � 5Fe2�

yMn2� � 4H2O � 5Fe3�

3. The end point occurs when the number ofequivalents of MnO4

� added equals thenumber of equivalents of Fe2� originallypresent in the reaction flask. One equivalentis the amount of reducing agent (or oxidizingagent) that can give (or accept) one mole ofelectrons. When all the Fe2� in the flask isoxidized, the next drop of MnO4

� remainsunreacted, and the solution in the flask turnslight purple, signaling the end point of thetitration.

4. Volume KMnO4 � Initial Volume � FinalVolume � 48.65 mL � 23.35 mL � 25.30 mL

Moles MnO4� � 0.02530 L� � �

0.020L�0 mol�

Moles MnO44� � 5.06 � 10�4 mol MnO44

�




% Fe in ore � �02

.

.19

43

18

gg

� � 100% � 4.80%

Vocabulary Review 201. oxidizing agent

2. oxidation-number-change method

3. reduction

4. half-reaction method

5. oxidation number

6. reducing agent

7. half-reaction

8. oxidation-reduction reaction

9. oxidation

10. redox reaction

Quiz for Chapter 201. b 4. b 7. c

2. a 5. a 8. d

3. b 6. a

Chapter 20 Test A

A. Matching1. d 5. i 9. c

2. j 6. g 10. h

3. f 7. e

4. a 8. b

B. Multiple Choice11. b 17. c 23. d

12. d 18. c 24. c

13. c 19. a 25. c

14. b 20. a 26. c

15. a 21. c

16. b 22. b

C. Questions27. a. Na oxidized, reducing agent; Br2 reduced,

oxidizing agent

b. S reduced, oxidizing agent; K oxidized,reducing agent

28. 2Cr � 3Br2y 2Cr3� � 6Br�

29. a. Li �1, Al �3, F �1

b. Na �1, O �2

c. S 0 (element)

30. Oxidation-number change method:

Fe2O3 � 3CO y 2Fe � 3CO2

Half-reaction method:

6H� � Fe2O3 � 6e�y 2Fe � 3H2O

3(H2O � CO y CO2 � 2H� � 2e�)

6H� � Fe2O3 � 6e� � 3H2O � 3CO y 2Fe � 3H2O � 3CO2 � 6H� � 6e�

Fe2O3 � 3CO y 2Fe � 3CO2

D. Essay31. An oxidation number is assigned to an

element in a compound according to a set ofarbitrary rules. The oxidation number of anelement in an uncombined state is zero. Theoxidation number of a monatomic ion is thesame in magnitude and sign as the ioniccharge. The sum of the oxidation numbers ofthe elements in a neutral compound is zero.In a polyatomic ion, however, the sum isequal to the charge on the ion. Oxidationnumbers help keep track of electrons in redoxreactions. An oxidation-number increase isoxidation. A decrease is reduction.

Chapter 20 Test B

A. Matching1. e 5. i 9. b

2. f 6. a 10. j

3. d 7. c

4. g 8. h

2 � (�3) � �6

�3 �2 �2 �2 0 �4 �2

Fe2O3 � CO y Fe � CO2

3 � (�2) � �6

Moles iron(II) � 5.06 � 10�4 mol MnO44�

��1

5m

mol

oMl F

neO

2�

4�� 2.53 � 10�3 mol Fe2�

Mass Fe � 2.53 � 10�3 mol Fe2�

� �515.

m85

olg

FFee2

2

�

�

� � 0.141 g

Answer Key 845

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B. Multiple Choice11. d 17. a 23. d

12. a 18. b 24. c

13. a 19. c 25. d

14. d 20. c 26. b

15. d 21. b 27. b

16. b 22. b 28. a

C. Questions29. a. K; I; I2; K

b. Na; H; H2O; Na

c. H; Cu; CuO; H2

d. Mg; Cu; Cu(NO3)2; Mg

30. a. K2SO4 � �1, �6, �2

b. Cu(NO3)2 � �2, �5, �2

c. HAsO3 � �1, �5, �2

d. MnO4� � �7, �2

31. a.

b.

32. a. HNO2 � HI y I2 � NO � H2O

H�(aq) � NO2�(aq) � H�(aq) � I�(aq) y I2(aq) � NO(g) � H2O(l)

Oxidation: 2I�(aq) y I2 � 2e�

Reduction: 2[2H�(aq) � NO22�(aq) � 1e�

y NO � H2O]

4H� � 2NO2� � 2e�y 2NO � 2H2O

4H� � 2I� � 2NO2�y I2 � 2NO � H2O

Final: 2HNO2 � 2HI y I2 � 2NO � 2H2O

b. K2Cr2O7 � FeCl2 � HCl yCrCl3 � KCl � FeCl3 � H2O

2K�(aq) � Cr2O72�(aq) � Fe2�(aq) � 2Cl�

2K�(aq) � H�(aq) � Cl�(aq) y Cr3�(aq) � 3Cl�(aq) � K�(aq) � Cl�(aq)y Cr3�(aq) � Fe3�(aq) � 3Cl�(aq) � H2O

Oxidation: 6[Fe2�y Fe2� � 1e�]

Reduction: 2Cr6� � 6e�y 2Cr3�

6Fe2� � 2Cr6�y 6Fe3� � 2Cr3�

Final: K2Cr2O7 � 6FeCl2 � 14HCl y 2CrCl3 � 2KCl � 6FeCl3 � 7H2O

D. Essay33. Since oxidation is the loss of electrons, it can

only occur in the presence of anothersubstance that will accept the lost electrons.The accepting substance gains electrons, andthus, undergoes reduction. In other words, aloss of electrons can only occur if a gain takesplace concurrently.


Section 20.3 Half Reactions,page 655

Analysis

1. Mg is most reactive because it bubbles mostvigorously. Cu did not react. The order ofreactivity is Mg Zn Fe Cu.

2. H2(g) is the gas produced all the reactions.

3. Mg(s) � 2HCl(aq) → H2(g) � MgCl2(aq)

Mg(s) � 2H�(aq) → H2(g) � Mg2�(aq)

Fe(s) � 2HCl(aq) → H2(g) � FeCl2(aq)

Fe(s) � 2H�(aq) → H2(g) � Fe2�(aq)

All are redox reactions because the oxidationnumber of reactants change.

4. Mg(s) → Mg2�(aq) � 2e�

Fe(s) → Fe2�(aq) � 2e�

5. 2H� � 2e�→ H2(g).

Mg(s) → Mg2�(aq) � 2e�

Figure A

HCl

Zn

Mg

Cu

Fe

Bubbles

Bubbles

No visiblereaction

No visiblereaction

No visiblereaction

Bubbles

Bubbles

Bubbles

Bubbles

Bubbles

Bubbles

Bubbles

HNO3 H2SO4

1 � (�2) � �2

Br2 � SO2 � 2H2O y H2SO4 � 2HBr

2 � (�1) � �2

3 � (�1) � �3

4HNO3 � 3Ag y 3AgNO3 � NO � 2H2O

1 � (�3) � �3




Mg(s) � 2H�(aq) → H2(g) � Mg2�(aq)

You’re the Chemist1. Add a drop of any acid to the damaged part

of the penny and notice that only the zincinterior reacts.

2. Many toilet-bowl cleaners and vinegardissolve metals. Keep products containingacids away from metal pipes and fixtures.

Section Review 21.1

Part A Completion1. electrochemical process

2. electrons

3. voltaic cells

4. salt bridge

5. ions

6. anode

7. cathode


9. AT 11. ST

Part C Matching12. g 15. b 17. e

13. f 16. c 18. a

14. d

Part D Problem19. The shorthand notation Mg(s) | MgSO4(aq) ||

PbSO4(aq) | Pb(s) represents a magnesium-lead voltaic cell. The single vertical linesindicate boundaries of phases that are incontact, and the double vertical linesrepresent the salt bridge that separates theanode compartment from the cathodecompartment. In this electrochemical cell,Mg is oxidized to Mg2� at the anode (thenegative electrode) and Pb2� is reduced to Pbat the cathode (the positive electrode).Electrons flow from the anode, through anexternal circuit (connected to a light bulb orvoltmeter), to the cathode. To complete thecircuit, sulfate (SO4

2�) anions move from thecathode compartment to the anodecompartment, and magnesium and sodiumcations move from the anode compartmentto the cathode compartment. Checkstudents’ diagrams.

Section Review 21.2

Part A Completion1. electric potential

2. electrons

3. cell potential

4. standard hydrogen electrode

5. 0.00 V

6. less

7. spontaneous


9. ST 11. NT

Part C Matching12. b 14. f 16. a

13. d 15. c 17. e

Part D Problem18. Oxidation: Mg yMg2� � 2e�

Reduction: 2e� � Cl2y 2Cl�

Redox: Mg � Cl2yMg2� � 2Cl�

E0cell � E0

red � E0oxid

E0cell � E0

Cl2� E0

Mg

E0cell � �1.36 V � (�2.37 V)

E0cell � �3.73 V

Section Review 21.3

Part A Completion1. electrolysis 5. electrolyte

2. electrolytic cell 6. hydrogen/oxygen

3. electrons 7. oxygen/hydrogen

4. battery 8. hydrogen gas


10. ST 12. AT


14. d 16. c

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Part D Questions and Problems18. In electrolytic cells, electrical energy is used

to bring about a normally nonspontaneouschemical reaction. In a voltaic cell, chemicalenergy is converted to electrical energy by aspontaneous redox reaction. Electrolytic cellsare used in electroplating, in refining metals,and in the production of substances such assodium hydroxide, aluminum, sodium, andchlorine. Voltaic cells are used in pacemakers,hearing aids, and cameras.

19.

Anode (oxidation): Ag(s) y Ag�(aq) � e�

Cathode (reduction): Ag�(aq) � e�y Ag(s)

Practice Problems

Section 21.21. a. Cl2(g) + Mg(s) y 2Cl�(aq) + Mg2�(aq)

E0cell � 1.36 V � (�2.37 V) � 3.73 V

cathode: Cl2(g) � 2e�y 2Cl�(aq)

b. 2Ag�(aq) � Ni(s) y Ni2�(aq) � 2Ag(s)

E0cell � 0.80 V � (�0.25 V) � 1.05 V

cathode: Ag�(aq) � e�y Ag(s)

c. 2MnO4�(aq) � 16H�(aq) � 5Cd(s)

y 5Cd2�(aq) � 2Mn2�(aq) � 8H2O(l)

E0cell � 1.51 V � (�0.40 V) � 1.91 V

cathode: MnO4�(aq) � 8H�(aq) � 5e�

yMn2�(aq) � 4H2O(l)

d. Br2 � 2Na(s) y 2Na�(aq) � 2Br�(aq)

E0cell � 1.07 V � (�2.71 V) � 3.78 V

cathode: Br2(l) � 2e�y 2Br�(aq)

e. MnO2(s) � 4H�(aq) � H2(g) y 2H�(aq) � Mn2�(aq) � 2H2O(l)

E0cell � 1.28 V � 0.00 V � 1.28 V

cathode: MnO2(s) + 4H�(aq) � 2e�

yMn2�(aq) � 2H2O(l)

2. a. E0cell � �0.14 V � (�2.90 V) � �2.76 V;

spontaneous

b. E0cell � �0.80 V � 1.36 V ��0.56 V;

nonspontaneous

c. E0cell � �2.87 V � (�0.76 V) ��3.63 V;

spontaneous

d. E0cell � �0.28 V � (�3.05 V) � �2.77 V;

spontaneous

e. E0cell � �2.93 V � 0.54 V � �3.47 V;

nonspontaneous

Interpreting Graphics 211. anode(�)

2. cathode(�)

3. electrorefining

4. a. The anode(�) of the electrolytic cellshould be connected to the positive(�)terminal of the battery. The cathode(�) ofthe electrolytic cell should be connectedto the negative(�) terminal of the battery.

b. The anode of the electrolytic cell isconnected to the cathode of the battery.The cathode of the electrolytic cell isconnected to the anode of the battery.

5. Oxidation occurs at the anode, labelednumber 1 in the diagram. Reduction occursat the cathode, labeled number 2 in thediagram.

6. Students should indicate the flow of electronsout of the anode(�) and into the cathode(�).

7. The voltage should be great enough to oxidizecopper metal at the anode and reducecopper(II) ions at the cathode, but not highenough to oxidize other metals at the anodeand reduce them at the cathode. The voltageshould be greater than 0.34 V but less than0.44 V.

8. a. gold, silver, and platinum

b. Zn2� and Fe2�

c. copper

Vocabulary Review 211. voltaic cell

2. fuel cell

3. electrochemical cell

4. electrochemical process

5. cathode

6. reduction potential

Solution: aluminum

Solution of silver ions

DC

AgAnode

e� e�

Cathode




Quiz for Chapter 211. NT 7. NT 13. NT

2. AT 8. AT 14. AT

3. NT 9. AT 15. NT

4. AT 10. ST 16. AT

5. ST 11. NT 17. AT

6. NT 12. AT 18. NT

Chapter 21 Test A

A. Matching1. b 5. j 9. f

2. h 6. g 10. a

3. d 7. i

4. c 8. e

B. Multiple Choice11. d 16. a 21. c

12. b 17. d 22. c

13. c 18. c 23. a

14. c 19. a 24. a

15. c 20. a 25. c

C. True-False26. NT 28. AT 30. NT

27. NT 29. NT

D. Question31.

E. Essay32. In both voltaic and electrolytic cells,

oxidation occurs at the anode and reductionoccurs at the cathode. In the voltaic cell, thecathode is positive and the anode is negative.In the electrolytic cell, the anode is positiveand the cathode is negative.

F. Additional Questions33. The negative value means that the tendency

for zinc ions to be reduced is less than that ofhydrogen ions to be reduced, so zinc metal isoxidized when paired with the standardhydrogen half-cell.

34. The reduction potential of a half-cell is ameasure of the tendency of a given half-reaction to occur as a reduction.

35. a. Ni2�(aq) � 2e�y Ni(s) E0 � �0.25 V

Zn2�(aq) � 2e�y Zn(s) E0 � �0.76 V

E0cell � E0

red � E0oxid

E0cell � �0.76 V � (�0.25 V)

E0cell � �0.51 V

This reaction is not spontaneous.

b. 2[Al3�(aq) � 3e�y Al(s)] E0 � �1.66 V

3[Co2�(aq) � 2e�y Co(s)] E0 � �0.28 V

E0cell � E0

red � E0oxid

E0cell � �1.66 V � (�0.28 V)

E0cell � �1.38 V

This reaction is not spontaneous.

Chapter 21 Test B

A. Matching1. h 5. c 9. b

2. d 6. g 10. a

3. f 7. j

4. i 8. e

B. Multiple Choice11. c 17. d 23. b

12. d 18. a 24. a

13. b 19. d 25. b

14. c 20. c 26. c

15. d 21. b 27. d

16. a 22. c

C. True-False28. NT 30. AT 32. AT

29. NT 31. AT 33. ST

Anode Cathode

Saltbridge

e�

e�

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D. Question34.

35. Cu(s) y Cu2�(aq) � 2e�

2Ag�(aq) � 2e�y 2Ag(s)

Net: Cu(s) � 2Ag�(aq) y Cu2�(aq) � 2Ag(s)

E0cell � E0

red � E0oxid

E0cell � �0.80 V � (�0.34 V)

E0cell � �0.46 V

E. Essay36. In both voltaic and electrolytic cells,

electrons flow from the anode to the cathodethrough the external circuit, reduction occursat the cathode, and oxidation occurs at theanode. However, while the flow of electronsin a voltaic cell is caused by a spontaneouschemical reaction, in an electrolytic cell theflow of electrons is being pushed by anoutside source such as a battery. Additionally,while the anode is the negative electrode andthe cathode is the positive electrode in avoltaic cell, the reverse is true in anelectrolytic cell—the anode in an electrolyticcell is the positive electrode and the cathodeis negative.

F. Additional Questions37. Since Al is above Pb in the reduction potential

table, Al is oxidized and Pb is reduced. Thus,the two half-cell reactions are as follows:

2[Al(s) y Al3�(aq) � 3e�]

3[Pb2�(aq) � 2e�y Pb(s)]

Net: 2Al(s) � 3Pb2�(aq) y 2Al3�(aq) � 3Pb(s)

E0cell � E0

red � E0oxid

E0cell � �0.13 V � (�1.66 V)

E0cell � �1.53 V

38. a. The half-reactions are:

Oxidation: Na(s) y Na�(aq) � e�

E0 � �2.71 V

Reduction: Cu2�(aq) � 2e�y Cu(s)

E0 � �0.34 V

E0cell � E0

red � E0oxid

E0cell � �0.34 V � (�2.71 V)

E0cell � �3.05 V

Since the standard cell potential ispositive, the redox reaction will bespontaneous.

b. The half-reactions are:

Oxidation: Ag(s) y Ag�(aq) � e�

E0 � �0.80 V

Reduction: Mg2�(aq) � 2e�yMg(s)

E0 � �2.37 V

E0cell � E0

red � E0oxid

E0cell � �2.37 V � (�0.80 V)

E0cell � �3.17 V

Since the standard cell potential isnegative, the redox reaction will benonspontaneous.

c. The half-reactions are:

Oxidation: Al(s) y Al3�(aq) � 3e�

E0 � �1.66 V

Reduction: Zn2�(aq) � 2e�y Zn(s)

E0 � �0.76 V

E0cell � E0

red � E0oxid

E0cell � �0.76 V � (�1.66 V)

E0cell � �0.90 V

Since the standard cell potential ispositive, the redox reaction will bespontaneous.

39. Reaction 1: since Z is oxidized and X2� isreduced, Z appears above X in the activityseries.

Reaction 2: since W is not oxidized in thepresence of X2�, W should appear below X.

Reaction 3: since Y is oxidized and Z2� isreduced, Y appears above Z.

The elements should be listed as follows: Y, Z, X, and W.

CuAnode

Cu(NO3)2 (aq) AgNO3 (aq)

AgCathode

Saltbridge

e�

e�





Section 21.3 Electrolysis of Water,page 684

Analysis

1. Pure water has too few ions to carry anelectric current.

2. Sodium sulfate is an electrolyte. Itdissociates into ions in solution, which carryan electric current through the solution.

3. The bubbles are H2(g) and the blue BTBsolution indicates the presence of OH� ions.

4. The bubbles are O2(g) and H� ions insolution impart the yellow color to the BTBsolution.

5. 2H2O � 2e��→ H2(g) � 2OH�

H2O → �12

� O2(g) � 2H� � 2e��

3H2O → H2(g) � �12

� O2(g) �2OH� � 2H�

3�H2O → H2(g) � �12

� O2(g) � 2H2O�H2O → H2(g) � �

12

� O2(g)

You’re the Chemist1.

The bubbles are H2(g) and the blue solutionindicates of the presence of OH� ions. Theyellow solution is I2(aq), which is black inthe presence of starch.

2. NaCl:Cathode: 2H2O � 2e� → H2(g) � 2OH�

Bubbles, blue BTBAnode: 2Cl� → Cl2(aq) � 2e�

Yellow solution.

KBr:Cathode: 2H2O � 2e� → H2(g) � 2OH�

Bubbles, blue BTBAnode: 2Br� → Br2(aq) � 2e�

Yellow solution.

CuSO4:Cathode: Cu2� � 2e� → H2(g) � 2OH�

Copper plates out.Anode: H2O → �

12

� O2(g) � 2H� � 2e�

Bubbles, yellow BTB

3. Eo

2H2O � 2e� → H2(g) � 2OH� 0.00 V

H2O → �12

�O2(g) � 2H� � 2e� �0.82 V

2I� → I2(aq) � 2e� �0.54 V

I� is more likely to oxidize (lose electrons)than H2O because it has a more favorable(more positive) Eo value.

Section Review 22.1

Part A Completion1. carbon 6. straight

2. organic 7. branches

3. hydrocarbons 8. alkyl

4. four 9. longest

5. single 10. parent


12. ST 14. ST

Part C Matching16. d 18. e 20. b

17. a 19. f 21. c

Part D Questions and Problems22. 2,2-dimethylbutane

23. a. 16 b. 16

24.

CH3 2 C 2 CH 2 CH2 2 CH 2 CH2 2 CH2 2 CH3

CH3

CH3

@

CH2@

CH3@

CH3@

@

Figure B

KI KI � starch KI � BTB

Bubbles atthe cathode.

Black solutionat the anode.

Bubbles atthe cathode.Yellow soln

at the anode.

Bubbles andblue soln atthe cathode.Yellow soln

at the anode.

Figure A

H2O

No visiblereaction

Bubbles atboth the

anode andcathode

Anode areaturns yellow,cathode areaturns blue

Na2SO4 Na2SO4 � BTB

Answer Key 851

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Section 22.2

Part A Completion1. unsaturated 6. alkane

2. double 7. -ene

3. triple 8. double bond

4. longest 9. -yne

5. double


11. ST 13. ST


15. d 17. b

Part D Questions and Problems18. 3-methyl-2-hexene

19. 2,3,4,5-tetramethylnonane

20. 4-methyl-1-hexene

21.

Section 22.3

Part A Completion1. molecular 7. trans

2. structures 8. stereoisomer

3. butane 9. asymmetric

4. properties 10. optical

5. Geometric 11. mirror

6. cis 12. superimposed


14. ST 16. AT

Part C Matching17. h 20. c 23. a

18. e 21. f 24. g

19. d 22. b

Part D Problems25.

26. Carbon 2 is the asymmetric carbon.

27.

Section 22.4

Part A Completion1. cyclic 7. Methylbenzene

2. aromatic 8. xylenes

3. six 9. ortho, o

4. hydrogen 10. meta, m

5. double 11. para, p

6. resonance

Part B True-False12. ST 14. ST 16. ST

13. AT 15. AT


18. a 20. b


21.

22. CH2 3 CH 2 CH 2 CH3@

2 CH2CH2CH3

CH3 CH2 CH2 CH3

C 3 CH H

%

%^

^

cis-3-hexene

C 3 CCH3 CH2

H

H

CH2 CH3

%

%^

^

trans-3-hexene

CH3 2 C 2 CH3

CH3

CH3

@

@

CH3 2 CH 2 CH2 2 CH3

CH3

@

CH3 2 CH2 2 CH2 2 CH2 2 CH3

CH3 2 CH 2 CH2 2 CH 2 C 4 C 2 CH 2 CH 2 CH 2 CH3

CH3@

CH3@

CH2@

CH3@

CH3@




23. a. cyclooctane

b. 1,3-diethylbenzene or m-diethylbenzene

Section 22.5

Part A Completion1. natural gas 6. boiling point

2. coal 7. lignite

3. methane 8. bituminous

4. aliphatic 9. anthracite

5. distillation 10. aromatic


12. NT 14. ST


16. d 18. c

Part D Problems20. 2C5H12(l) � 11O2(g) → 10CO(g) � 12H2O(g)

21. 2C6H6 � 15O2 → 12CO2 � 6H2O

Practice ProblemsSection 22.1

1. 5-ethyl-3,3,5-trimethyloctane

2. 3-ethyl-2,3,5,5-tetramethylheptane

3. a.

b.

c.

4. heptane: CH3—CH2—CH2—CH2—CH2—CH2—CH3

octane:CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH3

5. 19

Section 22.21. 2,4-dimethyl-2-hexene

2. 3,4-dimethyl-1-pentyne

3. 1-pentyne: CH�C—CH2—CH2—CH3

2-pentyne: CH3—C�C—CH2—CH3

3-methyl-1-butyne:

4. a.

b.

c.

Section 22.31. cis-2-pentene

2. trans-6-methyl-3-heptene

3.

4. a, d

5. Carbon 3 is the asymmetric carbon.

6. a, c

Section 22.41. 1-ethyl-3-methylbenzene

2. 5-phenyl-2-hexene

3. a.

b.

c. 2 CH2 2 CH3

CH3 2 CH2 2

H HC 3 C

%

%^

^

^% %

^ %

@

@ @

CH

CH3

CH2

CH2

CH2 CH3

CH

CH2

C 3 CCH3

H

H

CH2 2 CH2 2 CH2 2 CH3

%

%^

^

CH3 @

CH 4 C 2 C 2 CH3 @

CH3

CH2 3 CH 2 CH2 2 CH 2 CH3 @

CH3

CH2 3 CH 2 CH 2 CH2 2 CH 2 CH3 @ @

CH3 CH3

CH 4 C 2 CH 2 CH3 @

CH3

CH3 2 CH 2 CH 2 CH 2 CH 2 CH2 2 CH2 2 CH2 2 CH3

CH3

@

CH3

@

CH3

@

CH3

@

CH3 2 CH2 2 CH 2 CH 2 CH2 2 CH3

CH2

@

CH3

@

CH2

@

CH3

@

CH3 2 CH 2 C 2 CH2 2 CH3

CH3

CH2CH3

@

@ @

CH3

@

Answer Key 853

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Section 22.51. 2C8H18 � 25O2y 16CO2 � 18H2O

2. The refining process yields fractions thatdiffer with respect to the length of the carbonchains. Natural gas contains mainly lowmolar mass, straight-chain alkanes—methane, ethane, propane, and butane.Gasoline is composed of alkanes with five totwelve carbon atoms. Kerosene is composedof alkanes with twelve to fifteen carbonatoms.

Interpreting Graphics 221. A

2. C

3. 3,3-dimethylpentane

4. D and F; cis-2-pentene and trans-2-pentene

5. E; 2-phenylbutane; C-2 is asymmetric

6. 22

7. 10

8. Compounds A (2-phenylpropane) and E (2-phenylbutane) are aromatic compounds.

9. Compounds D and F are geometric isomers.Like all isomers, they are differentcompounds with different properties. Thus,the boiling points of compounds D and F arenot expected to be the same. (In fact, theboiling points of trans-2-pentene and cis-2-pentene are 36.3C and 36.9C respectively, asmall but measurable difference.)

Vocabulary Review 221. alkynes

2. substituent

3. cis configuration

4. homologous series

5. cracking

6. arene

7. stereoisomers

8. saturated compounds

Solution: hydrocarbons

Chapter 22 Quiz1. ST 5. NT 8. AT

2. NT 6. NT 9. ST

3. NT 7. NT 10. AT

4. AT

11. 2,5,7-trimethyl-3-octene

12.

Chapter 22 Test A

A. Matching1. g 5. d 8. e

2. i 6. c 9. j

3. a 7. f 10. b

4. h


12. c 17. c 21. c

13. a 18. d 22. c

14. b 19. a 23. b

15. d

C. Problems24. 2-methyl-2-phenylbutane

25.

26.

27. 2C6H14(l) � 19O2(g) → 12CO2(g) � 14H2O(l)

D. Essay28. 1. Find the root word (ending in -ane) in the

hydrocarbon name. Then draw the longestcarbon chain to form the parent structure. 2.Number the carbons in the chain. 3. Identifythe substituent groups. Attach thesubstituents to the numbered chain at theproper positions. 4. Add hydrogens as neededso that each carbon added has four bonds.

E. Additional Problems29. Carbon 3 is asymmetric.

CH3 2 C 3 CH 2 C 2 CH2 2 CH 2 CH3

@

CH3@

CH2

@

CH3

@

CH3

H 2 C 2 C 2 C 2 C 2 H and H 2 C 2 C 2 C 2 H

H H H H

H H H

@

H@

H@

@

@

@

@ @@ @

@

CH3

@

H@

H@

H@

H@

2 C 2 C 2 C 2 C 2 C 2 H

HH H

H H

@

@

H@

H@

@ @

H@

@

@

@

@

@




30.

F. True-False31. AT 33. AT

32. AT 34. NT

Chapter 22 Test B

A. Matching1. b 5. h 8. c

2. i 6. a 9. j

3. f 7. d 10. e

4. g

B. Multiple Choice11. d 17. b 23. c

12. c 18. d 24. d

13. b 19. d 25. a

14. b 20. d 26. d

15. c 21. b 27. b

16. d 22. c 28. d


30. AT 34. AT 37. NT

31. NT 35. AT 38. ST

32. AT

D. Problems39. a.

b.

c.

40. a. 3,5-diethyl-4-methylheptaneb. 3-ethyl-2,4,4-trimethyl-2-pentenec. 6,7-diethyl-2,8-dimethyl-5-propyl-3-

decyne

41. a. CH3—CH2—CH2—CH2—CH3 pentane

b.

2-methylbutane

c.

2,2-dimethylpropane

42.

E. Essay43. a. hexane C6H14

CH3 —CH2—CH2—CH2—CH2—CH3

b. 2-hexene C6H12

CH3—CH�CH—CH2—CH2—CH3

c. 2-hexyne C6H10

CH3—C�C—CH2—CH2—CH3

The number of hydrogen atoms decreaseswhen carbon atoms form double or triplebonds in the alkene and alkyne, respectively.The number of hydrogen atoms is at amaximum in the unsaturated alkane.

F. Additional Problems44. Carbon 3 is asymmetric because there are

four different groups attached to it—amethyl, an ethyl, propyl groups, plus a 3-carbon branched chain.

45. a. ethylbenzene

b. 2,3-dimethyl-3-phenylhexane

C 3 CCH3 H

H CH3%

^

%

^

trans-2-butene

C 3 CCH3 CH3

H H%

^

%

^

cis-2-butene

CH3 2 C 2 CH3

CH3

CH3

@

@

CH3 2 CH 2 CH2 2 CH3

CH3

@

CH3 2 C 4 C 2 CH 2 C 2 CH 2 CH 2 CH3

CH3

CH2

@

CH2@

CH3@

CH3@

@

CH2

@

CH3

@

CH3 2 C 3 C 2 C 2 CH2 2 CH3

CH3

CH2

@

CH3@

@

CH3

@

CH3 2 CH 2 CH2 2 CH2 2 CH3

CH3@

CH3@

C 3 CCH3

CH3

H

H

CH 2 CH2 2 CH3@

%

%^

^

C 3 CCH3

CH3

H H

CH 2 CH2 2 CH3

@

%

%^

^

Answer Key 855

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Chapter 22 Small-Scale Labs

Section 22.3 Hydrocarbon Isomers,page 708

Analyze1.

2. To find the number of hydrogen atoms onany carbon of a line-angle formula, countthe number of lines drawn to any point andsubtract from four.

3.

You’re the Chemist1.

2.

Section Review 23.1

Part A Completion1. functional 6. substitution

2. reactive/functional 7. hydrogen

3. alkenes 8. bromine

4. alkynes 9. alcohol

5. Halocarbons 10. salt


12. NT 14. AT

C 3 C 2 CH3

H%

^

2-ethyl-1-propene

H

@

CH2CH3

C 3 CH

H H%

^

%

^

3-methyl-1-butene

CCH3

CH3%

^

H^

C 3 CCH3 H

CH3CH2 H%

^

%

^

2-methyl-1-butene

C 3 CCH3 CH2CH3

H H%

^

%

^

cis-2-pentene

C 3 CCH3 H

H CH2CH3%

^

%

^

trans-2-pentene

C 3 CH

H CH2CH2CH3%%

^

1-pentene

H@

2,2-dimethylbutane

2,3-dimethylbutane

3-methylpentane

2-methylpentane

hexane

CH3CHCH3

methylpropane

CH3

butane

CH3CH2CH2CH3

H 2 C 2 C 2 C 2 C 2 C 2 H

H

H@

@

H

H@

@

H

H@

@

H

H@

@

H

H@

@

H 2 C 2 C 2 C 2 C 2 H

H 2 C 2 H

H

H

H@

H@

@

@

@

@

@H

H@

@

H

H@

@

H 2 C 2 C 2 C 2 H

H 2 C 2 H

H 2 C 2 H

H

H

@

@

@

@

pentane

2-methylbutane

2, 2-dimethylpropane




Part C Matching15. d 17. a

16. b 18. c

Part D Problems19. a. hydroxyl

b. carbonyl

c. carbonyl

d. carboxyl

20. a. b.

Section Review 23.2

Part A Completion1. Alcohols 7. hydration

2. primary 8. water

3. secondary 9. hydrogenation

4. tertiary 10. alkane

5. secondary 11. ethers

6. hydrogen bonding 12. lower

Part B True-False13. AT 15. AT

14. NT 16. NT

Part C Matching17. e 19. d 21. c

18. a 20. b

Part D Problems22. a. tertiary

b. primary

23. a.

b.

Section Review 23.3

Part A Completion1. oxygen

2. double

3. ketones/carboxylic acids

4. ketones/carboxylic acids

5. aldehyde

6. carboxylic acid

7. formaldehyde

8. carboxylic acids

9. esters

10. propanol

11. oxidation–reduction

12. potassium dichromate

Part B True-False14. AT 16. NT

15. NT 17. AT

Part C Matching18. c 20. e 22. a

19. d 21. b

Part D Questions and Problems23.

24. 3–hexanone

Section Review 23.4

Part A Completion1. polymer 4. Polyethylene

2. Addition 5. polyesters

3. Condensation 6. length

Part B True-False7. ST 9. NT 11. AT

8. AT 10. AT

R 2 C 2 H uuuy R 2 C 2 H uuuy R 2 C 2 OH

OH

H

O O

alcohol aldehyde carboxylicacid

oxidation–2H

R 2 C 2 R uuuy R 2 C 2 R

OH

H

O

alcohol ketone

oxidation–2H

K2 Cr2 O7

H2 SO4

#@

@

@

@ # #

CH3 2 C 2 C 2 H

OH H

H H

@ @

@ @

H

H

CH3 2 C 2 C 2 CH3

H

H

@ @

@ @

CH3 C CH2 CH2 CH3

Cl

Cl

@

@

ClCH3

^

@

Answer Key 857

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reserv

ed.



13. c 15. e

Part D Questions and Problems17.

18. Polyesters are polymers consisting of manyrepeating units of dicarboxylic acids anddihydroxy alcohols joined by ester bonds.Dacron™ is one example of a polyester.

Practice Problems

Section 23.11. a. ether c. halogen

b. carboxyl d. hydroxyl

2. They are all halocarbons.

a. m-bromobenzene

b. 1-bromo-1-chloroethane

c. chloroethene (vinyl chloride)

3. a. � NaBr

b. CCl4 � 4HCl

4. a.

b.

Section 23.21. a. 2-butanol; secondary

b. ethylphenyl ether

c. 3-methyl-1-butanol; primary

d. 1-pentanol; primary

2. dipropyl ether:

2-methyl-1-butanol:

2,3-butanediol:

2,3-butanediol is expected to be most solubledue to its two 2OH groups, which can formhydrogen bonds with water.

3. a.

b.

4. a. addition b. substitution

Section 23.31. a. benzaldehyde

b. 2-butanone

c. 3-methylpentanoic acid

d. ethyl butanoate

e. 3-phenyl-2-propenal

2. a. ethanal (acetaldehyde)

b. propane

c. 1-butanol

3. a. 2-pentanone

b. octanoic acid

c. 1-butene or 2-butene

4. a.

b.

5. a. esterification

b. oxidation-reduction; 1–butanol isoxidized to butanoic acid.

O

OH

K2

H2SO4

Cr2O7CH3 CH2 CH2 CH2 OH uuuy CH3 CH2 CH2 C#

%

H�

O

CH3

CH3

CH 2 OH � CH3 CH2 CH2 CCH3

CH3 OH

O

uy CH3 CH2 CH2 CO 2 CH � H2O

#

#

%

%

%

%^

^

� Br2 uuuy � HBrcatalyst

Br^

Cl

CH3 CH2 CH 3 CH2 � HCl uy CH3 CH2 CH CH3 @

OH

CH3 2 CH 2 CH 2 CH3

OH@ @

CH3 CH2 C CH2 OH

CH3

H

@

@

CH3 CH2 CH2 2 O 2 CH2 CH2 CH3

CH3

CH3 CH2 CH2 2 CH 2 CH 2 CH2 CH2 Br

CH3

@ @

CH2 CH3

Br

@

^

CH 2 CH2 2 CH2 2 OHCH3

CH3

%

^

H

H

H

H xx C 3 C uuy H CH2 2 CH2 H

%

^

^

%




Section 23.41. propene (propylene)

polypropylene

tetrafluoroethene

polytetrafluoroethene (PTFE)

Polypropylene is used extrensively in utensilsand containers. Polytetrafluoroethene, alsoknown as Teflon™, is used to coat nonstickcookware and to make bearings and bushingsin chemical reactors.

2.

Polyethylene terephthalate (PET) is formedfrom the condensation of terephthalic acidand ethylene glycol. One molecule of water islost for each bond formed. Because therepeating units are joined by ester bonds, PETis a polyester.

Interpreting Graphics 231. Only primary and secondary alcohols are

oxidized by dichromate ion. Tertiary alcohols,such as 2-methyl-2-propanol, are notexpected to react. Table 1 shows a change inabsorbance values with time, which indicatesa reaction between ethanol, a primaryalcohol, and the oxidizing agent. The data inTable 2 show no change even after fiveminutes. (The slight fluctuation is due torandom electronic noise in the instrument.)

2. This tertiary alcohol serves as a negativecontrol Investigators use negative andpositive controls to check that a chemicalassay is functioning properly.

3. a. CH3CH2OH CH3CHO

b. Rate ��[CH3

�

CtH2OH]�� k � [CH3CH2OH]

4. a. 0.000; 0.140; 0.304; 0.465; 0.627; 0.766

b.

c. Answers will vary slightly.slope � 0.157 min�1

0.157 min�1 � �2.3

k03�

k � 0.362 min�1

Vocabulary Review 231. functional group

2. aryl halides

3. substitution reaction

4. alcohols

5. hydration reactions

6. hydrogenation

7. ketones

Solution: aspirin

Quiz for Chapter 231.

2. NT 7. NT 12. AT

3. AT 8. NT 13. AT

4. ST 9. AT 14. AT

5. AT 10. NT 15. AT

6. AT 11. ST 16. ST

Chapter 23 Test A

A. Matching1. e 5. d 8. j

2. i 6. h 9. a

3. f 7. c 10. b

4. g

� I2 y � HI

I@

H@

1.000

0.600

0.200

0.800

0.400

0.0000 1 2 3 4 5

Time (min)

Log (absorbance)

K2 Cr2 O7

H2 SO4uuuuy

O

C 2 O 2 CH2 CH2 2 O

O

C 2 2x

# #

CF2 2 CF2 x

xCF2 3 CF2

CH2 2 CH

CH3

x

@

xCH2 3 CH

CH3@

Answer Key 859

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B. Multiple Choice11. b 17. c 23. d

12. c 18. d 24. c

13. a 19. d 25. c

14. b 20. c 26. c

15. d 21. c 27. c

16. a 22. a

C. Problems28. a.

b.

29. a. 2-butanol

b. chlorobenzene/phenyl chloride

30. a. R — X

b. R — O — R

c.

d.

D. Essay31. Polymers are large, chain-like molecules

formed by the covalent bonding of repeatingsmaller molecules, called monomers. Inaddition polymerization, unsaturatedmonomers, such as alkenes, are joined to oneanother. In condensation polymerization,monomers with two functional groups, suchas dicarboxylic acids and dihydroxy alcohols,react in a head-to-tail fashion. Because oftheir malleability, high strength-to-weightratio, and durability, polymers have manycommercial uses such as packaging,insulation, and synthetic fibers.

Chapter 23 Test B

A. Matching1. i 5. j 8. d

2. a 6. b 9. e

3. g 7. f 10. c

4. h

B. Multiple Choice11. c 17. d 23. d

12. a 18. d 24. a

13. b 19. c 25. c

14. c 20. b 26. c

15. c 21. c 27. c

16. c 22. b 28. b

C. Problems29. a. 3–chloro–2–methylpentane

b. 2,3–dimethyl–2–butanol

c. butanal

d. 2–hexanone

e. propyl ethanoate

30. a. aldehyde d. carboxylic acid

b. ester e. ether

c. alcohol f. ketone

31. a. CH3CH2I � KOH → CH3CH2OH � KIethanol

b. CH3 — CH2—CH � CH—CH2—CH2—CH3 � HBr

→ CH3—CH2—CHBr—CH2—CH2—CH2—CH3

3–bromohexane

c.

iodobenzene

D. Essay32. Ethylene glycol is an alcohol with both a high

boiling point and a low freezing point due tointermolecular hydrogen bonding. Ethyleneglycol is soluble in water. When ethyleneglycol is added to the water in a car radiator,the resulting mixture boils at a temperaturehigher than water alone, and freezes at atemperature lower than water alone. Thus,ethylene glycol protects against boiling insummer and freezing in winter.

� I2 y � HI

I^

R 2 C 2 OH

O#

R 2 C 2 O 2 R

O#

� Cl2 uuy � HClcatalyst

Cl@

H 2 C 2 C 3 C 2 C 2 H � HOH

H HH H

H

H H

@

@

H@

@ @ @

H HH H@ @ @ @

@ @

OH H@ @

y H 2 C 2 C 2 C 2 C 2 H`





Section 23.4 Polymers, page 753

Analysis1. The polymer is a gel-like liquid which is very

viscous. It will not hold its shape like a solidand will flow slowly if left to stand. It wiggles,squirms, and oozes.

2–3.

4.

5.

You’re the Chemist1. Through experimentation, students are able

to produce an amazing variety of polymerswith different properties.

2. The chain is like a polymer because itcontains many repeating units linked end toend. The rings that link two chains togetherare like the borate ion that cross linkspolymer chains.

Section Review 24.1

Part A Completion1. prokaryotic/eukaryotic

2. prokaryotic/eukaryotic

3. bacteria

4. green plants

5. organelles

6. Mitochondria

7. lysosomes

8. nucleus

9. Sunlight

10. Photosynthesis

11. oxygen

Part B True-False12. NT 14. AT 16. NT

13. AT 15. AT

Part C Matching17. b 19. a

18. c 20. d

Part D Question21. Chloroplasts contain the biological

molecules necessary for the conversion ofsolar energy into chemical energy. Plantsstore the excess chemical energy in carboncompounds. Like animals, they meet theirenergy demands by breaking down thesestored compounds. These oxidation reactionstake place in mitochondria.

Section Review 24.2

Part A Completion1. Carbohydrates 6. polysaccharide

2. energy 7. starch

3. cellulose 8. glucose

4. monosaccharides 9. Glycogen

5. disaccharides 10. liver


12. AT 14. AT

Part C Matching16. e 18. c 20. d

17. b 19. a

Part D Questions21. Starches are a source of energy for plants.

Cellulose is used by plants to construct cellwalls that are hard and rigid.

22. glucose and fructose

23. The hydroxyl group, �OH.

O O O O

O O O O

B B

B� 4R OH 4HOHHO OH

HO OHB�

RO OR

RO OR� �

OH OH OH OH

Answer Key 861

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Section Review 24.3

Part A Completion1. amino acid 6. water

2. side-chain group 7. protein

3. side-chain group 8. catalysts

4. peptide 9. enzymes

5. peptide


11. NT 13. AT

Part C Matching14. b 16. c

15. a

Part D Problem17.

Section Review 24.4

Part A Completion1. lipid

2. not soluble/insoluble

3. triglycerides

4. Triglycerides

5. Saponification

6. glycerol

7. Phospholipids

8. hydrophilic/polar

9. hydrophobic/nonpolar

10. lipid bilayer

11. Cell membranes

Part B True-False12. AT 14. AT 16. ST

13. NT 15. AT


18. b 20. a

Part D Questions21. Wax coats on the surface of plant leaves

protect against water loss and attack bymicroorganisms. In animals, waxes coat theskin, hair, and feathers, which keep thesestructures pliable and waterproof.

22. The molecules of both types of lipids havehydrophilic and hydrophobic ends. Thus,both types of lipids can interact with polarand nonpolar phases simultaneously. Thecleansing action of soaps relies on thisphysical property.


1. nucleotide

2. deoxyribonucleic acid

3. ribonucleic acid

4. proteins

5. nitrogen base

6. adenine, guanine, thymine, or cytosine




10. uracil

11. double helix

12. hydrogen bonds

13. thymine

14. cytosine


16. AT 18. AT

Part C Matching19. b 21. c 23. d

20. e 22. a

Part D Question24. Mutations are random changes in the

sequence of nucleotides in a DNA molecule.Mutations may arise from additions,deletions, or substitutions of one or more ofthe nucleotides. When a mutation occurswithin a gene, it may stop production of thespecified protein or cause production of aprotein with an altered amiono acidsequence. Sometimes the change isbeneficial; more often, the ability of the

H2N 2 C 2 C 2 NH 2 C 2 C 2 O 2 CH3

H H

CH2

@

@

@

CH2

COOH

@

O#

O#

@

@

1 1

3

4

2

5




protein to function is seriously impaired. Forexample, a mutation in the peptide chain ofhemoglobin reduces its ability to transportoxygen. People with this mutation have amolecular disease called sickle cell anemia,which is named for the distorted shape of thedefective red blood cells.

Section Review 24.6

Part A Completion1. adenosine triphosphate

2. adenosine diphosphate -ADP

3. oxidation

4. 30.5 kJ

5. nonspontaneous

6. catabolic or anabolic

7. catabolic or anabolic

8. metabolism

9. catabolism

10. anabolism

Part B True-False11. AT 13. NT 15. NT

12. AT 14. AT


17. b 19. a

Part D Question21. The free energy of ATP hydrolysis is used to

drive many nonspontaneous biologicalreactions. ATP hydrolysis provides the extraenergy needed to shift the equilibrium of anonspontaneous reaction in favor of theproducts. ATP captures energy fromcatabolism reactions to drive anabolismreactions.

Interpreting Graphics 24

Part A1. cytoplasm 4. nucleus

2. cell membrane 5. cytoplasm

3. cell wall 6. cell membrane

Part B1. The cell in panel a represents a prokaryotic

cell. It lacks a nucleus and organelles, whichare present in eukaryotic cells such as the onedepicted in panel b. Prokaryotic cells are thecells of bacteria. The cells of all otherorganisms are eukaryotic.

2. Figure 1a: 0.0001 to 0.001 mm, 0.1 to 1.0 �m,100 to 1000 nm.

Figure 1b: 0.001 to 0.01 mm, 1.0 to 10.0 �m,1000 to 10,000 nm.

3. All of the organelle types labeled in Figure 1bare found in a typical plant cell. Plant cells areeukaryotic.

4. Plant cells contain chloroplasts, structuresthat enable plants to produce carbohydratesthrough photosynthesis. Plant cells have cellwalls.

5. ATP is produced in the mitochondrion andtransported out to the cytoplasm, where it isused to fuel nonspontaneous processes.

6. Mitochondria produce energy needed forcellular activities. Muscle cells are highlyactive cells, which require manymitochondria to meet their energy demands.Skin cells are less active. They containsignificantly fewer mitochondria.

7. Carbohydrates are found in the cytoplasmand are attached to the extracellular surfacesof membrane-bound proteins. Theycomprise the cell walls, which providestructure and rigidity to plant cells. Becauseproteins catalyze metabolic reactions, theyare found throughout the cell. Proteinsembedded in cell membranes help totransport molecules and ions across thisbarrier. Lipids are found mainly in cell andorganelle membranes where they form abarrier to the free flow of ions and moleculesinto and out of the membrane-enclosedcompartments. DNA, a molecule that storesthe information needed to make proteins, isfound primarily in the nucleus of eukaryoticcells and in the cytoplasm of prokaryoticcells. RNA, a molecule that participates in thetransfer of information between DNA andprotein, is found in the cytoplasm of all cells.

Answer Key 863

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Vocabulary Review 241. c 5. k 9. i

2. f 6. h 10. d

3. b 7. e 11. a

4. g 8. j

Quiz for Chapter 241. NT 4. AT 7. b

2. AT 5. ST 8. b

3. NT 6. d 9. d

Chapter 24 Test A

A. Matching1. e 5. c 8. d

2. b 6. f 9. h

3. g 7. a 10. j

4. i

B. Multiple Choice11. d 16. d 21. d

12. b 17. a 22. d

13. b 18. c 23. c

14. d 19. c 24. a

15. b 20. b 25. c

C. True-False26. AT 30. ST 33. NT

27. AT 31. AT 34. AT

28. NT 32. NT 35. AT

29. ST

D. Questions and Problems36. Trp-Arg-Ala-Leu-Asn-end

37. a. ATP � H2O → ADP � Pi

b. Keq � �[A

[DA

PT

]P[]Pi]� 1

c. spontaneous

38. Phosphorus is essential for the synthesis ofphospholipids, nucleic acids, and energy-richmolecules such as ATP. Phosphorus is foundin the bloodstream as HPO4

2� and H2PO4�,

which together form an important buffer.Nitrogen is required for the synthesis ofamino acids and nitrogen-containing basesof nucleic acids.

39. Test the aqueous solubility of the substance.Many carbohydrates are soluble in water,whereas lipids are not.

40. Eukaryotic cells contain a nucleus and othermembrane-enclosed structures calledorganelles. Prokaryotic cells do not contain anucleus or organelles. Eukaryotic cells aretypically much larger than prokaryotic cells.

41. a. A membrane protein that acts as achannel must have contacts inside andoutside the cell membrane. To span theentire bilayer, membrane proteins musthave dimensions similar to the observedthickness of the lipid bilayer. The lengthalong the transmembrane axis of theprotein must be approximately 5 to 10 nm.

b. Because the lipid bilayer is composed oftwo sheets of phospholipid moleculesarranged tail to tail, each phospholipidmolecule must be approximately 2.5 to 5 nm long from head to tail.

E. Essay42. Nucleic acids are polymers found primarily in

cell nuclei. They are composed of nucleotidesthat contain a phosphate group, a five-carbon sugar, and a nitrogen-containingbase. There are two types: DNA and RNA.DNA stores the information needed to makeproteins. DNA governs the reproduction andgrowth of cells. RNA has a key role in thetransmission of the information stored inDNA.

Chapter 24 Test B

A. Matching1. c 5. b 8. e

2. d 6. a 9. h

3. g 7. f 10. j

4. i

B. Multiple Choice11. c 16. d 21. d

12. d 17. d 22. b

13. b 18. c 23. c

14. a 19. c 24. c

15. d 20. b 25. d




C. True-False26. NT 30. ST 33. AT

27. AT 31. AT 34. AT

28. AT 32. AT 35. NT

29. ST

D. Questions and Problems36. More than one answer is possible due to the

redundancy of the genetic code. Onepossibility: ACAGTTGGTACT

37. An enzyme catalyzes the conversion of asubstrate to product. Doubling and triplingthe number of enzyme molecules in thereaction mixture is equivalent to doublingand tripling the number of active sites towhich substrate can bind. Thus, when allother conditions are kept the same, the rate atwhich product is formed will increase withthe number of enzyme molecules present inthe reaction system.

38. Test the aqueous solubility of the substance.Most proteins are soluble in water, whereaslipids are not.

39.

40. Two dipeptides are possible. One possibility:

41. The extent to which the physical properties ofa cell membrane are altered by a substancemay depend on the solubility of thesubstance in the lipid bilayer. Beacuase theinterior of the lipid bilayer is a hydrophobicenvironment, nonpolar substances have thegreatest chance of becoming incorporatedinto this protion of the cell membrane.

E. Essay42. Enzymes are proteins that, like act as

biological catalysts. They reduce the timerequired for a chemical reaction to reachequilibrium, but do not change the normalposition of the equilibrium. Enzymes are notchanged by the reactions they promote. Themolecules on which an enzyme acts are

called substrates. In an enzyme-catalyzedreaction the substrate binds to the active siteon the enzyme form an enzyme-substratecomplex. Next, bond-breaking and bond-making occur at the active site to produce theproducts of the reaction. Finally, the productsdissociate from the enzyme leaving theenzyme free to bind new substrate and begina second reaction cycle.


Section 24.3 The Egg: ABiochemicalStorehouse, page 774

AnalysisSample answers are given.

mass � 62.42 g

1. Shape Index � �45

.

.59

50

cc

m�m�

� � 100 � 77.1

2. V � 0.5236 lw2 � (0.5236)(5.90 cm)(4.55 cm)2

V � 64.0 cm3

M � 0.5632lw2 � (0.5632)(5.90 cm)(4.55 cm)2

M � 68.8 g

A � 3.138 [(5.90 cm)(4.55 cm)2]2/3

A � 77.3 cm2

3. The measured mass of 62.42 g is less thanthe calculated mass of 68.8 g by 6.4 g. Theegg may have lost water over time.

4. d � �mv� � �

6642.0.4

c2mg

3� � 0.975 g/cm3

This is less than the density of a freshly laidegg.

width = 4.55 cm

length = 5.90 cm

H3N 2 C 2 C 2 OH � H2N 2 C 2 C 2 OH JJJH

CH3

H

O#

O#@

@

H

H

@

@

H3N 2 C 2 C 2 N 2 C 2 C 2 OH � H2O

CH3

H

O#

O#@

@

H

H

@

@

H@

2000 kJ� � �1 m

30o.5l A

kTJ�

P� � �

5107

m.2ol

g�AATTPP

� � �110k3

gg�

�

� 33.3 kg ATP

Answer Key 865

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You’re the Chemist1. Weigh the egg once each day for two or three

days. Typical eggs will loose 0.2 to 0.5 gramsper day depending on the temperature atwhich they are stored.

2. Assume Step 1 reveals the egg loses 0.20 gper day.

Age of egg � (68.8 g� � 62.42 g�) � �01.2d0ay

g��

Age of egg � 32 days old

3. Measure the volume of water displaced bythe egg.

4. The larger the egg, the smaller the shapeindex. Extra large eggs tend to be moreoblong than small eggs, which are rounder.

5. Extra large eggs are usually more than 70grams, medium eggs are less than 50 grams.

6. HCl produces bubbles at the surface of theegg shell. 2HCl � CaCO3 → CO2 � H2O � CaCl2

7. Powdered milk � NaOH � CuSO4 producesa violet color, a positive test for protein. Eggshell � NaOH � CuSO4 produces a violetcolor, a positive test for protein.

8. Weigh an egg once a day for three days andstore in a refrigerator between weighings.Weigh it once a day for three more days andstore it at room temperature. The warmerthe temperature the greater the mass loss.


1. radioactive 9. electrons

2. radioisotopes 10. metal foil

3. nuclei 11. Gamma

4. stable 12. mass

5. energy 13. Lead

6. beta 14. concrete

7. Alpha 15. stop

8. helium

Part B True-False16. ST 18. AT 20. AT

17. NT 19. NT


22. a 24. e

Part D Problems26. a. 84

218Po → 42He � 214

82Pb

b. 82210Pb → 210

83Bi � �10e


1. band of stability 7. billions

2. beta 8. transmutation

3. positron 9. radioactive decay

4. rate 10. atomic numbers

5. half-life 11. synthesized

6. radioactive


13. NT 15. ST

Part C Matching17. c 19 b 21. d

18. e 20. a

Part D Questions

22. �61

05

hh

rr

� � 4 half-lives; After 4 half-lives

1/2 � 1/2 � 1/2 � 1/2 � 1/16 of the originalmass will remain. 1/16 � 18.0 g � 1.13 g

23. 2.0 g � �12

� � �12

� � �12

� � 0.25 g or 3 half-lives

42 days 3 � 14 days

Section Review 25.3

Part A Completion1. fission 7. fusion

2. neutrons 8. mass

3. fissionable atom 9. energy

4. energy 10. hydrogen

5. moderation 11. helium

6. absorption


13. NT 15. NT

Part C Matching16. a 18. e 20. d

17. c 19. b




Part D Questions and Problems21. a. 3 b. 4

22. a. slow fast-moving neutrons so they can beabsorbed by the fuel atoms

b. decrease the number of slow-movingneutrons and slow the chain reaction

Section Review 25.4

Part A Completion1. ionizing 6. scintillation

2. electrons 7. all

3. senses 8. iodine-131

4. Geiger 9. phosphorus-32

5. gas 10. neutron activation


12. NT 14. AT


16. e 18. a

Part D Questions20. b

21. Neutron activation analysis is used to detecttrace amounts of elements in samples.Radioisotopes are used to study chemicalreactions and molecular structures.Radioisotopes are used to diagnose and treatdiseases such as cancer.


Section 25.11. The atomic number increases by one; the

mass number remains the same.

2. The atomic number decreases by two; themass number decreases by four.

3. a. 28 protons and 36 neutrons

b. 53 protons and 83 neutrons

c. 79 protons and 116 neutrons

4. a. 714N c. 0

�1e

b. 93237Np

Section 25.21. a. 87

208Fr → 42He � 204

85At

b. 74Be � �1

0e → 73Li

c. 3718Ar → 37

19K � �10e

d. 917F → 17

8O � �10e

2. a. 31H d. 58

144Ce

b. 9236Kr e. 94

239Pu

c. 3015P

3. It takes five half-lives, or 820 s.8.0 g→ 4.0 g→ 2.0 g→ 1.0 g→ 0.50 g→ 0.25 g

4. 16 g → 8 g → 4 g → 2.0 g → 1.0 gFour half-lives � 4 � 17 days � 68 days

5. �55.11

mm

iinn

� � 10 half-lives. The mass would

decrease by a factor of more than 1000.

��12

��10

� �10

124�

6. The mass decreases by a factor of 1/8, orthree half-lives. The half-life is 5.49/3 � 183 s

Section 25.31. a. 2 ([1 � 235] � [90 � 144] � 2)

b. 3 ([1 � 235] � [87 � 146] � 3)c. 4 ([1 � 235] � [72 � 160] � 4)

2. 2.0 � 107 kcal 8.0 kcal/g � 2.5 � 106 g

3. 42He

Section 25.41. Radioisotopes replace non-radioactive

isotopes in the structure of a compoundwithout changing its chemical properties.Tracing the pathways of radioactive isotopesallows scientists to study reactionmechanisms and reaction rates.

2. Teletherapy is the use of gamma radiation todestroy cancerous tissue.

Interpreting Graphics 251. 8

2. 6; 23892U y 206

82Pb � 8 42He � 6 0�1e

3. 8 days � 2 half-lives; 20 � �12

� � �12

� � 5.0 g

4. 20 minutes � 1 half-life; 1.0 mol �6.0 � 1023 � �

12

� � 3.0 � 1023 atoms

5. lead-210; The half-life of polonium-214 is insignificant compared to the half-life ofbismuth-214.

6. Three half-lives � 15 days; 16 g y 8 g y 4 gy 2.0 g.

Answer Key 867

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7. For heavier isotopes, such as lead-206, thestability ratio is about 1.5 neutrons to 1proton. 124 n 82 p � 1.5

8. Uranium-238 has the longest half-life (4.5 � 109 yr); polonium-210 has the shortesthalf-life (1.6 � 10�4 s).

Vocabulary Review 251. b 5. i 9. k

2. l 6. d 10. f

3. g 7. e 11. h

4. a 8. c 12. j

Quiz for Chapter 251. d 5. a 8. b

2. b 6. d 9. a

3. c 7. c 10. a

4. b

Chapter 25 Test A

A. Matching1. a 5. j 8. f

2. i 6. e 9. h

3. g 7. d 10. b

4. c

B. Multiple Choice11. b 17. b 22. a

12. c 18. d 23. b

13. b 19. b 24. a

14. b 20. c 25. d

15. a 21. a 26. b

16. c

C. Problems27. a. 42

19K → �10e � 42

20Ca

b. 92235U → 4

2He � 23190Th

28. If one-eighth of the sample remains, theisotope decayed through 3 half-lives.Three half-lives is 252 days, so one half-lifeperiod � 84 days.The half-life of scandium-42 is 84 days.

29. 40 days � 5 half-lives.

�312� of the original sample remains

� 0.13 gram remaining.

D. Essay30. The energy released from the sun is the

result of a nuclear fusion, or thermonuclearreaction. Fusion occurs when two lightnuclei combine to produce a nucleus ofheavier mass. In solar fusion, hdrogen nuclei(protons) are fused to make helium nuclei.The reaction requires two beta particles.

411H � 2�1

0e → 42He � energy

Chapter 25 Test B

A. Matching1. j 5. e 8. h

2. i 6. g 9. b

3. c 7. a 10. f

4. d

B. Multiple Choice11. a 16. d 21. d

12. d 17. b 22. d

13. c 18. a 23. d

14. b 19. a 24. a

15. a 20. c 25. d

C. Problems26. a. 88

226Ra → 22286Rn � 4

2He

b. 23491Pa → 234

92U � �10e

c. 90234Th → 91

234Pa � �10e

27. 27.0 h/6.75 h � 4.00 half-lives

Thus, 12.0 g � �12

� � �12

� � �12

� � �12

� � 0.750 g

28. � �312� of the sample remains

Since �312� represents ��

12

��5, or 5 half-lives,

�571

h.a5lfy-eli

avress

� � 14.3 years/half-life.

D. Essay

29. a. Chemical reactions occur in an effort toattain stable electron configurations.Nuclear reactions occur in an effort toobtain stable nuclear configurations.

b. Nuclear reactions release far more energythan typical exothermic chemicalreactions.

0.125 g��4.00 g�




c. Unlike chemical reactions, nuclearreactions are unaffected by changes intemperature, pressure, or the presence ofa catalyst.


Section 25.2 Radioactivity and HalfLives, page 809

AnalysisSample data are provided.

1.

2. The rate of disappearance of heads is non-linear. The rate decreases over time.

3. For each flip the probability of a head is 0.50.

4. Each trial represents one half-life becausethe number of heads approximately halvesfor each trial.

You’re the Chemist1. Count the total number of even numbers that

result in 100 rolls of the die. Roll the die againa number of times equal to the numberobtained in the first trial. Do trials until thenumber of events equals zero. Plot number ofevens vs. trial.

2. After 3.8 days, half the sample remains. After7.6 days, one-fourth remains, and, after 11.4days, one-eighth remains.

3. This time period is two half-lives (11,460years/5730 � 2) of carbon-14. After two half-lives, one-fourth of the sample remains.

Trial

Flip

s

100

80

60

40

20

00 2 4 61 3 5 7

1

Trial#

Number offlips

Number ofheads

2

3

5

6

7

4

100

42

20

5

3

1

9

42

20

9

3

1

0

5

Figure A

Answer Key 869

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arson

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