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Copyright © by The McGraw-Hill Companies, Inc.All rights reserved. Permission is granted to reproduce the material contained hereinon the condition that such material be reproduced only for classroom use; be providedto students, teachers, and families without charge; and be used solely in conjunctionwith the Chemistry: Matter and Change program. Any other reproduction, for use orsale, is prohibited without prior written permission of the publisher.

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ISBN 0-07-824533-8Printed in the United States of America.1 2 3 4 5 6 7 8 9 10 045 09 08 07 06 05 04 03 02 01

A Glencoe Program

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Challenge Problems Chemistry: Matter and Change iii

To the Teacher . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

Chapter 1 Production of Chlorofluorocarbons, 1950–1992 . . . . . . . . . 1

Chapter 2 Population Trends in the United States . . . . . . . . . . . . . . . . 2

Chapter 3 Physical and Chemical Changes . . . . . . . . . . . . . . . . . . . . . 3

Chapter 4 Isotopes of an Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Chapter 5 Quantum Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Chapter 6 Döbereiner’s Triads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Chapter 7 Abundance of the Elements . . . . . . . . . . . . . . . . . . . . . . . . 7

Chapter 8 Comparing the Structures of Atoms and Ions . . . . . . . . . . . 8

Chapter 9 Exceptions to the Octet Rule . . . . . . . . . . . . . . . . . . . . . . . . 9

Chapter 10 Balancing Chemical Equations . . . . . . . . . . . . . . . . . . . . . 10

Chapter 11 Using Mole-Based Conversions . . . . . . . . . . . . . . . . . . . . 11

Chapter 12 Mole Relationships in Chemical Reactions . . . . . . . . . . . . 12

Chapter 13 Intermolecular Forces and Boiling Points . . . . . . . . . . . . . 13

Chapter 14 A Simple Mercury Barometer . . . . . . . . . . . . . . . . . . . . . . 14

Chapter 15 Vapor Pressure Lowering . . . . . . . . . . . . . . . . . . . . . . . . . 15

Chapter 16 Standard Heat of Formation . . . . . . . . . . . . . . . . . . . . . . . 16

Chapter 17 Determining Reaction Rates . . . . . . . . . . . . . . . . . . . . . . . 17

Chapter 18 Changing Equilibrium Concentrations in a Reaction . . . . . 18

Chapter 19 Swimming Pool Chemistry . . . . . . . . . . . . . . . . . . . . . . . . 19

Chapter 20 Balancing Oxidation–Reduction Equations . . . . . . . . . . . . 20

Chapter 21 Effect of Concentration on Cell Potential . . . . . . . . . . . . . 21

Chapter 22 Structural Isomers of Hexane . . . . . . . . . . . . . . . . . . . . . . 22

Chapter 23 Boiling Points of Organic Families . . . . . . . . . . . . . . . . . . 23

Chapter 24 The Chemistry of Life . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Chapter 25 The Production of Plutonium-239 . . . . . . . . . . . . . . . . . . . 25

Chapter 26 The Phosphorus Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T27

CHALLENGE PROBLEMS

Contents

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iv Chemistry: Matter and Change Challenge Problems

Students can take their learning one step beyond their textbooks withChallenge Problems. The worksheets in this supplement to Chemistry:Matter and Change challenge students to apply their knowledge ofchemistry to new situations, analyze and interpret those situations, andsynthesize responses. Whether analyzing experimental results or investigatinga hypothetical situation, these worksheets encourage students to use chemicalconcepts along with their critical thinking skills to solve problems.

To the Teacher

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Challenge Problems Chemistry: Matter and Change • Chapter 1 1

Production ofChlorofluorocarbons, 1950–1992Production ofChlorofluorocarbons, 1950–1992

Chlorofluorocarbons (CFCs) were first produced in the laboratory in the late 1920s. They did not

become an important commercial product until sometime later. Eventually, CFCs grew in popularity untiltheir effect on the ozone layer was discovered in the1970s. The graph shows the combined amounts of twoimportant CFCs produced between 1950 and 1992.Answer the following questions about the graph.

CHALLENGE PROBLEMSCHAPTER 1

1. What was the approximate amount of CFCs produced in 1950? In 1960? In 1970?

2. In what year was the largest amount of CFCs produced? About how much was producedthat year?

3. During what two-year period did the production of CFCs decrease by the greatestamount? By about how much did their production decrease?

4. During what two-year period did the production of CFCs increase by the greatestamount? What was the approximate percent increase during this period?

5. How confident would you feel about predicting the production levels of CFCs during theodd numbered years 1961, 1971, and 1981? Explain.

6. Could the data in the graph be presented in the form of a circle graph? Explain.

Year

Am

ou

nt

of

CFC

s(b

illio

n k

ilog

ram

s)

050

100150200250300350400

1950 1960 1970 1980 1990

Use with Chapter 1,Section 1.1

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2 Chemistry: Matter and Change • Chapter 2 Challenge Problems

Population Trends in theUnited StatesPopulation Trends in theUnited States



1. By how much did the total U.S. population increase between 1990 and 2000? What wasthe percent increase during this period?

2. Calculate the total population for each of the five groups for 1990 and 2000.

3. Make a bar graph that compares the population for the five groups in 1990 and 2000. Inwhat ways is the bar graph better than the circle graphs? In what way is it less useful?

U.S. Population Distribution

(Percentages may not add up to 100% due to rounding.)

Caucasian71.4%Native American

0.70%

Asian American3.8%

Hispanic American11.8%

African American12.2%

Caucasian75.7%

Native American0.70%

Asian American2.8%

Hispanic American9.0%

African American11.8%

20001990

The population of the United States is becoming more diverse. The circle graphs below show thedistribution of the U.S. population among five ethnic groups in 1990 and 2000. The estimated

total U.S. population for those two years was 2.488 � 108 in 1990 and 2.754 � 108 in 2000.

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Physical and ChemicalChangesPhysical and ChemicalChanges

Physical and chemical changes occur all around us. One of the many places in which physical and chemical changes occur is the kitchen. For example, cooking spaghetti in a

pot of water on the stove involves such changes. For each of the changes described below, tell(a) whether the change that occurs is physical or chemical, and (b) how you made your choicebetween these two possibilities. If you are unable to decide whether the change is physical orchemical, tell what additional information you would need in order to make a decision.


1. As the water in the pot is heated, its temperature rises.

2. As more heat is added, the water begins to boil and steam is produced.

3. The heat used to cook is produced by burning natural gas in the stove burner.

4. The metal burner on which the pot rests while being heated becomes red as its temperature rises.

5. After the flame has been turned off, a small area on the burner has changed in color fromblack to gray.

6. A strand of spaghetti has fallen onto the burner, where it turns black and begins tosmoke.

7. When the spaghetti is cooked in the boiling water, it becomes soft.


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Isotopes of an ElementIsotopes of an Element

A mass spectrometer is a device for separatingatoms and molecules according to their

mass. A substance is first heated in a vacuum andthen ionized. The ions produced are acceleratedthrough a magnetic field that separates ions of dif-ferent masses. The graph below was producedwhen a certain element (element X) was analyzedin a mass spectrometer. Use the graph to answerthe questions below.



1. How many isotopes of element X exist?

2. What is the mass of the most abundant isotope?

3. What is the mass of the least abundant isotope?

4. What is the mass of the heaviest isotope?

5. What is the mass of the lightest isotope?

6. Estimate the percent abundance of each isotope shown on the graph.

7. Without performing any calculations, predict the approximate atomic mass for elementX. Explain the basis for your prediction.

8. Using the data given by the graph, calculate the weighted average atomic mass of element X. Identify the unknown element.

0

5

10

15

20

25

30

196194192190 198 200 202 204 206 208 210Atomic mass (amu)

Perc

ent

abu

nd

ance

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Quantum NumbersQuantum Numbers


The state of an electron in an atom can be completely described by four quantum numbers,designated as n, �, m�, and ms. The first, or principal, quantum number, n, indicates the

electron’s approximate distance from the nucleus. The second quantum number, �, describesthe shape of the electron’s orbit around the nucleus. The third quantum number, m�, describesthe orientation of the electron’s orbit compared to the plane of the atom. The fourth quantumnumber, ms, tells the direction of the electron’s spin (clockwise or counterclockwise).

The Schrödinger wave equation imposes certain mathematical restrictions on the quantumnumbers. They are as follows:

n can be any integer (whole number),

� can be any integer from 0 to n � 1,

m� can be any integer from �� to ��, and

ms can be � or �

As an example, consider electrons in the first energy level of an atom, that is, n � 1. Inthis case, � can have any integral value from 0 to (n � 1), or 0 to (1 � 1). In other words,� must be 0 for these electrons. Also, the only value that m� can have is 0. The electrons in

this energy level can have values of � or � for ms. These restrictions agree with the

observation that the first energy level can have only two electrons. Their quantum numbers

are 1, 0, 0, � and 1, 0, 0 � .

Use the rules given above to complete the table listing the quantum numbers for eachelectron in a boron atom. The correct quantum numbers for one electron in the atom is provided as an example.

1�2

1�2

1�2

1�2

1�2

1�2


Electron n � m� ms

1 1 0 0 �

2

3

4

5

1�2

Boron (B)

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Döbereiner’s TriadsDöbereiner’s Triads

One of the first somewhat successful attempts to arrange the elements in a systematic way was made by the German chemist Johann Wolfgang Döbereiner (1780–1849). In 1816,

Döbereiner noticed that the then accepted atomic mass of strontium (50) was midway betweenthe atomic masses of calcium (27.5) and barium (72.5). Note that the accepted atomic massesfor these elements today are very different from their accepted atomic masses at the timeDöbereiner made his observations. Döbereiner also observed that strontium, calcium, and bar-ium showed a gradual gradation in their properties, with the values of some of strontium’sproperties being about midway between the values of calcium and barium. Döbereiner eventu-ally found four other sets of three elements, which he called triads, that followed the same pat-tern. In each triad, the atomic mass of the middle element was about midway between theatomic masses of the other two elements. Unfortunately, because Döbereiner’s system did notturn out to be very useful, it was largely ignored.

Had Döbereiner actually discovered a way of identifying trends among the elements?Listed below are six three-element groups in which the elements in each group are consecutivemembers of the same group in the periodic table. The elements in each set show a gradation intheir properties. Values for the first and third element in each set are given. Determine the miss-ing value in each set by calculating the average of the two given values. Then, compare the val-ues you obtained with those given in the Handbook of Chemistry and Physics. Record theactual values below your calculated values. Is the value of the property of the middle elementin each set midway between the values of the other two elements in the set?



Element Melting Point (°C)

Fluorine �219.6

Chlorine Calculated:

Actual:

Bromine �7.2

Set 1

Element Atomic Mass

Lithium 6.941

Sodium Calculated:

Actual:

Potassium 39.098

Set 2

Element Boiling Point (°C)

Magnesium 1107

Calcium Calculated:

Actual:

Strontium 1384

Set 3


Krypton �153

Xenon Calculated:

Actual:

Radon �62

Set 4

Element Melting Point (°C)

Germanium 937

Tin Calculated:

Actual:

Lead 327

Set 5


Beryllium 1285

Magnesium Calculated:

Actual:

Calcium 851

Set 6

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Abundance of the ElementsAbundance of the Elements

The abundance of the elements differs significantly in various parts of the universe. The table below lists the abundance of some elements in various

parts of the universe. Use the table to answer the following questions.


1. What percent of all atoms in the universe are either hydrogen or helium? What percent ofall atoms in the solar system are either hydrogen or helium?

2. Explain the relatively high abundance of hydrogen and helium in the universe comparedto their relatively low abundance on Earth.

3. Only the top four most abundant elements on Earth and in Earth’s crust are shown in thetable. Name two additional elements you would expect to find among the top ten ele-ments both on Earth and in Earth’s crust. Explain your choices.

4. Name at least three elements in addition to those shown in the table that you wouldexpect to find in the list of the top ten elements in the human body. Explain your choices.


Abundance (Number of atoms per 1000 atoms)*

Element Universe Solar System Earth Earth’s Crust Human Body

Hydrogen 927 863 30 606

Helium 71.8 135

Oxygen 0.510 0.783 500 610 257

Nitrogen 0.153 0.0809 24

Carbon 0.0811 0.459 106

Silicon 0.0231 0.0269 140 210

Iron 0.0139 0.00320 170 19

* An element is not abundant in a region that is left blank.

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Comparing the Structures ofAtoms and IonsComparing the Structures ofAtoms and Ions

The chemical properties of an element depend primarily on its number of valence electrons in its atoms. The noble gas elements, for example, all have similar chemical properties

because the outermost energy levels of their atoms are completely filled. The chemical propertiesof ions also depend on the number of valence electrons. Any ion with a complete outermostenergy level will have chemical properties similar to those of the noble gas elements. The fluo-ride ion (F�), for example, has a total of ten electrons, eight of which fill its outermost energylevel. F� has chemical properties, therefore, similar to those of the noble gas neon.

Shown below are the Lewis electron dot structures for five elements: sulfur (S), chlorine (Cl),argon (Ar), potassium (K), and calcium (Ca). Answer the questions below about these structures.



1. Write the atomic number for each of the five elements shown above.

2. Write the electron configuration for each of the five elements.

3. Which of the above Lewis electron dot structures is the same as the Lewis electron dotstructure for the ion S2�? Explain your answer.

4. Which of the above Lewis electron dot structures is the same as that for the ion Cl�?Explain your answer.

5. Which of the above Lewis electron dot structures is like that for the ion K�? Explainyour answer.

6. Name an ion of calcium that has chemical properties similar to those of argon. Explainyour answer.

S Cl Ar K Ca

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Exceptions to the Octet RuleExceptions to the Octet Rule

The octet rule is an important guide to understanding how most compounds are formed. However, there are a number of cases in which the octet rule does not apply. Answer the

following questions about exceptions to the octet rule.


1. Draw the Lewis structure for the compound BeF2.

2. Does BeF2 obey the octet rule? Explain.

3. Draw the Lewis structure for the compound NO2.

4. Does NO2 obey the octet rule? Explain.

5. Draw the Lewis structure for the compound N2F2.

6. Does N2F2 obey the octet rule? Explain.

7. Draw the Lewis structure for the compound IF5.

8. Does IF5 obey the octet rule? Explain.


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Balancing ChemicalEquationsBalancing ChemicalEquations



Each chemical equation below contains at least one error. Identify the error or errors and then write the correct chemical equation for the reaction.

1. K(s) � 2H2O(l) 0 2KOH(aq) � H2(g)

2. MgCl2(aq) � H2SO4(aq) 0 Mg(SO4)2(aq) � 2HCl(aq)

3. AgNO3(aq) � H2S(aq) 0 Ag2S(aq) � HNO3(aq)

4. Sr(s) � F2(g) 0 Sr2F

5. 2NaHCO3(s) � 2HCl(aq) 0 2NaCl(s) � 2CO2(g)

6. 2LiOH(aq) � 2HBr(aq) 0 2LiBr(aq) � 2H2O

7. NH4OH(aq) � KOH(aq) 0 KOH(aq) � NH4OH(aq)

8. 2Ca(s) � Cl2(g) 0 2CaCl(aq)

9. H2SO4(aq) � 2Al(NO3)3(aq) 0 Al2(SO4)3(aq) � 2HNO3(aq)

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Using Mole-BasedConversionsUsing Mole-BasedConversions

The diagram shows three containers, each of which holds a certain mass of the substance indicated. Complete the table below for each of the three substances.


1. Compare and contrast the number of representative particles and the mass of UF6 withthe number of representative particles and mass of CCl3CF3. Explain any differences you observe.

2. UF6 is a gas used in the production of fuel for nuclear power plants. How many moles ofthe gas are in 100.0 g of UF6?

3. CCl3CF3 is a chlorofluorocarbon responsible for the destruction of the ozone layer inEarth’s atmosphere. How many molecules of the liquid are in 1.0 g of CCl3CF3?

4. Lead (Pb) is used to make a number of different alloys. What is the mass of lead presentin an alloy containing 0.15 mol of lead?

UF6 (g)

225.0 g

CCl3CF3 (l)

200.0 g

Pb (s)

250.0 g


Molar Mass Number of Number of Representative Substance Mass (g) (g/mol) Moles (mol) Particles

UF6(g)

CCl3CF3(l)

Pb(s)

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Mole Relationships inChemical ReactionsMole Relationships inChemical Reactions

The mole provides a convenient way of finding the amounts of the substances in a chemical reaction. The diagram below shows how this concept can be applied to the reaction

between carbon monoxide (CO) and oxygen (O2), shown in the following balanced equation.

2CO(g) � O2(g) 0 2CO2(g)

Use the equation and the diagram to answer the following questions.



1. What information is needed to make the types of conversions shown by double-arrow 1in the diagram?

2. What conversion factors would be needed to make the conversions represented by double-arrow 2 in the diagram for CO? By double-arrow 6 for CO2?

3. What information is needed to make the types of conversions represented by double-arrows 3 and 7 in the diagram?

4. What conversion factors would be needed to make the conversions represented by double-arrow 3 in the diagram for CO?

5. Why is it not possible to convert between the mass of a substance and the number of representative particles, as represented by double-arrow 4 of the diagram?

6. Why is it not possible to use the mass of one substance in a chemical reaction to find the massof a second substance in the reaction, as represented by double-arrow 5 in the diagram?

Moles ofCO

Grams ofCO

Moles ofCO2

Grams ofCO2

Particles ofCO

Particles ofCO2

1

5

2

3

46

7

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Intermolecular Forces andBoiling PointsIntermolecular Forces andBoiling Points


1. How do the boiling points of the group 4A hydrides change as the molecular masses ofthe hydrides change?

2. What are the molecular structure and polarity of the four group 4A hydrides?

3. Predict the strength of the forces between group 4A hydride molecules. Explain howthose forces affect the boiling points of group 4A hydrides.

4. How do the boiling points of the group 6A hydrides change as the molecular masses ofthe hydrides change?

5. What are the molecular structure and polarity of the four group 6A hydrides?

6. Use Table 9-4 in your textbook to determine the difference in electronegativities of thebonds in the four group 6A hydrides.

100

0

�100

H2O

H2S

CH4

SiH4

GeH4

SnH4

H2Se

H2Te

Group 6Ahydrides

Group 4Ahydrides

Molecular mass

Bo

ilin

g p

oin

t (°

C)

00 50 100 150


The boiling points of liquids depend partly on the mass of the particles of which they are made. The greater the mass of

the particles, the more energy is needed to convert a liquid to agas, and, thus, the higher the boiling point of the liquid. This pat-tern may not hold true, however, when there are significant forcesbetween the particles of a liquid. The graph plots boiling pointversus molecular mass for group 4A and group 6A hydrides. Ahydride is a binary compound containing hydrogen and one otherelement. Use the graph to answer the following questions.

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A Simple Mercury BarometerA Simple Mercury Barometer

In Figure 1, a simple mercury barometer is made by filling a long glass tube with mercury and then inverting the open end of the

tube into a bowl of mercury. Answer the following questions aboutthe simple mercury barometer shown here.



1. What occupies the space above the mercury column in thebarometer’s glass tube?

2. What prevents mercury from flowing out of the glass tube into the bowl of mercury?

3. When the barometer in Figure 1 is moved to a higher elevation, such as an altitude of5000 meters, the column of mercury changes as shown in Figure 2. Why is the mercurycolumn lower in Figure 2 than in Figure 1?

4. Suppose the barometer in Figure 1 was carried into an open mine 500 meters below sealevel. How would the height of the mercury column change? Explain why.

5. Suppose the liquid used to make the barometer was water instead of mercury. How wouldthis substitution affect the barometer? Explain.

6. Suppose a tiny crack formed at the top of the barometer’s glass tube. How would thisevent affect the column of mercury? Explain why.

Glass tube

Mercury column

Bowl of mercury

At sea level At 500 metersabove sea level

Figure 1 Figure 2

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Vapor Pressure LoweringVapor Pressure Lowering

You have learned that adding a nonvolatile solute to a solvent lowers the vapor pressure of that solvent. The amount by

which the vapor pressure is lowered can be calculated by means of a relationship discovered by the French chemist François MarieRaoult (1830–1901) in 1886. According to Raoult’s law, the vaporpressure of a solvent (P) is equal to the product of its vapor pressurewhen pure (P0) and its mole fraction (X) in the solution, or

P � P0X

The solution shown at the right was made by adding 75.0 g ofsucrose (C12H22O11) to 500.0 g of water at a temperature of 20°C.Answer the following questions about this solution.


1. Why do the sugar molecules in the solution lower the vapor pressure of the water?

2. What is the number of moles of sucrose in the solution?

3. What is the number of moles of water in the solution?

4. What is the mole fraction of water in the solution?

5. What is the vapor pressure of the solution if the vapor pressure of pure water at 20°C is17.54 mm Hg?

6. How much is the vapor pressure of the solution reduced from that of water by the addition of the sucrose?

Watermolecule

Solution

Sucrosemolecule


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Standard Heat of FormationStandard Heat of Formation

Hess’s law allows you to determine the standard heat of formation of a compound

when you know the heats of reactions that leadto the production of that compound. The firstdiagram on the right shows how Hess’s law canbe used to calculate the heat of formation ofCO2 by knowing the heats of reaction of twosteps leading to the production of CO2. Use thisdiagram to help you answer the questions belowabout the second diagram.



The equations below show how NO2 can be formed in two ways: directly from the elements or in two steps.

N2(g) � O2(g) 0 NO2(g) �H � 33 kJ/mol

or

N2(g) � O2(g) 0 NO(g) �H � 91 kJ/mol

NO(g) � O2(g) 0 NO2(g) �H � �58 kJ/mol

1. On the diagram at the right, draw arrowheadsto show the directions in which the three lineslabeled 1, 2, and 3 should point.

2. Write the correct reactants and/or products oneach of the lines labeled A, B, and C.

3. Write the correct enthalpy change next toeach number on the diagram.

1�2

1�2

1�2

1�2

CO(g) � O2(g)

�H � �110 kJ/mol

C(s) � O2(g)

�H � �393 kJ/mol

�H � �283 kJ/molEnth

alp

y

CO2(g)

12

NO2(g)

�H � �58 kJ/mol

NO(g) � 1/2 O2(g)

�H � 91 kJ/mol

Enth

alp

y

�H � 33 kJ/mol

1

2

3

A

B

C

1/2 N2(g) � O2(g)

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Determining Reaction RatesDetermining Reaction Rates

Dinitrogen pentoxide decomposes to produce nitrogen dioxide and oxygen as represented

by the following equation.

2N2O5(g) 0 4NO2(g) � O2(g)

The graph on the right represents the concen-tration of N2O5 remaining as the reaction proceedsover time. Answer the following questions aboutthe reaction.


1. What is the concentration of N2O5 at the beginning of the experiment? After 1 hour?After 2 hours? After 10 hours?

2. By how much does the concentration of N2O5 change during the first hour of the reaction? Calculate the percentage of change the concentration undergoes during the first hour of the reaction.

3. The instantaneous rate of reaction is defined as the change in concentration of reactantduring some specified time period, or instantaneous rate of reaction = [N2O5]/t. What isthe instantaneous rate of reaction for the decomposition of N2O5 for the time periodbetween the first and second hours of the reaction? Between the second and third hours?Between the sixth and seventh hours?

4. What is the instantaneous rate of reaction for the decomposition of N2O5 between the sec-ond and fourth hours of the reaction? Between the third and eighth hours of the reaction?

5. How long does it take for 0.10 mol of N2O5 to decompose during the tenth hour of the reaction?

6. What is the average rate of reaction for the decomposition of N2O5 overall?

Time (h)

Co

nce

ntr

atio

n (

mo

l/L)

0.2

0

0.4

0.6

0.8

1.0

1.2

1.4

1.6

10 2 3 4 5 6 7 8 9 10


87654321

10

0 2 3 4 5Time (sec)

Co

nce

ntr

atio

n (

mo

l/L)

6 7 8 9 10

SO2

SO3

SO3

O2O2

SO2

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Changing EquilibriumConcentrations in a Reaction

Reversible reactions eventually reach an equilibrium condition in which the concentrations of all reactants

and products are constant. Equilibrium can be disturbed,however, by the addition or removal of either a reactant orproduct. The graph on the right shows how the concentra-tions of the reactants and product of a reaction changewhen equilibrium is disturbed. Use the graph to answer thefollowing questions.



1. Write the equation for the reaction depicted in the graph.

2. Write the equilibrium constant expression for the reaction.

3. Explain the shapes of the curves for the three gases during the first 2 minutes of the reaction.

4. At approximately what time does the reaction reach equilibrium? How do you knowequilibrium has been reached?

5. What are the concentrations of the three gases at equilibrium?

6. Calculate the value of Keq for the reaction.

7. Describe the change made in the system 4 minutes into the reaction. Tell how you knowthe change was made.

8. At what time does the system return to equilibrium?

Changing EquilibriumConcentrations in a Reaction

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Swimming Pool ChemistrySwimming Pool Chemistry

The presence of disease-causing bacteria in swimming pools is a major health concern. Chlorine gas is added to the water in some large commercial swimming pools to kill

bacteria. However, in most home swimming pools, either solid calcium hypochlorite(Ca(OCl)2) or an aqueous solution of sodium hypochlorite (NaOCl) is used to treat thewater. Both compounds dissociate in water to form the weak acid hypochlorous acid(HOCl). Hypochlorous acid is a highly effective bactericide. By contrast, the hypochloriteion (OCl�) is not a very effective bactericide. Use the information above to answer the following questions about the acid-base reactions that take place in swimming pools.


1. Write an equation that shows the reaction between hypochlorous acid and water. Identifythe acid, base, conjugate acid, and conjugate base in this reaction.

2. Write an equation that shows the reaction that occurs when the hypochlorite ion (OCl�),in the form of calcium hypochlorite or sodium hypochlorite, is added to water. Name theacid, base, conjugate acid, and conjugate base in this reaction.

3. What effect does the addition of hypochlorite ion have on the pH of swimming pool water?

4. The effectiveness of hypochlorite ion as a bactericide depends on pH. How does high pHaffect the equilibrium reaction described in question 2? What effect would high pH haveon the bacteria?

5. In the presence of sunlight, hypochlorite ion decomposes to form chloride ion and oxygen gas. Write an equation for this reaction and tell how it affects the safety of pool water.


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Balancing Oxidation–Reduction EquationsBalancing Oxidation–Reduction Equations

Scientists have developed a number of methods for protecting metals from oxidation. One such method involves the use of a

sacrificial metal. A sacrificial metal is a metal that is more easily oxidized than the metal it is designed to protect. Galvanized iron, forexample, consists of a piece of iron metal covered with a thin layer of zinc. When galvanized iron is exposed to oxygen, it is the zinc,rather than the iron, that is oxidized.

Water heaters often contain a metal rod that is made by coating a heavy steel wire with magnesium or aluminum. In this case, themagnesium or aluminum is the sacrificial metal, protecting the ironcasing of the heater from corrosion.

The diagram shows a portion of a water heater containing a sacrificial rod. Answer the following questions about the diagram.



1. In the absence of a sacrificial metal, oxygen dissolved in water may react with the ironcasing of the heater. One product formed is iron(II) hydroxide (Fe(OH)2). Which elementis oxidized and which is reduced in this reaction?

2. Balance the oxidation–reduction equation for this reaction:Fe(s) � O2(aq) � H2O 0 Fe(OH)2(aq)

3. Write the two half-reactions for this example of corrosion.

4. Suppose the sacrificial rod in the diagram above is coated with aluminum metal. Writethe balanced equation for the reaction of aluminum with oxygen dissolved in the water.(Hint: The product formed is aluminum hydroxide (Al(OH)3).

5. Write the two half-reactions for this example of corrosion.

6. Suppose that some iron in the casing of the water heater is oxidized, as shown in theequation of question 2 above. The sacrificial metal (aluminum, in this case) immediatelyrestores the Fe2� ions to iron atoms. Write two half-reactions that represent this situation.

Ironcasing

Steel wire

Sacrificialmetal

Water

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Effect of Concentration onCell PotentialEffect of Concentration onCell Potential

In a voltaic cell where all ions have a concentration of 1M, the cell potential is equal to the standard potential. For cells in which ion concentrations are greater or

less than 1M, as shown below, an adjustment must be made to calculate cell potential.That adjustment is expressed by the Nernst equation:

Ecell � E0cell � �log

In this equation, n is the number of moles of electrons transferred in the reaction,and x and y are the coefficients of the product and reactant ions, respectively, in thebalanced half-cell reactions for the cell.

[product ion]x��[reactant ion]y

0.0592�n


1. Write the two half-reactions and the overall cell reaction for the cell shown above.

2. Use Table 21-1 in your textbook to determine the standard potential of this cell.

3. Write the Nernst equation for the cell.

4. Calculate the cell potential for the ion concentrations shown in the cell.

Voltmeter

Ag Cu

Cu2�

1.0 � 10�3MAg�

1.0 � 10�2M


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Structural Isomers of HexaneStructural Isomers of Hexane

The structural formula of an organic compound can sometimes be written in a variety of ways, but sometimes structural formulas that appear similar can

represent different compounds. The structural formulas below are ten ways of representing compounds having the molecular formula C6H14.


Use with Chapter 22,Sections 22.1 and 22.3

1. In the spaces provided, write the correct name for each of the structural formulas, labeleda–j, above.

a. e. i.

b. f. j.

c. g.

d. h.

2. How many different compounds are represented by the structural formulas above? Whatare their names?

CH3

CH2 CH2 CH2 CH2 CH3

a.

CH3

CH3

CH CH2 CH2 CH3

b.

CH3

CH3

CH3 CH CH CH3

c.

CH3

CH3

CH3 C CH2 CH3

d.

CH3

CH CH2

CH2

CH3

CH3e.

CH3 CH CH CH3

CH3 CH3

f.

CH2

CH2 CH3

CH CH3

CH3

g.

CH3

CH2

CH3 CH CH2

CH3

h.

CH2

CH3

CH2

CH2 CH2

CH3i.

CH3

CH2 CH CH2

CH3 CH3j.

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Boiling Points of OrganicFamiliesBoiling Points of OrganicFamilies

The most important factor determining the boiling point of a substance is its atomic or molecular mass. In general,

the larger the atomic or molecular mass of the substance, themore energy is needed to convert the substance from the liquidphase to the gaseous phase. As an example, the boiling pointof ethane (molecular mass � 30; boiling point � �89°C) ismuch higher than the boiling point of methane (molecularmass � 16; boiling point � �161°C).

Intermolecular forces between the particles of a liquid alsocan affect the liquid’s boiling point. The graph shows trends inthe boiling points of four organic families: alkanes, alcohols,aldehydes, and ethers. Use the graph and your knowledge ofintermolecular forces to answer the following questions.


1. For any one family, what is the relationship between molecular mass and boiling point?

2. For compounds of similar molecular mass, which family of the four shown in the graphhas the lowest boiling points? Which family has the highest boiling points?

3. Find and list the boiling points for ethanol (molecular mass � 46) and dimethyl ether(molecular mass � 46) on the graph. Why would you expect these two compounds tohave relatively similar boiling points?

4. Find the aldehyde with a molecular mass of about 58. Name that aldehyde and write itschemical formula.

5. Can this aldehyde form hydrogen bonds? Can other aldehydes form hydrogen bonds?Explain.

30 40 50 60Molecular mass

� alkane� alcohol

Bo

ilin

g p

oin

t (°

C)

70 80

�50

0

50

100

� aldehyde� ether


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The Chemistry of LifeThe Chemistry of Life

Proteins are synthesized when RNA molecules translate the DNA language of nitrogen bases

into the protein language of amino acids using agenetic code. The genetic code is found in RNA mole-cules called messenger RNA (mRNA), which are syn-thesized from DNA molecules. The genetic codeconsists of a sequence of three nitrogen bases in themRNA, called a codon. Most codons code for specificamino acids. A few codons code for a stop in the syn-thesis of proteins. The table shows the mRNA codonsthat make up the genetic code. To use the table, readthe three nitrogen bases in sequence. The first base isshown along the left side of the table. The second baseis shown along the top of the table. The third base isshown along the right side of the table. For example,the sequence CAU codes for the amino acid histidine(His). The table gives abbreviations for the aminoacids. Answer the following questions about thegenetic code.



1. What amino acid is represented by each of the following codons?

a. CUG b. UCA

2. Write the sequence of amino acids for which the following mRNA sequence codes.

-C-A-U-C-A-C-C-G-G-U-C-U-U-U-U-C-U-U-

3. Errors sometimes occur when mRNA molecules are synthesized from DNA molecules.Nitrogen bases may be omitted, an extra nitrogen base may be added, or a nitrogen basemay be changed during synthesis. The two mRNA sequences shown below are examplesof such errors. In each case, tell how the mRNA sequence shown differs from the correctmRNA sequence given in question 2.

a. -C-A-U-C-A-C-C-G-G-U-U-C-U-U-U-U-C-U-U-

b. -C-A-U-U-A-C-C-G-G-U-C-U-U-U-U-C-U-U-

4. Write the amino acid sequence for each of the mRNA sequences shown in question 3.

a.

b.

UUU UCU UAU UGU UUUC UCC UAC UGC CUUA UCA UAA UGA AUUG UCG UAG UGG GCUU CCU CAU CGU UCUC CCC CAC CGC CCUA CCA CAA CGA ACUG CCG CAG CGG GAUU ACU AAU AGU UAUC ACC AAC AGC CAUA ACA AAA AGA AAUG ACG AAG AGG GGUU GCU GAU GGU UGUC GCC GAC GGC CGUA GCA GAA GGA AGUG GCG GAG GGG G

Second base

Firs

t b

ase Th

ird b

ase

}} Phe

Leu

Ile

Met

Leu

Val

Pro

Ser

Ala

Thr

}

}}

}}

}}

His

Gln

Tyr

StopStop

Asn

Lys

Asp

Glu

}

}}

Arg

Gly

Cys

StopTrp

Ser

Arg

U

C

A

G

U C A G

The Genetic Code

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The Production ofPlutonium-239The Production ofPlutonium-239

When nuclear fission was first discovered, only two isotopes, uranium-233 and uranium-235, were

known of being capable of undergoing this nuclear change.Scientists later discovered a third isotope, plutonium-239,also could undergo nuclear fission. Plutonium-239 does notoccur in nature but can be made synthetically in nuclearreactors and particle accelerators.

The diagram shows the process by which plutonium-239is made in nuclear reactors. Answer the questions about thediagram.


1. Identify the isotope whose nucleus is labeled A in the

diagram.

2. Name the type of nuclear reaction that occurs when a

neutron strikes nucleus A.

3. Identify the isotope whose nucleus is labeled B.

4. Besides fragmented nuclei, what else is produced when a neutron strikes nucleus A?

5. Identify the isotope whose nucleus is labeled C.

6. Write the nuclear equation for the reaction that occurs when a neutron strikes nucleus C.Identify the product D formed in the reaction.

7. Write the nuclear equation for the decay of nucleus D. Identify isotope E formed in thereaction.

8. Write a balanced nuclear equation for the decay of nucleus E. Identify isotope F formedin the reaction.

9. Name the type of nuclear reaction that occurs when a neutron strikes nucleus F.

10. Write the nuclear equation for the reaction that occurs when a neutron strikes nucleus F.Identify isotope G formed in the reaction.

0–1

0–1

00

10n

10n

10n

10n

45p75n

48p77n

92p143n92p

143n

92p146nSource

ofneutrons

B

A

C

D

E

G

F


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The Phosphorus CycleThe Phosphorus Cycle

Phosphorus is an important element both in organisms and in the lithosphere. In organisms, phosphorus occurs in DNA and RNA molecules, cell membranes, bones

and teeth, and in the energy–storage compound adenosine triphosphate (ATP). In the litho-sphere, phosphorus occurs primarily in the form of phosphates, as a major constituent ofmany rocks and minerals. Phosphate rock is mined to produce many commercial products,such as fertilizers and detergents. When these products are used, phosphates are returned tothe lithosphere and hydrosphere. Thus, phosphorus—like carbon and nitrogen—cycles in theenvironment. Use the diagram of the phosphorus cycle to answer the questions below.



1. By what methods does phosphorus get into soil?

2. By what method do plants obtain the phosphorus they need?

3. By what method do animals obtain the phosphorus they need?

4. In what way is the phosphorus cycle different from the carbon and nitrogen cycles youstudied in the textbook?

5. The phosphorus cycle has both short-term and long-term parts. Use different colored pencils to show each part on the diagram.

Phosphate rocks

Phosphaterocks

Geological uplift

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CHALLENGE PROBLEMSAnswer Key

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T28 Chemistry: Matter and Change Challenge Problems Answer Key

Nam

eD

ate

Cla

ss

2C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

2C

hal

len

ge

Pro

ble

ms

Popu

lati

on T

rend

s in

the

Uni

ted

Stat

esPo

pula

tion

Tre

nds

in t

heU

nite

d St

atesCH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

2

Use

wit

h Ch

apte

r 2,

Sect

ion

2.4

1.

By

how

muc

h di

d th

e to

tal U

.S. p

opul

atio

n in

crea

se b

etw

een

1990

and

200

0? W

hat w

asth

e pe

rcen

t inc

reas

e du

ring

this

per

iod?

Tota

l in

crea

se �

2.75

4 �

108

�2.

488

�10

8�

2.66

�10

7 ;

per

cen

tag

e in

crea

se �

2.66

�10

7�

2.48

8 �

108

�10

.7%

.

2.

Cal

cula

te th

e to

tal p

opul

atio

n fo

r ea

ch o

f th

e fi

ve g

roup

s fo

r 19

90 a

nd 2

000.

For

1990

, to

tal �

per

cen

t fo

r g

rou

p �

2.48

8 �

108 ,

or

Cau

casi

an: 1

.88

�10

8 ;

Afr

ican

Am

eric

an: 2

.94

�10

7 ; H

isp

anic

Am

eric

an: 2

.2 �

107 ;

Asi

an A

mer

ican

:

7.0

�10

6 ; N

ativ

e A

mer

ican

: 1.7

�10

6 ; f

or

2000

, to

tal �

per

cen

t fo

r g

rou

p �

2.75

4 �

108 ,

or

Cau

casi

an: 1

.97

�10

8 ; A

fric

an A

mer

ican

: 3.4

�10

7 ; H

isp

anic

Am

eric

an: 3

.25

�10

7 ; N

ativ

e A

mer

ican

: 1.9

�10

6 ; A

sian

Am

eric

an: 1

.0 �

107 .

3.

Mak

e a

bar

grap

h th

at c

ompa

res

the

popu

latio

n fo

r th

e fi

ve g

roup

s in

199

0 an

d 20

00. I

nw

hat w

ays

is th

e ba

r gr

aph

bette

r th

an th

e ci

rcle

gra

phs?

In

wha

t way

is it

less

use

ful?

The

bar

gra

ph

may

mak

e th

e ch

ang

es f

rom

199

0 to

200

0 sh

ow

up

mo

re c

lear

ly.

Un

less

th

e n

um

eric

al v

alu

es f

or

each

bar

are

act

ual

ly w

ritt

en o

n t

he

gra

ph

, th

ey

may

be

mo

re d

iffi

cult

to

est

imat

e th

an t

he

valu

es g

iven

on

th

e ci

rcle

gra

ph

. Th

e

larg

e d

iffe

ren

ces

in p

op

ula

tio

ns

for

Cau

casi

an a

nd

no

n-C

auca

sian

gro

up

s m

ake

it

dif

ficu

lt t

o s

ho

w d

ata

for

the

latt

er g

rou

ps

clea

rly.

U.S

. Pop

ulat

ion

Dis

trib

utio

n

(Per

cen

tag

es m

ay n

ot

add

up

to

100

% d

ue

to r

ou

nd

ing

.)Cau

casi

an71

.4%

Nat

ive

Am

eric

an0.

70%

Asi

an A

mer

ican

3.8%

His

pan

ic A

mer

ican

11.8

%

Afr

ican

Am

eric

an12

.2%

Cau

casi

an75

.7%

Nat

ive

Am

eric

an0.

70%

Asi

an A

mer

ican

2.8%

His

pan

ic A

mer

ican

9.0%A

fric

an A

mer

ican

11.8

%

2000

1990

The

pop

ulat

ion

of th

e U

nite

d St

ates

is b

ecom

ing

mor

e di

vers

e. T

he c

ircl

e gr

aphs

bel

ow s

how

the

dist

ribu

tion

of t

he U

.S. p

opul

atio

n am

ong

five

eth

nic

grou

ps in

199

0 an

d 20

00. T

he e

stim

ated

tota

l U.S

. pop

ulat

ion

for

thos

e tw

o ye

ars

was

2.4

88 �

108

in 1

990

and

2.75

4 �

108

in 2

000.

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

11

Prod

ucti

on o

fCh

loro

fluo

roca

rbon

s,19

50–1

992

Prod

ucti

on o

fCh

loro

fluo

roca

rbon

s,19

50–1

992

Chl

orof

luor

ocar

bons

(C

FCs)

wer

e fi

rst p

rodu

ced

in

the

labo

rato

ry in

the

late

192

0s. T

hey

did

not

beco

me

an im

port

ant c

omm

erci

al p

rodu

ct u

ntil

som

etim

e la

ter.

Eve

ntua

lly,C

FCs

grew

in p

opul

arity

unt

ilth

eir

effe

ct o

n th

e oz

one

laye

r w

as d

isco

vere

d in

the

1970

s. T

he g

raph

sho

ws

the

com

bine

d am

ount

s of

two

impo

rtan

t CFC

s pr

oduc

ed b

etw

een

1950

and

199

2.A

nsw

er th

e fo

llow

ing

ques

tions

abo

ut th

e gr

aph.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

1

1.

Wha

t was

the

appr

oxim

ate

amou

nt o

f C

FCs

prod

uced

in 1

950?

In

1960

? In

197

0?

1950

: les

s th

an 1

0 b

illio

n k

g; 1

960:

ab

ou

t 50

bill

ion

kg

; 197

0: a

bo

ut

240

bill

ion

kg

2.

In w

hat y

ear

was

the

larg

est a

mou

nt o

f C

FCs

prod

uced

? A

bout

how

muc

h w

as p

rodu

ced

that

yea

r?

1988

; ab

ou

t 37

5 b

illio

n k

g

3.

Dur

ing

wha

t tw

o-ye

ar p

erio

d di

d th

e pr

oduc

tion

of C

FCs

decr

ease

by

the

grea

test

amou

nt?

By

abou

t how

muc

h di

d th

eir

prod

uctio

n de

crea

se?

Bet

wee

n 1

988

and

199

0; t

he

dec

reas

e am

ou

nte

d t

o a

bo

ut

140

bill

ion

kg

.

4.

Dur

ing

wha

t tw

o-ye

ar p

erio

d di

d th

e pr

oduc

tion

of C

FCs

incr

ease

by

the

grea

test

amou

nt?

Wha

t was

the

appr

oxim

ate

perc

ent i

ncre

ase

duri

ng th

is p

erio

d?

Bet

wee

n 1

970

and

197

2; t

he

per

cen

t in

crea

se w

as a

pp

roxi

mat

ely

(305

bill

ion

kg

�24

0 b

illio

n k

g) /2

40 b

illio

n k

g �

27%

.

5.

How

con

fide

nt w

ould

you

fee

l abo

ut p

redi

ctin

g th

e pr

oduc

tion

leve

ls o

f C

FCs

duri

ng th

eod

d nu

mbe

red

year

s 19

61,1

971,

and

1981

? E

xpla

in.

The

pro

du

ctio

n o

f C

FCs

incr

ease

d r

egu

larl

y fr

om

195

0 to

197

4 (e

xcep

t in

195

8)

and

dec

reas

ed r

egu

larl

y af

ter

1988

. Th

us,

on

e m

igh

t b

e ab

le t

o p

red

ict

accu

rate

ly

the

pro

du

ctio

n le

vels

du

rin

g 1

961

and

197

1 fr

om

dat

a in

th

e g

rap

h. H

ow

ever

, th

e

pro

du

ctio

n le

vels

wer

e le

ss r

egu

lar

bet

wee

n 1

974

and

198

8. T

her

efo

re, p

red

icti

on

s

reg

ard

ing

th

e p

rod

uct

ion

leve

l du

rin

g 1

981

mig

ht

be

less

acc

ura

te.

6.

Cou

ld th

e da

ta in

the

grap

h be

pre

sent

ed in

the

form

of

a ci

rcle

gra

ph?

Exp

lain

.

A c

ircl

e g

rap

h c

ou

ld n

ot

be

use

d b

ecau

se it

sh

ow

s h

ow

a t

ota

l am

ou

nt

is d

ivid

ed

into

cat

ego

ries

, no

t h

ow

a v

aria

ble

ch

ang

es o

ver

tim

e.

Yea

r

Amount of CFCs(billion kilograms)

050100

150

200

250

300

350

400 19

5019

6019

7019

8019

90

Use

wit

h Ch

apte

r 1,

Sect

ion

1.1

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.

Challenge Problems Answer Key Chemistry: Matter and Change T29

Nam

eD

ate

Cla

ss

4C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

4C

hal

len

ge

Pro

ble

ms

Isot

opes

of

an E

lem

ent

Isot

opes

of

an E

lem

ent

Am

ass

spec

trom

eter

is a

dev

ice

for

sepa

ratin

gat

oms

and

mol

ecul

es a

ccor

ding

to th

eir

mas

s. A

sub

stan

ce is

fir

st h

eate

d in

a v

acuu

m a

ndth

en io

nize

d. T

he io

ns p

rodu

ced

are

acce

lera

ted

thro

ugh

a m

agne

tic f

ield

that

sep

arat

es io

ns o

f di

f-fe

rent

mas

ses.

The

gra

ph b

elow

was

pro

duce

dw

hen

a ce

rtai

n el

emen

t (el

emen

t X)

was

ana

lyze

din

a m

ass

spec

trom

eter

. Use

the

grap

h to

ans

wer

the

ques

tions

bel

ow.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

4

Use

wit

h Ch

apte

r 4,

Sect

ion

4.3

1.

How

man

y is

otop

es o

f el

emen

t X e

xist

?

2.

Wha

t is

the

mas

s of

the

mos

t abu

ndan

t iso

tope

?

3.

Wha

t is

the

mas

s of

the

leas

t abu

ndan

t iso

tope

?

4.

Wha

t is

the

mas

s of

the

heav

iest

isot

ope?

5.

Wha

t is

the

mas

s of

the

light

est i

soto

pe?

6.

Est

imat

e th

e pe

rcen

t abu

ndan

ce o

f ea

ch is

otop

e sh

own

on th

e gr

aph.

196:

less

th

an 1

%; 1

98: a

bo

ut

10%

; 199

: ab

ou

t 17

%; 2

00: a

bo

ut

23%

;

201:

ab

ou

t 13

%; 2

02: a

bo

ut

30%

; 204

: ab

ou

t 7%

7.

With

out p

erfo

rmin

g an

y ca

lcul

atio

ns,p

redi

ct th

e ap

prox

imat

e at

omic

mas

s fo

r el

emen

tX

. Exp

lain

the

basi

s fo

r yo

ur p

redi

ctio

n.

The

ato

mic

mas

s o

f el

emen

t X

is b

etw

een

200

an

d 2

02 a

mu

, th

e m

asse

s o

f

the

two

mo

st a

bu

nd

ant

iso

top

es.

8.

Usi

ng th

e da

ta g

iven

by

the

grap

h,ca

lcul

ate

the

wei

ghte

d av

erag

e at

omic

mas

s of

el

emen

t X. I

dent

ify

the

unkn

own

elem

ent.

Mas

s o

f X

�(0

.001

4)(1

96 a

mu

) �

(0.1

0)(1

98 a

mu

) �

(0.1

68)(

199

amu

)

�(0

.231

)(20

0 am

u)

�(0

.132

)(20

1) �

(0.2

98)(

202

amu

) �

(0.0

68)(

204

amu

)

Mas

s o

f X

�0.

27 a

mu

�19

.8 a

mu

�33

.4 a

mu

�46

.2 a

mu

�26

.5 a

mu

�60

.2 a

mu

�13

.9 a

mu

Mas

s o

f X

�20

0.3

amu

The

elem

ent

in t

he

per

iod

ic t

able

wit

h a

n a

tom

ic m

ass

clo

sest

to

th

e ca

lcu

late

d

valu

e o

f 20

0.3

amu

is m

ercu

ry. E

lem

ent

X is

pro

bab

ly m

ercu

ry, w

hic

h h

as a

n

ato

mic

mas

s o

f 20

0.59

am

u.

196

amu

204

amu

196

amu

202

amu

7

051015202530

196

194

192

190

198

200

202

204

206

208

210

Ato

mic

mas

s (a

mu

)

Percent abundance

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

33

Phys

ical

and

Che

mic

alCh

ange

sPh

ysic

al a

nd C

hem

ical

Chan

ges

Phy

sica

l and

che

mic

al c

hang

es o

ccur

all

arou

nd u

s. O

ne o

f th

e m

any

plac

es in

whi

ch

phys

ical

and

che

mic

al c

hang

es o

ccur

is th

e ki

tche

n. F

or e

xam

ple,

cook

ing

spag

hetti

in a

pot o

f w

ater

on

the

stov

e in

volv

es s

uch

chan

ges.

For

eac

h of

the

chan

ges

desc

ribe

d be

low

,tel

l(a

) w

heth

er th

e ch

ange

that

occ

urs

is p

hysi

cal o

r ch

emic

al,a

nd (

b) h

ow y

ou m

ade

your

cho

ice

betw

een

thes

e tw

o po

ssib

ilitie

s. I

f yo

u ar

e un

able

to d

ecid

e w

heth

er th

e ch

ange

is p

hysi

cal o

rch

emic

al,t

ell w

hat a

dditi

onal

info

rmat

ion

you

wou

ld n

eed

in o

rder

to m

ake

a de

cisi

on.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

3

1.

As

the

wat

er in

the

pot i

s he

ated

,its

tem

pera

ture

ris

es.

Phys

ical

ch

ang

e; t

he

com

po

siti

on

of

the

wat

er d

oes

no

t ch

ang

e.

2.

As

mor

e he

at is

add

ed,t

he w

ater

beg

ins

to b

oil a

nd s

team

is p

rodu

ced.

Phys

ical

ch

ang

e; t

he

wat

er c

han

ges

fro

m li

qu

id t

o g

as, b

ut

ther

e is

no

ch

ang

e in

its

com

po

siti

on

.

3.

The

hea

t use

d to

coo

k is

pro

duce

d by

bur

ning

nat

ural

gas

in th

e st

ove

burn

er.

Ch

emic

al c

han

ge;

th

e co

mp

osi

tio

n o

f th

e n

atu

ral g

as c

han

ges

as

it b

urn

s.

4.

The

met

al b

urne

r on

whi

ch th

e po

t res

ts w

hile

bei

ng h

eate

d be

com

es r

ed a

s its

te

mpe

ratu

re r

ises

.

Phys

ical

ch

ang

e; t

he

app

eara

nce

of

the

met

al c

han

ges

as

it is

hea

ted

, bu

t it

s

com

po

siti

on

do

es n

ot.

Wh

en t

he

met

al c

oo

ls, i

t w

ill n

o lo

ng

er b

e re

d.

5.

Aft

er th

e fl

ame

has

been

turn

ed o

ff,a

sm

all a

rea

on th

e bu

rner

has

cha

nged

in c

olor

fro

mbl

ack

to g

ray.

It is

no

t p

oss

ible

to

kn

ow

wh

eth

er t

his

is a

ph

ysic

al c

han

ge

or

a ch

emic

al c

han

ge

wit

ho

ut

com

par

ing

th

e co

mp

osi

tio

n o

f th

e g

ray

mat

eria

l wit

h t

he

ori

gin

al b

lack

mat

eria

l.

6.

A s

tran

d of

spa

ghet

ti ha

s fa

llen

onto

the

burn

er,w

here

it tu

rns

blac

k an

d be

gins

tosm

oke.

Ch

emic

al c

han

ge;

th

e fo

rmat

ion

of

smo

ke a

nd

a p

erm

anen

t co

lor

chan

ge

are

evid

ence

of

a ch

emic

al c

han

ge.

7.

Whe

n th

e sp

aghe

tti is

coo

ked

in th

e bo

iling

wat

er,i

t bec

omes

sof

t.

Phys

ical

ch

ang

e; t

he

com

po

siti

on

of

the

spag

het

ti h

as n

ot

chan

ged

.

Use

wit

h Ch

apte

r 3,

Sect

ion

3.2

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

6C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

6C

hal

len

ge

Pro

ble

ms

Döb

erei

ner’

s Tr

iads

Döb

erei

ner’

s Tr

iads

One

of

the

firs

t som

ewha

t suc

cess

ful a

ttem

pts

to a

rran

ge th

e el

emen

ts in

a s

yste

mat

ic w

ay

was

mad

e by

the

Ger

man

che

mis

t Joh

ann

Wol

fgan

g D

öber

eine

r (1

780–

1849

). I

n 18

16,

Döb

erei

ner

notic

ed th

at th

e th

en a

ccep

ted

atom

ic m

ass

of s

tron

tium

(50

) w

as m

idw

ay b

etw

een

the

atom

ic m

asse

s of

cal

cium

(27

.5)

and

bari

um (

72.5

). N

ote

that

the

acce

pted

ato

mic

mas

ses

for

thes

e el

emen

ts to

day

are

very

dif

fere

nt f

rom

thei

r ac

cept

ed a

tom

ic m

asse

s at

the

time

Döb

erei

ner

mad

e hi

s ob

serv

atio

ns. D

öber

eine

r al

so o

bser

ved

that

str

ontiu

m,c

alci

um,a

nd b

ar-

ium

sho

wed

a g

radu

al g

rada

tion

in th

eir

prop

ertie

s,w

ith th

e va

lues

of

som

e of

str

ontiu

m’s

prop

ertie

s be

ing

abou

t mid

way

bet

wee

n th

e va

lues

of

calc

ium

and

bar

ium

. Döb

erei

ner

even

tu-

ally

fou

nd f

our

othe

r se

ts o

f th

ree

elem

ents

,whi

ch h

e ca

lled

tria

ds,t

hat f

ollo

wed

the

sam

e pa

t-te

rn. I

n ea

ch tr

iad,

the

atom

ic m

ass

of th

e m

iddl

e el

emen

t was

abo

ut m

idw

ay b

etw

een

the

atom

ic m

asse

s of

the

othe

r tw

o el

emen

ts. U

nfor

tuna

tely

,bec

ause

Döb

erei

ner’

s sy

stem

did

not

turn

out

to b

e ve

ry u

sefu

l,it

was

larg

ely

igno

red.

Had

Döb

erei

ner

actu

ally

dis

cove

red

a w

ay o

f id

entif

ying

tren

ds a

mon

g th

e el

emen

ts?

Lis

ted

belo

w a

re s

ix th

ree-

elem

ent g

roup

s in

whi

ch th

e el

emen

ts in

eac

h gr

oup

are

cons

ecut

ive

mem

bers

of

the

sam

e gr

oup

in th

e pe

riod

ic ta

ble.

The

ele

men

ts in

eac

h se

t sho

w a

gra

datio

n in

thei

r pr

oper

ties.

Val

ues

for

the

firs

t and

thir

d el

emen

t in

each

set

are

giv

en. D

eter

min

e th

e m

iss-

ing

valu

e in

eac

h se

t by

calc

ulat

ing

the

aver

age

of th

e tw

o gi

ven

valu

es. T

hen,

com

pare

the

val-

ues

you

obta

ined

with

thos

e gi

ven

in th

e H

andb

ook

of C

hem

istr

y an

d P

hysi

cs. R

ecor

d th

eac

tual

val

ues

belo

w y

our

calc

ulat

ed v

alue

s. I

s th

e va

lue

of th

e pr

oper

ty o

f th

e m

iddl

e el

emen

tin

eac

h se

t mid

way

bet

wee

n th

e va

lues

of

the

othe

r tw

o el

emen

ts in

the

set?

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

6

Use

wit

h Ch

apte

r 6,

Sect

ion

6.2

No

, on

ly s

om

e o

f th

e p

rop

erti

es o

f so

me

of

the

mid

dle

ele

men

ts o

f tr

iad

s ar

e

mid

way

in v

alu

e.

Elem

ent

Mel

ting

Poi

nt (

°C)

Flu

ori

ne

�21

9.6

Ch

lori

ne

Cal

cula

ted

:�

113.

4

Act

ual

: �

100.

98

Bro

min

e�

7.2

Set

1

Elem

ent

Ato

mic

Mas

s

Lith

ium

6.94

1

Sod

ium

Cal

cula

ted

: 23

.019

Act

ual

:22

.990

Pota

ssiu

m39

.098

Set

2

Elem

ent

Boili

ng P

oint

(°C

)

Mag

nes

ium

1107

Cal

ciu

mC

alcu

late

d:

1246

Act

ual

: 48

4

Stro

nti

um

1384

Set

3

Elem

ent

Boili

ng P

oint

(°C

)

Kry

pto

n�

153

Xen

on

Cal

cula

ted

:�

108

Act

ual

:�

108

Rad

on

�62

Set

4

Elem

ent

Mel

ting

Poi

nt (

°C)

Ger

man

ium

937

Tin

Cal

cula

ted

:63

2

Act

ual

: 23

2

Lead

327

Set

5

Elem

ent

Boili

ng P

oint

(°C

)

Ber

ylliu

m12

85

Mag

nes

ium

Cal

cula

ted

:10

68

Act

ual

:65

0

Cal

ciu

m85

1

Set

6

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

55

Qua

ntum

Num

bers

Qua

ntum

Num

bers

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

5

The

sta

te o

f an

ele

ctro

n in

an

atom

can

be

com

plet

ely

desc

ribe

d by

fou

r qu

antu

m n

umbe

rs,

desi

gnat

ed a

s n,

�,m

�,an

d m

s. T

he f

irst

,or

prin

cipa

l,qu

antu

m n

umbe

r,n,

indi

cate

s th

eel

ectr

on’s

app

roxi

mat

e di

stan

ce f

rom

the

nucl

eus.

The

sec

ond

quan

tum

num

ber,

�,de

scri

bes

the

shap

e of

the

elec

tron

’s o

rbit

arou

nd th

e nu

cleu

s. T

he th

ird

quan

tum

num

ber,

m�,

desc

ribe

sth

e or

ient

atio

n of

the

elec

tron

’s o

rbit

com

pare

d to

the

plan

e of

the

atom

. The

fou

rth

quan

tum

num

ber,

ms,

tells

the

dire

ctio

n of

the

elec

tron

’s s

pin

(clo

ckw

ise

or c

ount

ercl

ockw

ise)

.

The

Sch

rödi

nger

wav

e eq

uatio

n im

pose

s ce

rtai

n m

athe

mat

ical

res

tric

tions

on

the

quan

tum

num

bers

. The

y ar

e as

fol

low

s:

nca

n be

any

inte

ger

(who

le n

umbe

r),

�ca

n be

any

inte

ger

from

0 to

n�

1,

m�

can

be a

ny in

tege

r fr

om �

�to

��,

and

ms

can

be �

or �

As

an e

xam

ple,

cons

ider

ele

ctro

ns in

the

firs

t ene

rgy

leve

l of

an a

tom

,tha

t is,

n�

1. I

nth

is c

ase,

�ca

n ha

ve a

ny in

tegr

al v

alue

fro

m 0

to (

n�

1),o

r 0

to (

1 �

1). I

n ot

her

wor

ds,

�m

ust b

e 0

for

thes

e el

ectr

ons.

Als

o,th

e on

ly v

alue

that

m�

can

have

is 0

. The

ele

ctro

ns in

this

ene

rgy

leve

l can

hav

e va

lues

of

�or

�fo

r m

s. T

hese

res

tric

tions

agr

ee w

ith th

e

obse

rvat

ion

that

the

firs

t ene

rgy

leve

l can

hav

e on

ly tw

o el

ectr

ons.

The

ir q

uant

um n

umbe

rs

are

1,0,

0,�

and

1,0,

0 �

.

Use

the

rule

s gi

ven

abov

e to

com

plet

e th

e ta

ble

listin

g th

e qu

antu

m n

umbe

rs f

or e

ach

elec

tron

in a

bor

on a

tom

. The

cor

rect

qua

ntum

num

bers

for

one

ele

ctro

n in

the

atom

is

prov

ided

as

an e

xam

ple.

1 � 21 � 2

1 � 21 � 2

1 � 21 � 2

Use

wit

h Ch

apte

r 5,

Sect

ion

5.2

Elec

tron

n�

m�

ms

11

00

�

2 3 4 5

1 � 2

Bo

ron

(B

)Fo

r b

oro

n, t

he

firs

t tw

oel

ectr

on

s h

ave

qu

antu

mn

um

ber

s 1,

0,0,

�1 /

2 an

d1,

0,0,

�1 /

2. T

he

seco

nd

tw

oel

ectr

on

s h

ave

qu

antu

mn

um

ber

s 2,

0,0,

�1 /

2 an

d2,

0,0,

�1 /

2. T

he

fift

h e

lect

ron

may

hav

e an

y o

ne

of

six

po

ssib

le q

uan

tum

nu

mb

ers:

2,1,

1,�

1 /2;

2,1

,1,�

1 /2;

2,1,

0,�

1 /2;

2,1

,0,�

1 /2;

2,1,

1,�

1 /2;

2,1

,1,�

1 /2.

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

8C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

8C

hal

len

ge

Pro

ble

ms

Com

pari

ng t

he S

truc

ture

s of

Ato

ms

and

Ions

Com

pari

ng t

he S

truc

ture

s of

Ato

ms

and

Ions

The

che

mic

al p

rope

rtie

s of

an

elem

ent d

epen

d pr

imar

ily o

n its

num

ber

of v

alen

ce e

lect

rons

in

its a

tom

s. T

he n

oble

gas

ele

men

ts,f

or e

xam

ple,

all h

ave

sim

ilar

chem

ical

pro

pert

ies

beca

use

the

oute

rmos

t ene

rgy

leve

ls o

f th

eir

atom

s ar

e co

mpl

etel

y fi

lled.

The

che

mic

al p

rope

rtie

sof

ions

als

o de

pend

on

the

num

ber

of v

alen

ce e

lect

rons

. Any

ion

with

a c

ompl

ete

oute

rmos

ten

ergy

leve

l will

hav

e ch

emic

al p

rope

rtie

s si

mila

r to

thos

e of

the

nobl

e ga

s el

emen

ts. T

he f

luo-

ride

ion

(F�

),fo

r ex

ampl

e,ha

s a

tota

l of

ten

elec

tron

s,ei

ght o

f w

hich

fill

its

oute

rmos

t ene

rgy

leve

l. F�

has

chem

ical

pro

pert

ies,

ther

efor

e,si

mila

r to

thos

e of

the

nobl

e ga

s ne

on.

Show

n be

low

are

the

Lew

is e

lect

ron

dot s

truc

ture

s fo

r fi

ve e

lem

ents

:sul

fur

(S),

chlo

rine

(C

l),

argo

n (A

r),p

otas

sium

(K

),an

d ca

lciu

m (

Ca)

. Ans

wer

the

ques

tions

bel

ow a

bout

thes

e st

ruct

ures

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

8

Use

wit

h Ch

apte

r 8,

Sect

ion

8.1

1.

Wri

te th

e at

omic

num

ber

for

each

of

the

five

ele

men

ts s

how

n ab

ove.

S: 1

6; C

l: 17

; Ar:

18;

K: 1

9; C

a: 2

0

2.

Wri

te th

e el

ectr

on c

onfi

gura

tion

for

each

of

the

five

ele

men

ts.

S: 1

s22s

2 2p

6 3s2

3p4 ;

Cl:

1s2 2

s22p

6 3s2

3p5 ;

Ar:

1s2

2s2 2

p6 3

s23p

6 ;

K: 1

s22s

2 2p

6 3s2

3p6 4

s1; C

a: 1

s22s

2 2p

6 3s2

3p6 4

s2

3.

Whi

ch o

f th

e ab

ove

Lew

is e

lect

ron

dot s

truc

ture

s is

the

sam

e as

the

Lew

is e

lect

ron

dot

stru

ctur

e fo

r th

e io

n S2�

? E

xpla

in y

our

answ

er.

The

Lew

is e

lect

ron

do

t st

ruct

ure

fo

r ar

go

n; b

oth

arg

on

an

d S

2 �h

ave

18 e

lect

ron

s,

eig

ht

of

wh

ich

are

val

ence

ele

ctro

ns.

4.

Whi

ch o

f th

e ab

ove

Lew

is e

lect

ron

dot s

truc

ture

s is

the

sam

e as

that

for

the

ion

Cl�

?E

xpla

in y

our

answ

er.

The

Lew

is e

lect

ron

do

t st

ruct

ure

fo

r ar

go

n; b

oth

arg

on

an

d C

l�h

ave

18 e

lect

ron

s,

eig

ht

of

wh

ich

are

val

ence

ele

ctro

ns.

5.

Whi

ch o

f th

e ab

ove

Lew

is e

lect

ron

dot s

truc

ture

s is

like

that

for

the

ion

K�

? E

xpla

inyo

ur a

nsw

er.

The

Lew

is e

lect

ron

do

t st

ruct

ure

fo

r ar

go

n; b

oth

arg

on

an

d K

�h

ave

18 e

lect

ron

s,

eig

ht

of

wh

ich

are

val

ence

ele

ctro

ns.

6.

Nam

e an

ion

of c

alci

um th

at h

as c

hem

ical

pro

pert

ies

sim

ilar

to th

ose

of a

rgon

. Exp

lain

your

ans

wer

.

Ca2

�; l

ike

arg

on

, Ca2

�h

as 1

8 el

ectr

on

s, e

igh

t o

f w

hic

h a

re v

alen

ce e

lect

ron

s.

SC

lA

rK

Ca

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

77

Abu

ndan

ce o

f th

e El

emen

tsA

bund

ance

of

the

Elem

ents

The

abu

ndan

ce o

f th

e el

emen

ts d

iffe

rs s

igni

fica

ntly

in v

ario

us p

arts

of

the

univ

erse

. The

tabl

e be

low

list

s th

e ab

unda

nce

of s

ome

elem

ents

in v

ario

uspa

rts

of th

e un

iver

se. U

se th

e ta

ble

to a

nsw

er th

e fo

llow

ing

ques

tions

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

7

1.

Wha

t per

cent

of

all a

tom

s in

the

univ

erse

are

eith

er h

ydro

gen

or h

eliu

m?

Wha

t per

cent

of

all a

tom

s in

the

sola

r sy

stem

are

eith

er h

ydro

gen

or h

eliu

m?

un

iver

se: 9

27 H

�71

.8 H

e �

1000

�10

0% �

99.9

%;

sola

r sy

stem

: 863

H �

135

He

�10

00 �

100%

�99

.8%

2.

Exp

lain

the

rela

tivel

y hi

gh a

bund

ance

of

hydr

ogen

and

hel

ium

in th

e un

iver

se c

ompa

red

to th

eir

rela

tivel

y lo

w a

bund

ance

on

Ear

th.

Hyd

rog

en a

nd

hel

ium

are

gas

es o

f lo

w d

ensi

ty t

hat

hav

e es

cap

ed f

rom

Ear

th’s

gra

vita

tio

nal

att

ract

ion

.

3.

Onl

y th

e to

p fo

ur m

ost a

bund

ant e

lem

ents

on

Ear

th a

nd in

Ear

th’s

cru

st a

re s

how

n in

the

tabl

e. N

ame

two

addi

tiona

l ele

men

ts y

ou w

ould

exp

ect t

o fi

nd a

mon

g th

e to

p te

n el

e-m

ents

bot

h on

Ear

th a

nd in

Ear

th’s

cru

st. E

xpla

in y

our

choi

ces.

Bas

ed o

n t

he

info

rmat

ion

in S

ecti

on

7.1

, alu

min

um

an

d c

alci

um

are

po

ssib

le

cho

ices

. Alu

min

um

is t

he

mo

st a

bu

nd

ant

met

al in

Ear

th’s

cru

st. C

alci

um

fo

rms

man

y ki

nd

s o

f ro

cks

wit

h o

xyg

en, s

ilico

n, a

nd

car

bo

n.

4.

Nam

e at

leas

t thr

ee e

lem

ents

in a

dditi

on to

thos

e sh

own

in th

e ta

ble

that

you

wou

ldex

pect

to f

ind

in th

e lis

t of

the

top

ten

elem

ents

in th

e hu

man

bod

y. E

xpla

in y

our

choi

ces.

Stu

den

ts’ a

nsw

ers

may

var

y. E

lem

ents

dis

cuss

ed in

Sec

tio

n 7

.1 a

s h

avin

g

imp

ort

ant

role

s in

th

e h

um

an b

od

y in

clu

de

sod

ium

(#8

in li

st o

f to

p t

en),

po

tass

ium

(#9

), c

hlo

rin

e (#

10),

an

d p

ho

sph

oru

s (#

7).

Use

wit

h Ch

apte

r 7,

Sect

ion

7.1

Ab

un

dan

ce (

Nu

mb

er

of

ato

ms

per

1000 a

tom

s)*

Elem

ent

Uni

vers

eSo

lar

Syst

emEa

rth

Eart

h’s

Crus

tH

uman

Bod

y

Hyd

rog

en92

786

330

606

Hel

ium

71.8

135

Oxy

gen

0.51

00.

783

500

610

257

Nit

rog

en0.

153

0.08

0924

Car

bo

n0.

0811

0.45

910

6

Silic

on

0.02

310.

0269

140

210

Iro

n0.

0139

0.00

320

170

19

* A

n e

lem

ent

is n

ot

abu

nd

ant

in a

reg

ion

th

at is

left

bla

nk.

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

10C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

10C

hal

len

ge

Pro

ble

ms

Bala

ncin

g Ch

emic

alEq

uati

ons

Bala

ncin

g Ch

emic

alEq

uati

ons

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

10

Use

wit

h Ch

apte

r 10

,Se

ctio

n 10

.1

Eac

h ch

emic

al e

quat

ion

belo

w c

onta

ins

at le

ast o

ne e

rror

. Ide

ntif

y th

e er

ror

or e

rror

s an

d th

en w

rite

the

corr

ect c

hem

ical

equ

atio

n fo

r th

e re

actio

n.

1.

K(s

) �

2H2O

(l) 0

2KO

H(a

q) �

H2(

g)Th

e eq

uat

ion

is n

ot

bal

ance

d. 2

K(s

) �

2H2O

(l) 0

2KO

H(a

q)

�H

2(g

)

2.

MgC

l 2(aq

) �

H2S

O4(

aq) 0

Mg(

SO4)

2(aq

) �

2HC

l(aq

)A

n in

corr

ect

form

ula

is g

iven

fo

r m

agn

esiu

m s

ulf

ate.

Th

eref

ore

, th

e eq

uat

ion

is

no

t b

alan

ced

. Mg

Cl 2

(aq

) �

H2S

O4(

aq) 0

Mg

SO4(

aq)

�2H

Cl(

aq)

3.

AgN

O3(

aq)

�H

2S(a

q) 0

Ag 2S

(aq)

�H

NO

3(aq

)Th

e eq

uat

ion

is n

ot

bal

ance

d. 2

Ag

NO

3(aq

) �

H2S

(aq

) 0

Ag

2S(a

q)

�2H

NO

3(aq

)

4.

Sr(s

) �

F 2(g)

0Sr

2FA

n in

corr

ect

form

ula

is g

iven

fo

r st

ron

tiu

m f

luo

rid

e. T

her

efo

re, t

he

equ

atio

n is

no

t b

alan

ced

. Als

o, t

he

ph

ysic

al s

tate

of

stro

nti

um

flu

ori

de

is n

ot

giv

en.

Sr(s

) �

F 2(g

) 0

SrF 2

(s)

5.

2NaH

CO

3(s)

�2H

Cl(

aq) 0

2NaC

l(s)

�2C

O2(

g)A

ter

m m

ust

be

mis

sin

g b

ecau

se t

her

e is

no

pro

du

ct c

on

tain

ing

hyd

rog

en. T

her

efo

re,

the

equ

atio

n is

no

t b

alan

ced

. 2N

aHC

O3(

s) �

2HC

l(aq

) 0

2NaC

l(s)

�2C

O2(

g)

�2H

2O(l)

6.

2LiO

H(a

q) �

2HB

r(aq

) 0

2LiB

r(aq

) �

2H2O

The

coef

fici

ents

are

no

t ex

pre

ssed

in lo

wes

t te

rms.

Als

o, t

he

ph

ysic

al s

tate

of

wat

er is

no

t g

iven

. LiO

H(a

q)

�H

Br(

aq) 0

LiB

r(aq

) �

H2O

(l)

7.

NH

4OH

(aq)

�K

OH

(aq)

0K

OH

(aq)

�N

H4O

H(a

q)

The

pro

du

cts

are

the

sam

e as

th

e re

acta

nts

, so

no

rea

ctio

n o

ccu

rred

an

d n

o

equ

atio

n c

an b

e w

ritt

en.

8.

2Ca(

s) �

Cl 2(

g) 0

2CaC

l(aq

)A

n in

corr

ect

form

ula

is g

iven

fo

r ca

lciu

m c

hlo

rid

e. C

a(s)

�C

l 2(g

) 0

CaC

l 2(a

q)

9.

H2S

O4(

aq)

�2A

l(N

O3)

3(aq

) 0

Al 2(

SO4)

3(aq

) �

2HN

O3(

aq)

The

equ

atio

n is

no

t b

alan

ced

. 3H

2SO

4(aq

) �

2Al(

NO

3)3(

aq) 0

Al 2

(SO

4)3(

aq)

�

6HN

O3(

aq)

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

99

Exce

ptio

ns t

o th

e O

ctet

Rul

eEx

cept

ions

to

the

Oct

et R

ule

The

oct

et r

ule

is a

n im

port

ant g

uide

to u

nder

stan

ding

how

mos

t com

poun

ds a

re f

orm

ed.

How

ever

,the

re a

re a

num

ber

of c

ases

in w

hich

the

octe

t rul

e do

es n

ot a

pply

. Ans

wer

the

follo

win

g qu

estio

ns a

bout

exc

eptio

ns to

the

octe

t rul

e.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

9

1.

Dra

w th

e L

ewis

str

uctu

re f

or th

e co

mpo

und

BeF

2.

2.

Doe

s B

eF2

obey

the

octe

t rul

e? E

xpla

in.

BeF

2d

oes

no

t o

bey

th

e o

ctet

ru

le b

ecau

se t

he

cen

tral

ato

m (

Be)

of

the

mo

lecu

le

is n

ot

surr

ou

nd

ed b

y fo

ur

elec

tro

n p

airs

.

3.

Dra

w th

e L

ewis

str

uctu

re f

or th

e co

mpo

und

NO

2.

4.

Doe

s N

O2

obey

the

octe

t rul

e? E

xpla

in.

NO

2d

oes

no

t o

bey

th

e o

ctet

ru

le b

ecau

se t

he

cen

tral

ato

m (

N)

of

the

mo

lecu

le is

no

t su

rro

un

ded

by

fou

r el

ectr

on

pai

rs.

5.

Dra

w th

e L

ewis

str

uctu

re f

or th

e co

mpo

und

N2F

2.

6.

Doe

s N

2F2

obey

the

octe

t rul

e? E

xpla

in.

N2F

2o

bey

s th

e o

ctet

ru

le b

ecau

se e

very

ato

m o

f th

e m

ole

cule

is s

urr

ou

nd

ed

by

fou

r el

ectr

on

s p

airs

.

7.

Dra

w th

e L

ewis

str

uctu

re f

or th

e co

mpo

und

IF5.

8.

Doe

s IF

5ob

ey th

e oc

tet r

ule?

Exp

lain

.IF

5d

oes

no

t o

bey

th

e o

ctet

ru

le b

ecau

se t

her

e ar

e 12

ele

ctro

ns

atta

ched

to

the

cen

tral

ato

m (

I).

Use

wit

h Ch

apte

r 9,

Sect

ion

9.3

F

Be

F

O

N

O

F N

N

F

F

FF

FF

�

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

12C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

12C

hal

len

ge

Pro

ble

ms

Mol

e R

elat

ions

hips

inCh

emic

al R

eact

ions

Mol

e R

elat

ions

hips

inCh

emic

al R

eact

ions

The

mol

e pr

ovid

es a

con

veni

ent w

ay o

f fi

ndin

g th

e am

ount

s of

the

subs

tanc

es in

a c

hem

ical

re

actio

n. T

he d

iagr

am b

elow

sho

ws

how

this

con

cept

can

be

appl

ied

to th

e re

actio

nbe

twee

n ca

rbon

mon

oxid

e (C

O)

and

oxyg

en (

O2)

,sho

wn

in th

e fo

llow

ing

bala

nced

equ

atio

n.

2CO

(g)

�O

2(g)

02C

O2(

g)

Use

the

equa

tion

and

the

diag

ram

to a

nsw

er th

e fo

llow

ing

ques

tions

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

12

Use

wit

h Ch

apte

r 12

,Se

ctio

n 12

.2

1.

Wha

t inf

orm

atio

n is

nee

ded

to m

ake

the

type

s of

con

vers

ions

sho

wn

by d

oubl

e-ar

row

1in

the

diag

ram

?

the

coef

fici

ents

of

the

sub

stan

ces

in t

he

bal

ance

d c

hem

ical

eq

uat

ion

2.

Wha

t con

vers

ion

fact

ors

wou

ld b

e ne

eded

to m

ake

the

conv

ersi

ons

repr

esen

ted

by

doub

le-a

rrow

2 in

the

diag

ram

for

CO

? B

y do

uble

-arr

ow 6

for

CO

2?28

.01

g C

O/1

mo

l CO

an

d 1

mo

l CO

/28.

01 g

CO

; 44.

01 g

CO

2/1

mo

l CO

2an

d

1 m

ol C

O2/

44.0

1 g

CO

2

3.

Wha

t inf

orm

atio

n is

nee

ded

to m

ake

the

type

s of

con

vers

ions

rep

rese

nted

by

doub

le-a

rrow

s 3

and

7 in

the

diag

ram

?

the

nu

mb

er o

f re

pre

sen

tati

ve p

arti

cles

in a

mo

le, o

r A

vog

adro

’s n

um

ber

:

6.02

�10

23

4.

Wha

t con

vers

ion

fact

ors

wou

ld b

e ne

eded

to m

ake

the

conv

ersi

ons

repr

esen

ted

by

doub

le-a

rrow

3 in

the

diag

ram

for

CO

?

6.02

�10

23p

arti

cles

CO

/1 m

ol C

O a

nd

1 m

ol C

O/6

.02

�10

23p

arti

cles

CO

5.

Why

is it

not

pos

sibl

e to

con

vert

bet

wee

n th

e m

ass

of a

sub

stan

ce a

nd th

e nu

mbe

r of

re

pres

enta

tive

part

icle

s,as

rep

rese

nted

by

doub

le-a

rrow

4 o

f th

e di

agra

m?

6.

Why

is it

not

pos

sibl

e to

use

the

mas

s of

one

sub

stan

ce in

a c

hem

ical

rea

ctio

n to

fin

d th

e m

ass

of a

sec

ond

subs

tanc

e in

the

reac

tion,

as r

epre

sent

ed b

y do

uble

-arr

ow 5

in th

e di

agra

m?

The

mas

ses

of

the

sub

stan

ces

do

no

t re

act

wit

h e

ach

oth

er in

def

init

e ra

tio

s.

Mo

les

of

CO

Gra

ms

of

CO

Mo

les

of

CO

2

Gra

ms

of

CO

2

Part

icle

s o

fC

OPa

rtic

les

of

CO

2

1 5

2

3

46

7

Mas

ses

of

dif

fere

nt

sub

stan

ces

con

tain

dif

fere

nt

nu

mb

ers

of

rep

rese

nta

tive

par

ticl

es.

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

1111

Usi

ng M

ole-

Base

dCo

nver

sion

sU

sing

Mol

e-Ba

sed

Conv

ersi

ons

The

dia

gram

sho

ws

thre

e co

ntai

ners

,eac

h of

whi

ch h

olds

a c

erta

in m

ass

of th

e su

bsta

nce

indi

cate

d. C

ompl

ete

the

tabl

e be

low

for

eac

h of

the

thre

e su

bsta

nces

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

11

1.

Com

pare

and

con

tras

t the

num

ber

of r

epre

sent

ativ

e pa

rtic

les

and

the

mas

s of

UF 6

with

the

num

ber

of r

epre

sent

ativ

e pa

rtic

les

and

mas

s of

CC

l 3CF 3.

Exp

lain

any

dif

fere

nces

yo

u ob

serv

e.Ev

en t

ho

ug

h t

he

mas

s o

f U

F 6is

gre

ater

th

an t

he

mas

s o

f C

Cl 3

CF 3

in t

hei

r re

spec

-

tive

co

nta

iner

s, t

her

e ar

e fe

wer

mo

lecu

les

of

UF 6

than

of

CC

l 3C

F 3. T

hat

is b

ecau

se

the

mo

lar

mas

s o

f U

F 6is

sig

nif

ican

tly

gre

ater

th

an t

hat

of

CC

l 3C

F 3. A

sin

gle

UF 6

mo

lecu

le h

as a

sig

nif

ican

tly

gre

ater

mas

s th

an a

sin

gle

CC

l 3C

F 3m

ole

cule

has

.

2.

UF 6

is a

gas

use

d in

the

prod

uctio

n of

fue

l for

nuc

lear

pow

er p

lant

s. H

ow m

any

mol

es o

fth

e ga

s ar

e in

100

.0 g

of

UF 6?

100.

0 g

UF 6

�1

mo

l UF 6

/352

.03

g U

F 6�

0.28

41 m

ol U

F 6

3.

CC

l 3CF 3

is a

chl

orof

luor

ocar

bon

resp

onsi

ble

for

the

dest

ruct

ion

of th

e oz

one

laye

r in

Ear

th’s

atm

osph

ere.

How

man

y m

olec

ules

of

the

liqui

d ar

e in

1.0

g o

f C

Cl 3C

F 3?

1.0

g C

Cl 3

CF 3

�1.

0 m

ol C

Cl 3

CF 3

/187

.37

g C

Cl 3

CF 3

�6.

02 �

1023

mo

lecu

les

CC

l 3C

F 3/

1.0

mo

l CC

l 3C

F 3 �

3.2

�10

21m

ole

cule

s C

Cl 3

CF 3

4.

Lea

d (P

b) is

use

d to

mak

e a

num

ber

of d

iffe

rent

allo

ys. W

hat i

s th

e m

ass

of le

ad p

rese

ntin

an

allo

y co

ntai

ning

0.1

5 m

ol o

f le

ad?

0.15

mo

l Pb

�20

7.2

g P

b/1

.0 m

ol P

b �

31 g

Pb

UF 6

(g

)

225.

0 g

CC

l 3C

F 3 (

l)

200.

0 g

Pb (

s)

250.

0 gUse

wit

h Ch

apte

r 11

,Se

ctio

n 11

.3

Mol

ar M

ass

Num

ber

of

Num

ber

of R

epre

sent

ativ

e Su

bsta

nce

Mas

s (g

)(g

/mol

)M

oles

(m

ol)

Part

icle

s

UF 6

(g)

225.

035

2.03

0.63

923.

848

�10

23m

ole

cule

s

CC

l 3C

F 3(l

)20

0.0

187.

371.

067

6.42

3 �

1023

mo

lecu

les

Pb(s

)25

0.0

207.

21.

207

7.26

6 �

1023

ato

ms

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

14C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

14C

hal

len

ge

Pro

ble

ms

A S

impl

e M

ercu

ry B

arom

eter

A S

impl

e M

ercu

ry B

arom

eter

In Fi

gure

1,a

sim

ple

mer

cury

bar

omet

er is

mad

e by

fill

ing

a lo

ng

glas

s tu

be w

ith m

ercu

ry a

nd th

en in

vert

ing

the

open

end

of

the

tube

into

a b

owl o

f m

ercu

ry. A

nsw

er th

e fo

llow

ing

ques

tions

abo

utth

e si

mpl

e m

ercu

ry b

arom

eter

sho

wn

here

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

14

Use

wit

h Ch

apte

r 14

,Se

ctio

n 14

.1

1.

Wha

t occ

upie

s th

e sp

ace

abov

e th

e m

ercu

ry c

olum

n in

the

baro

met

er’s

gla

ss tu

be?

The

spac

e is

occ

up

ied

on

ly b

y sm

all a

mo

un

ts o

f

Hg

vap

or

du

e to

th

e p

rese

nce

of

a va

cuu

m a

bo

ve

the

mer

cury

co

lum

n in

th

e g

lass

tu

be.

2.

Wha

t pre

vent

s m

ercu

ry f

rom

flo

win

g ou

t of

the

glas

s tu

be in

to th

e bo

wl o

f m

ercu

ry?

Air

pre

ssu

re p

ush

ing

on

th

e su

rfac

e o

f th

e m

ercu

ry in

th

e b

ow

l kee

ps

the

mer

cury

in t

he

gla

ss t

ub

e.

3.

Whe

n th

e ba

rom

eter

in F

igur

e 1

is m

oved

to a

hig

her

elev

atio

n,su

ch a

s an

alti

tude

of

5000

met

ers,

the

colu

mn

of m

ercu

ry c

hang

es a

s sh

own

in F

igur

e 2.

Why

is th

e m

ercu

ryco

lum

n lo

wer

in F

igur

e 2

than

in F

igur

e 1?

Bec

ause

air

pre

ssu

re a

t th

e h

igh

er e

leva

tio

n in

Fig

ure

2 is

less

th

an t

he

air

pre

ssu

re a

t th

e lo

wer

ele

vati

on

in F

igu

re 1

, th

ere

is le

ss p

ush

ing

on

th

e su

rfac

e

of

the

mer

cury

in t

he

bo

wl i

n F

igu

re 2

.

4.

Supp

ose

the

baro

met

er in

Fig

ure

1 w

as c

arri

ed in

to a

n op

en m

ine

500

met

ers

belo

w s

eale

vel.

How

wou

ld th

e he

ight

of

the

mer

cury

col

umn

chan

ge?

Exp

lain

why

.

The

hei

gh

t o

f th

e m

ercu

ry c

olu

mn

wo

uld

incr

ease

in t

he

min

e b

ecau

se m

ore

air

pre

ssu

re w

ou

ld b

e p

ush

ing

on

th

e su

rfac

e o

f th

e m

ercu

ry in

th

e b

ow

l.

5.

Supp

ose

the

liqui

d us

ed to

mak

e th

e ba

rom

eter

was

wat

er in

stea

d of

mer

cury

. How

wou

ldth

is s

ubst

itutio

n af

fect

the

baro

met

er?

Exp

lain

.

The

colu

mn

of

liqu

id in

th

e tu

be

wo

uld

be

lon

ger

. Bec

ause

wat

er is

mu

ch le

ss

den

se t

han

mer

cury

, air

pre

ssu

re is

ab

le t

o s

up

po

rt a

mu

ch lo

ng

er c

olu

mn

of

wat

er t

han

of

mer

cury

.

6.

Supp

ose

a tin

y cr

ack

form

ed a

t the

top

of th

e ba

rom

eter

’s g

lass

tube

. How

wou

ld th

isev

ent a

ffec

t the

col

umn

of m

ercu

ry?

Exp

lain

why

.

Air

wo

uld

en

ter

the

gla

ss t

ub

e, c

ausi

ng

th

e m

ercu

ry c

olu

mn

to

flo

w o

ut

of

the

tub

e. A

ir p

ress

ure

exe

rted

on

th

e m

ercu

ry in

th

e tu

be

wo

uld

eq

ual

th

e ai

r

pre

ssu

re e

xert

ed o

n t

he

mer

cury

in t

he

bo

wl.

Gla

ss t

ub

e

Mer

cury

co

lum

n

Bo

wl o

f m

ercu

ry

At

sea

leve

lA

t 50

0 m

eter

sab

ove

sea

leve

l

Figu

re 1

Figu

re 2

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

1313

Inte

rmol

ecul

ar F

orce

s an

dBo

iling

Poi

nts

Inte

rmol

ecul

ar F

orce

s an

dBo

iling

Poi

nts

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

13

1.

How

do

the

boili

ng p

oint

s of

the

grou

p 4A

hyd

ride

s ch

ange

as

the

mol

ecul

ar m

asse

s of

the

hydr

ides

cha

nge?

As

the

mo

lecu

lar

mas

ses

of

the

hyd

rid

es in

crea

se, s

o d

o t

he

bo

ilin

g p

oin

ts o

f th

e

hyd

rid

es.

2.

Wha

t are

the

mol

ecul

ar s

truc

ture

and

pol

arity

of

the

four

gro

up 4

A h

ydri

des?

All

gro

up

4A

hyd

rid

e m

ole

cule

s ar

e re

gu

lar

tetr

ahed

ron

s an

d a

re n

on

po

lar.

3.

Pred

ict t

he s

tren

gth

of th

e fo

rces

bet

wee

n gr

oup

4A h

ydri

de m

olec

ules

. Exp

lain

how

thos

e fo

rces

aff

ect t

he b

oilin

g po

ints

of

grou

p 4A

hyd

ride

s.

Bec

ause

gro

up

4A

hyd

rid

es a

re n

on

po

lar,

the

rela

tive

ly w

eak

inte

rmo

lecu

lar

forc

es b

etw

een

th

eir

mo

lecu

les

hav

e lit

tle

effe

ct o

n t

hei

r b

oili

ng

po

ints

.

4.

How

do

the

boili

ng p

oint

s of

the

grou

p 6A

hyd

ride

s ch

ange

as

the

mol

ecul

ar m

asse

s of

the

hydr

ides

cha

nge?

Wit

h t

he

exce

pti

on

of

wat

er, a

s th

e m

ole

cula

r m

asse

s o

f th

e h

ydri

des

incr

ease

, so

do

th

e b

oili

ng

po

ints

of

the

hyd

rid

es. W

ater

, ho

wev

er, h

as a

mu

ch h

igh

er b

oili

ng

po

int

than

an

y o

ther

gro

up

6A

hyd

rid

e.

5.

Wha

t are

the

mol

ecul

ar s

truc

ture

and

pol

arity

of

the

four

gro

up 6

A h

ydri

des?

All

gro

up

6A

hyd

rid

e m

ole

cule

s ar

e b

ent,

po

lar

mo

lecu

les.

6.

Use

Tab

le 9

-4 in

you

r te

xtbo

ok to

det

erm

ine

the

diff

eren

ce in

ele

ctro

nega

tiviti

es o

f th

ebo

nds

in th

e fo

ur g

roup

6A

hyd

ride

s.H

—O

bo

nd

: �en

�1.

2; H

—S

bo

nd

: �en

�0.

4; H

—Se

bo

nd

: �en

�0.

4;

H—

Te b

on

d: �

en�

0.1

100 0

�10

0

H2O

H2S CH

4SiH

4

GeH

4

SnH

4

H2S

e

H2T

e

Gro

up

6A

hyd

rid

es

Gro

up

4A

hyd

rid

es

Mo

lecu

lar

mas

s

Boiling point (°C)

0 050

100

150

Use

wit

h Ch

apte

r 13

,Se

ctio

n 13

.3

The

boi

ling

poin

ts o

f liq

uids

dep

end

part

ly o

n th

e m

ass

of th

e pa

rtic

les

of w

hich

they

are

mad

e. T

he g

reat

er th

e m

ass

ofth

e pa

rtic

les,

the

mor

e en

ergy

is n

eede

d to

con

vert

a li

quid

to a

gas,

and,

thus

,the

hig

her

the

boili

ng p

oint

of

the

liqui

d. T

his

pat-

tern

may

not

hol

d tr

ue,h

owev

er,w

hen

ther

e ar

e si

gnif

ican

t for

ces

betw

een

the

part

icle

s of

a li

quid

. The

gra

ph p

lots

boi

ling

poin

tve

rsus

mol

ecul

ar m

ass

for

grou

p 4A

and

gro

up 6

A h

ydri

des.

Ahy

drid

e is

a b

inar

y co

mpo

und

cont

aini

ng h

ydro

gen

and

one

othe

rel

emen

t. U

se th

e gr

aph

to a

nsw

er th

e fo

llow

ing

ques

tions

.

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

16C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

16C

hal

len

ge

Pro

ble

ms

Stan

dard

Hea

t of

For

mat

ion

Stan

dard

Hea

t of

For

mat

ion

Hes

s’s

law

allo

ws

you

to d

eter

min

e th

e st

anda

rd h

eat o

f fo

rmat

ion

of a

com

poun

dw

hen

you

know

the

heat

s of

rea

ctio

ns th

at le

adto

the

prod

uctio

n of

that

com

poun

d. T

he f

irst

diag

ram

on

the

righ

t sho

ws

how

Hes

s’s

law

can

be u

sed

to c

alcu

late

the

heat

of

form

atio

n of

CO

2by

kno

win

g th

e he

ats

of r

eact

ion

of tw

ost

eps

lead

ing

to th

e pr

oduc

tion

of C

O2.

Use

this

diag

ram

to h

elp

you

answ

er th

e qu

estio

ns b

elow

abou

t the

sec

ond

diag

ram

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

16

Use

wit

h Ch

apte

r 16

,Se

ctio

n 16

.4

The

equ

atio

ns b

elow

sho

w h

ow N

O2

can

be

form

ed in

two

way

s:di

rect

ly f

rom

the

elem

ents

or

in tw

o st

eps.

N2(

g) �

O2(

g) 0

NO

2(g)

�H

�33

kJ/

mol

or

N2(

g) �

O2(

g) 0

NO

(g)

�H

�91

kJ/

mol

NO

(g)

�O

2(g)

0N

O2(

g)�

H �

�58

kJ/

mol

1.

On

the

diag

ram

at t

he r

ight

,dra

w a

rrow

head

sto

sho

w th

e di

rect

ions

in w

hich

the

thre

e lin

esla

bele

d 1,

2,an

d 3

shou

ld p

oint

.

2.

Wri

te th

e co

rrec

t rea

ctan

ts a

nd/o

r pr

oduc

ts o

nea

ch o

f th

e lin

es la

bele

d A

,B,a

nd C

.

3.

Wri

te th

e co

rrec

t ent

halp

y ch

ange

nex

t to

each

num

ber

on th

e di

agra

m.

1 � 2

1 � 21 � 21 � 2

CO

(g)

�

O2(

g)

�H

� �

110

kJ/m

ol

C(s

) �

O2(

g)

�H

� �

393

kJ/m

ol

�H

� �

283

kJ/m

ol

Enthalpy

CO

2(g

)

1 2

NO

2(g

)

�H

� �

58 k

J/m

ol

NO

(g)

� 1

/2 O

2(g

)

�H

� 9

1 kJ

/mo

l

Enthalpy

�H

� 3

3 kJ

/mo

l

1

2

3

A

B

C

1 /2

N2(

g)

� O

2(g

)

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

1515

Vapo

r Pr

essu

re L

ower

ing

Vapo

r Pr

essu

re L

ower

ing

You

hav

e le

arne

d th

at a

ddin

g a

nonv

olat

ile s

olut

e to

a s

olve

nt

low

ers

the

vapo

r pr

essu

re o

f th

at s

olve

nt. T

he a

mou

nt b

y w

hich

the

vapo

r pr

essu

re is

low

ered

can

be

calc

ulat

ed b

y m

eans

of

a re

latio

nshi

p di

scov

ered

by

the

Fren

ch c

hem

ist F

ranç

ois

Mar

ieR

aoul

t (18

30–1

901)

in 1

886.

Acc

ordi

ng to

Rao

ult’s

law

,the

vap

orpr

essu

re o

f a

solv

ent (

P)

is e

qual

to th

e pr

oduc

t of

its v

apor

pre

ssur

ew

hen

pure

(P

0 ) a

nd it

s m

ole

frac

tion

(X)

in th

e so

lutio

n,or

P�

P0 X

The

sol

utio

n sh

own

at th

e ri

ght w

as m

ade

by a

ddin

g 75

.0 g

of

sucr

ose

(C12

H22

O11

) to

500

.0 g

of

wat

er a

t a te

mpe

ratu

re o

f 20

°C.

Ans

wer

the

follo

win

g qu

estio

ns a

bout

this

sol

utio

n.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

15

1.

Why

do

the

suga

r m

olec

ules

in th

e so

lutio

n lo

wer

the

vapo

r pr

essu

re o

f th

e w

ater

?

The

pre

sen

ce o

f su

gar

mo

lecu

les

at t

he

surf

ace

of

the

solu

tio

n r

edu

ces

the

nu

mb

er o

f w

ater

mo

lecu

les

at t

he

surf

ace,

th

us

red

uci

ng

th

e n

um

ber

of

wat

er

mo

lecu

les

able

to

esc

ape

into

th

e va

po

r p

has

e.

2.

Wha

t is

the

num

ber

of m

oles

of

sucr

ose

in th

e so

lutio

n?

75.0

g s

ucr

ose

�1

mo

l su

cro

se/3

42.3

g s

ucr

ose

�0.

219

mo

l su

cro

se

3.

Wha

t is

the

num

ber

of m

oles

of

wat

er in

the

solu

tion?

500.

0 g

wat

er �

1 m

ol w

ater

/18.

02 g

wat

er �

27.7

5 m

ol w

ater

4.

Wha

t is

the

mol

e fr

actio

n of

wat

er in

the

solu

tion?

Mo

le f

ract

ion

of

wat

er �

mo

les

solv

ent /(

mo

les

solv

ent

�m

ole

s so

lute

)

�27

.75

mo

l /(27

.75

mo

l �0.

219

mo

l) �

0.99

2

5.

Wha

t is

the

vapo

r pr

essu

re o

f th

e so

lutio

n if

the

vapo

r pr

essu

re o

f pu

re w

ater

at 2

0°C

is17

.54

mm

Hg?

P�

P0 X

; P�

(17.

54 m

m H

g)(

0.99

2) �

17.4

mm

Hg

6.

How

muc

h is

the

vapo

r pr

essu

re o

f th

e so

lutio

n re

duce

d fr

om th

at o

f w

ater

by

the

addi

tion

of th

e su

cros

e?

red

uct

ion

�17

.54

mm

Hg

�17

.4 m

m H

g �

0.1

mm

Hg

Wat

erm

ole

cule

Solu

tio

n

Sucr

ose

mo

lecu

le

Use

wit

h Ch

apte

r 15

,Se

ctio

n 15

.3

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


8 7 6 5 4 3 2 1

10

02

34

5Ti

me

(sec

)

Concentration (mol/L)

67

89

10

SO2

SO3

SO3

O2

O2

SO2

Nam

eD

ate

Cla

ss

18C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

18C

hal

len

ge

Pro

ble

ms

Chan

ging

Equ

ilibr

ium

Conc

entr

atio

ns in

a R

eact

ion

Rev

ersi

ble

reac

tions

eve

ntua

lly r

each

an

equi

libri

um

cond

ition

in w

hich

the

conc

entr

atio

ns o

f al

l rea

ctan

tsan

d pr

oduc

ts a

re c

onst

ant.

Equ

ilibr

ium

can

be

dist

urbe

d,ho

wev

er,b

y th

e ad

ditio

n or

rem

oval

of

eith

er a

rea

ctan

t or

prod

uct.

The

gra

ph o

n th

e ri

ght s

how

s ho

w th

e co

ncen

tra-

tions

of

the

reac

tant

s an

d pr

oduc

t of

a re

actio

n ch

ange

whe

n eq

uilib

rium

is d

istu

rbed

. Use

the

grap

h to

ans

wer

the

follo

win

g qu

estio

ns.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

18

Use

wit

h Ch

apte

r 18

,Se

ctio

n 18

.1

1.

Wri

te th

e eq

uatio

n fo

r th

e re

actio

n de

pict

ed in

the

grap

h.2S

O2(

g)

�O

2(g

) 3

2SO

3(g

)

2.

Wri

te th

e eq

uilib

rium

con

stan

t exp

ress

ion

for

the

reac

tion.

Keq

�[S

O3]

2 /[SO

2]2 [

O2]

3.

Exp

lain

the

shap

es o

f th

e cu

rves

for

the

thre

e ga

ses

duri

ng th

e fi

rst 2

min

utes

of

the

reac

tion.

The

con

cen

trat

ion

s o

f th

e tw

o r

eact

ants

, SO

2an

d O

2, d

ecre

ase

as t

hey

are

use

d

up

to

pro

du

ce S

O3;

th

e co

nce

ntr

atio

n o

f SO

3in

crea

ses.

4.

At a

ppro

xim

atel

y w

hat t

ime

does

the

reac

tion

reac

h eq

uilib

rium

? H

ow d

o yo

u kn

oweq

uilib

rium

has

bee

n re

ache

d?

Equ

ilib

riu

m is

rea

ched

at

abo

ut

3 m

inu

tes;

th

e co

nce

ntr

atio

ns

of

all t

hre

e

gas

es b

eco

me

con

stan

t at

th

at t

ime.

5.

Wha

t are

the

conc

entr

atio

ns o

f th

e th

ree

gase

s at

equ

ilibr

ium

?[S

O2]

eq�

6.0

mo

l/L;

[O

2]eq

�4.

5 m

ol/

L; [

SO3]

eq�

1.0

mo

l/L

6.

Cal

cula

te th

e va

lue

of K

eqfo

r th

e re

actio

n.

Keq

�[1

.0]2

/[6.0

]2[4

.5]

�1.

0 /(3

6)(4

.5)

�0.

0062

7.

Des

crib

e th

e ch

ange

mad

e in

the

syst

em 4

min

utes

into

the

reac

tion.

Tel

l how

you

kno

wth

e ch

ange

was

mad

e.SO

2w

as a

dd

ed t

o t

he

syst

em b

ecau

se t

he

con

cen

trat

ion

of

SO2

incr

ease

d

sud

den

ly a

t th

at t

ime.

8.

At w

hat t

ime

does

the

syst

em r

etur

n to

equ

ilibr

ium

?

at 6

min

ute

s

Chan

ging

Equ

ilibr

ium

Conc

entr

atio

ns in

a R

eact

ion

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

1717

Det

erm

inin

g R

eact

ion

Rat

esD

eter

min

ing

Rea

ctio

n R

ates

Din

itrog

en p

ento

xide

dec

ompo

ses

to p

rodu

ce

nitr

ogen

dio

xide

and

oxy

gen

as r

epre

sent

edby

the

follo

win

g eq

uatio

n.

2N2O

5(g)

04N

O2(

g) �

O2(

g)

The

gra

ph o

n th

e ri

ght r

epre

sent

s th

e co

ncen

-tr

atio

n of

N2O

5re

mai

ning

as

the

reac

tion

proc

eeds

over

tim

e. A

nsw

er th

e fo

llow

ing

ques

tions

abo

utth

e re

actio

n.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

17

1.

Wha

t is

the

conc

entr

atio

n of

N2O

5at

the

begi

nnin

g of

the

expe

rim

ent?

Aft

er 1

hou

r?A

fter

2 h

ours

? A

fter

10

hour

s?

1.50

mo

l/L;

1.2

0 m

ol/

L; 0

.90

mo

l/L;

0.1

8 m

ol/

L

2.

By

how

muc

h do

es th

e co

ncen

trat

ion

of N

2O5

chan

ge d

urin

g th

e fi

rst h

our

of th

e re

actio

n? C

alcu

late

the

perc

enta

ge o

f ch

ange

the

conc

entr

atio

n un

derg

oes

duri

ng th

e fi

rst h

our

of th

e re

actio

n.

1.50

mo

l/L

�1.

20 m

ol/

L �

0.30

mo

l/L;

(1.5

0 m

ol/

L �

1.20

mo

l/L)

/1.5

0 m

ol/

L �

.20

�10

0% �

20%

3.

The

inst

anta

neou

s ra

te o

f re

actio

n is

def

ined

as

the

chan

ge in

con

cent

ratio

n of

rea

ctan

tdu

ring

som

e sp

ecif

ied

time

peri

od,o

r in

stan

tane

ous

rate

of

reac

tion

= [

N2O

5]/t

. Wha

t is

the

inst

anta

neou

s ra

te o

f re

actio

n fo

r th

e de

com

posi

tion

of N

2O5

for

the

time

peri

odbe

twee

n th

e fi

rst a

nd s

econ

d ho

urs

of th

e re

actio

n? B

etw

een

the

seco

nd a

nd th

ird

hour

s?B

etw

een

the

sixt

h an

d se

vent

h ho

urs?

(1.2

0 m

ol/

L �

0.09

mo

l/L)

/1 h

�0.

30 m

ol/

L/h

; (0.

90 m

ol/

L �

0.70

mo

l/L)

/1 h

�

0.20

mo

l/L/

h; (

0.35

mo

l/L

�0.

30 m

ol/

L)/1

h �

0.05

0 m

ol/

L/h

4.

Wha

t is

the

inst

anta

neou

s ra

te o

f re

actio

n fo

r th

e de

com

posi

tion

of N

2O5

betw

een

the

sec-

ond

and

four

th h

ours

of

the

reac

tion?

Bet

wee

n th

e th

ird

and

eigh

th h

ours

of

the

reac

tion?

(0.9

0 m

ol/

L �

0.55

mo

l/L)

/2 h

�0.

18 m

ol/

L/h

;

(0.7

0 m

ol/

L �

0.25

mo

l/L)

/5 h

�0.

09 m

ol/

L/h

5.

How

long

doe

s it

take

for

0.1

0 m

ol o

f N

2O5

to d

ecom

pose

dur

ing

the

tent

h ho

ur o

f th

e re

actio

n?

0.20

mo

l/L/

h �

0.10

mo

l/L/

x

x�

(0.1

0 m

ol/

L)/(0

.20

mo

l/L/

h)

�5

ho

urs

6.

Wha

t is

the

aver

age

rate

of

reac

tion

for

the

deco

mpo

sitio

n of

N2O

5ov

eral

l?

(1.5

0 m

ol/

L �

0.18

mo

l/L)

10 h

�0.

13 m

ol/

L/h

Tim

e (h

)

Concentration (mol/L)

0.2 0

0.4

0.6

0.8

1.0

1.2

1.4

1.6

10

23

45

67

89

10

Use

wit

h Ch

apte

r 17

,Se

ctio

n 17

.1

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

20C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

20C

hal

len

ge

Pro

ble

ms

Bala

ncin

g O

xida

tion

–R

educ

tion

Equ

atio

nsBa

lanc

ing

Oxi

dati

on–

Red

ucti

on E

quat

ions

Sci

entis

ts h

ave

deve

lope

d a

num

ber

of m

etho

ds f

or p

rote

ctin

g m

etal

s fr

om o

xida

tion.

One

suc

h m

etho

d in

volv

es th

e us

e of

asa

crif

icia

l met

al. A

sac

rifi

cial

met

al is

a m

etal

that

is m

ore

easi

ly

oxid

ized

than

the

met

al it

is d

esig

ned

to p

rote

ct. G

alva

nize

d ir

on,f

orex

ampl

e,co

nsis

ts o

f a

piec

e of

iron

met

al c

over

ed w

ith a

thin

laye

r of

zin

c. W

hen

galv

aniz

ed ir

on is

exp

osed

to o

xyge

n,it

is th

e zi

nc,

rath

er th

an th

e ir

on,t

hat i

s ox

idiz

ed.

Wat

er h

eate

rs o

ften

con

tain

a m

etal

rod

that

is m

ade

by c

oatin

g a

heav

y st

eel w

ire

with

mag

nesi

um o

r al

umin

um. I

n th

is c

ase,

the

mag

nesi

um o

r al

umin

um is

the

sacr

ific

ial m

etal

,pro

tect

ing

the

iron

casi

ng o

f th

e he

ater

fro

m c

orro

sion

.

The

dia

gram

sho

ws

a po

rtio

n of

a w

ater

hea

ter

cont

aini

ng

a sa

crif

icia

l rod

. Ans

wer

the

follo

win

g qu

estio

ns a

bout

the

diag

ram

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

20

Use

wit

h Ch

apte

r 20

,Se

ctio

n 20

.3

1.

In th

e ab

senc

e of

a s

acri

fici

al m

etal

,oxy

gen

diss

olve

d in

wat

er m

ay r

eact

with

the

iron

casi

ng o

f th

e he

ater

. One

pro

duct

for

med

is ir

on(I

I) h

ydro

xide

(Fe

(OH

) 2). W

hich

ele

men

tis

oxi

dize

d an

d w

hich

is r

educ

ed in

this

rea

ctio

n?

Iro

n is

oxi

diz

ed, a

nd

oxy

gen

is r

edu

ced

.

2.

Bal

ance

the

oxid

atio

n–re

duct

ion

equa

tion

for

this

rea

ctio

n:Fe

(s)

�O

2(aq

) �

H2O

0Fe

(OH

) 2(aq

)

2Fe(

s) �

O2(

aq)

+ 2

H2O

02F

e(O

H) 2

(aq

)

3.

Wri

te th

e tw

o ha

lf-r

eact

ions

for

this

exa

mpl

e of

cor

rosi

on.

Fe(s

) 0

Fe2 �

(aq

) �

2e�

;

O2(

aq)

�2H

2O(l

) �

4e�0

4OH

�(a

q)

4.

Supp

ose

the

sacr

ific

ial r

od in

the

diag

ram

abo

ve is

coa

ted

with

alu

min

um m

etal

. Wri

teth

e ba

lanc

ed e

quat

ion

for

the

reac

tion

of a

lum

inum

with

oxy

gen

diss

olve

d in

the

wat

er.

(Hin

t:T

he p

rodu

ct f

orm

ed is

alu

min

um h

ydro

xide

(A

l(O

H) 3)

.4A

l(s)

�3O

2(aq

) �

6H2O

(l) 0

4Al(

OH

) 3(a

q)

5.

Wri

te th

e tw

o ha

lf-r

eact

ions

for

this

exa

mpl

e of

cor

rosi

on.

Al(

s) 0

Al3

�(a

q)

�3e

�; O

2(aq

) �

2H2O

(l)

�4e

�0

4OH

�(a

q)

6.

Supp

ose

that

som

e ir

on in

the

casi

ng o

f th

e w

ater

hea

ter

is o

xidi

zed,

as s

how

n in

the

equa

tion

of q

uest

ion

2 ab

ove.

The

sac

rifi

cial

met

al (

alum

inum

,in

this

cas

e) im

med

iate

lyre

stor

es th

e Fe

2�io

ns to

iron

ato

ms.

Wri

te tw

o ha

lf-r

eact

ions

that

rep

rese

nt th

is s

ituat

ion.

Al(

s) 0

Al3

�(a

q)

�3e

�; F

e2�

(aq

) �

2e�0

Fe(s

)

Iro

nca

sin

g

Stee

l wir

e

Sacr

ific

ial

met

al

Wat

er

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

1919

Swim

min

g Po

ol C

hem

istr

ySw

imm

ing

Pool

Che

mis

try

The

pre

senc

e of

dis

ease

-cau

sing

bac

teri

a in

sw

imm

ing

pool

s is

a m

ajor

hea

lth c

once

rn.

Chl

orin

e ga

s is

add

ed to

the

wat

er in

som

e la

rge

com

mer

cial

sw

imm

ing

pool

s to

kill

bact

eria

. How

ever

,in

mos

t hom

e sw

imm

ing

pool

s,ei

ther

sol

id c

alci

um h

ypoc

hlor

ite(C

a(O

Cl)

2) o

r an

aqu

eous

sol

utio

n of

sod

ium

hyp

ochl

orite

(N

aOC

l) is

use

d to

trea

t the

wat

er. B

oth

com

poun

ds d

isso

ciat

e in

wat

er to

for

m th

e w

eak

acid

hyp

ochl

orou

s ac

id(H

OC

l). H

ypoc

hlor

ous

acid

is a

hig

hly

effe

ctiv

e ba

cter

icid

e. B

y co

ntra

st,t

he h

ypoc

hlor

iteio

n (O

Cl�

) is

not

a v

ery

effe

ctiv

e ba

cter

icid

e. U

se th

e in

form

atio

n ab

ove

to a

nsw

er th

e fo

llow

ing

ques

tions

abo

ut th

e ac

id-b

ase

reac

tions

that

take

pla

ce in

sw

imm

ing

pool

s.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

19

1.

Wri

te a

n eq

uatio

n th

at s

how

s th

e re

actio

n be

twee

n hy

poch

loro

us a

cid

and

wat

er. I

dent

ify

the

acid

,bas

e,co

njug

ate

acid

,and

con

juga

te b

ase

in th

is r

eact

ion.

HO

Cl(

aq)

�H

2O(l

) 3

H3O

�(a

q)

�O

Cl�

(aq

)

The

acid

is H

OC

l, an

d t

he

bas

e is

H2O

. Th

e co

nju

gat

e ac

id is

H3O

�, a

nd

th

e

con

jug

ate

bas

e is

OC

l�.

2.

Wri

te a

n eq

uatio

n th

at s

how

s th

e re

actio

n th

at o

ccur

s w

hen

the

hypo

chlo

rite

ion

(OC

l�),

in th

e fo

rm o

f ca

lciu

m h

ypoc

hlor

ite o

r so

dium

hyp

ochl

orite

,is

adde

d to

wat

er. N

ame

the

acid

,bas

e,co

njug

ate

acid

,and

con

juga

te b

ase

in th

is r

eact

ion.

OC

l�(a

q)

�H

2O(l

) 3

HO

Cl(

aq)

�O

H�

The

acid

is H

2O, a

nd

th

e b

ase

is O

Cl�

. Th

e co

nju

gat

e ac

id is

HO

Cl,

and

th

e

con

jug

ate

bas

e is

OH

�.

3.

Wha

t eff

ect d

oes

the

addi

tion

of h

ypoc

hlor

ite io

n ha

ve o

n th

e pH

of

swim

min

g po

ol w

ater

?

The

add

itio

n o

f h

ypo

chlo

rite

ion

incr

ease

s th

e co

nce

ntr

atio

n o

f O

H�

and

ther

efo

re in

crea

ses

pH

.

4.

The

eff

ectiv

enes

s of

hyp

ochl

orite

ion

as a

bac

teri

cide

dep

ends

on

pH. H

ow d

oes

high

pH

affe

ct th

e eq

uilib

rium

rea

ctio

n de

scri

bed

in q

uest

ion

2? W

hat e

ffec

t wou

ld h

igh

pH h

ave

on th

e ba

cter

ia?

A h

igh

pH

val

ue

ind

icat

es a

hig

h c

on

cen

trat

ion

of

OH

�, w

hic

h t

end

s to

dri

ve t

he

reac

tio

n t

o t

he

left

. As

a re

sult

, th

ere

will

be

rela

tive

ly le

ss H

OC

l an

d r

elat

ivel

y

mo

re O

Cl�

in s

olu

tio

n, m

akin

g it

eas

ier

for

bac

teri

a to

su

rviv

e.

5.

In th

e pr

esen

ce o

f su

nlig

ht,h

ypoc

hlor

ite io

n de

com

pose

s to

for

m c

hlor

ide

ion

and

oxyg

en

gas.

Wri

te a

n eq

uatio

n fo

r th

is r

eact

ion

and

tell

how

it a

ffec

ts th

e sa

fety

of

pool

wat

er.

2OC

l�(a

q) 0

2Cl�

(aq

) �

O2(

g)

Rem

ova

l of

OC

l�d

rive

s th

e re

acti

on

des

crib

ed in

qu

esti

on

2 t

o t

he

left

, cau

sin

g

mo

re H

OC

l to

bre

ak d

ow

n, l

eavi

ng

less

HO

Cl t

o k

ill b

acte

ria.

Use

wit

h Ch

apte

r 19

,Se

ctio

n 19

.2

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

22C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

22C

hal

len

ge

Pro

ble

ms

Stru

ctur

al Is

omer

s of

Hex

ane

Stru

ctur

al Is

omer

s of

Hex

ane

The

str

uctu

ral f

orm

ula

of a

n or

gani

c co

mpo

und

can

som

etim

es b

e w

ritte

n in

a

vari

ety

of w

ays,

but s

omet

imes

str

uctu

ral f

orm

ulas

that

app

ear

sim

ilar

can

repr

esen

t dif

fere

nt c

ompo

unds

. The

str

uctu

ral f

orm

ulas

bel

ow a

re te

n w

ays

of

repr

esen

ting

com

poun

ds h

avin

g th

e m

olec

ular

for

mul

a C

6H14

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

22

Use

wit

h Ch

apte

r 22

,Se

ctio

ns 2

2.1

and

22.3

1.

In th

e sp

aces

pro

vide

d,w

rite

the

corr

ect n

ame

for

each

of

the

stru

ctur

al f

orm

ulas

,lab

eled

a–j,

abov

e.

a.

e.

i.

b.

f.j.

c.g

.

d.

h.

2.

How

man

y di

ffer

ent c

ompo

unds

are

rep

rese

nted

by

the

stru

ctur

al f

orm

ulas

abo

ve?

Wha

tar

e th

eir

nam

es?

five

; hex

ane,

2-m

eth

ylp

enta

ne,

2,3

-dim

eth

ylb

uta

ne,

2,2

-dim

eth

ylb

uta

ne,

and

3-m

eth

ylp

enta

ne

2-m

eth

ylp

enta

ne

2,2-

dim

eth

ylb

uta

ne

3-m

eth

ylp

enta

ne

2,3-

dim

eth

ylb

uta

ne

3-m

eth

ylp

enta

ne

2,3-

dim

eth

ylb

uta

ne

2-m

eth

ylp

enta

ne

hex

ane

3-m

eth

ylp

enta

ne

hex

ane

CH

3

CH

2C

H2

CH

2C

H2

CH

3

a.

CH

3

CH

3

CH

CH

2C

H2

CH

3

b.

CH

3

CH

3

CH

3C

HC

HC

H3

c.

CH

3

CH

3

CH

3C

CH

2C

H3

d.

CH

3

CH

CH

2

CH

2

CH

3

CH

3e.

CH

3C

HC

HC

H3

CH

3C

H3

f.

CH

2

CH

2C

H3

CH

CH

3

CH

3

g.

CH

3

CH

2

CH

3C

HC

H2

CH

3

h.

CH

2

CH

3

CH

2

CH

2C

H2

CH

3i.

CH

3

CH

2C

HC

H2

CH

3C

H3

j.

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

2121

Effe

ct o

f Co

ncen

trat

ion

onCe

ll Po

tent

ial

Effe

ct o

f Co

ncen

trat

ion

onCe

ll Po

tent

ial

In a

volta

ic c

ell w

here

all

ions

hav

e a

conc

entr

atio

n of

1M

,the

cel

l pot

entia

l is

equa

l to

the

stan

dard

pot

entia

l. Fo

r ce

lls in

whi

ch io

n co

ncen

trat

ions

are

gre

ater

or

less

than

1M

,as

show

n be

low

,an

adju

stm

ent m

ust b

e m

ade

to c

alcu

late

cel

l pot

entia

l.T

hat a

djus

tmen

t is

expr

esse

d by

the

Ner

nst e

quat

ion:

Ece

ll�

E0 ce

ll�

log

In th

is e

quat

ion,

nis

the

num

ber

of m

oles

of

elec

tron

s tr

ansf

erre

d in

the

reac

tion,

and

xan

d y

are

the

coef

fici

ents

of

the

prod

uct a

nd r

eact

ant i

ons,

resp

ectiv

ely,

in th

eba

lanc

ed h

alf-

cell

reac

tions

for

the

cell.

[pro

duct

ion]

x�

�[r

eact

ant i

on]y

0.05

92�

n

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

21

1.

Wri

te th

e tw

o ha

lf-r

eact

ions

and

the

over

all c

ell r

eact

ion

for

the

cell

show

n ab

ove.

Ag

�(a

q)

�1e

�0

Ag

(s)

Cu

(s) 0

Cu

2 �(a

q)

�2e

�

2Ag

�(a

q)

�C

u(s

) 0

2Ag

(s)

�C

u2 �

(aq

)

2.

Use

Tab

le 2

1-1

in y

our

text

book

to d

eter

min

e th

e st

anda

rd p

oten

tial o

f th

is c

ell.

3.

Wri

te th

e N

erns

t equ

atio

n fo

r th

e ce

ll.E c

ell

�0.

46 V

�0.

0592

/2�l

og

1.0

�10

�3 /(

1.0

�10

�2 )

2 ; n

�2

in t

his

cas

e b

ecau

se

two

mo

les

of

elec

tro

ns

are

tran

sfer

red

in t

he

ove

rall

cell

reac

tio

n (

See

answ

er t

o

qu

esti

on

1.)

4.

Cal

cula

te th

e ce

ll po

tent

ial f

or th

e io

n co

ncen

trat

ions

sho

wn

in th

e ce

ll.

0.16

V

0.46

V

Vo

ltm

eter

Ag

Cu

Cu

2�

1.0

� 1

0�3 M

Ag

�

1.0

� 1

0�2 M

Use

wit

h Ch

apte

r 21

,Se

ctio

n 21

.1

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

24C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

24C

hal

len

ge

Pro

ble

ms

The

Chem

istr

y of

Lif

eTh

e Ch

emis

try

of L

ife

Pro

tein

s ar

e sy

nthe

size

d w

hen

RN

A m

olec

ules

tr

ansl

ate

the

DN

A la

ngua

ge o

f ni

trog

en b

ases

into

the

prot

ein

lang

uage

of

amin

o ac

ids

usin

g a

gene

tic c

ode.

The

gen

etic

cod

e is

fou

nd in

RN

A m

ole-

cule

s ca

lled

mes

seng

er R

NA

(m

RN

A),

whi

ch a

re s

yn-

thes

ized

fro

m D

NA

mol

ecul

es. T

he g

enet

ic c

ode

cons

ists

of

a se

quen

ce o

f th

ree

nitr

ogen

bas

es in

the

mR

NA

,cal

led

a co

don.

Mos

t cod

ons

code

for

spe

cifi

cam

ino

acid

s. A

few

cod

ons

code

for

a s

top

in th

e sy

n-th

esis

of

prot

eins

. The

tabl

e sh

ows

the

mR

NA

cod

ons

that

mak

e up

the

gene

tic c

ode.

To

use

the

tabl

e,re

adth

e th

ree

nitr

ogen

bas

es in

seq

uenc

e. T

he f

irst

bas

e is

show

n al

ong

the

left

sid

e of

the

tabl

e. T

he s

econ

d ba

seis

sho

wn

alon

g th

e to

p of

the

tabl

e. T

he th

ird

base

issh

own

alon

g th

e ri

ght s

ide

of th

e ta

ble.

For

exa

mpl

e,th

e se

quen

ce C

AU

cod

es f

or th

e am

ino

acid

his

tidin

e(H

is).

The

tabl

e gi

ves

abbr

evia

tions

for

the

amin

oac

ids.

Ans

wer

the

follo

win

g qu

estio

ns a

bout

the

gene

tic c

ode.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

24

Use

wit

h Ch

apte

r 24

,Se

ctio

n 24

.4

1.

Wha

t am

ino

acid

is r

epre

sent

ed b

y ea

ch o

f th

e fo

llow

ing

codo

ns?

a.

CU

G

b.

UC

A

2.

Wri

te th

e se

quen

ce o

f am

ino

acid

s fo

r w

hich

the

follo

win

g m

RN

A s

eque

nce

code

s.

-C-A

-U-C

-A-C

-C-G

-G-U

-C-U

-U-U

-U-C

-U-U

-

-His

-His

-Arg

-Ser

-Ph

e-Le

u-

3.

Err

ors

som

etim

es o

ccur

whe

n m

RN

A m

olec

ules

are

syn

thes

ized

fro

m D

NA

mol

ecul

es.

Nitr

ogen

bas

es m

ay b

e om

itted

,an

extr

a ni

trog

en b

ase

may

be

adde

d,or

a n

itrog

en b

ase

may

be

chan

ged

duri

ng s

ynth

esis

. The

two

mR

NA

seq

uenc

es s

how

n be

low

are

exa

mpl

esof

suc

h er

rors

. In

eac

h ca

se,t

ell h

ow th

e m

RN

A s

eque

nce

show

n di

ffer

s fr

om th

e co

rrec

tm

RN

A s

eque

nce

give

n in

que

stio

n 2.

a.

-C-A

-U-C

-A-C

-C-G

-G-U

-U-C

-U-U

-U-U

-C-U

-U-

An

ext

ra U

(u

raci

l) h

as b

een

ad

ded

at

po

siti

on

#10

.

b.

-C-A

-U-U

-A-C

-C-G

-G-U

-C-U

-U-U

-U-C

-U-U

-

C (

cyto

sin

e) a

t p

osi

tio

n #

4 h

as b

een

ch

ang

ed t

o U

(u

raci

l).

4.

Wri

te th

e am

ino

acid

seq

uenc

e fo

r ea

ch o

f th

e m

RN

A s

eque

nces

sho

wn

in q

uest

ion

3.

a.

b.

-His

-Tyr

-Arg

-Ser

-Ph

e-Le

u-

-His

-His

-Arg

-Ph

e-Ph

e-Se

r-

Seri

ne

Leu

cin

e

UU

UU

CU

UA

UU

GU

UU

UC

UC

CU

AC

UG

CC

UU

AU

CA

UA

AU

GA

AU

UG

UC

GU

AG

UG

GG

CU

UC

CU

CA

UC

GU

UC

UC

CC

CC

AC

CG

CC

CU

AC

CA

CA

AC

GA

AC

UG

CC

GC

AG

CG

GG

AU

UA

CU

AA

UA

GU

UA

UC

AC

CA

AC

AG

CC

AU

AA

CA

AA

AA

GA

AA

UG

AC

GA

AG

AG

GG

GU

UG

CU

GA

UG

GU

UG

UC

GC

CG

AC

GG

CC

GU

AG

CA

GA

AG

GA

AG

UG

GC

GG

AG

GG

GG

Seco

nd

bas

e

First base

Third base

}}Phe

Leu

Ile Met

Leu

Val

Pro

Ser

Ala

Thr

} }} }} }}His

Gln

Tyr

Sto

pSt

op

Asn

Lys

Asp

Glu

} }}Arg

Gly

Cys

Sto

pTr

p

Ser

Arg

U C A G

UC

AG

The

Gen

etic

Cod

e

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

2323

Boili

ng P

oint

s of

Org

anic

Fam

ilies

Boili

ng P

oint

s of

Org

anic

Fam

ilies

The

mos

t im

port

ant f

acto

r de

term

inin

g th

e bo

iling

poi

nt o

f a

subs

tanc

e is

its

atom

ic o

r m

olec

ular

mas

s. I

n ge

nera

l,th

e la

rger

the

atom

ic o

r m

olec

ular

mas

s of

the

subs

tanc

e,th

em

ore

ener

gy is

nee

ded

to c

onve

rt th

e su

bsta

nce

from

the

liqui

dph

ase

to th

e ga

seou

s ph

ase.

As

an e

xam

ple,

the

boili

ng p

oint

of e

than

e (m

olec

ular

mas

s �

30; b

oilin

g po

int �

�89

°C)

ism

uch

high

er th

an th

e bo

iling

poi

nt o

f m

etha

ne (

mol

ecul

arm

ass

�16

; boi

ling

poin

t ��

161°

C).

Inte

rmol

ecul

ar f

orce

s be

twee

n th

e pa

rtic

les

of a

liqu

id a

lso

can

affe

ct th

e liq

uid’

s bo

iling

poi

nt. T

he g

raph

sho

ws

tren

ds in

the

boili

ng p

oint

s of

fou

r or

gani

c fa

mili

es:a

lkan

es,a

lcoh

ols,

alde

hyde

s,an

d et

hers

. Use

the

grap

h an

d yo

ur k

now

ledg

e of

inte

rmol

ecul

ar f

orce

s to

ans

wer

the

follo

win

g qu

estio

ns.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

23

1.

For

any

one

fam

ily,w

hat i

s th

e re

latio

nshi

p be

twee

n m

olec

ular

mas

s an

d bo

iling

poi

nt?

As

the

mo

lecu

lar

mas

s in

crea

ses,

so

do

es t

he

bo

ilin

g p

oin

t.

2.

For

com

poun

ds o

f si

mila

r m

olec

ular

mas

s,w

hich

fam

ily o

f th

e fo

ur s

how

n in

the

grap

hha

s th

e lo

wes

t boi

ling

poin

ts?

Whi

ch f

amily

has

the

high

est b

oilin

g po

ints

?

The

alka

nes

hav

e th

e lo

wes

t b

oili

ng

po

ints

; th

e al

coh

ols

hav

e th

e h

igh

est

bo

ilin

g

po

ints

.

3.

Find

and

list

the

boili

ng p

oint

s fo

r et

hano

l (m

olec

ular

mas

s �

46)

and

dim

ethy

l eth

er(m

olec

ular

mas

s �

46)

on th

e gr

aph.

Why

wou

ld y

ou e

xpec

t the

se tw

o co

mpo

unds

toha

ve r

elat

ivel

y si

mila

r bo

iling

poi

nts?

Bo

ilin

g p

oin

t o

f et

han

ol:

abo

ut

78°C

; bo

ilin

g p

oin

t o

f d

imet

hyl

eth

er: a

bo

ut

�25

°C; t

he

two

co

mp

ou

nd

s w

ou

ld b

e ex

pec

ted

to

hav

e si

mila

r b

oili

ng

po

ints

bec

ause

th

ey h

ave

the

sam

e m

ole

cula

r m

ass.

4.

Find

the

alde

hyde

with

a m

olec

ular

mas

s of

abo

ut 5

8. N

ame

that

ald

ehyd

e an

d w

rite

its

chem

ical

for

mul

a.Th

e al

deh

yde

is p

rop

anal

, CH

3CH

2CH

O.

5.

Can

this

ald

ehyd

e fo

rm h

ydro

gen

bond

s? C

an o

ther

ald

ehyd

es f

orm

hyd

roge

n bo

nds?

Exp

lain

.

Nei

ther

pro

pan

al n

or

any

oth

er a

ldeh

yde

can

fo

rm h

ydro

gen

bo

nd

s b

ecau

se

ald

ehyd

es la

ck �

OH

gro

up

s.

3040

5060

Mo

lecu

lar

mas

s

� a

lkan

e�

alc

oh

ol

Boiling point (°C)70

80

�50050100

� a

ldeh

yde

� e

ther

Use

wit

h Ch

apte

r 23

,Se

ctio

n 23

.3

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.


Nam

eD

ate

Cla

ss

26C

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

26C

hal

len

ge

Pro

ble

ms

The

Phos

phor

us C

ycle

The

Phos

phor

us C

ycle

Pho

spho

rus

is a

n im

port

ant e

lem

ent b

oth

in o

rgan

ism

s an

d in

the

litho

sphe

re. I

n or

gani

sms,

phos

phor

us o

ccur

s in

DN

A a

nd R

NA

mol

ecul

es,c

ell m

embr

anes

,bon

esan

d te

eth,

and

in th

e en

ergy

–sto

rage

com

poun

d ad

enos

ine

trip

hosp

hate

(A

TP)

. In

the

litho

-sp

here

,pho

spho

rus

occu

rs p

rim

arily

in th

e fo

rm o

f ph

osph

ates

,as

a m

ajor

con

stitu

ent o

fm

any

rock

s an

d m

iner

als.

Pho

spha

te r

ock

is m

ined

to p

rodu

ce m

any

com

mer

cial

pro

duct

s,su

ch a

s fe

rtili

zers

and

det

erge

nts.

Whe

n th

ese

prod

ucts

are

use

d,ph

osph

ates

are

ret

urne

d to

the

litho

sphe

re a

nd h

ydro

sphe

re. T

hus,

phos

phor

us—

like

carb

on a

nd n

itrog

en—

cycl

es in

the

envi

ronm

ent.

Use

the

diag

ram

of

the

phos

phor

us c

ycle

to a

nsw

er th

e qu

estio

ns b

elow

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

26

Use

wit

h Ch

apte

r 26

,Se

ctio

n 26

.4

1.

By

wha

t met

hods

doe

s ph

osph

orus

get

into

soi

l?

An

imal

s ex

cret

e p

ho

sph

ates

. Dea

d o

rgan

ism

s re

leas

e p

ho

sph

ates

as

they

dec

ay.

Rai

nfa

ll le

ach

es p

ho

sph

ates

fro

m r

ock

s.

2.

By

wha

t met

hod

do p

lant

s ob

tain

the

phos

phor

us th

ey n

eed?

by

extr

acti

ng

ph

osp

hat

es f

rom

th

e so

il

3.

By

wha

t met

hod

do a

nim

als

obta

in th

e ph

osph

orus

they

nee

d?

by

eati

ng

pla

nts

or

oth

er a

nim

als

4.

In w

hat w

ay is

the

phos

phor

us c

ycle

dif

fere

nt f

rom

the

carb

on a

nd n

itrog

en c

ycle

s yo

ust

udie

d in

the

text

book

?

The

ph

osp

ho

rus

cycl

e h

as n

o a

tmo

sph

eric

co

mp

on

ent.

5.

The

pho

spho

rus

cycl

e ha

s bo

th s

hort

-ter

m a

nd lo

ng-t

erm

par

ts. U

se d

iffe

rent

col

ored

pe

ncils

to s

how

eac

h pa

rt o

n th

e di

agra

m.

Sho

rt-t

erm

cyc

le: a

nim

als

excr

etin

g p

ho

sph

ates

; dec

om

po

sin

g o

rgan

ism

s; a

nim

als

eati

ng

pla

nts

; pla

nts

ext

ract

ing

ph

osp

hat

es f

rom

th

e so

il; m

inin

g; u

se o

f fe

rtili

z-er

s, d

eter

gen

ts, a

nd

oth

er s

ynth

etic

pro

du

cts

con

tain

ing

ph

osp

ho

rus.

Lo

ng

-ter

mcy

cle:

geo

log

ical

up

lift,

ph

osp

hat

es le

ach

ing

fro

m r

ock

s, p

ho

sph

ate

rock

fo

rmin

gu

nd

er t

he

oce

ans

Pho

sph

ate

rock

s

Pho

sph

ate

rock

s

Geo

log

ical

up

lift

Nam

eD

ate

Cla

ss

Ch

alle

ng

e Pr

ob

lem

sC

hem

istr

y: M

atte

r an

d C

han

ge

• C

hap

ter

2525

The

Prod

ucti

on o

fPl

uton

ium

-239

The

Prod

ucti

on o

fPl

uton

ium

-239

Whe

n nu

clea

r fi

ssio

n w

as f

irst

dis

cove

red,

only

two

isot

opes

,ura

nium

-233

and

ura

nium

-235

,wer

ekn

own

of b

eing

cap

able

of

unde

rgoi

ng th

is n

ucle

ar c

hang

e.Sc

ient

ists

late

r di

scov

ered

a th

ird

isot

ope,

plut

oniu

m-2

39,

also

cou

ld u

nder

go n

ucle

ar f

issi

on. P

luto

nium

-239

doe

s no

toc

cur

in n

atur

e bu

t can

be

mad

e sy

nthe

tical

ly in

nuc

lear

reac

tors

and

par

ticle

acc

eler

ator

s.

The

dia

gram

sho

ws

the

proc

ess

by w

hich

plu

toni

um-2

39is

mad

e in

nuc

lear

rea

ctor

s. A

nsw

er th

e qu

estio

ns a

bout

the

diag

ram

.

CH

ALLEN

GE P

RO

BLEM

SCH

AP

TER

25

1.

Iden

tify

the

isot

ope

who

se n

ucle

us is

labe

led

A in

the

diag

ram

.

2.

Nam

e th

e ty

pe o

f nu

clea

r re

actio

n th

at o

ccur

s w

hen

a

neut

ron

stri

kes

nucl

eus

A.

3.

Iden

tify

the

isot

ope

who

se n

ucle

us is

labe

led

B.

4.

Bes

ides

fra

gmen

ted

nucl

ei,w

hat e

lse

is p

rodu

ced

whe

n a

neut

ron

stri

kes

nucl

eus

A?

5.

Iden

tify

the

isot

ope

who

se n

ucle

us is

labe

led

C.

6.

Wri

te th

e nu

clea

r eq

uatio

n fo

r th

e re

actio

n th

at o

ccur

s w

hen

a ne

utro

n st

rike

s nu

cleu

s C

.Id

entif

y th

e pr

oduc

t D f

orm

ed in

the

reac

tion.

1 0n �

238 92U

00 0

�23

9 92U

; Th

e p

rod

uct

nu

cleu

s is

ura

niu

m-2

39.

7.

Wri

te th

e nu

clea

r eq

uatio

n fo

r th

e de

cay

of n

ucle

us D

. Ide

ntif

y is

otop

e E

for

med

in th

ere

actio

n.23

9 92U

0�

10 �

239 93N

p; n

eptu

niu

m-2

39

8.

Wri

te a

bal

ance

d nu

clea

r eq

uatio

n fo

r th

e de

cay

of n

ucle

us E

. Ide

ntif

y is

otop

e F

form

edin

the

reac

tion.

239 93N

p 0

�10

�23

9 94Pu

; plu

ton

ium

-239

9.

Nam

e th

e ty

pe o

f nu

clea

r re

actio

n th

at o

ccur

s w

hen

a ne

utro

n st

rike

s nu

cleu

s F.

nu

clea

r fi

ssio

n

10.

Wri

te th

e nu

clea

r eq

uatio

n fo

r th

e re

actio

n th

at o

ccur

s w

hen

a ne

utro

n st

rike

s nu

cleu

s F.

Iden

tify

isot

ope

G f

orm

ed in

the

reac

tion.

1 0n �

239 94Pu

012

5 48C

d �

115 46Pd

; pal

lad

ium

-115

ura

niu

m-2

38 (

238 92U

)

neu

tro

ns

silv

er-1

14 (

114 47A

g)

nu

clea

r fi

ssio

n

ura

niu

m-2

35 (

235 92U

)

0 –1

0 –10 0�

1 0n

1 0n

1 0n

1 0n

45p

75n

48p

77n

92p

143n

92p

143n

92p

146n

Sou

rce

of

neu

tro

ns

B

A

C

D

E

G

F

Use

wit

h Ch

apte

r 25

,Se

ctio

n 25

.4

Cop

yrig

ht ©

Gle

ncoe

/McG

raw

-Hill

,a d

ivis

ion

of th

e M

cGra

w-H

ill C

ompa

nies

,Inc

.

Challenge Problems Chemistry: Matter and Change T41

CREDITS

Art CreditsNavta Associates: 1, 2, 4, 11, 12, 13, 15, 16, 17, 18, 20, 21, 23, 24; MacArt Design: 14, 25, 26

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