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10: Energy as a Driving Force

# 10: Energy as a Driving Force

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A22 Solutions to Self-Check Exercises

Step 2 We must first determine whether Pb2ϩ or SO42Ϫ is the limiting reactant by calculating the moles of Pb2ϩ and SO42Ϫ ions

present. Because 0.0500 M Pb(NO3)2 contains 0.0500 M Pb2ϩ ions,

we can calculate the moles of Pb2ϩ ions in 1.25 L of this solution

as follows:

0.0500 mol Pb2ϩ

1.25 L ϫ

ϭ 0.0625 mol Pb2ϩ

L

The 0.0250 M Na2SO4 solution contains 0.0250 M SO42Ϫ ions, and

the number of moles of SO42Ϫ ions in 2.00 L of this solution is

0.0250 mol SO42Ϫ

2.00 L ϫ

ϭ 0.0500 mol SO42Ϫ

L

Step 3 Pb2ϩ and SO42Ϫ react in a 1:1 ratio, so the amount of SO42Ϫ

ions is limiting because SO42Ϫ is present in the smaller number of

moles.

Step 4 The Pb2ϩ ions are present in excess, and only 0.0500 mole

of solid PbSO4 will be formed.

Step 5 We calculate the mass of PbSO4 by using the molar mass

of PbSO4 (303.3 g).

303.3 g PbSO4

0.0500 mol PbSO4 ϫ

ϭ 15.2 g PbSO4

1 mol PbSO4

Self-Check Exercise 15.9

Step 1 Because nitric acid is a strong acid, the nitric acid solution

contains Hϩ and NO3Ϫ ions. The KOH solution contains Kϩ and

OHϪ ions. When these solutions are mixed, the Hϩ and OHϪ react

to form water.

H ϩ (aq) ϩ OHϪ(aq)SH2O(l)

Step 2 The number of moles of OHϪ present in 125 mL of 0.050 M

KOH is

1L

0.050 mol OH Ϫ

125 mL ϫ

ϫ

ϭ 6.3 ϫ 10 Ϫ3 mol OH Ϫ

1000 mL

L

Step 3 Hϩ and OHϪ react in a 1:1 ratio, so we need 6.3 ϫ 10Ϫ3 mole

of Hϩ from the 0.100 M HNO3.

Step 4 6.3 ϫ 10Ϫ3 mole of OHϪ requires 6.3 ϫ 10Ϫ3 mole of Hϩ to

form 6.3 ϫ 10Ϫ3 mole of H2O.

Therefore,

0.100 mol H ϩ

ϭ 6.3 ϫ 10 Ϫ3 mol H ϩ

L

where V represents the volume in liters of 0.100 M HNO3 required.

Solving for V, we have

6.3 ϫ 10 Ϫ3 mol H ϩ

ϭ 6.3 ϫ 10 Ϫ2 L

0.100 mol H ϩ

L

1000 mL

Ϫ2

ϭ 6.3 ϫ 10 L ϫ

ϭ 63 mL

L

Self-Check Exercise 15.10

From the definition of normality, N ϭ equiv/L, we need to calculate (1) the equivalents of KOH and (2) the volume of the solution

in liters. To find the number of equivalents, we use the equivalent

weight of KOH, which is 56.1 g (see Table 15.2).

1 equiv KOH

23.6 g KOH ϫ

ϭ 0.421 equiv KOH

56.1 g KOH

Next we convert the volume to liters.

1L

755 mL ϫ

ϭ 0.755 L

1000 mL

Finally, we substitute these values into the equation that defines

normality.

equiv

0.421 equiv

Normality ϭ

ϭ

ϭ 0.558 N

L

0.755 L

Self-Check Exercise 15.11

To solve this problem, we use the relationship

Nacid ϫ Vacid ϭ Nbase ϫ Vbase

where

equiv

Nacid ϭ 0.50

L

Vacid ϭ ?

equiv

Nbase ϭ 0.80

L

Vbase ϭ 0.250 L

We solve the equation

Nacid ϫ Vacid ϭ Nbase ϫ Vbase

for Vacid by dividing both sides by Nacid.

Nacid ϫ Vacid

Nbase ϫ Vbase

ϭ

Nacid

Nacid

equiv

(0.80

) ϫ (0.250 L)

Nbase ϫ Vbase

L

Vacid ϭ

ϭ

equiv

Nacid

0.50

L

Vacid ϭ 0.40 L

Therefore, 0.40 L of 0.50 N H2SO4 is required to neutralize 0.250 L

of 0.80 N KOH.

Chapter 16

Self-Check Exercise 16.1

The conjugate acid–base pairs are

H2O,

H3O ϩ

Base

Conjugate acid

HC2H3O2,

C2H3O2Ϫ

and

Acid

Conjugate base

The members of both pairs differ by one Hϩ.

Self-Check Exercise 16.2

Because [Hϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14, we can solve for [Hϩ].

1.0 ϫ 10 Ϫ14

1.0 ϫ 10 Ϫ14

3HϪ 4 ϭ

ϭ

ϭ 5.0 ϫ 10 Ϫ13 M

Ϫ

3OH 4

2.0 ϫ 10 Ϫ2

This solution is basic: [OHϪ] ϭ 2.0 ϫ 10Ϫ2 M is greater than [Hϩ] ϭ

5.0 ϫ 10Ϫ13 M.

Self-Check Exercise 16.3

a. Because [Hϩ] ϭ 1.0 ϫ 10Ϫ3 M, we get pH ϭ 3.00 because pH ϭ

Ϫlog[Hϩ] ϭ Ϫlog[1.0 ϫ 10Ϫ3] ϭ 3.00.

b. Because [OHϪ] ϭ 5.0 ϫ 10Ϫ5 M, we can find [Hϩ] from the Kw

expression.

Kw

1.0 ϫ 10 Ϫ14

ϭ 2.0 ϫ 10 Ϫ10 M

3H ϩ 4 ϭ

Ϫ ϭ

3OH 4

5.0 ϫ 10 Ϫ5

pH ϭ Ϫlog[Hϩ] ϭ Ϫlog[2.0 ϫ 10Ϫ10] ϭ 9.70

Self-Check Exercise 16.4

pOH ϩ pH ϭ 14.00

pOH ϭ 14.00 Ϫ pH ϭ 14.00 Ϫ 3.5

pOH ϭ 10.5

Self-Check Exercise 16.5

Step 1 pH ϭ 3.50

Step 2 ϪpH ϭ Ϫ3.50

Solutions to Self-Check Exercises

Step 3

inv

log

Ϫ3.50 ϭ 3.2 ϫ 10Ϫ4

Self-Check Exercise 16.6

Step 1 pOH ϭ 10.50

Step 2 ϪpOH ϭ Ϫ10.50

inv

log

Self-Check Exercise 17.7

Ϫ11

Ϫ10.50 ϭ 3.2 ϫ 10

[OHϪ] ϭ 3.2 ϫ 10Ϫ11 M

Because HCl is a strong acid, it is completely dissociated:

5.0 ϫ 10Ϫ3 M HCl S 5.0 ϫ 10Ϫ3 M Hϩ and 5.0 ϫ 10Ϫ3 M ClϪ

so 3 Hϩ 4 ϭ 5.0 ϫ 10Ϫ3 M.

pH ϭ Ϫlog(5.0 ϫ 10Ϫ3 ) ϭ 2.30

Chapter 17

Applying the law of chemical equilibrium gives

Coefficient

Coefficient

of NO2

of H2O

T

T

3 NO2 4 4 3 H2O4 6

3 NH3 4 4 3O2 4 7

d Coefficient

of O2

c

Coefficient

of NH3

Self-Check Exercise 17.2

a. K ϭ [O2]3

The solids are not included.

b. K ϭ [N2O][H2O]2 The solid is not included. Water is gaseous

in this reaction, so it is included.

1

c. K ϭ

The solids are not included.

3 CO2 4

1

3 SO3 4

Water and H2SO4 are pure liquids and so

are not included.

Self-Check Exercise 17.3

When rain is imminent, the concentration of water vapor in the

air increases. This shifts the equilibrium to the right, forming

CoCl2 и 6H2O(s), which is pink.

Self-Check Exercise 17.4

a. No change. Both sides of the equation contain the same number of gaseous components. The system cannot change its

pressure by shifting its equilibrium position.

b. Shifts to the left. The system can increase the number of

gaseous components present, and so increase the pressure, by

shifting to the left.

c. Shifts to the right to increase the number of gaseous components and thus its pressure.

Self-Check Exercise 17.5

a.

b.

c.

d.

PbCrO4(s) Δ Pb2ϩ(aq) ϩ CrO42Ϫ(aq)

Ksp ϭ [Pb2ϩ][CrO42Ϫ] ϭ 2.0 ϫ 10Ϫ16

[Pb2ϩ] ϭ x

[CrO42Ϫ] ϭ x

Ksp ϭ 2.0 ϫ 10Ϫ16 ϭ x2

x ϭ [Pb2ϩ] ϭ [CrO42Ϫ] ϭ 1.4 ϫ 10Ϫ8

Chapter 18

Self-Check Exercise 17.1

d. K ϭ

(3.9 ϫ 10Ϫ5)2 ϭ 1.5 ϫ 10Ϫ9 ϭ Ksp

Self-Check Exercise 17.8

Self-Check Exercise 16.7

Self-Check Exercise 17.6

a. BaSO4(s) Δ Ba2ϩ(aq) ϩ SO42Ϫ(aq); Ksp ϭ [Ba2ϩ][SO42Ϫ]

b. Fe(OH)3(s) Δ Fe3ϩ(aq) ϩ 3OHϪ(aq); Ksp ϭ [Fe3ϩ][OHϪ]3

c. Ag3PO4(s) Δ 3Agϩ(aq) ϩ PO43Ϫ(aq); Ksp ϭ [Agϩ]3[PO43Ϫ]

[Hϩ] ϭ 3.2 ϫ 10Ϫ4 M

Step 3

A23

Shifts to the right away from added SO2.

Shifts to the right to replace removed SO3.

Shifts to the right to decrease its pressure.

Shifts to the right. Energy is a product in this case, so a decrease in temperature favors the forward reaction (which produces energy).

Self-Check Exercise 18.1

a. CuO contains Cu2ϩ and O2Ϫ ions, so copper is oxidized

(Cu S Cu2ϩ ϩ 2eϪ) and oxygen is reduced (O ϩ 2eϪ S O2Ϫ).

b. CsF contains Csϩ and FϪ ions. Thus cesium is oxidized

(Cs S Csϩ ϩ eϪ) and fluorine is reduced (F ϩ eϪ S FϪ).

Self-Check Exercise 18.2

a. SO3

We assign oxygen first. Each O is assigned an oxidation state of

Ϫ2, giving a total of Ϫ6 (3 ϫ Ϫ2) for the three oxygen atoms.

Because the molecule has zero charge overall, the sulfur must

have an oxidation state of ϩ6.

Check: ϩ6 ϩ 3(Ϫ2) ϭ 0

b. SO42Ϫ

As in part a, each oxygen is assigned an oxidation state of Ϫ2,

giving a total of Ϫ8 (4 ϫ Ϫ2) on the four oxygen atoms. The

anion has a net charge of Ϫ2, so the sulfur must have an oxidation state of ϩ6.

Check: ϩ6 ϩ 4(Ϫ2) ϭ Ϫ2

SO42Ϫhas a charge of Ϫ2, so this is correct.

c. N2O5

We assign oxygen before nitrogen because oxygen is more

electronegative. Thus each O is assigned an oxidation state of

Ϫ2, giving a total of Ϫ10 (5 ϫ Ϫ2) on the five oxygen atoms.

Therefore, the oxidation states of the two nitrogen atoms

must total ϩ10 because N2O5 has no overall charge. Each N is

assigned an oxidation state of ϩ5.

Check: 2(ϩ5) ϩ 5(Ϫ2) ϭ 0

d. PF3

First we assign the fluorine an oxidation state of Ϫ1, giving a

total of Ϫ3 (3 ϫ Ϫ1) on the three fluorine atoms. Thus P must

have an oxidation state of ϩ3.

Check: ϩ3 ϩ 3(Ϫ1) ϭ 0

e. C2H6

In this case, it is best to recognize that hydrogen is always ϩ1

in compounds with nonmetals. Thus each H is assigned an

oxidation state of ϩ1, which means that the six H atoms account for a total of ϩ6 (6 ϫ ϩ1). Therefore, the two carbon

atoms must account for Ϫ6, and each carbon is assigned an

oxidation state of Ϫ3.

Check: 2(Ϫ3) ϩ 6(ϩ1) ϭ 0

A24 Solutions to Self-Check Exercises

Self-Check Exercise 18.3

We can tell whether this is an oxidation–reduction reaction by

comparing the oxidation states of the elements in the reactants

and products:

N2 ϩ 3H2S2NH3

c

0

Oxidation states:

c

0

c

Ϫ3

a

ϩ1 (each H)

Nitrogen goes from 0 to Ϫ3. Thus it gains three electrons and is reduced. Each hydrogen atom goes from 0 to ϩ1 and is thus oxidized, so this is an oxidation–reduction reaction. The oxidizing

agent is N2 (it takes electrons from H2). The reducing agent is H2 (it

gives electrons to N2).

N2 ϩ 3H2S2NH3

6eϪ

Self-Check Exercise 18.4

The unbalanced equation for this reaction is

Cu(s) ϩ HNO3(aq)SCu(NO3 )2(aq) ϩ H2O(l) ϩ NO(g)

Copper

metal

Nitric acid

Aqueous

copper(II)

nitrate

(contains Cu2ϩ)

Water

Nitrogen

monoxide

¡

¡

¡

¡

¡

¡

¡

Step 1 The oxidation half-reaction is

Cu ϩ HNO3SCu(NO3 )2

ϩ1 ϩ5 Ϫ2 ϩ2 ϩ5 Ϫ2

(each 0)

(each 0)

The copper goes from 0 to ϩ2 and thus is oxidized. This reduction

reaction is

HNO3SNO

¡

¡

¡

¡

¡

Oxidation states: 0

ϩ1 ϩ5 Ϫ2 ϩ2 Ϫ2

(each 0)

In this case, nitrogen goes from ϩ5 in HNO3 to ϩ2 in NO and so is

reduced. Notice two things about these reactions:

1. The HNO3 must be included in the oxidation half-reaction to

supply NO3Ϫ in the product Cu(NO3)2.

2. Although water is a product in the overall reaction, it does not

need to be included in either half-reaction at the beginning.

It will appear later as we balance the equation.

Oxidation states:

Step 2 Balance the oxidation half-reaction.

Cu ϩ HNO3SCu(NO3 )2

a. Balance nitrogen first.

Cu ϩ 2HNO3SCu(NO3 )2

b. Balancing nitrogen also caused oxygen to balance.

c. Balance hydrogen using Hϩ.

Cu ϩ 2HNO3SCu(NO3 )2 ϩ 2Hϩ

d. Balance the charge using eϪ.

Cu ϩ 2HNO3SCu(NO3 )2 ϩ 2Hϩ ϩ 2eϪ

This is the balanced oxidation half-reaction.

Balance the reduction half-reaction.

HNO3SNO

a. All elements are balanced except hydrogen and oxygen.

b. Balance oxygen using H2O.

HNO3SNO ϩ 2H2O

c. Balance hydrogen using Hϩ.

3Hϩ ϩ HNO3SNO ϩ 2H2O

d. Balance the charge using eϪ.

3eϪ ϩ 3Hϩ ϩ HNO3SNO ϩ 2H2O

This is the balanced reduction half-reaction.

Step 3 We equalize electrons by multiplying the oxidation halfreaction by 3:

3 ϫ 3Cu ϩ 2HNO3SCu(NO3 )2 ϩ 2Hϩ ϩ 2eϪ 4

gives

3Cu ϩ 6HNO3S3Cu(NO3 )2 ϩ 6Hϩ ϩ 6eϪ

Multiplying the reduction half-reaction by 2:

2 ϫ 3 3eϪ ϩ 3Hϩ ϩ HNO3SNO ϩ 2H2O4

gives

6eϪ ϩ 6Hϩ ϩ 2HNO3S2NO ϩ 4H2O

Step 4 We can now add the balanced half-reactions, which both

involve a six-electron change.

3Cu ϩ 6HNO3S3Cu(NO3 )2 ϩ 6Hϩ ϩ 6eϪ

6eϪ ϩ 6Hϩ ϩ 2HNO3S2NO ϩ 4H2O

6eϪ ϩ 6H ϩ ϩ 3Cu ϩ 8HNO3S3Cu(NO3 )2 ϩ 2NO ϩ 4H2O ϩ 6H ϩ ϩ 6eϪ

Canceling species common to both sides gives the balanced overall equation:

3Cu(s) ϩ 8HNO3(aq)S3Cu(NO3 )2(aq) ϩ 2NO(g) ϩ 4H2O(l)

Step 5 Check the elements and charges.

Elements

Charges

3Cu, 8H, 8N, 24O S 3Cu, 8H, 8N, 24O

0S0

Chapter 19

Self-Check Exercise 19.1

a. An alpha particle is a helium nucleus, 42He. We can initially

represent the production of an ␣ particle by 226

88 Ra as follows:

226

4

A

88 RaS2He ϩ Z X

Because we know that both A and Z are conserved, we can

write

A ϩ 4 ϭ 226 and Z ϩ 2 ϭ 88

Solving for A gives 222 and for Z gives 86, so AZX is 222

86 X.

Because Rn has Z ϭ 86, AZX is 222

86 Rn. The overall

balanced equation is

226

4

222

88 RaS2He ϩ 86 Rn

Check: Z ϭ 88

Z ϭ 86 ϩ 2 ϭ 88

S

A ϭ 226

A ϭ 222 ϩ 4 ϭ 226

b. Using a similar strategy, we have

214

0

A

82 PbSϪ1 e ϩ Z X

Because Z Ϫ 1 ϭ 82, Z ϭ 83, and because A ϩ 0 ϭ 214, A ϭ 214.

Therefore, AZX ϭ 214

83 Bi. The balanced equation is

214

0

214

82 PbSϪ1e ϩ 83 Bi

Check: Z ϭ 82

Z ϭ 83 Ϫ 1 ϭ 82

S

A ϭ 214

A ϭ 214 ϩ 0 ϭ 214

Self-Check Exercise 19.2

a. The missing particle must be 42H (an ␣ particle), because

222

218

4

86 RnS 84 Po ϩ 2He

is a balanced equation.

Check: Z ϭ 86

Z ϭ 84 ϩ 2 ϭ 86

S

A ϭ 222

A ϭ 218 ϩ 4 ϭ 222

b. The missing species must be 157X or 157N, because the balanced

equation is

15

15

0

8 OS 7 N ϩ 1e

Check: Z ϭ 8

Zϭ7Ϫ1ϭ8

S

A ϭ 15

A ϭ 15 ϩ 0 ϭ 15

Solutions to Self-Check Exercises

Self-Check Exercise 19.3

Let’s do this problem by thinking about the number of half-lives

required to go from 8.0 ϫ 10Ϫ7 mole to 1.0 ϫ 10Ϫ7 mole of 228

88 Ra.

8.0 ϫ 10Ϫ7 mol ¡ 4.0 ϫ 10Ϫ7 mol ¡

First

Second

half-life

half-life

2.0 ϫ 10Ϫ7 mol ¡ 1.0 ϫ 10Ϫ7 mol

Third

half-life

A25

It takes three half-lives, then, for the sample to go from

Ϫ7

8.0 ϫ 10Ϫ7 mole of 228

mole of 228

88 Ra to 1.0 ϫ 10

88 Ra. From Table

19.3, we know that the half-life of 228

Ra

is 6.7 years. Therefore, the

88

elapsed time is 3(6.7 years) ϭ 20.1 years, or 2.0 ϫ 101 years when

we use the correct number of significant figures.

ANSWERS TO EVEN-NUMBERED END-OF-CHAPTER

QUESTIONS AND EXERCISES

Chapter 1

2.

4.

6.

8.

10.

12.

14.

16.

The answer depends on the student’s experiences.

Answers will depend on the student’s responses.

Answers will depend on the student’s choices.

Recognize the problem and state it clearly; propose possible solutions or explanations; decide which solution/explanation is

best through experiments.

Answers will depend on student responses. A quantitative observation must include a number, such as “There are three

windows in this room.” A qualitative observation could include something like “The chair is blue.”

The answer depends on the student’s responses/examples.

Chemistry is not just a set of facts that have to be memorized.

To be successful in chemistry, you have to be able to apply what

you have learned to new situations, new phenomena, new experiments. Rather than just learning a list of facts or studying

someone else’s solution to a problem, your instructor hopes

you will learn how to solve problems yourself, so that you will be

able to apply what you have learned in future circumstances.

In real-life situations, the problems and applications likely to

be encountered are not simple textbook examples. You must

be able to observe an event, hypothesize a cause, and then

test this hypothesis. You must be able to carry what has been

learned in class forward to new, different situations.

Chapter 2

2. “Scientific notation” means we have to put the decimal point

after the first significant figure, and then express the order of

magnitude of the number as a power of 10. So we want to put

the decimal point after the first 2:

2421 S 2.421 ϫ 10to some power

To be able to move the decimal point three places to the left

in going from 2421 to 2.421 means you will need a power of

103 after the number, where the exponent 3 shows that you

moved the decimal point three places to the left:

2421 S 2.421 ϫ 10to some power ϭ 2.421 ϫ 103

4. (a) 104; (b) 10Ϫ3; (c) 102; (d) 10Ϫ30

6. (a) negative; (b) zero; (c) positive; (d) negative

8. (a) 2789; (b) 0.002789; (c) 93,000,000; (d) 42.89;

(e) 99,990; (f) 0.00009999

10. (a) three places to the left; (b) one place to the left;

(c) five places to the right; (d) one place to the left;

(e) two places to the right; (f) two places to the left

12. (a) 6244; (b) 0.09117; (c) 82.99; (d) 0.0001771;

(e) 545.1; (f) 0.00002934

14. (a) 3.1 ϫ 103; (b) 1 ϫ 106; (c) 1 or 1 ϫ 100;

(d) 1.8 ϫ 10Ϫ5; (e) 1 ϫ 107; (f) 1.00 ϫ 106;

(g) 1.00 ϫ 10Ϫ7; (h) 1 ϫ 101

16. Answer depends on the student’s examples.

18. about 1⁄4 pound

20. about an inch

22. 2-liter bottle

24. the woman

26. (a) centimeter; (b) meter; (c) kilometer

28. d

30. Typically we read the scale on measuring devices to 0.1 unit

of the smallest scale division on the device. We estimate this

final significant figure, which makes the final significant figure in the measurement uncertain.

32. The scale of the ruler is marked to the nearest tenth of a centimeter. Writing 2.850 would imply that the scale was marked

to the nearest hundredth of a centimeter (and that the zero in

the thousandths place had been estimated).

34. (a) three: the relationship is exact; (b) two; (c) five;

(d) probably two

36. It is better to round off only the final answer and to carry

through extra digits in intermediate calculations. If there are

enough steps to the calculation, rounding off in each step

may lead to a cumulative error in the final answer.

38. (a) 4.18 ϫ 10Ϫ6; (b) 3.87 ϫ 104; (c) 9.11 ϫ 10Ϫ30;

(d) 5.46 ϫ 106

40. (a) 8.8 ϫ 10Ϫ4; (b) 9.375 ϫ 104; (c) 8.97 ϫ 10Ϫ1;

(d) 1.00 ϫ 103

42. The total mass would be determined by the number of decimal

places available on the readout of the scale/balance. For example, if a balance whose readout is to the nearest 0.01 g were

used, the total mass would be reported to the second decimal

place. For example, 42.05 g ϩ 29.15 g ϩ 31.09 g would be reported as 102.29 to the second decimal place. Even though

there are only four significant figures in each of the measurements, there are five significant figures in the answer because

we look at the decimal place when adding (or subtracting)

numbers.

44. Most calculators would say 0.66666666. If the 2 and 3 were experimentally determined numbers, this quotient would imply

far too many significant figures.

46. none

48. (a) 2.3; (b) 9.1 ϫ 102; (c) 1.323 ϫ 103;

(d) 6.63 ϫ 10Ϫ13

50. (a) one; (b) four; (c) two; (d) three

52. (a) 2.045; (b) 3.8 ϫ 103; (c) 5.19 ϫ 10Ϫ5;

(d) 3.8418 ϫ 10Ϫ7

54. an infinite number, a definition

2.54 cm

1 in.

1 lb

;

56.

58.

1 in.

2.54 cm

\$0.79

60. (a) 50.5 in.; (b) 3.11 ft; (c) 452 mm; (d) 76.12 cm;

(e) 1.32 qt; (f) 8.42 pt; (g) 13.7 lb; (h) 28.0 oz

62. (a) 1.03598 atm; (b) 3.13 qt; (c) 0.510 kg; (d) 1.007 cal;

(e) 8617 ft; (f) 9.04 qt; (g) 262 g; (h) 1.76 qt

64. 4117 km

66. 1 ϫ 10Ϫ8 cm; 4 ϫ 10Ϫ9 in.; 0.1 nm

68. freezing/melting

70. 273

72. Fahrenheit (F)

74. (a) 195 K; (b) 502 °C; (c) 216 °C; (d) 297 K

76. (a) 173 °F; (b) 104 °F; (c) Ϫ459 °F; (d) 90. °F

78. (a) 2 °C; (b) 28 °C; (c) Ϫ5.8 °F (Ϫ6 °F); (d) Ϫ40 °C (Ϫ40

is where both temperature scales have the same value)

80. g/cm3 (g/mL)

82. 100 in.3

84. Density is a characteristic property of a pure substance.

86. copper

88. (a) 22 g/cm3; (b) 0.034 g/cm3; (c) 0.962 g/cm3;

(d) 2.1 ϫ 10Ϫ5 g/cm3

90. 2.94 ϫ 103 g; 159 mL

92. float

94. 11.7 mL

96. (a) 966 g; (b) 394 g; (c) 567 g; (d) 135 g

98. (a) 301,100,000,000,000,000,000,000;

(b) 5,091,000,000; (c) 720; (d) 123,400; (e) 0.000432002;

(f) 0.03001; (g) 0.00000029901; (h) 0.42

100. (a) cm; (b) m; (c) km; (d) cm; (e) mm

A27

A28 Answers to Even-Numbered End-of-Chapter Questions and Exercises

102.

104.

110.

114.

116.

118.

120.

122.

124.

130.

134.

136.

138.

140.

142.

144.

146.

148.

150.

154.

156.

158.

(a) 5.07 ϫ 104 kryll; (b) 0.12 blim; (c) 3.70 ϫ 10Ϫ5 blim2

20. in.

106. \$1.33

108. °X ϭ 1.26 °C ϩ 14

3.50 g/L (3.50 ϫ 10Ϫ3 g/cm3)

112. 959 g

(a) negative; (b) negative; (c) positive; (d) zero;

(e) negative

(a) 2, positive; (b) 11, negative; (c) 3, positive;

(d) 5, negative; (e) 5, positive; (f) 0, zero;

(g) 1, negative; (h) 7, negative

(a) 1, positive; (b) 3, negative; (c) 0, zero;

(d) 3, positive; (e) 9, negative

(a) 0.0000298; (b) 4,358,000,000; (c) 0.0000019928;

(d) 602,000,000,000,000,000,000,000; (e) 0.101;

(f) 0.00787; (g) 98,700,000; (h) 378.99; (i) 0.1093;

(j) 2.9004; (k) 0.00039; (l) 0.00000001904

(a) 1 ϫ 10Ϫ2; (b) 1 ϫ 102; (c) 5.5 ϫ 10Ϫ2; (d) 3.1 ϫ 109;

(e) 1 ϫ 103; (f) 1 ϫ 108; (g) 2.9 ϫ 102; (h) 3.453 ϫ 104

kelvin, K

126. centimeter

128. 0.105 m

1 kg

132. 10

2.8 (the hundredths place is estimated)

(a) 0.000426; (b) 4.02 ϫ 10Ϫ5; (c) 5.99 ϫ 106;

(d) 400.; (e) 0.00600

(a) 2149.6; (b) 5.37 ϫ 103; (c) 3.83 ϫ 10Ϫ2;

(d) Ϫ8.64 ϫ 105

(a) 7.6166 ϫ 106; (b) 7.24 ϫ 103; (c) 1.92 ϫ 10Ϫ5;

(d) 2.4482 ϫ 10Ϫ3

1 yr 12 mo

;

12 mo 1 yr

(a) 25.7 kg; (b) 3.38 gal; (c) 0.132 qt;

(d) 1.09 ϫ 104 mL; (e) 2.03 ϫ 103 g; (f) 0.58 qt

for exactly 6 gross, 864 pencils

(a) 352 K; (b) Ϫ18 °C; (c) Ϫ43 °C; (d) 257 °F

78.2 g

152. 0.59 g/cm3

(a) 23 °F; (b) 32 °F; (c) Ϫ321 °F; (d) Ϫ459 °F;

(e) 187 °F; (f) Ϫ459 °F

1 mile

ϭ 62 miles, or about 60 miles,

(a) 100 km ϫ

1.6093 km

taking significant figures into account.

2.2046 lbs

ϭ 49,200 lbs. of fuel was

(b) 22,300 kg ϫ

1 kg

needed; 22,300 lbs. were added, so 26,900 additional

pounds were needed.

10Ϫ8 g

3.7854 L

1 lb.

ϫ

ϫ

Ϸ 10Ϫ11 lb/gal

L

1 gallon

453.59 g

Chapter 3

2. forces among the particles in the matter

4. liquids

6. gaseous

8. stronger

10. Because gases are mostly empty space, they can be compressed

easily to smaller volumes. In solids and liquids, most of the

sample’s bulk volume is filled with the molecules, leaving little empty space.

12. chemical

14. malleable; ductile

16. c

18. (a) physical; (b) chemical; (c) chemical; (d) chemical;

(e) physical; (f) physical; (g) chemical; (h) physical;

(i) physical; (j) physical; (k) chemical

20. Compounds consist of two or more elements combined together chemically in a fixed composition, no matter what their

source may be. For example, water on earth consists of molecules containing one oxygen atom and two hydrogen atoms.

Water on Mars (or any other planet) has the same composition.

22. compounds

24. In general, the properties of a compound are very different

from the properties of its constituent elements. For example,

the properties of water are altogether different from the properties of the elements (hydrogen gas and oxygen gas) that

make it up.

26. no; heating causes a reaction to form iron(II) sulfide, a pure

substance

28. Heterogeneous mixtures: salad dressing, jelly beans, the

change in my pocket; solutions: window cleaner, shampoo,

rubbing alcohol

30. (a) primarily a pure compound, but fillers and anti–caking

agents may have been added; (b) mixture; (c) mixture;

(d) pure substance

32. Concrete is a mixture. It consists of sand, gravel, water, and

cement (which consists of limestone, clay, shale, and gypsum). The composition of concrete can vary.

34. Consider a mixture of salt (sodium chloride) and sand. Salt is

soluble in water; sand is not. The mixture is added to water

and stirred to dissolve the salt, and is then filtered. The salt solution passes through the filter; the sand remains on the filter. The water can then be evaporated from the salt.

36. Each component of the mixture retains its own identity during the separation.

38. compound

40. physical

42. far apart

44. chemical

46. physical

48. electrolysis

50. (a) heterogeneous; (b) heterogeneous; (c) homogeneous

(if no lumps!); (d) heterogeneous (although it may appear

homogeneous); (e) heterogeneous

52. Answers depend on student responses.

54. physical; chemical

56. O2 and P4 are both still elements, even though the ordinary

forms of these elements consist of molecules containing

more than one atom (but all atoms in each respective molecule are the same). P2O5 is a compound, because it is made up

of two or more different elements (not all the atoms in the

P2O5 molecule are the same).

58. Assuming there is enough water present in the mixture to

have dissolved all the salt, filter the mixture to separate out

the sand from the mixture. Then distill the filtrate (consisting

of salt and water), which will boil off the water, leaving the

salt.

60. The most obvious difference is the physical states: water is a

liquid under room conditions, hydrogen and oxygen are both

gases. Hydrogen is flammable. Oxygen supports combustion.

Water does neither.

Chapter 4

2. Robert Boyle

4. 116 elements are presently known; 88 occur naturally; the remainder are manmade. Table 4.1 lists the most common elements on the earth.

6. (a) Trace elements are elements that are present in tiny

amounts. Trace elements in the body, while present in small

amounts, are essential.

(b) Answers will vary. For example, chromium assists in the

metabolism of sugars and cobalt is present in vitamin B12.

8. Answer depends on student choices/examples.

10. (a) 9; (b) 6; (c) 8; (d) 12; (e) 11; (f) 13; (g) 3;

(h) 5; (i) 4; (j) 2

12. zirconium; Cs; selenium; Au; cerium

14. B: barium, Ba; berkelium, Bk; beryllium, Be; bismuth, Bi;

bohrium, Bh; boron, B; bromine, Br

N: neodymium, Nd; neon, Ne; neptunium, Np; nickel, Ni;

niobium, Nb; nitrogen, N; nobelium, No

P: palladium, Pd; phosphorus, P; platinum, Pt; plutonium,

Pu; polonium, Po; potassium, K; praseodymium, Pr; promethium, Pm; protactinium, Pa

S: samarium, Sm; scandium, Sc; seaborgium, Sg;

selenium, Se; silicon, Si; silver, Ag; sodium, Na; strontium,

Sr; sulfur, S

16. (a) Elements are made of tiny particles called atoms. (b) All

atoms of a given element are identical; (c) The atoms of a

Answers to Even-Numbered End-of-Chapter Questions and Exercises

18.

20.

22.

24.

26.

30.

32.

34.

36.

given element are different from those of any other element;

(d) A given compound always has the same numbers and types

of atoms; (e) Atoms are neither created nor destroyed in

chemical processes. A chemical reaction simply changes the

way the atoms are grouped together.

According to Dalton, all atoms of the same element are identical; in particular, every atom of a given element has the same

mass as every other atom of that element. If a given compound always contains the same relative numbers of atoms of

each kind, and those atoms always have the same masses,

then the compound made from those elements always contains the same relative masses of its elements.

(a) CO2; (b) CO; (c) CaCO3; (d) H2SO4; (e) BaCl2;

(f) Al2S3

(a) False; Rutherford’s bombardment experiments with metal

foil suggested that the ␣ particles were being deflected by

coming near a dense, positively charged atomic nucleus;

(b) False; the proton and the electron have opposite charges,

but the mass of the electron is much smaller than the mass of

the proton; (c) True

The protons and neutrons are found in the nucleus. The protons are positively charged; the neutrons have no charge. The

protons and neutrons each weigh approximately the same.

neutron; electron

28. The electrons; outside the nucleus

The atomic number represents the number of protons in the

nucleus of the atom and makes the atom a particular element.

The mass number represents the total number of protons and

neutrons in the nucleus of an atom and distinguishes one isotope of an element from another.

Neutrons are uncharged and contribute only to the mass.

Atoms of the same element (atoms with the same number of

protons in the nucleus) may have different numbers of neutrons, and so will have different masses.

Z

Symbol

Name

14

Si

silicon

54

Xe

xenon

79

Au

gold

56

Ba

barium

53

I

iodine

50

Sn

tin

48

Cd

38. (a) 54

(b) 56

(c) 57

(d) 147 N; (e) 157 N; (f) 157 N

26 Fe;

26 Fe;

26 Fe;

40. Researchers have found that the concentrations of hydrogen-2

(deuterium) and oxygen-18 in drinking water vary significantly

from region to region in the United States. By collecting hair

samples around the country, they have also found that 86% of

the variations in the hair samples’ hydrogen and oxygen isotopes result from the isotopic composition of the local water.

Atomic

Mass

Number of

42.

Name

Symbol

Number

Number

Neutrons

oxygen

17

8O

8

17

oxygen

17

8O

8

17

9

neon

20

10 Ne

10

20

10

iron

56

26 Fe

26

56

30

plutonium

244

94 Pu

94

244

150

mercury

202

80 Hg

80

202

122

cobalt

59

27 Co

27

59

32

nickel

56

28 Ni

28

56

28

fluorine

19

9F

9

19

10

chromium

50

24 Cr

24

50

26

44. vertical; groups

A29

46. Metallic elements are found toward the left and bottom of the

periodic table; there are far more metallic elements than nonmetals.

48. nonmetallic gaseous elements: oxygen, nitrogen, fluorine,

chlorine, hydrogen, and the noble gases; There are no metallic gaseous elements at room conditions

50. A metalloid is an element that has some properties common

to both metallic and nonmetallic elements. The metalloids

are found in the “stair-step” region marked on most periodic

tables.

52. (a) fluorine, chlorine, bromine, iodine, astatine; (b) lithium,

sodium, potassium, rubidium, cesium, francium; (c) beryllium, magnesium, calcium, strontium, barium, radium;

(d) helium, neon, argon, krypton, xenon, radon

54. Arsenic is a metalloid. Other elements in the same group (5A)

include nitrogen (N), phosphorus (P), antinomy (Sb), and bismuth (Bi).

56. Most elements are too reactive to be found in the uncombined form in nature and are found only in compounds.

58. These elements are found uncombined in nature and do not

readily react with other elements. Although these elements

were once thought to form no compounds, this now has been

shown to be untrue.

60. diatomic gases: H2, N2, O2, F2, Cl2; monatomic gases: He, Ne,

Kr, Xe, Rn, Ar

62. chlorine

64. diamond

66. electrons

68. 3ϩ

70. -ide

72. nonmetallic

74. (a) 36; (b) 36; (c) 21; (d) 36; (e) 80; (f) 27

76. (a) two electrons gained; (b) three electrons gained;

(c) three electrons lost; (d) two electrons lost;

(e) one electron lost; (f) two electrons lost

78. (a) P3Ϫ; (b) Ra2ϩ; (c) AtϪ; (d) no ion; (e) Csϩ; (f) Se2Ϫ

80. Sodium chloride is an ionic compound, consisting of Naϩ and

ClϪ ions. When NaCl is dissolved in water, these ions are set

free and can move independently to conduct the electric current. Sugar crystals, although they may appear similar visually,

contain no ions. When sugar is dissolved in water, it dissolves

as uncharged molecules. No electrically charged species are

present in a sugar solution to carry the electric current.

82. The total number of positive charges must equal the total

number of negative charges so that the crystals of an ionic

compound have no net charge. A macroscopic sample of compound ordinarily has no net charge.

84. (a) CsI, BaI2, AlI3; (b) Cs2O, BaO, Al2O3; (c) Cs3P, Ba3P2,

AlP; (d) Cs2Se, BaSe, Al2Se3; (e) CsH, BaH2, AlH3

86. (a) 7, halogens; (b) 8, noble gases; (c) 2, alkaline earth

elements; (d) 2, alkaline earth elements; (e) 4; (f) 6;

(g) 8, noble gases; (h) 1, alkali metals

Atomic

88.

Element

Symbol

Number

Group 3

boron

aluminum

gallium

indium

B

Al

Ga

In

5

13

31

49

Group 5

nitrogen

phosphorus

arsenic

antimony

N

P

As

Sb

7

15

33

51

Group 6

oxygen

sulfur

selenium

tellurium

O

S

Se

Te

8

16

34

52

Group 8

helium

neon

argon

krypton

He

Ne

Ar

Kr

2

10

18

36

9

A30 Answers to Even-Numbered End-of-Chapter Questions and Exercises

90. Most of an atom’s mass is concentrated in the nucleus: the

protons and neutrons that constitute the nucleus have similar

masses and are each nearly 2000 times more massive than

electrons. The chemical properties of an atom depend on the

number and location of the electrons it possesses. Electrons are

found in the outer regions of the atom and are involved in interactions between atoms.

92. C6H12O6

94. (a) 29 electrons, 34 neutrons, 29 electrons;

(b) 35 protons, 45 neutrons, 35 electrons;

(c) 12 protons, 12 neutrons, 12 electrons

96. The chief use of gold in ancient times was as ornamentation,

whether in statuary or in jewelry. Gold possesses an especially

beautiful luster; since it is relatively soft and malleable, it can

be worked finely by artisans. Among the metals, gold is inert

to attack by most substances in the environment.

98. (a) I; (b) Si; (c) W; (d) Fe; (e) Cu; (f) Co

100. (a) Br; (b) Bi; (c) Hg; (d) V; (e) F; (f) Ca

102. (a) osmium; (b) zirconium; (c) rubidium;

(d) radon; (e) uranium; (f) manganese;

(g) nickel; (h) bromine

104. (a) CO2; (b) AlCl3; (c) HClO4; (d) SCl6

106. (a) 136 C; (b) 136 C; (c) 136 C; (d) 44

(e) 41

(f) 35

19 K;

20 Ca;

19 K

Number

Number

Mass

108. Symbol

of Protons

of Neutrons

Number

41

20 Ca

55

25 Mn

109

47 Ag

45

21 Sc

20

21

41

25

30

55

47

62

109

21

24

45

24. An oxyanion is a polyatomic ion containing a given element

and one or more oxygen atoms. The oxyanions of chlorine

and bromine are given below:

26.

28.

30.

32.

34.

36.

38.

40.

42.

44.

46.

48.

Chapter 5

2. A binary chemical compound contains only two elements;

the major types are ionic (compounds of a metal and a nonmetal) and nonionic or molecular (compounds between two

nonmetals). Answers depend on student responses.

4. cation (positive ion)

6. Some substances do not contain molecules; the formula we

write reflects only the relative number of each type of atom

present.

8. Roman numeral

10. (a) lithium chloride; (b) barium fluoride;

(c) calcium oxide; (d) aluminum iodide;

(e) magnesium sulfide; (f) rubidium oxide

12. (a) correct; (b) incorrect, copper(I) oxide;

(c) incorrect; potassium oxide; (d) correct;

(e) incorrect, rubidium sulfide

14. (a) copper(II) chloride; (b) chromium(III) oxide;

(c) mercury(II) chloride; (d) mercury(I) oxide;

(e) gold(III) bromide; (f) manganese(IV) oxide

16. (a) cobaltic chloride; (b) ferrous bromide;

(c) plumbic oxide; (d) stannic chloride;

(e) mercuric iodide; (f) ferrous sulfide

18. (a) chlorine pentafluoride; (b) xenon dichloride;

(c) selenium dioxide; (d) dinitrogen trioxide;

(e) diiodine hexachloride; (f) carbon disulfide

20. (a) lead(IV) sulfide, plumbic sulfide;

(b) lead(II) sulfide, plumbous sulfide;

(c) silicon dioxide;

(d) tin(IV) fluoride, stannic fluoride;

(e) dichlorine heptoxide;

(f) cobalt(III) sulfide, cobaltic sulfide

22. (a) barium fluoride; (b) radium oxide;

(c) dinitrogen oxide; (d) rubidium oxide;

(e) diarsenic pent(a)oxide; (f) calcium nitride

50.

52.

54.

56.

58.

60.

62.

64.

Oxyanion

Name

Oxyanion

Name

ClOϪ

ClO2Ϫ

ClO3Ϫ

ClO4Ϫ

hypochlorite

chlorite

chlorate

perchlorate

BrOϪ

BrO2Ϫ

BrO3Ϫ

BrO4Ϫ

hypobromite

bromite

bromate

perbromate

hypo- (fewest); per- (most)

IOϪ, hypoiodite; IO2Ϫ, iodite; IO3Ϫ, iodate; IO4Ϫ, periodate

(a) NO3Ϫ; (b) NO2Ϫ; (c) NH4ϩ; (d) CNϪ

CNϪ, cyanide; CO32Ϫ, carbonate; HCO3Ϫ, hydrogen carbonate; C2H3O2Ϫ, acetate

(a) ammonium ion; (b) dihydrogen phosphate ion;

(c) sulfate ion; (d) hydrogen sulfite ion (bisulfite ion);

(e) perchlorate ion; (f) iodate ion

(a) sodium permanganate; (b) aluminum phosphate;

(c) chromium(II) carbonate, chromous carbonate;

(d) calcium hypochlorite; (e) barium carbonate;

(f) calcium chromate

oxygen

(a) hypochlorous acid; (b) sulfurous acid; (c) bromic acid;

(d) hypoiodous acid; (e) perbromic acid; (f) hydrosulfuric

acid; (g) hydroselenic acid; (h) phosphorous acid

(a) MgF2; (b) FeI2; (c) HgS; (d) Ba3N2; (e) PbCl2;

(f) SnF4; (g) Ag2O; (h) K2Se

(a) N2O; (b) NO2; (c) N2O4; (d) SF6; (e) PBr3; (f) CI4;

(g) OCl2

(a) NH4C2H3O2; (b) Fe(OH)2; (c) Co2(CO3)3; (d) BaCr2O7;

(e) PbSO4; (f) KH2PO4; (g) Li2O2; (h) Zn(ClO3)2

(a) HCN; (b) HNO3; (c) H2SO4; (d) H3PO4;

(e) HClO or HOCl; (f) HBr; (g) HBrO2; (h) HF

(a) Ca(HSO4)2; (b) Zn3(PO4)2; (c) Fe(ClO4)3; (d) Co(OH)3;

(e) K2CrO4; (f) Al(H2PO4)3; (g) LiHCO3; (h) Mn(C2H3O2)2;

(i) MgHPO4; (j) CsClO2; (k) BaO2; (l) NiCO3

A moist paste of NaCl would contain Naϩ and ClϪ ions in solution and would serve as a conductor of electrical impulses.

H S Hϩ (hydrogen ion) ϩ eϪ; H ϩ eϪ S HϪ (hydride ion)

ClO4Ϫ, HClO4; IO3Ϫ, HIO3; ClOϪ, HClO; BrO2Ϫ, HBrO2;

ClO2Ϫ, HClO2

(a) gold(III) bromide (auric bromide); (b) cobalt(III) cyanide

(cobaltic cyanide); (c) magnesium hydrogen phosphate;

(d) diboron hexahydride (common name diborane);

(e) ammonia; (f) silver(I) sulfate (usually called silver

sulfate); (g) beryllium hydroxide

(a) ammonium carbonate; (b) ammonium hydrogen

carbonate, ammonium bicarbonate; (c) calcium phosphate;

(d) sulfurous acid; (e) manganese(IV) oxide; (f) iodic acid;

(g) potassium hydride

(a) M(C2H3O2)4; (b) M(MnO4)4; (c) MO2; (d) M(HPO4)2;

(e) M(OH)4; (f) M(NO2)4

Mϩ compounds: MD, M2E, M3F; M2ϩ compounds: MD2, ME,

M3F2; M3ϩ compounds: MD3, M2E3, MF

66.

Ca(NO3)2

Sr(NO3)2

NH4NO3

Al(NO3)3

Fe(NO3)3

Ni(NO3)2

AgNO3

Au(NO3)3

KNO3

Hg(NO3)2

Ba(NO3)2

CaSO4

SrSO4

(NH4)2SO4

Al2(SO4)3

Fe2(SO4)3

NiSO4

Ag2SO4

Au2(SO4)3

K2SO4

HgSO4

BaSO4

Ca(HSO4)2

Sr(HSO4)2

NH4HSO4

Al(HSO4)3

Fe(HSO4)3

Ni(HSO4)2

AgHSO4

Au(HSO4)3

KHSO4

Hg(HSO4)2

Ba(HSO4)2

Ca(H2PO4)2

CaO

Sr(H2PO4)2

SrO

NH4H2PO4 (NH4)2O

Al(H2PO4)3 Al2O3

Fe(H2PO4)3 Fe2O3

Ni(H2PO4)2

NiO

AgH2PO4

Ag2O

Au(H2PO4)3 Au2O3

KH2PO4

K2O

Hg(H2PO4)2 HgO

Ba(H2PO4)2

BaO

CaCl2

SrCl2

NH4Cl

AlCl3

FeCl3

NiCl2

AgCl

AuCl3

KCl

HgCl2

BaCl2

Answers to Even-Numbered End-of-Chapter Questions and Exercises

68. helium

70. F2, Cl2 (gas); Br2 (liquid); I2, At2 (solid)

72. 1Ϫ

74. 1Ϫ

76. (a) Al(13e) S Al3ϩ(10e) ϩ 3eϪ; (b) S(16e) ϩ 2eϪ S S2Ϫ(18e);

(c) Cu(29e) S Cuϩ(28e) ϩ eϪ; (d) F(9e) ϩ eϪ S FϪ(10e);

(e) Zn(30e) S Zn2ϩ(28e) ϩ 2eϪ; (f) P(15e) ϩ 3eϪ S P3Ϫ(18e)

78. (a) Na2S; (b) KCl; (c) BaO; (d) MgSe; (e) CuBr2;

(f) AlI3; (g) Al2O3; (h) Ca3N2

80. (a) silver(I) oxide or just silver oxide; (b) correct;

(c) iron(III) oxide; (d) plumbic oxide; (e) correct

82. (a) stannous chloride; (b) ferrous oxide; (c) stannic oxide;

(d) plumbous sulfide; (e) cobaltic sulfide; (f) chromous

chloride

84. (a) iron(III) acetate; (b) bromine monofluoride;

(c) potassium peroxide; (d) silicon tetrabromide;

(e) copper(II) permanganate; (f) calcium chromate

86. (a) CO32Ϫ; (b) HCO3Ϫ; (c) C2H3O2Ϫ; (d) CNϪ

88. (a) carbonate; (b) chlorate; (c) sulfate; (d) phosphate;

(e) perchlorate; (f) permanganate

90. Answer depends on student choices.

92. (a) NaH2PO4; (b) LiClO4; (c) Cu(HCO3)2; (d) KC2H3O2;

(e) BaO2; (f) Cs2SO3

Chapter 6

2. Most of these products contain a peroxide, which decomposes and releases oxygen gas.

4. Bubbling takes place as the hydrogen peroxide chemically decomposes into water and oxygen gas.

6. The appearance of the black color actually signals the breakdown of starches and sugars in the bread to elemental carbon.

You may also see steam coming from the bread (water produced by the breakdown of the carbohydrates).

8. atoms

10. Balancing an equation ensures that no atoms are created

or destroyed during the reaction. The total mass after the

reaction must be the same as the total mass before the reaction.

12. Solid, (s); liquid, (l); gas, ( g)

14. H2O2(aq) S H2(g) ϩ O2(g)

16. N2H4(l) S N2(g) ϩ H2(g)

18. C3H8(g) ϩ O2(g) S CO2(g) ϩ H2O(g);

C3H8(g) ϩ O2(g) S CO(g) ϩ H2O(g)

20. CaCO3(s) ϩ HCl(aq) S CaCl2(aq) ϩ H2O(l) ϩ CO2(g)

22. SiO2(s) ϩ C(s) S Si(s) ϩ CO(g)

24. Fe(s) ϩ H2O(l) S FeO(s) ϩ H2(g)

26. SO2(g) ϩ H2O(l) S H2SO3(aq); SO3(g) ϩ H2O(l) S H2SO4(aq)

28. NO(g) ϩ O3(g) S NO2(g) ϩ O2(g)

32. Xe(g) ϩ F2(g) S XeF4(s)

30. P4(s) ϩ O2(g) S P2O5(s)

34. NH3(g) ϩ O2(g) S HNO3(aq) ϩ H2O(l)

36. To balance a chemical equation we must have the same number of each type of atom on both sides of the equation. In addition, we must balance the equation we are given, that is, we

are not to change the nature of the substances.

For example, the equation 2H2O2(aq) S 2H2O(l) ϩ O2(g) can

be represented as

؉

؉

؉

The equation H2O2(aq) S H2(g) ϩ O2(g) can be represented as

؉

38. (a) Zn(s) ϩ CuO(s) S ZnO(s) ϩ Cu(l); (b) P4(s) ϩ 6F2(g) S

4PF3(g); (c) Xe( g) ϩ 2F2(g) S XeF4(s); (d) 2NH4Cl(g) ϩ

Mg(OH)2(s) S 2NH3(g) ϩ 2H2O(g) ϩ MgCl2(s); (e) 2SiO(s) ϩ

4Cl2(g) S 2SiCl4(l) ϩ O2(g); (f) Cs2O(s) ϩ H2O(l) S

2CsOH(aq); (g) N2O3(g) ϩ H2O(l) S 2HNO2(aq);

(h) Fe2O3(s) ϩ 3H2SO4(l) S Fe2(SO4)3(s) ϩ 3H2O(g)

A31

40. (a) Na2SO4(aq) ϩ CaCl2(aq) S CaSO4(s) ϩ 2NaCl(aq);

(b) 3Fe(s) ϩ 4H2O(g) S Fe3O4(s) ϩ 4H2(g);

(c) Ca(OH)2(aq) ϩ 2HCl(aq) S CaCl2(aq) ϩ 2H2O(l);

(d) Br2(g) ϩ 2H2O(l) ϩ SO2(g) S 2HBr(aq) ϩ H2SO4(aq);

(e) 3NaOH(s) ϩ H3PO4(aq) S Na3PO4(aq) ϩ 3H2O(l);

(f) 2NaNO3(s) S 2NaNO2(s) ϩ O2(g); ( g) 2Na2O2(s) ϩ

2H2O(l) S 4NaOH(aq) ϩ O2(g); (h) 4Si(s) ϩ S8(s) S 2Si2S4(s)

42. (a) 4NaCl(s) ϩ 2SO2(g) ϩ 2H2O(g) ϩ O2(g) S 2Na2SO4(s) ϩ

4HCl(g);

(b) 3Br2(l) ϩ I2(s) S 2IBr3(s);

(c) Ca(s) ϩ 2H2O(g) S Ca(OH)2(aq) ϩ H2(g);

(d) 2BF3(g) ϩ 3H2O(g) S B2O3(s) ϩ 6HF(g);

(e) SO2(g) ϩ 2Cl2(g) S SOCl2(l) ϩ Cl2O(g);

(f) Li2O(s) ϩ H2O(l) S 2LiOH(aq);

(g) Mg(s) ϩ CuO(s) S MgO(s) ϩ Cu(l);

(h) Fe3O4(s) ϩ 4H2(g) S 3Fe(l) ϩ 4H2O(g)

44. (a) Ba(NO3)2(aq) ϩ Na2CrO4(aq) S BaCrO4(s) ϩ 2NaNO3(aq);

(b) PbCl2(aq) ϩ K2SO4(aq) S PbSO4(s) ϩ 2KCl(aq); (c)

C2H5OH(l) ϩ 3O2(g) S 2CO2(g) ϩ 3H2O(l); (d) CaC2(s) ϩ

2H2O(l) S Ca(OH)2(s) ϩ C2H2(g); (e) Sr(s) ϩ 2HNO3(aq) S

Sr(NO3)2(aq) ϩ H2(g); (f) BaO2(s) ϩ H2SO4(aq) S BaSO4(s) ϩ

H2O2(aq); (g) 2AsI3(s) S 2As(s) ϩ 3I2(s); (h) 2CuSO4(aq) ϩ

4KI(s) S 2CuI(s) ϩ I2(s) ϩ 2K2SO4(aq)

46. Na(s) ϩ O2(g) S Na2O2(s); Na2O2(s) ϩ H2O(l) S NaOH(aq) ϩ

O2(g)

48. C12H22O11(aq) ϩ H2O(l) S 4C2H5OH(aq) ϩ 4CO2(g)

50. 2Al2O3(s) ϩ 3C(s) S 4Al(s) ϩ 3CO2(g)

52. 2Li(s) ϩ S(s) S Li2S(s); 2Na(s) ϩ S(s) S Na2S(s); 2K(s) ϩ S(s)

S K2S(s); 2Rb(s) ϩ S(s) S Rb2S(s); 2Cs(s) ϩ S(s) S Cs2S(s);

2Fr(s) ϩ S(s) S Fr2S(s)

54. BaO2(s) ϩ H2O(l) S BaO(s) ϩ H2O2(aq)

56. 2KClO3(s) S 2KCl(s) ϩ 3O2(g)

58. NH3(g) ϩ HCl(g) S NH4Cl(s)

60. The senses we call “odor” and “taste” are really chemical reactions of the receptors in our body with molecules in the

food we are eating. The fact that the receptors no longer detect the “fishy” odor or taste suggests that adding the lemon

juice or vinegar has changed the nature of the amines in the

fish.

62. Fe(s) ϩ S(s) S FeS(s)

64. K2CrO4(aq) ϩ BaCl2(aq) S BaCrO4(s) ϩ 2KCl(aq)

66. 2NaCl(aq) ϩ 2H2O(l) S Cl2(g) ϩ H2(g) ϩ 2NaOH(aq, s)

2NaBr(aq) ϩ 2H2O(l) S Br2(l) ϩ H2(g) ϩ 2NaOH(aq, s)

2NaI(aq) ϩ 2H2O(l) S I2(s) ϩ H2(g) ϩ 2NaOH(aq, s)

68. CaC2(s) ϩ 2H2O(l) S Ca(OH)2(s) ϩ C2H2(g)

70. CuO(s) ϩ H2SO4(aq) S CuSO4(aq) ϩ H2O(l)

72. Na2SO3(aq) ϩ S(s) S Na2S2O3(aq)

74. (a) Cl2(g) ϩ 2KI(aq) S 2KCl(aq) ϩ I2(s); (b) CaC2(s) ϩ

2H2O(l) S Ca(OH)2(s) ϩ C2H2(g); (c) 2NaCl(s) ϩ H2SO4(l)

S Na2SO4(s) ϩ 2HCl(g); (d) CaF2(s) ϩ H2SO4(l) S CaSO4(s)

ϩ 2HF(g); (e) K2CO3(s) S K2O(s) ϩ CO2(g); (f) 3BaO(s) ϩ

2Al(s) S Al2O3(s) ϩ 3Ba(s); (g) 2Al(s) ϩ 3F2(g) S 2AlF3(s);

(h) CS2(g) ϩ 3Cl2(g) S CCl4(l) ϩ S2Cl2(g)

76. (a) Pb(NO3)2(aq) ϩ K2CrO4(aq) S PbCrO4(s) ϩ 2KNO3(aq);

(b) BaCl2(aq) ϩ Na2SO4(aq) S BaSO4(s) ϩ 2NaCl(aq);

(c) 2CH3OH(l) ϩ 3O2(g) S 2CO2(g) ϩ 4H2O(g);

(d) Na2CO3(aq) ϩ S(s) ϩ SO2(g) S CO2(g) ϩ Na2S2O3(aq);

(e) Cu(s) ϩ 2H2SO4(aq) S CuSO4(aq) ϩ SO2(g) ϩ 2H2O(l);

(f) MnO2(s) ϩ 4HCl(aq) S MnCl2(aq) ϩ Cl2(g) ϩ 2H2O(l);

(g) As2O3(s) ϩ 6KI(aq) ϩ 6HCl(aq) S 2AsI3(s) ϩ 6KCl(aq) ϩ

3H2O(l); (h) 2Na2S2O3(aq) ϩ I2(aq) S Na2S4O6(aq) ϩ 2NaI(aq)

Chapter 7

2. Driving forces are types of changes in a system that pull a reaction in the direction of product formation; driving forces include formation of a solid, formation of water, formation of a

gas, and transfer of electrons.

A32 Answers to Even-Numbered End-of-Chapter Questions and Exercises

4. A reactant in aqueous solution is indicated with (aq); formation of a solid is indicated with (s).

6. There are twice as many chloride ions as magnesium ions.

8. The simplest evidence is that solutions of ionic substances

conduct electricity.

10. Answer depends on student choices.

12. (a) soluble; Rule 3; (b) soluble; Rule 2; (c) soluble; Rule 2;

(d) insoluble; Rule 5; (e) soluble; Rule 2; (f) soluble; Rule 1;

(g) soluble; Rule 4; (h) insoluble; Rule 6

14. (a) Rule 6; (b) Rule 6; (c) Rule 6; (d) Rule 3; (e) Rule 4

16. (a) MnCO3, Rule 6; (b) CaSO4, Rule 4; (c) Hg2Cl2, Rule 3;

(d) no precipitate, most sodium and nitrate salts are soluble;

(e) Ni(OH)2, Rule 5; (f) BaSO4, Rule 4

18. (a) Na2CO3(aq) ϩ CuSO4(aq) S Na2SO4(aq) ϩ CuCO3(s)

20.

22.

24.

26.

28.

30.

32.

34.

36.

38.

40.

42.

44.

46.

48.

(b) HCl(aq) ϩ AgC2H3O2(aq) S HC2H3O2(aq) ϩ AgCl(s)

(c) no precipitate

(d) 3(NH4)2S(aq) ϩ 2FeCl3(aq) S 6NH4Cl(aq) ϩ Fe2S3(s)

(e) H2SO4(aq) ϩ Pb(NO3)2(aq) S 2HNO3(aq) ϩ PbSO4(s)

(f) 2K3PO4(aq) ϩ 3CaCl2(aq) S 6KCl(aq) ϩ Ca3(PO4)2(s)

(a) CaCl2(aq) ϩ 2AgNO3(aq) S Ca(NO3)2(aq) ϩ 2AgCl(s);

(b) 2AgNO3(aq) ϩ K2CrO4(aq) S Ag2CrO4(s) ϩ 2KNO3(aq);

(c) BaCl2(aq) ϩ K2SO4(aq) S BaSO4(s) ϩ 2KCl(aq)

(a) Na2CO3(aq) ϩ K2SO4(aq) S no precipitate; all

combinations are soluble (b) CuCl2(aq) ϩ (NH4)2CO3(aq) S

2NH4Cl(aq) ϩ CuCO3(s) (c) K3PO4(aq) ϩ AlCl3(aq) S

3KCl(aq) ϩ AlPO4(s)

Spectator ions are ions that remain in solution during a

precipitation/double-displacement reaction. For example, in

the reaction BaCl2(aq) ϩ K2SO4(aq) S BaSO4(s) ϩ 2KCl(aq),

the Kϩ and ClϪ ions are spectator ions.

(a) Ca2ϩ(aq) ϩ SO42Ϫ(aq) S CaSO4(s); (b) Ni2ϩ(aq) ϩ

2OHϪ(aq) S Ni(OH)2(s); (c) 2Fe3ϩ(aq) ϩ 3S2(aq) S Fe2S3(s)

Agϩ(aq) ϩ ClϪ(aq) S AgCl(s); Pb2ϩ(aq) ϩ 2ClϪ(aq) S

PbCl2(s); Hg22ϩ(aq) ϩ 2ClϪ(aq) S Hg2Cl2(s)

Co2ϩ(aq) ϩ S2Ϫ(aq) S CoS(s); 2Co3ϩ(aq) ϩ 3S2Ϫ(aq) S

Co2S3(s); Fe2ϩ(aq) ϩ S2Ϫ(aq) S FeS(s); 2Fe3ϩ(aq) ϩ 3S2Ϫ(aq) S

Fe2S3(s)

The strong bases are those hydroxide compounds that dissociate fully when dissolved in water. The strong bases that are

highly soluble in water (NaOH, KOH) are also strong electrolytes.

acids: HCl (hydrochloric), HNO3 (nitric), H2SO4 (sulfuric);

bases: hydroxides of Group 1A elements: NaOH, KOH, RbOH,

CsOH

A salt is the ionic product remaining in solution when an acid

neutralizes a base. For example, in the reaction HCl(aq) ϩ

NaOH(aq) S NaCl(aq) ϩ H2O(l), sodium chloride is the salt

produced by the neutralization reaction.

RbOH(s) S Rbϩ(aq) ϩ OHϪ(aq); CsOH(s) S Csϩ(aq) ϩ OHϪ(aq)

(a) H2SO4(aq) ϩ 2KOH(aq) S K2SO4(aq) ϩ 2H2O(l)

(b) HNO3(aq) ϩ NaOH(aq) S NaNO3(aq) ϩ H2O(l)

(c) 2HCl(aq) ϩ Ca(OH)2(aq) S CaCl2(aq) ϩ 2H2O(l)

(d) 2HClO4(aq) ϩ Ba(OH)2(aq) S Ba(ClO4)2(aq) ϩ 2H2O(l)

Answer depends on student choice of example: Na(s) ϩ Cl2(g)

S 2NaCl(s) is an example.

The metal loses electrons, the nonmetal gains electrons.

Each magnesium atom would lose two electrons. Each oxygen

atom would gain two electrons (so the O2 molecule would

gain four electrons). Two magnesium atoms would be required to react with each O2 molecule. Magnesium ions are

charged 2ϩ, oxide ions are charged 2Ϫ.

Each potassium atom loses one electron. The sulfur atom

gains two electrons. So two potassium atoms are required to

react with one sulfur atom.

2 ϫ (K S Kϩ ϩ eϪ)

S ϩ 2eϪ S S2Ϫ

50. (a) P4(s) ϩ 5O2(g) S P4O10(s); (b) MgO(s) ϩ C(s) S Mg(s) ϩ

CO(g); (c) Sr(s) ϩ 2H2O(l) S Sr(OH)2(aq) ϩ H2(g);

(d) Co(s) ϩ 2HCl(aq) S CoCl2(aq) ϩ H2(g)

52. The reaction includes aluminum metal as a reactant and

products that contain aluminum ions. For this reaction,

electrons must be transferred. That is, to make an aluminum

cation, electrons must be removed from the metal. An oxidation reduction is one that involves transfers of electrons.

54. (a) oxidation–reduction; (b) oxidation–reduction;

(c) acid–base; (d) acid–base, precipitation; (e) precipitation; (f) precipitation; (g) oxidation–reduction;

(h) oxidation–reduction; (i) acid–base

56. oxidation–reduction

58. A decomposition reaction is one in which a given compound

is broken down into simpler compounds or constituent

elements. The reactions CaCO3(s) S CaO(s) ϩ CO2(g) and

2HgO(s) S 2Hg(l) ϩ O2(g) represent decomposition reactions. Such reactions often may be classified in other ways.

For example, the reaction of HgO(s) is also an oxidation–reduction reaction.

60. (a) C3H8(g) ϩ 5O2(g) S 3CO2(g) ϩ 4H2O(g);

(b) C2H4(g) ϩ 3O2(g) S 2CO2(g) ϩ 2H2O(g);

(c) 2C8H18(l) ϩ 25O2(g) S 16CO2(g) ϩ 18H2O(g)

62. Answer depends on student selection.

64. (a) 8Fe(s) ϩ S8(s) S 8FeS(s); (b) 4Co(s) ϩ 3O2(g) S

2Co2O3(s); (c) Cl2O7(g) ϩ H2O(l) S 2HClO4(aq)

66. (a) 2Al(s) ϩ 3Br2(l) S 2AlBr3(s)

(b) Zn(s) ϩ 2HClO4(aq) S Zn(ClO4)2(aq) ϩ H2(g)

(c) 3Na(s) ϩ P(s) S Na3P(s)

(d) CH4(g) ϩ 4Cl2(g) S CCl4(l) ϩ 4HCl(g)

(e) Cu(s) ϩ 2AgNO3(aq) S Cu(NO3)2(aq) ϩ 2Ag(s)

68. (a) silver ion: Agϩ(aq) ϩ ClϪ(aq) S AgCl(s); lead(II) ion:

Pb2ϩ(aq) ϩ 2ClϪ(aq) S PbCl2(s); mercury(I) ion: Hg22ϩ(aq) ϩ

2ClϪ(aq) S Hg2Cl2(s); (b) sulfate ion: Ca2ϩ(aq) ϩ SO42Ϫ(aq)

S CaSO4(s); carbonate ion: Ca2ϩ(aq) ϩ CO32Ϫ(aq) S

CaCO3(s); phosphate ion: 3Ca2ϩ(aq) ϩ 2PO43Ϫ(aq) S

Ca3(PO4)2(s); (c) hydroxide ion: Fe3ϩ(aq) ϩ 3OHϪ(aq) S

Fe(OH)3(s); sulfide ion: 2Fe3ϩ(aq) ϩ 3S2Ϫ(aq) S Fe2S3(s);

phosphate ion: Fe3ϩ(aq) ϩ PO43Ϫ(aq) S FePO4(s);

(d) barium ion: Ba2ϩ(aq) ϩ SO42Ϫ(aq) S BaSO4(s); calcium

ion: Ca2ϩ(aq) ϩ SO42Ϫ(aq) S CaSO4(s); lead(II) ion:

Pb2ϩ(aq) ϩ SO42Ϫ(aq) S PbSO4(s); (e) chloride ion:

Hg22ϩ(aq) ϩ 2ClϪ(aq) S Hg2Cl2(s); sulfide ion: Hg22ϩ(aq) ϩ

S2Ϫ(aq) S Hg2S(s); carbonate ion: Hg22ϩ(aq) ϩ CO32Ϫ(aq) S

Hg2CO3(s); (f) chloride ion: Agϩ(aq) ϩ ClϪ(aq) S AgCl(s);

hydroxide ion: Agϩ(aq) ϩ OHϪ(aq) S AgOH(s); carbonate

ion: 2Agϩ(aq) ϩ CO32Ϫ(aq) S Ag2CO3(s)

70. (a) HNO3(aq) ϩ KOH(aq) S H2O(l) ϩ KNO3(aq);

(b) H2SO4(aq) ϩ Ba(OH)2(aq) S BaSO4(s) ϩ 2H2O(l);

(c) HClO4(aq) ϩ NaOH(aq) S H2O(l) ϩ NaClO4(aq);

(d) 2HCl(aq) ϩ Ca(OH)2(aq) S CaCl2(aq) ϩ H2O(l)

72. (a) soluble (Rule 2: most potassium salts are soluble);

(b) soluble (Rule 2: most ammonium salts are soluble);

(c) insoluble (Rule 6: most carbonate salts are only slightly

soluble); (d) insoluble (Rule 6: most phosphate salts are

only slightly soluble); (e) soluble (Rule 2: most sodium salts

are soluble); (f) insoluble (Rule 6: most carbonate salts are

only slightly soluble); (g) soluble (Rule 3: most chloride

salts are soluble)

74. (a) AgNO3(aq) ϩ HCl(aq) S AgCl(s) ϩ HNO3(aq);

(b) CuSO4(aq) ϩ (NH4)2CO3(aq) S CuCO3(s) ϩ (NH4)2SO4(aq);

(c) FeSO4(aq) ϩ K2CO3(aq) S FeCO3(s) ϩ K2SO4(aq);

(d) no reaction; (e) Pb(NO3)2(aq) ϩ Li2CO3(aq) S PbCO3(s)

ϩ 2LiNO3(aq); (f) SnCl4(aq) ϩ 4NaOH(aq) S Sn(OH)4(s) ϩ

4NaCl(aq)

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