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10: Energy as a Driving Force
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
0.0500 mol Pb2ϩ
1.25 L ϫ
ϭ 0.0625 mol Pb2ϩ
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
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
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
0.050 mol OH Ϫ
125 mL ϫ
ϭ 6.3 ϫ 10 Ϫ3 mol OH Ϫ
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.
0.100 mol H ϩ
ϭ 6.3 ϫ 10 Ϫ3 mol H ϩ
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
ϭ 6.3 ϫ 10 L ϫ
ϭ 63 mL
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.
755 mL ϫ
ϭ 0.755 L
Finally, we substitute these values into the equation that defines
ϭ 0.558 N
Self-Check Exercise 15.11
To solve this problem, we use the relationship
Nacid ϫ Vacid ϭ Nbase ϫ Vbase
Nacid ϭ 0.50
Vacid ϭ ?
Nbase ϭ 0.80
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
) ϫ (0.250 L)
Nbase ϫ Vbase
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.
Self-Check Exercise 16.1
The conjugate acid–base pairs are
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 5.0 ϫ 10 Ϫ13 M
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
1.0 ϫ 10 Ϫ14
ϭ 2.0 ϫ 10 Ϫ10 M
3H ϩ 4 ϭ
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
Ϫ3.50 ϭ 3.2 ϫ 10Ϫ4
Self-Check Exercise 16.6
Step 1 pOH ϭ 10.50
Step 2 ϪpOH ϭ Ϫ10.50
Self-Check Exercise 17.7
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
Applying the law of chemical equilibrium gives
3 NO2 4 4 3 H2O4 6
3 NH3 4 4 3O2 4 7
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.
c. K ϭ
The solids are not included.
3 CO2 4
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
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
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
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
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
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.
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
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
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
N2 ϩ 3H2S2NH3
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
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)
Step 1 The oxidation half-reaction is
Cu ϩ HNO3SCu(NO3 )2
ϩ1 ϩ5 Ϫ2 ϩ2 ϩ5 Ϫ2
The copper goes from 0 to ϩ2 and thus is oxidized. This reduction
Oxidation states: 0
ϩ1 ϩ5 Ϫ2 ϩ2 Ϫ2
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.
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.
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
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
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.
3Cu, 8H, 8N, 24O S 3Cu, 8H, 8N, 24O
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:
88 RaS2He ϩ Z X
Because we know that both A and Z are conserved, we can
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
Because Rn has Z ϭ 86, AZX is 222
86 Rn. The overall
balanced equation is
88 RaS2He ϩ 86 Rn
Check: Z ϭ 88
Z ϭ 86 ϩ 2 ϭ 88
A ϭ 226
A ϭ 222 ϩ 4 ϭ 226
b. Using a similar strategy, we have
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
82 PbSϪ1e ϩ 83 Bi
Check: Z ϭ 82
Z ϭ 83 Ϫ 1 ϭ 82
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
86 RnS 84 Po ϩ 2He
is a balanced equation.
Check: Z ϭ 86
Z ϭ 84 ϩ 2 ϭ 86
A ϭ 222
A ϭ 218 ϩ 4 ϭ 222
b. The missing species must be 157X or 157N, because the balanced
8 OS 7 N ϩ 1e
Check: Z ϭ 8
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
8.0 ϫ 10Ϫ7 mol ¡ 4.0 ϫ 10Ϫ7 mol ¡
2.0 ϫ 10Ϫ7 mol ¡ 1.0 ϫ 10Ϫ7 mol
It takes three half-lives, then, for the sample to go from
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
is 6.7 years. Therefore, the
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.
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ANSWERS TO EVEN-NUMBERED END-OF-CHAPTER
QUESTIONS AND EXERCISES
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.
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
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)
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.
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
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
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.
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
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
A28 Answers to Even-Numbered End-of-Chapter Questions and Exercises
(a) 5.07 ϫ 104 kryll; (b) 0.12 blim; (c) 3.70 ϫ 10Ϫ5 blim2
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;
(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
128. 0.105 m
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
152. 0.59 g/cm3
(a) 23 °F; (b) 32 °F; (c) Ϫ321 °F; (d) Ϫ459 °F;
(e) 187 °F; (f) Ϫ459 °F
ϭ 62 miles, or about 60 miles,
(a) 100 km ϫ
taking significant figures into account.
ϭ 49,200 lbs. of fuel was
(b) 22,300 kg ϫ
needed; 22,300 lbs. were added, so 26,900 additional
pounds were needed.
Ϸ 10Ϫ11 lb/gal
2. forces among the particles in the matter
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.
14. malleable; ductile
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.
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
28. Heterogeneous mixtures: salad dressing, jelly beans, the
change in my pocket; solutions: window cleaner, shampoo,
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.
42. far apart
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
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.
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
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;
(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.
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.
38. (a) 54
(d) 147 N; (e) 157 N; (f) 157 N
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.
44. vertical; groups
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
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
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
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.
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
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:
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
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
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
(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;
(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;
(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
Answers to Even-Numbered End-of-Chapter Questions and Exercises
70. F2, Cl2 (gas); Br2 (liquid); I2, At2 (solid)
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
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
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).
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)
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
(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
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
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)
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
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)
(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
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,
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
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