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6: The Wave Mechanical Model of the Atom

6: The Wave Mechanical Model of the Atom

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A42 Answers to Even-Numbered End-of-Chapter Questions and Exercises

82. Strong hydrogen-bonding forces are present in an ice crystal,

while only the much weaker London forces exist in the crystal of a nonpolar substance like oxygen.

84. Ice floats on liquid water; water expands when it is frozen.

86. Although they are at the same temperature, steam at 100 ЊC

contains a larger amount of energy than hot water, equal to

the heat of vaporization of water.

88. Hydrogen bonding is a special case of dipole–dipole interactions that occur among molecules containing hydrogen

atoms bonded to highly electronegative atoms such as fluorine, oxygen, or nitrogen. The bonds are very polar, and the

small size of the hydrogen atom (compared to other atoms)

allows the dipoles to approach each other very closely. Examples: H2O, NH3, HF.

90. Evaporation and condensation are opposite processes. Evaporation is an endothermic process; condensation is an

exothermic process. Evaporation requires an input of energy

to provide the increased kinetic energy possessed by the

molecules when they are in the gaseous state. It occurs when

the molecules in a liquid are moving fast enough and randomly enough that molecules are able to escape from the

surface of the liquid and enter the vapor phase.

Chapter 15

2. A nonhomogeneous mixture may differ in composition in

various places in the mixture, whereas a solution (a homogeneous mixture) has the same composition throughout.

Examples of nonhomogeneous mixtures include spaghetti

sauce, a jar of jelly beans, and a mixture of salt and sugar.

4. solid

6. “Like dissolves like.” The hydrocarbons in oil have intermolecular forces that are very different from those in water,

so the oil spreads out rather than dissolving in the water.

8. Carbon dioxide is somewhat soluble in water, especially if

pressurized (otherwise, the soda you may be drinking while

studying chemistry would be “flat”). Carbon dioxide’s solubility in water is approximately 1.5 g/L at 25 °C under a pressure of approximately 1 atm. The carbon dioxide molecule

overall is nonpolar, because the two individual COO bond

dipoles cancel each other due to the linearity of the molecule. However, these bond dipoles are able to interact with

water, making CO2 more soluble in water than nonpolar

molecules such as O2 or N2, which do not possess individual

bond dipoles.

10. unsaturated

12. large

14. 100.

16. (a) 0.116%; (b) 0.0116%; (c) 10.4%; (d) 10.4%

18. (a) 20.5 g FeCl3; 504.5 g (505 g) water; (b) 26.8 g sucrose;

198.2 g (198 g) water; (c) 181.3 g (181 g) NaCl; 1268.7

(1.27 ϫ 103) g water; (d) 95.9 g KNO3; 539.1 (539) g water

20. 957 g Fe; 26.9 g C; 16.5 g Cr

22. 19.6% CaCl2

24. 7.81 g KBr

26. approximately 71 g

28. 9.5 g

30. 0.110 mol; 0.220 mol

32. b

34. (a) 3.35 M; (b) 1.03 M; (c) 0.630 M; (d) 4.99 M

36. (a) 0.403 M; (b) 0.169 M; (c) 0.629 M; (d) 0.829 M

38. 4.03 g KBr

40. 0.0902 M

42. 0.479 M

44. (a) 0.00130 mol; (b) 0.00609 mol; (c) 0.0184 mol;

(d) 0.0356 mol

46. (a) 0.235 g; (b) 0.593 g; (c) 2.29 g; (d) 2.61 g

48. 9.51 g

50. (a) 4.60 ϫ 10Ϫ3 mol Al3ϩ, 1.38 ϫ 10Ϫ2 mol ClϪ;

(b) 1.70 mol Naϩ, 0.568 mol PO43Ϫ;

(c) 2.19 ϫ 10Ϫ3 mol Cu2ϩ, 4.38 ϫ 10Ϫ3 mol ClϪ;

(d) 3.96 ϫ 10Ϫ5 mol Ca2ϩ, 7.91 ϫ 10Ϫ5 mol OHϪ

52. 1.33 g

54. half

56. (a) 0.0717 M; (b) 1.69 M; (c) 0.0426 M;

(d) 0.625 M

58. 0.541 L (541 mL)

60. Dilute 48.3 mL of the 1.01 M solution to a final volume of

325 mL.

62. 10.3 mL

64. 31.2 mL

66. 0.523 g

68. 0.300 g

70. 378 mL

72. 1.8 ϫ 10Ϫ4 M

74. (a) 63.0 mL; (b) 2.42 mL; (c) 50.1 mL;

(d) 1.22 L

76. 1 N

78. 1.53 equivalents OHϪ ion. By definition, one equivalent of

OHϪ ion exactly neutralizes one equivalent of Hϩ ion.

80. (a) 0.277 N; (b) 3.37 ϫ 10Ϫ3 N; (c) 1.63 N

82. (a) 0.134 N; (b) 0.0104 N; (c) 13.3 N

84. 7.03 ϫ 10Ϫ5 M, 1.41 ϫ 10Ϫ4 N

86. 22.2 mL, 11.1 mL

88. 0.05583 M, 0.1117 N

90. Molarity is defined as the number of moles of solute contained in 1 liter of total solution volume (solute plus solvent

after mixing). In the first example, the total volume after

mixing is not known and the molarity cannot be calculated.

In the second example, the final volume after mixing is

known and the molarity can be calculated simply.

92. 3.3%

94. 12.7 g NaHCO3

96. 56 mol

98. 1.12 L HCl at STP

100. 26.3 mL

102. 2.56 M

104. (a) 6.3% KNO3; (b) 0.25% KNO3;

(c) 11% KNO3; (d) 18% KNO3

106. 4.7% C, 1.4% Ni, 93.9% Fe

108. 28 g Na2CO3

110. 9.4 g NaCl, 3.1 g KBr

112. (a) 4.0 M; (b) 1.0 M; (c) 0.73 M; (d) 3.6 M

114. 0.812 M

116. 0.026 M

118. (a) 0.446 mol, 33.3 g; (b) 0.00340 mol, 0.289 g;

(c) 0.075 mol, 2.7 g; (d) 0.0505 mol, 4.95 g

120. (a) 0.938 mol Naϩ, 0.313 mol PO43Ϫ; (b) 0.042 mol Hϩ,

0.021 mol SO42Ϫ; (c) 0.0038 mol Al3ϩ, 0.011 mol ClϪ;

(d) 1.88 mol Ba2ϩ, 3.75 mol ClϪ

122. (a) 0.0909 M; (b) 0.127 M; (c) 0.192 M;

(d) 1.6 M

124. 0.90 M

126. 50. mL

128. 35.0 mL

130. (a) 0.822 N HCl; (b) 4.00 N H2SO4;

(c) 3.06 N H3PO4

132. 0.083 M NaH2PO4, 0.17 N NaH2PO4

134. 9.6 ϫ 10Ϫ2 N HNO3

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


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)


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)

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.

A44 Answers to Even-Numbered End-of-Chapter Questions and Exercises

84. pH

86. weak acid

88. (a) H2O and OHϪ are a conjugate acid–base pair (H2O is

the acid, having one more proton than the base, OHϪ);

(b) H2SO4 and SO42Ϫ are not a conjugate acid–base pair (they

differ by two protons). The conjugate base of H2SO4 is HSO4Ϫ;

the conjugate acid of SO42Ϫ is also HSO4Ϫ; (c) H3PO4 and

H2PO4Ϫ are a conjugate acid–base pair (H3PO4 is the acid, having one more proton than the base, H2PO4Ϫ); (d) HC2H3O2

and C2H3O2Ϫ are a conjugate acid–base pair (HC2H3O2 is the

acid, having one more proton than the base, C2H3O2Ϫ)

90. (a) NH4ϩ; (b) NH3; (c) H3Oϩ; (d) H2O

92. (a) CH3CH2COOH ϩ H2O 3

4 CH3CH2COOϪ ϩ H3Oϩ;


(b) NH4 ϩ H2O 3

4 NH3 ϩ H3Oϩ;

(c) H2SO4 ϩ H2O S HSO4Ϫ ϩ H3Oϩ;

(d) H3PO4 ϩ H2O 3

4 H2PO4Ϫ ϩ H3Oϩ

94. (a) [Hϩ(aq)] ϭ 2.4 ϫ 10Ϫ12 M, solution is basic;

(b) [Hϩ(aq)] ϭ 9.9 ϫ 10Ϫ2 M, solution is acidic;

(c) [Hϩ(aq)] ϭ 3.3 ϫ 10Ϫ8 M, solution is basic;

(d) [Hϩ(aq)] ϭ 1.7 ϫ 10Ϫ9 M, solution is basic

96. (a) [OHϪ(aq)] ϭ 0.0000032 M;

(b) [OHϪ(aq)] ϭ 1.54 ϫ 10Ϫ8 M;

(c) [OHϪ(aq)] ϭ 4.02 ϫ 10Ϫ7 M

98. (a) pH ϭ 8.15; solution is basic; (b) pH ϭ 5.97; solution is

acidic; (c) pH ϭ 13.34; solution is basic; (d) pH ϭ 2.90;

solution is acidic

100. (a) [OHϪ(aq)] ϭ 1.8 ϫ 10Ϫ11 M, pH ϭ 3.24, pOH ϭ 10.76;

(b) [Hϩ(aq)] ϭ 1.1 ϫ 10Ϫ10 M, pH ϭ 9.95, pOH ϭ 4.05;

(c) [OHϪ(aq)] ϭ 3.5 ϫ 10Ϫ3 M, pH ϭ 11.54, pH ϭ 2.46;

(d) [Hϩ(aq)] ϭ 1.4 ϫ 10Ϫ7 M, pH ϭ 6.86, pOH ϭ 7.14

102. (a) [Hϩ] ϭ 3.9 ϫ 10Ϫ6 M; (b) [Hϩ] ϭ 1.1 ϫ 10Ϫ2 M;

(c) [Hϩ] ϭ 1.2 ϫ 10Ϫ12 M; (d) [Hϩ] ϭ 7.8 ϫ 10Ϫ11 M

104. (a) [Hϩ(aq)] ϭ 1.4 ϫ 10Ϫ3 M, pH ϭ 2.85;

(b) [Hϩ(aq)] ϭ 3.0 ϫ 10Ϫ5 M, pH ϭ 4.52;

(c) [Hϩ(aq)] ϭ 5.0 ϫ 10Ϫ2 M, pH ϭ 1.30;

(d) [Hϩ(aq)] ϭ 0.0010 M, pH ϭ 3.00

Chapter 17

2. Four COH bonds and four ClOCl bonds must be broken;

four COCl and four HOCl bonds must form.

4. Ea represents the activation energy for the reaction, which is

the minimum energy needed for the reaction to be able to


6. Enzymes are biochemical catalysts that speed up the complicated reactions that would be too slow to sustain life at

normal body temperatures.

8. A state of equilibrium is attained when two opposing

processes are exactly balanced. The development of a vapor

pressure above a liquid in a closed container is an example

of a physical equilibrium. Any chemical reaction that appears to “stop” before completion is an example of a chemical equilibrium.

10. A system has reached equilibrium when no more product

forms, even though significant amounts of all the needed reactants are present. This lack of further product creation indicates that the reverse process is now occurring at the same

rate as the forward process—that is, every time a product

molecule forms in the system, another product molecule reacts to give back the original reactants elsewhere in the system. Reactions that come to equilibrium are indicated by a

double arrow.

12. The two curves come together when a state of chemical

equilibrium has been reached, after which point the forward

and reverse reactions are occurring at the same rate so that

there is no further net change in concentration.

14. The equilibrium constant is a ratio of the concentration of

products to the concentration of reactants, all at equilibrium. Depending on the amount of reactant that was originally present, different amounts of reactants and products

will be present at equilibrium, but their ratio will always be

the same for a given reaction at a given temperature. For example, the ratios 4/2 and 6/3 involve different numbers, but

each of these ratios has the value 2.

16. (a) K ϭ

(c) K ϭ

18. (a) K ϭ

(c) K ϭ



; (b) K ϭ










(b) K ϭ






20. K ϭ 0.734

22. K ϭ 4.85 ϫ 10Ϫ6

24. Equilibrium constants represent ratios of the concentrations

of products and reactants present at the point of equilibrium. The concentration of a pure solid or a pure liquid is constant and is determined by the density of the solid or liquid.

26. (a) K ϭ

28. (a) K ϭ

(c) K ϭ



; (b) K ϭ






(c) K ϭ

; (b) K ϭ








30. [CO2] increases; K does not change

32. If heat is applied to an endothermic reaction (the temperature is raised), the equilibrium is shifted to the right. More

product will be present at equilibrium than if the temperature had not been increased. The value of K increases.

34. (a) shift right; (b) no change; (c) shift left

36. (a) no change (B is solid); (b) shift right;

(c) shift left; (d) shift right

38. The forward reaction would be favored. The reaction is endothermic, and raising the temperature means heat is being

added to the system.

40. For an endothermic reaction, an increase in temperature will

shift the position of equilibrium to the right (toward products).

42. Add more CO(g); add more H2(g); decrease the volume of the


44. A small equilibrium constant implies that not much product

forms before equilibrium is reached. The reaction would not

be a good source of the products unless Le Châtelier’s principle can be used to force the reaction to the right.

46. K ϭ 8.63 ϫ 10Ϫ7

48. [H2] ϭ 0.119 M

50. [O2(g)] ϭ 8.0 ϫ 10Ϫ2 M

52. 5.4 ϫ 10Ϫ4 M

54. solubility product, Ksp

56. only the temperature

z 2Bi3ϩ(aq) ϩ 3S2Ϫ(aq);

58. (a) Bi2S3(s) y

Ksp ϭ [Bi3ϩ(aq)]2[S2Ϫ(aq)]3;

z Ca2ϩ(aq) ϩ 2OHϪ(aq);

(b) Ca(OH)2(s) y

Ksp ϭ [Ca2ϩ(aq)][OHϪ(aq)]2;

z Co3ϩ(aq) ϩ 3OHϪ(aq);

(c) Co(OH)3(s) y

Ksp ϭ [Co3ϩ(aq)][OHϪ(aq)]3;

z 2Cuϩ(aq) ϩ S2Ϫ(aq); Ksp ϭ [Cuϩ(aq)]2[S2Ϫ(aq)]

(d) Cu2S(s) y

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



















1.9 ϫ 10Ϫ4 M; 0.016 g/L

7.4 ϫ 10Ϫ4 g/L

Ksp ϭ 2.27 ϫ 10Ϫ4

Ksp ϭ 1.23 ϫ 10Ϫ15

Ksp ϭ 1.9 ϫ 10Ϫ4; 10. g/L

4 ϫ 10Ϫ17 M, 4 ϫ 10Ϫ15 g/L

increase in temperature increases the fraction of molecules

with energy Ͼ Ea



reaction is still taking place, but in opposing directions, at

the same speeds



In an exothermic process, heat is a product of the reaction.

So adding heat (increasing the temperature) fights against

the forward process.

An equilibrium reaction may come to many positions of

equilibrium, but the numerical value of the equilibrium

constant is fulfilled at each possible position. If different experiments vary the amounts of reactant, the absolute

amounts of reactants and products present at the point of

equilibrium will differ from one experiment to another, but

the ratio that defines the equilibrium constant will remain

the same.

9.0 ϫ 10Ϫ3 M

z Ba2ϩ(aq) ϩ CO32Ϫ(aq); 7.1 ϫ 10Ϫ5 M

BaCO3(s) y

z Cd2ϩ(aq) ϩ CO32Ϫ(aq); 2.3 ϫ 10Ϫ6 M

CdCO3(s) y


CaCO3(s) y Ca2ϩ(aq) ϩ CO32Ϫ(aq); 5.3 ϫ 10Ϫ5 M

z Co2ϩ(aq) ϩ CO32Ϫ(aq); 3.9 ϫ 10Ϫ7 M

CoCO3(s) y

Although a small solubility product generally implies a

small solubility, comparisons of solubility based directly on

Ksp values are valid only if the salts produce the same numbers of positive and negative ions per formula when they

dissolve. For example, the solubilities of AgCl(s) and NiS(s)

can be compared directly using Ksp, since each salt produces

one positive and one negative ion per formula when dissolved. AgCl(s) cannot be directly compared with a salt such

as Ca3(PO4)2, however.

At higher temperatures, the average kinetic energy of the

reactant molecules is larger, as is the probability that a collision between molecules will be energetic enough for reaction to take place. On a molecular basis, a higher temperature means a given molecule will be moving faster.

96. (a) K ϭ

(c) K ϭ



(b) K ϭ






98. K ϭ 1.2 ϫ 10Ϫ3


100. (a) K ϭ



(c) K ϭ


(b) K ϭ






102. An exothermic reaction liberates energy as heat. Increasing

the temperature (adding heat) for such a reaction is fighting

against the reaction’s own tendency to liberate heat. The net

effect of raising the temperature will be a shift to the left and

a decrease in the amount of product. To increase the amount

of products in an exothermic reaction, heat must be removed

from the system. Changing the temperature does change the

numerical value of the equilibrium constant for a reaction.

104. The reaction is exothermic. An increase in temperature (addition of heat) will shift the reaction to the left (toward



106. [NH3(g)] ϭ 1.1 ϫ 10Ϫ3 M

z Cu2ϩ(aq) ϩ 2OHϪ(aq);

108. (a) Cu(OH)2(s) y

Ksp ϭ [Cu (aq)][OHϪ(aq)]2;

z Cr3ϩ(aq) ϩ 3OHϪ(aq);

(b) Cr(OH)3(s) y

Ksp ϭ [Cr (aq)][OHϪ(aq)]3;

z Ba2ϩ(aq) ϩ 2OHϪ(aq);

(c) Ba(OH)2(s) y

Ksp ϭ [Ba (aq)][OHϪ(aq)]2;

z Sn2ϩ(aq) ϩ 2OHϪ(aq);

(d) Sn(OH)2(s) y

Ksp ϭ [Sn (aq)][OHϪ(aq)]2

110. Ksp ϭ 3.9 ϫ 10Ϫ11

112. Ksp ϭ 1.4 ϫ 10Ϫ8

114. The activation energy is the minimum energy that two colliding molecules must possess for the collision to result in a


116. Once a system has reached equilibrium, the net concentration of product no longer increases because molecules of

product already present react to form the original reactants.

118. (a) K ϭ [H2O][CO2]; (b) K ϭ [CO2];


(c) K ϭ


Chapter 18

2. Oxidation is a loss of one or more electrons by an atom or

ion. Reduction is the gaining of one or more electrons by an

atom or ion. Equations depend on student responses.

4. (a) boron is oxidized; oxygen is reduced; (b) nitrogen is oxidized; oxygen is reduced; (c) carbon is oxidized; hydrogen

is reduced; (d) magnesium is oxidized; copper is reduced

6. (a) sulfur is oxidized, oxygen is reduced; (b) phosphorus is

oxidized, oxygen is reduced; (c) hydrogen is oxidized, carbon is reduced; (d) boron is oxidized, hydrogen is reduced

8. Oxidation numbers represent a “relative charge” one atom

has compared to another in a compound. In an element, all

the atoms are equivalent.

10. Because fluorine is the most electronegative element, its oxidation state is always negative relative to other elements;

because fluorine gains only one electron to complete its outermost shell, its oxidation number in compounds is always

Ϫ1. The other halogen elements are almost always more

electronegative than the atoms to which they bond, and almost always have Ϫ1 oxidation numbers. However, in an interhalogen compound involving fluorine and some other

halogen, since fluorine is the most electronegative element

of all, the other halogens in the compound will have positive oxidation states relative to fluorine.

12. Ϫ3

14. (a) Cr, ϩ3; Cl, Ϫ1; (b) Ni, ϩ2; O, Ϫ2; H, ϩ1;

(c) H, ϩ1; S, Ϫ2; (d) C, ϩ4; S, Ϫ2

16. (a) 0; (b) Ϫ3; (c) ϩ4; (d) ϩ5

18. (a) ϩ2; (b) ϩ7; (c) ϩ4; (d) ϩ3

20. (a) Ca, ϩ2; O, Ϫ2; (b) Al, ϩ3; O, Ϫ2; (c) P, ϩ3; F, Ϫ1;

(d) P, ϩ5; O, Ϫ2

22. (a) H, ϩ1; S, ϩ6; O, Ϫ2; (b) Mn, ϩ7; O, Ϫ2;

(c) Cl, ϩ5; O, Ϫ2; (d) Br, ϩ7; O, Ϫ2

24. Electrons are negative; when an atom gains electrons, it

gains one negative charge for each electron gained. For example, in the reduction reaction Cl ϩ eϪ S ClϪ, the oxidation state of chlorine decreases from 0 to Ϫ1 as the electron

is gained.

26. Answer depends on student selection of example.

28. An antioxidant is a substance that prevents oxidation of

some molecule in the body. It is not certain how all antioxidants work, but one example is by preventing oxygen molecules and other substances from stripping electrons from

A46 Answers to Even-Numbered End-of-Chapter Questions and Exercises














cell membranes, which leaves them vulnerable to destruction by the immune system.

(a) aluminum is oxidized; sulfur is reduced; (b) carbon is

oxidized; oxygen is reduced; (c) carbon is oxidized; iron is

reduced; (d) chlorine is oxidized; chromium is reduced

(a) carbon is oxidized, chlorine is reduced; (b) carbon is oxidized, oxygen is reduced; (c) phosphorus is oxidized, chlorine is reduced; (d) calcium is oxidized, hydrogen is reduced

Iron is reduced [ϩ3 in Fe2O3(s), 0 in Fe(l)]; carbon is oxidized

[ϩ2 in CO( g), ϩ4 in CO2(g)]. Fe2O3(s) is the oxidizing agent;

CO(g) is the reducing agent.

(a) chlorine is reduced, iodine is oxidized; chlorine is the oxidizing agent, iodide ion is the reducing agent; (b) iron is

reduced, iodine is oxidized; iron(III) is the oxidizing agent,

iodide ion is the reducing agent; (c) copper is reduced, iodine is oxidized; copper(II) is the oxidizing agent, iodide ion

is the reducing agent

Oxidation–reduction reactions are often more complicated

than “regular” reactions; the coefficients necessary to balance the number of electrons transferred are often large


Under ordinary conditions it is impossible to have “free”

electrons that are not part of some atom, ion, or molecule.

Thus, the total number of electrons lost by the species being

oxidized must equal the total number of electrons gained by

the species being reduced.

(a) 2ClϪ(aq) S Cl2(g) ϩ 2eϪ; (b) Fe2ϩ(aq) S Fe3ϩ(aq) ϩ eϪ;

(c) Fe(s) S Fe3ϩ(aq) ϩ 3eϪ; (d) Cu2ϩ(aq) ϩ eϪ S Cuϩ(aq)

(a) 4Hϩ ϩ 4eϪ ϩ O2 S 2H2O;

(b) 4Hϩ ϩ 2eϪ ϩ SO42Ϫ S H2SO3 ϩ H2O;

(c) 2Hϩ ϩ 2eϪ ϩ H2O2 S 2H2O;

(d) H2O ϩ NO2Ϫ S NO3Ϫ ϩ 2Hϩ ϩ 2eϪ

(a) 2Al ϩ 6Hϩ S 2Al3ϩ ϩ 3H2;

(b) 8Hϩ ϩ 2NO3Ϫ ϩ 3S2Ϫ S 3S ϩ 2NO ϩ 4H2O;

(c) 6H2O ϩ I2 ϩ 5Cl2 S 2IO3Ϫ ϩ 2Hϩ ϩ 10HCl;

(d) 2Hϩ ϩ AsO4Ϫ ϩ S2Ϫ S S ϩ AsO3Ϫ ϩ H2O

Cu(s) ϩ 2HNO3(aq) ϩ 2Hϩ(aq) S Cu2ϩ(aq) ϩ 2NO2(g) ϩ

2H2O(l); Mg(s) ϩ 2HNO3(aq) S Mg(NO3)2(aq) ϩ H2(g)

A salt bridge typically consists of a U-shaped tube filled

with an inert electrolyte (one involving ions that are not

part of the oxidation–reduction reaction). A salt bridge

completes the electrical circuit in a cell. Any method that allows transfer of charge without allowing bulk mixing of the

solutions may be used (another common method is to set up

one half-cell in a porous cup, which is then placed in the

beaker containing the second half-cell).

Reduction takes place at the cathode and oxidation takes

place at the anode.

Salt bridge


























Pb2ϩ(aq) ion is reduced; Zn(s) is oxidized. The anode reaction

is Zn(s) S Zn2ϩ(aq) ϩ 2eϪ. The cathode reaction is Pb2ϩ(aq) ϩ

2eϪ S Pb(s).

56. Cd ϩ 2OHϪ S Cd(OH)2 ϩ 2eϪ (oxidation);

NiO2 ϩ 2H2O ϩ 2eϪ S Ni(OH)2 ϩ 2OHϪ (reduction)

58. Aluminum is a very reactive metal when freshly isolated in

the pure state. Upon standing for even a relatively short period of time, aluminum metal forms a thin coating of Al2O3


on its surface from reaction with atmospheric oxygen. This

Al2O3 coating is much less reactive than the metal and protects the metal’s surface from further attack.

Chromium protects stainless steel by forming a thin coating

of chromium oxide on the surface of the steel, which prevents oxidation of the iron in the steel.

The main recharging reaction for the lead storage battery is

2PbSO4(s) ϩ 2H2O(l) S Pb(s) ϩ PbO2(s) ϩ 2H2SO4(aq). A major side reaction is the electrolysis of water, 2H2O(l) S

2H2(g) ϩ O2(g), which produces an explosive mixture of hydrogen and oxygen that accounts for many accidents during

the recharging of such batteries.

The balanced equation is 2H2O(l) S 2H2(g) ϩ O2(g). Oxygen

is oxidized (going from Ϫ2 oxidation state in water to zero

oxidation state in the free element). Hydrogen is reduced

(going from ϩ1 oxidation state in water to zero oxidation

state in the free element). Heat is produced by burning the

hydrogen gas produced by the electrolysis. Because energy

must be applied to water to electrolyze it, energy is released

when hydrogen gas produced by the electrolysis and oxygen

gas combine to form water in the fireplace.

loss; oxidation state


An oxidizing agent is an atom, molecule, or ion that causes the

oxidation of some other species, while itself being reduced.


separate from


In an electrolysis reaction, an ordinarily nonspontaneous reaction is forced to occur by the application of an electric current of sufficient voltage. For example, water may be electrolyzed into its elements: 2H2O(l) S 2H2(g) ϩ O2(g).

hydrogen; oxygen


(a) 4Fe(s) ϩ 3O2(g) S 2Fe2O3(s); iron is oxidized, oxygen is reduced; (b) 2Al(s) ϩ 3Cl2(g) S 2AlCl3(s); aluminum is oxidized, chlorine is reduced; (c) 6Mg(s) ϩ P4(s) S 2Mg3P2(s);

magnesium is oxidized, phosphorus is reduced

(a) Al is oxidized (0 S ϩ3); H is reduced (ϩ1 S 0);

(b) H is reduced (ϩ1 S 0); I is oxidized (Ϫ1 S 0);

(c) Cu is oxidized (0 S ϩ2); H is reduced (ϩ1 S 0)

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

(b) CO(g) ϩ 2H2(g) S CH3OH(l);

(c) SnO2(s) ϩ 2C(s) S Sn(s) ϩ 2CO(g);

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

(a) sodium is oxidized, oxygen is reduced; (b) iron is oxidized, hydrogen is reduced; (c) oxygen (O2Ϫ) is oxidized,

aluminum (Al3ϩ) is reduced; (d) magnesium is oxidized,

nitrogen is reduced

(a) H, ϩ1; N, Ϫ3; (b) C, ϩ2; O, Ϫ2;

(c) C, ϩ4; O, Ϫ2; (d) N, ϩ3; F, Ϫ1

(a) Mn, ϩ4; O, Ϫ2; (b) Ba, ϩ2; Cr, ϩ6; O, Ϫ2;

(c) H, ϩ1; S, ϩ4; O, Ϫ2; (d) Ca, ϩ2; P, ϩ5; O, Ϫ2

(a) Bi, ϩ3; O, Ϫ2; (b) P, ϩ5; O, Ϫ2;

(c) N, ϩ3; O, Ϫ2; (d) Hg, ϩ1

(a) oxygen is oxidized, chlorine is reduced; (b) germanium

is oxidized, oxygen is reduced; (c) carbon is oxidized, chlorine is reduced; (d) oxygen is oxidized, fluorine is reduced

(a) SiO2(s) ϩ 4Hϩ(aq) ϩ 4eϪ S Si(s) ϩ 2H2O(l);

(b) S(s) ϩ 2Hϩ(aq) ϩ 2eϪ S H2S(g);

(c) NO3Ϫ(aq) ϩ 3Hϩ(aq) ϩ 2eϪ S HNO2(aq) ϩ H2O(l);

(d) NO3Ϫ(aq) ϩ 4Hϩ(aq) ϩ 3eϪ S NO(g) ϩ 2H2O(l)

(a) 16Hϩ(aq) ϩ 2MnO4Ϫ(aq) ϩ 5C2O42Ϫ(aq) S 2Mn2ϩ(aq) ϩ

8H2O(l) ϩ 10CO2(g); (b) 8Hϩ(aq) ϩ MnO4Ϫ(aq) ϩ

5Fe2ϩ(aq) S Mn2ϩ(aq) ϩ 4H2O(l) ϩ 5Fe3ϩ(aq);

(c) 16Hϩ(aq) ϩ 2MnO4Ϫ(aq) ϩ 10ClϪ(aq) S 2Mn2ϩ(aq) ϩ

8H2O(l) ϩ 5Cl2(g)

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

104. A galvanic cell is a battery. A spontaneous oxidation–

reduction reaction is separated physically into the two halfreactions, and the electrons being transferred between the

two half-cells are made available as an electric current.

Chapter 19

2. The radius of a typical atomic nucleus is on the order of

10Ϫ13 cm, which is roughly 100,000 times smaller than the

radius of an atom overall.

4. The mass number represents the total number of protons

and neutrons in a nucleus.

6. The atomic number (Z) is written as a left subscript, while

the mass number (A) is written as a left superscript. That is,

the general symbol for a nuclide is AZX. As an example, consider the isotope of oxygen with 8 protons and 8 neutrons:

its symbol would be 168O.

8. electron

10. Emission of a neutron, 01 n, does not change the atomic number of the parent nucleus, but it causes the mass number of

the parent nucleus to decrease by one unit.

12. Gamma rays are high-energy photons of electromagnetic radiation; they are not normally considered to be particles.

When a nucleus produces only gamma radiation, the atomic

number and mass number of the nucleus do not change.

14. Electron capture occurs when one of the inner-orbital electrons is pulled into, and becomes part of, the nucleus.

16. The fact that the average atomic mass of potassium is only

slightly above 39 amu reflects the fact that the isotope of

mass number 39 predominates.


Number of Neutrons



20 neutrons



21 neutrons



22 neutrons

18. Based on the predominance of Mg-24, but with significant amounts of the other isotopes, one would expect the

average atomic molar mass to be slightly higher than 24

(24.31 g).

20. (a) electron; (b) positron; (c) neutron; (d) proton


22. (a) 12

Mg; (b) 35Li; (c) 24He


24. (a) 86 Rn; (b) 10e (positron); (c) 137

56 Ba





26. (a) 234

(b) 222

92U S 2He ϩ 90Th;

86Rn S 2He ϩ 84Po;




(c) 75Re S 2He ϩ 73Ta





28. (a) 212

(b) 212

82Pb S Ϫ1e ϩ 83Bi;

81Tl S Ϫ1e ϩ 82Pb;




(c) 88Ra S Ϫ1e ϩ 89Ac

30. In a nuclear bombardment process, a target nucleus is bombarded with high-energy particles (typically subatomic particles or small atoms) from a particle accelerator. This may

result in the transmutation of the target nucleus into some

other element. For example, nitrogen-14 may be transmuted

into oxygen-17 by bombardment with ␣ particles.



32. 12

Mg ϩ 12H S 11

Na ϩ 24He

34. The half-life of a nucleus is the time required for one-half of

the original sample of nuclei to decay. A given isotope of an

element always has the same half-life, although different isotopes of the same element may have greatly different halflives. Nuclei of different elements have different half-lives.


36. 226

88Ra is the most stable (longest half-life);

88Ra is the

“hottest” (shortest half-life).

38. With a half-life of 2.6 hours, strontium-87 is the hottest;

with a half-life of 45.1 days, iron-59 is the most stable to



40. Four half-lives; 1/16 (0.54) remains

42. For an administered dose of 100 ␮g, 0.39 ␮g remains after

2 days. The fraction remaining is 0.39͞100 ϭ 0.0039; on a

percentage basis, less than 0.4% of the original radioisotope


44. Carbon-14 is produced in the upper atmosphere by the

bombardment of nitrogen with neutrons from space:



ϩ 10nS146C ϩ 11H

46. We assume that the concentration of C-14 in the atmosphere

is effectively constant. A living organism is constantly replenishing C-14 through the processes of either metabolism

(sugars ingested in foods contain C-14) or photosynthesis

(carbon dioxide contains C-14). When a plant dies, it no

longer replenishes itself with C-14 from the atmosphere. As

the C-14 undergoes radioactive decay, its amount decreases

with time.

48. 1 day is about 13 half-lives for 189F. If we begin with

6.02 ϫ 1023 atoms (1 mol), then after 13 half-lives, 7.4 ϫ 1019

atoms of 189F will remain.

50. fission, fusion, fusion, fission




52. 10n ϩ 235

92 US 56 Ba ϩ 36Kr ϩ 30n

54. A critical mass of a fissionable material is the amount

needed to provide a high enough internal neutron flux to

sustain the chain reaction (production of enough neutrons

to cause the continuous fission of further material). A sample with less than a critical mass is still radioactive, but cannot sustain a chain reaction.

56. An actual nuclear explosion, of the type produced by a nuclear weapon, cannot occur in a nuclear reactor because the

concentration of the fissionable materials is not sufficient to

form a supercritical mass.

58. Some advantages are that the fuel is available domestically,

is relatively “clean,” and does not produce the greenhouse

gases that fossil fuel plants produce. Some disadvantages are

safety, waste disposal, and cost.

60. In one type of fusion reactor, two 21H atoms fuse to produce


2 He. Because the hydrogen nuclei are positively charged, extremely high energies (temperatures of 40 million K) are

needed to overcome the repulsion between the nuclei as

they are shot into each other.

62. protons (hydrogen), helium

64. Somatic damage is damage directly to the organism itself,

causing nearly immediate sickness or death to the organism.

Genetic damage is damage to the genetic machinery of the

organism, which will be manifested in future generations of


66. Gamma rays penetrate long distances, but seldom cause ionization of biological molecules. Because they are much heavier, although less penetrating, alpha particles ionize biological molecules very effectively and leave a dense trail of

damage in the organism. Isotopes that decay by releasing alpha particles can be ingested or breathed into the body,

where the damage from the alpha particles will be more


68. Most reactor waste is still in “temporary” storage. Various

suggestions have been made for a more permanent solution,

such as casting the spent fuel into glass bricks to contain it

and then storing the bricks in corrosion-proof metal containers deep underground.

70. radioactive

72. mass

74. neutron; proton

76. radioactive decay

78. mass number

80. transuranium

82. half-life

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