Download Chapter 1: Matter and Measurements

Chapter 1: Matter and Measurements
22. Refer to Section 1.2.
(a) 4020.6 mL = 4.0206 x 103 mL
(b) 1.006 g (This is already in proper scientific notation.)
(c) 100.1°C = 1.001 x 102°C
32. Refer to Section 1.2 and Table 1.2.
Convert one of the numbers to the units of the other. Once the numbers are expressed in
common units, they can be compared directly.
(a) 37.12 g x
1k g
= 0.03712 k g , thus: 37.12 g < 0.3712 kg.
1000 g
3
⎛ 100 cm ⎞
6
3
⎟ = 28 ×10 cm , thus: 28 m3 > 28 x 102 cm3.
⎝ 1m ⎠
(b) 28 m 3 × ⎜
(c) 525 mm ×
1m
1 nm
×
= 525 × 106 nm ,
−9
1000 mm
1 × 10 m
thus 525 mm = 525 x 106 nm
48. Refer to Section 1.3.
Calculate the volume of the object from the change in
the volume of the graduated cylinder, then calculate the
density.
42.3 mL
35.0 mL
Volume of object = 42.3 mL – 35.0 mL = 7.3 mL
density =
mass (g)
11.33 g
=
= 1.6 g/mL
volume (mL) 7.3 mL
1
Chapter 2: Atoms, Molecules, and Ions
16. Refer to Section 2.2.
(a) Na-21 has 11 protons and 21 – 11 = 10 neutrons. Adding one neutron makes the number
22, thus the symbol is 22
11 Na .
(b) An isobar of Na-21 with 10 protons would have an atomic number of 10, thus the
21
element symbol would be Ne. The nuclear symbol is 10
Ne .
(c) A nucleus with 11 protons is sodium. The mass number is 11 + 12 = 23. The nuclear
symbol is 23
11 Na . This is indeed an isotope of Na-21.
38. Refer to Section 2.5.
The sum of the charges must equal zero; if not, add extras of the needed ion to “balance”
the charges.
(a) Ba2+ and I- give BaI2 (+2 + 2(-1) = 0)
Ba2+ and N3- give Ba3N2 (3(+2) + 2(-3) = 0)
(b) O2- and Fe2+ give FeO (-2 + (+2) = 0)
O2- and Fe3+ give Fe2O3 (3(-2) + 2(+3) = 0)
52. Refer to Sections 2.4 and 2.5.
(a) “Compounds containing carbon atoms are molecular” is usually true. Carbon, being a
nonmetal can combine with other non-metals to form molecular compounds such as those
mentioned in Section 2.4 (indeed, a whole field of chemistry, called organic chemistry, is
dedicated to the study of such compounds). Carbon can, however, also form ionic
compounds such as calcium carbide (used in old miners lamps).
(b) “A molecule is made up of nonmetal atoms” is always true, by definition.
(c) “An ionic compound has at least one metal atom” is usually true. Most ionic
compounds do contain a metal, but there are ionic compounds in which the cation
(positive ion) is not a metal, such as in ammonium chloride, NH4Cl.
2
Chapter 3: Mass Relations in Chemistry; Stoichiometry
26. Refer to Section 3.2 and Example 3.3.
Multiply the number of each atom in the molecule by the atomic mass of that atom, and then
add up the masses to get the molar mass of the molecule.
(a) Ga: 1(69.723) = 69.723 g/mol.
.
(b) CaSO4 ½ H2O: 1(40.078) + 1(32.066) + 4(15.9994) + ½ [2(1.00794) + 1(15.9994)]
= 136.1416 + ½ (18.01528) =145.149 g/mol.
(c) C14H10O4: 14(12.011) + 10(1.0079) + 4(15.9994)
= 168.154 + 10.0794 + 63.9976 = 242.231 g/mol.
38. Refer to Section 3.3.
Mass of 1 mol. C7H5BiO4 = 7(12.01) + 5(1.008) + 209.0 + 4(16.00) = 362.1 g
346 mg Bi x
1 mol. C 7 H 5 BiO 4
1g
362.1 g
x
x
= 0.599 g C 7 H 5 BiO 4
1000 mg
209.0 g Bi
1 mol. C 7 H 5 BiO 4
0.599 g C 7 H 5 BiO 4
x 100% = 39.9%
1.500 g Pepto − Bismal
49. Refer to Section 3.3.
68. Refer to Section 3.4 and Example 3.9.
(a) First, the mass of the sample of tin, Sn, must be determined. Note: the thickness is given
in units of millimeters and must first be converted into cm.
3
0.600 mm ×
cm
= 0.0600 cm
10 mm
7.28 g Sn
= 77.5 g Sn
cm3
1 mol. Sn 1 mol. SnO 2 150.71 g SnO 2
77.5 g Sn ×
×
×
= 98.4 g SnO 2
118.71 g Sn 1 mol. Sn
1 mol. SnO 2
(8.25 cm × 21.5 cm × 0.0600 cm) ×
(b) 98.4 g SnO 2 ×
31.999 g O 2
L O2
100 L air
×
×
= 76 L air
150.71 g SnO 2 1.309 g O 2 21 L O 2
4
Chapter 4: Reactions in Aqueous Solution
4. Refer to Section 4.1.
⎛ 1 mol ⎞
⎟
15.00 g ScI 3 ⎜⎜
425.7 g ⎟⎠
mol
⎝
(a) M =
=
= 0.0538 M
L
0.655 L
0.0538 M Sc3+ (aq)
0.0538 M * 3 = 0.161 M I- (aq)
⎛ 1 mol ⎞
⎟
15.00 g Na 2 CO 3 ⎜⎜
106.0 g ⎟⎠
mol
⎝
(b) M =
=
= 0.216 M
L
0.655 L
0.216 M Na+ (aq)
0.216 M CO32+ (aq)
⎛ 1 mol ⎞
⎟
15.00 g Mg 3 ( PO 4 ) 2 ⎜⎜
262.84 g ⎟⎠
mol
⎝
(c) M =
=
= 0.0871 M
L
0.655 L
0.0871 M * 3 = 0.261 M Mg2+ (aq)
0.0871 M * 2 = 0.174 M PO43- (aq)
⎛ 1 mol ⎞
⎟⎟
15.00 g K 2 O⎜⎜
94
.
20
g
mol
⎠ = 0.243 M
⎝
(d) M =
=
0.655 L
L
0.243 M * 2 = 0.486 M K+ (aq)
0.243 M O2- (aq)
18. Refer to Sections 2.6, 4.1, and 4.2, Figures 4.3 and 4.4, and Example 4.4.
Recall that soluble salts ionize when dissolved. Write the reactions for the ionizations. Look
at the resulting ions. If there are pairs that would result in insoluble salts, these salts would
form and precipitate from solution.
5
(a) Ions present: Cu2+, SO42-, Na+, ClPossible precipitates: CuCl2, Na2SO4
Both compounds are soluble, thus no precipitate forms.
(b) Ions present: Mn2+, NO3-, Na+, OHPossible precipitates: Mn(OH)2, NaNO3
NaNO3 is soluble, but Mn(OH)2 is not.
Mn2+(aq) + 2OH-(aq) → Mn(OH)2(s)
(c) Ions present: Ag+, NO3-, H+, ClPossible precipitates: AgCl, HNO3
HNO3 is soluble, but AgCl is not.
Ag+(aq) + Cl-(aq) → AgCl(s)
(d) Ions present: Co2+, SO42-, Ba2+, OHPossible precipitates: Co(OH)2, BaSO4
Both compounds are insoluble.
Co2+(aq) + 2OH-(aq) → Co(OH)2(s)
Ba2+(aq) + SO42-(aq) → BaSO4(s)
(e) Ions present: NH4+, K+, OH-, CO32Possible precipitates: NH4OH, K2CO3
Both compounds are soluble, thus no precipitate forms.
36. Refer to Sections 2.6 and 4.3 and Table 4.2.
For the net ionic equation OH-(aq) + HB(aq) → B-(aq) + H2O(l) to be correct, the reactants
must be a weak acid and a strong base.
(a) The equation is not correct.
Hydrochloric acid (HCl) is a strong acid, the reacting species is H+.
Pyridine (C5H5N) is a weak base, the reacting species is C5H5N.
H+(aq) + C5H5N(aq) → C5H5NH+(aq)
(b) The equation is not correct.
Sulfuric acid (H2SO4) is a strong acid, the reacting species is H+.
Rubidium hydroxide (RbOH) is a strong base, the reacting species is OH-.
H+(aq) + OH-(aq) → H2O(l)
(c) The equation is correct.
Hydrofluoric acid (HF) is a weak acid, the reacting species is HF.
Potassium hydroxide (KOH ) is a strong base, the reacting species is OH-.
OH-(aq) + HF(aq) → F-(aq) + H2O(l)
(d) The equation is not correct.
Hydroiodic acid (HI) is a strong acid, the reacting species is H+.
Ammonia (NH3) is a weak base, the reacting species is NH3.
H+(aq) + NH3(aq) → NH4+(aq)
6
(e) The equation is correct.
Hydrocyanic acid (HCN) is a weak acid, the reacting species is HCN.
Strontium hydroxide (Sr(OH)2 ) is a strong base, the reacting species is OH-.
OH-(aq) + HCN(aq) → CN-(aq) + H2O(l)
56. Refer to Section 4.4 and Example 4.9.
Begin by determining the oxidation numbers of the element being reduced or oxidized.
Balance the element being oxidized or reduced and then balance the oxidation number by
adding electrons. Balance the charge by adding H+ (if acidic) or OH- (if basic), and H2O to
balance H and O.
(a) ClO-(aq) → Cl-(aq)
[basic]
+
Cl → Cl
ClO-(aq) +2e- → Cl-(aq)
ClO-(aq) +2e- → Cl-(aq) + 2OH-(aq)
ClO-(aq) +2e- + H2O(l) → Cl-(aq) + 2OH-(aq)
Reduction
Cl and electrons balanced
charges balanced
O and H balanced
(b) NO3-(aq) → NO(g)
[acidic]
N5+ → N2+
NO3-(aq) + 3e- → NO(g)
NO3-(aq) + 3e- + 4H+(aq)→ NO(g)
NO3-(aq) + 3e- + 4H+(aq)→ NO(g) + 2H2O(l)
Reduction
N and electrons balanced
charges balanced
O and H balanced
(c) Ni2+(aq) → Ni2O3(s)
[basic]
Ni2+ → Ni3+
2Ni2+(aq) → Ni2O3(s)
2Ni2+(aq) → Ni2O3(s) + 2e2Ni2+(aq) + 6OH-(aq) → Ni2O3(s) + 2e2Ni2+(aq) + 6OH-(aq) → Ni2O3(s) + 2e- + 3H2O(l)
Oxidation
Ni balanced
electrons balanced
charges balanced
O and H balanced
(d) Mn2+(aq) → MnO2(s)
[acidic]
2+
4+
Mn → Mn
Mn2+(aq) → MnO2(s) + 2eMn2+(aq) → MnO2(s) + 2e- + 4H+(aq)
Mn2+(aq) + 2H2O(l) → MnO2(s) + 2e- + 4H+(aq)
Oxidation
Mn and electrons balanced
charges balanced
O and H balanced
7
Chapter 5: Gases
12. Refer to Section 5.1 and Example 5.3.
V1 = 25.0 mL = 0.0250 L
T1 = 23oC = 23 + 273 = 296 K
P1 = 745 mm Hg ×
1 atm
= 0.980 atm
760.0 mm Hg
Since volume increased by 8%, multiply V1 by 1.08 to get V2.
V2 = (0.0250 L)(1.08) = 0.0270 L
T2 = 82oC = 82 + 273 = 355 K
P2 = ?
P1V1 P2V2
=
T1
T2
(0.980 atm)(0.0250 L) ( P2 )(0.0270 L)
=
(296 K)
355 K
P2 = 1.09 atm = 828 mm Hg
Yes, the pressure increased.
20. Refer to Section 5.3 and Example 5.3.
Start by converting all units to those used in the ideal gas law. Calculated moles using the
ideal gas equation if P, V and T are given, otherwise use the molar mass and the given mass.
Then apply the ideal gas law to fill in any other missing data.
Pressure
22.7 atm
0.895 atm
433 mm Hg
1.74 bar
Volume
1.75 L
6.17 L
92.4 mL
8.66 L
Temperature
19°C
6°C
1.62 K
98°F
Moles
1.66
0.241
0.395
0.585
T = 19°C + 273 = 292 K
P=
nRT (1.66 mol.)(0.0821 L ⋅ atm/mol. ⋅ K)(292 K)
=
= 22.7 atm
V
(1.75 L)
8
Grams
96.5
14.0
23.0
34.0
1.66 mol. C 4 H10 ×
58.12 g
= 96.5 g
1 mol.
T = 6°C + 273 = 279 K
14.0 g ×
V=
1 mol.
= 0.241 mol.
58.12 g
nRT (0.241 mol.)(0.0821 L ⋅ atm/mol. ⋅ K)(279 K)
=
= 6.17 L
P
(0.895 atm)
433 mm Hg x
1 atm
= 0.570 atm
760 mm Hg
(0.570 atm)(0.0924 L)
PV
=
= 1.62 K
nR (0.0821 L ⋅ atm/mol. ⋅ K)(0.395 mol.)
1 atm
1.74 bar x
= 1.72 atm
1 .013 bar
T=
T=
98 D F - 32
+ 273 = 310 K
1.8
n=
(1.72 atm)(8.66 L)
PV
=
= 0.585 mol.
RT (0.0821 L ⋅ atm/mol. ⋅ K)(310 K)
0.585 mol. x
58.12 g
= 34.0 g
1 mol.
36. Refer to Section 5.4 and Example 5.5.
(a) This equation can be balanced by inspection.
NH4NO3(s) → N2O(g) + 2H2O(l)
(b) T = 250°C + 273 = 523 K
V=
nRT (0.0625 mol.)(0.0821 L ⋅ atm/mol.⋅ K)(298 K)
=
= 2.7 L
P
1.0 atm
52. Refer to Section 5.6.
(a) P (H2) = P (Total) – P (H2O) = 748 mm Hg – 23.8 mm Hg = 724 2 mm Hg
9
⎤
⎡
⎢ 23.8 mm Hg ⎥
⎥(0.250 L)
⎢
⎢ 760 mm Hg ⎥
⎥⎦
⎢ 1 atm
PV
(b) n =
= ⎣
= 0.000320 moles H2O
RT (0.08216 L • atm )( 298 K )
mol • K
⎤
⎡
⎢ 724.2 mm Hg ⎥
⎥(0.250 L)
⎢
⎢ 760 mm Hg ⎥
⎥⎦
⎢
1 atm
PV
(c) n =
= ⎣
= 0.00973 moles H2
RT (0.08216 L • atm )( 298 K )
mol • K
⎛ 1 mol ⎞
⎟⎟ = 0.00465 mol He are added to the flask
⎝ 4.0026 g ⎠
(d) 0.0186 g He ⎜⎜
New Total Pressure:
P=
nRT
V
(0.000320 + 0.00973 + 0.00465 moles)(0.08216
P=
0.250 L
P = 1.44 atm or 1094 mm Hg
P (He) = X (He) · P (Total) =
⎛
⎞
(.00465 moles)
⎜⎜
⎟⎟ (1094 mm Hg)
⎝ (0.000320 + 0.00973 + .00465 moles) ⎠
= 346 mm Hg
(e) P (Total) = 1.44 atm or 1094 mm Hg (from above)
10
L • atm
)(298 K )
mol • K
Chapter 6: Electronic Structure and the Periodic Table
8. Refer to Section 6.1 and Example 6.2.
Calculate the amount of energy in one photon. Use that and the total amount of energy to
calculate the number of photons emitted.
λ = 633 nm x
E=
hc
=
λ
12 kJ ×
1m
= 6.33 x 10 -7 m
1 x 10 9 nm
(6.626 x 10 -34 J ⋅ s)(2.998 x 10 8 m/s)
= 3.14 x 10 -19 J
-7
6.33 x 10 m
1000 J
1 photon
×
= 3.8 x 10 22 photons
1 kJ 3.14 x 10-19 J
22. Refer to Section 6.3.
The type of orbital is determined by the A value.
(a) A = 1, therefore: p-orbital.
(b) A = 0, therefore: s-orbital.
(c) A = 2, therefore: d-orbital
48. Refer to Section 6.6.
The only main group metals in the 4th period are K, Ca and Ga, the remainder of the elements
are transition metals, nonmetals or metalloids.
.
K
Ca
Ga
4s
(↑ )
(↑↓)
(↑↓)
4p
( )( )( )
( )( )( )
(↑ )( )( )
(a) Ca
(b) K and Ga
(c) none
(d) none
11
54. Refer to Section 6.8, Figures 6.13 and 6.15, and Example 6.11.
(a) Atomic radius increases from right to left across a period, therefore:
S < Si < Na
(b) Ionization energy increases from left to right across a period, therefore:
Na < Si < S
(c) Electronegativity decreases from right to left across a period, therefore:
S > Si > Na
12
Chapter 7: Covalent Bonding
12. Refer to Section 7.1 and Example 7.1.
2C: 2 x 4 valence electrons
5H: 5 x 1 valence electrons
2O: 2 x 6 valence electrons
N: 5 valence electrons
total: 30 electrons
H
H
H
O
N
C
C
O
From the information given, the structure must be that on
the left.
H
H
H
H
H
O
N
C
C
O
Note that this structure does not have a complete octet
around one of the carbon atoms. To complete the octet,
move a pair of electrons from the terminal oxygen to
form a C=O double bond.
H
H
H
H
H
O
N
C
C
O
This structure provides octets for all the atoms and
provides that all formal charges are zero.
H
H
34. Refer to Section 7.2, Examples 7.5 and 7.6, Figures 7.4 and 7.5, and Table 7.3.
Draw the Lewis structure of the compound and determine the number of bonded groups and
the number of electron pairs. Then use Table 7.3 to assign the geometry.
(a)
NNO ⇒
N N O
2 bonded groups, no electron pairs, thus AX2 and linear.
(b)
ONCl ⇒
O N Cl
2 bonded groups, one electron pair, thus AX2E and bent.
(c)
NH4 ⇒
+
H
+
H N H
H
13
4 bonded groups, no electron pairs, thus AX4 and tetrahedron.
O3 ⇒
(d)
O O O
2 bonded groups, one electron pair, thus AX2E and bent.
52. Refer to Section 7.4, Table 7.4, Example 7.9, and Problem 34 (above).
Recall that the total number of groups (bonded atoms and electron pairs) around the central
atom is equal to the number of orbitals that hybridized. Furthermore, the sum of the
superscripts in the hybrid orbital notation gives the total number of hybrid orbitals.
(a)
NNO
AX2
2 groups
sp
(b)
ONCl
AX2E
3 groups
sp2
(c)
NH4+
AX4
4 groups
sp3
(d)
O3
AX2E
3 groups
sp2
14
Chapter 8: Thermochemistry
8. Refer to Problem 1.16, Section 8.1, and Example 8.1.
(a) In the solution process, heat is absorbed. Thus, it is not exothermic, it is endothermic.
(b) Heat is absorbed from the water. Thus, heat exits the water and the process for the water
is exothermic. Thus, q H 2O = −27.6 kJ .
(c) 250.0 mL ×
− 27.6 kJ ×
1.00 g
= 250.0 g
1 mL
1000 J
= −27600 J
1 kJ
q = mcΔt
-27600 J = (250.0 g)( 4.18 J/g°C) Δt
Δt = -26.4°C
Δt = t f - t i
-26.4°C = tf – 30.0°C
tf = 3.6°C
(d) t°F = 1.8t°C + 32
ti = 30.0°C
t°F = 1.8(30.0°C) + 32
ti = 86.0°F
tf = 3.6°C
t°F = 1.8(3.6°C) + 32
tf = 38.5°F
16. Refer to Section 8.3.
Calculate qreaction for 5.00 g caffeine using 1 mol. caffeine per 4.96 x 103 kJ as a conversion
factor. Then use qreaction to calculate Ccal.
Since heat is evolved, q is negative. qreaction = -4.96 x 103 kJ
5.00 g caffeine x
1 mol. caffeine
- 4.96 x 10 3 kJ
= - 128 kJ
x
194.20 g caffeine 1 mol. caffeine
qreaction = -Ccal x Δt
-128 kJ = -Ccal x 11.37°C
Ccal = 11.2 kJ/°C = 1.12 x 104 J/°C
15
28. Refer to Sections 8.1 - 8.4.
Calculate the amount of heat needed to raise the temperature of the water 3°C. Recall that the
heat absorbed by the water will be equal to the heat released by the fat. Finally, use the
enthalpy of the reaction to calculate the amount of fat needed to produce the required heat.
C57H104O6(s) + 80O2(g) → 57CO2(g) + 52H2O(l)
100.0 mL H2O = 100.0 g H2O
ΔH = -3.022 x 104 kJ/mol.
qH2O = (100.0 g)(4.18 J/g°C)(25.00°C – 22.00°C) = 1.25 x 103 J
qfat = -qH2O = -1.25 x 103 J
− 1.25 ×103 J x
1 kJ
1 mol. fat
885.4 g fat
x
x
= 3.67 x 10-2 g fat
4
1000 J - 3.022 x 10 kJ 1 mol. fat
16
Chapter 9: Liquids and Solids
8. Refer to Section 9.1 and Example 9.2.
⎛ P ⎞ + Δ H vap
(a) ln⎜⎜ 2 ⎟⎟ =
R
⎝ P1 ⎠
⎛1
1
⎜⎜ −
⎝ T1 T2
⎞
⎟⎟
⎠
+ Δ H vap ⎛ 1
1 ⎞
⎛ 760 atm ⎞
−
ln⎜
⎟=
⎜
⎟
⎝ 203 atm ⎠ 8.31 J/mol. ⋅ K ⎝ 308 K 337.8 K ⎠
1.32 =
+ Δ H vap
8.31 J/mol. ⋅ K
(2.86 x 10
-4
K -1 )
ΔHvap = 3.84 x 104 J/mol.
ΔHvap = 38.4 kJ/mol.
⎛ P ⎞ + Δ H vap
(b) ln⎜⎜ 2 ⎟⎟ =
R
⎝ P1 ⎠
⎛1
1
⎜⎜ −
⎝ T1 T2
⎞
⎟⎟
⎠
⎛ 760 atm ⎞ 3.84 × 10 4 J/mol. ⎛
1
1 ⎞
⎟⎟ =
−
ln⎜⎜
⎜
⎟
8.31 J/mol. ⋅ K ⎝ 313.2 K 337.8 K ⎠
⎝ P1 ⎠
⎛ 760 atm ⎞
⎟⎟ = 4.62 × 103 K (2.32 × 10 − 4 K -1 )
ln⎜⎜
P
1
⎝
⎠
ln(760 atm) – lnP1 = 1.07
5.56 = lnP1
P1 = 260 mm Hg
28. Refer to Section 9.3, Examples 9.4 and 9.6, and Chapter 7.
All molecules have dispersion forces. One can consider each molecule to be roughly
spherical. Those that have an even distribution of charge over the surface (b and c) will not
have a dipole, while those that have an uneven distribution (a and d) will have a dipole.
(a) PH3:
Dispersion and dipole. The electrons are not symmetrically distributed, so
this molecule has a dipole. Each H bears a slight negative charge and the P
bears a slight positive charge. Recall from Chapter 7 that PH3 is trigonal
pyramid in shape.
(b) N2:
Dispersion. The electrons are symmetrically distributed, so this molecule does
not have a dipole. Each N has identical, zero, charge.
17
(c) CH4:
Dispersion. The electrons are symmetrically distributed, so this molecule does
not have a dipole. Each H bears an identical, slight positive charge, thus, the
surface of the "sphere" has an even distribution of charge.
(d) H2O:
Dispersion and dipole. The electrons are not symmetrically distributed, so
this molecule has a dipole. The O bears a slight negative charge, while the H's
each bear a slight positive charge. Recall from Chapter 7 that water is bent.
48. Refer to Section 9.4.
(a) Graphite is a network covalent compound, so the C atoms are held together in an
extended network of covalent bonds.
(b) Silicon carbide is a network covalent compound, so the Si and C atoms are held together
in an extended network of covalent bonds.
(c) FeCl2 is an ionic solid. The structural units are composed of individual Fe2+ and Cl- ions.
(d) Acetylene (C2H2) is a molecular solid, composed of individual C2H2 molecules held
together by dispersion forces.
18
Chapter 10: Solutions
10. Refer to Section 10.1 and Examples 10.4 and 10.5.
Molality
Mass Percent
of Solvent
Ppm Solute
Mole Fraction
of Solvent
2.577
86.58%
1.340 x 105
0.9556
(b)
20.4
45.0
5.50 x 10
(c)
0.07977
99.5232%
4768
(a)
(d)
12.6
4.30 x 10
57.0%
5
0.731
0.9986
5
0.815
Since these are aqueous solutions, the solvent is water. The solute is urea, CO(NH2)2.
Molar mass of urea: (12.01) + 16.00 + 2(14.01 + 2(1.008)) = 60.06 g/mol.
(a) 2.577 m indicates that there are 2.577 mol. of solute per 1000 g of solvent and the
calculations are based on that ratio.
2.577 mol. x
1 kg x
60.06 g urea
= 154.8 g urea
1 mol. urea
1000 g 1 mol. H 2 O
x
= 55.49 mol. H 2 O
1 kg
18.02 g H 2 O
mass % =
1000 g
1000 g
x 100% =
= 86.58%
(1000 g + 154.8 g)
1155 g
ppm solute =
X H 2O =
154.8 g
x 10 6 = 1.340 x 10 5
(1000 g + 154.8 g)
55.49 mol.
= 0.9556
(55.49 mol. + 2.577 mol.)
(b) 45.0 mass % of solvent means that 45.0 g of solvent are present for each 100.0 g of
solution. Calculations are then based on this ratio.
45.0 g x
1 mol. H 2 O
= 2.50 mol. H 2 O
18.02 g
mass of solute = 100.0 g (total) – 45.0 g (water) = 55.0 g CO(NH2)2
55.0 g urea x
1 mol. urea
= 0.916 mol. urea
60.06 g urea
19
molality =
0.916 mol. urea
= 20.4 m (remember to convert mass of solvent to kg)
0.0450 kg H 2 O
ppm solute =
X H 2O =
(c) ppm =
55.0 g urea
x 106 = 5.50 x 105
100 g
2.50 mol.
= 0.731
(2.50 mol. + 0.916 mol.)
mass of solute
x 10 6 = 4768 g
mass of solution
If we assume a total mass of 106 g, then the mass of solute = 4768 g by definition. Any
assumption for mass is valid here, this one was chosen for simplicity.
4768 g urea x
1 mol. urea
= 79.39 mol. urea
60.06 g urea
10 6 g - 4768 g = 995232 g; 995232 g H 2 O x
X H 2O =
1 mol.
= 55244.0 mol. H 2 O
18.0152 g
55244.0 mol.
= 0.9986
(55244.0 mol. + 79.39 mol.)
mass % =
9.95232 x 10 5 g
x 100% = 99.5232%
10 6 g
molality =
79.39 mol.
= 0.07977 m
995.232 kg
(d) Xsolvent = 0.815 indicates that there are 0.815 moles H2O per 1 mole of solvent and solute
combined. Consequently, there must be 0.185 mol. (1-0.815) urea.
0.185 mol. urea x
60.06 g urea
= 11.1 g urea
1 mol. urea
0.815 mol. H 2 O x
18.02 g
= 14.7 g H 2 O
1 mol.
0.185 mol. urea
= 12.6 m (remember to convert mass of solvent to kg)
0.0147 kg H 2 O
Mass % =
mass of solvent
14.7 g
x 100% =
x 100% = 57.0%
total mass
(11.1 g + 14.7 g)
20
ppm =
mass of solute
11.1 g urea
x 10 6 =
x 10 6 = 4.30 x 10 5
mass of solution
(11.1 g + 14.7 g)
26. Refer to Section 10.2 and Chapter 8.
(a) For the solution process, NaOH(s) → Na+(aq) + OH-(aq)
ΔH = [(1 mol.)(-240.1kJ/mol.) + (1 mol.)(-230.0kJ/mol.)] – [(1 mol.)(-425.6kJ/mol.)]
ΔH = -44.5kJ
(b) Since the solution process is exothermic (ΔH < 0), an increase in temperature will
decrease the solubility of NaOH.
52. Refer to Section 10.3 and Example 10.8.
4.60 mm Hg x
1 atm
= 6.05 x 10 -3 atm
760 mm Hg
π = MRT
6.05 x 10-3 atm = (M)(0.0821 L⋅atm/mol.⋅K)(293 K)
M = 2.52 x 10-4 mol./L
0.200 L x
2.52 x 10 -4 mol.
= 5.03 x 10 -5 mol.
1L
3.27 g hemoglobin
= 6.50 x 10 4 g/mol.
-5
5.03 x 10 mol.
21
Chapter 11: Rate of Reaction
6. Refer to Section 11.1.
rate =
(a)
- Δ[Br - ] - Δ[BrO3- ] - Δ[H + ] Δ[Br2 ] Δ[H 2 O]
=
=
=
=
5Δ t
6Δ t
3Δ t
3Δ t
Δt
Δ[Br2 ]
Δ[Br2 ] 0.039 mol./L ⋅ s
= 0.039 mol./L ⋅ s ⇒
=
Δt
3Δ t
3
Δ[Br2 ] Δ[H 2 O] Δ[H 2 O] 0.039 mol./L ⋅ s
=
⇒
=
3Δ t
3Δ t
3Δ t
3
Δ[H 2 O] 3(0.039 mol./L ⋅ s)
=
= 0.039 mol./L ⋅ s
Δt
3
- Δ[Br - ] Δ[Br2 ]
=
(b)
5Δ t
3Δ t
- Δ[Br - ] 0.039 mol./L ⋅ s
=
5Δ t
3
- Δ[Br - ] 5(0.039 mol./L ⋅ s)
=
= 0.065 mol./L ⋅ s
3
Δt
(c)
- Δ[H + ] Δ[Br2 ] - Δ[H + ] 0.039 mol./L ⋅ s
=
⇒
=
6Δ t
3Δ t
6Δ t
3
- Δ[H + ] 6(0.039 mol./L ⋅ s)
=
= 0.078 mol./L ⋅ s
3
Δt
16. Refer to Section 11.2.
(a) Zero order
(b) rate = k[Y]0
(c) rate = k[Y]0
22
0.045
M
= k(1)
hr
k = 0.045
M
hr
26. Refer to Section 11.1 and Example 11.2.
(a) To determine the reaction order in [BF3], select two experiments in which [NH3] is
constant (such as experiments 1 and 4, as used below).
rate1 k[BF3 ]1m [NH 3 ]1n
=
rate 4 k[BF3 ]m4 [NH 3 ]n4
⇒
rate1 ⎛ [BF3 ]1 ⎞
⎟
=⎜
rate 4 ⎜⎝ [BF3 ]4 ⎟⎠
m
⎛ [NH 3 ]1 ⎞
⎟⎟
⎜⎜
⎝ [NH 3 ]4 ⎠
0.0341 mol./L ⋅ s ⎛ 0.100 mol./L ⎞ ⎛ 0.100 mol./L ⎞
=⎜
⎟ ⎜
⎟
0.102 mol./L ⋅ s ⎝ 0.300 mol./L ⎠ ⎝ 0.100 mol./L ⎠
m
n
n
0.334 = (0.333)m(1)n
0.334 = (0.333)m
m = 1 (The reaction is 1st order in BF3.)
To determine the reaction order in [NH3], select two experiments in which [BF3] is
constant (such as experiments 2 and 3) and repeat the process above.
0.159 mol./L ⋅ s ⎛ 0.200 mol./L ⎞ ⎛ 0.233 mol./L ⎞
=⎜
⎟ ⎜
⎟
0.0512 mol./L ⋅ s ⎝ 0.200 mol./L ⎠ ⎝ 0.750 mol./L ⎠
m
n
0.311 = (1)m(0.311)n
0.311 = (0.311)n
n = 1 (The reaction is 1st order in NH3.)
Overall reaction order = m + n = 1 + 1 = 2
(b) Rate = k[BF3][NH3]
(c) Substitute values from any of the experiments into the rate equation and solve for k.
0.0341 M/s = k(0.100 M)(0.100 M)
0.0341 M/s = k(0.01 M2)
k = 3.41 s-1·M-1 = 3.41 L/mol.·s
(d) Rate = k[BF3][NH3]
Rate = (3.41 L/mol.·s)(0.553 M)(0.300 M)
Rate = 0.566 M/s
23
Chapter 12: Gaseous Chemical Equilibrium
4. Refer to Section 12.1.
Time
(min)
PA (atm)
PB (atm)
PC (atm)
0
1
2
3
4
5
6
1.000
0.400
0.000
0.778
0.326
0.148
0.580
0.260
0.280
0.415
0.205
0.390
0.355
0.185
0.430
0.325
0.175
0.450
0.325
0.175
0.450
1 min: 1.000 atm A – 0.788 atm A = 0.222 atm A
0.222 atm A ×
1B
= 0.0740 atm B
3A
0.400 atm B – 0.0740 atm B = 0.326 atm B
0.222 atm A ×
2C
= 0.148 atm C
3A
2 min: 0.400 atm B – 0.260 atm B = 0.140 atm B
0.140 atm B ×
3A
= 0.420 atm A
1B
1.000 atm A – 0.420 atm A = 0.580 atm A
0.140 atm B ×
2C
= 0.280 atm C
1B
3 min: 0.390 atm C ×
1B
= 0.195 atm B
2C
0.400 atm B – 0.195 atm B = 0.205 atm B
0.390 atm C ×
3A
= 0.585 atm A
2C
1.000 atm A – 0.585 atm A = 0.415 atm A
4 min: 0.400 atm B – 0.185 atm B = 0.215 atm B
0.215 atm B ×
3A
= 0.645 atm A
1B
1.000 atm A – 0.645 atm A = 0.355 atm A
0.215 atm B ×
2C
= 0.430 atm C
1B
24
5 min: 1.000 atm A – 0.325 atm A = 0.675 atm A
0.675 atm A x
1B
= 0.225 atm B
3A
0.400 atm B – 0.225 atm B = 0.175 atm B
0.675 atm A x
2C
= 0.450 atm C
3A
6 min: Since PB has not changed from 5 min, PA and PC will also equal the values from 5
min.
6. Refer to Section 12.2 and Examples 12.1 and 12.2.
Recall that liquids and solids do not show up in equilibrium expressions.
(a) K =
( PH 2 ) 3 ( PCO )
( PCH 4 )
(b) K =
(d) K =
(c) K = PCO 2
( PNO ) 4 ( PH 2O ) 6
( PNH3 ) 4 ( PO 2 ) 5
1
( PNH 3 )( PHCl )
14. Refer to Section 12.2.
(a) K1 =
[O 2 ][SO 2 ] 2
[SO 3 ] 2
1
2
(b) K2 =
[O 2 ] [SO 2 ]
[SO 3 ]
(c) K2 = K1½
24. Refer to Section 12.2 and Chapter 5.
(a) CH4(g) + 2H2S(g) ⇌ CS2(g) + 4H2(g)
(b)
Use the ideal gas law equation to calculate the pressure of each gas, then
substitute those values into the equation for the equilibrium constant.
PCH 4 =
n
0.00142 mol.
RT =
(0.0821 L ⋅ atm/mol. ⋅ K)(1123 K) = 0.131 atm
V
1L
25
PH 2 S =
n
6.14 x 10 -4 mol.
RT =
(0.0821 L ⋅ atm/mol. ⋅ K)(1123 K) = 0.0566 atm
V
1L
PCS2 =
0.00266 mol.
n
RT =
(0.0821 L ⋅ atm/mol. ⋅ K)(1123 K) = 0.245 atm
1L
V
PH 2 =
n
0.00943 mol.
RT =
(0.0821 L ⋅ atm/mol. ⋅ K)(1123 K) = 0.869 atm
V
1L
(PH 2 ) 4 (PCS 2 )
(0.869 atm) 4 (0.245 atm)
K=
=
= 333
(PCH 4 )(PH 2S ) 2 (0.131 atm)(0.0566 atm)2
38. Refer to Section 12.4.
Ptotal = PNO + PNO 2 + PO 2 ⇒ 1.25 = PNO + PNO 2 + 0.515
0.735 atm = PNO + PNO 2
let x = PNO; then 0.735 – x = PNO2
K=
( PNO ) 2 ( PO 2 )
( PNO 2 ) 2
1.69 =
1.3 =
( x) 2 (0.515)
⇒ 0.87 =
(0.735 - x) 2
( x) 2
(0.735 - x) 2
x
⇒ x = 0.42
0.735 − x
PNO 2 = 0.32 atm
PNO = 0.42 atm
26
Chapter 13: Acids and Bases
6. Refer to Section 13.1.
Write the equation for the dissociation of the species (HB) into H+ and B-. The resulting Bwill be the conjugate base of the acid HB.
(a) HCO3- → H+ + CO32(b) [Cu(H2O)(OH)3]- → H+ + [Cu(OH)4]2(c) HNO2 → H+ + NO2(d) (CH3)2NH2+ → H+ + (CH3)2NH
(e) H2SO3 → H+ + HSO338. Refer to Section 13.4.
(a) The stronger the acid, the greater the degree of ionization and thus the larger the Ka.
C<D<B<A
(b) The largest Ka corresponds to the smallest pKa.
A has the smallest pKa.
44. Refer to Section 13.4.
pH of 2.642 M HB is 5.32
[H+] = x = 10-pH = 4.79 x 10-6 M
Ka =
(4.79 x 10 −6 ) 2
[H + ][B−]
x2
=
=
= 1.78 x 10-10
[HB]
0.129 − x
0.129 − 4.79 x 10 −6
27
Chapter 14: Equilibria in Acid-Base Solutions
4. Refer to Chapters 4 and 13 and Problem 2 above.
Write the reactions, eliminating the spectator ions. Bear in mind that weak acids and bases
(Table 13.2) do not ionize significantly, thus they exist in solution as the undissociated acid
or base.
(a) NH4NO3(aq) + OH-(aq) ⇌ NH3(aq) + H2O(l) + NO3-(aq)
NH4+(aq) + NO3-(aq) + OH-(aq) ⇌ NH3(aq) + H2O(l) + NO3-(aq)
NH4+(aq) + OH-(aq) ⇌ NH3(aq) + H2O(l)
+
(b) NaH2PO4(aq) + OH-(aq) ⇌ Na (aq) + HPO42-(aq) + H2O(l)
+
+
Na (aq) + H2PO4-(aq) + OH-(aq) ⇌ Na (aq) + HPO42-(aq) + H2O(l)
H2PO4-(aq) + OH-(aq) ⇌ HPO42-(aq) + H2O(l)
(c) [Al(H2O)6]3+(aq) + OH-(aq) ⇌ [Al(H2O)5OH]2+(aq) + H2O(l)
26. Refer to Section 14.1, Example 14.2 and Table 13.2.
Calculate the moles of the acid and base, then calculate the [H+] and pH.
(a) Calculate [H+] from pH, then calculate the acid/base ratio of the buffer.
[H+] = 10-pH = 10-3.0 = 1 x 10-3 M
[H + ] = K a ×
[HB]
[HCHO 2 ]
⇒ 1 × 10-3 = 1.9 × 10 -4 ×
[B ]
[CHO 2 ]
[HCHO 2 ]
=5
[CHO 2 ]
(b) NaCHO2 will dissociate to Na+ and CHO2-. Use this fact and the equation from part (a) to
calculate [HCHO2].
[HCHO 2 ]
[HCHO 2 ]
=5⇒
=5
0.139
[CHO 2 ]
[HCHO2] = 0.7 M, thus 0.7 mol. would need to be added to the liter of solution.
(c)
[HCHO 2 ]
0.159
=5⇒
=5
[CHO 2 ]
[CHO 2 ]
[CHO2-] = 0.03 M
28
−
0.03 mol. CHO 2 1 mol. NaCHO2 68.01 g NaCHO2
0.3500 L ×
×
×
= 0.7 g NaCHO2
−
1L
1 mol. NaCHO2
1 mol. CHO 2
[HCHO 2 ]
[HCHO 2 ]
(d)
=5⇒
=5
0.500
[CHO 2 ]
[HCHO2] = 2 M, since the volume of the formate solution is one liter, 2 mol of HCHO2 is
needed.
2 mol. HCHO 2 ×
1L
= 1× 101 L
0.236 mol. HCHO 2
44. Refer to Section 14.3.
(a) MBVB = MAVA
(MA)(0.02500 L) = (0.117 M)(0.03974 L)
MA = 0.186 M
(b) At the equivalence point
H2O + CHO2- (aq) ⇌ HCHO2 (aq) + OH- (aq)
Kb=Kw/Ka=1x10-14/1.9x10-4=5.26x10-11
[OH ][HCHO2 ]
x
= 5.26x10−11 =
0.0783- x
[CHO2 ]
X=1.94x10-6
-
2
[HCHO2]=1.94x10-6
[CHO2-]=7.83x10-2
[OH-]=1.94x10-6
[K+]=7.83x10-2
(c) pH = 8.29
29
Chapter 15: Complex Ions
4. Refer to Section 15.1.
Put the metal and the ligands together and add up the charges to get the overall charge.
(a) Pt2+ + 2NH3 + C2O42- → [Pt(NH3)2(C2O4)]
(b) Pt2+ + 2NH3 + SCN- + Br- → [Pt(NH3)2(SCN)Br]
(c) Pt2+ + en + 2NO2- → [Pt(en)(NO2)2]
* en = ethylenediamine
22. Refer to Section 15.2, Example 15.3 and Figures 15.5 and 15.6.
(a) As in Figure 15.5, there are four of one ligand (SCN-) and two of another (NH3). There
are two possible geometric isomers (NH3 cis and NH3 trans).
SCN
SCN
H3N
Cr
Cr
SCN
H3N
SCN
H3N
NH3
NCS
SCN
SCN
SCN
cis
trans
(b) As in Figure 15.6, there are three of one ligand (NO2-), and three of another (NH3). There
are two possible geometric isomers.
NO2
H3N
Co
H3N
NO2
NO2
H3N
NO2
H3N
NH3
(c)
NO2
NH3
NO2
This complex has three NH3, two H2O and one OH-.
OH
Mn
OH2
H3N
NH3
OH2
OH2
OH2
H3N
Co
H3N
OH
H3N
NH3
H3N
Mn
H3N
Mn
OH2
NH3
OH
OH2
36. Refer to Section 15.3 and Problem 34.
For weak field ligands, Δ0 is small, resulting in high spin complexes.
30
(a)
Rh3+: [Kr]4d6
(b)
Mn3+: [Ar]3d4
(c)
Ag+: [Kr]4d10
(d)
Pt4+: [Xe]5d6
(e)
Au3+: [Xe]5d8
E
4 unpaired electrons
E
4 unpaired electrons
E
0 unpaired electrons
E
4 unpaired electrons
E
2 unpaired electrons
52. Refer to Chapter 15.
(a) Fals(e) en occupies two coordination sites. Thus, the coordination number is 6.
(b) Fals(e) [Ni(CN)6]4- is a low spin complex with a large Δo, thus it absorbs at a short
wavelength. [Ni(NH3)6]2+ is a high spin complex with a small Δo, thus it absorbs
at a long wavelength.
(c) Tru(e) Cr3+ has a partially filled d orbital. Thus there will be crystal field splitting
energy and thus its complexes will be colored. Zn2+ has a d10 configuration.
Thus there is no crystal field splitting energy and its complexes are not colored.
(d) Fals(e) Ions with eight or more d electrons cannot form both high- and low-spin
octahedral complexes.
31
Chapter 16: Precipitation Equilibria
10. Refer to Section 16.1.
(a)
(b)
(c)
Compound
BaCO2O4
Cr(OH)3
Pb3(PO4)2
[cation]
1.3 x 10-3
2.7 x 10-8
1 x 10-8
[anion]
1.3 x 10-3
2.9 x 10-8
8 x 10-6
Ksp
1.6 x 10-6
6.3 x 10-31
1 x 10-34
30. Refer to Section 16.2.
0.00865 g
100.0 mL
⎛ 1 mol ⎞
⎜⎜
⎟⎟
⎝ 86.0 g ⎠
⎛ 1000 mL ⎞
⎜⎜
⎟⎟ = 1.000x10-3 M=[Cr2+]
1
L
⎝
⎠
pH = 8.5 so pOH = 5.5 or [OH-]=1x10-5.5
Ksp = [Cr2+] [OH-]2 =1.00x10-14
40. Refer to Section 16.2, Table 16.2, and Example 16.9.
(a) Cu(OH)2(s) + 4NH3(aq) → Cu(NH3)42+(aq) + 2OH-(aq)
(b) Cd2+(aq) + 4NH3(aq) → Cd(NH3)42+(aq)
(c) The lead must be combining with an anion to give an electrically neutral precipitate. A
likely candidate is OH- formed by the reaction of ammonia and water.
Pb2+(aq) + 2NH3(aq) + 2H2O(l) → Pb(OH)2(s) + 2NH4+(aq)
32
Chapter 17: Spontaneity of Reaction
8. Refer to Section 17.2.
(a) O2 (g) at 758 mm Hg
(b) glucose (s)
(c) Hg (l)
30. Refer to Section 17.4 and Example 17.5.
Determine ΔH fD from the table in Appendix 1. Write the equation for the formation of the
compounds from their elements. Calculate ΔS° and then ΔGfD .
3
(a) N2(g) + 2H2(g) + 2 O2(g) → NH4NO3(s)
ΔHreaction = ΔH fD = -365.6 kJ
ΔS° = Σ S°products - Σ S°reactants
ΔS° = [(1 mol.)(0.1511 kJ/mol.·K)] - [(1 mol.)(0.1915 kJ/mol.·K)
3
+ (2 mol.)(0.1306 kJ/mol.·K) + ( 2 mol.)(0.2050 kJ/mol.·K)]
ΔS° = -0.6091 kJ/K
ΔGreaction = ΔGfD = ΔH° - TΔS°
ΔGfD = -365.6 kJ - (298 K)(-0.6091 kJ/K) = -184.1 kJ
Note that ΔGfD is defined for 1 mole, thus ΔGfD = -184.1 kJ/mol.
(b) C(s) +
1
2
H2(g) +
3
2
Cl2(g) → CHCl3(l)
ΔHreaction = ΔH fD = -134.5 kJ
ΔS° = Σ S°products - Σ S°reactants
ΔS° = [(1 mol.)(0.2017 kJ/mol.·K)] - [(1 mol.)(0.0057 kJ/mol.·K)
3
+ ( 12 mol.)(0.1306 kJ/mol.·K) + ( 2 mol.)(0.2230 kJ/mol.·K)]
ΔS° = -0.2038 kJ/K
ΔGreaction = ΔGfD = ΔH° - TΔS°
ΔGfD = -134.5 kJ - (298 K)(-0.2038 kJ/K)
ΔGfD = -73.8 kJ
33
Note that ΔGfD is defined for 1 mole, thus ΔGfD = -73.8 kJ/mol.
(c) K(s) +
1
2
Cl2(g) → KCl(s)
ΔHreaction = ΔH fD = -436.7 kJ
ΔS° = Σ S°(prod) - Σ S°(react.)
ΔS° = [(1 mol.)(0.0826 kJ/mol.·K)]
- [(1 mol.)(0.0642 kJ/mol.·K) + ( 12 mol.)(0.2230 kJ/mol.·K)]
ΔS° = -0.0931 kJ/K
ΔGreaction = ΔGfD = ΔH° - TΔS°
ΔGfD = -436.7 kJ - (298 K)(-0.0931 kJ/K)
ΔGfD = -409.0 kJ
Note that ΔGfD is defined for 1 mole, thus ΔGfD = -409.0 kJ/mol.
66. Refer to Section 17.5.
(a) ΔG° = -RT ln K
ΔG° = -(0.00831 kJ/mol.·K)(721 K)ln(50.0)
ΔG° = -23.4 kJ/mol.
(b) ΔG° = Σ ΔGf°products - Σ ΔGf°reactants
ΔG° = (2 mol.)(1.7 kJ/mol.) - [(1 mol.)(0.0 kJ/mol.) + (1 mol.)(19.4 kJ/mol.)]
ΔG° = -16 kJ/mol.
ΔG° = -RT ln K
-16 kJ = -(0.00831 kJ/mol.·K)(298 K)ln K
ln K = 6.46
K = 6.4 x 102
34
Chapter 18: Electrochemistry
4. Refer to Section 18.1 and Example 18.1.
(a) Sn(s) is oxidized at the anode to Sn2+(aq).
Ag+(aq) is reduced at the cathode to Ag(s).
e-
Sn
anode
Ag
cathode
anions cations
Sn2+
Ag+
(b) H2(g) is oxidized at the anode to H+(aq).
Hg22+(Hg2Cl2) is reduced at the cathode to Hg(l).
Pt
anode
Pt
cathode
e-
H2(g)
anions cations
H
+
ClHg, Hg2Cl2
(c) Pb(s) is oxidized at the anode to Pb2+(PbSO4(s)).
Pb4+(PbO2(s)) is reduced at the cathode to Pb2+(PbSO4(s)).
35
Pb
anode
PbO2
cathode
e-
anions cations
2+
Pb , PbSO4
2+
Pb , PbSO4
16. Refer to Section 18.1.
(a) Co3+ (aq)
(b) Mn (s)
(c) Co3+ (aq)
(d) Mn2+ (aq)
(e) Yes
(f) No
(g) Fe3+ (aq) and Co3+ (aq)
(h) Pb (s), Cd (s), and Mn (s)
74. Refer to Sections 18.5, 18.6, and Example 18.8.
Use the equation on page 502 for the reaction of a lead storage battery to determine the moles of
e- per mole of Pb.
(a) 12.0 lb Ca ×
453.6 g
1 mol.
2 mol. e −
×
×
= 272 mol. e −
1 lb
40.078 g Ca 1 mol. Ca
272 mol. e − ×
3.2 V =
9.648 ×10 4 C
= 2.62 ×107 C
−
1 mol. e
J
2.62 × 107 C
J = (3.2 V)(2.62 × 107 C) = 8.4 × 107 J
36
(b) $ 0.09 = 1 kWh = 3.600 x 106 J
8.4 × 107 J ×
1 kWh
$ 0.09
×
= $ 2.1
6
3.600 × 10 J 1 kWh
37
Chapter 19: Nuclear Chemistry
4. Refer to Section 19.1.
(a)
52
26 Fe
0
→ 52
25 Mn + 1 e
(b)
228
88 Ra
(c)
236
95
→ 228
89 Ac +
0 −1 e
4
Am → 232
93 Np + 2 He
26. Refer to Sections 19.1 and 19.2.
36
0
36
17Cl → -1e + 18Ar
(Note there is 1 β-particle per 1
36
17 Cl
decay.)
1 hr
1d
1 yr
2.3 x 10-6
x
x
x
= 4.4 x 10 -12 /min
365 d 24 hr 60 min
yr
1.00 mg Cl - 36 x
1g
1 mol. Cl - 36
x
= 2.78 x 10-5 mol. Cl - 36
1000 mg
36.0 g
2.78 x 10-5 mol. Cl - 36 x
1 mol. β - particles 6.022 x 10 23
x
= 1.67 x 1019 β - particles
36
1 mol.
1 mol. Cl
A = kN = (4.4 x 10-12 /min)(1.67 x 1019 β-particles) = 7.3 x 107 particles/min
7.3 × 107 particles 1 min
1 Ci
×
×
= 3.3 × 10-5 Ci
10
1 min
60 s 3.700 × 10 particles/s
38. Refer to Section 19.3 and Example 19.4.
(a)
90
38
Sr → 2 -10 e +
90
40
Zr
(See problem 4a)
(b) Δm = 2(0.00055 g/mol.) + 89.8824 g/mol. – 89.8869 g/mol. = -0.0034 g/mol.
(c) Calculate ΔE for one mole, then convert to kJ/g.
ΔE = 9.00 x 1010 kJ/g x Δm = (9.00 x 1010 kJ/g)(-0.0034 g/mol.) = -3.1 x 108 kJ/mol.
6.50 mg ×
1g
1 mol. Sr - 3.1 x 108 kJ
×
×
= - 2.2 x 10 4 kJ
1000 mg 89.8869 g
1 mol.
38
Chapter 20: Chemistry of the Metals
10. Refer to Section 20.1 and Chapter 18.
one metric ton = 1 x 103 kg
1.000 x 103 kg x
1000 g 1 mol. Zn
x
= 1.529 x 104 mol. Zn
65.39 g
1 kg
1.529 x 10 4 mol. Zn x
2 mol. e - 9.648 x 10 4 C
x
= 2.950 x 10 9 C
1 mol. Zn
1 mol. e
J = VC = (3.0 V)(2.950 x 109 C) = 8.9 x 109 J
8.9 x 109 J x
1 kWh
= 2.5 x 103 kWh
6
3.600 x 10 J
16. Refer to Section 20.2 and Table 20.1.
(a) Na2O2(s) + 2H2O(l) → 2Na+(aq) + 2OH-(aq) + H2O2(aq)
sodium ion, hydroxide ion, hydrogen peroxide.
(b) 2Ca(s) + O2(g) → 2CaO(s)
calcium oxide.
(c) Rb(s) + O2(g) → RbO2(s)
rubidium superoxide.
(d) SrH2(s) + 2H2O(l) → Sr2+(aq) + 2OH-(aq) + 2H2(g)
strontium ion, hydroxide ion, hydrogen gas.
24. Refer to Section 20.3 and Chapter 4.
(a) Fe0 → Fe3+ + 3e5+
-
Oxidation half-reaction
4+
N +e → N
2H+(aq) + NO3-(aq) + e- → NO2(g) + H2O(l)
Reduction half-reaction
Balancing electrons gives:
6H+(aq) + Fe(s) + 3NO3-(aq) → 3NO2(g) + Fe3+(aq) + 3H2O(l)
(b) Cr3+ → Cr6+ + 3eOxidation half-reaction
Cr(OH)3(s) → CrO42-(aq) + 3e5 OH-(aq) + Cr(OH)3(s) → CrO42-(aq) + 3e5 OH-(aq) + Cr(OH)3(s) → CrO42-(aq) + 3e- + 4H2O(l)
39
O0 + 2e- → O2O2(g) + 4e- → 2H2O(l)
O2(g) + 4e- → 2H2O(l) + 4 OH-(aq)
4H2O(l) + O2(g) + 4e- → 2H2O(l) + 4OH-(aq)
2H2O(l) + O2(g) + 4e- → 4 OH-(aq)
Reduction half-reaction
Balancing electrons gives:
20 OH-(aq) + 4Cr(OH)3(s) + 6H2O(l) + 3O2(g) + 12e→ 12 OH-(aq) + 4CrO42-(aq) + 12e- + 16H2O(l)
8 OH (aq) + 4Cr(OH)3(s) + 3O2(g) → 4CrO42-(aq) + 10H2O(l)
40
Chapter 21: Chemistry of the Nonmetals
20. Refer to Section 21.1 and Table 18.1.
(a) 2Br-(aq) → Br2(l) + 2e-
D
E ox
= -1.077 V
Cl2(g) + 2e- → 2Cl-(aq)
Cl2(g) + 2Br-(aq) → 2Cl-(aq) + Br2(l)
D
E red
= +1.360 V
E° is (+) thus reaction occurs.
(b) 2Cl-(aq) → Cl2(g) + 2e-
D
E ox
= -1.360 V
I2(s) + 2e- → 2I-(aq)
I2(s) + Cl-(aq) → N.R.
D
E red
= +0.534 V
E° is (-) thus no reaction occurs.
(c) 2Br-(aq) → Br2(l) + 2e-
D
E ox
= -1.077 V
I2(s) + 2e- → 2I-(aq)
I2(s) + Br-(aq) → N.R.
D
E red
= +0.534 V
E° is (-) thus no reaction occurs.
(d) 2Cl-(aq) → Cl2(g) + 2e-
D
E ox
= -1.360 V
Br2(l) + 2e- → 2Br-(aq)
Br2(l) + Cl-(aq) → N.R.
D
E red
= +1.077 V
E° is (-) thus no reaction occurs.
32. Refer to Section 21.4 and Chapters 7 and 13.
-
O
(a)
NO3-
O
N
O
-
O
(b)
HSO4-
H
O
S
This Lewis structure has a -1 formal
charge on the most electronegative
atom (O).
O
O
-
O
(c)
H2PO4-
H
O
P
O
H
O
41
This Lewis structure has a -1 formal
charge on the most electronegative
atom (O).
70. Refer to Sections 21.3 and 21.4.
(a) Increasing oxidation number corresponds to an increase in oxygens atoms around the
central atom. These oxygen atoms stabilize the negative charge of the resulting conjugate
base by i) delocalizing the negative charge (consider resonance structures) and ii) by
the electronegative oxygens pulling electron density away from the atom bearing the
charge.
(b) NO2 has an odd number of electrons and will thus have an unpaired electron.
(c) In general, the reduction reactions of the oxoanions (oxoanions acting as oxidizing
agents) involve H+ as a reactant. Therefore higher [H+] (lower pH) causes the reaction to
be more spontaneous.
(d) The sugar is oxidized to carbon, which is black.
42
Chapter 22: Organic Chemistry
12. Refer to Section 22.2.
The structures are printed in Appendix 6, p670. Thus, this space will be used to explain the
answer.
(a) “pent” = five carbon chain
“-yne” = triple bond
“2-” = triple bond starts at carbon 2
(b) “pent” = five carbon chain
“-yne” = triple bond
“2-” = triple bond starts at carbon 2
“4-methyl” = methyl group attached to carbon 4
(c) “hex” = six carbon chain
“-yne” = triple bond
“3-” = triple bond starts at carbon 3
“2-methyl” = methyl group attached to carbon 2
(d) “but” = four carbon chain
“-yne” = triple bond
“1-” = triple bond starts at carbon 1
“3,3-dimethyl” = two methyl groups attached to carbon 3
Note the nomenclature priority system. The multiple bond has a lower number than the
functional groups.
32. Refer to Section 22.5 and Example 22.9.
This type of problem is best approached by systematically moving the C1 atoms, first
keeping two to a carbon, then moving them one at a time. Be aware that the isomer with both
C1’s on the 1st carbon is identical to that with both on the last carbon.
CI
│
CI2CH—CH2—CH3 CH3—CCI2—CH3
CICH2—CH—CH3
CICH 2—CH2—CH2CI
46. Refer to Section 22.5 and Example 22.11.
The chiral carbons (denoted with an *) are those that have four different groups attached.
43
44
Chapter 23: Organic Polymers, Natural and Synthetic
4. Refer to Section 23.1 and Example 23.2.
This problem is solved the same way as Example 23.2(a) by replacing CH3 with a phenyl
group (abbreviated as Ph)
14. Refer to Section 23.2 and Example 23.4.
Condensation polymers are made by removing OH from one monomer and H from the other.
To identify the monomer, locate the bond forming the polymer linkage and add OH to one
end of the bond and H to the other.
26. Refer to Table 23.3.
Calculate the possible combinations of the three amino acids. This can be done mathematically
or by inspection.
Thus, there are 6 possible tripeptides.
45