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Solutions Manual For Chemistry 4th Canadian Edition By John Olmsted, Gregory Williams, Robert Burk (All Chapters 1-22, 100% Original Verified, A+ Grade) All Chapters Arranged
Reverse: 22-1
This is The Original Solutions Manual For 4th Canadian Edition, All other Files in The Market are Fake/Old/Wrong Edition. 1 / 4
Chemistry, 4e Instructor Solutions Manual Chapter 22 Chapter 22 Environmental Chemistry Solutions to Problems in Chapter 22 22.1
10 11 1 Q P Q 5.15 10 kg 365days thus,P 2.8 10 kg y 66days 1year - t = ´
= = ´ = ´
t 22.2 10 12 1 1
4.39 10
0.00696
6.30 10
365 2.5
Q kg y P kg y d y d t - - ´ = = = ´ ´ =
22.3 From the data we can calculate the mass of COS in the atmosphere:
8 -1 9
Q P = τ Q = Pτ = 5.0×10 kg y (7.0 y) = 3.51×10 kg And because volume is an intensive property, the volume of COS is calculated using its total
mass and density:
9
- 3
-3 Q 3.51×10 kg V = = = 2.14×10 m ρ 1.64kg m 22.4 (a) Difluoromethane. C1H2F2Cl0, thus X=0, Y=3, Z=2, thus CFC-32 (b) Fluoromethane. C1H3F1Cl0, thus X=1, Y=4, Z=1, thus CFC-041 (c) 1,1,2,2-tetrachloro-1,2-difluoroethane. C2H0F2Cl4, thus CFC-112 22.5(a) 1,1,1,2-tetrafluoroethane. C2H2F4Cl0, thus CFC-134 (b) 1,1-dichloro-2,2,2-trifluoroethane. C2HF3Cl2, thus CFC-123 (c) Chlorofluoromethane. CH2FCl, thus CFC-031 22.6 (a) (b)(c)
© 2021 John Wiley and Sons Canada, Ltd. 22-1 2 / 4
Chemistry, 4e Instructor Solutions Manual Chapter 22 22.7 (a)
- 2
CH ClF OH CHClF H O · ·
+ ® +
(b) h
- 2 2
CF Cl CF Cl Cl · · u
¾¾® +
22.8
11 12 2
2 kgSO 1000g 1mol 1.2 10 1.9 10 mol SO y kg 64.1g
æ öæ ö
´ = ´ ç ÷ç ÷
è ø è ø
This will produce an equivalent amount of H2SO4, 1.9 × 10 12 mol.From Appendix E, the first hydrolysis of H2SO4 is “strong”, i.e. it fully hydrolyzes. The second is much smaller, so we ignore it. H2SO4(aq) therefore hydrolyzes according to: H2SO4(aq) + H2O(l) H3O + (aq) + HSO4 - (aq) Thus the moles of hydronium ion are also 1.9 × 10 12 mol.The volume of rainwater is 1.7 × 10 14 m 2 × 1 m = 1.7 × 10 14 m 3
(= 1.7 × 10
17 L).The hydronium concentration it therefore 12 5
- (aq)17
1.9 10 mol [H O ] 1.12 10 mol L
1.7 10 L
+ - ´
= = ´
´
5 10 3 (aq) 10 pH log [H O ] log (1.12 10 ) 4.95
- -
= - = - ´ =
22.9 If the initial pH is 4.00, and the final pH is 5.00, then the initial [H3O + (aq)] = 10 -4 mol L -1
and the final [H3O + (aq)] = 10 -5 mol L -1 . The amount of H3O + (aq) neutralized must therefore
be:
4 5 1 8 4
(10 10 mol L ) 5.23 10 L 4.71 10 mol
- - -
- ´ ´ = ´
The neutralization reaction is 2 H3O + (aq) + Ca(OH)2(s) Ca 2+ (aq) + 2 H2O(l). The required amount of lime is therefore 42 2
- (aq) 26
- 2
1molCa(OH) 74.1g Ca(OH) 1ton 4.71 10 mol H O 1.75tonsCa(OH) 2mol H O 1molCa(OH) 10 g + + ´ =
æ öæ öæ ö
ç ÷ç ÷ç ÷
è øè øè ø
22.10 We see from the reactions in Example 22-4 that producing 1 mol of nickel metal also produces 1 mol of SO2, which in turn will produce 1 mol of H2SO4, which in turn produces two moles of H3O + . The amount of H3O + produced per year is therefore © 2021 John Wiley and Sons Canada, Ltd. 22-2 3 / 4
Chemistry, 4e Instructor Solutions Manual Chapter 22 6 2
82 4 3 3
- 2 4
- mol H SO 2mol H O molH O
10 tons Ni 365d 1molNi 1mol SO10 g d ton 1y 58.69 g Ni 1molNi
1.24 10
- mol SO 1 mol H SO y
- +
´
´ = ´
æ öæ öæ öæ öæ ö
ç ÷ ç ÷ç ÷ç ÷ç ÷
è ø è øè øè øè ø
æ öæ ö
ç ÷ç ÷
è øè ø
The deposition rate of hydronium is therefore:
- 3
- 1 1
3 molH O
1.24 10
y 1.24 10 molH O ha y 100,000ha +
- - -
´ = ´
This is larger than the maximum tolerable, so the trees will likely suffer irreversible damage.
22.11 -6 -6 NO (g) (aq) (aq) H,NO NO
-5 -3 -1
-6 -3 -9 -1 p =1.0 ×10 (1.0bar)=10 bar = 0.10Pa NO NO [NO ]=K p =1.9 ×10 molm Pa (0.10Pa) =1.9 ×10 molm =1.9 ×10 molL ƒ 1
- +
- +
- (aq) 3 (aq)
3(aq) 2 3 (aq) 3 (aq)
a 3(aq) -9
2 -9 -8
-8 1/2 -4 +
- (aq)
-4 10
HNO +H O NO +H O
[NO ][H O ]
K =
[HNO ]
x(x) thus, 30.0 =
1.9×10
x = 30.0(1.9×10 )= 5.70 ×10 x = (5.70 ×10 ) = 2.39 ×10 (=[H O ]) thus,pH= -log [2.39×10 ]= 3.62 ƒ 22.12 © 2021 John Wiley and Sons Canada, Ltd. 22-3
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