Page 344 - Theory and Problems of BEGINNING CHEMISTRY

PRACTICE QUIZZES 333

Chapter 13
1. According to Graham’slaw,

v Ne MM Ar 39.9 amu
= = = 1.41
v Ar MM Ne 20.2 amu
The neon molecules are on average 1.41 times faster than those of argon.
2. Only (c) “The average kinetic energies of their molecules must be the same,” must be true. The average speeds and their
effusion rates depend on the molar masses (Graham’s law). Their pressures and numbers of moles (and thus molecules)
are governed by the ideal gas law.

Chapter 14
1. (a) +5 (b) −1 (a peroxide) (c) +6 (d) −1

−
+
2. 34 H + 5Cr 2 O 7 2− + 3Cl 2 −→ 6 ClO 3 + 10 Cr 3+ + 17 H 2 O

Chapter 15
1. Assume that there is 1.000 mol total of alcohol plus water. Then there is 0.150 mol of alcohol and 0.850 mol of water.

18.0gH O
2
0.850 mol H 2 O = 15.3g H O = 0.0153 kg H O
2
2
1 mol H 2 O
0.150 mol alcohol
= 9.80 m alcohol
0.0153 kg H O
2
2. Assume 100.0 g of solution total. Then there are 20.0 g CH 4 O and 80.0 g H 2 O.
1 mol CH 4 O

20.0gCH O = 0.625 mol CH 4 O
4
32.0gCH O
4
1 mol H 2 O

80.0gH O = 4.44 mol H 2 O
2
18.0gH O
2
0.625 mol methanol
X methane = = 0.123 X water = 0.877
5.07 mol total
Chapter 16
[CO] 2
1. K = (Note that solids are not included in the expression.)
[O 2 ]
2. All in mol/L: N 2 + 2O 2 −− 2NO 2
−−
Initial: 0.0700 0.100 0.000
Change: x/2 x x
Final: 0.0700 −x/2 0.100−x x
[NO 2 ] 2 x 2
K = = = 6.0 × 10 −6
[N 2 ][O 2 ] 2 (0.0700)(0.100) 2
2
x = 4.2 × 10 −9
x = 6.5 × 10 −5 mol/L = [NO 2 ]
[N 2 ] = 0.0700 mol/L

[O 2 ] = 0.100 mol/L