5.13 see textbook p606.
ANS
1.00 atm = 760. mmHg x 13.6 = 10336 mmH2O = 10.3 mH2O
1.5 km = 1.5 x 103 m, then (1.5 x 103 m)/(10.3 mH2O) = 145.6 atm = 150 atm = 15 ×
10 atm.
ANS
PV = nRT
(89 x 103/101325 atm) × (158 L) = n × (0.08206 atm×L/mol×K) × (273 + 14) K
n = 5.893 mol = 5.9 mol
PV = nRT
(a) 1.0 atm × 1.4 L = n × (0.08206 atm×L/mol×K) x 273 K
n = 0.0625 mol = 0.063 mol
mass = 0.063 mol × 64.0g/mol = 4.0 g.
(b) 1.0 atm × 3.5 x 105 L = n × (0.08206 atm×L/mol×K) x 273 K
n = 15623 = 1.6 x 104 mol
mass = 1.6 x 104 mol × 44.0g/mol = 704000 g = 7.0 x 102 kg
5.35 see textbook, page 607.
ANS
(800.0/760.0 atm) × V = n ×(0.08206 atm×L/mol×K) ×(287 K), V/n = 23.153 L/mol =
23.1 L/mol
T = PV/nR = (35/14.7) atm ×(23.15 mol/L)/(0.08206 atm×L/mol×K) = 670.2 K = 670
K
5.47 see textbook, page 607.
ANS
PH2O = (14.0/18.0) × (0.08206 atm×L/mol×K) × 295.5 K/ (3.00L) = 6.287 atm =
6.29 atm
PO2 = (11.5/32.0) × (0.08206 atm×L/mol×K) × 295.5 K/ (3.00L) = 2.905 atm = 2.91
atm.
PN2 = (37.3/28.0) × (0.08206 atm×L/mol×K) × 295.5 K/ (3.00L) = 10.768 atm =
10.8 atm
PTotal = 6.29 + 2.91 + 10.8 = 20.00 atm = 20.0 atm
nO2 = (620x103g)/(97.4g/mol)×(3/2) = 9548 mol = 9550 mol = 955 × 10 mol
V = nRT/P = (9550 mol)×(0.08206 atm×L/mol×K)×(307.0 K)/(0.977 atm) = 246251 L
=
2.46 x 105 L.
5Mg + N2 + O2 Æ Mg3N2 + 2MgO
Mg = 3.11/24.3 = 0.1279 mol = 0.128 mol
Mass of Mg3N2 = 0.128 × (100.9)/5 = 2.583 g = 2.58 g.
Mass of MgO = 0.128 × (40.3) × 2/5 = 2.063 g = 2.06 g.
5.65
ANS
Ni(s) + 4 CO(g) Æ Ni(CO)4(g)
Before reaction, nCO = PV/RT = (748/760 atm)×(2.00 L)/((0.08206 atm×L/mol×K) ×
(350.0 K) = 0.06854 mol = 0.0685 mol.
Ni = 5.00/58.69 = 0.08519 = 0.0852 mol, then CO is limiting reagent.
The final pressure is from Ni(CO)4, whose n = 0.0685/4 = 0.0171 mol.
PNi(CO)4 = nNi(CO)4RT/V = (0.0171 mol)×(0.08206 atm×L/mol×K)×(350.0 K)/(2.00L) =
0.2456 atm = 0.246 atm (or 187 torr).
5.74
ANS
V = (2RT/M)0.5, thus heavier gas moves more slowly than lighter gas.
5.78
ANS
(a)
PV = nRT,
P = nRT/V
= (11.4g/64.0g/mol)×(0.08206 atm×L/mol×K)×(298.0K)/(24.0 x 10-3 L) = 181.4 atm =
181 atm.
(b)
(P +a×(n/V)2)×(V-nb) = nRT
P = nRT/(V-nb) - a×(n/V)2
= (11.4g/64.0g/mol)×(0.08206 atm×L/mol×K)×(298.0K)/ (24.0 x 10-3 L – 0.05636×
11.4/64.0)-6.714×[(11.4/64.0)/(24.0 x 10-3 L)]2 = -57.83 atm = -57.8 atm.
The van der Waals equation underestimates the pressure of SO2 in this condition.
5.86
ANS
150 L of liquid O2 = (150 L)×(1.14 g/mL)×(1000 mL/L)/(32.0 g/mol) = 5344 mol =
5340 mol.
V = nRT/P = (5340 mol)/(0.21)×(0.08206 atm×L/mol×K)×(298 K)/(750/760 atm) =
6.3 x 105 L.
ANS
PoriginVorigin = noriginRT = (33.7/28.0 mol) × (0.08206 atm×L/mol×K) × (277.4 K) =
27.39 atm×L = 27.4 atm×L
Then, 27.4 atm×L = PnewVnew = nnewRT = (33.7/28.0 mol)×(100.0-30.1)/(100.0%)×
(0.08206 atm×L/mol×K) × T
Thus T = 397.1 K = 397 K ( or 123 ℃).