# Answers to Chapter 05 Study Questions

1. STP:  $$\mathrm{P_T = 1\: atm}$$; $$\mathrm{T = 273\: K}$$;  $$\mathrm{P_{O_2}=\left(\dfrac{n_{O_2}}{n_T}\right)P_T}$$;  $$\mathrm{n_T = 0.039 + 0.010 + 0.001 = 0.050\: moles}$$
$$\mathrm{P_{O_2}=\left(\dfrac{n_{O_2}}{n_T}\right)P_T}$$; $$\mathrm{P_{O_2}=\left(\dfrac{0.010}{0.050}\right)(1.00\:atm)=0.20\: atm}$$
2. $$\mathrm{V = ?}$$;  STP:  $$\mathrm{T = 273\: K}$$, $$\mathrm{P_T = 1\: atm}$$;  $$\mathrm{n_T = 0.050\: mol}$$;  $$\mathrm{PV = nRT}$$
$$\mathrm{V=\dfrac{nRT}{P}=\dfrac{(0.050\:mol)(0.08206)(273\:K)}{1\:atm}=1.1\:L}$$

1. $$\mathrm{P_T = P_{H_2} + P_{H_2O}}$$; Find $$\mathrm{P_{H_2O}}$$ in Table from lab report; at 19°C, $$\mathrm{P_{H_2O} = 16\: mmHg}$$
$$\mathrm{P_{H_2} = P_T - P_{H_2O} = 756 - 16 = 740.\: mmHg}$$
2. $$\mathrm{740\: mmHg\times\dfrac{1\:atm}{760\:mmHg}=0.974\: atm}$$

1. $$\mathrm{V_1 = 600.\: cm^3}$$;  $$\mathrm{T_1 = 25^\circ C = 298\: K}$$;  $$\mathrm{P_1 = 750.\: mmHg}$$

$$\mathrm{V_2 = 480.\: cm^3}$$;  $$\mathrm{T_2 = 41^\circ C = 314\: K}$$;  $$\mathrm{P_2 = ?}$$

$$\mathrm{P_2=P_1\times\dfrac{V_1}{V_2}\times\dfrac{T_2}{T_1}=750\:mmHg\times\dfrac{600\:cm^3}{480\:cm^3}\times\dfrac{314\:K}{298\:K}=988\: mmHg}$$

1. $$\mathrm{density =\dfrac{molar\:mass}{molar\:volume}=\dfrac{4.00\:g}{22.4\:L}=0.178\: g/L}$$

1. $$\mathrm{2C_4H_{10}(g) + 13O_2(g) \rightarrow 10H_2O(l) + 8CO_2(g)}$$
2. $$\mathrm{2.0\: L\: CO_2\times\dfrac{13\:L\:O_2}{8\:L\:CO_2}=3.2\: L\: O_2}$$
3. $$\mathrm{11.6\: g\: C_4H_{10}\times\dfrac{1\:mol\:C_4H_{10}}{58.0\:g\:C_4H_{10}}\times\dfrac{8\:mol\:CO_2}{2\:mol\:C_4H_{10}}\times\dfrac{22.4\:L\:CO_2}{1\:mol\:CO_2}=17.9\: L\: CO_2}$$
4. $$\mathrm{5.6\: L\: C_4H_{10}\times\dfrac{1\:mol\:C_4H_{10}}{22.4\:L}\times\dfrac{10\:mol\:H_2O}{2\:mol\:C_4H_{10}}\times\dfrac{6.02\times10^{23}\:molecules}{1\:mol\:H_2O}=7.5\times10^{23}\:molecules}$$

1. $$\mathrm{n = 1\: mole}$$;  $$\mathrm{T = 68^\circ C = 341\: K}$$;  $$\mathrm{P = 2.00\: atm}$$;  $$\mathrm{V = ?}$$

$$\mathrm{V=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(341\:K)}{2.00\:atm}=14.0\: L}$$

1. $$\mathrm{8.00\:g\:CH_4\times\dfrac{1\:mol\:CH_4}{16.0\:g\:CH_4}\times\dfrac{22.4\:L}{1\:mol}=11.2\: L}$$

1. $$\mathrm{d=\dfrac{mm}{mV}}$$; $$\mathrm{mV=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(546\:K)}{4.00\:atm}=11.2\:L}$$;  $$\mathrm{d=\dfrac{44.0\:g}{11.2\:L}=3.93\: g/L}$$

1. Find molar volume at 710 mmHg and 36°C and then use conversion factors:

T = 36 + 273 = 309 K;  $$\mathrm{P = 710\: mmHg\times\dfrac{1\:atm}{760\:mmHg}=0.934\:atm}$$

$$\mathrm{V=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(309\:K)}{0.934\:atm}=27.1\: L}$$

$$\mathrm{6.52\: g\: Cu S\times\dfrac{1\:mol\:CuS}{95.6\:g\:CuS}\times\dfrac{2\:mol\:O_2}{1\:mol\:CuS}\times\dfrac{27.1\:L\:O_2}{1\:mol\:O_2}=3.70\: L\: O_2}$$

1. $$\mathrm{molar\: mass =\dfrac{mass}{moles}}$$;  so use $$\mathrm{PV = nRT}$$ to calculate $$\mathrm{n}$$;  $$\mathrm{T = 29 + 273 = 302\: K}$$;  $$\mathrm{P = 1\: atm}$$

$$\mathrm{n=\dfrac{PV}{RT}=\dfrac{(1\:atm)(6.20\:L)}{(0.08206)(302\:K)}=0.250\:moles}$$;  $$\mathrm{molar\:mass=\dfrac{7.00\:g}{0.250\:mol}=28.0\: g/mole}$$

1. $$\mathrm{15.0\: g\: CO_2\times\dfrac{1\:mol\:CO_2}{44.0\:g\:CO_2}=0.341\: mol\: CO_2}$$;   $$\mathrm{12.0\:g\:CH_4\times\dfrac{1\:mol\:CH_4}{16.0\:g\:CH_4}=0.750\: mole\: CH_4}$$;

At constant T and P, $$\mathrm{\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2}}$$;   $$\mathrm{V_2=V_1\times\dfrac{n_2}{n_1}=7.16\:L\times\dfrac{0.750\:mol}{0.341\:mol}=15.7\:L}$$

1. $$\mathrm{P_{H_2}= P_T - P_{H_2O}}$$; At 22°C, $$\mathrm{P_{H_2O} = 20\: mm Hg}$$; $$\mathrm{P_{H_2} = 750\: mmHg - 20\: mmHg = 730\: mmHg}$$

$$\mathrm{P_{H_2} = 730\: mmHg \times\dfrac{1\:atm}{760\:mmHg} = 0.960\: atm}$$;  T = 22 + 273 = 295 K;

$$\mathrm{7.78\: g\: Zn \times\dfrac{1\:mol\:Zn}{65.4\:g\:Zn}\times\dfrac{1\:mol\:H_2}{1\:mol\:Zn}=0.119\: mol\: H_2 = n}$$

$$\mathrm{V=\dfrac{nRT}{P}=\dfrac{(0.119\:mol)(0.08206)(295\:K)}{0.960\:atm}=3.00\: L}$$