chem test

Chapter 12 Homework Study Notes

Problem 1: Partial Pressure of O2

• Given: O2 gas collected over water at 23°C, total pressure = 772 torr
• Vapor pressure of water at 23°C: 21.2 torr
• Calculation for partial pressure:
  $\text{Partial pressure of O2} = \text{Total pressure} - \text{Vapor pressure of water}$
  $\text{Partial pressure of O2} = 772 \text{ torr} - 21.2 \text{ torr} = 750.8 \text{ torr}$

Problem 2: Partial Pressure of CH4

• Given: CH4 gas collected over water at 29°C, total pressure = 0.750 atm
• Convert total pressure to torr:
  $0.750 \text{ atm} \times 760 \text{ torr/atm} = 570.0 \text{ torr}$
• Vapor pressure of water at 29°C: 30.0 torr
• Calculation for partial pressure:
  $\text{Partial pressure of CH4} = 570.0 \text{ torr} - 30.0 \text{ torr} = 540.0 \text{ torr}$

Problem 3: Total Pressure of a Gas Mixture

• Components of mixture:
  • H2 gas: 600.0 torr
  • N2 gas: 200.0 torr
  • O2 gas: 300.0 torr
• Total pressure calculation:
  $\text{Total pressure} = 600.0 \text{ torr} + 200.0 \text{ torr} + 300.0 \text{ torr} = 1100.0 \text{ torr}$

Problem 4: Gas Pressure Change

• Given: 775 mL of NO2 gas at STP.
• Volume change: 615 mL
• Temperature changes to: 25.0 °C = 298 K
• Using Ideal Gas Law:$\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$
  • At STP:
  • $P<em>1 = 1.0 \text{ atm}$, $V</em>1 = 775 \text{ mL}$, $T_1 = 273 \text{ K}$
  • Calculating new pressure P2:
    $\frac{(1.0 \text{ atm})(775 \text{ mL})}{273 \text{ K}} = \frac{P_2(615 \text{ mL})}{298 K}$
  • Solve for $P<em>2$: $P</em>2 = \frac{(1.0)(775)(298)}{(273)(615)} = 1.095 \text{ atm}$

Problem 5: New Temperature Calculation

• Initial conditions:
  • Volume = 21.5 L
  • Pressure = 725 mmHg
  • Temperature = 25°C
• New conditions:
  • Volume = 24.9 L
  • Pressure = 755 mmHg
• Using the combined gas law:
  $\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$
• Convert temperature to Kelvin:
  • Initial: 25°C = 298 K
• Computing new temperature T2:
  $\frac{(725 \text{ mmHg})(21.5 \text{ L})}{298 K} = \frac{(755 \text{ mmHg})(24.9 \text{ L})}{T_2}$
• Solving for T2 gives: $T_2 = 351.29 ext{ K}$ or $78.14 ext{ °C}$

Problem 6: Volume Calculation After Changes

• Initial conditions:
  • Volume = 348 mL
  • Pressure = 675 torr
  • Temperature = 13°C
• New conditions:
  • Pressure = 582 torr
  • Temperature = 11°C
• Convert temperatures to Kelvin:
  • Initial: 13°C = 286 K
  • New: 11°C = 284 K
• Using the combined gas law:
  $\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$
• Calculating new volume V2:$V<em>2 = \frac{V</em>1 \cdot P<em>1 \cdot T</em>2}{P<em>2 \cdot T</em>1}$
  • Put in the values:
    $V_2 = \frac{(348 \text{ mL}) \cdot (675 \text{ torr}) \cdot (284 K)}{(582 \text{ torr}) \cdot (286 K)} = 351.78 mL$

Problem 7: Temperature Calculation for Xe Gas

• Given: 25.25 moles of Xe, volume = 645 L, pressure = 732 torr
• Using Ideal Gas Law:
  $PV = nRT$
• Solve for temperature T:
  $T = \frac{PV}{nR}$
• Conversion of pressure to atm:
  $\frac{732 \text{ torr}}{760 \text{ torr/atm}} = 0.9645 \text{ atm}$
• Substituting values into the equation:
  $T = \frac{(0.9645 \text{ atm})(645 \text{ L})}{25.25 \text{ moles} imes 0.0821 \text{ L atm/(K mol)}} = 233.2 K$

Problem 8: Volume at STP for O2

• Given: 12.5 g of O2 gas
• Molar mass of O2: 32 g/mol
• Moles of O2:
  $\text{moles} = \frac{12.5 \text{ g}}{32 \text{ g/mol}} = 0.390625 \text{ mol}$
• At STP (standard temperature and pressure):
  • 1 mol of gas occupies 22.4 L
• Volume at STP:
  $0.390625 \text{ mol} \times 22.4 \text{ L/mol} = 8.75 L$

Problem 9: Grams of C3H8

• Given: 945 mL at STP
• At STP: 1 mol = 22.4 L
• Moles of C3H8:
  $\text{moles} = \frac{945 \text{ mL}}{22400 \text{ mL/mol}} = 0.04215 \text{ mol}$
• Molar mass of C3H8: 44.10 g/mol
• Calculation of grams:
  $\text{mass} = 0.04215 \text{ mol} \times 44.10 \text{ g/mol} = 1.86 g$

Problem 10: Mole Calculation for CO2

• Given: 10.5 L of CO2, pressure = 713 mmHg, temperature = 56.7°C
• Convert temperature to Kelvin:
  $56.7 + 273.15 = 329.85 K$
• Convert pressure to atm:
  $\frac{713 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.937 ext{ atm}$
• Using Ideal Gas Law:
  $PV = nRT$
• Calculation of moles n:
  $n = \frac{PV}{RT} \rightarrow n = \frac{(0.937 \text{ atm})(10.5 \text{ L})}{(0.0821 \text{ L atm/(K mol)})(329.85 K)} = 0.3887\text{ mol}$
• Using Avogadro's number for molecules:
  $\text{molecule count} = n \times 6.022 \times 10^{23} = 0.3887 \times 6.022 \times 10^{23} = 2.34 \times 10^{23} \text{ molecules}$

Problem 11: Molar Mass of a Gas

• Given: 2.50 g of gas occupies 0.875 L, pressure =