chem test

Chapter 12 Homework Study Notes

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}

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}

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}

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{P1V1}{T1} = \frac{P2V2}{T2}
- At STP:
- $P1 = 1.0 \text{ atm}$, $V1 = 775 \text{ mL}$, $T_1 = 273 \text{ K}$
- 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 $P2$: P2 = \frac{(1.0)(775)(298)}{(273)(615)} = 1.095 \text{ atm}

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{P1V1}{T1} = \frac{P2V2}{T2}
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}

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{P1V1}{T1} = \frac{P2V2}{T2}
Calculating new volume V2: V2 = \frac{V1 \cdot P1 \cdot T2}{P2 \cdot T1}
- 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

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

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

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

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}