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The problem in the image involves a 9.00 L tank at 7.58 °C containing sulfur tetrafluoride (SF₄) and chlorine pentafluoride (ClF₅). The mass of sulfur tetrafluoride is 10.3 g, and the mass of chlorine pentafluoride is 18.3 g. The goal is to calculate the mole fractions, partial pressures of each gas, and the total pressure in the tank, assuming ideal gas behavior.

Let me walk you through the calculations:

### Given dataThe problem in the image involves a 9.00 L tank at 7.58 °C containing sulfur tetrafluoride (SF₄) and chlorine pentafluoride (ClF₅). The mass of sulfur tetrafluoride is 10.3 g, and the mass of chlorine pentafluoride is 18.3 g. The goal is to calculate the mole fractions, partial pressures of each gas, and the total pressure in the tank, assuming ideal gas behavior.

Let me walk you through the calculations:

### Given 
- Volume of the tank (V) = 9.00 L
- Temperature (T) = 7.58 °C = 280.73 K
- Mass of SF₄ = 10.3 g
- Mass of ClF₅ = 18.3 g
- Molar mass of SF₄ = 108.07 g/mol
- Molar mass of ClF₅ = 130.45 g/mol
- Ideal gas constant (R) = 0.0821 L·atm/(mol·K)

### 1. Moles of each gas
To calculate the moles of each gas, use the formula:
\[
n = \frac{\text{mass}}{\text{molar mass}}
\]

For sulfur tetrafluoride (SF₄):
\[
n_{\text{SF}_4} = \frac{10.3}{108.07} = 0.0953 \, \text{mol}
\]

For chlorine pentafluoride (ClF₅):
\[
n_{\text{ClF}_5} = \frac{18.3}{130.45} = 0.1403 \, \text{mol}
\]

### 2. Total moles of gas
\[
n_{\text{total}} = n_{\text{SF}_4} + n_{\text{ClF}_5} = 0.0953 + 0.1403 = 0.2356 \, \text{mol}
\]

### 3. Mole fraction of each gas
Mole fraction is calculated using:
\[
\text{Mole fraction of SF₄} = \frac{n_{\text{SF}_4}}{n_{\text{total}}} = \frac{0.0953}{0.2356} = 0.4046
\]
\[
\text{Mole fraction of ClF₅} = \frac{n_{\text{ClF}_5}}{n_{\text{total}}} = \frac{0.1403}{0.2356} = 0.5954
\]

### 4. Total pressure in the tank using the ideal gas law
\[
P_{\text{total}} = \frac{n_{\text{total}} \cdot R \cdot T}{V}
\]
Substitute the known values:
\[
P_{\text{total}} = \frac{0.2356 \cdot 0.0821 \cdot 280.73}{9.00} = 0.6056 \, \text{atm}
\]

### 5. Partial pressures of each gas
Partial pressure is calculated as:
\[
P_{\text{SF}_4} = \text{mole fraction of SF₄} \times P_{\text{total}} = 0.4046 \times 0.6056 = 0.245 \, \text{atm}
\]
\[
P_{\text{ClF}_5} = \text{mole fraction of ClF₅} \times P_{\text{total}} = 0.5954 \times 0.6056 = 0.360 \, \text{atm}
\]

### Results:
- Mole fraction of SF₄ = 0.4046
- Partial pressure of SF₄ = 0.245 atm
- Mole fraction of ClF₅ = 0.5954
- Partial pressure of ClF₅ = 0.360 atm
- Total pressure in the tank = 0.606 atm

These answers differ slightly from the image results due to rounding.

Would you like more details on any step, or have any questions?

### Related Questions:
1. How do mole fractions affect partial pressures in gas mixtures?
2. How is the ideal gas law adjusted for non-ideal gas behavior?
3. What would happen to the total pressure if the temperature increased?
4. Can you explain how Dalton’s Law of Partial Pressures applies here?
5. How can we use the van der Waals equation for real gases in this scenario?

### Tip:
Always check the units used for temperature in gas law calculations (Kelvin instead of Celsius).

A 9.00 L tank at 7.58 °C is filled with 10.3 g of sulfur tetrafluoride gas and 18.3 g of chlorine pentafluoride gas. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank.

Learn how to calculate the mole fraction and partial pressure of sulfur tetrafluoride and chlorine pentafluoride in a 9.00 L tank at 7.58 °C. This example covers key concepts such as the Ideal Gas Law, mole fractions, and Dalton's Law of Partial Pressures.

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