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To solve this problem, we need to calculate the **mole fraction** and **partial pressure** of each gas in the mixture, and finally, the **total pressure** in the tank.

### Given:
- Volume of the tank, \( V = 9.00 \, \text{L} \)
- Temperature, \( T = 5.26 \, ^\circ\text{C} = 278.41 \, \text{K} \) (Convert to Kelvin by adding 273.15)
- Mass of chlorine pentafluoride, \( \text{ClF}_5 = 16.2 \, \text{g} \)
- Mass of sulfur tetrafluoride, \( \text{SF}_4 = 7.73 \, \text{g} \)
- Ideal gas constant, \( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)

### Step 1: Calculate moles of each gas
We need the molar masses of the two gases:
- Molar mass of chlorine pentafluoride (\( \text{ClF}_5 \)):  
  \[
  \text{ClF}_5 = 1 \times \text{Cl} (35.45 \, \text{g/mol}) + 5 \times \text{F} (5 \times 19.00 \, \text{g/mol}) = 130.45 \, \text{g/mol}
  \]
- Molar mass of sulfur tetrafluoride (\( \text{SF}_4 \)):  
  \[
  \text{SF}_4 = 1 \times \text{S} (32.07 \, \text{g/mol}) + 4 \times \text{F} (4 \times 19.00 \, \text{g/mol}) = 108.07 \, \text{g/mol}
  \]

Now, calculate the moles of each gas:
\[
n_{\text{ClF}_5} = \frac{16.2 \, \text{g}}{130.45 \, \text{g/mol}} = 0.1242 \, \text{mol}
\]
\[
n_{\text{SF}_4} = \frac{7.73 \, \text{g}}{108.07 \, \text{g/mol}} = 0.0715 \, \text{mol}
\]

### Step 2: Calculate total moles
The total moles in the tank are:
\[
n_{\text{total}} = n_{\text{ClF}_5} + n_{\text{SF}_4} = 0.1242 + 0.0715 = 0.1957 \, \text{mol}
\]

### Step 3: Calculate mole fractions
The mole fraction of each gas is the ratio of its moles to the total moles:
\[
X_{\text{ClF}_5} = \frac{n_{\text{ClF}_5}}{n_{\text{total}}} = \frac{0.1242}{0.1957} = 0.6346
\]
\[
X_{\text{SF}_4} = \frac{n_{\text{SF}_4}}{n_{\text{total}}} = \frac{0.0715}{0.1957} = 0.3654
\]

### Step 4: Use the ideal gas law to calculate total pressure
The ideal gas law is:
\[
PV = nRT
\]
Solve for \( P \) (total pressure):
\[
P = \frac{n_{\text{total}} RT}{V} = \frac{(0.1957 \, \text{mol}) (0.0821 \, \text{L} \cdot \text{atm}/\text{mol} \cdot \text{K}) (278.41 \, \text{K})}{9.00 \, \text{L}} = 0.496 \, \text{atm}
\]

### Step 5: Calculate partial pressures
The partial pressure of each gas is the product of its mole fraction and the total pressure:
\[
P_{\text{ClF}_5} = X_{\text{ClF}_5} \times P_{\text{total}} = 0.6346 \times 0.496 \, \text{atm} = 0.315 \, \text{atm}
\]
\[
P_{\text{SF}_4} = X_{\text{SF}_4} \times P_{\text{total}} = 0.3654 \times 0.496 \, \text{atm} = 0.181 \, \text{atm}
\]

### Final answers:
- **Mole fraction of chlorine pentafluoride (ClF₅):** 0.6346
- **Partial pressure of chlorine pentafluoride (ClF₅):** 0.315 atm
- **Mole fraction of sulfur tetrafluoride (SF₄):** 0.3654
- **Partial pressure of sulfur tetrafluoride (SF₄):** 0.181 atm
- **Total pressure in the tank:** 0.496 atm

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Would you like further details or clarifications?

### Relative questions:
1. How do you calculate the molar mass of a compound?
2. Why do we convert temperature from Celsius to Kelvin in gas law calculations?
3. What would happen to the pressure if the temperature increased?
4. Can the partial pressure of one gas be higher than the total pressure?
5. How does the ideal gas law differ from real gas behavior?

### Tip:
Always ensure the units in gas law calculations (L, atm, K, mol) are consistent to avoid errors in results.

A 9.00 L tank at 5.26 °C is filled with 16.2 g of chlorine pentafluoride gas and 7.73 g of sulfur tetrafluoride gas. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Assume both gases behave as ideal gases.

Learn how to calculate the mole fraction and partial pressure of chlorine pentafluoride and sulfur tetrafluoride in a 9.00 L tank at 5.26 °C. Using the ideal gas law and mole fraction equations, this detailed solution provides the step-by-step process for finding the total pressure and gas contributions.

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