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Let's solve this step-by-step.

### Given dataLet's solve this step-by-step.

### Given 
- Volume of the tank \(V = 8.00 \, \text{L}\)
- Temperature \(T = 7.43^\circ C = 280.58 \, \text{K}\) (converted to Kelvin)
- Mass of dinitrogen monoxide (N₂O) \(m_{\text{N₂O}} = 6.72 \, \text{g}\)
- Mass of sulfur hexafluoride (SF₆) \(m_{\text{SF₆}} = 7.23 \, \text{g}\)

We are asked to calculate:
1. The **mole fraction** of each gas.
2. The **partial pressure** of each gas.
3. The **total pressure** in the tank.

### Step 1: Calculate the moles of each gas
To find the number of moles (\(n\)) of each gas, use the formula:

\[
n = \frac{m}{M}
\]
where \(m\) is the mass of the gas, and \(M\) is the molar mass.

- Molar mass of N₂O (dinitrogen monoxide):
  \[
  M_{\text{N₂O}} = 2(14.01) + 16.00 = 44.01 \, \text{g/mol}
  \]
- Molar mass of SF₆ (sulfur hexafluoride):
  \[
  M_{\text{SF₆}} = 32.06 + 6(18.998) = 146.06 \, \text{g/mol}
  \]

Now calculate the moles:
- Moles of N₂O:
  \[
  n_{\text{N₂O}} = \frac{6.72 \, \text{g}}{44.01 \, \text{g/mol}} = 0.1527 \, \text{mol}
  \]
- Moles of SF₆:
  \[
  n_{\text{SF₆}} = \frac{7.23 \, \text{g}}{146.06 \, \text{g/mol}} = 0.0495 \, \text{mol}
  \]

### Step 2: Calculate the mole fraction of each gas
The mole fraction (\(X\)) of a gas is given by:

\[
X = \frac{n_{\text{gas}}}{n_{\text{total}}}
\]
where \(n_{\text{total}} = n_{\text{N₂O}} + n_{\text{SF₆}}\).

- Total moles:
  \[
  n_{\text{total}} = 0.1527 \, \text{mol} + 0.0495 \, \text{mol} = 0.2022 \, \text{mol}
  \]
- Mole fraction of N₂O:
  \[
  X_{\text{N₂O}} = \frac{0.1527}{0.2022} = 0.7552
  \]
- Mole fraction of SF₆:
  \[
  X_{\text{SF₆}} = \frac{0.0495}{0.2022} = 0.2448
  \]

### Step 3: Calculate the total pressure using the ideal gas law
The ideal gas law is:

\[
PV = nRT
\]
where:
- \(P\) is the total pressure (in atm),
- \(V\) is the volume (in L),
- \(n\) is the total moles of gas,
- \(R = 0.0821 \, \text{L·atm/mol·K}\) is the ideal gas constant,
- \(T = 280.58 \, \text{K}\).

Rearranging for \(P\):

\[
P = \frac{nRT}{V}
\]

Substitute the known values:

\[
P = \frac{(0.2022 \, \text{mol}) (0.0821 \, \text{L·atm/mol·K}) (280.58 \, \text{K})}{8.00 \, \text{L}}
\]
\[
P = 0.5827 \, \text{atm}
\]

### Step 4: Calculate the partial pressures
The partial pressure of each gas is given by Dalton’s Law of Partial Pressures:

\[
P_{\text{gas}} = X_{\text{gas}} \times P_{\text{total}}
\]

- Partial pressure of N₂O:
  \[
  P_{\text{N₂O}} = 0.7552 \times 0.5827 = 0.4401 \, \text{atm}
  \]
- Partial pressure of SF₆:
  \[
  P_{\text{SF₆}} = 0.2448 \times 0.5827 = 0.1426 \, \text{atm}
  \]

### Final Answers:
1. Mole fraction of N₂O: **0.755**
2. Mole fraction of SF₆: **0.245**
3. Partial pressure of N₂O: **0.440 atm**
4. Partial pressure of SF₆: **0.143 atm**
5. Total pressure: **0.583 atm**

Let me know if you need further clarification on any part of this process!

Here are some related questions:
1. How does the ideal gas law change if temperature is expressed in Celsius instead of Kelvin?
2. How would the partial pressures change if the tank's volume doubled?
3. What would happen to the total pressure if another gas were added to the tank?
4. Can you explain Dalton’s Law in more detail for non-ideal gas behavior?
5. How would you calculate the pressure if one of the gases became a liquid at this temperature?

**Tip**: Always make sure your units are consistent, especially when using the ideal gas law. Always convert Celsius to Kelvin!

A 8.00 L tank at 7.43 °C is filled with 6.72 g of dinitrogen monoxide gas and 7.23 g of sulfur hexafluoride 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 dinitrogen monoxide and sulfur hexafluoride in an 8.00 L tank at 7.43 °C, assuming ideal gas behavior. This problem involves key concepts such as the ideal gas law and Dalton's Law of Partial Pressures.

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