3. Suppose 2.94g of H2 and 2.34g of He are trapped in 2.65L container at standard temperature. Determine the.... a. Pressure of the Hydrogen gas in atm b. Pressure of the Helium gas in atm c. Total Pressure of the container using parts a and b.

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please help with question 3 a-c
### Educational Transcription of Gas Laws Problems

#### Problem 2
With the given information, follow these steps:
1. **Identify the gas law applicable.**
2. **Solve for the unknown.**

**a.** If a 425mL of Helium (He) gas is cooled from 76°C to 22°C, what is the new volume of the gas?

- **Gas Law:** Charles’s Law
- **Equation:** \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)
- **Solution:**
  - \(T_1 = 76°C + 273K = 349K\)
  - \(T_2 = 22°C + 273K = 295K\)
  - Solve for \(V_2\): \(V_2 = \frac{425mL \times 295K}{349K} = 359.24mL\)

**b.** Determine the new pressure of a gas when it expands to 1.25L from an original volume of 235mL at a pressure of 5.34atm.

- **Gas Law:** Boyle’s Law
- **Equation:** \(P_1V_1 = P_2V_2\)
- **Solution:**
  - Solve for \(P_2\): \(P_2 = \frac{2.35 \times 5.34}{1.25} = 1.004atm\)

**c.** For 23.4 grams of Argon with a volume of 1.35L, find the volume for the same mass of Cl2 gas, assuming other conditions remain constant.

- **Gas Law:** Avogadro’s Law
- **Equation:** \(\frac{V_1}{n_1} = \frac{V_2}{n_2}\)
- **Solution:**
  - Molar Mass of Argon = 39.948g/mol
  - Moles of Argon = \(\frac{23.4g}{39.948g/mol} = 0.5857mol\)
  - Molar Mass of Cl2 = 70.906g/mol
  - Moles of Cl2 = \(\frac{23.4g}{70.906g/mol} = 0.330mol\)
  - Solve for \(
Transcribed Image Text:### Educational Transcription of Gas Laws Problems #### Problem 2 With the given information, follow these steps: 1. **Identify the gas law applicable.** 2. **Solve for the unknown.** **a.** If a 425mL of Helium (He) gas is cooled from 76°C to 22°C, what is the new volume of the gas? - **Gas Law:** Charles’s Law - **Equation:** \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) - **Solution:** - \(T_1 = 76°C + 273K = 349K\) - \(T_2 = 22°C + 273K = 295K\) - Solve for \(V_2\): \(V_2 = \frac{425mL \times 295K}{349K} = 359.24mL\) **b.** Determine the new pressure of a gas when it expands to 1.25L from an original volume of 235mL at a pressure of 5.34atm. - **Gas Law:** Boyle’s Law - **Equation:** \(P_1V_1 = P_2V_2\) - **Solution:** - Solve for \(P_2\): \(P_2 = \frac{2.35 \times 5.34}{1.25} = 1.004atm\) **c.** For 23.4 grams of Argon with a volume of 1.35L, find the volume for the same mass of Cl2 gas, assuming other conditions remain constant. - **Gas Law:** Avogadro’s Law - **Equation:** \(\frac{V_1}{n_1} = \frac{V_2}{n_2}\) - **Solution:** - Molar Mass of Argon = 39.948g/mol - Moles of Argon = \(\frac{23.4g}{39.948g/mol} = 0.5857mol\) - Molar Mass of Cl2 = 70.906g/mol - Moles of Cl2 = \(\frac{23.4g}{70.906g/mol} = 0.330mol\) - Solve for \(
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