QuestionAugust 7, 2025

Approximately 1.000 g each of four gasses H_(2) Ne, Ar, and Kr are placed in a sealed container all under 5 atm of pressure. Assuming ideal behavior, determine the partial pressure of the H_(2) and Ne? View Available Hint(s) 1.279 atm H_(2) and 0.1267 atm Ne 0.06394 atm H_(2) and 0.03051 atm Ne 0.1267 atm H_(2) and 1.279 atm Ne 0.03051 atm H_(2) and 0.06394 atm Ne

Approximately 1.000 g each of four gasses H_(2) Ne, Ar, and Kr are placed in a sealed container all under 5 atm of pressure. Assuming ideal behavior, determine the partial pressure of the H_(2) and Ne? View Available Hint(s) 1.279 atm H_(2) and 0.1267 atm Ne 0.06394 atm H_(2) and 0.03051 atm Ne 0.1267 atm H_(2) and 1.279 atm Ne 0.03051 atm H_(2) and 0.06394 atm Ne
Approximately 1.000 g each of four gasses H_(2) Ne, Ar, and Kr are placed in a sealed container all under 5 atm of pressure. Assuming ideal behavior, determine the partial pressure of the H_(2) and Ne? View Available Hint(s) 1.279 atm H_(2) and 0.1267 atm Ne 0.06394 atm H_(2) and 0.03051 atm Ne 0.1267 atm H_(2) and 1.279 atm Ne 0.03051 atm H_(2) and 0.06394 atm Ne

Solution
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Answer

None of the provided options match the calculated partial pressures. Explanation 1. Calculate moles of each gas Use the formula n = \frac{m}{M}, where m is mass and M is molar mass. For H_2, M = 2.02 \, \text{g/mol}; for Ne, M = 20.18 \, \text{g/mol}. Thus, n_{H_2} = \frac{1.000}{2.02} \approx 0.495 \, \text{mol} and n_{Ne} = \frac{1.000}{20.18} \approx 0.0495 \, \text{mol}. 2. Calculate total moles Sum moles of all gases: n_{total} = n_{H_2} + n_{Ne} + n_{Ar} + n_{Kr}. Assume similar calculations for Ar and Kr, resulting in n_{Ar} \approx 0.025 \, \text{mol} and n_{Kr} \approx 0.012 \, \text{mol}. Therefore, n_{total} \approx 0.495 + 0.0495 + 0.025 + 0.012 = 0.5815 \, \text{mol}. 3. Calculate partial pressures Use Dalton's Law: P_i = X_i \cdot P_{total}, where X_i = \frac{n_i}{n_{total}}. For H_2, X_{H_2} = \frac{0.495}{0.5815} \approx 0.851, so P_{H_2} = 0.851 \times 5 \approx 4.255 \, \text{atm}. For Ne, X_{Ne} = \frac{0.0495}{0.5815} \approx 0.085, so P_{Ne} = 0.085 \times 5 \approx 0.425 \, \text{atm}.

Explanation

1. Calculate moles of each gas<br /> Use the formula $n = \frac{m}{M}$, where $m$ is mass and $M$ is molar mass. For $H_2$, $M = 2.02 \, \text{g/mol}$; for Ne, $M = 20.18 \, \text{g/mol}$. Thus, $n_{H_2} = \frac{1.000}{2.02} \approx 0.495 \, \text{mol}$ and $n_{Ne} = \frac{1.000}{20.18} \approx 0.0495 \, \text{mol}$.<br /><br />2. Calculate total moles<br /> Sum moles of all gases: $n_{total} = n_{H_2} + n_{Ne} + n_{Ar} + n_{Kr}$. Assume similar calculations for Ar and Kr, resulting in $n_{Ar} \approx 0.025 \, \text{mol}$ and $n_{Kr} \approx 0.012 \, \text{mol}$. Therefore, $n_{total} \approx 0.495 + 0.0495 + 0.025 + 0.012 = 0.5815 \, \text{mol}$.<br /><br />3. Calculate partial pressures<br /> Use Dalton's Law: $P_i = X_i \cdot P_{total}$, where $X_i = \frac{n_i}{n_{total}}$. For $H_2$, $X_{H_2} = \frac{0.495}{0.5815} \approx 0.851$, so $P_{H_2} = 0.851 \times 5 \approx 4.255 \, \text{atm}$. For Ne, $X_{Ne} = \frac{0.0495}{0.5815} \approx 0.085$, so $P_{Ne} = 0.085 \times 5 \approx 0.425 \, \text{atm}$.
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