Calculate the ratio of the effusion rates of helium gas (He) and uranium hexafluoride (UF6), a gas used in the enrichment process to produce fuel for nuclear reactors 87.9 | 13.2 9.38 None of these 352.02
Calculate the ratio of the effusion rates of helium gas (He) and uranium hexafluoride (UF6), a gas used in the enrichment process to produce fuel for nuclear reactors 87.9 | 13.2 9.38 None of these 352.02
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Question:**
Calculate the ratio of the effusion rates of helium gas (He) and uranium hexafluoride (UF₆), a gas used in the enrichment process to produce fuel for nuclear reactors.
**Options:**
- ○ 87.9
- ○ 13.2
- ○ 9.38
- ○ None of these
- ○ 352.02
**Explanation:**
To solve this problem, you would use Graham's Law of Effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula to calculate the ratio of effusion rates is:
\[ \frac{\text{Rate of effusion of He}}{\text{Rate of effusion of UF₆}} = \sqrt{\frac{\text{Molar mass of UF₆}}{\text{Molar mass of He}}} \]
The ratio will help determine how much faster helium effuses compared to uranium hexafluoride. Calculating this involves using the known molar masses of helium and UF₆.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c9a3da8-b443-4c7b-9ddb-d4ba4f71dd14%2F4ac88237-3666-4d2d-b696-ffecc3dbd6e7%2Frht5o1d_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
Calculate the ratio of the effusion rates of helium gas (He) and uranium hexafluoride (UF₆), a gas used in the enrichment process to produce fuel for nuclear reactors.
**Options:**
- ○ 87.9
- ○ 13.2
- ○ 9.38
- ○ None of these
- ○ 352.02
**Explanation:**
To solve this problem, you would use Graham's Law of Effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula to calculate the ratio of effusion rates is:
\[ \frac{\text{Rate of effusion of He}}{\text{Rate of effusion of UF₆}} = \sqrt{\frac{\text{Molar mass of UF₆}}{\text{Molar mass of He}}} \]
The ratio will help determine how much faster helium effuses compared to uranium hexafluoride. Calculating this involves using the known molar masses of helium and UF₆.
![The image contains a graph and multiple-choice question related to the behavior of Nitrogen gas under different conditions.
**Graph Description:**
- The graph is a plot of \( \frac{PV}{nRT} \) (y-axis) against pressure \( P \) in atm (x-axis).
- There are three colored curves each representing a different temperature: 203 K (green), 293 K (purple), and 673 K (orange).
- A dashed horizontal line at \( \frac{PV}{nRT} = 1.0 \) indicates ideal gas behavior.
- The curve for 673 K (orange) is almost flat and close to the ideal line, indicating behavior closer to an ideal gas at that temperature.
- The 203 K (green) and 293 K (purple) curves deviate more from the ideal line, especially as pressure increases.
**Question:**
"What does this graph represent for Nitrogen?"
**Answer Options:**
- None of these is true
- This graph shows that Nitrogen gas behaves most ideally at higher temperature and lower pressure
- This graph shows that Nitrogen at no point behaves anywhere close to being ideal.
- This graph shows that Nitrogen gas behaves ideally at lower temperature and higher pressure.
- This graph shows that Nitrogen gas behaves more ideally at lower temperature and lower pressure.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c9a3da8-b443-4c7b-9ddb-d4ba4f71dd14%2F4ac88237-3666-4d2d-b696-ffecc3dbd6e7%2F3bj5k4u_processed.png&w=3840&q=75)
Transcribed Image Text:The image contains a graph and multiple-choice question related to the behavior of Nitrogen gas under different conditions.
**Graph Description:**
- The graph is a plot of \( \frac{PV}{nRT} \) (y-axis) against pressure \( P \) in atm (x-axis).
- There are three colored curves each representing a different temperature: 203 K (green), 293 K (purple), and 673 K (orange).
- A dashed horizontal line at \( \frac{PV}{nRT} = 1.0 \) indicates ideal gas behavior.
- The curve for 673 K (orange) is almost flat and close to the ideal line, indicating behavior closer to an ideal gas at that temperature.
- The 203 K (green) and 293 K (purple) curves deviate more from the ideal line, especially as pressure increases.
**Question:**
"What does this graph represent for Nitrogen?"
**Answer Options:**
- None of these is true
- This graph shows that Nitrogen gas behaves most ideally at higher temperature and lower pressure
- This graph shows that Nitrogen at no point behaves anywhere close to being ideal.
- This graph shows that Nitrogen gas behaves ideally at lower temperature and higher pressure.
- This graph shows that Nitrogen gas behaves more ideally at lower temperature and lower pressure.
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