The rate of effusion of Ar gas through a porous barrier is observed to be 1.27 x 10-4 mol/h. Under the same conditions, the rate of effusion of 03 gas would be mol/h.
The rate of effusion of Ar gas through a porous barrier is observed to be 1.27 x 10-4 mol/h. Under the same conditions, the rate of effusion of 03 gas would be mol/h.
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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![**Effusion Rate Comparison of Gases**
The rate of effusion of argon (Ar) gas through a porous barrier is observed to be:
\[ 1.27 \times 10^{-4} \, \text{mol/h}. \]
Under the same conditions, the rate of effusion of ozone (O\(_3\)) gas would be:
\[ \boxed{ \hspace{20mm} } \, \text{mol/h}. \]
**Understanding Rates of Effusion:**
This scenario involves comparing the effusion rates of different gases under the same conditions using Graham's law of effusion. Generally, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. To find the effusion rate of ozone given that of argon, we can use:
\[ \frac{\text{Rate of effusion of Ar}}{\text{Rate of effusion of O}_3} = \sqrt{\frac{\text{Molar mass of O}_3}{\text{Molar mass of Ar}}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20a0d4d5-788e-4de6-b89d-c8e91545f77f%2Fd43dc7dd-6969-4088-917f-3e15df8890b4%2F310vv6w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Effusion Rate Comparison of Gases**
The rate of effusion of argon (Ar) gas through a porous barrier is observed to be:
\[ 1.27 \times 10^{-4} \, \text{mol/h}. \]
Under the same conditions, the rate of effusion of ozone (O\(_3\)) gas would be:
\[ \boxed{ \hspace{20mm} } \, \text{mol/h}. \]
**Understanding Rates of Effusion:**
This scenario involves comparing the effusion rates of different gases under the same conditions using Graham's law of effusion. Generally, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. To find the effusion rate of ozone given that of argon, we can use:
\[ \frac{\text{Rate of effusion of Ar}}{\text{Rate of effusion of O}_3} = \sqrt{\frac{\text{Molar mass of O}_3}{\text{Molar mass of Ar}}} \]
![A mixture of argon and krypton gases, in a 9.30 L flask at 46 °C, contains 6.46 grams of argon and 45.4 grams of krypton. The partial pressure of krypton in the flask is [blank] atm and the total pressure in the flask is [blank] atm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20a0d4d5-788e-4de6-b89d-c8e91545f77f%2Fd43dc7dd-6969-4088-917f-3e15df8890b4%2Fx8bl18_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A mixture of argon and krypton gases, in a 9.30 L flask at 46 °C, contains 6.46 grams of argon and 45.4 grams of krypton. The partial pressure of krypton in the flask is [blank] atm and the total pressure in the flask is [blank] atm.
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