It takes 228 s for 4.85 × 10-5 mol of gas with molar mass M = 213 g/mol to effuse through a %D tiny hole. Under the same conditions, how long will it take for 4.85 × 10-5 mol Ar to effuse? Answer: time(s) =
It takes 228 s for 4.85 × 10-5 mol of gas with molar mass M = 213 g/mol to effuse through a %D tiny hole. Under the same conditions, how long will it take for 4.85 × 10-5 mol Ar to effuse? Answer: time(s) =
Chemistry
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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 Time Calculation for Gases**
*Concept Overview:*
This example examines the time it takes for a specific amount of gas to effuse through a tiny hole. Effusion time can be calculated using the molar mass of gases and comparing them under consistent conditions.
*Problem Statement:*
It takes 228 seconds for 4.85 × 10⁻⁵ moles of a gas with a molar mass (M) of 213 g/mol to effuse through a tiny hole. Under the same conditions, how long will it take for 4.85 × 10⁻⁵ moles of Argon (Ar) to effuse?
*Solution Approach:*
Use Graham's Law of Effusion:
\[
\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}}
\]
Where \(\text{rate}_1\) and \(\text{rate}_2\) are the effusion rates of gas 1 and gas 2, respectively, and \(M_1\) and \(M_2\) are their molar masses.
Given:
- Initial gas: M = 213 g/mol
- Argon (Ar) molar mass: 39.95 g/mol
- Time for initial gas effusion = 228 s
\[
\text{time (s)} = \ldots
\]
Apply this equation to find the effusion time for Argon.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe3925343-d6c3-4a25-992b-202f02b2b634%2Fc662ad76-2bd8-4c9b-8929-6e82b122a6ef%2Fk2n5p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Effusion Time Calculation for Gases**
*Concept Overview:*
This example examines the time it takes for a specific amount of gas to effuse through a tiny hole. Effusion time can be calculated using the molar mass of gases and comparing them under consistent conditions.
*Problem Statement:*
It takes 228 seconds for 4.85 × 10⁻⁵ moles of a gas with a molar mass (M) of 213 g/mol to effuse through a tiny hole. Under the same conditions, how long will it take for 4.85 × 10⁻⁵ moles of Argon (Ar) to effuse?
*Solution Approach:*
Use Graham's Law of Effusion:
\[
\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}}
\]
Where \(\text{rate}_1\) and \(\text{rate}_2\) are the effusion rates of gas 1 and gas 2, respectively, and \(M_1\) and \(M_2\) are their molar masses.
Given:
- Initial gas: M = 213 g/mol
- Argon (Ar) molar mass: 39.95 g/mol
- Time for initial gas effusion = 228 s
\[
\text{time (s)} = \ldots
\]
Apply this equation to find the effusion time for Argon.
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