A sample of O3 gas is observed to effuse through a porous barrier in 4.08 minutes. Under the same conditions, the same number of moles of an unknown gas requires 3.89 minutes to effuse through the same barrier. The molar mass of the unknown gas is g/mol.
A sample of O3 gas is observed to effuse through a porous barrier in 4.08 minutes. Under the same conditions, the same number of moles of an unknown gas requires 3.89 minutes to effuse through the same barrier. The molar mass of the unknown gas is g/mol.
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...
Related questions
Question
Transcribed Image Text:#### Effusion of Gases
A sample of \( O_3 \) gas is observed to effuse through a porous barrier in 4.08 minutes. Under the same conditions, the same number of moles of an unknown gas requires 3.89 minutes to effuse through the same barrier.
The molar mass of the unknown gas is _____ g/mol.
![### Problem Statement: Effusion Rates of Gases
#### Given Data:
The rate of effusion of \( \text{NO}_2 \) gas through a porous barrier is observed to be:
\[ 3.05 \times 10^{-4} \, \text{mol/h} \]
#### Objective:
Determine the rate of effusion of \( \text{CO}_2 \) gas under the same conditions.
#### To Calculate:
The rate of effusion of \( \text{CO}_2 \) gas in \( \text{mol/h} \).
### Explanation:
Effusion refers to the process by which gas particles pass through a tiny hole. According to Graham's law of effusion, the rate of effusion (\( r \)) of a gas is inversely proportional to the square root of its molar mass (\( M \)):
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]
Where:
- \( r_1 \) and \( r_2 \) are the rates of effusion of gases 1 and 2.
- \( M_1 \) and \( M_2 \) are the molar masses of gases 1 and 2.
**Note**: There is a blank box in the image to be filled with the calculated value for the rate of effusion of \( \text{CO}_2 \) gas.
### Steps to Solve:
1. Identify the molar masses of \( \text{NO}_2 \) and \( \text{CO}_2 \) gases.
2. Using Graham's law, solve for the rate of effusion of \( \text{CO}_2 \).
This will provide students with a practical application of Graham's Law in determining the unknown rate of effusion for \( \text{CO}_2 \) based on known data.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84435934-315d-4f3f-bf89-3ba66c6d068b%2Fcadc2b14-a9a1-484c-a3a4-e5beb09f3534%2F549d1yk_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement: Effusion Rates of Gases
#### Given Data:
The rate of effusion of \( \text{NO}_2 \) gas through a porous barrier is observed to be:
\[ 3.05 \times 10^{-4} \, \text{mol/h} \]
#### Objective:
Determine the rate of effusion of \( \text{CO}_2 \) gas under the same conditions.
#### To Calculate:
The rate of effusion of \( \text{CO}_2 \) gas in \( \text{mol/h} \).
### Explanation:
Effusion refers to the process by which gas particles pass through a tiny hole. According to Graham's law of effusion, the rate of effusion (\( r \)) of a gas is inversely proportional to the square root of its molar mass (\( M \)):
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]
Where:
- \( r_1 \) and \( r_2 \) are the rates of effusion of gases 1 and 2.
- \( M_1 \) and \( M_2 \) are the molar masses of gases 1 and 2.
**Note**: There is a blank box in the image to be filled with the calculated value for the rate of effusion of \( \text{CO}_2 \) gas.
### Steps to Solve:
1. Identify the molar masses of \( \text{NO}_2 \) and \( \text{CO}_2 \) gases.
2. Using Graham's law, solve for the rate of effusion of \( \text{CO}_2 \).
This will provide students with a practical application of Graham's Law in determining the unknown rate of effusion for \( \text{CO}_2 \) based on known data.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY