Determine what the unknow gas is when it has a rate of diffusion which is 1.897 times faster than N,o, Will it be CHe, CO, NO, CO,, N,O, or C,H,
Determine what the unknow gas is when it has a rate of diffusion which is 1.897 times faster than N,o, Will it be CHe, CO, NO, CO,, N,O, or C,H,
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter10: Gases And Their Properties
Section: Chapter Questions
Problem 83GQ: Carbon dioxide, CO2, was shown lo effuse through a porous plate at the rate of 0.033 mol/min. The...
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![**Determine the Unknown Gas**
To identify the unknown gas, given that it has a rate of diffusion which is 1.897 times faster than \( N_2O_5 \), we are considering one of the following gases: \( CH_4 \), \( CO \), \( NO \), \( CO_2 \), \( N_2O_4 \), or \( C_2H_6 \).
To solve this problem, you can use Graham's Law of Effusion, which states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass:
\[ \frac{\text{Rate}_{\text{Gas}_1}}{\text{Rate}_{\text{Gas}_2}} = \sqrt{\frac{M_{2}}{M_{1}}} \]
Where:
- \( \text{Rate}_{\text{Gas}_1} \) is the rate of diffusion of the unknown gas.
- \( \text{Rate}_{\text{Gas}_2} \) is the rate of diffusion of \( N_2O_5 \).
- \( M_{1} \) is the molar mass of the unknown gas.
- \( M_{2} \) is the molar mass of \( N_2O_5 \).
Given that the rate of the unknown gas is 1.897 times faster than \( N_2O_5 \), we can set up the equation:
\[ 1.897 = \sqrt{\frac{M_{N_2O_5}}{M_{\text{unknown}}}} \]
You can then square both sides to get:
\[ (1.897)^2 = \frac{M_{N_2O_5}}{M_{\text{unknown}}} \]
\[ 3.598 = \frac{M_{N_2O_5}}{M_{\text{unknown}}} \]
To rearrange for \( M_{\text{unknown}} \):
\[ M_{\text{unknown}} = \frac{M_{N_2O_5}}{3.598} \]
The molar mass of \( N_2O_5 \) is approximately 108 g/mol.
\[ M_{\text{unknown}} = \frac{108}{3.598} \approx 30.02 \text{ g/mol} \]
Now, comparing the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4963a1b7-231e-4415-aa15-591e301cddd8%2F2e7df6cb-dc53-4bdc-8c31-d197f855b9a0%2Fgusd4qe.jpeg&w=3840&q=75)
Transcribed Image Text:**Determine the Unknown Gas**
To identify the unknown gas, given that it has a rate of diffusion which is 1.897 times faster than \( N_2O_5 \), we are considering one of the following gases: \( CH_4 \), \( CO \), \( NO \), \( CO_2 \), \( N_2O_4 \), or \( C_2H_6 \).
To solve this problem, you can use Graham's Law of Effusion, which states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass:
\[ \frac{\text{Rate}_{\text{Gas}_1}}{\text{Rate}_{\text{Gas}_2}} = \sqrt{\frac{M_{2}}{M_{1}}} \]
Where:
- \( \text{Rate}_{\text{Gas}_1} \) is the rate of diffusion of the unknown gas.
- \( \text{Rate}_{\text{Gas}_2} \) is the rate of diffusion of \( N_2O_5 \).
- \( M_{1} \) is the molar mass of the unknown gas.
- \( M_{2} \) is the molar mass of \( N_2O_5 \).
Given that the rate of the unknown gas is 1.897 times faster than \( N_2O_5 \), we can set up the equation:
\[ 1.897 = \sqrt{\frac{M_{N_2O_5}}{M_{\text{unknown}}}} \]
You can then square both sides to get:
\[ (1.897)^2 = \frac{M_{N_2O_5}}{M_{\text{unknown}}} \]
\[ 3.598 = \frac{M_{N_2O_5}}{M_{\text{unknown}}} \]
To rearrange for \( M_{\text{unknown}} \):
\[ M_{\text{unknown}} = \frac{M_{N_2O_5}}{3.598} \]
The molar mass of \( N_2O_5 \) is approximately 108 g/mol.
\[ M_{\text{unknown}} = \frac{108}{3.598} \approx 30.02 \text{ g/mol} \]
Now, comparing the
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