#### Effusion of GasesA 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.

Since you have asked multiple questions, we will solve the first question for you. If you want any…

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...

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Concept explainers

Kinetic-Theory of Gases

Kinetic theory of gases is the model or method which was developed in the 19th century by Maxwell and Boltzmann. This is possible as the interatomic force which is of short-range forces that are important for solids and liquids can be neglected for gases. This theory or gas law gives a molecular interpretation of the pressure and temperature of a gas and is consistent with gas laws and Avogadro’s hypothesis. This theory is very helpful to correctly explain the specific heat capacities of many gases, kinetic energy and it also explains the transport qualities such as viscosity, thermal conductivity, and diffusion with molecular parameters. As stated in the kinetic-molecular theory, the temperature of a substance or gas particle is related to the average kinetic energy of the particles of that substance. When a substance is heated, some of the absorbed energy is stored within the particles, while some of the energy increases the motion of the particles, called average kinetic energy.

Question

#### 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.

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