A sample of oxygen gas was found to effuse at a rate equal to two times that of an unknown gas. The molecular weight of the unknown gas is g/mol. A 128 64 16 D) 8.0 E 8

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Question:**

A sample of oxygen gas was found to effuse at a rate equal to two times that of an unknown gas. The molecular weight of the unknown gas is _______ g/mol.

- **A)** 128
- **B)** 64
- **C)** 16
- **D)** 8.0
- **E)** 8

This question presents a problem based on effusion, which can be understood using Graham's Law of Effusion. The law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. By comparing oxygen gas with an unknown gas, you can determine the molecular weight of the unknown gas using the given rate ratio.
Transcribed Image Text:**Question:** A sample of oxygen gas was found to effuse at a rate equal to two times that of an unknown gas. The molecular weight of the unknown gas is _______ g/mol. - **A)** 128 - **B)** 64 - **C)** 16 - **D)** 8.0 - **E)** 8 This question presents a problem based on effusion, which can be understood using Graham's Law of Effusion. The law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. By comparing oxygen gas with an unknown gas, you can determine the molecular weight of the unknown gas using the given rate ratio.
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