Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 4, Problem 46E
Interpretation Introduction

Interpretation:

The condition of STP is to be interpreted.

Concept introduction:

STP stands for standard temperature and pressure conditions.

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What volume (in L) does a 1.34 g  sample of chlorine gas occupy at SATP?
This relationship is known as Graham’s Law of Effusion. Since both gases are at the same temperature, they must have the same average kinetic energy (½ mv2), where m is mass and v is velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ mLvL2 = ½ mHvH2. Multiplying both sides by 2 gives you mLvL2 = mHvH2. Rearranging the equation to get both masses on the same side of the equation will give you mL/mH = VH2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, VH2/VL2 = ¼. If the light gas was Ne, what would be a reasonable identity for the heavy gas? Explain.
c) This relationship is known as Graham's Law of Effusion. Since both gases are at te same temperature, they must have the same average kinetic energy (½ mv²), where m is mass and v is velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ muvL2 = v ². Multiplying both sides by 2 gives you m v 2 y ². Rearranging the equation to get H H LL H H 2 m = m both masses on the same side of the equation will give you mu/mH = V 2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, V 2/VL2 = ¼. How many times heavier is the heavy gas compared to the light gas? d) If the light gas was Ne, what would be a reasonable identity for the heavy gas?

Chapter 4 Solutions

Introductory Chemistry: An Active Learning Approach

Ch. 4 - Prob. 11ECh. 4 - The density of liquid oxygen is about 1.4 g/cm3....Ch. 4 - Prob. 13ECh. 4 - Prob. 14ECh. 4 - Prob. 15ECh. 4 - Prob. 16ECh. 4 - Prob. 17ECh. 4 - Prob. 18ECh. 4 - Prob. 19ECh. 4 - Prob. 20ECh. 4 - Prob. 21ECh. 4 - Prob. 22ECh. 4 - Prob. 23ECh. 4 - Prob. 24ECh. 4 - Prob. 25ECh. 4 - Prob. 26ECh. 4 - Prob. 27ECh. 4 - Many common liquids have boiling points that are...Ch. 4 - Hydrogen cyanide is the deadly gas used in some...Ch. 4 - Prob. 30ECh. 4 - Prob. 31ECh. 4 - Prob. 32ECh. 4 - Prob. 33ECh. 4 - Prob. 34ECh. 4 - Prob. 35ECh. 4 - Prob. 36ECh. 4 - Prob. 37ECh. 4 - A sample of carbon dioxide gas at a pressure of...Ch. 4 - Prob. 39ECh. 4 - A sample of krypton gas at a pressure of 905 torr...Ch. 4 - Prob. 41ECh. 4 - A sample of krypton gas occupies a volume of 6.68...Ch. 4 - Prob. 43ECh. 4 - Prob. 44ECh. 4 - Prob. 45ECh. 4 - Prob. 46ECh. 4 - If 1 cubic foot-28.3 L-of air at common room...Ch. 4 - Prob. 48ECh. 4 - Prob. 49ECh. 4 - Prob. 50ECh. 4 - Prob. 51ECh. 4 - Prob. 52ECh. 4 - A container with a volume of 56.2 L holds helium...Ch. 4 - At STP, a sample of neon fills a 4.47-L container....Ch. 4 - Prob. 55ECh. 4 - Prob. 56ECh. 4 - Prob. 57ECh. 4 - Prob. 58ECh. 4 - Prob. 59ECh. 4 - Prob. 60ECh. 4 - Prob. 61ECh. 4 - Air in a steel cylinder is heated from 19 0C to 42...Ch. 4 - A gas storage tank is designed to hold a fixed...Ch. 4 - Prob. 64ECh. 4 - If 1.62 m3 of air at 120C and 738 torr is...Ch. 4 - Prob. 66ECh. 4 - The compression ratio in an automobile engine is...Ch. 4 - Target check For each of the macroscopic...Ch. 4 - Prob. 4.2TCCh. 4 - Prob. 4.3TCCh. 4 - Prob. 4.4TCCh. 4 - Prob. 1CLECh. 4 - Prob. 2CLECh. 4 - Prob. 3CLECh. 4 - Prob. 4CLECh. 4 - Prob. 5CLECh. 4 - Prob. 1PECh. 4 - Prob. 2PECh. 4 - Prob. 3PECh. 4 - Prob. 4PECh. 4 - Prob. 5PECh. 4 - Prob. 6PE
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