Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
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Question
Chapter 10.2, Problem 32LC
Interpretation Introduction
Interpretation: The mass of the given collected gas should be calculated and the reason for the importance of mol for this calculation needs to be explained.
Concept Introduction: A factor that uses one set of units to change it to another set of units is said to be a conversion factor.
Expert Solution & Answer
Explanation of Solution
Given information:
Chemist collects 2.94 L of carbon monoxide gas at STP.
The molar volume of any compound at STP is 22.4 L for 1 mol of the compound.
22.4 L = 1 mol
In the given information, carbon monoxide, CO gas is considered an ideal gas. So, the number of moles can be determined using the following conversion factor:
1 mol22.4 L×2.94 L
So,
1 mol22.4 L×2.94 L = 0.13 mol
Now, the mass of the gas can be calculated using the number of moles and molar mass of gas as:
mass = moles × molar mass
Substituting the values as:
mass = 0.13 mol × 28 g/molmass = 3.6 g
Since the number of moles is intermediate for the determination of the mass of the gas, so, it is important to calculate the number of moles of the gas.
Chapter 10 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
Ch. 10.1 - Prob. 1SPCh. 10.1 - Prob. 2SPCh. 10.1 - Prob. 3SPCh. 10.1 - Prob. 4SPCh. 10.1 - Prob. 5SPCh. 10.1 - Prob. 6SPCh. 10.1 - Prob. 7SPCh. 10.1 - Prob. 8SPCh. 10.1 - Prob. 9LCCh. 10.1 - Prob. 10LC
Ch. 10.1 - Prob. 11LCCh. 10.1 - Prob. 12LCCh. 10.1 - Prob. 13LCCh. 10.1 - Prob. 14LCCh. 10.1 - Prob. 15LCCh. 10.2 - Prob. 16SPCh. 10.2 - Prob. 17SPCh. 10.2 - Prob. 18SPCh. 10.2 - Prob. 19SPCh. 10.2 - Prob. 20SPCh. 10.2 - Prob. 21SPCh. 10.2 - Prob. 22SPCh. 10.2 - Prob. 23SPCh. 10.2 - Prob. 24LCCh. 10.2 - Prob. 25LCCh. 10.2 - Prob. 26LCCh. 10.2 - Prob. 27LCCh. 10.2 - Prob. 28LCCh. 10.2 - Prob. 29LCCh. 10.2 - Prob. 30LCCh. 10.2 - Prob. 31LCCh. 10.2 - Prob. 32LCCh. 10.3 - Prob. 33SPCh. 10.3 - Prob. 34SPCh. 10.3 - Prob. 35SPCh. 10.3 - Prob. 36SPCh. 10.3 - Prob. 37SPCh. 10.3 - Prob. 38SPCh. 10.3 - Prob. 39SPCh. 10.3 - Prob. 40SPCh. 10.3 - Prob. 41SPCh. 10.3 - Prob. 42SPCh. 10.3 - Prob. 43LCCh. 10.3 - Prob. 44LCCh. 10.3 - Prob. 45LCCh. 10.3 - Prob. 46LCCh. 10.3 - Prob. 47LCCh. 10.3 - Prob. 48LCCh. 10.3 - Prob. 49LCCh. 10 - Prob. 50ACh. 10 - Prob. 51ACh. 10 - Prob. 52ACh. 10 - Prob. 53ACh. 10 - Prob. 54ACh. 10 - Prob. 55ACh. 10 - Prob. 56ACh. 10 - Prob. 57ACh. 10 - Prob. 58ACh. 10 - Prob. 59ACh. 10 - Prob. 60ACh. 10 - Prob. 61ACh. 10 - Prob. 62ACh. 10 - Prob. 63ACh. 10 - Prob. 64ACh. 10 - Prob. 65ACh. 10 - Prob. 66ACh. 10 - Prob. 67ACh. 10 - Prob. 68ACh. 10 - Prob. 69ACh. 10 - Prob. 70ACh. 10 - Prob. 71ACh. 10 - Prob. 72ACh. 10 - Prob. 73ACh. 10 - Prob. 74ACh. 10 - Prob. 75ACh. 10 - Prob. 76ACh. 10 - Prob. 77ACh. 10 - Prob. 78ACh. 10 - Prob. 79ACh. 10 - Prob. 80ACh. 10 - Prob. 81ACh. 10 - Prob. 82ACh. 10 - Prob. 83ACh. 10 - Prob. 84ACh. 10 - Prob. 85ACh. 10 - Prob. 86ACh. 10 - Prob. 87ACh. 10 - Prob. 88ACh. 10 - Prob. 89ACh. 10 - Prob. 90ACh. 10 - Prob. 91ACh. 10 - Prob. 92ACh. 10 - Prob. 93ACh. 10 - Prob. 94ACh. 10 - Prob. 95ACh. 10 - Prob. 96ACh. 10 - Prob. 97ACh. 10 - Prob. 98ACh. 10 - Prob. 99ACh. 10 - Prob. 100ACh. 10 - Prob. 101ACh. 10 - Prob. 102ACh. 10 - Prob. 103ACh. 10 - Prob. 104ACh. 10 - Prob. 105ACh. 10 - Prob. 106ACh. 10 - Prob. 108ACh. 10 - Prob. 109ACh. 10 - Prob. 110ACh. 10 - Prob. 111ACh. 10 - Prob. 112ACh. 10 - Prob. 113ACh. 10 - Prob. 114ACh. 10 - Prob. 115ACh. 10 - Prob. 116ACh. 10 - Prob. 117ACh. 10 - Prob. 118ACh. 10 - Prob. 119ACh. 10 - Prob. 120ACh. 10 - Prob. 121ACh. 10 - Prob. 122ACh. 10 - Prob. 123ACh. 10 - Prob. 124ACh. 10 - Prob. 125ACh. 10 - Prob. 126ACh. 10 - Prob. 127ACh. 10 - Prob. 128ACh. 10 - Prob. 1STPCh. 10 - Prob. 2STPCh. 10 - Prob. 3STPCh. 10 - Prob. 4STPCh. 10 - Prob. 5STPCh. 10 - Prob. 6STPCh. 10 - Prob. 7STPCh. 10 - Prob. 8STPCh. 10 - Prob. 9STPCh. 10 - Prob. 10STPCh. 10 - Prob. 11STPCh. 10 - Prob. 12STPCh. 10 - Prob. 13STPCh. 10 - Prob. 14STP
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