Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 13, Problem 94A
Solution

(a)

Interpretation:

The volume of given amount of carbon dioxide is to be calculated.

Concept introduction:

Avogadro’s Law is used to find out the result.

The volume of carbon dioxide is 4.48L

According to Avogadro’s Law, equal numbers of molecules of all gases occupy 22.4 liters volume at STP.

44gCO2occupiesvolume=22.4L8.80gwilloccupy=8.80×22.444=4.48L

(b)

Interpretation:

The volume of the gas is to be calculated when pressure, temperature and amount of the gas are given.

Concept introduction:

Ideal gas equation helps to calculate volume of a gas at given temperature and pressure. According to this law,

PV=nRT

Where

P is pressure of the gas

V is volume of the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

The volume of the gas is 2.37L

Ideal gas equation helps to calculate volume of a gas at given temperature and pressure. According to this law,

PV=nRT

Or

V=nRTP

Given data is:

m=8.80g

P=3atm

R=0.0821LatmK-1mol-1

T=160°C=160+273=433K

n=mMn=8.8044=0.2mol

Now substituting the values into formula,

V=nRTP=0.2×0.0821×4333=2.37L

Therefore volume of the gas is 2.37L

(c)

Interpretation:

The volume of the gas is to be calculated when pressure, temperature and amount of the gas are given.

Concept introduction:

Ideal gas equation helps to calculate volume of a gas at given temperature and pressure. According to this law,

PV=nRT

Where

P is pressure of the gas

V is volume of the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

The volume of the gas is 4.06L

Ideal gas equation helps to calculate volume of a gas at given temperature and pressure. According to this law,

PV=nRT

Or

V=nRTP

Given data is:

m=8.80g

P=118kPa

R=8.314LkPaK-1mol-1

T=288K

n=mMn=8.8044=0.2mol

Now substituting the values into formula,

V=nRTP=0.2×8.314×288118=4.06L

Therefore volume of the gas is 4.06L

Chapter 13 Solutions

Chemistry: Matter and Change

Ch. 13.1 - Prob. 11PPCh. 13.1 - Prob. 12PPCh. 13.1 - Prob. 13PPCh. 13.1 - Prob. 14SSCCh. 13.1 - Prob. 15SSCCh. 13.1 - Prob. 16SSCCh. 13.1 - Prob. 17SSCCh. 13.1 - Prob. 18SSCCh. 13.1 - Prob. 19SSCCh. 13.2 - Prob. 20PPCh. 13.2 - Prob. 21PPCh. 13.2 - Prob. 22PPCh. 13.2 - Prob. 23PPCh. 13.2 - Prob. 24PPCh. 13.2 - Prob. 25PPCh. 13.2 - Prob. 26PPCh. 13.2 - Prob. 27PPCh. 13.2 - Prob. 28PPCh. 13.2 - Prob. 29PPCh. 13.2 - Prob. 30PPCh. 13.2 - Prob. 31SSCCh. 13.2 - Prob. 32SSCCh. 13.2 - Prob. 33SSCCh. 13.2 - Prob. 34SSCCh. 13.2 - Prob. 35SSCCh. 13.2 - Prob. 36SSCCh. 13.2 - Prob. 37SSCCh. 13.3 - Prob. 38PPCh. 13.3 - Prob. 39PPCh. 13.3 - Prob. 40PPCh. 13.3 - Prob. 41PPCh. 13.3 - Prob. 42PPCh. 13.3 - Prob. 43PPCh. 13.3 - Prob. 44PPCh. 13.3 - Prob. 45PPCh. 13.3 - Prob. 46SSCCh. 13.3 - Prob. 47SSCCh. 13.3 - Prob. 48SSCCh. 13.3 - Prob. 49SSCCh. 13 - Prob. 50ACh. 13 - Prob. 51ACh. 13 - Prob. 52ACh. 13 - Prob. 53ACh. 13 - Prob. 54ACh. 13 - Prob. 55ACh. 13 - Prob. 56ACh. 13 - Prob. 57ACh. 13 - Prob. 58ACh. 13 - Prob. 59ACh. 13 - Prob. 60ACh. 13 - Prob. 61ACh. 13 - Prob. 62ACh. 13 - Prob. 63ACh. 13 - Prob. 64ACh. 13 - Prob. 65ACh. 13 - Prob. 66ACh. 13 - Prob. 67ACh. 13 - Prob. 68ACh. 13 - Prob. 69ACh. 13 - Prob. 70ACh. 13 - Prob. 71ACh. 13 - Prob. 72ACh. 13 - Prob. 73ACh. 13 - Prob. 74ACh. 13 - Prob. 75ACh. 13 - Prob. 76ACh. 13 - Prob. 77ACh. 13 - Prob. 78ACh. 13 - Prob. 79ACh. 13 - Prob. 80ACh. 13 - Prob. 81ACh. 13 - Prob. 82ACh. 13 - Prob. 83ACh. 13 - Prob. 84ACh. 13 - Prob. 85ACh. 13 - Prob. 86ACh. 13 - Prob. 87ACh. 13 - Prob. 88ACh. 13 - Prob. 89ACh. 13 - Prob. 90ACh. 13 - Prob. 91ACh. 13 - Prob. 92ACh. 13 - Prob. 93ACh. 13 - Prob. 94ACh. 13 - Oxygen Consumption If 5.00 L of hydrogen...Ch. 13 - Prob. 96ACh. 13 - If 2.33 L of propane at 24°C and 67.2 kPa is...Ch. 13 - Prob. 98ACh. 13 - Prob. 99ACh. 13 - Prob. 100ACh. 13 - Prob. 101ACh. 13 - Apply Calculate the pressure of (4.671022)...Ch. 13 - Analyze When nitroglycerin (C3H5N3O9) explodes,...Ch. 13 - Prob. 104ACh. 13 - Prob. 105ACh. 13 - Prob. 106ACh. 13 - Prob. 107ACh. 13 - Prob. 108ACh. 13 - Prob. 109ACh. 13 - Prob. 110ACh. 13 - Prob. 111ACh. 13 - Prob. 112ACh. 13 - Prob. 113ACh. 13 - Prob. 114ACh. 13 - Prob. 115ACh. 13 - Prob. 116ACh. 13 - Prob. 117ACh. 13 - Prob. 118ACh. 13 - Prob. 119ACh. 13 - Prob. 1STPCh. 13 - Prob. 2STPCh. 13 - Prob. 3STPCh. 13 - Prob. 4STPCh. 13 - Prob. 5STPCh. 13 - Prob. 6STPCh. 13 - Prob. 7STPCh. 13 - Prob. 8STPCh. 13 - Prob. 9STPCh. 13 - Prob. 10STPCh. 13 - Prob. 11STPCh. 13 - Prob. 12STPCh. 13 - Prob. 13STPCh. 13 - Prob. 14STP
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