EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
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Chapter 12, Problem 4PE

(a)

Interpretation Introduction

Interpretation:

Conversion of 649 torr pressure into kPa has to be determined.

Concept Introduction:

Barometer is used for measurement of pressure of gas in atmospheric units (atm). Pressure can also be represented in many units such as torr, mm Hg, and pascal (Pa). These units can be related to one another as follows:

  1 atm=760 torr=760 mm Hg=101325 Pa=101.325 kPa

(a)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 649 torr in kPa units is 86.5 kPa.

Explanation of Solution

Conversion factor to convert torr to kPa is as follows:

  (101.325 kPa760 torr)

Pressure 649 torr can be converted to kPa as follows:

  Pressure(kPa)=(649 torr)(101.325 kPa760 torr)=86.5 kPa

Hence 649 torr in kPa units is 86.5 kPa.

(b)

Interpretation Introduction

Interpretation:

Conversion of 5.07 kPa pressure into atm has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 5.07 kPa in atm units is 0.0500 atm.

Explanation of Solution

Conversion factor to convert kPa to atm is as follows:

  (1 atm101.325 kPa)

Pressure 5.07 kPa can be converted to kPa as follows:

  Pressure(atm)=(5.07 kPa)(1 atm101.325 kPa)=0.0500 atm

Hence 5.07 kPa in atm units is 0.0500 atm.

(c)

Interpretation Introduction

Interpretation:

Conversion of 3.64 atm pressure into mm Hg has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 3.64 atm in mm Hg units is 2.77×103 mm Hg.

Explanation of Solution

Conversion factor to convert atm to mm Hg is as follows:

  (760 mm Hg1 atm)

Pressure 3.64 atm can be converted to mm Hg as follows:

  Pressure(mm Hg)=(3.64 atm)(760 mm Hg1 atm)=2.77×103 mm Hg

Hence 3.64 atm in mm Hg units is 2.77×103 mm Hg.

(d)

Interpretation Introduction

Interpretation:

Conversion of 803 torr pressure into atm has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 803 torr in atm units is 1.06 atm.

Explanation of Solution

Conversion factor to convert torr to atm is as follows:

  (1 atm760 torr)

Pressure 803 torr can be converted to atm as follows:

  Pressure(atm)=(803 torr)(1 atm760 torr)=1.06 atm

Hence 803 torr in atm units is 1.06 atm.

(e)

Interpretation Introduction

Interpretation:

Conversion of 1.08 atm pressure into cm Hg has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 1.08 atm in cm Hg units is 82.1 cm Hg.

Explanation of Solution

Conversion factor to convert atm to mm Hg is as follows:

  (760 mm Hg1 atm)

Pressure 1.08 atm can be converted to cm Hg as follows:

  Pressure(cm Hg)=(1.08 atm)(760 mm Hg1 atm)(1 cm Hg10 mm Hg)=82.1 cm Hg

Hence 1.08 atm in cm Hg units is 82.1 cm Hg.

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Chapter 12 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 12.8 - Prob. 12.11PCh. 12.9 - Prob. 12.12PCh. 12.9 - Prob. 12.13PCh. 12 - Prob. 1RQCh. 12 - Prob. 2RQCh. 12 - Prob. 3RQCh. 12 - Prob. 4RQCh. 12 - Prob. 5RQCh. 12 - Prob. 6RQCh. 12 - Prob. 7RQCh. 12 - Prob. 8RQCh. 12 - Prob. 9RQCh. 12 - Prob. 10RQCh. 12 - Prob. 11RQCh. 12 - Prob. 12RQCh. 12 - Prob. 13RQCh. 12 - Prob. 14RQCh. 12 - Prob. 15RQCh. 12 - Prob. 16RQCh. 12 - Prob. 17RQCh. 12 - Prob. 18RQCh. 12 - Prob. 19RQCh. 12 - Prob. 20RQCh. 12 - Prob. 21RQCh. 12 - Prob. 22RQCh. 12 - Prob. 23RQCh. 12 - Prob. 24RQCh. 12 - Prob. 25RQCh. 12 - Prob. 26RQCh. 12 - Prob. 1PECh. 12 - Prob. 2PECh. 12 - Prob. 3PECh. 12 - Prob. 4PECh. 12 - Prob. 5PECh. 12 - Prob. 6PECh. 12 - Prob. 7PECh. 12 - Prob. 8PECh. 12 - Prob. 9PECh. 12 - Prob. 10PECh. 12 - Prob. 11PECh. 12 - Prob. 12PECh. 12 - Prob. 13PECh. 12 - Prob. 14PECh. 12 - Prob. 15PECh. 12 - Prob. 16PECh. 12 - Prob. 17PECh. 12 - Prob. 18PECh. 12 - Prob. 19PECh. 12 - Prob. 20PECh. 12 - Prob. 21PECh. 12 - Prob. 22PECh. 12 - Prob. 23PECh. 12 - Prob. 24PECh. 12 - Prob. 25PECh. 12 - Prob. 26PECh. 12 - Prob. 27PECh. 12 - Prob. 28PECh. 12 - Prob. 29PECh. 12 - Prob. 30PECh. 12 - Prob. 31PECh. 12 - Prob. 32PECh. 12 - Prob. 33PECh. 12 - Prob. 34PECh. 12 - Prob. 35PECh. 12 - Prob. 36PECh. 12 - Prob. 37PECh. 12 - Prob. 38PECh. 12 - Prob. 39PECh. 12 - Prob. 40PECh. 12 - Prob. 41PECh. 12 - Prob. 42PECh. 12 - Prob. 43PECh. 12 - Prob. 44PECh. 12 - Prob. 45PECh. 12 - Prob. 46PECh. 12 - Prob. 47PECh. 12 - Prob. 48PECh. 12 - Prob. 49PECh. 12 - Prob. 50PECh. 12 - Prob. 51PECh. 12 - Prob. 52PECh. 12 - Prob. 53PECh. 12 - Prob. 54PECh. 12 - Prob. 55AECh. 12 - Prob. 56AECh. 12 - Prob. 57AECh. 12 - Prob. 58AECh. 12 - Prob. 59AECh. 12 - Prob. 60AECh. 12 - Prob. 61AECh. 12 - Prob. 62AECh. 12 - Prob. 63AECh. 12 - Prob. 64AECh. 12 - Prob. 65AECh. 12 - Prob. 66AECh. 12 - Prob. 67AECh. 12 - Prob. 68AECh. 12 - Prob. 69AECh. 12 - Prob. 70AECh. 12 - Prob. 71AECh. 12 - Prob. 72AECh. 12 - Prob. 73AECh. 12 - Prob. 74AECh. 12 - Prob. 75AECh. 12 - Prob. 76AECh. 12 - Prob. 77AECh. 12 - Prob. 78AECh. 12 - Prob. 79AECh. 12 - Prob. 80AECh. 12 - Prob. 81AECh. 12 - Prob. 82AECh. 12 - Prob. 83AECh. 12 - Prob. 84AECh. 12 - Prob. 85AECh. 12 - Prob. 86AECh. 12 - Prob. 87AECh. 12 - Prob. 88AECh. 12 - Prob. 89AECh. 12 - Prob. 90AECh. 12 - Prob. 91AECh. 12 - Prob. 92AECh. 12 - Prob. 93AECh. 12 - Prob. 94CECh. 12 - Prob. 95CECh. 12 - Prob. 96CECh. 12 - Prob. 97CECh. 12 - Prob. 98CECh. 12 - Prob. 99CECh. 12 - Prob. 100CECh. 12 - Prob. 101CE
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