Foundations of College Chemistry, Binder Ready Version
Foundations of College Chemistry, Binder Ready Version
15th Edition
ISBN: 9781119083900
Author: Morris Hein, Susan Arena, Cary Willard
Publisher: WILEY
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Chapter 2, Problem 22PE

(a)

Interpretation Introduction

Interpretation:

The quantity 35.6m has to be converted into foot.

(a)

Expert Solution
Check Mark

Answer to Problem 22PE

The quantity 35.6m in foot is 117ft.

Explanation of Solution

The given quantity is 35.6m.

Meters is converted into foot using the conversion factor,

  (100cm1m)(1in.2.54cm)(1ft12in.)

The quantity 35.6m is converted as,

  ft=(35.6m)(100cm1m)(1in.2.54cm)(1ft12in.)ft=117ft

The quantity 35.6m in foot is 117ft.

(b)

Interpretation Introduction

Interpretation:

The quantity 16.5km has to be converted into miles.

(b)

Expert Solution
Check Mark

Answer to Problem 22PE

The quantity 16.5km in miles is 10.3 mi.

Explanation of Solution

The given quantity is 16.5km.

Kilometers is converted into miles using the conversion factor,

  (1mi1.609km)

The quantity 16.5km is converted as,

  km=(16.5km)(1mi1.609km)km=10.3km

The quantity 16.5km in miles is 10.3 mi.

(c)

Interpretation Introduction

Interpretation:

The quantity 4.5in.3 has to be converted into millimeter cube (mm3).

(c)

Expert Solution
Check Mark

Answer to Problem 22PE

The quantity 4.5in.3 in millimeter cube is 7.4×104mm3.

Explanation of Solution

The given quantity is 4.5in.3.

Cubic inches is converted into millimeter cube using the conversion factor,

  (2.54cm1in.)3(10mm1cm)3

The quantity 4.5in.3 is converted as,

  mm3=(4.5in.3)(2.54cm1in.)3(10mm1cm)3mm3=7.4×104mm3

The quantity 4.5in.3 in millimeter cube is 7.4×104mm3.

(d)

Interpretation Introduction

Interpretation:

The quantity 95lb has to be converted into grams.

(d)

Expert Solution
Check Mark

Answer to Problem 22PE

The quantity 95lb in grams is 4.3×104g.

Explanation of Solution

The given quantity is 95lb.

Pounds is converted into grams using the conversion factor,

  (453.6 g1lb)

The quantity 95lb is converted as,

  g=(95lb)(453.6 g1lb)g=4.3×104g

The quantity 95lb in grams is 4.3×104g.

(e)

Interpretation Introduction

Interpretation:

The quantity 20.0 gal has to be converted into liters.

(e)

Expert Solution
Check Mark

Answer to Problem 22PE

The quantity 20.0 gal in liters is 75.7L.

Explanation of Solution

The given quantity is 20.0 gal.

Gallons is converted into liters using the conversion factor,

  (4qt1gal)(0.946mL1qt)

The quantity 20.0 gal is converted as,

  L=(20.0gal)(4qt1gal)(0.946mL1qt)L=75.7 L

The quantity 20.0 gal in liters is 75.7L.

(f)

Interpretation Introduction

Interpretation:

The quantity 4.5×104ft3 has to be converted into cubic meter.

(f)

Expert Solution
Check Mark

Answer to Problem 22PE

The quantity 4.5×104ft3 in cubic meter is 1.3×103m3.

Explanation of Solution

The given quantity is 4.5×104ft3.

Cubic feet is converted into meter cube using the conversion factor,

  (12in.1ft)3(2.54cm1in.)3(1m100cm)3

The quantity 4.5×104ft3 is converted as,

  m3=(4.5×104ft3)(12in.1ft)3(2.54cm1in.)3(1m100cm)3m3=1.3×103m3

The quantity 4.5×104ft3 in cubic meter is 1.3×103m3.

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Chapter 2 Solutions

Foundations of College Chemistry, Binder Ready Version

Ch. 2.6 - Prob. 2.11PCh. 2.6 - Prob. 2.12PCh. 2.6 - Prob. 2.13PCh. 2.6 - Prob. 2.14PCh. 2.6 - Prob. 2.15PCh. 2.7 - Prob. 2.16PCh. 2.7 - Prob. 2.17PCh. 2.7 - Prob. 2.18PCh. 2.7 - Prob. 2.19PCh. 2.8 - Prob. 2.20PCh. 2.8 - Prob. 2.21PCh. 2.9 - Prob. 2.22PCh. 2.9 - Prob. 2.23PCh. 2 - Prob. 1RQCh. 2 - Prob. 2RQCh. 2 - Prob. 3RQCh. 2 - Prob. 4RQCh. 2 - Prob. 5RQCh. 2 - Prob. 6RQCh. 2 - Prob. 7RQCh. 2 - Prob. 8RQCh. 2 - Prob. 9RQCh. 2 - Prob. 10RQCh. 2 - Prob. 11RQCh. 2 - Prob. 12RQCh. 2 - Prob. 13RQCh. 2 - Prob. 14RQCh. 2 - Prob. 15RQCh. 2 - Prob. 16RQCh. 2 - Prob. 17RQCh. 2 - Prob. 18RQCh. 2 - Prob. 19RQCh. 2 - Prob. 20RQCh. 2 - Prob. 21RQCh. 2 - Prob. 1PECh. 2 - Prob. 2PECh. 2 - Prob. 3PECh. 2 - Prob. 4PECh. 2 - Prob. 5PECh. 2 - Prob. 6PECh. 2 - Prob. 7PECh. 2 - Prob. 8PECh. 2 - Prob. 9PECh. 2 - Prob. 10PECh. 2 - Prob. 11PECh. 2 - Prob. 12PECh. 2 - Prob. 13PECh. 2 - Prob. 14PECh. 2 - Prob. 15PECh. 2 - Prob. 16PECh. 2 - Prob. 17PECh. 2 - Prob. 18PECh. 2 - Prob. 19PECh. 2 - Prob. 20PECh. 2 - Prob. 21PECh. 2 - Prob. 22PECh. 2 - Prob. 23PECh. 2 - Prob. 24PECh. 2 - Prob. 25PECh. 2 - Prob. 26PECh. 2 - Prob. 27PECh. 2 - Prob. 28PECh. 2 - Prob. 29PECh. 2 - Prob. 30PECh. 2 - Prob. 31PECh. 2 - Prob. 32PECh. 2 - Prob. 33PECh. 2 - Prob. 34PECh. 2 - Prob. 35PECh. 2 - Prob. 36PECh. 2 - Prob. 37PECh. 2 - Prob. 38PECh. 2 - Prob. 39PECh. 2 - Prob. 40PECh. 2 - Prob. 41PECh. 2 - Prob. 42PECh. 2 - Prob. 43PECh. 2 - Prob. 44PECh. 2 - Prob. 45PECh. 2 - Prob. 46PECh. 2 - Prob. 47PECh. 2 - Prob. 48PECh. 2 - Prob. 49PECh. 2 - Prob. 50PECh. 2 - Prob. 51PECh. 2 - Prob. 52PECh. 2 - Prob. 53PECh. 2 - Prob. 54PECh. 2 - Prob. 55PECh. 2 - Prob. 56PECh. 2 - Prob. 57PECh. 2 - Prob. 58PECh. 2 - Prob. 59PECh. 2 - Prob. 60PECh. 2 - Prob. 61PECh. 2 - Prob. 62PECh. 2 - Prob. 63PECh. 2 - Prob. 64PECh. 2 - Prob. 65PECh. 2 - Prob. 66PECh. 2 - Prob. 67PECh. 2 - Prob. 68PECh. 2 - Prob. 69PECh. 2 - Prob. 70PECh. 2 - Prob. 71AECh. 2 - Prob. 72AECh. 2 - Prob. 73AECh. 2 - Prob. 74AECh. 2 - Prob. 75AECh. 2 - Prob. 76AECh. 2 - Prob. 77AECh. 2 - Prob. 78AECh. 2 - Prob. 79AECh. 2 - Prob. 80AECh. 2 - Prob. 81AECh. 2 - Prob. 82AECh. 2 - Prob. 83AECh. 2 - Prob. 84AECh. 2 - Prob. 85AECh. 2 - Prob. 86AECh. 2 - Prob. 87AECh. 2 - Prob. 88AECh. 2 - Prob. 89AECh. 2 - Prob. 90AECh. 2 - Prob. 91AECh. 2 - Prob. 92AECh. 2 - Prob. 93AECh. 2 - Prob. 94AECh. 2 - Prob. 95AECh. 2 - Prob. 96AECh. 2 - Prob. 97AECh. 2 - Prob. 98AECh. 2 - Prob. 99AECh. 2 - Prob. 100AECh. 2 - Prob. 101AECh. 2 - Prob. 102AECh. 2 - Prob. 103AECh. 2 - Prob. 104AECh. 2 - Prob. 105AECh. 2 - Prob. 106CECh. 2 - Prob. 108CECh. 2 - Prob. 109CECh. 2 - Prob. 110CECh. 2 - Prob. 111CECh. 2 - Prob. 112CE
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