Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
15th Edition
ISBN: 9781119231318
Author: Morris Hein
Publisher: Wiley (WileyPLUS Products)
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Chapter 2, Problem 13PE

(a)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for 12.62+1.5+0.25 and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Significant figures:  The digits that are measured in a number including one estimated digit are called as significant figures.

The answers for those calculations that involve addition and subtraction will have the same number of decimal places as the original number with the fewest decimal places.

(a)

Expert Solution
Check Mark

Answer to Problem 13PE

The answer for the sum of 12.62+1.5+0.25 is 14.4.

Explanation of Solution

The calculation for 12.62+1.5+0.25 is,

  12.621.50.25_14.37

The number with the least precision is 0.25.  Therefore, the answer is rounded off to the nearest tenth: 14.4.  The answer for the sum of 12.62+1.5+0.25 is 14.4.

(b)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for (2.25×103)(4.80×104) and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Significant figures:  The digits that are measured in a number including one estimated digit are called as significant figures.

The answers for those calculations that involve multiplication and division will have the same number of significant figures as the original number with the fewest number of significant figures.

(b)

Expert Solution
Check Mark

Answer to Problem 13PE

The answer for the product of (2.25×103)(4.80×104) is 1.08×108.

Explanation of Solution

The product of (2.25×103)(4.80×104) is 10.8×107.  The answer should have three significant figures because both the numbers ((2.25×103)(4.80×104)), with the fewest significant figures, has only three significant figures.

The decimal point is placed between 1 and 0.  Since it’s shifted towards left, the number 10.8 is multiplied by 108.  The answer for the product of (2.25×103)(4.80×104) is 1.08×108.

(c)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for (452)(6.2)14.3 and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Refer to part (b).

(c)

Expert Solution
Check Mark

Answer to Problem 13PE

The answer for (452)(6.2)14.3 is 2.0×102.

Explanation of Solution

The answer for (452)(6.2)14.3 is 195.97.  The answer should have only two significant figures, since the number 6.2, with the fewest significant figures, has only two significant figures.

In the number 195.97, the last three digits are dropped and one is added to the next digit 9.  The answer for (452)(6.2)14.3 is 2.0×102.

(d)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for (0.0394)(12.8) and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Refer to part (b).

(d)

Expert Solution
Check Mark

Answer to Problem 13PE

The answer for (0.0394)(12.8) is 0.504.

Explanation of Solution

The product of (0.0394)(12.8) is 0.50432.  The answer should have only three significant figures, since the number 12.8, with fewest significant figures, has only three significant figures.

In the number 0.50432, the last two digits 3,2 are dropped.  The answer for (0.0394)(12.8) is 0.504.

(e)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for 0.427859.6 and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Refer to part (b).

(e)

Expert Solution
Check Mark

Answer to Problem 13PE

The answer for 0.427859.6 is 7.18×10-3.

Explanation of Solution

The answer for 0.427859.6 is 0.007177.  The answer should have only three significant figures since the number 59.6, with fewest significant figures, have only three significant figures.

In the number 0.007177, the last digit is dropped and one is added to the next digit 7.  The rounded off three significant number is 0.00718.  The answer for 0.427859.6 is 7.18×10-3.

(f)

Interpretation Introduction

Interpretation:

Calculations have to be carried out for 10.4+(3.75)(1.5×104) and the answer has to be provided using proper number of significant figures.

Concept Introduction:

Refer to part (b).

(f)

Expert Solution
Check Mark

Answer to Problem 13PE

The answer for 10.4+(3.75)(1.5×104) is 5.6×104.

Explanation of Solution

The answer for 10.4+(3.75)(1.5×104) is 56260.4.  The answer should have only three significant figures, since 1.5×104, with the fewest significant figures, has only three significant figures.

The decimal point is shifted between 5 and 6, the last few digits are dropped and the number is multiplied by 104.  The answer for 10.4+(3.75)(1.5×104) is 5.6×104.

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Chapter 2 Solutions

Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card

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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