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
Question
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Chapter 2.3, Problem 43PP

(a)

Interpretation Introduction

Interpretation:

The answer of given problem needs to be round off to the correct significant figures.

4.84m÷2.4s

Concept introduction:

In order to determine the number of significant figures, the following rules should be applied,

1. The non-zero digits are always significant.

2. In a number, the zeros to the left of the first non-zero digit are not significant.

3. Zeros between non zero digits are significant

4. Zeros to the right of the decimal place are significant

5. If a number ends in zero that are not to the right of a decimal, the zeros may or may not be significant.

To round off a number, look at the digit up to which number needs to be rounded off. If the digit right to it is less than 5 simply add 1 to the previous digit and remove all the digits after it from the number. If the digit right to the digit up to which rounding needs to be done is less than 5, just remove all the digits after it from the number.

Now, if the digit is equal to 5 then 1 is added to the previous digit if it is an odd number. If the previous digit is an even number, simply remove all the digits after it from the number.

(a)

Expert Solution
Check Mark

Answer to Problem 43PP

The final answer of the given problem is correct to 2.0m/s.

Explanation of Solution

When dividing the numbers, answer must have the same number of the significant figures as the measurement with the lowest number of significant figures. The lowest number of the significant figure from the given two values is 2. Hence, the solution must contain 2 significant figures.

4.84m÷2.4s=4.84m2.4s=2.01666m/s=2.0m/s

Hence, the correct value is 2.0m/s.

(b)

Interpretation Introduction

Interpretation:

The answer of given problem needs to be round off to the correct significant figures.

60.2m÷20.1s

Concept introduction:

In order to determine the number of significant figures, the following rules should be applied,

1. The non-zero digits are always significant.

2. In a number, the zeros to the left of the first non-zero digit are not significant.

3. Zeros between non zero digits are significant

4. Zeros to the right of the decimal place are significant

5. If a number ends in zero that are not to the right of a decimal, the zeros may or may not be significant.

To round off a number, look at the digit up to which number needs to be rounded off. If the digit right to it is less than 5 simply add 1 to the previous digit and remove all the digits after it from the number. If the digit right to the digit up to which rounding needs to be done is less than 5, just remove all the digits after it from the number.

Now, if the digit is equal to 5 then 1 is added to the previous digit if it is an odd number. If the previous digit is an even number, simply remove all the digits after it from the number.

(b)

Expert Solution
Check Mark

Answer to Problem 43PP

The final answer of the given problem is correct to 3.00m/s.

Explanation of Solution

When dividing the numbers, answer must have the same number of the significant figures as the measurement with the lowest number of significant figures. The lowest number of the significant figure from the given two values is 3. Hence, the solution must contain 3 significant figures.

60.2m÷20.1s=60.2m20.1s=2.9950m/s=3.00m/s

Hence, the correct value is 3.00m/s.

(c)

Interpretation Introduction

Interpretation:

The answer of given problem needs to be round off to the correct significant figures.

102.4m÷51.2s

Concept introduction:

In order to determine the number of significant figures, the following rules should be applied,

1. The non-zero digits are always significant.

2. In a number, the zeros to the left of the first non-zero digit are not significant.

3. Zeros between non zero digits are significant

4. Zeros to the right of the decimal place are significant

5. If a number ends in zero that are not to the right of a decimal, the zeros may or may not be significant.

To round off a number, look at the digit up to which number needs to be rounded off. If the digit right to it is less than 5 simply add 1 to the previous digit and remove all the digits after it from the number. If the digit right to the digit up to which rounding needs to be done is less than 5, just remove all the digits after it from the number.

Now, if the digit is equal to 5 then 1 is added to the previous digit if it is an odd number. If the previous digit is an even number, simply remove all the digits after it from the number.

(c)

Expert Solution
Check Mark

Answer to Problem 43PP

The final answer of the given problem is correct to 2.00m/s.

Explanation of Solution

When dividing the numbers, answer must have the same number of the significant figures as the measurement with the lowest number of significant figures. The lowest number of the significant figure from the given two values is 3. Hence, the solution must contain 3 significant figures.

102.4m÷51.2s=102.4m51.2s=2.00m/s

Hence, the correct value is 2.00m/s.

(d)

Interpretation Introduction

Interpretation:

The answer of given problem needs to be rounded off to the correct significant figures.

168m÷58s

Concept introduction:

In order to determine the number of significant figures, the following rules should be applied,

1. The non-zero digits are always significant.

2. In a number, the zeros to the left of the first non-zero digit are not significant.

3. Zeros between non zero digits are significant

4. Zeros to the right of the decimal place are significant

5. If a number ends in zero that are not to the right of a decimal, the zeros may or may not be significant.

To round off a number, look at the digit up to which number needs to be rounded off. If the digit right to it is less than 5 simply add 1 to the previous digit and remove all the digits after it from the number. If the digit right to the digit up to which rounding needs to be done is less than 5, just remove all the digits after it from the number.

Now, if the digit is equal to 5 then 1 is added to the previous digit if it is an odd number. If the previous digit is an even number, simply remove all the digits after it from the number.

(d)

Expert Solution
Check Mark

Answer to Problem 43PP

The final answer of the given problem is correct to 2.9m/s.

Explanation of Solution

When dividing the numbers, answer must have the same number of the significant figures as the measurement with the lowest number of significant figures. The lowest number of the significant figure from the given two values is 2. Hence, the solution must contain 2 significant figures.

168m÷58s=168m58s=2.8965m/s=2.9m/s

Hence, the correct value is 2.9m/s.

Chapter 2 Solutions

Chemistry: Matter and Change

Ch. 2.2 - Prob. 11PPCh. 2.2 - Prob. 12PPCh. 2.2 - Prob. 13PPCh. 2.2 - Prob. 14PPCh. 2.2 - Prob. 15PPCh. 2.2 - Prob. 16PPCh. 2.2 - Prob. 17PPCh. 2.2 - Prob. 18PPCh. 2.2 - Prob. 19PPCh. 2.2 - Prob. 20PPCh. 2.2 - Prob. 21PPCh. 2.2 - Prob. 22PPCh. 2.2 - Prob. 23PPCh. 2.2 - Prob. 24SSCCh. 2.2 - Prob. 25SSCCh. 2.2 - Prob. 26SSCCh. 2.2 - Prob. 27SSCCh. 2.2 - Prob. 28SSCCh. 2.2 - Prob. 29SSCCh. 2.2 - Prob. 30SSCCh. 2.2 - Prob. 31SSCCh. 2.3 - Prob. 32PPCh. 2.3 - Prob. 33PPCh. 2.3 - Prob. 34PPCh. 2.3 - Prob. 35PPCh. 2.3 - Prob. 36PPCh. 2.3 - Prob. 37PPCh. 2.3 - Prob. 38PPCh. 2.3 - Prob. 39PPCh. 2.3 - Prob. 40PPCh. 2.3 - Prob. 41PPCh. 2.3 - Prob. 42PPCh. 2.3 - Prob. 43PPCh. 2.3 - Prob. 44PPCh. 2.3 - Prob. 45SSCCh. 2.3 - Prob. 46SSCCh. 2.3 - Prob. 47SSCCh. 2.3 - Prob. 48SSCCh. 2.3 - Prob. 49SSCCh. 2.3 - Prob. 50SSCCh. 2.3 - Prob. 51SSCCh. 2.4 - Prob. 52SSCCh. 2.4 - Prob. 53SSCCh. 2.4 - Prob. 54SSCCh. 2.4 - Prob. 55SSCCh. 2.4 - Prob. 56SSCCh. 2.4 - Prob. 57SSCCh. 2.4 - Prob. 58SSCCh. 2 - Prob. 59ACh. 2 - Prob. 60ACh. 2 - Prob. 61ACh. 2 - Prob. 62ACh. 2 - Prob. 63ACh. 2 - Prob. 64ACh. 2 - Prob. 65ACh. 2 - Prob. 66ACh. 2 - Prob. 67ACh. 2 - Prob. 68ACh. 2 - Prob. 69ACh. 2 - Prob. 70ACh. 2 - Prob. 71ACh. 2 - Prob. 72ACh. 2 - Prob. 73ACh. 2 - Prob. 74ACh. 2 - Prob. 75ACh. 2 - Prob. 76ACh. 2 - Prob. 77ACh. 2 - Prob. 78ACh. 2 - Prob. 79ACh. 2 - Prob. 80ACh. 2 - Prob. 81ACh. 2 - Prob. 82ACh. 2 - Prob. 83ACh. 2 - Prob. 84ACh. 2 - Prob. 85ACh. 2 - Prob. 86ACh. 2 - Prob. 87ACh. 2 - Prob. 88ACh. 2 - Prob. 89ACh. 2 - Prob. 90ACh. 2 - Prob. 91ACh. 2 - Prob. 92ACh. 2 - Prob. 93ACh. 2 - Prob. 94ACh. 2 - Prob. 95ACh. 2 - Prob. 96ACh. 2 - Prob. 97ACh. 2 - Prob. 98ACh. 2 - Prob. 99ACh. 2 - Prob. 100ACh. 2 - Prob. 101ACh. 2 - Prob. 102ACh. 2 - Prob. 103ACh. 2 - Prob. 104ACh. 2 - Prob. 105ACh. 2 - Prob. 106ACh. 2 - Prob. 107ACh. 2 - Prob. 108ACh. 2 - Prob. 109ACh. 2 - Prob. 110ACh. 2 - Prob. 111ACh. 2 - Prob. 112ACh. 2 - Prob. 113ACh. 2 - Prob. 114ACh. 2 - Prob. 115ACh. 2 - Prob. 116ACh. 2 - Prob. 117ACh. 2 - Prob. 118ACh. 2 - Prob. 119ACh. 2 - Prob. 120ACh. 2 - Prob. 121ACh. 2 - Prob. 122ACh. 2 - Prob. 123ACh. 2 - Prob. 124ACh. 2 - Prob. 125ACh. 2 - Prob. 126ACh. 2 - Prob. 127ACh. 2 - Prob. 128ACh. 2 - Prob. 129ACh. 2 - Prob. 130ACh. 2 - Prob. 1STPCh. 2 - Which value is NOT equivalent to the others? A....Ch. 2 - Prob. 3STPCh. 2 - Prob. 4STPCh. 2 - Prob. 5STPCh. 2 - Prob. 6STPCh. 2 - Prob. 7STPCh. 2 - Prob. 8STPCh. 2 - Prob. 9STPCh. 2 - Prob. 10STPCh. 2 - Prob. 11STPCh. 2 - Prob. 12STPCh. 2 - Prob. 13STPCh. 2 - Prob. 14STPCh. 2 - Prob. 15STPCh. 2 - Prob. 16STPCh. 2 - Prob. 17STPCh. 2 - Prob. 18STPCh. 2 - Prob. 19STPCh. 2 - Prob. 20STP

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