CHEM 212:CHEMISTSRY V 2
CHEM 212:CHEMISTSRY V 2
8th Edition
ISBN: 9781260304503
Author: SILBERBERG
Publisher: MCGRAW-HILL CUSTOM PUBLISHING
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Chapter 1, Problem 1.59P

(a)

Interpretation Introduction

Interpretation:

The following calculation is to be solved to the correct number of significant figures.

2.420 g+15.6 g4.8 g

Concept introduction:

Significant figures of a number are the digits which carry meaningful contribution to its measurement resolution. The rightmost digit of the quantity is the most uncertain digit. The number of certain and uncertain digit in a quantity is considered as significant figures. The digit with a higher number of significant figures has a higher certainty of measurement.

To determine the number of significant figures in a quantity following steps is followed.

1. The quantity must has a decimal point.

2. Start counting from the left and proceed towards the right until the first nonzero digit is encountered. All nonzero digit and the zeroes between two nonzero digits are considered as significant figures. For example, 0.0000765 has three significant figures and 7009 has four significant figures.

3. Zeroes after a decimal point are significant figures. For example, 42.0 have three significant figures.

4. Trailing zeroes that do nothing but are used to set a decimal point are non-significant figures. However, exponential notation can be used to avoid confusion. For example, 4300 has 3 significant figures. It can be expressed in scientific notation as 4.30×103 or

4.300×103. The number of significant figures in 4.30×103 and 4.300×103 is 3 and 4 respectively.

5. Zeroes present before a trailing decimal point are significant figures. For example, 3200 has only two significant figures but 3200. has 4 significant figures.

Rules to determine significant figures in calculations are as follows:

(1) In multiplication and division operations the result carries the same number of significant figures as the operand or measurement with the fewest significant figures.

(2) In addition and subtraction operations, the result carries the same number of decimal places as the operand or measurement with fewest decimal places.

(3) Exact numbers do not affect the number of significant digits in the final answer.

In mathematical expression which involves mixed operations the result of each intermediate step with proper significant figures. Avoid rounding of the result at intermediate steps. Round off the final answer of the calculation. The rules to round off are as follows:

(1) If the last dropped digit is greater than 5 then increase the preceding digit by 1.

(2) If the last dropped digit is less than 5 then the preceding digit does not change.

(3) If the last digit dropped is 5, then the preceding digit is increased by 1 if it is odd and remains the same if it is even. Also, if 5 is followed by zeroes only then rule (3) is applicable and if it is followed by non-zero digit then rule (1) is applicable.

(a)

Expert Solution
Check Mark

Answer to Problem 1.59P

The answer of the calculation to a correct number of significant figures is 3.8 g.

Explanation of Solution

The given expression is,

2.420 g+15.6 g4.8 g

The least significant digit in each number is underlined as follows:

2.420_ g+15.6_ g4.8_ g

Add 2.420_ g and 15.6_ g in the numerator as:

2.420_ g+15.6_ g=18.0_2 g

The result of the addition operation must carry the same number of decimal places as the measurement with fewest decimal places. Measurement 15.6_ g has one digit after the decimal point. Therefore, the least significant figure in the result is 0.

Divide 18.02 g by 4.8 g as:

18.02 g4.8 g=3.7_54 g=3.8 g

The result of the division operation must carry the same number of significant figures as the measurement with the fewest significant figures. 18.02 g has 4 significant figures and 4.8 g has 2 significant figures. The result must have 2 significant figures. Therefore, the final answer is rounded off to the 2 significant figures.

The last digit to be dropped is 5 and the preceding digit is odd thus it is increased by 1.

Conclusion

The final answer is rounded off to 2 significant figures. The answer of the calculation to a correct number of significant figures is 3.8 g.

(b)

Interpretation Introduction

Interpretation:

The following calculation is to be solved to the correct number of significant figures.

7.87 mL16.1 mL8.44 mL

Concept introduction:

Significant figures of a number are the digits which carry meaningful contribution to its measurement resolution. The rightmost digit of the quantity is the most uncertain digit.

The number of certain and uncertain digit in a quantity is considered as significant figures. The digit with a higher number of significant figures has a higher certainty of measurement.

To determine the number of significant figures in a quantity following steps is followed.

1. The quantity must has a decimal point.

2. Start counting from the left and proceed towards the right until the first nonzero digit is encountered. All nonzero digit and the zeroes between two nonzero digits are considered as significant figures. For example, 0.0000765 has three significant figures and 7009 has four significant figures.

3. Zeroes after a decimal point are significant figures. For example, 42.0 have three significant figures.

4. Trailing zeroes that do nothing but are used to set a decimal point are non-significant figures. However, exponential notation can be used to avoid confusion. For example, 4300 has 3 significant figures. It can be expressed in scientific notation as 4.30×103 or

4.300×103. The number of significant figures in 4.30×103 and 4.300×103 is 3 and 4 respectively.

5. Zeroes present before a trailing decimal point are significant figures. For example, 3200 has only two significant figures but 3200. has 4 significant figures.

Rules to determine significant figures in calculations are as follows:

(1) In multiplication and division operations the result carries the same number of significant figures as the operand or measurement with the fewest significant figures.

(2) In addition and subtraction operations, the result carries the same number of decimal places as the operand or measurement with fewest decimal places.

(3) Exact numbers do not affect the number of significant digits in the final answer.

In mathematical expression which involves mixed operations the result of each intermediate step with proper significant figures. Avoid rounding of the result at intermediate steps. Round off the final answer of the calculation. The rules to round off are as follows:

(1) If the last dropped digit is greater than 5 then increase the preceding digit by 1.

(2) If the last dropped digit is less than 5 then the preceding digit does not change.

(3) If the last digit dropped is 5, then the preceding digit is increased by 1 if it is odd and remains the same if it is even. Also, if 5 is followed by zeroes only then rule (3) is applicable and if it is followed by non-zero digit then rule (1) is applicable.

(b)

Expert Solution
Check Mark

Answer to Problem 1.59P

The answer of the calculation to a correct number of significant figures is 1.0.

Explanation of Solution

The given expression is,

7.87 mL16.1 mL8.44 mL

The least significant digit in each number is underlined as follows:

7.87_ mL16.1_ mL8.44_ mL

Subtract 8.44 mL from 16.1 mL in the denominator as,

16.1_ mL8.44_ mL=7.6_6 mL

The result of the subtraction operation must carry the same number of decimal places as the measurement with fewest decimal places. Measurement 16.1 mL has one digit after the decimal point. Therefore, the result must also contain only one digit after the decimal point. The least significant figure in the result is 6.

Divide 7.87 mL by 7.66 mL.

7.87_ mL7.6_6 mL=1.0_274=1.0

The result of the division operation must carry the same number of significant figures as the measurement with the fewest significant figures. 7.6_6 mL has 2 significant figures and 7.87_ mL has 3 significant figures. The result must have 2 significant figures. Therefore, the final answer must be rounded off to the 2 significant figures. The last digit to be dropped is 2. 2 is less than 5, therefore, the preceding digit remains the same.

Conclusion

The final answer is rounded off to 2 significant figures. The answer of the calculation to a correct number of significant figures is 1.0.

(c)

Interpretation Introduction

Interpretation:

The following calculation is to be solved to the correct number of significant figures.

V=πr2h

Concept introduction:

Significant figures of a number are the digits which carry meaningful contribution to its measurement resolution. The rightmost digit of the quantity is the most uncertain digit.

The number of certain and uncertain digit in a quantity is considered as significant figures. The digit with a higher number of significant figures has a higher certainty of measurement.

To determine the number of significant figures in a quantity following steps is followed.

1. The quantity must has a decimal point.

2. Start counting from the left and proceed towards the right until the first nonzero digit is encountered. All nonzero digit and the zeroes between two nonzero digits are considered as significant figures. For example, 0.0000765 has three significant figures and 7009 has four significant figures.

3. Zeroes after a decimal point are significant figures. For example, 42.0 have three significant figures.

4. Trailing zeroes that do nothing but are used to set a decimal point are non-significant figures. However, exponential notation can be used to avoid confusion. For example, 4300 has 3 significant figures. It can be expressed in scientific notation as 4.30×103 or

4.300×103. The number of significant figures in 4.30×103 and 4.300×103 is 3 and 4 respectively.

5. Zeroes present before a trailing decimal point are significant figures. For example, 3200 has only two significant figures but 3200. has 4 significant figures.

Rules to determine significant figures in calculations are as follows:

(1) In multiplication and division operations the result carries the same number of significant figures as the operand or measurement with the fewest significant figures.

(2) In addition and subtraction operations, the result carries the same number of decimal places as the operand or measurement with fewest decimal places.

(3) Exact numbers do not affect the number of significant digits in the final answer.

In mathematical expression which involves mixed operations the result of each intermediate step with proper significant figures. Avoid rounding of the result at intermediate steps. Round off the final answer of the calculation. The rules to round off are as follows:

(1) If the last dropped digit is greater than 5 then increase the preceding digit by 1.

(2) If the last dropped digit is less than 5 then the preceding digit does not change.

(3) If the last digit dropped is 5, then the preceding digit is increased by 1 if it is odd and remains the same if it is even. Also, if 5 is followed by zeroes only then rule (3) is applicable and if it is followed by non-zero digit then rule (1) is applicable.

(c)

Expert Solution
Check Mark

Answer to Problem 1.59P

The answer of the calculation to a correct number of significant figures is 565 cm3.

Explanation of Solution

The value of r is 6.23 cm and the value of h is 4.630 cm.

The formula to calculate the volume is,

V=πr2h

Substitute 6.23 cm for r, 4.630 cm for h and 3.14159 for π in the above equation.

V=(3.14159_)(6.23_ cm)2(4.630_ cm)=564_.556 cm3=565 cm3

The result of the multiplication and division operation must carry the same number of significant figures as the measurement with the fewest significant figures. 6.23 cm has 3 significant figures, 4.630 cm has 4 significant figures and 3.14159 has 6 significant figures. The result must have 3 significant figures. Therefore, the final answer must be rounded off to the 3 significant figures. The last digit to be dropped is 5 thus the preceding digit is increased by 1.

Conclusion

The final answer is rounded off to 3 significant figures. The answer of the calculation to a correct number of significant figures is 565 cm3.

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Chapter 1 Solutions

CHEM 212:CHEMISTSRY V 2

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