The absorbance detection limit of the sample has to be calculated. Concept Introduction:- Detection limit :- The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank. The minimum detectable concentration y dl can be defined as, signal detection limit y dl = y blank + 3s Where, s = Standard deviation. m = Slope of linear calibration curve.

Question

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Answer 1

Question

Chapter 5, Problem 5.AE

a)

Interpretation Introduction

Interpretation:

The absorbance detection limit of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The minimum detectable concentration $ydl$ can be defined as,

$signal detection limit$ $ydl= yblank+ 3s$

Where,

s = Standard deviation.

m = Slope of linear calibration curve.

a)

Expert Solution

Answer to Problem 5.AE

The absorbance detection limit of the sample has to be calculated as $0.003112$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The calculated standard deviation of $9$ samples is0.000644.

$ydl= yblank+ 3s$

$ydl= 0.00118+ 3(0.00688)=0.003112$

Conclusion

The absorbance detection limit of the sample has to be calculated as $0.003112$ .

b)

Interpretation Introduction

Interpretation:

The minimum detectable concentration of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$minimum detectable concentration = 3sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

b)

Expert Solution

Answer to Problem 5.AE

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$minimum detectable concentration = 3sm =3(0.000644)2.24×104 M-1=8.6×10-8 M$

Conclusion

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

c)

Interpretation Introduction

Interpretation:

The lower limit quantitation of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$lower limit of quantitation = 10sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Lower limit of quantitation:-

In analytical chemistry, the smallest amount of sample which can be measured with reasonable accuracy.

$lower limit of quantitation = 10sm$

c)

Expert Solution

Answer to Problem 5.AE

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$lower limit of quantitation = 10sm =10(0.000644)2.24×104 M-1=2.9×10-7 M$

Conclusion

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

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Answer 2

Question

Chapter 5, Problem 5.AE

a)

Interpretation Introduction

Interpretation:

The absorbance detection limit of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The minimum detectable concentration $ydl$ can be defined as,

$signal detection limit$ $ydl= yblank+ 3s$

Where,

s = Standard deviation.

m = Slope of linear calibration curve.

a)

Expert Solution

Answer to Problem 5.AE

The absorbance detection limit of the sample has to be calculated as $0.003112$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The calculated standard deviation of $9$ samples is0.000644.

$ydl= yblank+ 3s$

$ydl= 0.00118+ 3(0.00688)=0.003112$

Conclusion

The absorbance detection limit of the sample has to be calculated as $0.003112$ .

b)

Interpretation Introduction

Interpretation:

The minimum detectable concentration of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$minimum detectable concentration = 3sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

b)

Expert Solution

Answer to Problem 5.AE

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$minimum detectable concentration = 3sm =3(0.000644)2.24×104 M-1=8.6×10-8 M$

Conclusion

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

c)

Interpretation Introduction

Interpretation:

The lower limit quantitation of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$lower limit of quantitation = 10sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Lower limit of quantitation:-

In analytical chemistry, the smallest amount of sample which can be measured with reasonable accuracy.

$lower limit of quantitation = 10sm$

c)

Expert Solution

Answer to Problem 5.AE

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$lower limit of quantitation = 10sm =10(0.000644)2.24×104 M-1=2.9×10-7 M$

Conclusion

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

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Answer 3

Question

Chapter 5, Problem 5.AE

a)

Interpretation Introduction

Interpretation:

The absorbance detection limit of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The minimum detectable concentration $ydl$ can be defined as,

$signal detection limit$ $ydl= yblank+ 3s$

Where,

s = Standard deviation.

m = Slope of linear calibration curve.

a)

Expert Solution

Answer to Problem 5.AE

The absorbance detection limit of the sample has to be calculated as $0.003112$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The calculated standard deviation of $9$ samples is0.000644.

$ydl= yblank+ 3s$

$ydl= 0.00118+ 3(0.00688)=0.003112$

Conclusion

The absorbance detection limit of the sample has to be calculated as $0.003112$ .

b)

Interpretation Introduction

Interpretation:

The minimum detectable concentration of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$minimum detectable concentration = 3sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

b)

Expert Solution

Answer to Problem 5.AE

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$minimum detectable concentration = 3sm =3(0.000644)2.24×104 M-1=8.6×10-8 M$

Conclusion

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

c)

Interpretation Introduction

Interpretation:

The lower limit quantitation of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$lower limit of quantitation = 10sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Lower limit of quantitation:-

In analytical chemistry, the smallest amount of sample which can be measured with reasonable accuracy.

$lower limit of quantitation = 10sm$

c)

Expert Solution

Answer to Problem 5.AE

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$lower limit of quantitation = 10sm =10(0.000644)2.24×104 M-1=2.9×10-7 M$

Conclusion

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

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Answer 4

Question

Chapter 5, Problem 5.AE

a)

Interpretation Introduction

Interpretation:

The absorbance detection limit of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The minimum detectable concentration $ydl$ can be defined as,

$signal detection limit$ $ydl= yblank+ 3s$

Where,

s = Standard deviation.

m = Slope of linear calibration curve.

a)

Expert Solution

Answer to Problem 5.AE

The absorbance detection limit of the sample has to be calculated as $0.003112$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The calculated standard deviation of $9$ samples is0.000644.

$ydl= yblank+ 3s$

$ydl= 0.00118+ 3(0.00688)=0.003112$

Conclusion

The absorbance detection limit of the sample has to be calculated as $0.003112$ .

b)

Interpretation Introduction

Interpretation:

The minimum detectable concentration of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$minimum detectable concentration = 3sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

b)

Expert Solution

Answer to Problem 5.AE

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$minimum detectable concentration = 3sm =3(0.000644)2.24×104 M-1=8.6×10-8 M$

Conclusion

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

c)

Interpretation Introduction

Interpretation:

The lower limit quantitation of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$lower limit of quantitation = 10sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Lower limit of quantitation:-

In analytical chemistry, the smallest amount of sample which can be measured with reasonable accuracy.

$lower limit of quantitation = 10sm$

c)

Expert Solution

Answer to Problem 5.AE

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$lower limit of quantitation = 10sm =10(0.000644)2.24×104 M-1=2.9×10-7 M$

Conclusion

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

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Answer 5

Chapter 5, Problem 5.AE

a)

Interpretation Introduction

Interpretation:

The absorbance detection limit of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The minimum detectable concentration $ydl$ can be defined as,

$signal detection limit$ $ydl= yblank+ 3s$

Where,

s = Standard deviation.

m = Slope of linear calibration curve.

a)

Expert Solution

Answer to Problem 5.AE

The absorbance detection limit of the sample has to be calculated as $0.003112$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The calculated standard deviation of $9$ samples is0.000644.

$ydl= yblank+ 3s$

$ydl= 0.00118+ 3(0.00688)=0.003112$

Conclusion

The absorbance detection limit of the sample has to be calculated as $0.003112$ .

b)

Interpretation Introduction

Interpretation:

The minimum detectable concentration of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$minimum detectable concentration = 3sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

b)

Expert Solution

Answer to Problem 5.AE

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$minimum detectable concentration = 3sm =3(0.000644)2.24×104 M-1=8.6×10-8 M$

Conclusion

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

c)

Interpretation Introduction

Interpretation:

The lower limit quantitation of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$lower limit of quantitation = 10sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Lower limit of quantitation:-

In analytical chemistry, the smallest amount of sample which can be measured with reasonable accuracy.

$lower limit of quantitation = 10sm$

c)

Expert Solution

Answer to Problem 5.AE

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$lower limit of quantitation = 10sm =10(0.000644)2.24×104 M-1=2.9×10-7 M$

Conclusion

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Answer 6

Chapter 5, Problem 5.AE

a)

Interpretation Introduction

Interpretation:

The absorbance detection limit of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The minimum detectable concentration $ydl$ can be defined as,

$signal detection limit$ $ydl= yblank+ 3s$

Where,

s = Standard deviation.

m = Slope of linear calibration curve.

a)

Expert Solution

Answer to Problem 5.AE

The absorbance detection limit of the sample has to be calculated as $0.003112$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The calculated standard deviation of $9$ samples is0.000644.

$ydl= yblank+ 3s$

$ydl= 0.00118+ 3(0.00688)=0.003112$

Conclusion

The absorbance detection limit of the sample has to be calculated as $0.003112$ .

Answer 7

Chapter 5, Problem 5.AE

a)

Interpretation Introduction

Interpretation:

The absorbance detection limit of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The minimum detectable concentration $ydl$ can be defined as,

$signal detection limit$ $ydl= yblank+ 3s$

Where,

s = Standard deviation.

m = Slope of linear calibration curve.

Answer 8

a)

Expert Solution

Answer to Problem 5.AE

The absorbance detection limit of the sample has to be calculated as $0.003112$

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The calculated standard deviation of $9$ samples is0.000644.

$ydl= yblank+ 3s$

$ydl= 0.00118+ 3(0.00688)=0.003112$

Answer 13

Conclusion

The absorbance detection limit of the sample has to be calculated as $0.003112$ .

Answer 14

b)

Interpretation Introduction

Interpretation:

The minimum detectable concentration of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$minimum detectable concentration = 3sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

b)

Expert Solution

Answer to Problem 5.AE

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$minimum detectable concentration = 3sm =3(0.000644)2.24×104 M-1=8.6×10-8 M$

Conclusion

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Answer 15

b)

Interpretation Introduction

Interpretation:

The minimum detectable concentration of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$minimum detectable concentration = 3sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Answer 16

b)

Expert Solution

Answer to Problem 5.AE

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Answer 17

b)

Expert Solution

Answer 18

b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$minimum detectable concentration = 3sm =3(0.000644)2.24×104 M-1=8.6×10-8 M$

Answer 21

Conclusion

The minimum detectable concentration of the sample has to be calculated as $8.6×10-8M$

Answer 22

c)

Interpretation Introduction

Interpretation:

The lower limit quantitation of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$lower limit of quantitation = 10sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Lower limit of quantitation:-

In analytical chemistry, the smallest amount of sample which can be measured with reasonable accuracy.

$lower limit of quantitation = 10sm$

c)

Expert Solution

Answer to Problem 5.AE

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$lower limit of quantitation = 10sm =10(0.000644)2.24×104 M-1=2.9×10-7 M$

Conclusion

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Answer 23

c)

Interpretation Introduction

Interpretation:

The lower limit quantitation of the sample has to be calculated.

Concept Introduction:-

Detection limit:-

The concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

$lower limit of quantitation = 10sm$

Where,

S = Standard deviation.

m = Slope of linear calibration curve.

Lower limit of quantitation:-

In analytical chemistry, the smallest amount of sample which can be measured with reasonable accuracy.

$lower limit of quantitation = 10sm$

Answer 24

c)

Expert Solution

Answer to Problem 5.AE

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Answer 25

c)

Expert Solution

Answer 26

c)

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

Given,

The low concentrations samples near the detection limit are.

$0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011$

Mean reading of nine blanks are $0.001189$

The slope of the calibration curve is $2.24×104M−1$

The calculated standard deviation of $9$ samples is0.000644.

$lower limit of quantitation = 10sm =10(0.000644)2.24×104 M-1=2.9×10-7 M$

Answer 29

Conclusion

The lower limit of quantitation of the sample has to be calculated as $2.9×10-7M$

Answer 30

Ch. 5.1 - Prob. 1TY Ch. 5.2 - Prob. 2TY Ch. 5.3 - Prob. 3TY Ch. 5.4 - Prob. 5TY Ch. 5 - Prob. 5.AE Ch. 5 - Prob. 5.BE Ch. 5 - Prob. 5.CE Ch. 5 - Prob. 5.DE Ch. 5 - Prob. 5.EE Ch. 5 - Prob. 5.FE

Answer to Problem 5.AE

Explanation of Solution

Answer to Problem 5.AE

Explanation of Solution

Answer to Problem 5.AE

Explanation of Solution

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Chapter 5 Solutions