You set out to determine the density of lead in the laboratory. Using a top loading balance to determine the mass and the water displacement method (Study Question 41) to determine the volume of a variety of pieces of lead, you calculate the following densities: 11.6 g/cm 3 , 11.8 g/cm 3 , 11.5 g/cm 3 , and 12.0 g/cm 3 . You consult a reference book and find that the accepted value for the density of lead is 11.3 g/cm 3 . Calculate your average value, percent error, and standard deviation of your results.

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Answer 1

Textbook Question

Chapter 1, Problem 62RGQ

You set out to determine the density of lead in the laboratory. Using a top loading balance to determine the mass and the water displacement method (Study Question 41) to determine the volume of a variety of pieces of lead, you calculate the following densities: 11.6 g/cm³, 11.8 g/cm³, 11.5 g/cm³, and 12.0 g/cm³. You consult a reference book and find that the accepted value for the density of lead is 11.3 g/cm³. Calculate your average value, percent error, and standard deviation of your results.

Expert Solution & Answer

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$Percent error = Error in measurementAccepted value× 100%$

$Error in measurement = Experimentally determined value - Accepted value.$

Answer to Problem 62RGQ

The average value is $11.725 g/cm3.$

The percent error obtained is $3.7%$

The obtained Standard deviation is $0.19g/cm3$

The average value, percent error and standard deviation is calculated as,

Explanation of Solution

Average density Value:

$Average density : 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.$

Hence, the average density value is obtained as shown above. Hence, the average value is $11.725 g/cm3.$

Percent error:

$Error in measurement = 11.725 g/cm3 - 11.3 g/cm3Percent error = 11.725 g/cm3 - 11.3 g/cm311.3 g/cm3× 100% = 3.7%$

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

$SD =Σ|x-μ|2Nwhere, x is values in the data set, μ is average of the data set (mean value), N is no.of data points.Given : Find mean value (μ):μ = 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.DeterminationMeasured valuesDifference between Measurement and Average(x-μ)Square of Difference|x-μ|2111.6-0.1250.016211.80.0750.006311.5-0.2250.051412.00.2750.076Σ|x-μ|2=0.016 + 0.006 + 0.051 + 0.076 = 0.149SD=0.1494= 0.19g/cm3$

The obtained Standard deviation is $0.19g/cm3$

Conclusion

The average value was calculated as $11.725 g/cm3.$

The percent error obtained was calculated as $3.7%$

The obtained Standard deviation was calculated as $0.19g/cm3$

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Answer 2

Textbook Question

Chapter 1, Problem 62RGQ

You set out to determine the density of lead in the laboratory. Using a top loading balance to determine the mass and the water displacement method (Study Question 41) to determine the volume of a variety of pieces of lead, you calculate the following densities: 11.6 g/cm³, 11.8 g/cm³, 11.5 g/cm³, and 12.0 g/cm³. You consult a reference book and find that the accepted value for the density of lead is 11.3 g/cm³. Calculate your average value, percent error, and standard deviation of your results.

Expert Solution & Answer

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$Percent error = Error in measurementAccepted value× 100%$

$Error in measurement = Experimentally determined value - Accepted value.$

Answer to Problem 62RGQ

The average value is $11.725 g/cm3.$

The percent error obtained is $3.7%$

The obtained Standard deviation is $0.19g/cm3$

The average value, percent error and standard deviation is calculated as,

Explanation of Solution

Average density Value:

$Average density : 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.$

Hence, the average density value is obtained as shown above. Hence, the average value is $11.725 g/cm3.$

Percent error:

$Error in measurement = 11.725 g/cm3 - 11.3 g/cm3Percent error = 11.725 g/cm3 - 11.3 g/cm311.3 g/cm3× 100% = 3.7%$

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

$SD =Σ|x-μ|2Nwhere, x is values in the data set, μ is average of the data set (mean value), N is no.of data points.Given : Find mean value (μ):μ = 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.DeterminationMeasured valuesDifference between Measurement and Average(x-μ)Square of Difference|x-μ|2111.6-0.1250.016211.80.0750.006311.5-0.2250.051412.00.2750.076Σ|x-μ|2=0.016 + 0.006 + 0.051 + 0.076 = 0.149SD=0.1494= 0.19g/cm3$

The obtained Standard deviation is $0.19g/cm3$

Conclusion

The average value was calculated as $11.725 g/cm3.$

The percent error obtained was calculated as $3.7%$

The obtained Standard deviation was calculated as $0.19g/cm3$

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Answer 3

Textbook Question

Chapter 1, Problem 62RGQ

You set out to determine the density of lead in the laboratory. Using a top loading balance to determine the mass and the water displacement method (Study Question 41) to determine the volume of a variety of pieces of lead, you calculate the following densities: 11.6 g/cm³, 11.8 g/cm³, 11.5 g/cm³, and 12.0 g/cm³. You consult a reference book and find that the accepted value for the density of lead is 11.3 g/cm³. Calculate your average value, percent error, and standard deviation of your results.

Expert Solution & Answer

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$Percent error = Error in measurementAccepted value× 100%$

$Error in measurement = Experimentally determined value - Accepted value.$

Answer to Problem 62RGQ

The average value is $11.725 g/cm3.$

The percent error obtained is $3.7%$

The obtained Standard deviation is $0.19g/cm3$

The average value, percent error and standard deviation is calculated as,

Explanation of Solution

Average density Value:

$Average density : 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.$

Hence, the average density value is obtained as shown above. Hence, the average value is $11.725 g/cm3.$

Percent error:

$Error in measurement = 11.725 g/cm3 - 11.3 g/cm3Percent error = 11.725 g/cm3 - 11.3 g/cm311.3 g/cm3× 100% = 3.7%$

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

$SD =Σ|x-μ|2Nwhere, x is values in the data set, μ is average of the data set (mean value), N is no.of data points.Given : Find mean value (μ):μ = 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.DeterminationMeasured valuesDifference between Measurement and Average(x-μ)Square of Difference|x-μ|2111.6-0.1250.016211.80.0750.006311.5-0.2250.051412.00.2750.076Σ|x-μ|2=0.016 + 0.006 + 0.051 + 0.076 = 0.149SD=0.1494= 0.19g/cm3$

The obtained Standard deviation is $0.19g/cm3$

Conclusion

The average value was calculated as $11.725 g/cm3.$

The percent error obtained was calculated as $3.7%$

The obtained Standard deviation was calculated as $0.19g/cm3$

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 1, Problem 62RGQ

You set out to determine the density of lead in the laboratory. Using a top loading balance to determine the mass and the water displacement method (Study Question 41) to determine the volume of a variety of pieces of lead, you calculate the following densities: 11.6 g/cm³, 11.8 g/cm³, 11.5 g/cm³, and 12.0 g/cm³. You consult a reference book and find that the accepted value for the density of lead is 11.3 g/cm³. Calculate your average value, percent error, and standard deviation of your results.

Expert Solution & Answer

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$Percent error = Error in measurementAccepted value× 100%$

$Error in measurement = Experimentally determined value - Accepted value.$

Answer to Problem 62RGQ

The average value is $11.725 g/cm3.$

The percent error obtained is $3.7%$

The obtained Standard deviation is $0.19g/cm3$

The average value, percent error and standard deviation is calculated as,

Explanation of Solution

Average density Value:

$Average density : 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.$

Hence, the average density value is obtained as shown above. Hence, the average value is $11.725 g/cm3.$

Percent error:

$Error in measurement = 11.725 g/cm3 - 11.3 g/cm3Percent error = 11.725 g/cm3 - 11.3 g/cm311.3 g/cm3× 100% = 3.7%$

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

$SD =Σ|x-μ|2Nwhere, x is values in the data set, μ is average of the data set (mean value), N is no.of data points.Given : Find mean value (μ):μ = 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.DeterminationMeasured valuesDifference between Measurement and Average(x-μ)Square of Difference|x-μ|2111.6-0.1250.016211.80.0750.006311.5-0.2250.051412.00.2750.076Σ|x-μ|2=0.016 + 0.006 + 0.051 + 0.076 = 0.149SD=0.1494= 0.19g/cm3$

The obtained Standard deviation is $0.19g/cm3$

Conclusion

The average value was calculated as $11.725 g/cm3.$

The percent error obtained was calculated as $3.7%$

The obtained Standard deviation was calculated as $0.19g/cm3$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$Percent error = Error in measurementAccepted value× 100%$

$Error in measurement = Experimentally determined value - Accepted value.$

Answer to Problem 62RGQ

The average value is $11.725 g/cm3.$

The percent error obtained is $3.7%$

The obtained Standard deviation is $0.19g/cm3$

The average value, percent error and standard deviation is calculated as,

Explanation of Solution

Average density Value:

$Average density : 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.$

Hence, the average density value is obtained as shown above. Hence, the average value is $11.725 g/cm3.$

Percent error:

$Error in measurement = 11.725 g/cm3 - 11.3 g/cm3Percent error = 11.725 g/cm3 - 11.3 g/cm311.3 g/cm3× 100% = 3.7%$

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

$SD =Σ|x-μ|2Nwhere, x is values in the data set, μ is average of the data set (mean value), N is no.of data points.Given : Find mean value (μ):μ = 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.DeterminationMeasured valuesDifference between Measurement and Average(x-μ)Square of Difference|x-μ|2111.6-0.1250.016211.80.0750.006311.5-0.2250.051412.00.2750.076Σ|x-μ|2=0.016 + 0.006 + 0.051 + 0.076 = 0.149SD=0.1494= 0.19g/cm3$

The obtained Standard deviation is $0.19g/cm3$

Conclusion

The average value was calculated as $11.725 g/cm3.$

The percent error obtained was calculated as $3.7%$

The obtained Standard deviation was calculated as $0.19g/cm3$

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$Percent error = Error in measurementAccepted value× 100%$

$Error in measurement = Experimentally determined value - Accepted value.$

Answer to Problem 62RGQ

The average value is $11.725 g/cm3.$

The percent error obtained is $3.7%$

The obtained Standard deviation is $0.19g/cm3$

The average value, percent error and standard deviation is calculated as,

Explanation of Solution

Average density Value:

$Average density : 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.$

Hence, the average density value is obtained as shown above. Hence, the average value is $11.725 g/cm3.$

Percent error:

$Error in measurement = 11.725 g/cm3 - 11.3 g/cm3Percent error = 11.725 g/cm3 - 11.3 g/cm311.3 g/cm3× 100% = 3.7%$

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

$SD =Σ|x-μ|2Nwhere, x is values in the data set, μ is average of the data set (mean value), N is no.of data points.Given : Find mean value (μ):μ = 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.DeterminationMeasured valuesDifference between Measurement and Average(x-μ)Square of Difference|x-μ|2111.6-0.1250.016211.80.0750.006311.5-0.2250.051412.00.2750.076Σ|x-μ|2=0.016 + 0.006 + 0.051 + 0.076 = 0.149SD=0.1494= 0.19g/cm3$

The obtained Standard deviation is $0.19g/cm3$

Conclusion

The average value was calculated as $11.725 g/cm3.$

The percent error obtained was calculated as $3.7%$

The obtained Standard deviation was calculated as $0.19g/cm3$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation: The average, percent error and standard deviation has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$Percent error = Error in measurementAccepted value× 100%$

$Error in measurement = Experimentally determined value - Accepted value.$

Answer 12

Answer to Problem 62RGQ

The average value is $11.725 g/cm3.$

The percent error obtained is $3.7%$

The obtained Standard deviation is $0.19g/cm3$

The average value, percent error and standard deviation is calculated as,

Answer 13

Explanation of Solution

Average density Value:

$Average density : 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.$

Hence, the average density value is obtained as shown above. Hence, the average value is $11.725 g/cm3.$

Percent error:

$Error in measurement = 11.725 g/cm3 - 11.3 g/cm3Percent error = 11.725 g/cm3 - 11.3 g/cm311.3 g/cm3× 100% = 3.7%$

From the above calculation, the percent error is obtained as shown above.

Standard deviation:

$SD =Σ|x-μ|2Nwhere, x is values in the data set, μ is average of the data set (mean value), N is no.of data points.Given : Find mean value (μ):μ = 11.6+ 11.8 + 11.5 + 12.04 = 11.725 g/cm3.DeterminationMeasured valuesDifference between Measurement and Average(x-μ)Square of Difference|x-μ|2111.6-0.1250.016211.80.0750.006311.5-0.2250.051412.00.2750.076Σ|x-μ|2=0.016 + 0.006 + 0.051 + 0.076 = 0.149SD=0.1494= 0.19g/cm3$