3. The following molar concentrations were obtained from replicate standardization of a solution: 0.5026, 0.5029, 0.5023, 0.5031, 0.5025, 0.5032, 0.5027 and 0.5026. Calculate the confidence interval at 95 and 99% level. t95% = 2.36; t99% = 5.30

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Answer question #3  of this problem 

Answer all questions completely.
1.
Replicate water samples are analyzed for water hardness with the following results:
102.2, 102.8, 103.1 and 102.3 ppm CaCO3. Calculate the a) average deviation b) standard
deviation c) range.
2. Given the following data on the establishment of the precision of a method:
22.23, 22.18, 22.25, 22.09 and 22.17%.
Is 22.09 % a valid measurement at 95% confidence level? Qtab = 0.710
3.
The following molar concentrations were obtained from replicate standardization of a
solution:
0.5026, 0.5029, 0.5023, 0.5031, 0.5025, 0.5032, 0.5027 and 0.5026. Calculate the confidence
interval at 95 and 99% level. t95% = 2.36; t99% = 5.30
Transcribed Image Text:Answer all questions completely. 1. Replicate water samples are analyzed for water hardness with the following results: 102.2, 102.8, 103.1 and 102.3 ppm CaCO3. Calculate the a) average deviation b) standard deviation c) range. 2. Given the following data on the establishment of the precision of a method: 22.23, 22.18, 22.25, 22.09 and 22.17%. Is 22.09 % a valid measurement at 95% confidence level? Qtab = 0.710 3. The following molar concentrations were obtained from replicate standardization of a solution: 0.5026, 0.5029, 0.5023, 0.5031, 0.5025, 0.5032, 0.5027 and 0.5026. Calculate the confidence interval at 95 and 99% level. t95% = 2.36; t99% = 5.30
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