Fall 09. Exam1 key

Fall 09. Exam1 key - CHEMISTRY 2101 Fall 2009 Dr. Philippe...

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CHEMISTRY 2101 Name___________________________ Fall 2009 Dr. Philippe Buhlmann Hour Exam I October 9, 2009 Page 1 (1) (20 points in total) Statistics The oxalate concentrations in five apples were determined as: 659; 660; 740; 757; 899 μg/g The oxalate concentrations in four pears were determined as: 230; 269; 278; 278 μg/g (a) 5 points Is the value of 899 μg/g an outlier? G = | 899 - mean| / standard deviation = (899 - 743) / 98.1 = 1.59 G at 95% for 5 observations is 1.672. The 899 μg/g value does not qualify as an outlier. (b) 15 points Is there a statistically relevant difference (at the 95% level) between the oxalate concentrations in these apples and pears? Ignore outliers if that is permissible. s apples = 98.1; s pears = 22.9; x apples = 743.0 ; x pears = 263.8 s pooled = s 1 2 n 1 - 1 ( ) + s 2 2 n 2 -1 ( ) n 1 + n 2 - 2 = s 1 2 5 - 1 ( ) + s 2 2 4 -1 ( ) 5 + 4 - 2 = 75.6 t = x 1 - x 2 s pooled n 1 n 2 n 1 + n 2 = 743.0 - 263.8 75.6 5 " 4 5 + 4 = 9.44 => The student t value for 7 degrees of freedom is 2.365 at the 95% confidence level. The calculated t value is larger than 2.365. Therefore, there is a significant difference.
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CHEMISTRY 2101 Name___________________________ Fall 2009 Dr. Philippe Buhlmann Hour Exam I October 9, 2009 Page 2 (2) (25 points) A pH Buffer Design a 100 mM buffer solution to have a pH = 2.75. You can assume that all activity coefficients are 1. Make all necessary calculations. Answers of the type “…and use a glass electrode to adjust the pH” will not get any credit; in the lab
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This note was uploaded on 10/02/2011 for the course CHEM 2101 taught by Professor Bulman during the Spring '11 term at Minnesota.

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Fall 09. Exam1 key - CHEMISTRY 2101 Fall 2009 Dr. Philippe...

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