Fall09. Final key

Fall09. Final key - CHEMISTRY 2101 Fall 2009 Dr. Philippe...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEMISTRY 2101 Name___________________________ Fall 2009 Dr. Philippe Buhlmann Final Exam December 19, 2009 Page 1 FINAL EXAM There are 8 pages with problems to this exam. Put your name (or initials on subsequent pages) at the top of EACH PAGE ! Show your work! Problem Points Possible Your Score 1 30 2 15 3 20 4 15 5 30 6 10 7 30 Total 150 We will correct everything that we can read—promised! Answer the whole question but be succinct! Manage your time! Move on if you get stuck with a problem. h = 6.6262 10 -34 J s 1 eV = 1.6022 10 -19 J At 25 ºC: ln(10) RT/F = 2.303 RT/F = 59.16 mV
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
CHEMISTRY 2101 Name___________________________ Fall 2009 Dr. Philippe Buhlmann Final Exam December 19, 2009 Page 2 1 (30 points) Statistics (a) (7 points) The As content of a sample was determined in four measurements as 3.102%, 3.076%, 3.024%, and 3.030%. What final answer should be given? For the associated uncertainty level choose the 95% confidence level. mean 3.058 standard deviation 0.0374166 t for 3 degrees of freedom and 95%: 3.182 Final result: 3.058 ±0.060 (b) (8 points) Using the Grubbs test, decide whether the value 0.195 should be rejected from the set of results 0.217, 0.224, 0.195, 0.221, 0.221, 0.223. G = (0.195-0.2168)/ 0.0109621= 1.99 G for 95% and 6 values is 1.822 This value may be rejected. (c) (15 points) The bromide concentrations in 4 tomatoes determined as: 777; 790; 759; 790 µ g/g The bromide concentrations in 4 potatoes determined as: 782; 773; 777; 764 µ g/g Is there a statistically relevant difference (at the 95% level) between the bromide concentrations in these tomatoes and potatoes? _ x = 1 n n i=1 x i s x = 1 n-1 n i=1 (x i - _ x) 2 _ x tom = 779; n = 4; s tom = 1 4-1 4 i=1 (x i - _ x) 2 = 14.674
Background image of page 2
CHEMISTRY 2101 Name___________________________ Fall 2009 Dr. Philippe Buhlmann Final Exam December 19, 2009 Page 3 _ x pot = 774 n = 4; s pot = 1 4-1 4 i=1 (x i - _ x) 2 = 7.6158 s pooled = (n 1 - 1) s 1 2 + ( n 2 - 1) s 2 2 n 1 + n 2 - 2 = (4 - 1) 14.67 2 + ( 4 - 1) 7.616 2 4 + 4 - 2 = 11.690 t = | _ x 1 - _ x 2 | s pooled n 1 n 2 n 1 + n 2 = | 779 - 774 | 11.690 4 4 4 + 4 = 0.60488 => The student t value for 6 degrees of freedom is 2.45 at the 95% confidence level. The calculated t value is smaller than 2.45. Therefore, There is no significant difference.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
CHEMISTRY 2101 Name___________________________ Fall 2009 Dr. Philippe Buhlmann Final Exam December 19, 2009 Page 4 2 (15 points) Spectroscopy (a) (5 points) What is an isosbestic point? Explain why an isosbestic point is observed. One or more isosbestic points are obtained by the overlap of the UV-Vis spectra of solutions containing two absorbing species if the sum of the concentrations of the two species in all the solutions is the same . The isosbestic point occurs at the wavelength(s) at which the two absorbing species have the same molar absorptivities, ε . (b) (5 points) 2.0% of the incident light passes through a sample. Calculate the absorbance, A, and transmittance, T. T = 0.020; A = - log 0.020 = 1.70 (c) (5 points) The ultraviolet absorbance at 345 nm for a 1.00 mM solution of compound A is 0.088 (1.000 cm cell). Calculate the molar absorbtivity of A. ε A ' = Absorbance / (b c) = 0.088 /{1.000 cm 1.00 10 -3 M} = 8.8 10 1 M -1 cm -1
Background image of page 4
CHEMISTRY 2101 Name___________________________ Fall 2009 Dr. Philippe Buhlmann Final Exam December 19, 2009 Page 5 3 (20 points) Elemental Analysis and Electrochemistry The combustion of an Eosin-related dye (4.321 mg; 9.74 x 10 -6 mol; Mr 443.63) with the molecular formula C 20 H 8 O 5 BrCl was performed in an oxygen atmosphere in a closed flask containing 20 mL of water. Upon vigorous shaking of the flask for a while, all the gases dissolved in the water. The solution
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/02/2011 for the course CHEM 2101 taught by Professor Bulman during the Spring '11 term at Minnesota.

Page1 / 15

Fall09. Final key - CHEMISTRY 2101 Fall 2009 Dr. Philippe...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online