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exam3key - Chemistry 241 Exam 3 June 9 2010 Honor Pledge I...

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Unformatted text preview: Chemistry 241 Exam 3 June 9, 2010 Honor Pledge: I have neither given or received help on this exam, nor have I seen anyone else do so. (Signed)___________________________ 1. 2. 3. 4. 5. 6. 7. 8. Total________ 15 10 10 15 10 10 15 15 100 SHOW YOUR WORK CIRCLE NUMERICAL ANSWERS 1. (15pts) An unknown sample of Cu 2+ gave an absorbance of 0.266 in an AA analysis. A 100.0 ml solution containing 1.00 ml of 100.0 ppm (=μg/mL) Cu 2+ and 95.0 ml of the unknown Cu 2+ sample gave an absorbance of 0.523. What is the molarity of the unknown Cu 2+ solution? In 100 ml solution - [Cu 2+ ] S = 100.0 ppm (1/100) = 1.00 ppm [Cu 2+ ] U = (95/100)[Cu 2+ ] I [Cu 2+ ] I /(1.00 ppm + .95[Cu 2+ ] I ) = 0.266/0.523 [Cu 2+ ] I = 0.984 ppm .984 μg/mL (1000mL/L)(1g/10 6 μg)(1mole/63.55g) = 1.55 x 10-5 M 2. (10 pts) If a molecule fluoresces at 565 nm and the excited triplet state is 0.2008 eV lower in energy than the excited singlet state, what wavelength would the molecule phosphoresce at?...
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This note was uploaded on 01/09/2011 for the course CHEM 241 taught by Professor Tiani during the Summer '08 term at UNC.

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exam3key - Chemistry 241 Exam 3 June 9 2010 Honor Pledge I...

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