Chem 161-2011 Exam I + solutions

# Chem 161-2011 Exam I + solutions - Dr Ed Tavss Chem...

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Dr. Ed Tavss Chem 161-2011 Exam I Burdge and Overby, Chapter III – Quantum Theory and Atomic Structure Light and matter 1. A laser produces a light of wavelength 573 nm. How many moles of photons of this light would have to strike a material for the material to absorb 84.8 kJ? A . 0.406 moles B. 0.127 moles C. 0.212 moles D. 0.872 moles E. 0.651 moles E photon = hc/λ E photon = ((6.626x10 -34 Js) x (3x10 8 m/s))/(573 nm x (1x10 -9 m/nm)) = 3.469x10 -19 J 84800J/(3.47 x 10 -19 J/photon) = 2.44 x 10 23 photons 2.44 x 10 23 photons x (1 mol/6.022x10 23 photons) = 0.405 moles 1

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Chem 161-2011 Exam I Burdge and Overby, Chapter 1 – Chemistry: Science of Change Enthalpy 2. Which of the following are intensive properties? 1. volume 2. boiling point 3. mass 4. density A. 1, 2, 3, and 4 B. 1 and 3 C . 2 and 4 D. 1, 3 and 4 E. 2 and 3 Extensive property: Depends on the quantity. e.g., volume is an extensive property. Intensive property: Doesn’t depend on the quantity; e.g., temperature, density. Go from 100 mL of a substance to 200 mL of that substance, what changes? 1. Volume. The volume increases. Therefore, volume is an extensive property. 2. Boiling point. The boiling point doesn’t change. Therefore, boiling point is an intensive property. 3. Mass. The mass increases. Therefore, mass is an extensive property. 4. Density. The density doesn’t change. Therefore, density is an intensive property. 2
Chem 161-2011 Exam I Burdge and Overby, Chapter I – Chemistry: Science of Change Significant figures/Precision/Accuracy 3. To how many significant figures can the answer to the following equation be reported? Note: None of the numbers is exact. (11.87-10.952) 10.75 x 1.56 A. 1 B . 2 C. 3 D. 4 E. 5 Addition & Subtraction: Smallest number of digits to the right of the decimal point. Then count SF’s. 11.87-10.952 = 0.918 This is limited to two digits to the right of the decimal point, which is two significant figures. 0.918/(10.75x1.56) = 0.0547406 This has two significant figures (0.918 is really two significant figures) divided by 4x3 SF’s, which is two significant figures, 0.055, or 5.5 x 10 -2 . 3

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Chem 161-2011 Exam I Burdge and Overby, Chapter III – Quantum Theory and Atomic Structure Aufbau 4. Select the atom/ion that has the incorrect configuration listed next to it. A. Ti : [Ar] 4s 2 3d 2 B. I : [Kr] 5s 2 4d 10 5p 6 C. Cu + : [Ar] 3d 10 D. N 3– : [He] 2s 2 2p 6 E . Ge 2+ : [Ar] 4s 2 3d 8 4p 2 A. Correct. 22 Ti: [ 18 Ar]4s 2 3d 2 B. Correct. 53
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Chem 161-2011 Exam I + solutions - Dr Ed Tavss Chem...

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