key_final - FINAL EXAM Mar. 19, 2001 CHEMISTRY 20A Winter...

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1 FINAL EXAM CHEMISTRY 20A Winter 01 SOLUTIONS Mar. 19, 2001 . . S t u d e n t N a m e Bobby N. Johnny S. Rebecca E. Circle Name of TA . . S t u d e n t I . D . N o . PLEASE WRITE YOUR NAME & Circle Name of your TA . Then PLEASE WRITE YOUR NAME and ADDRESS ON THE BACK OF THE EXAM in the appropriate address label. If you want the exam returned, then PLACE $0.76 POSTAGE in the appropriate place. READ DIRECTIONS on THIS PAGE, but DO NOT TURN PAGE UNTIL TOLD TO DO SO . DIRECTIONS . All work and answers should be written on the exam. Please write legibly and clearly. Wherever applicable, show your work, justify approximations, and explain what you are doing. It is usually useful, when possible, to work problem out algebraically first. Partial credit will be given; correct statements will count in your favor, incorrect ones will count against you, unintelligible or illegible ones will not count at all. Allocate your time between problems judiciously. This is a closed book exam, but you may make use of a single 8.5" by 11" sheet of handwritten notes (both sides). This exam consists of 16 pages. NOTE that each part of each question can be answered independently, whether or not you are able to answer an earlier part; however, the answers to later parts may depend upon statements included in the previous parts. LEAVE OUT ONE OF THE 41 POINT PROBLEMS. MARK CLEARLY BELOW WHICH ONE YOU HAVE LEFT OUT. Problem. Max Grade Grade 1 27 2 27 Class average on Final. 3 34 4 41 YOUR GRADE for THE COURSE: 5 41 6 41 7 41 8 41 . Total 252
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2 Some possibly useful data: Molecular weights can be obtained from periodic table h = 6.63x10 -34 J s |e| = 1.60 x 10 -19 C. c = 3.00x10 8 m s -1 m electron = 9.11x10 -31 kg N av = 6.02x10 23 particles mol -1 m proton • m neutron = 1.67x10 -27 kg 1 eV = 1.60x10 -19 J Some possibly useful formulas. E = h ν KE = (1/2)mv 2 λ m = h/mv c = λ ν F = ma = -dU/dr (minus slope of U) U elect = ± e 2 4 πε o 1 r . For convenient numerical use U elect = 1.3 eV ( nm r ) = 1,250 kJ mol -1 ( nm r ) nm = nanometer (length) E n = - Z 2 R H n 2 = - (1/2) Z 2 e 4 m e (4 πε o h) 2 1 n 2 = -13 eV Z 2 n 2 = - 1,250 kJ Z 2 n 2
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3 1. When 14.8 g of a compound containing only carbon, hydrogen and oxygen was burned in excess oxygen, 18.0 g of H 2 O and 35.2 g of CO 2 were formed. What is the empirical formula of the compound, i.e. , in the formula C k H m O n what are the simplest integral values for k,n,m? 18.0 g of H 2 O is 1 mol, i.e ., 2 mol of H. 35.2 g of CO 2 is 0.8 mol , i.e . 0.8 mol of C. 2 mol of H is 2.0 g. 0.8 mol of C is 9.6 g. Thus the mass of H and C in the compound is 2.0 + 9.6 g = 11.6 g. Thus the mass of oxygen in the compound is 14.8 g - 11.6 g = 3.2 g. 3.2 g of O is 0.2 mol.
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This note was uploaded on 05/17/2009 for the course CHEM chem20a taught by Professor Neuhauser during the Spring '09 term at UCLA.

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key_final - FINAL EXAM Mar. 19, 2001 CHEMISTRY 20A Winter...

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