101E3_2006a - Chemistry 101 Examination 3, June 30, 2006...

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Chemistry 101 Examination 3, June 30, 2006 page 1 of 13 PRINT NAME______________________________ SIGNATURE_________________________ Last First Recitation Section_______ Recitation Instructor_________________________________ There are 2 questions in Part I and 18 multiple choice questions in Part II, in addition to one page of equations and a Periodic Table with constants, for a total of 12 pages . Be sure you have a complete copy before you begin. The backs of pages 3-12 can be used as scrap paper. Turn in pages 1 and 2. Part I. Answer each question in the space provided on the page. You must show your work for credit. Part II: For each question choose the one best answer. Indicate the letter corresponding to this answer in the proper space in the above box. It is best to use pencil and erase completely any answers you want to change. Questions with more than one answer marked will be graded as incorrect. Do not leave any answers blank - guess if you don't know. Part I 1. __________(5) 2. __________(5) Part II __________(90) Total __________(100) version A A B C D E 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 x 15 x 16 x 17 x 18 x 19 x 20 x A B C D E
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Chemistry 101 Examination 3 June 30, 2006 page 2 of 13 Part I. NAME______________________________ 1. (5 points) Draw the Lewis structure, including resonance forms, if necessary, for HCO 2 1- (C in center, all others connected to it). C O O H C O O H 2. (5 points) The apparatus shown consists of three bulbs, connected by stopcocks. Bulb A contains CO 2 at a pressure of 0.213 atm and has a volume of 15.0 L. Bulb B is empty and has a volume of 10.0 L. Bulb C contains Ar at a pressure of 0.115 atm and has a volume of 20.0 L. What is the total pressure inside the system after the stopcocks are opened and the gases freely mix? Assume the lines connecting the bulbs have zero volume. A B C for CO 2 : P = (0.213 atm)(15.0 L)/(15.0 + 10.0 + 20.0 L) = 0.0710 atm for Ar: P = (0.115 atm)(20.0 L)/(15.0 + 10.0 + 20.0 L) = 0.0511 atm P (total) = 0.0710 atm + 0.0511 atm = 0.1221 atm
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Chemistry 101 Examination 3 June 30, 2006 page 3 of 13 Part II. Each question is worth 5 points. 3. Assuming an MO diagram with orbital energies like that of O 2 what is the bond order and number of unpaired electrons for the molecule NO? (a) bond order 2, 2 unpaired electrons
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This note was uploaded on 06/05/2008 for the course CHE 101 taught by Professor Churchhill during the Summer '08 term at SUNY Buffalo.

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101E3_2006a - Chemistry 101 Examination 3, June 30, 2006...

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