Test II - ViewAttempt1of1 Title Exam2 Started...

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View Attempt 1 of 1  Title: Exam 2 Started: March 30, 2011 5:16 PM Submitted: March 30, 2011 5:59 PM Time spent: 00:42:40  Total score: 17/20 = 85%   Total score adjusted by 0.0  Maximum possible score: 20  1.   The highly toxic substance, disulfur decafluoride, decomposes as shown:  S 2 F 10  <=> SF 4  + SF 6 In a study, S 2 F 10  was placed in a 2.0 L flask and heated to 100  o C.  At equilibirum, [S 2 F 10 ] = 0.50 M.  More S 2 F 10  was added and, when equilibrium was reattained, [S 2 F 10 ] was 4.5 M.  How did [SF 4 ] change from the original  equilibrium position to the new one, after the addition of more S 2 F 10 ?   Student Response Value Correct Answer A.  new [SF 4 ] = 1/3 x old [SF 4 ]    B.  new [SF 4 ] = 1/9 x old [SF 4 ]    C.  Cannot be determined  without knowing the value  of K c .    D.  new [SF 4 ] = 3 x old [SF 4 ]    E.  new [SF 4 ] = 9 x old [SF 4 ] 0%     Score: 0/1    2.
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  The nuclide  11 C undergoes spontaneous positron decay to  11 B. What is the maximum kinetic energy of the  positron in MeV? (1 amu = 931.5 MeV).  Note that masses (in amu) are: electron = 0.00054857990; proton = 1.00727647; neutron = 1.00866490;
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This note was uploaded on 04/21/2011 for the course CHEM 1332 taught by Professor Shiv during the Spring '08 term at University of Houston.

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Test II - ViewAttempt1of1 Title Exam2 Started...

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