ex1s11 - NAME: Natalie Portman 2011 10 Circle Section...

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NAME: Natalie Portman CHEMISTRY 444, SPRING, 2011 Circle Section Number: 10 11 80 81 Examination 1, March 5, 2011 Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY understand your work; only work on the exam sheet will be considered. Write answers, where appropriate, with reasonable numbers of significant figures. You may use only the "Student Handbook," a calculator, and a straight edge . 1. (15 points) What is the total collisional frequency (Z Ar,Kr ) at 298.15 K for a collection of argon and krypton molecules confined to a volume of exactly 1 cm 3 , in which the partial pressure of argon is 0.300 bar and the partial pressure of krypton Is 0.400 bar? [HINT: Both argon and krypton are noble gases.] To begin, one needs the number density of each gas. Since these are noble gases, one may use the ideal-gas law to find these quantities. 3 24 1 1 23 5 * 3 24 1 1 23 5 * 10 7172 . 9 ) 15 . 298 )( 3144349 . 8 ( ) / 10 02211415 . 6 )( 10 400 . 0 ( 10 2879 . 7 ) 15 . 298 )( 3144349 . 8 ( ) / 10 02211415 . 6 )( 10 300 . 0 ( m molecules K mole K J mole molecules Pa RT L P n m molecules K mole K J mole molecules Pa RT L P n Kr Kr Ar Ar One also needs the collision cross section. This is found from the gas diameters of the two atoms in Table 8.2 in your Handbook: m nm nm d d r Kr Ar 10 10 666 . 3 3666 . 0 2 ) 3827 . 0 3504 . 0 ( 2 2 19 2 10 2 10 222 . 4 ) 10 666 . 3 ( m m r One also needs the average relative speed of the krypton and argon atoms, which requires one to calculate the reduced mass of the two atoms: kg kg kg kg M M L M M m m m m Kr Ar Kr Ar Kr Ar Kr Ar 26 23 10 4921 . 4 ) 083798 . 0 039948 . 0 )( 10 02211415 . 6 ( ) 083798 . 0 )( 039948 . 0 ( ) ( With this reduced mass, one can calculate the average relative speed of the two atoms: 1 26 23 , 06 . 483 ) 10 4921 . 4 ( ) 15 . 298 )( / 10 3806505 . 1 ( 8 8 s m kg K K J kT v Kr Ar  The almost-final step is to calculate the total collision frequency per unit volume: 1 3 34 3 24 3 24 1 2 19 * * , , , 10 4425 . 1 ) 10 7172 . 9 )( 10 2879 . 7 )( 06 . 483 ( 10 222 . 4 s m m m s m m n n v Z Kr Ar Kr Ar Kr Ar Kr Ar Finally multiply by the volume to find out the number of collisions in a second: 1 28 3 6 1 3 34 10 4425 . 1 ) 10 1 ( 10 4425 . 1 s m s m frequency collision Total DO NOT WRITE IN THIS SPACE p. 1_______/15 p. 2_______/10 p. 3_______/10 p. 4_______/15 p. 5_______/15 p. 6_______/10 p. 7_______/10 p. 8_______/15 ============= p. 9 _______/5 (Extra credit) ============= TOTAL PTS /100
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Score for Page NAME: CHEM 444, Exam 1, Spring, 2011, page 2 2. (10 points) Nucleosome particles are composed of DNA wrapped around a core protein. For a particular type of
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ex1s11 - NAME: Natalie Portman 2011 10 Circle Section...

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