Quiz 4 Key - 1 mol = 6.022 10 23 atoms 1 pm = 10 10 cm...

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Chemistry 201X: Quiz #4 September 21, 2007 Name:____KEY___________________________ (1) (6 pts) Using the picture below of a cubic unit cell of a superconducting material made up of Nb and Ge used in numerous applications, write down the empirical formula of the substance and calculate its density (in g/cm 3 ). (HINT: how many atoms of each type are in the unit cell?) a = 514.4 pm # Nb atoms = 12/2 = 6 atoms (all in faces) # Ge atoms = 1 + 8/8 = 2 atoms In unit cell: Nb 6 Ge 2 = Nb 3 Ge Mass = (6AW Nb + 2AW Ge ) / 6.022 × 10 23 = 1.1670 × 10 21 g Volume = (514.4 pm)(10 10 cm/1 pm) 3 = 1.3611 × 10 22 cm 3 Large circles = Ge Small circles (in faces) = Nb AW(Ge) = 72.64 g/mol AW(Nb) = 92.91 g/mol
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Unformatted text preview: 1 mol = 6.022 10 23 atoms 1 pm = 10 10 cm Density = Mass/Volume = 8.574 g/cm 3 Ans. __ Nb 3 Ge; 8.574 g/cm 3 ___ (2) (2 pts) Calculate the scattering angle , 2 , where you would see the Bragg reflection for the d-spacing equal to a , if Mo X-radiation is used ( = 71.069 pm). (Braggs law: n = 2 d sin ) (1)(71.069 pm) = 2(514.4 pm) sin sin = 0.06908 = 3.961 2 = 7.922 Ans. ___ 2 = 7.922 ______ (3) (2 pts) When one mole of this substance melts, what is the heat change for the melting process? Circle your choice (a) + H vap (b) H vap (c) + H fus (d) H fus (e) None of these....
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This homework help was uploaded on 03/27/2008 for the course CHEM 201 taught by Professor Miller during the Fall '07 term at Iowa State.

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