Solution4

# Solution4 - HW 4 Solution 5-13 (a) c 20% 25% = 2000 at %...

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HW 4 Solution 5-13 (a) 2000 0025 . 0 % 25 % 20 = = Δ Δ cm x c at % Zn/cm (b)We need to determine the wt % of zinc in each portion. wt % = 46 . 20 100 ) 54 . 63 )( 80 ( ) 38 . 65 )( 20 ( / 38 . 65 20 = × + × mol g wt %= 54 . 25 100 ) 54 . 63 )( 75 ( ) 38 . 65 )( 25 ( / 38 . 65 25 = × + × mol g % 2032 0025 . 0 % 54 . 25 % 46 . 20 wt cm x c = = Δ Δ Zn/cm (c) Find the number of atoms per cm 3 . ( ) 24 3 8 10 0167 . 0 10 63 . 3 ) 2 . 0 )( / 4 ( × = × cell atoms ( ) 24 3 8 10 0209 . 0 10 63 . 3 ) 25 . 0 )( / 4 ( × = × cell atoms 68 . 1 0025 . 0 10 0209 . 0 10 0167 . 0 24 24 = × × = Δ Δ cm x c ×10 24 Zn atoms/cm 3 -cm

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5-33 s cm D / 10 019 . 3 1473 987 . 1 900 , 32 exp 23 . 0 2 6 1200 × = × = s cm D / 10 034 . 3 1223 987 . 1 900 , 32 exp 23 . 0 2 7 950 × = × = 1200 1200 950 950 t D t D = h t 1 1200 = h D t D t 95 . 9 10 034 . 3 10 019 . 3 / 7 6 950 1200 1200 950 = × × = = 5-36 Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.
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## This note was uploaded on 06/09/2009 for the course MAE 614106 taught by Professor Marcmeyers during the Spring '09 term at UCSD.

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Solution4 - HW 4 Solution 5-13 (a) c 20% 25% = 2000 at %...

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