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edugen.wileyplus.com ber 96 Made in a Chicago Hood waeypuusu MywileyPLus I Help I Contact Us I Log allister s Rethwisch, Fundamentals of Materials Science and Engineering: An Integrated Approach, Se MATERIALS SCI (CHEM E) (ENGR:2720) NEXT I Problem S-38 (Go Tutorial) Gold forms a substtutional solid solution with silver compute the number of gold atoms per abic centimeter for a siver gold alloy that contains 16 wt% Au and 84 wit% Ag. The denseles of pure gold and silver are 19.32 and 10.49 gom respectively. The a weight Au atoms/cm3 the tolerance is +/-2.0% LINK TO TEXT By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Question Attempts o of 3 used SAVE FOR LATER SUBMITANSwER copyright o2000-20iy by sohn waey Ssoni mnc or related comoares reserved Rights Reserved A Division of zahn wilerA Sons,

Explanation / Answer

Q5.38

number of gold atoms in cm3 of silver-gold mix...

Density of mix = 0.16*19.32 + (1-0.16)*10.49 = 11.9028 g/m3

for 1 cm3 = 11.9028 g

so

0.16 is gold --> 11.9028*0.16 = 1.904448 g of Gold

Au mol = mass/MW = (1.904448)/(196.97) = 0.009668 mol

number of atoms = 0.009668*(6.022*10^23) = 5.8220*10^21 atoms of gold per cm3

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