3) what is the composition, in atom percent and in weight percent, of an alloy that contains 33 g cu and 47 g zn?

composition in weight percent = 41.25% Cu and 58.75% Zn

Composition in Atom percent = 40.54% Cu and 59.46% Zn

Explanation:

Starting with the composition in weight percent.

Let Mcu= mass of cu and Mzn= mass of zn

%W= weight percent

%Wcu = (Mcu/(Mcu + Mzn)) × 100%

By substituting the values

%Wcu = (33/(33+47)) × 100%

%Wcu = (33/80) × 100%

%Wcu = 41.25%

%Wzn = (Mzn/(Mcu + Mzn)) × 100%

= 47/80 × 100%

%Wzn = 58.75%

Then we can proceed to composition in atom percent

Atomic mass of Cu= Ac = 63.5g/mol

For zn = Az = 65.4g/mol

Atom percent cu = {(%Wcu×Ac)/[(%Wcu×Ac)+(%Wzn×Az)]} × 100%

Substituting the values

= {2619.375/6461.625}×100%

=40.54% Cu

Atom percent zn = {(%Wzn×Az)/[(%Wcu×Ac)+(%Wzn×Az)]} × 100%

={3842.25/6461.625}×100%

=59.46% Zn

Thanks...

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