a) mCu(NO3)2=64u+2*14u+6*16u = 188u%Cu = (64/188)*100% = 34%b) mAlCl3=27u+3*35,5u=133,5u%Al = (27/133,5)*100% = 20,2%
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a) mCu(NO3)2=64u+2*14u+6*16u = 188u
%Cu = (64/188)*100% = 34%
b) mAlCl3=27u+3*35,5u=133,5u
%Al = (27/133,5)*100% = 20,2%