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a. glinu w azotanie V glinu
m Al(NO3)3 = 27u + 3* 14u + 9* 16u = 27u + 42u + 144u = 213u
%Al = 27u/213u * 100% = 12,68 %
b. miedzi w siarczanie VI miedzi II
m CuSO4 = 64 u + 32 u + 4*16 u = 160 u
%Cu = 64 u/160u * 100% = 40 %
213 -100 %
27 - x
x = 12,68 % Al
masa molowa CuSO₄ = 159 g/mol
159 - 100 %
63 - x
x = 39,62 % Cu