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masa Al = 27g
213g = 100%
27g = x%
x = 12,68% Al w Al(NO3)3
b) masa CuSO4 = 159,5g
masa Cu = 63,5g
159,5g = 100%
63,5g = x%
x = 39,81% Cu w CuSO4
Masa molowa Al(NO3)3=1*27g+3*14g+9*16=213g/mol
213g----100%
27g------x%
x=27*100/213
x=12,67% Al
b)miedzi w siarczanie(VI) miedzi(II)
Masa molowa CuSO4=1*64g+1*32g+4*16g=160g/mol
160g---100%
64g-----x%
x=64*100/160
x=40% Cu