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Masa Al2O3 = (27ux2) + (3 x 16u) = 102 u <-- to jest nasze 100%
Masa Al = 27u
Masa O = 16u
54 u Al - x%
102 u Al2O3 - 100%
x= 52,94 %
48 u O ------ y%
102 u Al2O3 -100%
y= 47,06 %
%Al = 52,94, a %O = 47,06 :)
zawartość % Al w związku = (54:102)x100%= 52,94%
zawartość % O w związku = (48:102)x100%= 47,06%