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m CaCO3 = 1 x 40u + 1 x 12u + 3 x 16u = 40u + 12u + 48u = 100u
%Ca= x 100% = 4000 : 100 = 40%
%C= x 100% = 1200 : 100 = 12%
%O= x 100% = 4800 : 100 = 48%
Liczę na naj ;)