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14:3:12 stosunek masowy
100%-(14+3+12)
%Fe-14
%Fe=1400/29=48,26%
100%-(14+3+12)
%C-3
%C=300/29=10,34%
%O=100%-48,26%-10,34%=41,4%
zad 8
M((NH4)2SO4+2NH4NO3)=132+2*80=292u
MN=28+4*14=84u
100%-292
%N-84
%N=8400/292=28,77%