Oblicz skład procentowy;
a) (NH4)3
b) Cr(NO2)3.
BARDZO PILNE!
a) N=14 H=1 więc (NH4)3=14*3+1*4*3=54
% N=14*3/54*100%=77,77%
% H=1*4*3/54*100%=22,22%
b) Cr=52 N=14 O=16 więc Cr(NO2)3=52+14*3+16*2*3=190
% Cr= 52/190 *100%=27,36%
% N= 14*3/190 *100%=22,10%
%O= 16*2*3/190*100%=50,52%
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a) N=14 H=1 więc (NH4)3=14*3+1*4*3=54
% N=14*3/54*100%=77,77%
% H=1*4*3/54*100%=22,22%
b) Cr=52 N=14 O=16 więc Cr(NO2)3=52+14*3+16*2*3=190
% Cr= 52/190 *100%=27,36%
% N= 14*3/190 *100%=22,10%
%O= 16*2*3/190*100%=50,52%