Oblicz zawartość procentową węgla w metanie, etylenie i acetylenie. Daje Naj! Prosze o pomoc !
w metanie (CH4):
m= 12+ 4= 16u
%C= 12/16 * 100% = 75%
w etylenie(C2H4):
m= 12*2+4= 28u
%C= 24/28 * 100% = ok. 86%
w acetylenie(C2H2):
m= 12*2+2= 26
%C= 24/26 * 100% = ok. 92%
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w metanie (CH4):
m= 12+ 4= 16u
%C= 12/16 * 100% = 75%
w etylenie(C2H4):
m= 12*2+4= 28u
%C= 24/28 * 100% = ok. 86%
w acetylenie(C2H2):
m= 12*2+2= 26
%C= 24/26 * 100% = ok. 92%