Odpowiedzi

2009-11-15T16:30:43+01:00
MH2CO3=2+12+48=62g/mol

62g------100%
2g--------x
x=3,22%H2

62g----100%
12g-----y
y=19,36%C

100-19,3-3,22=77,42%O
TLENU JEST 77,42% WĘGLA 19,36% A WODORU 3,22%%
2009-11-15T16:32:16+01:00
MH2CO3=1u*2+12u+3*16u=2u+12u+48u=62u
%H=2u/62u * 100% =3,23%
%C=12u/62u * 100 % =19,35%
%O=48u/62u * 100 %=77,42%