1.Oblicz stosunek masy węgla do masy tlenu i do masy wodoru w cząsteczce etanolu.
C2H5OH
2mC=2*12u=24u
6mH=6*1u=6u
mO=16u
C : H: O
24:6:16
12:3:8
mC2H5OH=24u+6u+16u=46u
mC : mH : mO = 24 : 6 : 16
mC : mH : mO = 12 : 3 : 8
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C2H5OH
2mC=2*12u=24u
6mH=6*1u=6u
mO=16u
C : H: O
24:6:16
12:3:8
mC2H5OH=24u+6u+16u=46u
mC : mH : mO = 24 : 6 : 16
mC : mH : mO = 12 : 3 : 8