Oblicz zawartość procentową węgla w metanolu i etanolu.
More Questions From This User See All
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
100% - 32u
x% - 12u
x% C = 37,5%
m C2H5OH =46u
100% - 46u
x% - 24u
x% c = 52,2%
mCH3OH=12u+4*1u+1*16u=32u
32g------100%
12g-------x%
x=12*100/32
x=37,5% węgla
Etanol
mC2H5OH=2*12u+6*1u+1*16u=46u
46g-------100%
24g--------x%
x=24*100/46
x=52,17% węgla