zad.Opisz skład procentowy (masowy) gliceryny.
C3H8O3
masa: (12*3)+(1*8)+(16*3)=92g
C: 36/92*100%=39,1%
H: 8/92*100%=8,7%
O: 48/92*100%=52,2%
Gliceryna C3H8O3
mC3H8O3=3*12u+8*1u+3*16u=92u
%C = (36/92)*100% = 39,1%
%H = (8/92)*100% = 8,7%
%O = (48/92)*100% = 52,2%
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C3H8O3
masa: (12*3)+(1*8)+(16*3)=92g
C: 36/92*100%=39,1%
H: 8/92*100%=8,7%
O: 48/92*100%=52,2%
Gliceryna C3H8O3
mC3H8O3=3*12u+8*1u+3*16u=92u
%C = (36/92)*100% = 39,1%
%H = (8/92)*100% = 8,7%
%O = (48/92)*100% = 52,2%