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2009-11-27T21:19:03+01:00
A)C4H8
mC4H8 = 4*12u + 8*1u = 48u + 8u =56u
%C = 48\56 * 100% = 85,71%
%H= 100% - 85,71% = 14,29%
b)C4H6
mC4H6 = 4*12u+ 6*1u = 48u + 6u =54u
%C= 48/54*100% = 88,89%
%H=100%-88,89%=11,11%
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2009-11-27T21:25:16+01:00
A) C4H8 buten
m= 4*12u+8*1u
m=56u
węgiel
x=48u*100%/56u
x=85,70%
wodór
x=8u*100%/56u
x=14,30%
b)C4H6 butyn
m=4*12u+6*1u
m=54u
węgiel
x=48u*100%/54u
x=88,90%
wodór
x=6u*100%/54u
x=11,10%
2 5 2
2009-11-27T21:38:13+01:00
Alken C4H8

%C=4*12 / 4*12+8 *100%
%C=48 / 56 *100%
%C=85,71

%H=100% - 85,71% = 14,29%

Alkin C4H6

%C=4*12 / 4*12+6 *100%
%C=48 / 54 *100%
%C=88,9

%H=100% - 88,89% = 11,11
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