Oblicz zawartość procentową węgla i wodoru w dekanie, etynie i etenie. Z góry dzięki za odpowiedź!
dekan C10H22
m = 142g --- 100%
120g --- %C
%C = 84,5%
%H = 100% - 84,5% = 15,5%
etyn C2H2
m = 26g --- 100%
24g --- %C
%C = 92,3%
%H = 100% - 92,3% = 7,7%
eten C2H4
m = 28g --- 100%
%C = 85,7%
%H = 100% - 85,7% = 14,3%
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dekan C10H22
m = 142g --- 100%
120g --- %C
%C = 84,5%
%H = 100% - 84,5% = 15,5%
etyn C2H2
m = 26g --- 100%
24g --- %C
%C = 92,3%
%H = 100% - 92,3% = 7,7%
eten C2H4
m = 28g --- 100%
24g --- %C
%C = 85,7%
%H = 100% - 85,7% = 14,3%