Oblicz mase etamu i procentowy udział węgla w tej cząsteczce.
Pomózcie !!! . Dzięki z góry :)
etanu :)
C₂H₆
m = 2 x 12u + 6 x 1u = 30u
mC = 24u
%C = 24u/30u x 100% = 80%
Pozdrawiam :)
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etanu :)
C₂H₆
m = 2 x 12u + 6 x 1u = 30u
mC = 24u
%C = 24u/30u x 100% = 80%
Pozdrawiam :)