Oblicz zawartość procentową tlenku w cząsteczce tlenku siarki VI
mSO3=32u+48u=80u
%O=(48/80)*100% = 60%
%S=(32/80)*100% = 40%
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
mSO3=32u+48u=80u
%O=(48/80)*100% = 60%
%S=(32/80)*100% = 40%