Oblicz masę cząsteczkową kwasu metanowego, oraz jego skład procentowy.
HCOOH - kwas mrówkowy (metanowy)mH=1u*2=2umO=16u*2=32umC=12umHCOOH=2u+32u+12u=46u%H=(2u*100%)/46u=4,35%%O=(32u*100%)/46u=69,57%%C=100%-(4,35%+69,57%)=100%-73,92%=26,08%
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HCOOH - kwas mrówkowy (metanowy)
mH=1u*2=2u
mO=16u*2=32u
mC=12u
mHCOOH=2u+32u+12u=46u
%H=(2u*100%)/46u=4,35%
%O=(32u*100%)/46u=69,57%
%C=100%-(4,35%+69,57%)=100%-73,92%=26,08%