Odpowiedź :
masa H₂CO₃ wynosi 2+12+48 = 62 g
H = 2 g
C = 12 g
O = 48g
%H = 2/62 = 3,23%
%C = 12/62 = 19,35%
%O = 48/62 = 77,42%
H = 2 g
C = 12 g
O = 48g
%H = 2/62 = 3,23%
%C = 12/62 = 19,35%
%O = 48/62 = 77,42%
procent masowy:
mH₂CO₃ = 2*1u +12u +3*16u= 62u
%H = (2*1u / 62u) *100%
%C= (12u/62u)*100%
%O= (3*16u / 62u) *100%
%H≈3,23%
%C ≈ 19,35%
%O≈77,42%
SPR. 77,42 +19,35 +3,23= 100% -> zgadza się
Pozdrawiam! :)
mH₂CO₃ = 2*1u +12u +3*16u= 62u
%H = (2*1u / 62u) *100%
%C= (12u/62u)*100%
%O= (3*16u / 62u) *100%
%H≈3,23%
%C ≈ 19,35%
%O≈77,42%
SPR. 77,42 +19,35 +3,23= 100% -> zgadza się
Pozdrawiam! :)