Odpowiedzi

2009-11-03T18:31:36+01:00
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%
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2009-11-03T18:33:55+01:00
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! :)
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