Odpowiedzi

2009-11-07T16:58:40+01:00
A. mc=2*1u+32u+3*16u=2u+32u+48u=82u
% S=32u*100% /82u =3200/82≈39%
% 0₃=48u*100%/82u=4800/82≈59%

c. mS/m0₂= 32/32 , 32:32

b. H₂ : S : 0₃
2u:32u:48u / 2
16u:24u
4 3 4
2009-11-07T18:15:24+01:00
A.

mH2SO3 =2*1u+32u+3*16u=2u+32u+48u=82u
% S=32u/82u*100%=39.02%
% 0₃=48u/82u*100%=58,54%
%H=2u/82u*100%=2,44%


b.mH/mS/mO
2u/32u/48u /:2
1u/16u/24
Stosunek masowy jest jak 1 do 16 i do 24

c.
mSO2=32u+32u
mS/mO
32u/32u /:32
1/1
Stosunek masowy jest jak 1 do 1

1/1
2 5 2