Odpowiedź:
1.
mMgSO4= 24u +32u +4*16u=24u + 32u +64u= 120u
%O=64u/120u*100%=53,3%
mMgCO3=24u +12u + 3*16u=24u +12u +48u = 84u
%O= 48u/84u *100%^= 57,1%
2
a. N2O3
mN; mO = 2*14u : 3*16u= 28u :48u
mN;mO = 7 :12
b. CO2
mC : mO = 12u :2*16u= 12u : 32u
mC: mO= 3:8
c. Al2O3
mAl : mO= 2*27u : 3*16u =54u :48u
mAl : mO = 9:8
d. H2O
mH :mO= 2*1u : 16u=2u:18u
mH : mO = 1:9
