Đáp án:
a, PTK $Fe_{2}$$O_{3}$ = 56. 2+ 16. 3= 160(đvC)
%Fe=$\frac{56.2}{160}$ .100= 70%
%O=$\frac{16.3}{160}$ .100= 30%
b, PTK $S_{}$$O_{3}$ = 32+ 16.3 =80(đvC)
%S= $\frac{32}{80}$ .100= 40%
%O= $\frac{16.3}{80}$ .100= 60%
c, PTK $H_{2}$$SO_{4}$ = 1.2+ 32+ 16.4= 98(đvC)
%H= $\frac{2}{98}$ .100= 2.04%
%S= $\frac{32}{98}$ .100= 32.65%
%O= $\frac{16.4}{80}$ .100= 65.31%
d, PTK $Cu_{}$$SO_{4}$ = 64+ 32+ 16.4= 160(đvC)
%Cu= $\frac{64}{160}$ .100=40%
%S= $\frac{32}{160}$ .100= 20%
%O= $\frac{16.4}{160}$ .100= 40%
e, PTK $P_{2}$$O_{5}$ = 31.2+16.5= 142(đvC)
%P= $\frac{31.2}{142}$ .100= 43,66%
%O= $\frac{16.5}{142}$ .100= 56,34%
f, PTK $Al_{}$$(OH)_{3}$ = 27+16.3+3.1= 78(đvC)
%Al=$\frac{27}{78}$ .100= 34,62%
%O=$\frac{16.3}{78}$ .100= 61.54%
%H=$\frac{3.1}{78}$ .100= 3.84%
Mk làm hơi cực, mong bạn cho 5*và 1 cảm ơn.
