а)Fe2O3
= Мr(Fe2O3)= Аr(Fe)·56×2+Ar(O)·16×3=160
W(fe)= 2*56/160 = 0.7 = 70%
W(O)= 3*16/160 = 0,3 = 30%
б)Fe3O4
=Mr(Fe3O4)= 56×3+16×4= 232
W(Fe)=3*56/232 ·100%= 168:232·100%=72.4%
W(O)= 4*16/232·100%=64:232·100%=27.5%
в)FeO
=Mr(FeO)= 56+16=72
W(Fe)=56/72·100%=77.7%
W(O)=16/72·100%= 22.2%
г)FeS=Mr(FeS)=56+32=88
W(Fe)=56/88·100%=63.3%
W(S)= 32/88·100%=36.3%
M(CnH2n₊₁OH)=144
12n+2n+1+16+1=144
14n+18=144
14n=126
n=9
С₉Н₁₉ОН ,нонанол
CuSO4 + Fe = Cu + FeSO4
m(Fe)=1,6*56/159,5=0,56г
m(Fe которое осталось)=5,2-0,56=4,64г
17.2
18.3
19.3
20.1
21.1
22.2
23.4
24.4
25.2
26.4
27.1
28.1
29.2
30.2
31.3
32. 3