Ответы на вопрос:
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%
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