Odpowiedź :
Odpowiedź:
masa tlenek żelaza (II) = mFeO= mFe + mO = 56u + 16u = 72u
%O = 16*100%/72 = 1600%/72 = 22%
masa tlenku żelaza (III) = mFe203 = 2mFe +3 mO = 2*56 + 3*16 = 112 + 48 = 160 (u)
%O= 3*16*100%/160 = 4800%/160 = 30%
Odpowiedź:
masa tlenek żelaza (II) = mFeO= mFe + mO = 56u + 16u = 72u
%O = 16*100%/72 = 1600%/72 = 22%
masa tlenku żelaza (III) = mFe203 = 2mFe +3 mO = 2*56 + 3*16 = 112 + 48 = 160 (u)
%O= 3*16*100%/160 = 4800%/160 = 30%