wypisz wszystkie możliwe tlenki Fe oraz podaj ich skład %
FeO
M = 56 + 16 = 72
%Fe = 56/72 * 100% = 77,78%
%O = 100% - 77,78% = 22,22%
Fe2O3
M = 2*56u + 3*16u = 112u + 48u = 160u
%Fe = 112/160 * 100% = 70%
%O = 30%
Fe3O4
M = 3* 56u + 4* 16u = 168u +64u = 232u
%Fe = 168/232 * 100% =72%
%O = 28%
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FeO
M = 56 + 16 = 72
%Fe = 56/72 * 100% = 77,78%
%O = 100% - 77,78% = 22,22%
Fe2O3
M = 2*56u + 3*16u = 112u + 48u = 160u
%Fe = 112/160 * 100% = 70%
%O = 30%
Fe3O4
M = 3* 56u + 4* 16u = 168u +64u = 232u
%Fe = 168/232 * 100% =72%
%O = 28%