1.oblicz zawartośc procentowa żelaza w syderycie zawierajacym FeCO3
2.oblicz stezenie molowe siarczanu VI chromu III zawierajacego 3,92g tej soli w 200 cm^3 roztworu
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1.
M FeCO3 = 116g/mol
%Fe = 56/116 * 100% = 48,28%
2.
M Cr2(SO4)3= 392g/mol
m = 3,92g
liczba moli - n = m/M = 3,92/392 = 0,01mol
V = 200cm3 = 0,2dm3
Cm = n/V = 0,01/0,02 = 0,5mol/dm3