Odpowiedzi

2010-04-14T16:14:48+02:00
M Fe2O3 = 56u*2 + 3*16u = 112u + 48u = 160u
m Ca(NO3)2 = 40u + 14u*2 + 6*16u = 40u + 28u + 96u = 164u
2010-04-14T16:17:15+02:00
Fe₂O₃ - 56*2 + 16*3 = 112 + 48 = 160u

Ca(NO₃)₂ - 40 + 14*2 +(16*3)*2= 40 + 28 + 96 = 164u