Odpowiedzi

2009-11-30T16:03:28+01:00
MCuCO3= 63u+12u+(16*3)=123u
mNi(NO₃)₂=59u +(14*2) +(16*6)=183u
m (NH₄)₃PO₄ = (14*3)+(1*12)+31u+(16*4)=149u
Najlepsza Odpowiedź!
2009-11-30T16:30:32+01:00
A) CuCO₃
Cu- 64u
C- 12u
O - 16u

CuCO₃ = 64u + 12u + 3*16u = 64u + 12u + 48u = 124u

b) Ni(NO₃)₂
Ni - 59u
N- 14u
O- 16u

Ni(NO₃)₂ = 59u + 2*14u + 6*16u = 59u + 28u + 96u = 183u


c) (NH₄)₃PO₄
N - 14u
H - 1u
P - 31u
O - 16u

(NH₄)₃PO₄ = 3*14u + 12*1u + 31u + 4*16u = 42u + 12u +31u + 64u = 149u