Răspuns:
Explicație:
BaCO3
r.a.=1 :1 :3
r.m.=137 : 12 : 48
M BaCO3=137 + 12 +48=197---->197g/moli
Comp. proc .prin masa moleculara
197g BaCO3-------137g Ba ------12g C------48g O
100g ------------------x------------------y------------z
x=100.137: 197=69,54 % Ba
y=100.12:197=6,09% C
z=100.48 :197=24, 37% O