482VP I think is the correct answer.
Lithium 6 would have 6 valence electrons in the outer orbital, while lithium 7 would have 7 in the outer orbital.
Answer:
4.56×10^-10
Explanation:
Co(OH)2(s)<------->Co^2+(aq) + 2OH^-(aq)
Co^2+(aq) + 6NH3(aq) ---------->[Co(NH3)6]^2+(aq)
Overall equation:
Co(OH)2(s)+ 6NH3(aq) ---------->[Co(NH3)6]^2+(aq) + 2OH^-(aq)
The increased solubility of the cobalt salt in the presence of ammonia is as a result of the formation of the hexammine cobalt II complex shown above.
Knet= Ksp×Kf
Ksp of Co(OH)2= 5.92 × 10^-15.
Kf= 7.7 x 10^4
Knet= 5.92 × 10^-15 × 7.7 x 10^4
Knet= 4.56×10^-10
Note Knet is the equilibrium constant of the overall reaction.
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