This is an incomplete question, here is a complete question.
Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.
CaCO₃, Ksp = 8.7 × 10⁻⁹
Answer : The solubility of CaCO₃ is, 
Explanation :
As we know that CaCO₃ dissociates to give
ion and
ion.
The solubility equilibrium reaction will be:

The expression for solubility constant for this reaction will be,
![K_{sp}=[Ca^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
Let solubility of CaCO₃ be, 's'




Therefore, the solubility of CaCO₃ is, 
It would be an physical change ; if you melt butter the butter goes from a solid to a liquid so therefore the physical state is changed.
Answer:
613 mg
Explanation:

Number of fargday's 
Here, I = 9.20 A
t = 10.5 min
= 10.5 x 60 seconds
So, 

= 0.0208 F
Here, 2e, 2F
2F = 1 mol of Ni

1 mol = 59 gm of Ni
0.0104 mol = 59 x0.0104 gm Ni
= 0.613 gm Ni
= (0.613 x 1000 ) mg of Ni
= 613 mg of Ni