I’m assuming you mean 5g of magnesium :)
As the temperature is lowered matter is more likely to exist in the solid state
<u>Answer:</u>
<u>For A:</u> The
for the given reaction is 
<u>For B:</u> The
for the given reaction is 1642.
<u>Explanation:</u>
The given chemical reaction follows:

The expression of
for the above reaction follows:

We are given:

Putting values in above equation, we get:

Hence, the
for the given reaction is 
Relation of
with
is given by the formula:

where,
= equilibrium constant in terms of partial pressure = 
= equilibrium constant in terms of concentration = ?
R = Gas constant = 
T = temperature = 500 K
= change in number of moles of gas particles = 
Putting values in above equation, we get:

Hence, the
for the given reaction is 1642.
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, 