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,
Answer is: <span>- delta G.
</span>The change in Gibbs free energy (ΔG), at constant temperature and pressure, is: <span>ΔG=ΔH−TΔS.
</span>ΔH<span> is the change in enthalpy.
</span>ΔS is change in entropy.
T is temperature of the system.
When ΔG is negative, a reaction (<span>occurs without the addition of external energy)</span><span> will be spontaneous (</span>exergonic).
Answer:
2.5×10⁶ s
Explanation:
From the question given above, the following data were obtained:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
The half-life of a first order reaction is given by:
Half-life (t½) = 0.693 / Rate constant (K)
t½ = 0.693 / K
With the above formula, we can obtain the half-life of the reaction as follow:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 2.8×10¯⁷
t½ = 2.5×10⁶ s
Therefore, the half-life of the reaction is 2.5×10⁶ s
