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,
Here is the chemical equation for gun powder, in it’s simple form:
<span>2 KNO3 + S + 3 C → K2S + N2 + 3 CO2.</span>
This is the same simplified formula, only balanced:
<span>10 KNO3 + 3 S + 8 C → 2 K2CO3 + 3 K2SO4 + 6 CO2 + 5 N2.</span>
Answer:
-105 kJ
Explanation:
The enthalpy change of a reaction is the sum of the energy of the bonds of the reactants and the products. The bonds at the reactants are being broken, so it's an endothermic reaction, so the bond energy must be positive.
The bonds at the products are being formed, so the process is exothermic, and the bond energy must be negative. There are being broken 1 N≡N bond and 3 H-H bonds, and are being formed 6 N-H bonds:
Reactants: 945 + 3*432 = 2241 kJ
Products: 6*(-391) = -2346 kJ
ΔH = 2241 - 2346
ΔH = -105 kJ
