Answer:
5.3 × 10⁻¹⁷ mol·L⁻¹
Explanation:
Let <em>s</em> = the molar solubility.
Cu₂S(s) ⇌ 2Cu⁺(aq) + S²⁻(aq); K_{sp} = 6.1 × 10⁻⁴⁹
E/mol·L⁻¹: 2<em>s</em> <em>s
</em>
K_{sp} =[Cu⁺]²[S²⁻] = (2<em>s</em>)²×<em>s</em> = 4s^3 = 6.1 × 10⁻⁴⁹

![s = \sqrt[3]{1.52 \times 10^{-49}} \text{ mol/L} = 5.3 \times 10^{-17} \text{ mol/L}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%5B3%5D%7B1.52%20%5Ctimes%2010%5E%7B-49%7D%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%205.3%20%5Ctimes%2010%5E%7B-17%7D%20%5Ctext%7B%20mol%2FL%7D)
Main sequence starts maybe
Answer:
its endothermic .
because your using a cooling process instead of a heating process if it was a heating process it would be exothermic