Explanation:
Generally, heat flows from a hot environment to a cold (lesser temperature) environment. In this case, the soup is the hot environment and the air is the cold temperature.
Heat would continue to flow from one environment to another until thermal equilibrium is reached. At this thermal equilibrium, both environments would have the same temperature.
As,
CuCO₃ ⇆ Cu²⁺ + CO₃²⁻
So,
Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]
Or,
Ksp = [Cu²⁺] [CO₃²⁻]
As,
Ksp = 1.4 × 10⁻¹⁰
So,
1.4 × 10⁻¹⁰ = [x] [x]
Or,
x² = 1.4 × 10⁻¹⁰
Or,
x = 1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
x = 1.18 × 10⁻⁵ mol/L × 123.526 g/mol
x = 1.45 × 10⁻³ g/L
<span>0.002 moles is the answer </span>
