The solubility of AgCl(s) in water at 25 °C is 1.33 × 10?? mol/L and its ?H° of solution is 65.7 kJ/mol. Part A What is its solubility at 54.3 °C? Express your answer using two significant figures. S_AgCl = 1.4 • 10?? M
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Step 1: We know that the solubility of a substance is related to the temperature by the Van't Hoff equation, which is given by: ln(K2/K1) = -ΔH°/R * (1/T2 - 1/T1) where K1 and K2 are the equilibrium constants at temperatures T1 and T2 respectively, ΔH° is the Show more…
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The solubility of $\operatorname{AgCl}(s)$ in water at $25^{\circ} \mathrm{C}$ is $1.33 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$ and its $\Delta H^{\circ}$ of solution is $65.7 \mathrm{~kJ} / \mathrm{mol}$. What is its solubility at $50.0^{\circ} \mathrm{C}$ ?
The solubility of $\mathrm{AgCl}(s)$ in water at $25^{\circ} \mathrm{C}$ is $1.33 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$ and its $\Delta H^{\circ}$ of solution is $65.7 \mathrm{~kJ} / \mathrm{mol}$. What is its solubility at $50.0^{\circ} \mathrm{C} ?$
The solubility of AgCl(s) in water at 25 C is 1.33 * 10-5 mol>L, and its H of solution is 65.7 k>mol. What is its solubility at 50.0 C?
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