= 5.4×10-6 g AgBr
AgBr(s) → Ag+(aq) + Br-(aq) Ksp = 3.3×10^-13
[Ag+][Br-] = Ksp = 3.3×10^-13 = X²
X = 5.7×10^-7 M = [Ag+] = mol/L AgBr that dissolve
1 mole of AgBr = 187.8 g
Therefore;
= 0.05 L × (5.7 × 10^-7 mol/L) × (187.8 g /mol)
= 5.4×10-6 g AgBr
The two factors that accounts for increased rate of chemical reaction when temperature is increased are:
More important factor; Energy factor is more important.