Under which condition will a precipitate of a sparingly soluble salt form when two solutions are mixed?
Correct answer A. When the ionic product exceeds the Ksp of the salt
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Correct answer A. When the ionic product exceeds the Ksp of the salt
Correct answer D. mol⁻² dm⁶
Correct answer B. There is no change in the equilibrium position
Correct answer B. H₂(g) + I₂(g) ⇌ 2HI(g)
Correct answer C. 49
Correct answer A. 1.8 × 10⁻⁹ mol dm⁻³
Correct answer C. It decreases
Correct answer D. About 450 °C, about 200 atm and an iron catalyst
Correct answer D. At lower temperatures the rate of reaction becomes too slow to be economical
Correct answer B. Ksp = [Ag⁺]²[CrO₄²⁻]
Correct answer B. 1.7 × 10⁻¹⁰ mol² dm⁻⁶
Correct answer C. 3.2 × 10⁻¹¹ mol³ dm⁻⁹
Correct answer B. Some AgCl precipitates because its solubility is lowered
Correct answer B. Ethanoic acid and sodium ethanoate
Correct answer A. NH₄OH and NH₄Cl
Correct answer B. It combines with CH₃COO⁻ ions to form undissociated CH₃COOH
Correct answer D. 5.74
Correct answer B. The forward and reverse reactions continue at equal rates
Correct answer C. Equilibrium is reached sooner but its position and K are unchanged
Correct answer B. The equilibrium shifts to the right because fewer gas molecules are on that side