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Answer:
Explanation:
AgCl ⇄ Ag⁺ + Cl⁻
m m m
If x mole of AgCl be dissolved in one litre .
[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰
m² = 1.6 x 10⁻¹⁰
m = 1.26 x 10⁻⁵ moles
So solubility of AgCl is 1.26 x 10⁻⁵ moles / L
Answer:
373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.
Explanation:
Mass of silver to be precipitated on ecah spoon = 0.500 g
Number of silver spoons = 250
Total mass of silver = 250 × 0.500 g = 125 g
Moles of AgCN = n =
Volume of AgCN solution =V
Molarity of the AgCN = 2.50 M
(1 L = 1000 mL)
373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.
Answer:
Explanation:
The equation for density is:
We plug in the given values: