A. Two electrodes separated by an electrolyte that can generate an electrical current.
Two or more different elements
PH of a solution will be <span>higher than 7
</span>
Ammonium cyanide is a salt formed by hydrogen cyanide and ammonia. Ammonia is a weak base and hydrogen cyanide is a weak acid.
NH₄CN + H₂O ⇒ NH₃ + HCN
NH₄⁺ + H₂O -----> H₃O⁺ + NH₃
CN⁻ + H₂O -----> HCN + OH⁻
Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:
Ka(HCN) = 4.9 x × 10⁻¹⁰
Kb(NH₃) = 1.8 × 10⁻⁵<span>
Kw= </span>1.0 × 10⁻¹⁴
Let's first calculate Ka for NH₄⁺:
Ka(NH₄⁺) x Kb(NH₃<span>) = pKw
</span>Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰
Then, Kb for CN⁻:
Kb(CN⁻) x Ka(HCN) = pKw
Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵
From this, we can see that the acid constant NH4⁺ is much lower than the base constant of CN⁻, which will say that the solution of NH₄CN will react slightly alkaline because of the higher presence of hydroxyl ions in solution.
Strong Acid cause it has a very low pH
The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C.
