Complete Question:
Ions to calculate the p-values: Na⁺, Cl⁻, and NH₄⁺
Answer:
pNa = 0.307
pCl = 0.093
pNH₄ = 0.503
Explanation:
The p-value is calculated by the antilog of the concentration of the substance of interest. For example, pH = -log[H⁺]. Thus, first, let's find the ions concentration.
Both substances are salts that solubilize completely, thus, by the solution reactions:
NaCl → Na⁺ + Cl⁻
NH₄Cl → NH₄⁺ + Cl⁻
So, for both reactions the stoichiometry is 1:1:1 and the concentration of the ions is equal to the concentration of the salts.
[Na⁺] = 0.493 M
[Cl⁻] = 0.493 + 0.314 = 0.807 M
[NH₄⁺] = 0.314 M
The p-values are:
pNa = -log[Na⁺] = -log(0.493) = 0.307
pCl = -log[Cl⁻] = -log(0.807) = 0.093
pNH₄ = -log[NH₄⁺] = -log(0.314) = 0.503
Your Answer Will Be Intensive Property
It is a.combination that is the correct answer i think
Answer:
Copper ions are reduced into copper atoms.
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
Explanation:
During electrolysis, the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.
At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)
Thus the half equation is as follows:
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
