Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Answer: The number of protons
Explanation:
Answer:
a. HCl + NH₃ ⇄ NH₄Cl
b. KOH + NH₄⁺ ⇄ K⁺ + NH₃ + H₂O
Explanation:
A buffer system is made by a weak base (ammonia, NH₃) and its conjugate acid (ammonium ion NH₄⁺ coming from ammonium chloride NH₄Cl). Its function is to resist abrupt changes in the pH when acids or bases are added.
a. When aqueous hydrochloric acid is added, it reacts with the base of the buffer. The corresponding equation is:
HCl + NH₃ ⇄ NH₄Cl
b. When aqueous potassium hydroxide is added, it reacts with the acid of the buffer. The corresponding equation is:
KOH + NH₄⁺ ⇄ K⁺ + NH₃ + H₂O