Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺
|| Ag⁺ | Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu
⇄ Cu²⁺ + 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺ + 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu ⇄ Cu²⁺ + 2e⁻
Ag⁺ + e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺ + 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu + 2Ag⁺ + 2e⁻⇄ Cu²⁺ + 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu + 2Ag⁺ ⇄ Cu²⁺ + 2Ag
Answer:
I believe it's A. to reduce air bubbles. Tbh, it's been a while
Answer:
Name ; Nickel(ii)Hydroxide
Formula;NI3O6
Explanation:
For [Ni(en)³]²⁺ which is purple, the crystal field splitting energy is greater than the complex ion, [Ni(H₂O)₆]²⁺ which is green in color.
When a Lewis base id attached to the metal ion by covalent bond, then the complex ion is formed and when these complex ions are present with other ions of opposite charge or neutral charge, they will make complex compounds.
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
