Explanation:
It is known that relation between
,
, and pH is as follows.
![E_{cell} = E^{o}_{cell} - (\frac{0.0591}{n}) \times log[H^{+}] ](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%28%5Cfrac%7B0.0591%7D%7Bn%7D%29%20%5Ctimes%20log%5BH%5E%7B%2B%7D%5D%0A)
Also, it is known that
for hydrogen is equal to zero.
Hence, substituting the given values into the above equation as follows.
![E_{cell} = E^{o}_{cell} - (\frac{0.0591}{n}) \times log[H^{+}] ](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%28%5Cfrac%7B0.0591%7D%7Bn%7D%29%20%5Ctimes%20log%5BH%5E%7B%2B%7D%5D%0A)
0.238 V = 0 - (\frac{0.0591}{1}) \times log[H^{+}]
[/tex]
= 4.03
= antilog 4.03
= 3.5
As, pH =
.
Thus, we can conclude that pH of the given unknown solution at 298 K is 3.5.
Answer:
(D) Neither I nor II
Explanation:
The reagent potassium hydrogen phthalate is found in pure form and it is not partially hydrated .
The pallet form of NaOH ( solid form ) absorbs carbon dioxide gas from the atmosphere but NaOH in solution form does not absorb carbon dioxide .
Hence option D ) is right choice .
Answer:
Pure Water
Explanation:
The common ion effect describes the effect on equilibrium that occurs when a common ion (an ion that is already contained in the solution) is added to a solution. The common ion effect generally decreases solubility of a solute(Khan Academy).
NaCl, AgNO3, KCl, BaCl2 solutions all have a common ion with AgCl. As a result of this, AgCl will be much less soluble in these solvents than it is in pure water.
Therefore, AgCl will have the highest solubility in pure water compared to all the solutions listed above.
Answer: 2) see below
3) 2AgCl → 2Ag + Cl₂
<u>Explanation:</u>
2)
Step 1: Crushing & Grinding CuFeS₂(s)→ CuFeS₂(s)
Step 2: Froth Flotation CuFeS₂(s)----> CuFeS₂ (l)
Step 3: Roasting 2CuFeS₂(l) + 3O₂(g) → 2FeO(s) + 2CuS(s) + 2SO₂(g)
Step 4: Converting matte to blister Cu₂S(l) + O₂(g) → 2Cu(l) + SO₂(g)
Step 5: Anode Casting Cu(l) → Cu(s)
Step 6: Electrolytic Refining Cu(s) → Cu(s)
Anode: Cu(s) → Cu₂ + (aq) + 2e-
Cathode: Cu2 + (aq) + 2e- → Cu(s)
<em>see diagram below for illustration of steps</em>
3) When silver chloride (solid) is exposed to sunlight, it decomposes into silver (solid) and chlorine (gas).
The equation is written as: 