At constant volume, q_V = C_VΔT = ΔU

At constant pressure, q_P = C_PΔT = ΔH

For a mole of an ideal gas,

ΔH = ΔU + Δ(PV)

= ΔU + Δ(RT)

= ΔU + RΔT

On putting the values of ΔH and ΔU, we have

C_PΔT = C_VΔT + RΔT

C_P = C_V + R

C_P − C_V = R

In 1 L solution of H₂SO₄,

Therefore, [H⁺] = 10^-2

pH = -log 10^-2 = 2

Hence, option (1) is correct.

All the other options are incorrect.

The conjugate base of H₂S is HS^-.

BF₃ is a Lewis acid.

Phenolphthalein is pink in basic medium.

According to Le Chatelier's principle, when a system at equilibrium is subjected to a change, the equilibrium shifts in the direction so as to undo the change. Therefore, if the concentration of product is decreased, the forward reaction takes place so as to increase the concentration.

An acidic buffer can be prepared by making a solution of an acid and its salt solution. So, an acidic buffer can be prepared by making a solution of formic acid and sodium formate.

Buffer solutions are those solutions which resist the change in pH on addition of small amount of acid or base.

Buffers solutions are the mixture of either a weak acid and its salt with a strong base or a weak base and its salt with a strong acid.

(a) NaCl + NaOH is not a buffer solution because NaOH is a strong base.

(c) CH₃COOH + CH₃COONH₄ is not a buffer solution because CH₃OONH₄ is a salt with a weak base.

(d) H₂SO₄ + CuSO₄ is not a buffer solution because H₂SO₄ is a strong acid.

(d) CH₃COOH + CH₃COONa is a buffer solution because CH₃COOH is a weak acid and CH₃COONa is its salt with a strong base.

If the total volume of a container is increased, this means that the total pressure is decreased. The equilibrium will then shift to the side with more moles of gases. In option 2, since CaO and CaCO₃ are solids, we can assume that their concentration are constant. But CO₂ is a gas. Therefore, in this case, the reaction will proceed in forward direction.

Be(OH)₂ ​has the lowest value of K_sp at ordinary temperature because Be²⁺ ion is smaller than the other metal ions in the group, which results in a tighter bond with the OH^– ions, thus much lower solubility.

The solubility of a hydroxides of group 2 elements increases down the group because as you go down the group, the size of metal increases, thereby increasing the bond length and decreasing the bond energy.

Let 's' be the solubility of salt M₂S which undergoes dissociation as follows:-

Hence, the solubility product (K_sp) = (s) × (2s)²

∴ K_sp = 4 × s³

∴ K_sp = 4 × (3.5 × 10^-6)³

∴ K_sp = 1.7 × 10^-16

Ba(OH)₂(s) →  Ba²⁺(aq) + 2OH^-(aq)

One mole of Ba(OH)₂ give 2 moles of OH^- ions.

It is given that the concentration of Ba(OH)₂ solution is 0.10 M.

Hence, the concentration of OH^- ions is 0.20 M

pOH = -log[OH^-]

= -log(0.20)

= 0.699

From pOH calcuate the pH of the solution as follows:

pH = 14 - pOH

= 14 - 0.699

= 13.3