Emission of a positron results in the conversion of a proton into neutron, along with the release of neutrino. Thus, the number of protons decreases and of neutron increases, i.e., n/p ratio increases and the nuclei becomes stable.

$\underset{\mathrm{Proton}}{{}_1{}^1\mathrm{H}} \to \underset{\mathrm{Neutron}}{{}_0{}^1\mathrm{n}} + \underset{\mathrm{Psitron}}{{}_{+1}\mathrm{e}^0} + \mathrm{v}$

Given,

    C + O₂ → CO₂ ; $∆\mathrm{H}°$ = -x kJ        ...(i)

2CO + O₂ → 2CO₂ ; $∆\mathrm{H}°$ = -y kJ      ...(ii)

Required equation as per question is C + $\frac{1}{2}$O₂ → CO, $∆{\mathrm{H}}_1^{°}$

On reversing equation (ii), we get

2CO₂ → 2CO + O₂; $∆{\mathrm{H}}_2^{°}$ = +y kJ      ...(iii)

Dividing equation (iii) by 2 gives the following reaction

CO₂ → CO + $\frac{1}{2}{\mathrm{O}}_2$ ; $∆H_3^{°}$ = + $\frac{\mathrm{y}}{2}\mathrm{kJ}$    ...(iv)

On adding equation (i) and (iv), we get 

C + $\frac{1}{2}{\mathrm{O}}_2$ → CO ; $$\begin{gathered} ∆{\mathrm{H}}_1^{°} = ∆\mathrm{H}° + ∆{\mathrm{H}}_3^{°} \\ = \left(-\mathrm{x} + \frac{\mathrm{y}}{2}\right) \\ = \frac{\mathrm{y} - 2\mathrm{x}}{2} \mathrm{kJ} \end{gathered}$$

The given statement is true as electrons are shifted more towards the more electronegative atom.

₃₂Ge⁷⁶ and ₃₄Se⁷⁶ have the same mass number but different atomic number, therefore, they are isobars.

Number of neutrons, in

 ₁₄Si³⁰ = 30 - 14 = 16

₁₆Si³² = 32 - 16 = 16

Because of the presence of same number pf neutrons, ₁₄Si³⁰ and ₁₆S³², they are known as isotones.

10^-4 M KOH = 10^-4 M [OH^-]

We know that,

[H⁺][OH^-] = 1 × 10^-14

[H⁺] × 10^-4 = 1 × 10^-14

[H⁺] = $\frac{1 \times {10}^{-14}}{1 \times {10}^{-4}}$ = 1 × 10^-10 M

$\therefore$ pH = - log [H⁺] = -log (1 × 10^-10 M) = 10

Number of atoms = $\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}$ × N_A × Number of atoms in 1 mole

$\therefore$Number of atoms in 8 gm

O₂ = $\frac{8}{32 } \times {\mathrm{N}}_{\mathrm{A}} \times 2 = \frac{{\mathrm{N}}_{\mathrm{A}}}{2}$

Number of atoms in 2 gm

H₂ = $\frac{2}{2} \times {\mathrm{N}}_{\mathrm{A}} \times 2 = 2{\mathrm{N}}_{\mathrm{A}}$

Number of atoms in 4 gm

He = $\frac{4}{4} \times {\mathrm{N}}_{\mathrm{A}} \times 1 = {\mathrm{N}}_{\mathrm{A}}$

Thus, 2 gm H₂ contains the maximum number of atoms among the given.

Given,

$∆$H = -400 kJ mol^-1 ; $∆$S = -20 kJ mol^-1 K^-1

Gibbs Helmholtez equation is

$∆\mathrm{G} = ∆\mathrm{H} - \mathrm{T}∆\mathrm{S}$

For a reaction to be spontaneous, $∆$G must be less than 0, i.e. it should be negative.

$$\begin{gathered} \therefore 0 \ge ∆\mathrm{H} - \mathrm{T}∆\mathrm{S} \\ \mathrm{T}∆\mathrm{S} \le ∆\mathrm{H} \\ \mathrm{T} \le \frac{∆\mathrm{H}}{∆\mathrm{S}} \le \frac{-400}{-20} \le 20 \end{gathered}$$

The value of $∆$H is

$∆$H = -nC_p$∆\mathrm{T}$

      = -2 × $\frac{5}{2}$R ( 225 - 125)

      = -5R (100)

      = -500 R

Therefore, $∆$H of -500 R is required for cooling 2 mole of an ideal monoatomic gas from 225°C to 125°C at constant pressure.

Supoose, the unknown radioactive substance is _xA^y.

_xA^y + ₆C¹² → ₉₈Cf²⁴⁶ + ₀n¹

On comparing atomic numbers, we get

x + 6 = 98 + 0

x = 98 - 6 = 92

On comparing mass numbers, we get

y + 12 = 246 + 1

y = 247 - 12 = 235

Since, uranium has the atomic number 92, thus, the given unknown substance was ₉₂U²³⁵.

Entropy of vapourisation ($∆$S) = $\frac{\mathrm{enthalpy} \mathrm{of} \mathrm{vaporisation} (∆\mathrm{H})}{\mathrm{boiling} \mathrm{point} (\mathrm{in}\hspace{0.17em}\mathrm{K})}$

Given, $∆$H = 24.64 kJ mol^-1 and boiling point = 35 + 273 = 308 K

$\therefore$ $$\begin{gathered} ∆\mathrm{S} = \frac{24.64 \times {10}^{3 }\mathrm{J} {\mathrm{mol}}^{-1}}{308 \mathrm{K}} \\ = 80 \mathrm{JK}{ }^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$