Solution:
To find the time required to coat the metal surface, we use Faraday's law of electrolysis: \[ m = \frac{M I t}{n F}, \] where:
\( m \) is the mass of the substance deposited,
\( M \) is the molar mass of the substance (nickel, 60 g/mol),
\( I \) is the current (2 A),
\( t \) is the time in seconds,
\( n \) is the valency of the substance (for nickel, \( n = 2 \)),
\( F \) is the Faraday constant (96500 C/mol).
We are given that the thickness of the nickel layer is 0.001 mm (0.0001 cm), so the volume of the nickel deposited is: \[ V = \text{Area} \times \text{Thickness} = 100 \, \text{cm}^2 \times 0.0001 \, \text{cm} = 0.01 \, \text{cm}^3. \] Using the density of nickel \( \rho_{\text{Ni}} = 10 \, \text{g/mL} \), we can calculate the mass of the deposited nickel: \[ m = \rho_{\text{Ni}} \times V = 10 \times 0.01 = 0.1 \, \text{g}. \] Now, substitute into Faraday’s equation: \[ 0.1 = \frac{60 \times 2 \times t}{2 \times 96500}. \] Solving for \( t \): \[ t = \frac{0.1 \times 2 \times 96500}{60 \times 2} = 160.83 \, \text{seconds}. \] Thus, the time required to coat the desired layer is approximately 161 seconds.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are

| Column I (Chemical reactions) | Column II (Enzymes used) | ||
| (i) | Glucose → CO2 + Ethanol | a | Pepsin |
| (ii) | Sucrose→Glucose + Fructose | b | Diastase |
| (iii) | Starch →Maltose | c | Zymase |
| (iv) | Protein→Amino acids | d | Invertase |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)