The correct reaction for the decomposition of lithium nitrate is:

4LiNO₃ → 2Li₂O + 4NO₂ + O₂

This reflects the breakdown of nitrate to oxide, nitrogen dioxide, and oxygen.

The -M effect of NH₂ is stabilizing the carbocation.

The correct order based on the acidity of the compounds is b > d > a > c.

The rate of SN1 reaction depends upon the stability of the carbocation which follows the order (b) > (d) > (c) > (a).

The correct sequence of reagents is Fe, H; Br₂(aq); HNO₂; H₃PO₂ to achieve the desired transformation.

The maximum number of electrons in the n = 4 shell is calculated by summing the maximum electrons each subshell can hold:

4s = 2, 4p = 6, 4d = 10, 4f = 14

Total = 2 + 6 + 10 + 14 = 32

The correct matches based on the hybridization of the central metal in each complex are:

• A - [Ni(CO)₄] uses sp³ hybridization.
• B - [Cu(NH₃)₄]²⁺ uses dsp² hybridization.
• C - [Fe(NH₃)₆]²⁺ uses d²sp³ hybridization.
• D - [Fe(H₂O)₆]²⁺ uses sp³d² hybridization.

Therefore, the correct matching is: A-I, B-II, C-IV, D-III

In the facial isomer of the octahedral complex [Co(NH₃)₃Cl₃], the Cl - Co - Cl bond angles are 90° due to the arrangement of the ligands around the cobalt center.

Assertion (A): The statement is false because OH groups are not typically reduced by Zn-Hg/HCl.

Reason (R): The statement is true as Zn-Hg/HCl is commonly used to reduce carbonyl groups to -CH₂- groups.

Therefore, Assertion A is false, but Reason R is true.

BeCl₂ has a covalent nature, making it soluble in organic solvents, unlike the other options which form ionic chlorides.

Step 1: Understand the mechanism of action.

Antiallergic drugs target histamine receptors, while antacid drugs act on gastric acid secretion, implying they function on different receptors.

Final Answer: Both A and R are true, and R correctly explains A.

Step 1: Identify the condition for a node.

At a radial node, Ψ = 0. Therefore:

2 - r₀/a₀ = 0 ⇒ r₀ = 2a₀.

Final Answer: The radial node is formed at r₀ = 2a₀.

Step 1: Oxidation in acidic medium.

In acidic solution:

2KMnO₄ + 10KI + 8H₂SO₄ → 2K₂SO₄ + 2MnSO₄ + 5I₂ + 8H₂O

Step 2: Oxidation in neutral/faintly alkaline medium.

In neutral or slightly alkaline solution:

2KMnO₄ + 10KI + 8H₂O → 2KOH + 2MnO₂ + 5I₂ + 8H₂O

However, more accurately, in neutral conditions, iodide can be oxidized to iodate (IO₃⁻):

2KMnO₄ + 10KI + 8H₂O → 2MnO₂ + 5I₂ + 4KOH + 8H₂O

Final Answer: I₂ in acidic solution and IO₃⁻ in neutral/faintly alkaline solution.

Step 1: Understand bond strength in group 16 hydrides.

The bond dissociation energy typically decreases as we move down the group due to increasing atomic size and decreasing bond strength.

Thus, H₂O > H₂S > H₂Se > H₂Te.

Final Answer: Bond dissociation energy decreases in the order A (O) > B (S) > C (Se) > D (Te).

Step 1: Understand the range of BOD values.

Biochemical Oxygen Demand (BOD) is an indicator of water quality. A BOD value < 5 mg/L indicates very clean water, while higher values signify increasing levels of pollution.

Final Answer: The pond water is very clean.

Step 1: Analyze the separation techniques and mixtures.

• A: CHCl₃ + C₆H₅NH₂ – These can be separated by simple distillation due to different boiling points. Hence, A-III.
• B: C₆H₁₄ + C₅H₁₂ – These hydrocarbons have similar boiling points and require fractional distillation for separation. Hence, B-IV.
• C: C₆H₅NH₂ + H₂O – Aromatic amines can be separated using steam distillation. Hence, C-I.
• D: Organic compound in H₂O – Organic compounds dissolved in water can be separated using differential extraction. Hence, D-II.

Therefore, the correct matching is: A-III, B-IV, C-I, D-II.

Step 1: Understand the relationship between current and force.

The magnetic force between two parallel current-carrying wires is given by:

F ∝ I₁I₂

Initially, both wires carry current I, so F₁ ∝ I².

After increasing the current to 2I in both wires, F₂ ∝ (2I)² = 4I².

Therefore, the ratio of forces F₁:F₂ = I² : 4I² = 1:4.

Final Answer: The ratio of magnetic forces per unit length is 1:4.

Step 1: Use the formula for elastic deformation.

Δl ∝ (F × l) / (A × E)

Where Δl is the elongation, F is the force, l is the length, A is the cross-sectional area, and E is Young's modulus (constant for both wires).

Step 2: Determine the changes for wire B.

• Length of wire B, l₂ = 2 × l₁
• Diameter of wire B, d₂ = 2.4 × d₁
• Cross-sectional area A₂ = π (d₂/2)² = π (1.2d₁)² = 1.44A₁

Step 3: Calculate the new elongation Δl₂.

Δl₂ = (F × 2l₁) / (1.44A₁ × E) = (2 / 1.44) × (F × l₁) / (A₁ × E) = (2 / 1.44) × Δl₁ ≈ 1.3889 × 0.2 mm ≈ 0.2778 mm ≈ 6.9 × 10^-2 mm

Final Answer: The elongation in wire B is 6.9 × 10^-2 mm.

Step 1: Use conservation of energy.

Potential energy at height R: U₁ = -G(Mm)/(2R)

Potential energy at Earth's surface: U₂ = -G(Mm)/R

Kinetic energy at Earth's surface: K = G(Mm)/(2R)

Thus, velocity v can be found using K = ½ mv²:

½ mv² = G(Mm)/(2R)

v² = 2G M / R = gR

v = √(gR)

Final Answer: The velocity is √(gR).

Step 1: Calculate the effective power.

P_eff = 100 W × 0.05 = 5 W

Step 2: Calculate the total intensity at 5 m.

I_total = P_eff / (4π r²) = 5 W / (4π × 25 m²) = 5 W / (100π m²) = 1 W/(20π m²)

Step 3: Determine the intensity of the electric field component.

In electromagnetic waves, the electric and magnetic field components carry equal energy, so the intensity due to the electric field is half of the total intensity.

I_E = I_total / 2 = (1 W/(20π m²)) / 2 = 1 W/(40π m²)

Final Answer: The intensity produced by the electric field component is 1 W/(40π m²).

Step 1: Identify the given data.

• q = 0 (adiabatic process).
• ΔU = w = -3000 J (work done on the system is negative).
• C_v = 20 J mol^-1K^-1.
• Initial temperature T₁ = 300 K.

Step 2: Use the formula for internal energy change.

ΔU = C_v × (T₂ - T₁)

Substitute the values:

-3000 = 20 × (T₂ - 300)

Simplify:

T₂ - 300 = -150 ⇒ T₂ = 150 K.

Final Answer: The final temperature is T₂ = 150 K.

Step 1: Use the density formula for a unit cell.

d = (Z × M) / (N_A × a³)

where:

• d = 4.0 g cm^-3 (density),
• M = 72 g mol^-1 (molar mass of FeO),
• a = 5.0 Å = 5.0 × 10^-8 cm (edge length),
• N_A = 6.0 × 10²³ mol^-1 (Avogadro's number),
• Z = number of units per unit cell.

Step 2: Substitute the values.

4 = (Z × 72) / (6.0 × 10²³ × (5.0 × 10^-8)³)

Simplify:

4 = (Z × 72) / (6.0 × 10²³ × 125 × 10^-24)

4 = (Z × 72) / (7.5 × 10^-1) ⇒ Z × 72 = 4 × 7.5 × 10^-1 ⇒ Z = (4 × 7.5 × 10^-1) / 72 = 0.04167 ≈ 4.1667

Rounding to the nearest integer, Z = 4.

Final Answer: Z = 4 FeO units per unit cell.

Step 1: Use the formula for first-order reactions.

The time for decomposition is given by:

t = (1/k) ln(a / (a - x))

where k is the rate constant.

For 60% decomposition:

t₁ = (1/k) ln(a / 0.4a) = (1/k) ln(2.5)

For 90% decomposition:

t₂ = (1/k) ln(a / 0.1a) = (1/k) ln(10)

Find the ratio of t₂ to t₁:

t₂ / t₁ = ln(10) / ln(2.5)

Given ln(10) = 2.3 and ln(2.5) ≈ 0.9163,

t₂ / t₁ ≈ 2.3 / 0.9163 ≈ 2.51

Calculate t₂:

t₂ ≈ 2.51 × 540 ≈ 1355.4 s

Rounding to the nearest integer, t₂ ≈ 1350 s.

Final Answer: 1350 s

Step 1: Use the freezing point depression formula.

ΔT_f = i × K_f × m

where:

• i = 2.67 (van't Hoff factor)
• K_f = 1.8 kg mol^-1
• m = molality

Step 2: Calculate molality.

Given 38% by weight H₂SO₄ in 1 kg of solution:

Mass of H₂SO₄ = 0.38 kg = 380 g

Molar mass of H₂SO₄ = 98 g mol^-1

Number of moles of H₂SO₄ = 380 g / 98 g mol^-1 ≈ 3.8776 mol

Mass of solvent (water) = 1 kg - 0.38 kg = 0.62 kg

Molality, m = 3.8776 mol / 0.62 kg ≈ 6.25 mol kg^-1

Step 3: Calculate freezing point depression.

ΔT_f = 2.67 × 1.8 × 6.25 ≈ 30.06 K

Step 4: Determine the freezing temperature.

Assuming pure water freezes at 273 K:

T_f = 273 K - 30.06 K ≈ 242.94 K

Rounding to the nearest integer, T_f ≈ 243 K.

Final Answer: 243 K

Step 1: Analyze each factor.

• A. Increasing temperature: The reaction is exothermic (ΔH = -190 kJ). Increasing temperature shifts the equilibrium towards the reactants, decreasing SO₃ yield.
• B. Increasing pressure: The reaction reduces the number of gas molecules (3 moles on reactant side to 2 moles on product side). Increasing pressure shifts the equilibrium towards the products, increasing SO₃ yield.
• C. Adding more SO₂: Increases the concentration of reactants, shifting the equilibrium towards the products, increasing SO₃ yield.
• D. Adding more O₂: Similarly increases the concentration of reactants, shifting equilibrium towards the products, increasing SO₃ yield.
• E. Addition of catalyst: Catalysts speed up the rate of reaching equilibrium but do not affect the position of equilibrium or the yield.

Step 2: Count the factors that increase SO₃ yield.

Factors B, C, and D increase the yield of SO₃.

Final Answer: 3 factors increase the yield of SO₃.

Step 1: Write the Freundlich adsorption isotherm.

log (x/m) = log K + (1/n) log p

The slope of the line is (1/n) and the intercept is log K.

Step 2: Determine the values of K and n.

The line is inclined at 45°, which means the slope (1/n) = 1 ⇒ n = 1.

The intercept is log K = 0.6020 ⇒ K = 10^0.6020 ≈ 4.

Step 3: Use the Freundlich equation to find x/m.

x/m = K × p^1/n = 4 × p

Given p = 0.4 atm,

x/m = 4 × 0.4 = 1.6

Expressed as 1.6 × 10^-1 × 10¹ = 16 × 10^-1

Final Answer: 16 × 10^-1

Step 1: Identify the compounds.

Compounds with phenolic –OH groups or carboxylic acid groups dissolve in NaOH and NaHCO₃ solutions, while those with ester or amide groups require hot NaOH for hydrolysis.

Step 2: Analyze the structures.

• Compound 1: Ester group, dissolves in hot NaOH.
• Compound 2: Phenolic group, dissolves in cold NaOH.
• Compound 3: Ester group, dissolves in hot NaOH.
• Compound 4: Phenolic group, dissolves in cold NaOH.
• Compound 5: Amide group, dissolves in hot NaOH.

Step 3: Count the compounds dissolving only in hot NaOH.

The compounds are 1, 3, and 5.

Final Answer: 3 compounds dissolve in hot NaOH only.

Step 1: Understand the structure of peptides.

A peptide linkage is formed between two amino acids. The number of linkages is one less than the total number of amino acids.

Step 2: Calculate the total number of amino acids.

Total amino acids = 3 (G) + 2 (L) + 2 (V) = 7.

Step 3: Calculate the number of peptide linkages.

Peptide linkages = Total amino acids - 1 = 7 - 1 = 6.

Final Answer: The peptide has 6 linkages.

Step 1: Define the volume strength.

A 50-volume solution releases 50 L of O₂ at STP per litre of H₂O₂.

Step 2: Calculate the molarity.

Molarity = Volume strength / 11.35 = 50 / 11.35 ≈ 4.405 M.

Step 3: Calculate the strength in g/L.

Strength = Molarity × Molar mass = 4.405 × 34 ≈ 149.77 g/L ≈ 150 g/L.

Final Answer: The strength of the solution is 150 g/L.

Step 1: Write the Nernst equation.

The cell potential is:

E_cell = E^°_cell - (0.06/n) log(Q)

where Q is the reaction quotient.

Step 2: Determine E^°_cell.

E^°_cell = E^°_Y²⁺/Y - E^°_X²⁺/X = 0.36 V - (-2.36 V) = 2.72 V.

Step 3: Substitute the values into the Nernst equation.

E_cell = 2.72 V - (0.06/2) log(0.001/0.01)

log(0.001/0.01) = log(0.1) = -1.

E_cell = 2.72 V - (0.03)(-1) = 2.72 V + 0.03 V = 2.75 V.

Expressed as 275 × 10^-2 V.

Final Answer: 275 × 10^-2 V.