Solution:

We are given that B is the adjoint of a matrix A and the determinant of A is |A| = 4.

The adjoint matrix B is related to the determinant of A as follows:

B = adj(A) = |A| × A⁻¹.

The elements of B are the cofactors of the corresponding elements in A. The element B₁₂ (α in this case) is the cofactor of A₁₂. Since B is given as the adjoint matrix and |A| = 4, the cofactor corresponding to A₁₂ is equal to 1.

Conclusion: Therefore, α = 1.

Solution:

The IUPAC name of an ether is derived by naming the two alkyl groups attached to the oxygen atom and adding the word "ether" at the end. In this case, the ether has a methoxy group (CH₃O-) and an ethane group (C₂H₅) attached to the oxygen atom.

The correct IUPAC name is "Methoxy ethane," where "methoxy" represents the CH₃ group, and "ethane" represents the C₂H₅ group.

Conclusion: The correct name is Methoxy ethane.

Solution:

The inverse of a 3x3 matrix A is given by the formula:

A⁻¹ = (1/|A|) × adj(A),

where |A| is the determinant of A and adj(A) is the adjoint matrix.

Step 1: Calculate the determinant of A.

|A| = 0 × | 2 3 | | 1 1 | - 1 × | 1 3 | | 3 1 | + 2 × | 1 2 | | 3 1 |.

|A| = 0 - 1 × (-8) + 2 × (-5) = 8 - 10 = -2.

Step 2: Calculate the adjoint of A.

After calculating cofactors, adj(A) = | -1 -1 -1 | | -8 6 -1 | | 5 -3 1 |.

Step 3: Compute A⁻¹.

A⁻¹ = (1/|A|) × adj(A) = (1/-2) × adj(A) = (1/2) | -1 -1 -1 | | -8 6 -1 | | 5 -3 1 |.

Conclusion: The correct inverse is:

A⁻¹ = (1/2) | -1 -1 -1 | | -8 6 -1 | | 5 -3 1 |.

Solution:

Clemmensen reduction involves the reduction of aldehydes or ketones to alkanes using zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl).

Reaction:

R-CO-R' + Zn(Hg) + HCl → R-CH₂-R'.

This method is suitable for compounds that are stable in acidic conditions.

Conclusion: Clemmensen reduction reduces aldehydes and ketones to alkanes using zinc amalgam and HCl.

Solution:

Zinc (Zn) in the 3d series shows the lowest oxidation state because of its fully filled 3d orbitals (d¹⁰ configuration).

Electronic Configuration:

Zn: [Ar] 3d¹⁰ 4s².

Since the d-orbitals are completely filled, zinc typically shows an oxidation state of +2 by losing two 4s electrons, which is lower than the variable oxidation states seen in other 3d elements.

Conclusion: Zinc has the lowest oxidation state in the 3d series.

Solution:

The pH of a buffer solution is calculated using the Henderson-Hasselbalch equation:

pH = pKₐ + log([salt]/[acid])

Here:

• pKₐ is the negative logarithm of the acid dissociation constant (Kₐ).
• [salt] is the concentration of the conjugate base (salt).
• [acid] is the concentration of the weak acid.

Conclusion: Use the given concentrations and pKₐ to calculate the pH using the Henderson-Hasselbalch equation.

Solution:

The pH of a solution is related to the concentration of hydrogen ions (H⁺) by the equation:

pH = -log[H⁺]

Given that the pH is 2.7, rearrange the equation to solve for [H⁺]:

[H⁺] = 10⁻²·⁷

Calculating:

[H⁺] ≈ 10⁻²·⁷ ≈ 1.99 × 10⁻³ M

Conclusion: The concentration of hydrogen ions is approximately 1.99 × 10⁻³ M.

Solution:

According to Henry's law, the solubility of a gas (S) in a liquid is directly proportional to the external pressure (P) at constant temperature.

Mathematically:

S ∝ P

Conclusion: The relationship between solubility and pressure is S ∝ P.

Solution:

In a Body-Centered Cubic (BCC) unit cell:

• There is 1 atom at the center of the cube.
• There are 8 atoms at the corners, each contributing 1/8th to the unit cell.

Total contributions:

Corner atoms = 8 × (1/8) = 1 atom

Center atom = 1 atom

Total = 1 + 1 = 2 atoms

Conclusion: The total number of unit particles in a BCC unit cell is 2.

Solution:

Pyridinium chlorochromate (PCC) is a mild oxidizing agent that selectively oxidizes primary alcohols (R-CH₂-OH) to aldehydes (R-CHO) without further oxidation to carboxylic acids.

Reaction:

R-CH₂-OH + PCC → R-CHO

Conclusion: The most suitable reagent for this conversion is PCC.

Solution:

For a Body-Centered Cubic (BCC) unit cell, the relation between edge length (a) and the atomic radius (r) is given by:

a = 4r/√3

This relationship is derived from the geometry of the BCC unit cell, where the body diagonal contains 2 atomic radii.

Conclusion: The edge length is 4r/√3.

Solution:

Nanoparticles are generally characterized using advanced techniques such as Transmission Electron Microscopy (TEM) or Dynamic Light Scattering (DLS). None of the provided options represent preliminary tests for nanoparticles.

Conclusion: The correct answer is None of these.

Solution:

The IUPAC name of a haloarene is derived by prefixing "halo-" (e.g., chloro, bromo, iodo) to the name of the parent hydrocarbon (e.g., benzene, toluene). The exact name depends on the specific halogen group present.

Conclusion: The name is based on the halogen and the parent hydrocarbon.

Solution:

The converse of an implication statement A ⇒ B is B ⇒ A.

For the given statement ((∼p) ∧ q) ⇒ r, its converse is r ⇒ ((∼p) ∧ q).

Rewriting r ⇒ ((∼p) ∧ q) using logical equivalence:

(∼r) ⇒ (p ∨ (∼q))

Conclusion: The converse of the statement is (p ∨ (∼q)) ⇒ (∼r).

Solution:

To find the negative of (p ∧ (∼q)) ∨ (∼p):

∼[(p ∧ (∼q)) ∨ (∼p)]

Using De Morgan's laws:

∼(p ∧ (∼q)) ∧ ∼(∼p)

(∼p ∨ q) ∧ p

Simplifying:

p ∧ q

Conclusion: The negative of (p ∧ (∼q)) ∨ (∼p) is equivalent to p ∧ q.

Solution:

The variance of a probability distribution is calculated as:

Variance = E(X²) - (E(X))²

Step 1: Calculate E(X):

E(X) = Σ x × P(X)

E(X) = (0 × 9/16) + (1 × 3/8) + (2 × 1/16)

E(X) = 0 + 3/8 + 2/16

E(X) = 4/8 = 1/2

Step 2: Calculate E(X²):

E(X²) = Σ x² × P(X)

E(X²) = (0² × 9/16) + (1² × 3/8) + (2² × 1/16)

E(X²) = 0 + 3/8 + 4/16

E(X²) = 5/8

Step 3: Calculate Variance:

Variance = E(X²) - (E(X))²

Variance = 5/8 - (1/2)²

Variance = 5/8 - 2/8

Variance = 3/8

Conclusion: The variance is 3/8.