An aqueous solution of sugar cannot conduct electricity because sugar does not dissociate into ions in water. While, sodium chloride (NaCl) dissociates into Na⁺ and Cl⁻ ions, which carry electric current.

OR

It is appropriate to use the term formula weight instead of molecular weight for sodium chloride (NaCl) because NaCl is an ionic compound, not a molecule. Formula weight refers to the sum of the atomic weights of the ions in the formula unit.

Hydrogen sulfide (H₂S) gas can be prepared in Kipp\'s apparatus.

Balanced Chemical Equation:

FeS + H₂SO₄ → FeSO₄ + H₂S ↑

In this reaction, iron(II) sulfide (FeS) reacts with sulfuric acid (H_₂SO_₄) to produce iron(II) sulfate (FeSO_₄) and hydrogen sulfide gas (H_₂S).

To obtain metallic zinc from zinc oxide, zinc oxide is reduced by carbon or carbon monoxide. The balanced chemical equation for the reduction of zinc oxide with carbon is:

ZnO + C → Zn + CO

OR

When a piece of zinc is added to an aqueous solution of CuSO₄, zinc displaces copper from the solution, resulting in the formation of zinc sulfate and metallic copper. The balanced chemical equation is:

Zn + CuSO₄​ → ZnSO₄​ + Cu

Electronic Theory Explanation:

Oxidation:

Zinc (Zn) loses electrons and is oxidized to Zn²⁺:

Zn → Zn ²⁺ + 2𝑒⁻

Reduction:

Copper ions (Cu²⁺) gain electrons and are reduced to metallic copper (Cu):

Cu²⁺ + 2e⁻ → Cu

Ethylene (C₂H₄) Structural Formula:

H₂​C=CH₂ ​

In ethylene, the two carbon atoms are connected by a double bond. The presence of the double bond indicates that ethylene is an unsaturated hydrocarbon.

OR

To prevent the usage of Ethanol as beverages, a small amount (approximately) of Methanol (CH₃OH) is added to it. This makes Ethanol unfit for drinking and is called denatured spirit.

Charle's law: At constant pressure, the volume of a given mass of a gas increases or decreases by 1/273 of its volume at 0°C for each one-degree rise or fall in temperature.

The mathematical form of Charle's law:

V_t=Vo(1+t/273)

V_o = volume at 0°C

V_t = volume at t°C

Balanced chemical equation:

(NH₄)_2​SO₄ + 2NaOH → 2NH₃  +Na₂​SO₄​ + 2H₂​O

Molecular mass of NH₃ = 17 g

Molecular mass of (NH₄)2​SO₄

= [2 × (14 + 4)] + [32 + (4 × 16)]

= 132 g

From the balanced equation, 2 moles of NH₃ are produced from 1 mole of (NH₄)2 ​SO₄. Therefore, the moles of ammonium sulfate required to produce 0.4 moles of NH₃ are:

Moles of (NH₄)2 ​SO₄ = 0.4/2 = 0.2 mol

Mass of (NH₄)2SO₄ = 0.2  × 132 = 26.4 g

26.4 g of ammonium sulfate will be required to prepare 6.8 g of ammonia.

The volume expansion coefficient of a gas at constant pressure is the fractional change in volume per unit change in temperature.

At 0°C (273 K), its value is approximately 0.00366 K⁻¹, meaning the gas volume increases by about 0.366% per 1°C increase in temperature at constant pressure.

OR

The two similarities between heat conduction and electrical conduction are

1. Both processes involve the transfer of energy.
2. Both require a medium for transfer.

Diamond is a non-metal with very high thermal conductivity.

The two laws of refraction of light:

1. The incident ray, the refracted ray and the normal at the point of incidence on the refracting surface lie in the same plane.
2. For a given pair of media and for a given colour of light, the sine of the angle of incidence bears a constant ratio to the sine of the angle of refraction.

Mathematical form of the second law

sin i/ sin r =constant

In an optical system, a ray close to and nearly parallel to the principal axis is known as a paraxial ray.

When a paraxial ray of light OA incidents on a concave of a very small aperture, then it gets reflected along AQ. If the principal focus and centre of curvature of the mirror be F and C respectively then AC being the radius of curvature will act as a normal to the reflecting surface at A.

Hence, at point A, the angle of incidence (i) is equal to the angle of reflection (r) and the reflected ray will meet the principal axis at the principal focus E ∴ OA || XY, AC is the transversal, ∠OAC = ∠FCA = i. Again, ∠FAC is also equal to i, hence ∠FCA =∠FAC ∴ FA = FC.

As the aperture of the mirror is very small and we are considering only the paraxial rays (very close to the principal axis) FA is nearly equal to FP Hence, FP = FC and hence CP = 2FP

∴ r = 2f

∴ radius of curvature of the mirror =2 × focal length of the mirror

OR

Let a ray of monochromatic light OP incidents at P on the refracting face AB at an angle of incidence i₁ and suffer refraction from air to glass, bend towards the normal making an angle of refraction equal to r₁ and finally travel along the path PQ inside the prism. Again the refracted ray PQ strikes the refracting surface AC of the prism at an angle of incidence r₂ and after suffering refraction from glass to air, it emerges out of the prism at an angle of emergence i₂.

Here, the angle between the direction of the incident ray and the emergent ray (∠LMQ) is called the angle of deviation (δ) of the ray of light after refraction through the prism and its value is

δ = ∠MPQ + ∠MQP = (i_₁- r₁)+(i₂ - r₂)

δ = (i_₁+ i₂) - (r₁ + r₂)  …(i)

Again, in the quadrilateral APNQ,

∠PAQ + ∠PNQ + 90° + 90° = 360°

or, A + ∠PNQ = 180° [where A = ∠PAQ]

∴ ∠PNQ=180° - A …(ii)

(ii) But in ΔPNQ,

r_₁ + r_₂ + ∠PNQ = 180°

or, ∠PNQ = 180° - (r_₁ + r_₂) …(ii)

From equation (ii) and (iii) we get,

A = r_₁ + r_₂  …(iv)

∴ From equation (i)and (iv) we get,

δ = i_₁+ i₂ - A (Hence Proved)

Faraday's law of electromagnetic induction:

1. Whenever there is a change in magnetic flux linked with a coil, an emf is induced, which lasts as long as the change in magnetic flux is linked with the coil changes.
2. The magnitude of the induced emf is directly proportional to the rate of change magnitude flux linked with the coil.

Alternating Current: An alternating current can be defined as a current that changes its magnitude and polarity at regular intervals of time.

Resistance (R) = 440 ohm

Voltage (V) = 220 V

Time (t) = 10 h

Calculation of power:

P = V^2\over R = 220^2/440 = 110 watts

Calculation of Electrical Energy:

E = P × t

= 110 watts × 10 h

= 1100 Wh

= 1.1 BOT

OR

Initial length = l

Initial area = a

Initial Resistance (R) = ρ l\over a = 6 ohm

New length = 2 × l

New Area = a \over 2

Final Resistance (R') = ρ 2 × l\over {a\over2} = 4 × ρ l\over a  = 4 × 6 ohm = 24 ohm

The final resistance of the stretched wire is 24 ohms.