Explore topic-wise InterviewSolutions in Current Affairs.

The answer is Zero order.

(i) It prevents froth formation by one sulphide ore and allowing the other to come into froth. 

(ii) If the ore contains mixture of ZnS and PbS, NaCN allows PbS to come with froth and ZnS to sink.

(iii) It helps to separate two sulphide ores in a mixture like ZnS and PbS.

The answer is Chemisorption.

The gas is evolved at cathode during the electrolysis of an aqueous solution of NaCl is Chlorine.

Half-life of a first order reaction is independent of the initial concern trution it remains constant.

(b) Negative

(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate

(c) Co₂

Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.

Given: θ₀ = 25ºC, θ₁ = 50ºC, θ₂ = 45ºC,

[\(\frac{d\theta}{dt}\)]₁ = 4ºC/,om

To find: Rate of cooling at θ₂ (\(\frac{d\theta}{dt}\))₂

Formula: \(\frac{d\theta}{dt}\) = K (θ-θ₀)

Calculation: Using formula, for θ₁ = 50ºC

4 = K (50-25)

∴ K = \(\frac{4}{25}\) min^-1

∴ θ₂ = 45ºC

(\(\frac{d\theta}{dt}\))₂ = \(\frac{4}{25}\) x (45-25)

= \(\frac{4}{25}\) x (20)

= \(\frac{16}{5}\)

= 3.2ºC/min

The rate of cooling at 45 ºC is 3.2ºC/min.

Formation of stationary waves by analytical method:

Consider two identical progressive waves of equal amplitude and frequency travelling along X axis in opposite direction. They are given by

y₁ = A sin \(\frac{2π}{λ}\) (vt - x) along positive X-axis     .........(1)

y₂ = A sin \(\frac{2π}{λ}\) (vt + x) along positive X-axis     .........(2)

The resultant displacement ‘y’ is given by the principle of superposition of waves,

Y = y₁ + y₂                    .....(3)

_{y =}  A sin \(\frac{2π}{λ}\) (vt - x) +  A sin \(\frac{2π}{λ}\) (vt + x)

By using,

sin C + sin D = 2sin [\(\frac{C+D}{2}\)] cos [\(\frac{C-D}{2}\)]

We get,

y = 2A sin [\(\frac{2π}{λ}\) \((\frac{vt-x+vt+x}{2})\)] cos  [\(\frac{2π}{λ}\) \((\frac{vt-x+vt+x}{2})\)]

= 2A sin \((\frac{2πvt}{λ})\) cos \((\frac{2π}{λ}{(-x)})\)

∴ y = 2Asin 2πnt cos \((\frac{2πx}{λ})\)     .....[∴ n = \(\frac{v}{λ}\) and cos (-θ) = cosθ]

∴ y = 2Asin \((\frac{2πx}{λ})\) sin 2πnt

Let R = 2Acos \((\frac{2πx}{λ})\)

∴ y = Rsin (2πnt)    ...........(4)

But, ω = 2πn

∴ y = R sin ωt        ............(5)

Equation (5) represents the equation of S.H.M. Hence, the resultant wave is a S.H.M. of amplitude R which varies with x.

The absence of x in equation (5) shows that the resultant wave is neither travelling forward nor backward. Therefore it is called as stationary wave.

Amplitude at node is minimum, i.e., 0.

∴ R_min = 0

Since R = 2A cos \((\frac{2πx}{λ})\)

∴ cos \((\frac{2πx}{λ})\) = 0

∴ \(\frac{2πx}{λ}\) = \(\frac{π}{2},\frac{3π}{2},\frac{5π}{2},.......\)

∴ x = \(\frac{λ}{4},\frac{3λ}{4},\frac{5λ}{4},.......\)

At antinodes: R = ± 2A

∴ cos \((\frac{2πx}{λ})\) = ± 1

∴ \((\frac{2πx}{λ})\) = 0, π, 2π,3π,.............nπ

∴ 0, \(\frac{λ}{2}\), λ, \(\frac{3λ}{2}........\)

∴ Distance between first node and adjacent

antinode = x_n - x_an = \(\frac{λ}{4}\)- 0 = \(\frac{λ}{4}\) 

Thus, the distance between a node and an adjacent antinode is \(\frac{λ}{4}\)

a. Optical activity (Chirality): The property of a substance to rotate the plane of a plane polarised light towards the right (clockwise) or towards the left (anticlockwise) is called optical activity. 

eg. Lactic acid, glucose, cane sugar, 2-chlorobutane, etc., shows optical activity.

b. The molecules or ions which are coordinated to the central atom or ion in the coordination compound are called ligands or donor groups. 

eg. 1. In [Ni(NH₃)₆]²⁺, central metal ion is Ni²⁺ and ligands are NH₃ molecules. 

2. In [Cr(H₂O)₆]³⁺, the central metal ion is Cr³⁺ and the ligands are H₂O molecules.

c. 1. Interstitial compounds are those which are formed when small atoms like H, C, N, B, etc., are trapped inside the crystal lattice of metals. 

2. Interstitial compounds have variable composition and are non-stoichiometric in nature. 

eg. TiC, TiH_1.73, Mn₄N, Fe₃H, VH_0.56, ZrH_1.92, etc.

Given \(v = k\sqrt{ gh}\)

Dimensional formula \(v = [LT^{-1}]\) 

Dimensional formula \(\sqrt {gh} = [LT^{-1} ]^\frac12 [L] ^\frac 12\) 

\(= \sqrt{[LT^{-2}] [L]}\)

\(= \sqrt{[L^2T^{^{-2}}]}\)

\(\sqrt{gh} = \sqrt{[LT^{-1}]^2}\)

\(\sqrt{gh} = [LT^{-1}]\)

Then 

\([LT^{-1}] = K [LT^{-1}]\)

\(k = \frac{[LT^{-1}]}{[LT^{-1}]}\)

\(k = 0\)

K is dimensionless.

dx² – y² and dz² orbitals.

The limiting molar conductivity of an electrolyte can be represented as the sum of the limiting molar conductivities of the individual cation and anion. 

Application: Kohlrausch law is applied to calculate the limiting molar conductivity of any electrolyte from the limiting molar conductivities λ⁰ of individual ions.

Faraday’s 1^st law states that:- The amount of chemical reaction which occurs at any electrode during electrolysis by a current is directly proportional to the quantity of electricity passed through the electrolyte.

Correct option is  A) letter carrier

Correct option is  B) clay

Smoke from fire looks white due to scattering from smoke molecules therefore for scattering molecule of smoke are bigger than λ of light, so option (1) is true.