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Class 12th Chemistry

Among Co³⁺, Ti²⁺, V²⁺ and Cr²⁺ ions, one if used as a reagent cannot liberate H₂ from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is _________ B.M. (Nearest integer)

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Vishal Baghel

Contributor-Level 10

$C o^{3 +}$ $^{}$ will not get oxidized, so will not liberate H2 gas upon reaction with acidic solution.

No. of unpaired electron = 4

Spin only magnetic moment = $\sqrt[]{4 (4 + 2)} B M = \sqrt[]{24} B M \approx 5 B M$

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Class 12th Chemistry

Consider the following metal complexes:

[Co(NH₃)₆]³⁺

[CoCl(NH₃)₅]²⁺

[Co(CN)₆]^3-

[Co(NH₃)₅(H₂O)]³⁺

The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is _____________. B.M (Nearest integer)

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Vishal Baghel

Contributor-Level 10

Complex absorbing minimum wavelength of light means the value of $Δ$ for it is maximum.

That complex will be ${[C o {(C N)}_{6}]}^{3 -}$ ${_{}}^{}$

$C o^{3 +} \Rightarrow 3 d^{6} 4 s^{0} \Rightarrow t_{2} g^{6} e g^{0}$

No. of unpaired electron = 0

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Class 12th Chemical Kinetics

The half life for the decomposition of gaseous compound A is 240s when the gaseous pressure was 500 Torr initially, When the pressure was 250 Torr, the half life was found to be 4.0 min. The order of the reaction is _________. (Nearest integer)

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Vishal Baghel

Contributor-Level 10

$\frac{{(t_{1 / 2})}_{l}}{{(t_{1 / 2})}_{I I}} = {(\frac{P_{I}}{P_{I I}})}^{1 - n} \Rightarrow n = 1$

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Class 12th Electrochemistry

The cell potential for $Z n | Z n^{2 +} (a q) ? S n^{* +} | S n$ is 0.801 V at 298 K. The reaction quotient of the above reaction is 10^-2. The number of electrons involved in the given electrochemical cell reaction is _________.

(Given: $E_{Z n^{2 +} | Z n}^{o} = - 0.763 V, E_{S n^{* +} | S n}^{o} = + 0.008 V a n d \frac{2.303 R T}{F} = 0.06 V)$

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Vishal Baghel

Contributor-Level 10

$A n o d e : (Z n \to Z n^{2 +} + 2 e) * x$

$C a t h o d e : \underline{(S n^{x +} + x e \to S n) * 2}$

Net call reaction : $x Z n + 2 S n^{x +} \to x Z n^{2 +} + 2 S n (n = 2 x)$

Using Nernst equation

No. of electron involved = 4

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5 months ago

Class 12th Communication Systems

The required height of a TV tower which can cover the population of 6.03 lakh is h. If the average population density is 100 per square km and the radius of earth is 6400 km, then the value of h will be ________ m.

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Vishal Baghel

Contributor-Level 10

$d = \sqrt[]{2 R h}$

$d = \sqrt[]{2 * 6400 * h * 1 0^{- 3}} (h i n m)$

Area = $π d^{2}$

$= (π * 2 * 6400 * h * 1 0^{- 3})$ km²

h = 150m

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Class 12th Physics

The energy band gap of semiconducting material to produce violet (wavelength = 4000 ) LED is ________ eV. (Round off to the nearest integer).

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Vishal Baghel

Contributor-Level 10

$E_{9} = \frac{h c}{λ} = \frac{1242}{λ (n m)} = \frac{1242}{400} = 3.105$

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5 months ago

Class 12th Physics

Eight copper wire of length $l$ and diameter d are joined in parallel to from a single composite conductor of resistance R. If a single copper wire of length 2 $l$ have the same resistance (R) then its diameter will be ________d.

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Vishal Baghel

Contributor-Level 10

Each wile has resistance = $\frac{ρ 4 l}{π d^{2}} = r$ $\frac{}{^{}}$

Eight wire in parallel, then equivalent resistance is

$\frac{r}{8} = \frac{ρ l}{2 π d^{2}}$

Single copper wire of length $2 l$ has resistance

$R = ρ \frac{2 l * 4}{π d_{1}^{2}} = \frac{ρ l}{2 π d^{2}}$

d1 = 4d

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Class 12th Electrostatic Potential and Capacitance

In the given figure, the value of V₀ will be ________ V.

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Vishal Baghel

Contributor-Level 10

K V L Loop 1

6 - l * 1 -l₁ * 1 - 4 = 0

2 = l + l₁ - (i)

K V L Loop 2

4 + l₁ * 1 - (l – l₁) * 1 – 2 = 0

$\Rightarrow 2 + l_{1} - l = 0$

l = 2 + 2l₁ &nb

...more

K V L Loop 1

6 - l * 1 -l₁ * 1 - 4 = 0

2 = l + l₁ - (i)

K V L Loop 2

4 + l₁ * 1 - (l – l₁) * 1 – 2 = 0

$\Rightarrow 2 + l_{1} - l = 0$

l = 2 + 2l₁ - (ii)

from (i) & (ii)

l₁ = 0 v₀ = 4

less

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5 months ago

Class 12th Physics

The volume charge density of a sphere of radius 6m is 2 $μ C c m^{- 3} .$ The number of lines of force per unit surface area coming out from the surface of the sphere is _________ * 10¹⁰ NC^-1.

[Given : Permittivity of vacuum $\in_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} - m^{- 2}]$

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Vishal Baghel

Contributor-Level 10

No. of electric field line per unit area = electric field

$E = \frac{ρ r}{3 \in_{0}} f o r r = R$

$E = 45 * 1 0^{10} N / C$

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6 months ago

Class 12th Maths

Let $S = {(x, y) \in N * N : 9 {(x - 3)}^{2} + 16 {(y - 4)}^{2} \leq 144}$ and $T = {(x, y) \in R * R : {(x - 7)}^{2} + {(y - 4)}^{2} \leq 36} .$ Then $n (S \cap T)$ is equal to

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Payal Gupta

Contributor-Level 10

$\frac{{(x - 3)}^{2}}{16} + \frac{{(y - 4)}^{2}}{9} \leq 1, x, y \in N, {(x - 7)}^{2} + {(y - 4)}^{2} \geq 36, x, y \in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

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