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6 months ago

Class 12th Polymers

${(C_{7} H_{5} O_{2})}_{2} \overset{h v}{\to} [X] \to 2 {\overset{•}{C}}_{6} H_{5} + 2 C O_{2}$

Consider the above reaction and identify the intermediate 'X'

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Payal Gupta

Contributor-Level 10

This is the formation of free radical mechanism, hence free radical will be formed

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A ball is spun with angular acceleration a = 6t² – 2t where t is in second and a is in rads^-¹. At t = 0, the ball has angular velocity of 10 rads^-¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

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alok kumar singh

Contributor-Level 10

a = 6t² – 2t, of t 20, w = 10 rad/sec

, q = 4 rad

$α = \frac{d ω}{d t}$

$\Rightarrow \int_{10}^{ω} d ω = \int_{t - 0}^{t} α d t$

$\Rightarrow ω - 10 = \int_{0}^{t} (6 t^{2} - 2 t) d t$

$ω = \frac{d θ}{d t}$

$d θ = ω d t$

$\int_{4}^{θ} d θ = \int_{0}^{t} (2 t^{3} - t^{2} + 10) d t$

= $\frac{t^{4}}{2} - \frac{t^{3}}{3} + 10 t + 4$

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The most common oxidation state of Lanthanoid elements is +3. Which of the following is likely to deviate easily from +3 oxidation state?

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Vishal Baghel

Contributor-Level 10

$C e^{3 +} \to [X e] 4 f^{1} 5 d^{0}$

$C e^{4 +} \to [X e] 4 f^{0} 5 d^{0}$ (Noble gas configuration)

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Which one of the following is an example of disproportionation reaction?

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Vishal Baghel

Contributor-Level 10

Disproportionation reaction is

$\overset{+ 6}{M g} O_{4}^{- -} + 4 H^{+} \to 2 \overset{+ 7}{M n} O_{4}^{-} + \overset{+ 4}{M n} O_{2} + 2 H_{2} O$

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The oxide which contains an odd electron at the nitrogen atom is

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Class 12th Biomolecules

Given below are two statements:

Statement I: In 'Lassaigne's Test', when both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed.

Statement II: If both nitrogen and sulphur are present in an organic compound, then the excess of sodium used in sodium fusion will decompose the sodium thiocyanate formed to given NaCN and Na₂S.

In the light of the above statements, choose the most appropriate answer from the option given below

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Payal Gupta

Contributor-Level 10

Regarding Lassaigne's Test, both statement are correct.

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

s-block element which cannot be qualitatively confirmed by the flame test is

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Vishal Baghel

Contributor-Level 10

Be does not perform in flame colouration due to highly ionization energy.

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Boiling of hard water is helpful in removing the temporary hardness by converting calcium hydrogen carbonate and magnesium hydrogen carbonate to

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Vishal Baghel

Contributor-Level 10

During removal of temporary hardness of water.

$\begin{array}{l} M g {(H C O_{3})}_{2} \overset{B o i l}{\to} M g {(O H)}_{2} + 2 C O_{2} ↑ \\ C a {(H C O_{3})}_{2} \overset{B o i l}{\to} C a C O_{3} + H_{2} O + C O_{2} \end{array}$

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The role depressants in 'Froth Floatation method' is to

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Vishal Baghel

Contributor-Level 10

In sulphide ore, depressants selectively prevent impurity from coming to the fourth.

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Class 12th physics ncert exemplar solutions class 12th chapter two

Velocity $(υ)$ and acceleration (a) in two system of units 1 and 2 are related as $υ_{2} = \frac{n}{m^{2}} υ_{1} a n d a_{2} = \frac{a_{1}}{m n}$ respectively. Here m and n are constants. The relations for distance and time in two system respectively are:

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alok kumar singh

Contributor-Level 10

Let, L₁ & L₂ one distance for system (1) & (2) respectively

T₁ & T₂ are time for system (1) & (2) respectively

Given : v2 = $\frac{n}{m^{2}} v_{1}$

$\Rightarrow (\frac{L_{2}}{T_{2}}) = \frac{n}{m_{2}} (\frac{4}{T_{1}})$ $[? v_{2} = \frac{L_{2}}{T_{2}}, v_{1} = \frac{L_{1}}{T_{1}}]$

$\Rightarrow \frac{L_{2}}{L_{1}} = (\frac{n}{m_{2}}) [\frac{T_{2}}{T_{1}}] - - (i)$

And a2 = $\frac{a_{1}}{m n} \Rightarrow \frac{v_{2}}{T_{2}} = \frac{v_{1}}{T_{1} m n}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \frac{v_{1}}{v_{2}} (\frac{1}{m n})$ $[? \frac{v_{1}}{v_{2}} = \frac{m^{2}}{n}]$

$\Rightarrow \frac{n^{2}}{m} T_{1} = T_{2}$

from (1)

$\frac{L^{2}}{L_{1}} = \frac{n}{m^{2}} [\frac{n^{2}}{m}] [? \frac{T_{2}}{T_{1}} = \frac{n^{2}}{m}]$

$\Rightarrow \frac{n^{3}}{m^{3}} L_{1} = L_{2}$

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