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Class 12th Chemistry

Which of the following will have maximum stabilization due to crystal field?

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Vishal Baghel

Contributor-Level 10

Among all CO³⁺ has the highest +ve charge density and CN^- is the strongest ligand among all which will give maximum CFSE.

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Class 12th Chemistry

Heating white phosphorous with conc. NaOH solution gives mainly:

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Vishal Baghel

Contributor-Level 10

P₄ + 3NaOH + 3H₂O -> 3NaH₂PO₂ + PH₃

(It's a disproportionation reaction)

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Class 12th Chemistry

Match the List-I with List-II

List – II List – II

(Si-Compounds) (Si-polymeric / Other Products)

(A) &n

...more

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Vishal Baghel

Contributor-Level 10

(CH₃)₄Si is called silane, while CH₃Si (OH)₃ produced 2- dimensional silicone after the removal of water, similarity ${(C H_{3})}_{2} S i {(O H)}_{2}$ produced chain form and ${(C H_{3})}_{3} S i (O H)$ produced dimer form after releasing water molecules.

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Class 12th The S-block Elements

BeCl₂ reacts with LiAlH₄ to give:

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Vishal Baghel

Contributor-Level 10

Reaction of BeCl₂ and LiAlH₄

2BeCl₂ + LiAlH₄ -> 2BeH₂ + LiCl + AlCl₃

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Class 12th Chemistry

Addition of H₂SO₄ to BaO₂ produces:

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Vishal Baghel

Contributor-Level 10

BaO₂ + H₂SO₄ BaSO₄ + H₂O₂

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Class 12th Chemistry

Math the List-I with List-II

List – I List – II

(A) Concentration of gold ore &nbs

...more

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Vishal Baghel

Contributor-Level 10

Cyanidation is applicable for gold extraction, leaching for NaOH, froth stabilizer is aniline and blister copper is related with SO₂.

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Class 12th Chemistry

Match the List-I with List-II

List – I List – II

(A) Lyophilic colloid

...more

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Vishal Baghel

Contributor-Level 10

Emulsion are liquid colloids, protective colloids are lyophilic, FeCl3 + H2O $\overset{Δ}{\to} (+ v e) l y$ charged colloids and FeCl3 + NaOH $\to_{}^{}$ (-ve)ly charged colloids.

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Class 12th Solutions

Given below are two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R)

Assertion (A): At 10°C, the density of a 5M solution of KCl [atomic masses of K % Cl are 39 & 35.5 g mol^-¹ respectively is 'x' g ml^-¹. The solution is cooled to -21°C. The molality of the solution will remain unchanged.

Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature

In the light of the above statements, choose the correct answer from the options given below.

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Vishal Baghel

Contributor-Level 10

Molality itself a strength representing terms for solution which does not depend upon the temperature.

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Class 12th Communication Systems

The height of a transmitting antenna at the top of a tower is 25m and that of receiving antenna is, 49m. The maximum distance between them, for satisfactory communication is LOS (Line-Of-Sight) is $K \sqrt[]{5} * 1 0^{2} m .$ The value of K is___________.

(Assume radius of Earth is 64 * 10⁺⁵m) [Calculate upto nearest integer value]

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Vishal Baghel

Contributor-Level 10

Given, hT = 25m, R = 6400 km

hR = 49m

dm (Maximum distance for satisfactory communication)

dm = $\sqrt[]{2 R h_{T}} + \sqrt[]{2 R h_{R}}$ $_{}_{}$

dm = $\sqrt[]{2 * 6400 * 1 0^{3} * 25} + \sqrt[]{2 * 6400 * 1 0^{3} * 49}$ $^{}^{}$

= $\sqrt[]{5 * 1 0^{4} * 6400} + \sqrt[]{6400 * 1 0^{3} * 49 * 2}$ $^{}^{}$

= $192 \sqrt[]{5} * 1 0^{2}$ $^{}$

= $k \sqrt[]{5} * 1 0^{2}$ $^{}$

k= 192

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Class 12th Physics

A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _________nJ.

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Vishal Baghel

Contributor-Level 10

Q = CV

= 50 * 10^-12 * 100

= 5 * 10^-12 * 10³

Q = 5 * 10^-9C

U_i (Initial Energy)

= $\frac{Q^{2}}{2 C} = \frac{25 * 1 0^{- 18}}{2 * 50 * 1 0^{- 12}}$

= $\frac{1}{4} * 1 0^{- 6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_{1}}{C} = \frac{Q_{2}}{C} = 0$

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2 Q_{1} = Q$

$Q_{1} = \frac{Q}{2}$

$u_{F} = \frac{Q_{1}^{2}}{2 C} + \frac{Q_{1}^{2}}{2 C}$

$= \frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(Q / 2)}^{2}}{2 C}$

$= 2 * \frac{Q^{2}}{4 * 2 C}$

...more

Q = CV

= 50 * 10^-12 * 100

= 5 * 10^-12 * 10³

Q = 5 * 10^-9C

U_i (Initial Energy)

= $\frac{Q^{2}}{2 C} = \frac{25 * 1 0^{- 18}}{2 * 50 * 1 0^{- 12}}$

= $\frac{1}{4} * 1 0^{- 6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_{1}}{C} = \frac{Q_{2}}{C} = 0$

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2 Q_{1} = Q$

$Q_{1} = \frac{Q}{2}$

$u_{F} = \frac{Q_{1}^{2}}{2 C} + \frac{Q_{1}^{2}}{2 C}$

$= \frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(Q / 2)}^{2}}{2 C}$

$= 2 * \frac{Q^{2}}{4 * 2 C}$

$U_{f} = \frac{Q_{2}}{4 * 2 C}$

Loss in energy = U_i = U_t

= $\frac{Q^{2}}{2 C} - \frac{Q^{2}}{4 C}$

= $\frac{Q^{2}}{4 C}$

= $\frac{1}{2} [\frac{Q^{2}}{2 C}]$

$\frac{1}{2} * 0.25 * 1 0^{- 6}$

= 125 * 10^-9

= 125 nJ

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