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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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Class 12th Physics

The current density in a cylindrical wire of radius 4mm is 4 * 10⁶Am^-². The current through the outer portion of the wire between radial distances $\frac{R}{2}$ and R is _________πA.

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Vishal Baghel

Contributor-Level 10

J (current density) = 4 * 10⁶ Am^-2

Area between radial distance $\frac{R}{2} t o R$

A = $π [R^{2} - \frac{R^{2}}{4}] = \frac{3 R^{2}}{4} π$

I = AJ

= $\frac{3 R^{2}}{4} π * 4 * 1 0^{6}$

= 3R² * 10⁶ πAmp

= 3 (4 * 10^-3)² * 10⁶ πAmp

= 3 * 16 * 10^-6 * 10⁶π Amp

= 48πA

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Class 12th Physics

A cell, shunted by a8 $Ω$ resistance, is balanced across a potentiometer wire of length 3m. The balancing length is 2m when cell is shunted by 4 $Ω$ resistance. The value of internal resistance of the cell will be__________ $Ω$ .

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Vishal Baghel

Contributor-Level 10

R shunt Resistance

Given

CD = 3m

CF = 2m

Current through potentiometer wire

I = $\frac{v}{R_{0}} = \frac{V A}{ρ L}$ $\frac{}{_{}} \frac{}{}$ (1)

A Area of potentiometer wire

$ρ \to$ Resistivity of the wire

L Length of the potentiometer wire

Case 1 When 8 $Ω$ shunt is bal aced at 3m length

$V_{C D} = I R_{C D} = \frac{V}{ρ L} A * \frac{ρ C F}{A}$

$V_{C D} = \frac{V l}{L} = \frac{V * 3}{L}$

VCD = VAB = E – I1r

= $- \frac{E}{R + r} r = \frac{V}{L} * 3$ $\frac{}{} \frac{}{}$

$\Rightarrow E [1 - \frac{r}{8 + r}] = \frac{V}{L} * 3$ (2)

Case 2 When 4 $Ω$ shunt is balanced at 2m length

$V_{C F} = I R_{C F} = \frac{V}{ρ L} A * \frac{ρ C F}{A} = \frac{V * 2}{L}$

$V_{C F} = V_{A B} = E - I_{2} r$

= $E - \frac{E}{R + r} r$ $\frac{}{}$

= $E [1 - \frac{r}{4 + r}]$ $\frac{}{}$

$\frac{}{} \frac{}{}$ $\Rightarrow E (1 - \frac{r}{4 + r}) = \frac{V * 2}{L}$ (3)

$(2) \div (3)$

$\frac{1 - \frac{r}{8 + r}}{1 - \frac{r}{4 + r}} = \frac{3}{2}$

$\Rightarrow (\frac{8}{8 + r}) * (\frac{4 + 8}{4}) = \frac{3}{2}$

$\Rightarrow 4 (4 + r) = 3 (8 + r)$

$\Rightarrow 16 + 4 r = 24 + 3 r$

r = 24 – 16

r = 8 $Ω$

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Class 12th Chemistry

A Chromatography column, packed with silica gel as stationary phase, was used to separate a mixture of compounds consisting of (a) benzanilide, (b) aniline and (c) acetophenone. When the column is eluted with a mixture of solvents, hexane; ethyl acetate $(20 : 80)$ , the sequence of obtained compounds is