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6 months ago

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

An acidified magnate solution undergoes disproportionation reaction. The spin- only magnetic moment value of the product having manganese in higher oxidation state is________ B.M. (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

$3 M n O_{4}^{2 -} + 4 H^{+} \to_{}^{} 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$

No. of unpaired electron in Mn⁷⁺ = 0

$μ = 0 B . M$

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6 months ago

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

The number of terminal oxygen atoms present in the product B obtained from the following reaction is_____________.

$F e C r_{2} O_{4} + N a_{2} C O_{3} + O_{2} \to A + F e_{2} O_{3} + C O_{2}$

$A + H^{+} \to B + H_{2} O + N a^{+}$

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eight

If the constant term in the expansion of ${(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10} i s 2^{k} . l,$ where $l$ is an odd integer, then the value of k is equal to:

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alok kumar singh

Contributor-Level 10

${(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10}$

General term $10! \frac{{(3 x^{3})}^{α} \cdot {(- 2^{x})}^{β} \cdot {(5 x^{- 5})}^{γ}}{α! β! γ!}$

$= 10! \frac{3^{α} \cdot {(- 2)}^{β} \cdot 5^{γ} \cdot x^{3 α + 2 β - 5 γ}}{α! β! γ!}$

Now for constant term 3 + 2 $- 5 γ = 0$ ………….(i)

$α + β + γ = 10$ …………(ii)

From equation (i) & (ii)

$3 α + 2 (10 - α - γ) = 5 γ$

$α + 20 = 7 γ$

= 1, $γ$ = 3, = 6

Constant term $10! \cdot \frac{3^{1} \cdot {(- 2)}^{6} \cdot 5^{3}}{3! 6!} = 2^{9} \cdot 3^{2} \cdot 5^{4} \cdot 7^{1}$

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6 months ago

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

The activation energy of one of the reactions is a biochemical process is 532611 J mol^-1 when the temperature falls from 310 K to 300 K, the change in rate constant observed is k₃₀₀ = x * 10^-3 k₃₁₀. The value of x is___________.

[Given: In 10 = 2.3 R = 8.3 JK^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

$l n (\frac{K_{2}}{K_{1}}) = \frac{E a}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l n (\frac{K_{2}}{K_{1}}) = \frac{532611}{8.3} * (\frac{10}{310 * 300})$

Where, K₂ is at 310 K and K₁ at 300K

$l n (\frac{K_{2}}{K_{1}}) = 6.9 = 3 * l n 10$

$l n (\frac{K_{2}}{K_{1}}) = l n 1 0^{3}$

K₂ = K₁ * 10³

K₁ = K₂ * 10^-3

$∴$ x = 1

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6 months ago

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

The domain of the function $c o s^{-} (\frac{2 s i n^{- 1} (\frac{1}{4 x^{2} - 1})}{π}) i s :$

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alok kumar singh

Contributor-Level 10

Case 1: 1 $\leq \frac{1}{4 x^{2} - 1} \leq 1$

4x2 – 1 $\geq 1 o r 4 x^{2} - 1 \leq - 1$

$x^{2} \geq \frac{2}{4} o r x^{2} \leq 0$

So x $\in (- \infty, - \frac{1}{\sqrt[]{2}}] \cup [\frac{1}{\sqrt[]{2}}, \infty) \cup {0}$

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6 months ago

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

A dilute solution of sulphuric acid is electrolyzed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is __________ cm³. (Nearest Integer).

[Given: Faraday constant F = 96500 C mol^-1 at STP, molar volume of an ideal gas is 22.7 L mol^-1]

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Vishal Baghel

Contributor-Level 10

At anode (oxidation)

2H₂O $\to_{}^{}$ O₂ (g) + 4H⁺ + 4e^-

At cathode (Reduction)

2H⁺ + 2e^- $\to_{}^{}$ H₂ (g)

No. of gm- equivalents = $\frac{i * t}{96500} = \frac{0.1 * 2 * 60 * 60}{96500} = 0.00746$

$V_{O_{2}} = \frac{0.00746}{4} * 22.7 = 0.0423 L$

$V_{H_{2}} = \frac{0.00746}{2} * 22.7 = 0.0846 L$ $V_{t o t a l} ≅ 127 m L$

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6 months ago

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

1.2 mL of acetic acid is dissolved in water to make 2.0L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is_____________. (Nearest Integer).

[Given Density of acetic acid is 1.02g mL^-1

Molar mass of acetic acid is 60g mol^-1]

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Vishal Baghel

Contributor-Level 10

Mass = d * v = 1.02 * 1.2 = 1.224gm

Moles of acetic acid = 0.0204 moles in 2L

So molality = 0.0102 mol/kg

$Δ T_{f} = i * K_{f} * m$

i = 1 + a for acetic acid

0.0198 = (1 + a) * 1.85 * 0.0102

α = 0.04928 = 5%

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6 months ago

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

17.0g of NH₃ completely vapourises at -33.42°C and 1 bar pressure and the enthalpy change the process is 23.4 kJ mol^-1. The enthalpy change for the vapourisation of 85g of NH₃ under the same conditions is_____ kJ.

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Vishal Baghel

Contributor-Level 10

85gm NH₃ = 5 moles of NH₃

Enthalpy change for 1 mol = 23.4 kJ

Then enthalpy change for 5 mol = 23.4 * 5 = 117 kJ

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Six

A wire of length 22m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is:

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alok kumar singh

Contributor-Level 10

Let perimeter of $Δ$ is x and that of square is 22 – x

now area $= \frac{\sqrt[]{3}}{4} {(\frac{x}{3})}^{2} + {(\frac{22 - x}{4})}^{2}$

for maximum or minimum, $\frac{d A}{d x} = 0$

x $= \frac{22 \sqrt[]{3}}{\sqrt[]{3} + 4}$

now side of a $Δ = \frac{x}{3}$

$= \frac{22 \sqrt[]{3}}{3 (\sqrt[]{3} + 4)}$

$= 66$ $9$ $+$ $4$

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6 months ago

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

Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL^-1 at 257°C and 100 mm Hg. The molar mass of geraniol is___________ g mol^-1 (Nearest Integer).

[Given R = 0.082 L atm K^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

Assuming ideal behaviour, $P = \frac{d R T}{M}$

$P = \frac{100}{760} a t m, T = 257 + 273 = 530 K$

d = 0.46gm/L

$M = \frac{0.46 * 0.082 * 530}{100} * 760 = 151.93 ≅ 152 g / m o l$

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