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Class 11th Chemistry

In the given reaction,

The number of p- electron present in the product 'p' in__________.

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Vishal Baghel

Contributor-Level 10

It is aldol condensation.

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Class 11th Chemistry

In the given reaction

The number of sp2 hybridised carbon(s) in compound 'X' is____________.

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Vishal Baghel

Contributor-Level 10

Number of sp² carbon = 8

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Class 11th Thermodynamics

2.2g of nitrous oxide (N₂O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1mL to 167.75mL. The change in thermal energy of the process, $Δ U$ is '-x' J. The value of 'x' is________. [Nearest Integer]

(Given : atomic mass of N = 14 g mol^-1 and of O = 16 g mol^-1.

Molar heat capacity of N₂O is 100 J K^-1 mol^-1)

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Vishal Baghel

Contributor-Level 10

Moles of N₂O= $\frac{2.2}{44} = \frac{1}{20}$

$Δ H = n C_{p} Δ T = \frac{1}{20} * 100 (- 40) = - 200 J$

$Δ U = q_{p} + w$

$w = - P_{e x t} . Δ V$

$w = - 1 \frac{(167.75 - 217.1)}{1000} * 101.3 J = + 5 J$

$Δ U = - 200 + 5 = 195 J$

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Class 11th Chemistry

A box contains 0.90g of liquid water in equilibrium with water vapour at 27°C. The equilibrium vapour pressure of water at 27°C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be________ litre. [Nearest Integer]

(Given : R = 0.082 L atm K^-1 mole^-1)

(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

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Vishal Baghel

Contributor-Level 10

$\frac{n R T}{P} = \frac{0.90 * 0.0821 * 300}{18 * 32} * 760 = 29.21 ≅ 29$

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Class 11th Chemistry Organic Chemistry

On complete combustion 0.30g of an organic compound gave 0.20g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is __________. (Nearest integer)

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alok kumar singh

Contributor-Level 10

Complete combustion of compound produces 0.2 gm CO₂

Hence wt of carbon in 0.2 gm CO₂

$= (\frac{12}{44} * 0.2) g m$

Therefore % of carbon in compound

$= \frac{w t o f c a r b o n * 100}{w t o f c o m p o u n d} = \frac{12 * 0.2 * 100}{44 * 0.3} = \frac{2400}{44 * 3} = 18.18 \approx 18 %$

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Class 11th Physics

An ideal fluid of density 800kgm^-³, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2} .$ The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt[]{x}}{6} m s^{- 1}$ for x = ________. (Given g = 10 ms^-²)

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Raj Pandey

Contributor-Level 9

$ρ = 800 k g / m^{3}$

P₁ – P₂ = 4100 Pa

Bernoulli's question b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

⇒ $\frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{v_{2}^{2}}{2 g} + 0$ -(1)

Equation of continuity

A₁v₁ = A₂v₂

v₂ = 2v₁ -(2)

from (1) & (2)

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{{(2 v_{1})}^{2}}{2 g}$

$\frac{4100}{800 * 10} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{4 v_{1}^{2}}{2 g}$

$\frac{12100}{8000} = \frac{3 v_{1}^{2}}{2 g}$

$\Rightarrow v_{1}^{2} = \frac{2 * 10}{3} * \frac{12100}{8000}$

$= \frac{2}{3} * \frac{121}{8}$

x = 363

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Class 11th Chemistry

A white precipitate was formed when BaCl₂ was added to water extract of an inorganic salt. Further, a gas 'X' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl. The anion present in the inorganic salt is

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Vishal Baghel

Contributor-Level 10

$B a C l_{2} + S O_{3}^{2 -} \overset{}{\to} \underset{W h i t e}{B a S O_{3}} ↓ \overset{d i l . H C l}{\to} \underset{\begin{array}{l} b u r n i n g s u l p h u r \\ l i k e s m e l l \end{array}}{S O_{2} ↑}$

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3 months ago

Class 11th Chemistry

50 mL of 0.1 M CH₃COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be_________* 10^-². (Nearest integer)

(Given : pKa (CH₃-COOH) = 4.76)

log 2 = 0.30

log 3 = 0.48

log 5 = 0.69

log 7 = 0.84

log 11 = 1.04

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alok kumar singh

Contributor-Level 10

Here, total meq of acetic acid = 50 * 0.1 = 5

And total meq of NaOH = 25 * 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$p H = p K a + l o g_{10} \frac{[S]}{[A]}$

$p H = 4.76 + l o g_{10} \frac{2.5}{2.5} = 4.76 = 476 * 1 0^{- 2}$

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Class 11th Chemistry

Kindly go through the questions

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Vishal Baghel

Contributor-Level 10

With AlCl₃, alkyl halide will form cabocation which will show rearrangement.

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3 months ago

Class 11th Chemistry

Which of the following carbocations is most stable?

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Vishal Baghel

Contributor-Level 10

In this carbocation +M effect of -OCH₃ group stabilizes the carbocation.

While in option (A) and (B), +M of -OCH₃ will not work but in option (C), +M of -OCH₃ works so due to more delocalization in option (D), it is more stable.

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