Tags Class 11th

Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

6 months ago

Class 11th Chemistry

Number of lone pair(s) of electrons on central atom and the shape of BrF₃ molecule respectively.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

0 0

New answer posted

6 months ago

Class 11th Physics Waves

The velocity of sound in a gas, in which two wavelengths 4.08m and 4.16m produce 40 beats in 12s, will be:

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

$| f_{1} - f_{2} | = \frac{40}{12} = \frac{10}{3}$

$\Rightarrow v (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = \frac{10}{3}$

$\Rightarrow V = \frac{10}{3} * \frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}} = \frac{10}{3} * \frac{4.08 * 4.16}{. 08} = \frac{10}{3} * \frac{408 * 4.16}{8} = 707.2 m / s$

0 0

New answer posted

6 months ago

Class 11th Chemistry

Amongst baking soda, caustic soda and washing soda, carbonate anion is present

0 Follower 1 View

alok kumar singh

Contributor-Level 10

Baking soda = NaHCO₃

Washing soda = Na₂CO₃. 10H₂O

Caustic soda = NaOH

0 0

New answer posted

6 months ago

Class 11th Chemistry Classification of Elements and Periodicity in Properties

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

Assertion (A) : The first ionization enthalpy for oxygen is lower than that of nitrogen

Reason (R) : The four electrons in 2p orbitals of oxygen experience more electron electorn repulsion.

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

1^st ionization energy of N>O in oxygen atom, 2 electrons out of 4 electron of 2p orbital resulting in an increased electron repulsion.

0 0

New answer posted

6 months ago

Class 11th physics ncert solutions class 11th

THE de Broglie wavelength for an electron and a photon are $λ_{e} a n d λ_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

$λ_{e} = \frac{h}{P_{e}} = \frac{h}{\sqrt[]{2 m_{e} K_{e}}} \Rightarrow λ_{e}^{2} = \frac{h^{2}}{2 m_{e} k_{e}} \Rightarrow K_{e} = \frac{h^{2}}{2 m_{e} λ_{e}^{2}}$ - (i)

$λ_{P} = \frac{h}{P_{P}} = \frac{h c}{K_{P}} \Rightarrow K_{P} = \frac{h c}{λ_{P}}$ - - (ii)

K_e = K_P

_{$\Rightarrow λ_{P} α λ_{e}^{2}$}

0 0

New answer posted

6 months ago

Class 11th Equilibrium

4.0 moles of argon and 5.0 moles of PCl₅ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K_P for the reaction is

[Given R = 0.082 L atm K^-¹ mol^-¹]

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$

$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$

$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 0 0

2.5 - P P P

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5

...more

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$

$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$

$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 0 0

2.5 - P P P

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5 atm

$K_{P} = \frac{1.5 * 1.5}{1} = 2.25$

less

0 0

New answer posted

6 months ago

Class 11th Chemistry Chemical Bonding and Molecular Structure

Consider the species CH₄, $N H_{4}^{+} a n d B H_{4}^{-} .$ Choose the correct option with respect to the there species

0 Follower 1 View

alok kumar singh

Contributor-Level 10

Species = CH₄ $N H_{4}^{+}$ $B H_{4}^{?}$

Electrons = 10 10

0 0

New answer posted

6 months ago

Class 11th Physics Thermodynamics

Given below are two statements:

Statement – I : When $μ$ moment of an ideal gas undergoes adiabatic change from state $(P_{1}, V_{1}, T_{1})$ to state $(P_{2}, V_{2}, T_{2})$ then work done is w = $\frac{μ R (T_{2} - T_{1})}{1 - γ},$ where $γ = \frac{C_{p}}{C_{v}} a n d$ R = universal gas constant.

Statement – II : In the above case, when work done on the gas, the temperature of the gas would rise.

Choose the correct answer from the options given below:

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

In case of adiabatic process -

$W o r k d o n e, W = \frac{η R (T_{2} - T_{1})}{1 - Y}$

$A s, Δ Q = Δ U + W$

for adiabatic process :- $Δ Q = 0$

$\Rightarrow Δ U = - W$

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

0 0

New answer posted

6 months ago

Class 11th Physics

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be:

0 Follower 4 Views

Raj Pandey

Contributor-Level 9

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 * \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 * 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 * 12 * 3.14 * 1 0^{- 4} = 2.8 * 1 0^{- 4} J$

0 0

New answer posted

6 months ago

Class 11th Physics

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

0 Follower 3 Views

Raj Pandey

Contributor-Level 9

According to kepler's third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4 r_{B}^{3}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 11th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience