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

2 months ago

Class 11th Thermodynamics

The only INCORRECT information is

0 Follower 26 Views

Vishal Baghel

Contributor-Level 10

q = 0, w = 0, U = 0 = f (V, T)

As volume changes, temperature must change to maintain constant internal energy.

0 0

New answer posted

2 months ago

Class 11th Chemistry

If Q is a colourless gas with rotten fish smell then select CORRECT statement.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Zero acidic hydrogen present in anionic part of (R)

0 0

New answer posted

2 months ago

Class 11th Equilibrium

An equilibrium mixture contains 2.0 moles of A(g) and 4.0 moles of B(g). Now, 2.0 moles of A(g) is added in this equilibrium mixture and the system is allowed to re-achieve equilibrium at constant volume and temperature. Now, the volume of system is doubled at constant temperature. What should be the moles of B(g) at new equilibrium? The reaction involved is

$A (g) \overset{}{?} 2 B (g)$ .

0 Follower 24 Views

Vishal Baghel

Contributor-Level 10

$A (g) ? 2 B (g) .$

$E_{m} \frac{2 m o l}{V l} \frac{4 m o l}{V l}$

$N e w {E_{q}}^{m} \frac{4 - x}{V} \frac{4 + 2 x}{V}$

$V * 2 \frac{4 - x}{2 V} \frac{4 - 2 x}{2 V}$

New Eqm $\frac{4 - x - y}{2 V} \frac{4 + 2 x + 2 y}{2 V}$

= $\frac{4 - Z}{2 V} = \frac{4 + 2 Z}{2 V} (Z = x + y)$

Now, KC = $\frac{^{}}{}$ $\frac{{[B]}^{2}}{[A]} =$ constant

$o r, \frac{{(\frac{4}{V})}^{2}}{(\frac{2}{V})} = \frac{(\frac{4 + 2 Z}{2 V})}{(\frac{4 - Z}{2 V})} \Rightarrow Z = 1.29$

mole of B at new Eqm = 4 + 2Z = 6.58

0 0

New question posted

2 months ago

Class 11th Structure of Atom

A sample of hydrogen atoms de-excites from 6th excited state to the ground state in one or more electronic transitions. If no spectral line is obtained in the Paschen and Brackett series, then the maximum number of spectral lines of different photon energies obtained will be-

0 Follower 3 Views

New answer posted

2 months ago

Class 11th Chemistry

The compressibility factor of a Vanderwaal gas is 0.5 at $27 ° C$ and 24 atm. The value of Vanderwaal constant 'a' (in atm-L²/mol²) is (b = 0.10 L/mol, R = 0.08 L-atm/K-mol)

0 Follower 15 Views

Vishal Baghel

Contributor-Level 10

^{$Z = \frac{P V_{m}}{R T} \Rightarrow 0.5 = \frac{24 * V_{m}}{24}$}

^{$V_{m} = 0.5 l / m o l e$}

^{$(P + \frac{a}{V_{m}^{2}}) (V_{m} - b) = R T$}

^{$(24 + \frac{a}{0 . 5^{2}}) (0.5 - 0.1) = 24$}

^{$24 + \frac{a}{0 . 5^{2}} = 60$}

^{$\frac{a}{0 . 5^{2}} = 36$}

^{$a = 36 * 0 . 5^{2} = 9 \frac{a t m - l^{2}}{m o l^{2}}$}

0 0

New answer posted

2 months ago

Class 11th Chemistry

Consider the following reactions of 1A metals.

M + $(x + y) N H_{3}$ $\overset{}{\to} {[M {(N H_{3})}_{x}]}^{+} + {[e {(N H_{3})}_{y}]}^{-}$

$M_{(a m)}^{+} + e^{-} + N H_{3} (l) \overset{}{\to} M N H_{2 (a m)} + \frac{1}{2} H_{2} (g)$

Which of the following statement is

INCORRECT?

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

The blue color of the solution is due to the ammoniated electron which absorbs energy in infrared region of light

0 0

New question posted

2 months ago

Class 11th Chemistry Chemical Bonding and Molecular Structure

Which of the following set contains specie that have interaction energy proportional to $\frac{1}{r^{3}} ?$

(r -> internuclear distance)

0 Follower 3 Views

New question posted

2 months ago

Class 11th Chemistry Hydrocarbon

The oxidation of toluene with hot KMnO₄ gives-

0 Follower 2 Views

New answer posted

2 months ago

Class 11th Chemistry

The sum of “number off filled orbitals” and “number of vacant orbitals” upto valence shell of Fe is

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

Kindly go through the answer

0 0

New answer posted

2 months ago

Class 11th Physics

The new white belt of a long horizontal conveyor is moving with a constant speed v = 3.0m/s. A small block of carbon is placed on the belt with zero initial velocity relative to the ground. The block will slip a bit before moving with the belt, leaving a black mark on the belt (figure). How long is that mark (in m) if the coefficient of kinetic friction between the carbon block and the belt is 0.20 and the coefficient of static friction is 0.30?

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

$0^{2} = v^{2} - 2 μ_{k} g S_{r e l}$

$\frac{9}{2 * 2} = S_{r e l} = 2.25 m$

0 0

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

On Shiksha, get access to

65k Colleges
1.2k Exams
678k 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