Tags Ncert Solutions Chemistry Class 11th

Ncert Solutions Chemistry Class 11th

Get insights from 2k questions on Ncert Solutions Chemistry Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Chemistry Class 11th

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

Given below are two statements:

Statement I : An allotrope of oxygen is an important intermediate in the formation of reducing smog.

Statement II : Gases such as oxides of nitrogen and sulphur present in troposphere contribute to the formation of photochemical smog.

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

Primary pollutants are oxides of N & C and classical smog is due to SO₂ & particulates of fossil

0 0

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

Which of the following compound is added to the sodium extract before addition of silver nitrate for testing of halogens?

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

NaCN converted into HCN and Na₂S converted into H₂S

$\begin{array}{l} N a C N + H N O_{3} \to N a N O_{3} + H C N \\ N a_{2} S + 2 H N O_{3} \to N a N O_{3} + H_{2} S \end{array}$

0 0

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

The correct order of bond dissociation enthalpy of halogens is

0 Follower 1 View

Payal Gupta

Contributor-Level 10

Bond dissociation enthalpy is the energy required to break 1 mole of bond according to experimental data

Bond energy : Cl₂ > Br₂ > F₂ > l₂

0 0

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

The solubility of Ca(OH)₂ in water is:

[Given : The solubility product of Ca(OH)₂ in water = 5.5 * 10^-6]

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

0 0

New question posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

Given below are two statements:

Statement I : α and β forms of sulphur can change reversibly between themselves with slow heating or slow cooling.

Statement II : At room temperature the stable crystalline form of sulphur is monoclinic sulphur.

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

0 Follower 3 Views

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

Using the provided information in the following paper chromatogram: Fig : Paper chromatography for compounds A and B. The calculated R_f value of A------------* 10^-1

0 Follower 9 Views

alok kumar singh

Contributor-Level 10

$R_{f} v a l u e = \frac{D i s t a n c e m o v e d b y s u b s t a n c e f r o m b a s e l i n e}{D i s t a n c e m o v e d b y t h e s o l v e n t f r o m b a s e l i n e} = \frac{2}{5} = 0.4 = 4.0 * 1 0^{- 1}$

0 0

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

The ionization enthalpy of Na⁺ formation from Na_(g)is 495.8 kJ mol^-1, while the electron gain enthalpy of Br is - 325.0 kJ mol^-1. Given the lattice enthalpy of NaBr is - 728.4 kJ mol^-1. The energy for the formation of NaBr ionic solid is (-) -------------- * 10^-1kJ mol^-1.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

Δ H_{s u b} o f N a m e t a l = 496 k J / m o l e

Bond dissociation energy of Br₂ = 228 kJ/mole

$Δ H_{f} = 495.8 + - 325 - 728.4 = - 557.6 k J / m o l e = - 5576 * 1 0^{- 1} k J / m o l e$

Note : In question is neglect $Δ_{a} H^{0} o f N a, Δ_{v a p} H^{O} o f B r_{2} (l) a n d Δ_{b o n d} H^{0} o f B r_{2} (g)$

0 0

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

Bond dissociation energy of Br₂ = 228 kJ/mole

$? H_{f} = 495.8 + ? 325 ? 728.4 = ? 557.6 k J / m o l e = ? 5576 * 1 0^{? 1} k J / m o l e$

Note : $?_{a} H^{0} ? ? o f ? ? N a, ?_{v a p} H^{O} o f ? ? B r_{2} (l) a n d ? ? ?_{b o n d} H^{0} ? ? o f ? ? B r_{2} (g)$ In question is neglected

0 0

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

A car tyre is filled with nitrogen gas at 35 psi at 27°C. It will burst if pressure exceeds 40 psi. The temperature in °C at which the car tyre will burst is--------------. (Rounded – off to the nearest integer)

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \Rightarrow \frac{35}{300} = \frac{40}{T_{2}}$

$∴ T_{2} = \frac{40 * 300}{35} = 342.85 K$

$T_{2} (° C) = 69.707 \approx 70 ° C$

0 0

New answer posted

3 months ago

Ncert Solutions Chemistry Class 11th Class 11th

The reaction of cyanamide, NH₂CN_(s) with oxygen was run in a bomb calorimeter and $Δ U$ was found to be -742.24 kJ mol^-1. The magnitude of $Δ H_{298}$ for the reaction

------------- kJ. (Rounded off to the nearest integer)

[Assume ideal gases and R = 8.314 J mol^-1 K^-1]

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

$? U = ? 742.24 ? k J / m o l e ? H_{298} = ?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

= $? 742.24 + \frac{1}{2} * \frac{8.314}{1000} * 298$

$? 742.24 + 1.238 = ? 741.002 ? ? 741 k J / m o l$

So, magnitude of $? H$ = 741 kJ/mol.

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
679k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Ncert Solutions Chemistry Class 11th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience