Ncert Solutions Chemistry Class 11th

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New answer posted•a year 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?

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Asked by Payal Gupta

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Payal Gupta | Contributor-Level 10

a year ago

NaCN converted into HCN and Na₂S converted into H₂S

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

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

The correct order of bond dissociation enthalpy of halogens is

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

Bond energy : Cl₂ > Br₂ > F₂ > l₂

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New answer posted•a year 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]

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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New answer posted•a year 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

16 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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New answer posted•a year 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.

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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

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New answer posted•a year 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.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Bond dissociation energy of Br₂ = 228 kJ/mole

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

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