Ncert Solutions Chemistry Class 11th

Lithium salts are extensively hydrated due to high hydration enthalpy of Li⁺

$L{i}^{+}>N{a}^{+}>K^{+}>R{b}^{+}>C{s}^{+}$ (order of polarizing power)

Moles of sodium = $\frac{8}{23}mol$  

Number of atoms = $\frac{8}{23}\times 6.02\times 10^{23}$  

$\approx 2\times 10^{23}$                                            

$B_2^{+}\to \sigma 1s^{2}s*1s^2\sigma 2s^2\sigma *2s^2\pi 2p^1$  

Here, number of unpaired electrons, n = 1

Spin only moment ;    $\mu =\sqrt[]{n\left(n+2\right)}B.M$

= 173 × 10^-2 B.M

$H_2\underset{hv}{\overset{high\text{?}\text{?}T}{?}}2H$

Zn + 2HCl -> ZnCl₂ + H₂

$NaO{H}_{\left(aq\right)}+Zn?N{a}_{2}Zn{O}_2+H_2$        

$H_2\stackrel{2000K}{?}2H$    

It is nearly 0.081%.

Ammonium salt in rain drop resulting wet deposition

$N{H}_4^{+}?Salt+H_{2}O?N{H}_{4}OH$

Oxides of N & S settle down on ground as dry deposition (SO₂).

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

(a) $\Delta G=\Delta G°+RT\mathcal{l}nQ$

$\Delta G°=\Delta H°-T\Delta S°..............\left(1\right)$                      

At equilibrium $\Delta G=0,Q=K_{eq}$

  $\therefore \Delta G°=-RT\mathcal{l}n{K}_{eq}.................\left(2\right)$                   

$\therefore \Delta H°-T\Delta S°=-RT\mathcal{l}n{K}_{eq}$                      

$\mathcal{l}nKeq=-\left(\frac{\Delta H°-T\Delta S°}{RT}\right)$         Views|Posted a year ago