Chemistry

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 = $\frac{8}{23}*6.02*10^{23}$  

$\approx 2*10^{23}$                                            

$Cl{O}_2^{-},C{l}_2,M{n}^{3+}$ can show disproportionation reaction while $Mn{O}_4^{-}$ cannot show disproportionation reaction is Mn is in +7 oxidation state.

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

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%.

$2N{O}_2\left(g\right)?N_2{O}_4\left(g\right)$

$\Delta S=-176J{K}^{-1}=-0.176kJ{K}^{-1}$  

$\Delta H=-57.8KJmo{l}^{-1}$               

 T = 298 K

Using ,   $\Delta G=\Delta H-T\Delta S$

$=-57.8+\left(298*0.176\right)KJmo{l}^{-1}$              

$=-5.3KJmo{l}^{-1}$               

Hence, magnitude of  $\Delta G$ in kJmol^-1 is 5 (nearest integers)

Kindly consider the following figure

Peptide contains four amino acid i.e. glycine, aspartic acid and histidine, so it will have three  peptide linkage.

Power = 50 watt = 50 J/sec

Energy emitted per second = 50 J

Wavelength ; $\lambda$  = 795 nm

Energy of one photon =   

$=0.025*10^{-17}J$             

Number of photons emitted per second = $\frac{50}{0.025*10^{-17}}$  

= 2 * 10²⁰

Kindly consider the following figure