Chemistry

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

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

Mass of CuSO₄. 5H₂O = 80 g

Volume of solution = 5L

Molar mass of CuSO₄.5H₂O = 249.54g/ml

$Concentration=\frac{moles}{volumeofsolution}$          

$\begin{array}{l}=\frac{0.32}{5}=0.064M\\ =64*10^{-3}M\end{array}$                

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).

$Zn{\left(OH\right)}_2?Z{n}^{2+}+2O{H}^{-}$  

S                           -            -

NaOHNa⁺ + OH^--

0.1 M    -            -

$\left[Z{n}^{2+}\right]=S$                

Here ; 2 * 10^-20 = S (0.1)²

So; S = 2 * 10^-18 M