Chemistry

N > O > Be > B

Due to half filled configuration N has more I.E than oxygen and due to fully filled configuration Be has more I.E than B.

(a) → (ii)                (b) → (iii)            (c) → (iv)            (d) → (i)

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

Calamine is the ore of zinc i.e. ZnCO₃.

Malachite is the ore of copper i.e. CuCO₃.Cu (OH)₂.

Electronic configuration of Fe is [Ar] 4s²3d⁶ and in +3 oxidation state it has [Ar] 4s⁰3d⁵ configuration.

K = 3.3 * 10^-4 s^-1

Time for 40% completion ; t

Using $K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$  

3.3 * 10^-4 =   $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$

$t=\frac{2.303}{3.3}*10^4*0.22\Rightarrow t=25.58mins$  

so; the nearest integer is 26.

Coordination no. of an atom in BCC is 8.

Using $\lambda =\frac{h}{\sqrt[]{2qVm}}$  

${\lambda }_{Li}=\frac{h}{\sqrt[]{2*3e*V*8.3m}}$                 m = mass of proton

${\lambda }_p=\frac{h}{\sqrt[]{2*e*V*m}}$  

$\frac{{\lambda }_{Li}}{{\lambda }_p}=\sqrt[]{\frac{2eVm}{2*24.9eVm}}$            

= 0.2 = 2 * 10^-1

So; x = 2

W = 4.5 g

M.W = 90

V = 250 ml

Using M = $\frac{W}{M.W*V}*1000$  

$M=\frac{4.5}{90*250}*1000=0.2$  

M = 2 * 10^-1 M

So, x = 2

The balanced equation is :

$S_8\left(s\right)+12O{H}^{-}\left(aq\right)\to 4S^{2-}\left(aq\right)+2S_2{O}_3^{2-}\left(aq\right)+6H_{2}O\left(\mathcal{l}\right)$           

Here ; a = 12

Mass of Na⁺ in 50 ml = 70 * 50 = 3500 mg

23000 mg of Na⁺ is present in 85000 mg of NaNO₃ (1 mole NaNO₃ contains 1 mole Na⁺)

$\therefore 3500$ mg Na⁺ will be present in   $\frac{85000}{23000}*3500$

= 12934.78 mg

= 12.93478 gm

$\cong 13$