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Given are the oxide of alkali and alkaline earth metals which are ionic in nature.

Simple oxide are Li₂O, CaO, MgO and K₂O.

Peroxide is Na₂O₂ and superoxide is KO₂.

All simple oxides are diamagnetic as it has no unpaired electron.

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Zn (30) = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$  

$Z{n}^{+}=\left[Ar\right]4s^{1}3d^{10}$      

Outermost electron is 4s electron,

$n=4,\mathcal{l}=0,m=0,s=\pm 1/2$         

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log k = 20.35 - $\frac{\left(2.47\times 10^3\right)}{T}$  

Comparing with,

$logk=logA-\frac{Ea}{2.303RT}$

$\frac{Ea}{2.303R}=2.47\times 10^3$

= 47.29 kJ/mole

Ans. = 47

  $\Delta T_b=i{K}_{b}m$ ${}_{}{}_{}$

$m=\frac{w\times 1000}{M\times W_{solvent}}$

For acetone solution,

$0.17=1\times 1.7\times \frac{1.22\times 1000}{M\times 100}$

For Benzene solution,

$2Acid\to {\left(Acid\right)}_2\therefore i=1/2$

$\Delta T_b=i\times K_b\times m$

$=\frac{1}{2}\times 2.6\times \frac{1.22\times 1000}{122\times 100}°C$

= $0.13°C=13\times 10^{-2}°C$ ${}^{}$

$\therefore x\times 10^{-2}=13\times 10^{-2}$

$\therefore x=13$

Effective number of atom in C.C.P

$\therefore =\frac{1}{8}\times 8+\frac{1}{2}\times 6$     

= 4

Number of octahedral void = 4

Number of cations = 4

Number of anion = 4

Formula of compound = A₄B₄

Empirical formula = AB

Ans. x = 1

$A_{max}=A_c+A_m$

$A_{min}=A_c-A_m$

$\frac{A_{min}}{A_{max}}=\frac{A_c-A_m}{A_c+A_m}=\frac{250-150}{250+150}$

$\Rightarrow \frac{100}{400}=\frac{50}{200}$  

50 ml of 1 (M) HCl + 30 ml of 1 (M) NaOH

NaOH    +            HCl  ->          NaCl + H₂O

30 × 1 mmol      50 × 1 mmol     

0 mmol               20 mmol

${\left[H^{+}\right]}_{mix}=\frac{20}{50+30}M=\frac{20}{80}M=\frac{1}{4}M=0.25M$

$x\times 10^{-4}=6021\times 10^{-4}$    

$\therefore$ x = 6021

Ans. = 6021

By work energy theorem

Work done = change in K.E.

Work done by friction work done by spring

$=0-\frac{1}{2}m{V}^2$              

As 90% of K.E. is losed by friction so that

$-\frac{90}{100}\left(\frac{1}{2}m{V}^2\right)-\frac{1}{2}K{x}^2=-\frac{1}{2}m{V}^2$                

-K -> -16 × 10⁵

K = 16 × 10⁵

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$x?\left(x\right)={\displaystyle \underset{}{\overset{}{\int }}\left(3t^2-2?\left(t\right)\right)dt,x>-2,By}$  Leibniz theorem

$?\left(x\right)+x?\text{'}\left(x\right)=3x^2-2?\text{'}\left(x\right)\Rightarrow ?\text{'}\left(x\right)+\frac{?\left(x\right)}{x+2}=\frac{3x^2}{x+2}..............\left(i\right)$

$\therefore I.F.=e^{{\displaystyle \int \frac{dx}{x+2}}}=x+2$

Multiplying (i) by I.F. we get,

$\therefore ?\left(x\right)=\frac{x^3+8}{x+2}$

$\therefore ?\left(2\right)=4$

$O_2^{-2}=8\times 2+2=18e^{-}$

${\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2={\pi }_{2py}^2{\pi }_{2px}^{*2}={\pi }_{2py}^{*2}$      

Number of unpaired e^- = 0

Ans. = 0

$l=\frac{m{\mathcal{l}}^2}{3}$  

Energy conservation Low

$mg\mathcal{l}=\frac{1}{2}\frac{m{\mathcal{l}}^2}{3}{\omega }^2.....\left(i\right)$                

And speed V =    $\omega r=\omega \mathcal{l}$

then    $V=\sqrt[]{6g\mathcal{l}}=\sqrt[]{6\times 10\times .6}$

->6 m/s

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P (E₁)= 0.9 $P\left({\overline{E}}_t\right)=0.1andP\left(E_2\right)=0.8,P\left({\overline{E}}_2\right)=0.2$

$\therefore RequiredprobabilityP=\frac{0.8\times 0.1}{0.8\times 0.1+0.2\times 0.1+0.9\times 0.2}=\frac{0.8}{2.8}=\frac{2}{7}$

$\therefore 98P=28$

for smooth surface

$a=gsin30°=\frac{g}{2}$

$S_1=ut+\frac{1}{2}a{t}^2$         

$S_1=\frac{1}{2}\frac{g}{2}{t}^2=\frac{g}{4}{t}^2.......\left(i\right)$               

for rough Surface

$S=\frac{1}{2}\frac{g}{2}\left(1-\mu \sqrt[]{3}\right){\alpha }^2{t}^2........\left(ii\right)$               

By (i) and (ii)

$\mu =\frac{1}{\sqrt[]{3}}\left(\frac{{\alpha }^2-1}{{\alpha }^2}\right)x=\sqrt[]{3}$