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${\overline{z}}_1=i{\overline{z}}_2,arg\left(\frac{z_1}{z_2}\right)=\pi$

$\Rightarrow z_1=-z_2,arg\left(\frac{z_1{z}_2}{{\left|z_2\right|}^2}\right)=\pi$

$\Rightarrow arg\left(z_1{z}_2\right)=\pi$

$20-\frac{\pi }{2}=\pi$

$\theta =\frac{3\pi }{4}$

$\theta -\frac{\pi }{2}=\frac{\pi }{4}$

Stannane is the tetrahedral shape of molecule.

Hybridisation is- sp³, shape and structure are tetrahedral

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$F-Be-F$ $\mu =0$      

BF₃             $\mu =0$            

H₂O         $\mu =0$

NH₃           $\mu =0$                          

CCl₄           $\mu =0$             

HCl             $\mu =0$            

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$T_1=27°C=300K$  T₂ = 45° C = 45 + 273 = 318 K

P₁ = 30 atm        P₂ =?

V is constant due to rigid tank

$\therefore \frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{30}{300}=\frac{P_2}{318}\therefore P_2=\frac{1}{10}\times 318=31.8atm\approx 32atm$    

Liquation refining process is applicable for the metal having low m.p, but containing impurities have higher m.p.

$D=0\Rightarrow b^2=15a,\alpha =\frac{2b\pm 0}{2a}=\frac{b}{a}$  

  $\frac{b}{b^{\frac{2}{15}}}=\frac{15}{b}$

$\alpha +\beta =2b{\alpha }^2+{\beta }^2=4b^2-42$

$\alpha \beta =21$

$a{\left(x-\alpha \right)}^2=a\left(x^2-2\alpha x+{\alpha }^2\right)=a{x}^2-2bx+15$

a = 3                    a = -3

b = 7                    a = -7

a² + b² = 9 + 49 = 58

$?$ 0.30 gm glycine is present in 100 gm protein

$?$ 75 gm glycine is present in $\frac{100}{0.30}\times 75=\frac{1000\cancel{0}}{3\cancel{0}}\times 75=25\times 10^{3}g/mol$

Higher adsorbtion of gas is corresponds to higher liquefaction and higher liquefaction is directly proportional to the higher critical temperature.

Dettol is a mixture of

(B) is Terpineol.

Total concentration of H+ after the mixing of HCl and H2SO4;

$\left[H^{+}\right]=\frac{\left(200\times 0.01\right)+\left(400\times 2\times 0.01\right)}{600}$

$\left[H^{+}\right]=\frac{1}{60}$

$\therefore pH=lo{g}_{10}\left(\frac{1}{60}\right)=1.78$

Kindly go through the solution

 $\begin{array}{l}-2<x+1<2\\ -3<x<1\end{array}|\begin{array}{cc}\hfill x-1\le -2orx-1\ge 2\hfill & \hfill x\le -1orx\ge 3\hfill \end{array}$

Enolisable carbonyl compound gives enamine with 2° amine.

Does not form enol thus not form enamine.

Mass of pure carbon in coal = 0.6 × 1000gm × $\frac{60}{100}=360gm$

Mass of carbon converted into CO = 6 $\frac{60}{100}\times 360=216gm\Rightarrow Mol=\frac{216}{12}=18$ mole of carbon

Mass of carbon converted into CO₂ = 360 – 216 = 144gm Mole = $\frac{144}{12}=12$ mole of carbon.

C (s) + O₂ (g) CO₂ (g) + 400KJ

Mole – 12 for CO₂ production, Hence total energy produced = 400 × 12 = 4800KJ

$C_{\left(s\right)}+\frac{1}{2}{O}_2\left(g\right)\underset{}{\overset{}{\to }}C{O}_{\left(g\right)}+100KJ$

Mole = 18 for CO production, Hence energy produced = 100 × 18 = 1800KJ

Total heat = 4800 + 1800 = 6600KJ

Kindly consider the following figure

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