Class 12th

Simple distillation can be applied for the separation of two liquids has boiling point difference greater than 20°C.

Boiling point of propanol > boiling point of propane (due to H-bonding effect)

Let $t=\frac{3}{2}{x}^2+\frac{5}{3}{x}^3+\frac{7}{4}{x}^4+......$

$=\left(2-\frac{1}{2}\right)x^2+\left(2-\frac{1}{3}\right)x^3+\left(2-\frac{1}{4}\right)x^4+......$   

= $2\left(x^2+x^3+x^4+.....\right)-\left(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+.....\right)$

$=2\frac{x^2}{1-x}+ln\left(1-x\right)+x=\frac{x^2+x}{1-x}+ln\left(1-x\right)=\frac{x\left(1+x\right)}{1-x}+ln\left(1-x\right)$              

$=2\frac{x^2}{1-x}+ln\left(1-x\right)+x=\frac{x^2+x}{1-x}+ln\left(1-x\right)=\frac{x\left(1+x\right)}{1-x}+ln\left(1-x\right)$

A4 is in the middle in terms of height.

$\sum P\left(x\right)=1\Rightarrow k+2k+2k+3k+k=1sok=\frac{1}{9}$

 Now, $P\left(\frac{1<x<4}{x\angle 3}\right)=P\frac{\left(x=2\right)}{P\left(x\angle 3\right)}=\frac{\frac{2k}{9k}}{\frac{k}{9k}+\frac{2k}{9k}}=\frac{2}{3}$  

$\Rightarrow P=\frac{2}{3}$          

So, $5P=\lambda kgives\frac{10}{3}=\lambda *\frac{1}{9}\Rightarrow \lambda =30$  

La⁺² (z = 57) = 5d¹

Ce⁺² (z = 58) = 4f²

Nd⁺² (z = 60) = 4f⁴

Yb⁺² (z = 70) = 4f¹⁴

Yb⁺² has no unpaired electrons thus diamagnetic in nature.

 Normal vector to the given plane be

$2\widehat{i}-\widehat{j}+3\widehat{k}so$                   

Equation of line QS :

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda$

So let P $\left(2\lambda +1,-\lambda +3,\lambda +4\right)$  

Now P lies on given plane so

$4\lambda +2+\lambda -3+8\lambda +4+3=0$  

So, S (-3, 5, 2)

also given R lies on given plane so

6 – 5 + $\gamma$ + 3 = 0 so   $\gamma$ = -4

So, R (3, 5, -4)

SR² = 72

Non-linear and symmetrical Cr-O-Cr bond due to resonance.

$A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]then{A}^2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -2k\hfill & \hfill 4k+2\hfill \\ \hfill 2k^2+k\hfill & \hfill -4k-1\hfill \end{array}\right]$              

Now, A(A³ + 3l) = 2l gives A³ + 3l = 2A^-1

$\Rightarrow -2k+3=\frac{1}{k}\Rightarrow 2k^2-3k+1=0$    

  $k=\frac{1}{2},1$            

& $4k+2=\frac{2}{k}or2k+1-\frac{1}{k}soor2k^2+k-1=0$

=> $k=\frac{1}{2}$

Compound react with S_N¹ mechanism in presence of polar protic solvent, which follows carbocation forming path.

Hence correct order is.