Class 12th

Hinsberg test for amines

Boiling points of chloroform is 334 K and aniline is 457 K. Here difference in b. pt is more so they can be separated by simple distillation.

Aniline is insoluble (immiscible in water), the mixture of insoluble substance in water boils close to but below 373 K. so, aniline boils below its b.pt

By truth table

So F₁ (A, B, C) is not a tautology

Now again by truth table

So      F₂ (A, B) be a tautology.

From option let it be isosceles where AB = AC then

$x=\sqrt[]{r^2-{\left(h-r\right)}^2}$            

=  $\sqrt[]{r^2-h^2-r^2+2rh}$

$x=\sqrt[]{2hr-h^2}........\left(i\right)$

 Now ar $\left(\Delta ABC\right)=\Delta =\frac{1}{2}BC*AL$

$\Delta =\frac{1}{2}*2\sqrt[]{2hr-a^2}*h$        

then  $x=\sqrt[]{2*\frac{3r}{2}*r-\frac{9r^2}{4}}=\frac{\sqrt[]{3}}{2}rfrom\left(i\right)$

$\Rightarrow BC=\sqrt[]{3}r$    

So $AB=\sqrt[]{h^2+x^2}=\sqrt[]{\frac{9r^2}{4}+\frac{3r^2}{4}}=\sqrt[]{3}r$ .

Hence $\Delta$ be equilateral having each side of length $\sqrt[]{3}r.$

$\frac{dq}{dt}=t=20t+8t^2$

$\Rightarrow {\displaystyle \underset{0}{\overset{q}{\int }}dq={\displaystyle \underset{0}{\overset{15}{\int }}\left(20t+8t^2\right)dt}}$

$\Rightarrow q=20*\frac{15^2}{2}+8.\frac{15^3}{3}=11250C$

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series

$A_1+A_2={\displaystyle {\int }_0^{\pi /2}cosxdx}$

$={\left(Sinx\right)}_0^{\pi /2}=1$
$A_1={\displaystyle {\int }_0^{\pi /4}\left(cosx-sinx\right)dx={\left(sinx+cosx\right)}_0^{\pi /4}}$

$=\frac{2}{\sqrt[]{2}}-1=\sqrt[]{2}-1$

$So{A}_2=1-\left(\sqrt[]{2}-1\right)=2-\sqrt[]{2}=\sqrt[]{2}\left(\sqrt[]{2}-1\right)$

$Now\frac{A_1}{A_2}=\frac{1}{\sqrt[]{2}}$             

$K_{2}C{r}_2{O}_7+H_{2}S{O}_4+\underset{GasX}{S{O}_2}\to \underset{\begin{array}{l}Green\\ Y\end{array}}{C{r}_2{\left(S{O}_4\right)}_3}+K_{2}S{O}_4+H_{2}O$