Class 12th

$Pb{l}_2?P{b}^{2+}+2l^{-}{K}_{sq}=8.0*10^{-9}$

Pb (NO₃)₂ -> Pb²⁺ + ^{$2N{O}_3^{-}$}

0.1 M-

-0.1 M    0.1 M

  $K_{sq}=8*10^{-9}Using{K}_{sq}=\left[P{b}^{2+}\right]{\left[I^{-}\right]}^2$          

$8*10^{-9}=0.1*{\left(2S\right)}^2$

S = 141 * 10^-6 M

$x\frac{dy}{dx}+y=b{x}^4$       

$\frac{dy}{dx}+\frac{y}{x}=b{x}^3$           

As per question

$\frac{32b}{25}+\left(10-b\right)ln2-\frac{b}{32}=\frac{62}{5}$           

$\left(10-b\right)ln2=\frac{62}{5}-\frac{31}{25}b=\frac{31}{5}\left(2-\frac{b}{5}\right)$           
$\therefore b=10$

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$

$K=\frac{2.303}{t}log\frac{a}{a-x}$

$\frac{kt}{2.303}=log\frac{a}{a-x}$

$\frac{ln2*9}{\frac{10}{3}*2.303}=log\left(\frac{1}{f}\right)$

$log\left(\frac{1}{f}\right)=81.24*10^{-2}=81*10^{-2}$

Gabriel phthalimide synthesis is used for 1° Aliphatic /alicyclic amine

1° Amine

CH₃ – CH₂ – NH₂

(A)  

tautology

           (B)            

                             tautology

Kindly consider the following figure

  $\left|\begin{array}{cc}\hfill f\left(X\right)\hfill & \hfill f\text{'}\left(x\right)\hfill \\ \hfill f\text{'}\left(x\right)\hfill & \hfill f\text{'}\text{'}\left(x\right)\hfill \end{array}\right|=0$         

$f\left(x\right)f\text{'}\text{'}\left(x\right)=f\text{'}\left(x\right)f\text{'}\left(x\right)$

$\frac{f\text{'}\text{'}\left(x\right)}{f\text{'}\left(x\right)}=\frac{f\text{'}\left(x\right)}{f\left(x\right)},$ Integrating on both sides

$\frac{f\text{'}\left(x\right)}{f\left(x\right)}=2,$ again integrating on both side

ln f(x) = 2^X + k

f(x) = e^{2x + k}

f(0) = e^k = e^k = 1-> k = 0

$\therefore f\left(x\right)=e^{2x}\left[\therefore e=2.718\right]$           

$\therefore e^2\in \left(6,9\right)$           

$\sim \left(\sim p\wedge \left(p\vee q\right)\right)=pV\left(\sim p\wedge \sim q\right)$

$\begin{array}{l}=\left(pv\sim p\right)\wedge \left(pv\sim q\right)\\ =pv\sim q\end{array}$

If we add -NO₂ group on ortho and para position, then electron deficiency increasing in benzene ring so, order of nucleophilic attack is

Clean water would have BOD Value less than 5ppm, highly polluted water could have BOD value of 17 ppm or more.