Class 12th

No. of moles =  $\frac{5}{92}\left(toluene\right)$

Moles of benzaldehyde = $\frac{5}{92}*\frac{92}{100}=5*10^{-2}$  

Mass of benzaldehyde = 5 * 10^-2 *106 = 5.3 gm

= 530 * 10^-2 gm

Fehling's solution is a complex of   $C{u}^{2+},C{u}^{2+}=3d^9$

No. of unpaired e^- = 1

MM =  $\sqrt[]{1\left(1+2\right)}=\sqrt[]{3}=1.73BM$

  $2KMn{O}_4+3H_2{O}_2\underset{}{\overset{Basicmedium}{\to }}2Mn{O}_2+3O_2+2H_{2}O+2KOH$

Bauxite = AlOx (OH)_3-2x (where O < x < 1)

Siderite = FeCO₃

Cuprite = Cu₂O

Calamine = ZnCO₃

Haemetite = Fe₂O₃

Kaolinite = Al₂ (OH)₄Si₂O₅

Malachite = CuCO₃ Cu (OH)₂

Magneite = Fe₃O₄

Sphalerite = ZnS

Limonite = Fe₂O₃.3H₂O

C = 500   $\mu F,$ V = 100 v, L = 50 mH                                                        

In this LC – oscillation

q = q₀ cos   $\omega t$

$i=\frac{-dq}{dt}=q_0\omega sin\omega t\omega =\frac{1}{\sqrt[]{2c}}=\frac{1}{\sqrt[]{50*10^{-3}*5*10^{-4}}}$

=  $\frac{1000}{5}=200$  

So,     i_max =   $q_0\omega =500*10^6*100*200$

= 10A

$K=\frac{0.693}{t_{1/2}}=\frac{0.693}{70*60}=\frac{6930}{7*6}*10^{-6}$

= 165 * 10^-6 s^-1

$\frac{P^0-P_S}{P^0}=\frac{n_{solute}}{n_{solvent}}$

$\frac{P^0-\frac{P^0}{2}}{P^0}=\frac{n_{solute}}{n_{solvent}}$

$n_{solute}=\frac{n_{solvent}}{2}=\frac{100}{18*2}=2.78mol$

  ${\lambda }_A=25\lambda ,$ ${\lambda }_B=16\lambda$      

At t = 0 Þ N_A = N_B = N₀

after t =  $\frac{1}{a\lambda }:-\frac{N_B}{N_A}=\frac{N_0{e}^{-16\lambda t}}{N_0{e}^{-25\lambda t}}=e^{\left(25\lambda -16\lambda \right)t}$  

->e =  $e^{\left(9\lambda \frac{1}{a\lambda }\right)}$  

$\Rightarrow \frac{9}{a}=1\Rightarrow a=9$      

$i_{\left(zener-max\right)}=25mA$                                                                

 20 – i_max R – 8 = 0

i_max R = 12At minimum zener current $\left(\mu A\right):-$  

$20-i_{min}R-i_{min}{R}_L=0$  

  $\frac{R}{R_L}=\frac{12}{8}=\frac{3}{2}$

${\mathcal{l}}_{min}R=12$

  $i_{min}{R}_L=8$

At max^m zener current –

$20-i_{max}R-8=0$  

$i_L=O\left\{as{i}_{z}ma{x}^m=25mA\right\}$             

i_maxR = 12v

25 * 10^-3 R = 12

$R=\frac{12*10^3}{25}=12*40=480\Omega$

$C_{eff}=\left[\frac{{\epsilon }_0\left(7*4\right)}{4/10}+\frac{5{\epsilon }_0\left(1*4\right)}{4/10}\right]*10^{-2}$

$C_{eff}=1.2{\epsilon }_0$

Energy = $\frac{1}{2}{C}_{eff}{V}^2$ $\frac{}{}{}_{}{}^{}$

$\frac{1}{2}\left(1.2{\epsilon }_0\right)\left(20\right)\left(20\right)=240{\epsilon }_0$