Class 11th

Funnel of option A is useful for the separation of liquid – liquid mixture of immiscible form.

Following are the component of photochemical smog,

$C{H}_4+O_3\stackrel{}{\to }HCHO+H_{2}O+C{H}_2\text{?}=\text{?}CH-CHO+C{H}_{3}COON{O}_3$

Ionic radius of isoelectronic species depends upon $\frac{Z}{e}$ $\frac{}{}$ ratio; Hence Ionic radius of $O^{--}>$ ${}^{}$ ionic radius of Mg+

For spontaneous process   $\Delta G<0$

For Isobaric process;    $\Delta P=0$

For Isothermal process;    $\Delta T=0$

$\Delta H$ reaction =  ${\displaystyle \sum _{}^{}B.E_{\left(R\right)}.-{{\displaystyle \sum _{}^{}B.E.}}_{\left(P\right)}}$

Kindly consider the following figure

Com

m1v + m2 * 0 = m1v1 + m2v2

75v = $45\frac{v}{3}+50*10$ $\frac{}{}$

$\Rightarrow 75*\frac{2v}{3}=50*10$

$\Rightarrow v=\frac{3*50*10}{75*2}=10m/sec$

Bernoulli's equation between (1) & (2)

$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v_2^2}{zg}+z_2$

$\frac{P_{atm}+mg/A}{\rho g}+0+40+10^{-2}$

$\frac{P_{atm}}{\rho g}+\frac{v_2^2}{2g}+0\left[v_1\approx 0\right]$

$\frac{mg}{A\rho g}+40*10^{-2}=\frac{v_2^2}{2g}$

$\frac{m}{A\rho }+40*10^{-2}=\frac{v_2^2}{2g}$

$\Rightarrow \frac{25}{0.5*10^3}+40*10^2=\frac{V_2^2}{2*10}$

$\Rightarrow V_2^2=\left(5*10^{-2}+40*10^{-2}\right)20$

$V_2^2=45*10^{-2}*20$

$V_2^2=90*10^{-1}$

$V_2=3m/sec$

$=300cm/sec$

T₂ = 200 k

$T_1=527°C+273=800K$      

w = 12000 kJ = 12 * 10⁶ J

Q₁ =?

  $x=1-\frac{T_2}{T_1}=\frac{w}{Q_1}=1-\frac{200}{800}=\frac{12*10^6}{Q_1}$              

$\Rightarrow \frac{3}{4}=\frac{12*10^6}{Q_1}$      

$\Rightarrow Q_1=16*10^{6}J$        

$V_T=2000c{m}^3=2000*10^{-6}{m}^3\left(Totalvolumeofmixture\right)$

T = 300 K

${}_{}{}^{}{}^{}$  $P_T=100kPa=100*10^{3}Pa=10^5$ Pa

mT = 8.3 J/K1 mol1

Now, using Ideal gas equation

$P_{T}V=\left(x_1+x_2\right)RT$

$\Rightarrow 10^5*200*10^{-6}=\left(x_1+x_2\right)*8.3*300$

$\frac{x_1}{x_2}=\frac{0.06}{0.02}=3:1$

  $\mathrm{}{\mathrm{C}\mathrm{N}}^{-}\Rightarrow 14$ electrons, so $\mathrm{B}.\mathrm{O}=3$ and diamagnetic.