Class 12th

Oxidation of HCHO (a) will produce HCOOH, which is produced by ants.

$\mathrm{\%}Opticalpurity=\frac{Observedrotationofmixture*100}{rotationofpureenautiomer}$

$=\frac{+12.6}{+30}*100=42\mathrm{\%}$

  $2KMn{O}_4+5H_2{\stackrel{+3}{C}}_2{O}_4+3H_{2}S{O}_4\left(dil\right)\to K_{2}S{O}_4+2MnS{O}_4+10C{\stackrel{+4}{O}}_2+8H_{2}O$

Oxidation state of carbon changes from +3 to +4

$\therefore$ change in O. S is 1

Complex is   $\left[Co{\left(N{H}_3\right)}_{4}C{l}_2\right]Cl$

Primary valency = No. of ionisable species = 1

Solubility of CaF₂ = S mol/L

$S=\frac{2.34*10^{-3}}{0.1*78}=\frac{2.34}{78}*10^{-2}=3*10^{-4}mol/L$          

$K_{sp}\left(Ca{F}_2\right)=4S^3=4{\left(3*10^{-4}\right)}^3=108*10^{-12}$        

= 0.0108 * 10^-8 (mol/L)³

On decreasing pressure of NO by a factor of '2' the rate of reaction decreases by a factor of '4'

$\therefore$ Order of reaction w.r.t 'NO' = 2

Meq of K₂Cr₂O₇ = Meq of Fe²⁺

-> (Molarity * Volume * nf) of K₂Cr₂O₇ = (molarity * volume * nf) of Fe²⁺

->0.02 * 20 * 6 = M * 10 * 1

M = 0.24 M

Molarity = 24 * 10^-2 M

The current lags the voltage by an angle of 90 Degree (? /2 radians). This is due to the inductor opposing the change in current.

Radius, R = R₀A^1/3

take z = x + iy

$z^2+\overline{z}=0$               

$\Rightarrow x^2-y^2+x+i2xy-yi=0$               

$\Rightarrow x^2-y^2+x=0andy\left(2x-1\right)=0$             

if y = 0 Þ x = 0, -1

$ifx=\frac{1}{2}\Rightarrow y=\pm \frac{\sqrt[]{3}}{2}$               

$\Sigma \left(Re\left(z\right)+lm\left(z\right)\right)=\left(0-1+\frac{1}{2}+\frac{1}{2}\right)+\left(0+0+\frac{\sqrt[]{3}}{2}-\frac{\sqrt[]{3}}{2}\right)=0$