Class 12th

Rate constant, k = 5.5 * 10^-14 s^-1.

$t=\frac{2.303}{k}log\frac{{\left[R\right]}_0}{\left[R\right]}$  

 = $\frac{2.303}{k}log3-(i)$  

$t_{50\mathrm{\%}}=\frac{2.303}{k}log2-(ii)$  

 From (i) & (ii)

$\frac{t_{67\mathrm{\%}}}{t_{50\mathrm{\%}}}=\frac{log3}{log2}$  

= 1.58 t_50%

So;         t_67% is 15.8 * 10^-1 times half life.

X = 16 (the nearest integer)

$C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$  

Here, 6F electricity is required to reduce 1 mol  $C{r}_2{O}_7^{2-}$ to Cr³⁺.

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$  

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$  

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$  

It is same as number of roots of  $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (-2, 2)

Using ; k_sp = 108 S⁵

1.1 * 10^-23 = 108 S⁵

S =  ${\left(\frac{110}{108}*10^{-25}\right)}^{1/5}M\approx 10^{-5}M$  

Specific conductance,

${\lambda }_m^0=\frac{\kappa }{5*1000}S{m}^{2}mo{l}^{-1}$  

= 3 * 10^-3 Sm² mol^-1

So; x = 3

Given a > b

Area common to x² + y² $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$  

is  $\pi a^2-\pi ab=30\pi ..............\left(i\right)$  

Similarly  $\pi ab-\pi b^2=18\pi .................\left(ii\right)$  

Equation (i) and equation (ii)  $\frac{a}{b}=\frac{5}{3}$  

Equation (i) + equation (ii)  $a^2-b^2=48$  

a² = 75, b² = 27

$P_A^0=50torr$  

$P_B^0=100torr$  

Mole fraction of A in liquid phase,           x_A = 0.3

Mole fraction of B in liquid phase,            x_B = 0.7

Now;     $P_A=P_A^0{x}_A$  

$P_B=P_B^0{x}_B$

= 100 * 0.7 = 70 torr

Mole fraction of B in vapour phase,   $y_B=\frac{P_B}{P_A+P_B}=\frac{70}{85}$  

$\frac{70}{85}=\frac{x}{17}$  

                             X = 14

$\underset{\begin{array}{l}Sodiumextract\\ ofsulphur\end{array}}{N{a}_{2}S}+N{a}_2\underset{Sodiumnitroprusside}{\left[Fe{\left(CN\right)}_{5}NO\right]}\to N{a}_4\underset{violet/purple}{\left[Fe{\left(CN\right)}_{5}NOS\right]}$

$\alpha -$  Helix (2°) structure of protein is stabilized by intermolecular H-bonding.

$\frac{dy}{dx}+\frac{2x}{x-1}\cdot y=\frac{1}{{\left(x-1\right)}^2}$

IF  $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$  

=  $e^{2x}\cdot {\left(x-1\right)}^2$  

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

y =  $\frac{e^{2x}}{2{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^2}$  

y(2) =  $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$  

y(3) =  $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$  

Data contradiction.

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$