Class 12th

 f (3x)- f (x) = x

Replace  $x\to \frac{x}{3}\Rightarrow f\left(x\right)-f\left(\frac{x}{3}\right)=\frac{x}{3}$  

Again replace  $x\to \frac{x}{3}\Rightarrow f\left(\frac{x}{3}\right)-f\left(\frac{x}{3^2}\right)-f\left(\frac{x}{3^2}\right)=\frac{x}{3^2}$

$\Rightarrow f\left(3x\right)-f\left(0\right)=\frac{3x}{2}puttingx=\frac{8}{3}\Rightarrow f\left(8\right)-f\left(0\right)=4\therefore f\left(0\right)=3$  

Also putting x =  $\frac{14}{3}$ in f (3x) – 3 = $\frac{3x}{2}\Rightarrow$ F (14) – 3 = 7 Þ f (14) = 10

$\left[Cu{\left(en\right)}_2{\left(SCN\right)}_2\right]$

More stable isomers = 3 (trans isomers)

t_1/2 = 0.301 min

              t = 2 min

              $K=\frac{2.303}{t}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{0.693}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{2.303*0.301}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\therefore 2=\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

              $\frac{Co}{Ct}=10^2=100$  

              Ans. 100

Ka for C₃H₇COOH = 2 * 10^-5

              $pKa=-\mathcal{l}og\left(2*10^{-5}\right)=5-\mathcal{l}og2$  

              =5 – 0.3 = 4.7

              pH of 0.2 (M) solution =

              $pH=\frac{pKa-\mathcal{l}ogC}{2}$  

              $=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

              $pH=27*10^{-1}$     

          $\therefore$    Ans 27

$m=\frac{w*1000}{molecularwt*w_{solvent}}=\frac{10.2*1000}{176*150}$

$=\frac{1020\cancel{0}}{176*15\cancel{0}}=0.386$

$\Delta T_f=K_f*m$

3.9 * 0.386

$\therefore x*10^{-1}=15.05*10^{-1}$

Ans = 15

$FeO+\underset{\left(Acidicflux\right)}{Si{O}_2}\to \underset{\left(Slag\right)}{FeSi{O}_3}$

IUPAC nomenclature of element with atomic no. 103 is uniltrium

Uses of catalyst is very specific for a particular reaction.