Chemistry

Kindly go through the solution 

 $\frac{0.02858*0.112}{0.5702}=0.0561$

A salt bridge is a U-shaped tube that connects the two halves (oxidation and reduction) of the cell with an electrochemical cell, such as a galvanic cell. It allows the flow of ions in both halves and maintains the electricity to stay neutral.

In electrochemistry, metallic and electrolytic are two types of conductors.

The two branches of electrochemistry are: Electricity generated by chemical reactions and Chemical reactions that generate electricity.

% of C in organic compound

              $=\frac{12}{14}*\frac{WC{O}_2}{W.O.C.}*100$  

              $=\frac{952.56}{21.648}=44\mathrm{\%}$  

              $\mathrm{\%}ofH=\frac{2}{18}*\frac{W{H}_{2}O}{W.O.C.}*100$  

              = $\frac{20}{18}*\frac{0.4428}{0.492}*100$  

              = $\frac{88.56}{8.856}=10\mathrm{\%}$  

              = 100 – 54 = 46%

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

More stable isomers = 3 (trans isomers)

$C\mathcal{l}{F}_3$ ® T-shaped (sp³d)

 IF₇ ® Pentagonal bipyramidal (sp³d³)

BrF₅ ® Square pyramidal (sp³d²)

BrF₃ ® T-Shaped (sp³d)

I₂Cl₆ ® Triangular bipyramidal (sp³d)

$I{F}_5\to$  Square Pyramidal (sp³d²)

ClF ® (sp³)

ClF₅ ® Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

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