Class 12th

Kindly consider the following figure

Meq of NH₃ = Meq of used H₂SO₄ = Meq of NaOH = 0.25 * 30 = 7.5

Millmoles of N = millimoles of NH₃ = 7.5 (As n factor = 1)

Mass of nitrogen = 7.5 * 14 * 10^-3 = 0.105 gm

% of Nitrogen = $\frac{0.105}{0.166}*100\mathrm{\%}=63.25\mathrm{\%}~63\mathrm{\%}$

Number of Co-Co Bond = X = 1

Number of terminal ligand = Y = 6

ax + by + cz = d

2a + 3b – 5c = d               2a + 3b – 5c = d …………….(v)

Putting (i) & (iv) in (v) we get a = -9d.

In conditions

$\left(2\widehat{i}+\widehat{j}-5\widehat{k}\right),\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

2b = d ………….(i) d > 0

2a + b – 5c = 0                  2a + 3b – 5c = d   $\left|a\right|,\left|b\right|,\left|c\right|,d\to g.c.d$           

=   $\frac{\alpha }{2}$

$\therefore \frac{\alpha }{2}=1\Rightarrow \alpha =2$

$\left(3\widehat{i}+5\widehat{j}-7\widehat{k}\right).\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

3a + 5b – 7c = 0) *   $\frac{4}{7}...................\left(iii\right)$

$\therefore a=-18,b=1$

c = -7, d = 2

(iii)…(ii) ® 2a + 3b   $\frac{-33c}{7}=0$

2a + 3b =  $\frac{33}{7}c$ $c=\frac{-7}{2}d.................\left(iv\right)$

Putting values

a + 7b + c + 20d = 22

Number of Co-Co Bond = X = 1

Number of terminal ligand = Y = 6

Kindly consider the following figure

$a=\left|\frac{4.2-3\left(-1\right)-21}{5}\right|$

$=\left|\frac{8+3-21}{5}\right|=2$

$\therefore L=4a=8$

$k=A.e^{-\left(\frac{Ea}{RT}\right)}$

$\mathcal{l}nk=\mathcal{l}nA-\frac{Ea}{RT}$

$\Rightarrow \mathcal{l}nk=\mathcal{l}nA+\left(-\frac{Ea}{1000R}\right).\frac{1000}{T}$

Slope = $\frac{-Ea}{1000R}=-18.5$  

=> Ea = 18.5 * 1000 * 8.31 = 153.735 * 10³ J = 154 KJ

Anode, H₂ (g) -> 2H⁺ + 2e^-

Cathode, Cu²⁺ + 2e -> Cu (s)

Net cell reaction H₂ + Cu²⁺ -> 2H⁺ Cu (s)

$E_{cell}^0=E_{C{u}^{2+}/Cu}^0-E_{H^{+}/H_2}^0$

= 0.34 V – 0

= 0.34 V

$E_{cell}=E_{cell}^0-\frac{0.06}{2}log\frac{{\left[H^{+}\right]}^2}{\left[C{u}^{2+}\right]}$

$\begin{array}{l}\Rightarrow 0.576=0.34-\frac{0.06}{2}log\frac{{\left[H^{+}\right]}^2}{0.01}\\ \Rightarrow -log\left[H^{+}\right]=4.93\\ \Rightarrow pHofsolution=4.93~5\end{array}$