Class 12th

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

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

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

$\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}-\widehat{k}\&\overrightarrow{b}=-2\widehat{i}+\alpha \widehat{j}+\widehat{k}$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\sqrt[]{15\left(a^2+4\right)}$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2=?$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$cos\theta =\frac{-1}{\left(a^2+5\right)}$ $\therefore \left|sin\theta \right|=\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\left|\stackrel{?}{a}\right|*\left|\overrightarrow{b}\right|*\left|sin\theta \right|$

$=\left(a^5+5\right)*\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}=\sqrt[]{15\left(a^2+4\right)}$

${\left(a^2+5\right)}^2-1=15\left(a^2+4\right)$

$\Rightarrow \left(\alpha =\pm 3\right)$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2$

= $2\left(a^2+5\right)-\left(a^2+5\right)$

$\left(a^2+5\right)=14$

For isotonic solution

${?}_{injector}={?}_{Blood}$

$?C*0.082*300=7.47$

$?C=0.3\text{?}\text{?}mole/\mathcal{l}$

$=0.3*180gm/\mathcal{l}=54gm/\mathcal{l}$

The value of crystal field spilitting energy ${}_{}$ $\left({\Delta }_0\right)$ increases down the group. 5d series member will have more ${}_{}$ ${\Delta }_0$ compared to 3d series or 4d series member.

Unit cell = Hexagonal close packing (hcp)

No. of X in one unit cell = 6

No. of tetrahedral voids in one unit cell = 12

No. of Y in one unit cell = $\frac{2}{3}*12=8$ $\frac{}{}$

Formula => X₆ Y₈ => X₃ Y₄

% of X in unit cell $=\frac{3}{7}*100\mathrm{\%}=43\mathrm{\%}$ $\frac{}{}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$e=\sqrt[]{1+\frac{b^2}{a^2}}$ $e\text{'}=\sqrt[]{1+\frac{a^2}{b^2}}$

$\mathcal{l}=\frac{2b^2}{a}$ $\mathcal{l}\text{'}=\frac{2a^2}{b}$

$\left(1+\frac{b^2}{a^2}\right)=\frac{11}{14}*\frac{2b^2}{a}$ $\left(1+\frac{a^2}{b^2}\right)=\frac{11}{8}*\frac{2a^2}{b}$

$7*\left\{{\left(\frac{7b}{4}\right)}^2+b^2\right\}=11*b^2*\frac{7b}{4}\Rightarrow a*77=65$

65b² = 44b³

65 = b * 44

$\therefore 77a+44b=65*2=130$

$N{i}^{2+}+2dmg\to \left[Ni{\left(dmg\right)}_2\right]\downarrow$ Rosy-red precipitate

In Hall Heroult process electrolysis is carried out using carbon anode. The liberated

O2 at anode reacts with anode.

Cathode : $A{l}^{3+}\left(melt\right)+3ek\to A\mathcal{l}\left(\mathcal{l}\right)$

Anode : $\begin{array}{l}C\left(s\right)+O^{2-}\left(Melt\right)\to CO\left(g\right)+2e\\ C\left(s\right)+2O^{2-}\left(Melt\right)\to C{O}_2\left(g\right)+4e\end{array}$

Net Reaction 2Al2O3 + 3C $4A\mathcal{l}+3C{O}_2$

Penicillin is not a broad spectrum artibiotic and is used to treat only certain infections caused by streptococci and staphylococci such as pneumonia.