Chemistry

(a) Nitric oxide formation -> Pt is used as catalyst

(b) Haber's process -> Fe is used as catalyst

(c) Hydrolysis of ester -> Acid (H₂SO₄) is used as catalyst

(d) SO₃ formation -> NO is used as catalyst

(a)  $Cd\left(s\right)+2Ni{\left(OH\right)}_3\left(s\right)\to CdO\left(s\right)+2Ni{\left(OH\right)}_2\left(s\right)+H_{2}O\left(\mathcal{l}\right)$ ${}_{}{}_{}{}_{}$

During discharging of secondary battery this reaction takes place.

(b) $Zn\left(Hg\right)+HgO\left(s\right)\to ZnO\left(s\right)+Hg\left(\mathcal{l}\right)$ $$

Primary battery mercury cell reaction

(c) $2PbS{O}_4\left(s\right)+2H_{2}O\left(\mathcal{l}\right)\to Pb\left(s\right)+Pb{O}_2\left(s\right)+2H_{2}S{O}_4\left(aq\right)$ ${}_{}{}_{}{}_{}{}_{}{}_{}$

During charging of secondary battery PbSO4 reacts and $H_{2}S{O}_4$ ${}_{}{}_{}$ generated

(d) $H_2\&O_2$ ${}_{}{}_{}$ reacts in fuel cell to form $H_{2}O\left(\mathcal{l}\right)$ ${}_{}$

$4HNO{I}_{3\left(\mathcal{l}\right)}+3KC{l}_{\left(s\right)}\underset{}{\overset{}{\to }}C{l}_{2\left(g\right)}+NOC{l}_{\left(g\right)}+2H_{2}O\left(g\right)+3KN{O}_3\left(g\right)$

$\downarrow$              

4 moles of HNO₃ produced 3 mol of KNO₃

Here mole of produced KNO₃ = $\frac{110}{101}$  

If 3 mol of KNO₃ produced by 4 moles of HNO₃

$\therefore$  1 mole of KNO₃ produced by  $\frac{4}{3}$ moles of HNO₃

and  $\frac{110}{101}$ mole of KNO₃ produced by $\frac{4*110}{3*101}$  moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃ = mole * mol.wt = 145 * 63 = 91.48 $\approx$ 91.5gm

$\Delta H=\Delta U+\Delta \left(PV\right)$

= $\Delta U+P\Delta V+V\Delta P$

= $\Delta U+P\Delta V$   (At constant pressure)

An atomic orbital is characterized by $n,\mathcal{l}$ $$ and m.

Xenon has 8 valence electrons. In ${\mathrm{X}\mathrm{e}\mathrm{F}}_2$ , it forms two bonds with fluorine, leaving three lone pairs. The steric number =2 bonds +3 lone pairs $=5$ .
As per the VSEPR Theory  the electron? pair geometry due to SN = 5 will be trigonal bipyramidal. The hybridization type will be ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ hybridisation. However, the three lone pairs occupy equatorial positions, resulting in a linear molecular geometry.

Total 3 streo- isomers

Mole of polyhydric alcohol =  $\frac{1.84*10^{-3}}{92}=2.0*10^{-5}mole.$

Mole of H₂ gas produced = $\frac{1.344}{22400}=6.0*10^{-5}mole.$  

No of -OH gp present =  $\frac{6.0*10^{-5}}{2.0*10^{-5}}=3$

Only H₂S₂O₈ has per-oxo bond.

Oxidation state of Co = 3

And co-ordination No = 6

Sum = 3 + 6 = 9