Chemistry

Higher adsorbtion of gas is corresponds to higher liquefaction and higher liquefaction is directly proportional to the higher critical temperature.

Total concentration of H⁺ after the mixing of HCl and H₂SO₄

$\left[H^{+}\right]=\frac{\left(200*0.01\right)+\left(400*2*0.01\right)}{600}$

$\left[H^{+}\right]=\frac{1}{60}$

$\therefore pH=lo{g}_{10}\left(\frac{1}{60}\right)=1.78$

Mass of pure carbon in coal = 0.6 * 1000gm * $\frac{60}{100}=360gm$  

Mass of carbon converted into CO = 6 $\frac{60}{100}*360=216gm\Rightarrow Mol=\frac{216}{12}=18$  mole of carbon

Mass of carbon converted into CO₂ = 360 – 216 = 144gm Þ Mole = $\frac{144}{12}=12$  mole of carbon.

C _(s) + O₂ (g) -> CO₂ (g) + 400KJ

Mole – 12 for CO₂ production, Hence total energy produced = 400 * 12 = 4800KJ

$C_{\left(s\right)}+\frac{1}{2}{O}_2\left(g\right)\underset{}{\overset{}{\to }}C{O}_{\left(g\right)}+100KJ$  

Mole = 18 for CO production, Hence energy produced = 100 * 18 = 1800KJ

Total heat = 4800 + 1800 = 6600KJ

Rb and Cs has nearly same ${\Delta }_{e.g}H=-46kJ/mole$ ${}_{}$

Ar and Kr has same ${\Delta }_{e.g.}H=+96kJ/mole$ ${}_{}$

Sequence of sub-energy level decided by the rule of  $\left(n+\mathcal{l}\right).$

(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)