Chemistry

In leaching suitable compound is used as solvent.

$A{\mathcal{l}}_2{O}_3+2NaOH+3H_{2}O\to 2Na\left[A\mathcal{l}{\left(OH\right)}_4\right]$

Stannane is the tetrahedral shape of molecule.

Hybridisation is- sp³, shape and structure are tetrahedral

Liquation refining process is applicable for the metal having low m.p, but containing impurities have higher m.p

The difference in boiling or melting point, even after nearly the same molecular geometry, is because opf the presence of hydrogen bond. HF forms intermolecular hydrogen bonds due to fluorine's high electronegativity. This intermolecular hydrogen bonding requires more energy due to break the hydrogen bonds and melt or boil. So, stronger intermolecular forces of HF result in a higher boiling point.

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  $\left(n+\mathcal{l}\right).$