Chemistry

In sulphide ore, depressants selectively prevent impurity from coming to the fourth.

Except Te, all are metals.

Invertase Cane sugar to glucose and fructose

Zymase Glucose to ethanol

Diastase Starch to Maltose

Maltase Maltose to glucose

HOCl produce in the stratospheric cloud, by the hydrolysis reaction of ClONO₂.

$ClON{O}_{2\left(g\right)}+H_2{O}_{\left(g\right)}\to HOC{l}_{\left(g\right)}+HN{O}_{3\left(vap\right)}$

In 4d orbital, n = 4 and $\mathcal{l}=2$

Radial nodes = $n-\mathcal{l}-1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes,  $\mathcal{l}=2$

$P_4+3NaOH+3H_{2}O\to P{H}_3+3Na{H}_{2}P{O}_2$

$RedP+alkali\to \underset{\left(NoP-Hbond\right)}{H_4{P}_2{O}_6}$

Due to H- bond in water, it has high melting point and melting point of other hydrides of the group are depending upon the molecular weight.

Due to high crystallity Be has the highest M.P.

Be = 1560 K

Mg = 925 K

Ca = 1120 K

Sr = 1062 K

During the electrolysis of dilute H₂SO₄

$2H_{2}S{O}_4\stackrel{\left(1\right)electrolysis}{\to }2HS{O}_4^{-}+2H^{+}$

$H_2{S}_2{O}_8+2H_{2}O\stackrel{hydrolysis}{\to }2H_{2}S{O}_4+\underset{\left(A\right)}{H_2{O}_2}$

$2H^{+}+2e\to H_2\uparrow$

In the solid form of $H_2{O}_2$ dihedral angle is equal to 90.2°.

According to Ellingham diagram, the metal oxide with lower $\Delta G^o$ is more stable than metal oxide of higher $\Delta G^o$ at same temperature.