Class 11th

$\frac{0.02858*0.112}{0.5702}=0.0561$

In an oxidation-reduction reaction or redox reaction, the process of oxidation (addition of electrons) and reduction (removal of electrons) happens at the same time.

Redox reaction does not have a fixed formula but is an equation that forms when oxidation and reduction occur simultaneously. For example, 2Fe + 3Cl? 2FeCl? (Fe loses electrons and Cl gains).

Redox reactions are of four kinds: Decomposition, Combination, Displacement, and disproportionation reactions.

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°.

In boron family lower O.S (+1) is more stabilize down the group due to inert pair effect.

According to question, we can write

$\mathcal{l}=2\pi r\Rightarrow r=\frac{\mathcal{l}}{2\pi }$

$I_1=\frac{m{\mathcal{l}}^2}{3},and{I}_2=\frac{m{r}^2}{2}=\frac{m{\mathcal{l}}^2}{8{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{m{\mathcal{l}}^2}{3}*\frac{8{\pi }^2}{m{\mathcal{l}}^2}=\frac{8{\pi }^2}{3}$