Ncert Solutions Chemistry Class 11th

NaOH is obtained by electrolysis of NaCl. Solution of NaCl using Hg cathode

$\begin{array}{l}Cathode:N{a}^{+}+e\underset{}{\overset{Hg}{\to }}Na-Amalgam\\ Anode:2C{l}^{-}\to C{l}_2+2e\\ 2Na-Amalgam+2H_{2}O\to 2NaOH+2Hg+H_2\end{array}$

Enamel has calcium hydroxyl apatite [Ca₁₀ (PO₄)₆ (OH)₂], CaCO₃, CaF₂ and Mg₃ (PO₄)₂

Ca²⁺, P⁵⁺ and F^- are present

NF₃ is stable due to high N-F bond energy.

Other halides of nitrogen (NCl₃ NBr₃, Nl₃) are explosive

B₂H₆ has 4 2 c -2e bonds and 2 3c-2e bonds.

Bridging (B-H) bonds have more value of bond- length then terminal (B – H) bonds

Bridging bonds are in one plane, but terminal bonds are in perpendicular plane.

Due to presence of (3c-2e) bonds, it behaves as electrons deficient and prone to get attached by lewis base.

$3NaB{H}_4+4B{F}_3\underset{}{\overset{\Delta }{\to }}3NaB{F}_4+2B_2{H}_6$                

Due to higher extent of polarization by Li⁺ and Mg²⁺, LiCl and Mgcl₂ have covalent character. Therefore they are soluble in ethanol.

Due to very high value of lattice energy, LiF is having very less solubility in water.

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having use in manufacturing of N-based fertizers)

Acidic oxide => Cl₂O₇

Neutral oxide => N₂O, NO

Basic oxide => Na₂O

Amphoteric oxide => As₂O₃

Degenerate orbitals must have same value of energy

Orbitals with same n and   values are degenerate orbitals.

$\pi =\frac{n}{v}RT$

$\pi =\frac{W}{M*v}RT$

$5.03*10^{-3}=\frac{2.5}{M*0.5}*0.083*300$

$M=\frac{2.5*0.083*300}{5.03*0.5}*10^{3}g/mol$

Molar mass of protein, M = 24751 g/mol

 Molar mass of glycine = 75 g/mol

So; number of glycine units =   $\frac{24751}{75}=330$