Chemistry

Melting point of monocarboxylic acids with even number of carbon is more than that with odd number of carbon due to lattice energy.

With increase in molar mass of monocarboxylic acids size of alkyl group (Hydrophobic portion) increases and therefore solubility in water decreases.

Bond order of   $C_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-4}{2}=3$

Bond order of $N_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-6}{2}=2$

Bond order of   $O_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-8}{2}=1$

N_B = No. of electrons in bonding molecular orbitals.

N_A = No. of electron is Anti bonding molecular orbitals

Kindly consider the following figure

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

Energy of 1 photon =  $\frac{hc}{\lambda }$

Energy of 1 mole of photon = $N_A*\frac{hc}{\lambda }$ ${}_{}\frac{}{}$

$=\frac{6.022*10^{23}*6.63*10^{-34}*3*10^8}{300*10^{-9}}$

= 0.399 * 106 J = 399 KJ

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)

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

              Number of moles of H = 2 * no. of moles of H₂O = $\left(\frac{270}{18}*2\right)$ moles

              Mass of C = $\frac{330}{44}*12$ gm = 90 gm

              Mass of H = $\frac{270}{18}*2*1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}*100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}*100\mathrm{\%}=25\mathrm{\%}\end{array}$