Chemistry

$T_{90\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{10}$

$T_{50\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{50}$

$\frac{T_{90\mathrm{\%}}}{T_{50\mathrm{\%}}}=\frac{log10}{log2}=\frac{1}{0.301}=3.32$

Kindly consider the following figure

Kindly consider the following figure

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