Class 11th

CH₄, O₃ & H₂O are green-house gases.

2O₃(g) -> 3O₂

t = 0      a moles               0

-0.5a mole     +0.75a mole

At Eq.    0.5a mole         0.75a mole

Total moles at eq = 0.5a + 0.75 a = 1.25a

$P_{O_3}=x_{O_3}*P_T=\frac{0.5a}{1.25a}*1=\frac{2}{5}atm$

$P_{O_2}=x_{O_2}*p_T=\frac{0.75a}{1.25a}*1=\frac{3}{5}atm$

$K_p=\frac{{\left(P_{O_2}\right)}_{eq}^3}{{\left(P_{O_3}\right)}_{eq}^3}=\frac{{\left(\frac{3}{5}\right)}^3}{{\left(2/5\right)}^2}=1.35atm$

$\Delta G^{°}=-RT\mathcal{l}nKP$

$\begin{array}{l}=-8.3*300\mathcal{l}n1.35\\ =-747J/mole\end{array}$

Phosphorous can expand its covalency beyond 4, as it has vacant 3d orbitals. Nitrogen can't expand its covelency beyond 4, as it does not posses vacant 3d- orbitals.

Fire extinguisher contains sodium Bicarbonate (NaHCO3), also known as baking soda.

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

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

$2y{e}^{x/y^2}dx+\left(y^2-4x{e}^{x/y^2}\right)dy=0$

  $\Rightarrow 2e^{2/y^2}\left(ydx-2xdy\right)+y^{2}dy=0$            

$2e^{x/y^2}d\left(\frac{x}{y^2}\right)+\frac{dy}{y}=0$

Integrating   $2e^{x/y^2}+Iny=c$

y = 1, n = 0  c = 2

  ^{$2e^{x/y^2}+Iny=2$}  

$\therefore y=e\Rightarrow 2e^{x/e^2}+1=2$

x = -e² In2

  $y=3?\left|x?\frac{1}{2}\right|=\left|x+1\right|$

Graph

Area =   $\left(\frac{1}{2}*\frac{3}{2}*\frac{3}{4}\right)*2+\frac{3}{2}*\frac{3}{2}=\frac{3}{2}*\frac{3}{2}*\frac{3}{2}=\frac{27}{8}$

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

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

Ca²⁺, P⁵⁺ and F^- are present