Chemistry

$logK=logA-\frac{Ea}{2.303RT}$

$Slope=\frac{Ea}{2.303R}=-10,000K$  

$\frac{Ea}{2.303R}=10^4$            

$log\left(10^{-5}\right)=logA-10^4*\frac{1}{500}$   (at 500 K temperature)

$\therefore T=\frac{10^4}{19}K=\frac{100}{19}*10^{2}K=5.2631*10^{2}K=526.31K\approx 526K$

(A) Ethane                   one $\left(C-C\right)\sigma$ bond

(B) Ethene                         one $\left(C-C\right)\sigma$ and one $\left(C-C\right)\pi$ bond

(C) $C_2$                                                       two $\left(C-C\right)\pi$ bonds

(D) Ethyne $H-C\equiv C-H$                       two $\left(C-C\right)\pi$ bonds and one $\left(C-C\right)\sigma$ bond

$E_n=-R_H\left(\frac{Z^2}{n^2}\right)J$

For ${}^{+}\left(n=1\right)$ ,

$E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4R_H$

$\therefore R_H=\frac{x}{4}$

For ${}^{3+}\left(n=2\right)$ ,

$E_n=-R_H\left(\frac{z^2}{n^2}\right)J$

$=-\frac{x}{4}*\left(\frac{4*4}{2*2}\right)=-xJ$

(A) Isothermal process $\Rightarrow$ Temperature is constant throughout the process

(B) Isochoric process $\Rightarrow$ Volume is constant throughout the process

(C) Isobaric process $\Rightarrow$ Pressure is constant throughout the process

(D) Adiabatic process $\Rightarrow$ No exchange of heat $\left(q\right)$ between system and surrounding

(1) 4 mol of $=4{}_A$ atoms

(2) 4 u of $=\frac{4u}{4u}=1$ atom

(3) 4 g of Helium $=\frac{4}{4}$ mole $=1$ mole $=N_A$ He atom

(4) $2.2710982$ of He at STP $=\frac{2.271}{22.710982}$ mole

$=0.1$

$=0.1{}_A$

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

(1) 4 mol of $=4{}_A$ atoms

(2) 4 u of $=\frac{4u}{4u}=1$ atom

(3) 4 g of Helium $=\frac{4}{4}$ mole $=1$ mole $=N_A$ He atom

(4) $2.2710982$ of He at STP $=\frac{2.271}{22.710982}$ mole

$=0.1$

$=0.1{}_A$

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

As $Q_c>K_c$ , so reaction has a tendency to move backward.

Nucleophilic addition of sodium hydrogen sulphite to aldehyde or ketone is a;

$NaHS{O}_3?N{a}^{+}+HS{O}_3^{?}$                

So; nucleopilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion.

Addition of hydrogen cyanide;

Final product is cyanohydrin.

All given oxide have nitrogen – nitrogen bond except N₂O₅ as ;