Chemistry

$2C\left(s\right)+2H_2\left(g\right)\text{?}\text{?}\to C_2{H}_6\left(g\right)\Delta H=\Delta H_f\left(C_2{H}_6\right)$  

$\Delta H_f\left(C_2{H}_6\right)=\left[2*\left(-394\right)\right]+\left[3*\left(-286\right)\right]-\left[1*\left(-1560\right)\right]KJ/mole$  

= (-788 – 858 + 1560) KJ/mole = -86 kJ/mole

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

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

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

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

= 0.399 * 10⁶ J = 399 KJ

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 =  $\frac{270}{18}*2*1gm=30gm$

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

Strength = 0.72 g/L

Using ;

$Molarity=\frac{Strength}{Molarmassofsolute}$              

$M=\frac{0.72}{180}M$              

= 4 * 10^-3 M

Ans. is 4.

Theoretically; 1 mol benzene gives 1 mol toluene

Moles of benzene = $\frac{10}{78}mol$

Moles of toluene (Theoretical) = $\frac{10}{78}mol$

Mole of toluene (observed) = $\frac{9.2}{92}=0.1mol$

% yield = $\frac{0.1}{\frac{10}{78}}*100$

= 78%

Using ; Nernst equation

$E_{cell}=E_{cell}^o-\frac{0.06}{n}lo{g}_{10}\frac{\left[C{u}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}$              

$E_{cell}=2.97-\frac{0.06}{2}lo{g}_{10}\frac{0.25}{10^{-6}}V$              

$=2.97-0.03\left(log2.5+5\right)V$              

$E_{cell}=2.8V$              

Ans. is 3 (the nearest integer)

Acyclic isomers of pentene are :-

So; total isomers are 6

Density, $\rho =7.62gc{m}^{-3}$  

Cell edge, a = 0.4518nm = 0.458 * 10^-7 cm

Effective number of atoms, Z = 4 for CCP.

Using ; $\rho =\frac{Z*A}{N_A*a^3}$  

$\rho =\frac{4*A}{6.022*10^{23}*{\left(0.458*10^{-7}\right)}^3}$              

$7.62=\frac{4*A}{6.022*{\left(0.4518\right)}^3*10^2}$              

A = 105.79 g/mol

Ans. is 106 (the nearest integers)

K_p = 47.9      T = 288 K

K_C =?                    R = 0.083 L bar/K mol

$N_2{O}_4\left(g\right)?2N{O}_2\left(g\right)$           

Using

$K_p=K_C{\left(RT\right)}^{\Delta ng}$              

Here ;

$\Delta ng=1$                

49.7 = K_C (0.083 * 288)¹

K_C = 2