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 = 399KJ

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

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

In disproportionation reaction, one element of a compound will simultaneously get reduced and oxidised. In ClO₄ -, oxidation number of Cl is +7 and it can not increase it further. So, ClO₄- will not get oxidised and so will not undergo disporportionation reaction.

Correct order is Be

InA-O-H, if EN of ' A' is 2.1 then it will be neutral, as X_A-X₀=X₀-X_H. (where X is EN)

Bond order in N $N_2$ and O $O_2^{2+}$ is 3 (calculated by energy level diagram)