Chemistry

$B_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1={\pi }_{2py}^1\to$ Paramagnetic

$\begin{array}{l}L{i}_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2\to Diamagnetic\\ C_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2\to Diamagnetic\\ C_2^{-}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^1\to Paramagnetic\end{array}$

$O_2^{-2}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*2}\equiv {\pi }_{2py}^{*2}$ Diamagnetic

$O_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^0\to$  Paramagnetic

$H{e}_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*1}\to$ Paramagnetic

$\therefore$ Paramagnetic molecules are ${}_{}{}_{}^{}{}_{}^{}{}_{}^{}$ $=B_2,C_2^{-},O_2^{+},H{e}_2^{+}$

B.C.C structure

a = 300 pm = 300 * 10^-12 m

d = 6g/cm³

z = 2

$d=\frac{Z*M}{a^3}$            

$6=\frac{2*A}{{\left(300*10^{-10}\right)}^3}=\frac{2*A}{27*10^{-24}}$            

$\therefore AtomsofM$ = 3.69 * 6.022 * 10²³

= 22.22 * 10²³

the nearest integer = 22

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$\therefore n_x=1$          

  $n_y=1$              

$n_z=\frac{0.05}{3}=0.0167$ here z is limiting reagent.

$?\frac{0.05}{3}$ mole z gives 1 mole xyz₃

$\therefore$ mass of xyz₃ = n * molecular mass

=  $\frac{0.05}{3}*\left(10+20+3*30\right)a.m.u.$

= 0.5 * 4 = 2g

Moles of Br₂= moles of C₅H₁₀= $\frac{5}{60+10}=\frac{5}{70}$

$\therefore \frac{w}{160}=\frac{5}{70}$

$\therefore w=\frac{5*160}{70}=\frac{80}{7}g$

$=1142.8*10^{-2}\approx 1143*10^{-2}g$

 ${\Delta }_t=\frac{hc}{\lambda }=\frac{6.63*10^{-34}*3.08*10^8}{600*10^{-9}}$

$=\frac{6.63*3.08*10^{-17}}{600}$ J

$=765.765*10^{-21}J$

$?766*10^{-21}$ J

$\frac{d\left[C\right]}{dt}=\left(\frac{20-10}{10}\right)=1mmold{m}^{-3}$

$+\frac{d\left[D\right]}{dt}=\left\{-\frac{d\left(B\right)}{dt}\right\}*1.5=\frac{3}{2}\left\{-\frac{d\left[B\right]}{dt}\right\}$

$\left\{-\frac{d\left[B\right]}{dt}\right\}=2*\left\{\frac{-d\left[A\right]}{dt}\right\}$

$\therefore \frac{1}{6}\left\{\frac{-d\left[B\right]}{dt}\right\}=\frac{1}{3}\left\{\frac{-d\left[A\right]}{dt}\right\}=\frac{1}{9}\left\{\frac{+d\left[D\right]}{dt}\right\}=\frac{d\left[C\right]}{dt}$

$\therefore$ rate of reaction = $\frac{+d\left[C\right]}{dt}$ = 1m.m dm^-3 S^-1

$F{e}^{+3}+l^{-}\to I_2+F{e}^{+2}$

Fe⁺² undergoes reduction & I₂ undergoes oxidation.

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

= (0.77 – 0.54) V = 0.23 V

= 23 * 10^-2 V

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2*0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}*1\right)}^2}{\left(\frac{0.5}{1.5}*1\right)}$

$=\frac{4}{3}=1.33$

$=-710.15J/mol$

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

= 0.00002 * 10³ mm

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium ; $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165*1000}{-550}K$

$\begin{array}{l}=3*100K\\ =300K\end{array}$