Chemistry

Blue cupric metaborate is reduced to cuprous metaborate in a luminous flame.

  $2Cu{\left(B{O}_2\right)}_2+2NaB{O}_2+C\underset{}{\overset{Luminous}{\to }}2CuB{O}_2+N{a}_2{B}_4{O}_7+CO$            

Cupric metaborate is obtained by heating boric anhydride & copper sulphate in a non-luminous flame as

$CuS{O}_4+B_2{O}_3\underset{}{\overset{Non-LuminiousFlame}{\to }}\underset{Blue}{Cu{\left(B{O}_2\right)}_2}+S{O}_3.$            

Al < Mg < Si < S < P

1^st I. E increase along period with exception on moving from group 2 to group 13 and group 15 to group 16

Cu is the only element of 3d – series whose M²⁺ / M value is positive because of fact that low hydration enthalpy and high sublimation & ionization enthalpies.

Correctly identifying isotopes and isobars requires knowing both the atomic and mass numbers. Relying on only one is a common error.

• Isotopes: Same element (atomic number), different mass.
• Isobars: Different elements (atomic number), same mass.

Rutherford's atomic model was a breakthrough, but it was flawed. It couldn't explain atomic stability, as orbiting electrons should lose energy and spiral into the nucleus. It also failed to account for the discrete line spectra observed from excited atoms.

α - sulphur & β - sulphur – Diamagnetic, S₂ – form is paramagnetic due to presence of unpaired electron in π* orbital like O₂.

  $\Delta T_f=k_f.m$

$T_f^0-T_f=5.12*\frac{\left(\frac{10}{58}\right)}{\left(\frac{200}{1000}\right)}$

    5.5 – T_f = $\frac{5.12*5*10}{58}$  

  $T_f=1.086°C=\left(1.086\right)°C\cong 1°C$             

$3CH\equiv CH\left(g\right)\to C_6{H}_6\left(\mathcal{l}\right)$       

$\Delta G^0=-RT\mathcal{l}nK.........\left(i\right)$

$\Delta G^0=\sum \Delta {G^0}_P-\sum \Delta {G^0}_R..........\left(ii\right)$    

Equating (i) & (ii)

-2.303 RTlogk = 4.88 * 10⁵

$logK=\frac{-4.88*10^5}{2.303*R*T}=\frac{-488000}{5705.848}=-85.52=-855*10^{-1}$

So; magnitude of log K = 855 * 10^-1

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

$E_1=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}{\left[\frac{1}{H^{+}}\right]}^8$  

[H⁺] = 1M

$E_2=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}+\frac{0.059}{5}log10^{-32}$   

So, difference in E₁ & E₂ is

= 0.3776 V

= 3776 * 10^-4 V