Ncert Solutions Chemistry Class 12th

Sphalerite is ore of zinc consists of ZnS during its concentration group 1 cyanides are used as depressants like NaCN

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

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

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.

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

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

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

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

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

$\frac{1.86}{93}=\frac{w}{135}$

W = 2.70 g

Mass produced actual = 2.70 $*\frac{90}{100}=2.43=243*10^{-2}g$

$Pb{l}_2?P{b}^{2+}+2l^{-}{K}_{sq}=8.0*10^{-9}$

Pb (NO₃)₂ -> Pb²⁺ + ^{$2N{O}_3^{-}$}

0.1 M-

-0.1 M    0.1 M

  $K_{sq}=8*10^{-9}Using{K}_{sq}=\left[P{b}^{2+}\right]{\left[I^{-}\right]}^2$          

$8*10^{-9}=0.1*{\left(2S\right)}^2$

S = 141 * 10^-6 M

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$

$K=\frac{2.303}{t}log\frac{a}{a-x}$

$\frac{kt}{2.303}=log\frac{a}{a-x}$

$\frac{ln2*9}{\frac{10}{3}*2.303}=log\left(\frac{1}{f}\right)$

$log\left(\frac{1}{f}\right)=81.24*10^{-2}=81*10^{-2}$