Class 12th

The best conductor of electricity is Silver (Ag) at room temperature. It is not used in normal wiring even though being the best conductor because of its high cost.

• Electrical Conductivity ($\sigma$): $6.3 \times 10^ {7} \ \text {S/m}$

• Electrical Resistivity ($\rho$): $1.59 \times 10^ {-8} \ \Omega \cdot \text {m}$ (the lowest among all metals)

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

  $\frac{2^{21}{\left(\frac{-1+i\sqrt[]{3}}{2}\right)}^{21}}{{\left(\sqrt[]{2}\right)}^{24}{\left(\frac{1-i}{\sqrt[]{2}}\right)}^{24}}+\frac{2^{21}{\left(\frac{1+\sqrt[]{3}i}{2}\right)}^{21}}{{\left(\sqrt[]{2}\right)}^{24}{\left(\frac{1+i}{\sqrt[]{2}}\right)}^{24}}=k$

=   ${\displaystyle \sum _{j=0}^5\left(j+5\right)\left(j+5-1\right)={\displaystyle \sum _{j=0}^5\left(j+5\right)\left(j+4\right)}}$

= ${\displaystyle \sum _{j=0}^5\left(j^2+9\widehat{j}+20\right)=\frac{5*6*11}{6}+\frac{9*5*6}{2}+20*6}$  

55 + 135 + 120 = 310

  $\Delta T_f=k_f.m$

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

    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

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

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

$x\frac{dy}{dx}+y=b{x}^4$       

$\frac{dy}{dx}+\frac{y}{x}=b{x}^3$           

As per question

$\frac{32b}{25}+\left(10-b\right)ln2-\frac{b}{32}=\frac{62}{5}$           

$\left(10-b\right)ln2=\frac{62}{5}-\frac{31}{25}b=\frac{31}{5}\left(2-\frac{b}{5}\right)$           
$\therefore b=10$

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

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

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

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

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