Chemistry

$C{H}_{3}C{H}_{2}C{H}_{2}Br\stackrel{Mg}{\to }C{H}_{3}C{H}_{2}C{H}_{2}MgBr\stackrel{C{O}_2}{\to }$

$C{H}_{3}C{H}_{2}C{H}_{2}COOMgBr\stackrel{H_3{O}^{+}}{\to }C{H}_{3}C{H}_{2}C{H}_{2}COOH\left(butanoicacid\right)$

O₂F₂ oxidises plutonium to PuF₆ and the reaction is used in removing plutonium as PuF₆ from spent nuclear fuel.

Ionic radii of cations is smaller than anions, also more the positive charge less be the ionic radii and more the negative charge more be the ionic radii. Hence correct order of ionic radii is, $p^{3-}>S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$

When red phosphorus is heated in a sealed tube at 803 K, a - black phosphorus is formed.

$A{g}^{+}\left(aq\right)+2N{H}_3\left(aq\right)?Ag{\left(N{H}_3\right)}_2^{+}\left(aq\right)$

t = 00.8 M $\left(\frac{a}{2}\right)M$                  

0 $=\infty$   5 * 10^-8 M   $\left(\frac{a}{2}-1.6\right)M$   0.8M

$k_f=\frac{\left[Ag{\left(N{H}_3\right)}_2^{+}\right]}{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2}$

$10^8=\frac{0.8}{\left(5*10^{-8}\right){\left(\frac{a}{2}-1.6\right)}^2}$

$\frac{a}{2}=2$

 Concentration of NH₃ added =   $\frac{a}{2}=2M$

 Volume of solution = 2L

 Moles of NH₃ added = 2 * 2 mol = 4 mol.

Millimoles of HCl = 200 * 0.2 = 40

Millimoles of NaOH = 300 * 0.1 = 30

Heat released =  $\frac{30}{1000}*57.1*1000J=1713J$

$\left[\begin{array}{l}\rho =\frac{m}{v}\\ m=\rho v\end{array}\right]$

Mass of solution = 500 * 1 g

= 500 g

Specific heat of water = 4.18 Jg^-1 K^-1

$\Delta T=\frac{q}{mc}$

$=\frac{1713}{500*4.18}°C$  

$=81.96*10^{-2}°C\approx 82*10^{-2}°C$

Ans. = 82

Let molarity of KMnO₄ = x

$KMn{O}_4+FeS{O}_4\to F{e}_2{\left(S{O}_4\right)}_3+M{n}^{2+}$

n = 5         n = 1         Ferric sulphate

equivalent of KMnO₄ = equivalent of FeSO₄

5 * x * 10 = 1 * 0.1 * 10

x = 0.02 M

Strength = (0.02 * 158) = 3.16g/L

= 316 * 10^-2 g/L

Ans. = 316