NEET Chemistry Solved Question Paper 2015

Bomb calorimeter commonly used to find the heat of combustion of organic substances, consists of a sealed combustion chamber, called a bomb.

The reaction in a bomb calorimeter proceeds at constant volume so, w = 0 and $∆$U = q.

Hence, $∆$U < 0, w = 0

Significant numbers in 0.0000135 are 3. It is because zeros to the left of the first non-zero digitin a number are not significant.

$\underset{1-0.6= 0.4}{\underset{1}{2{\mathrm{SO}}_3 \left(\mathrm{g}\right)}} \rightleftharpoons \underset{0.6}{\underset{0}{ 2{\mathrm{SO}}_2 \left(\mathrm{g}\right)}} + \underset{0.3}{\underset{0}{{\mathrm{O}}_2 \left(\mathrm{g}\right)}} \underset{\mathrm{at} \mathrm{equilibrium} \mathrm{moles}}{\underset{\mathrm{initial} \mathrm{moles}}{{}_{} }}$

K_c = $\frac{[{\mathrm{SO}}_2{]}^2 \left[{\mathrm{O}}_2\right]}{[{\mathrm{SO}}_3{]}^2} = \frac{(0.6{)}^2 {\left(0.3\right)}^2}{{\displaystyle {\left(0.4\right)}^2}} = 0.675$

$\mathrm{\alpha }$ particles can be detected by using zinc sulphide screen. It emits a flash of light upon  $\mathrm{\alpha }$-particle. It was first used by Rutherford.

Al and Be show similar properties due to their diagonal relationship.

In a decay process, when ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$ changes into ${}_{\mathrm{Z}-1}{}^{\mathrm{A}}\mathrm{Y}$, the process is known as ${\mathrm{\beta }}^{+}$ decay. It can be explained as-

${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X} \to {}_{\mathrm{Z}-1}{}^{\mathrm{A}}\mathrm{Y} + {}_{+1}{}^0\mathrm{e} (\mathrm{or} {\mathrm{\beta }}^{+})$

Bond order of N₂ = 3, bond order of O₂ = 2. Higher the bond order, shorter is the bond length and higher is the bond dissociation energy, i.e., higher stability or lesser reactivity. Thus, N₂ is less reactive than O₂.

[Ba²⁺] = 0.5 mol L^-1

[S${\mathrm{O}}_4^{2-}$] = 1mol L^-1

Ba²⁺ (aq) + S${\mathrm{O}}_4^{2-}$ (aq) $\rightleftharpoons$ BaSO₄ (s)

0.5 mol of Ba²⁺ would react with 0.5 mol of S${\mathrm{O}}_4^{2-}$ ions to form 0.5 mol of BaSO₄.

For a given value of n, l = 0 to n - 1;

For a given value of l, m = -l to +l including zero;

For a given value of m, s = $+\frac{1}{2}, -\frac{1}{2}$

Therefore, when n = 5, then l = 0, 1, 2, 3, 4;

when l = 2, then m = -2, -1, 0, +1, +2

The value of s can be $\pm \frac{1}{2}$

Hence, the arrangement, n = 5, l = 2, m = 2, s = +$\frac{1}{2}$ is possible for an electron.

Due to high electronegativity and electron withdrawing nature of F-atom, the lone pair of N-atom in NF₃ molecule is not easily available for donation. However, in N(CH₃)₃, -CH₃ group is an electron releasing group, thus lone pair of N-atom in N(CH₃)₃ molecule can be ligated easily. Except for nitrogen fluoride, all other trihalides hydrolyse in water.