NEET Chemistry Solved Question Paper 2007

Acidic salt (NaHCO₃) and base (NaOH) cannot exist together in solution.

In liquid $\rightleftharpoons$ vapour, equilibrium the rate at which molecules escape from the liquid surface to gaseous phase equals to rate at which molecules in the gas phase strike the surface and becomes the part of liquid. Hence, the molecules in the two phases have equal kinetic energy.

Homogenous catalysis - When the reactants and catalyst are in the same phase, the catalysis is known as homogenous catalysis.
Example: $2{\mathrm{KClO}}_3\left(\mathrm{s}\right) \stackrel{{\mathrm{MnO}}_2\left(\mathrm{s}\right)}{\to }2\mathrm{KCl}\left(\mathrm{s}\right) + 3{\mathrm{O}}_2\left(\mathrm{g}\right)$

Equilibrium is  affected by temperature and pressure due to change in heat as well as change in volume of substances.

$$\begin{gathered} 4.6 \times {10}^{22} \mathrm{atoms} \mathrm{weigh} = 13.8 \mathrm{g} \\ \mathrm{Hence}, \\ 6.02 \times {10}^{13} \mathrm{atoms} \mathrm{will} \mathrm{weigh} \\ =\frac{13.8 \times 6.02\times {10}^{23}}{4.6 \times {10}^{22}}= 180.6\mathrm{g} (\mathrm{molar} \mathrm{mass}) \end{gathered}$$

(a) 0.1 mole of CO₂

$\left(\mathrm{b}\right)\hspace{0.17em}\frac{11.2}{22.4}$ = 0.5 mole of CO₂

(c) $\frac{22}{44}$ = 0.5 mole of CO₂

(d) $\frac{22.4 \times {10}^3}{22400}$= 1 mole of CO₂

Hence, the number of moles of CO₂ is least ie, 0.1.

Li^-= 1s², 2s² (In it all sub-shells are saturated so, it is stable).

Be^- = 1s², 2s², 2p¹(very much less stable).

B^- = 1s^2, 2s², 2p² (less stable)

C^- = 1s², 2s², 2p³ (stable due to presence of half-filled 2p-subshell).

$$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{atoms} \mathrm{in} 24\mathrm{g} \mathrm{of} \\ \mathrm{C} =\frac{24}{12}\times 6.02 \times {10}^{23} = 2 \times 6.02\times {10}^{23} \\ \mathrm{No}. \mathrm{of} \mathrm{atoms} \mathrm{in} 56 \mathrm{g} \mathrm{of} \\ \mathrm{Fe}= \frac{56}{56}\times 6.02 \times {10}^{23} \\ \mathrm{No}. \mathrm{of} \mathrm{atoms} \mathrm{in} 26 \mathrm{g} \mathrm{of} \mathrm{Al}= \frac{26}{27} \times 6.02\times {10}^{23}\approx 6.02\times {10}^{23} \\ \mathrm{No}. \mathrm{of} \mathrm{atoms} \mathrm{in} 108 \mathrm{g} \mathrm{of} \\ \mathrm{Ag} =\frac{108}{108}\times 6.02 \times {10}^{23} =6.02\times {10}^{23} \end{gathered}$$

In SO₂ molecule, S is sp²-hybridised

S(16)  = 1s², 2s², 2p⁶, 3s², 3p⁴

The number of lone pair in XeOF₄ is one (1). The structure of XeOF₄ is given as follows :