NEET Chemistry Solved Question Paper 2008

In the equation,

Zn + 2OH^- → ZnO${}_2^{2-}$ + 2H⁺ + 2e^-

NO${}_3^{-}$ + 8H⁺ + 8e^- → OH^- + 2H₂O + NH₃

Now, from the equation, we get

8 moles of electron absorbed by 85 gm of NaNO₃.

$\therefore$ 1 mole of electron absorbed by $\frac{85}{8}\mathrm{g}$ of NaNO₃ = 10.625 gm

Normality = n × molarity (M)

where, n = the number of moles of H⁺ per mole of the compound that a solute is capable of releasing [acid] on reacting with base.

In case of HCl, n = 1

So, 2N HCl solution ≈ 2 M HCl solution

And in case of H₂SO₄, n = 2

So, 4.0 N H₂SO₄ solution ≈ 2 M H₂SO₄ solution

The order of screening effect is s > p > d > f. Electrons experience the greatest effective nuclear charge when in s-orbital, then p-orbital and so on. Ionisation energy increases with an increase in penetration power. 

A catalyst changes the activation energy of the reaction. As a result, the reaction follows an alternate path and the rate of reaction wlll change.

Due to presence of intermolecular hydrogen bonding in H₂O, its molecules are associated with each other, which results unusual high boiling point of water.

According to the question, oxygen contents in element oxide is 20% by weight.

Hence, element contents in element oxide is 80% by weight.

Then, equivalent weight of unknown element = $\frac{80}{20}\times 8$

$\therefore$ Equivalent weight of unknown element = 32

Addition of sodium acetate in acetic acid solution, due to common ion NH${}_4^{+}$, The supressed, ionisation of acetic acid so concentration of [H⁺] decreases. Hence, pH of solution increases.

From the equation given below,

PbS + 4H₂O₂ → PbSO₄ + 4H₂O

$\because$ 1 mole of PbS requires 4 moles of H₂O₂.

$\therefore$ 0.01 moles of PbS requires 0.04 mole of H₂O₂.

Now, 

Weight of 0.04 mole H₂O₂ = 1.36 gm

And, 10 volume of H₂O₂ means,

1 mL of such solution of H₂O₂ on decomposition by heat produces 10 mL of oxygen at NTP.

H₂O₂ decomposes itself as,

2H₂O₂ → 2H₂O + O₂

Thus, 1 mL of 10 volume of H₂O₂ solution contains

= $\frac{68}{22400}$ × 10 gm of H₂O₂ = 0.03035 gm H₂O₂

$\because$ 0.03035 gm of H₂O₂ is present in 1 mL of 10 volume of H₂O₂

$\therefore$ 1.36 gm of H₂O₂ present in $\frac{1}{0.03035}$ × 1.36 mL of 10 volume of H₂O₂

= 44.81 mL

The de-Broglie relation is

$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$

where, λ = de-Broglie wavelength; h = Planck's constant; m = mass of particle; v = velocity of particle

Electronic configuration of Copper (Cu) is 1s², 2p², 2p⁶, 3s², 3p⁶, 4s¹, 3d¹⁰ and electronic configuration of Cu²⁺ is 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁹. Hence, the given configuration represents metallic cation.