VITEEE 2019 :: Chemistry :: General questions

• Chemistry - General questions

The wavelengths of the two photons are 2000Å and 4000Å respectively. What is the ratio of their energies?

Answer:

Explanation:

$E_{1}=h.\frac{c}{\lambda_{1}}$; $E_{2}=h.\frac{c}{\lambda_{2}}$

$\frac{E_{1}}{E_{2}} = \frac{hc}{\lambda_{1}}\times\frac{\lambda_{2}}{hc}$

= $\frac{\lambda_{2}}{\lambda_{1}}$ = $\frac{4000}{2000}$

= 2

Deep-sea divers used to respirate a mixture of

Answer:

Explanation:

An oxygen-helium mixture is used for artificial respiration in deep-sea diving instead of air because nitrogen present in the air dissolves in blood under high pressure when sea diver goes into the deep sea. When he comes to the surface nitrogen bubbles out of the blood due to a decrease in pressure, causing pain. This disease is called "bends".

The electrode potentials for Cu²⁺ (aq) + e^- →  Cu⁺ (aq) and Cu⁺ (aq) + e^- → Cu(s) are + 0. 15 V and + 0.50 V, respectively The value of E°_Cu²⁺/Cu will be: 

Answer:

Explanation:

Cu²⁺  + 1e^- →  Cu⁺    : ΔG^°₁ = -n₁E°₁F

Cu⁺  +  1e^- → Cu       :  ΔG₂^° = -n₂E°₂F

$Cu^{2+} + 2e^{-}\rightarrow Cu$ : $ΔG^{\circ} = ΔG_1^\circ + ΔG_2^\circ$

$-nE^{\circ}F = -1n_{1}$

$E_1^\circ F + (-1)n_{2}E_2^\circ F$

$ -nE^{\circ}F= -F(n_{1} E_1^\circ+n_{2}E_2^\circ F)$

$E^{\circ} =\frac{n_{1} E_1^\circ + n_{2}E_2^\circ}{n} $

= $\frac{0.15\times1 + 0.50\times1}{2}$ = 0.325V

The value of K_C for the reaction: A+3B $\rightleftharpoons$ 2C at 400ºC is 0.5.Calculate the value of K_P

Answer:

Explanation:

K_P = K_C [RT]^Δng    Δn = 2-4 = -2

T = 673K,  K_C = 0.5,

R=0.082L.atm.mol^-1 K^-1

K_P = 0.5×[0.082×673]^-2

= $1.6\times 10^{-4}$

For the reaction N₂ + 3H₂ → 2NH_{3 ,} If $\frac{Δ[NH_{3}]}{Δt} = 2\times 10^{-4} mol/ls$  then value of $\frac{-Δ\left[H_{2}\right]}{Δt}$ would be

Answer:

Explanation:

 N₂ + 3H₂ → 2NH₃

$\frac{-Δ[N_{2}]}{Δt}$ = $\frac{-1}{3}\frac{Δ\left[H_{2}\right]}{Δt}$

= $\frac{1}{2}\frac{Δ[NH_{3}]}{Δt}$

$\frac{-Δ\left[H_{2}\right]}{Δt}$ = $\frac{3}{2}$× $\frac{Δ[NH_{3}]}{Δt}$

=$\frac{3}{2}$× $2\times 10^{-4} mol/ls$

= $3\times 10^{-4}mol L^{-1}S^{-1}$

• Chemistry - General questions