Chemistry Test-35

\(\quad \quad \quad \quad \quad \quad N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)\)

Initial moles \(\quad \quad 1\quad \quad \quad\quad\quad 0\)

Moles at equil. \(\quad(1-\alpha) \quad \quad 2 \alpha\)

\((\alpha=\text { degree of dissociation })\)

Total number of moles at equil.

\(=(1-\alpha)+2 \alpha\)

\(=(1+\alpha)\)

\(P_{N_{2} O _{4}}=\frac{(1-\alpha)}{(1+\alpha)} \times P\)

\(P_{ NO _{2}}=\frac{2 \alpha}{(1+\alpha)} \times P\)

\(K_{P}=\frac{\left(P_{ NO _{2}}\right)^{2}}{P_{N_{2} O_{4}}}=\frac{\left(\frac{2 \alpha}{(1+\alpha)} \times P\right)^{2}}{\left(\frac{1-\alpha}{1+\alpha}\right) \times P}=\frac{4 \alpha^{2} P}{1-\alpha^{2}}\)

Given, \(K_{P}=2, P=0.5 atm\)

\(\therefore \quad K_{P}=\frac{4 \alpha^{2} P}{1-\alpha^{2}}\)

\(=\frac{4 \alpha^{2} \times 0.5}{1-\alpha^{2}}\)

\(\alpha=0.707=0.71\)

\(\therefore\) Percentage dissociation

\(=0.71 \times 100=71\)

Corrosion is an example of oxidation reaction.

Oxidation is defined as a corrosion reaction in which the corroded metal forms an oxide. Usually, this reaction is with a gas containing oxygen, such as air.

Corrosion is a natural process that converts a refined metal into a more chemically stable oxide. It is the gradual deterioration of materials by chemical or electrochemical reaction with their environment.

Paramagnetic species contain unpaired electrons in their molecular orbital electronic configuration. The molecular orbital configuration of the given species is as \(\mathrm{CO}~(6+8=14)\)

\(=\sigma 1 s^{2}, \stackrel{\star}{1} s^{2}, \sigma 2 s^{2}, \stackrel{\star}{2} s^{2}, \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \sigma 2 p_{z}^{2}\)

(All the electrons are paired so, it is diamagnetic).

\(O_{2}^{-}(8+8+1=17)\)

\(=\sigma 1 s^{2}, \stackrel{\star}{1} s^{2}, \sigma 2 s^{2}, \stackrel{\star}{2} s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2},\stackrel{\star}{\pi} 2 p_{x}^{2} \approx \hat{2} p_{y}^{1}\)

(It contains one unpaired electron so, it is paramagnetic).

\(C N^{-}(6+7+1=14)=\) same as \(\mathrm{CO}\)

\(N O^{+}(7+8-1=14)=\) same as \(\mathrm{CO}\)

Thus, among the given species only \(O_{2}^{-}\) is paramagnetic.

Given,

Decay constant of \(A, \lambda_{A}=10 \lambda\)

Decay constant of \(B, \lambda_{B}=\lambda\)

Initially, Nuclei of \(A=\) Nuclei of \(B\)

After \(t\) time ratio of their nuclei is, \(\frac{N_{A}}{N_{B}}=\frac{1}{e}\)

The decay constant \(\lambda\) represents the average probability per nucleus of decay occurring per unit time. The rate of radioactive decay is proportional to the number of each type of radioactive nuclei present in a given sample.

According to the radioactive decay formula,

\(N=N_{0} e^{-\lambda t}\)

Where \(N_{0}\) be the initial number of nuclei in both \(A\) and \(B\)

For a radioactive material \(A , N _{ A }= N _{0} e ^{-\lambda_{A} t }\)

Where \(N_{A}\) be the number of nuclei in \(A\)

For a radioactive material \(B , N _{ B }= N _{0} e ^{-\lambda_{B} t }\)

Where \(N_{B}\) be the number of nuclei in \(B\).

Thus, their ratio will be

\(\frac{ N _{ A }}{ N _{ B }}=\frac{ N _{0} e^{-\lambda_{A} t }}{ N _{0} e ^{-\lambda_{B} t }}\)

\(\frac{ N _{ A }}{ N _{ B }}= e ^{-\lambda_{A} t {+\lambda_{B} t }}\)

\(\frac{ N _{ A }}{ N _{ B }}= e ^{\left(\lambda_{ B }-\lambda_{ A }\right) t}\)

\(\frac{1}{ e }= e ^{\left(\lambda_{ B }-\lambda_{ A }\right) t}\)

\(e^{-1}= e ^{\left(\lambda_{ B }-\lambda_{ A }\right) t}\)

\(\left(\lambda_{ B }-\lambda_{ A }\right) t =-1\)

\(\left(\lambda_{ A }-\lambda_{ B }\right) \times t =1\)

\(t =\frac{1}{\left(\lambda_{ A }-\lambda_{ B }\right)}\)

\(t=\frac{1}{10 \lambda-\lambda}=\frac{1}{9 \lambda}\)

Thus, the ratio of the number of nuclei of \(A\) to that of \(B\) will be \(1 / e\) after a time \(\frac{1}{9 \lambda}\)

Drugs are chemicals of low molecular masses (~100 – 500u). These interact with macromolecular targets and produce a biological response. When the biological response is therapeutic and useful, these chemicals are called medicines and are used in diagnosis, prevention and treatment of diseases. Most of the drugs used as medicines are potential poisons, if taken in doses higher than those recommended. Use of chemicals for therapeutic effect is called chemotherapy.

If the adsorption is due to the weak van der waals forces between adsorbent and adsorbate, it is termed as Physisorption.

Physisorption is adsorption by van der Waals force, which is a weak intermolecular attraction that takes place below the critical temperature of the adsorbate and can result in the development of a monolayer or multilayer.

Adsorption of gas on a solid is a spontaneous exothermic reaction. Amount of heat liberated when a unit mass of a gas is adsorbed on the surface is called heat of adsorption. The process of removal of adsorbent from the surface of adsorbate is known as Desorption.

Adsorption is defined as the deposition of molecular species onto the surface. The molecular species that gets adsorbed on the surface is known as Adsorbent and the surface on which adsorption occurs is known as Adsorbate.

CO₂is a linear molecule because of sp-hybridization around the carbon atom.

Among the given antibiotics, only chloramphenicol i.e., option D contains a nitro group attached to aromatic ring. Its structure is as follows:

Given value in this solution:

Power consumed in electrolysis \(=100 W\)

Current inflow in this solution, \(i =\frac{100}{125}=0.8 A\)

According to the first law of Faraday's law \(M=Z Q\) and \(Q=\) it

So, \(M = Zit\)

\(Z=0.367 \times 10^{-6} ; i=0.8 A ; t =60 s\)

\(M =0.367 \times 10^{-6} \times 0.8 \times 60\)

\(=17.6 \times 10^{-6} kg\)

\(=17.6 mg\)

Let after \(t s\) amount of the \(A _{1}\) and \(A _{2}\) will become equal in the mixture.

As \(N=N_{0}\left(\frac{1}{2}\right)^{n}\) where \(n\) is the number of half-lives

For \(A_{1}, N_{1}=N_{01}\left(\frac{1}{2}\right)^{t / 20}\)

For \(A_{2}, N_{2}=N_{02}\left(\frac{1}{2}\right)^{t / 10}\)

According to question, \(N_{1}=N_{2}\)

\(\frac{40}{2^{t / 20}}=\frac{160}{2^{t / 10}} \)

\(2^{t / 10}=4\left(2^{t / 20}\right) \text { or } 2^{t / 10}=2^{2} 2^{t / 20}\)

\(\Rightarrow 2^{t / 10}=2^{\left(\frac{t}{20}+2\right)}\)

\(\frac{t}{10}=\frac{t}{20}+2 \text { or } \frac{t}{10}-\frac{t}{20}=2\)

\(\frac{t}{20}=2 \text { or } t=40 \) second