Chemistry Test-36

'Enzyme is non-sensitive for the changes of temperature and pH'is not true for enzymes.

Enzymes are complex biomolecules made up of proteins. Few RNA molecules are catalytic in nature. They are called as ribozymes but mostly enzymes are macromolecules made up of proteins. Enzymes are biological catalysts and function at physiological pH and temperature. They are inactive at extremes of temperature and pH. Enzymes are biological catalysts and catalyze metabolic reactions in living systems. Enzymes can be regulated by specific molecules called as allosteric modulators, the allosteric modulators bind on separate allosteric sites to enzymes. Those enzymes having allosteric site are called as allosteric enzymes.

We have been given, Bond length of \(\mathrm{HCl}\) bond \(=2.29 \times 10^{-10} \mathrm{~m}\)

Measured dipole moment \(=6.226 \times 10^{-30} C-m\)

We have to find out : Percentage ionic character of \(\mathrm{HCl}\)

Now, the percentage ionic character:

It may be defined as the ratio of observed dipole moment to calculated dipole moment multiplied by \(100 .\) Percentage ionic character \(=\frac{\text { Observed dipole moment }}{\text { Calculated dipole moment }} \times 100\)

The observed dipole moment is given to us while the calculated one is alsoneeded.

So, let us calculate that one first.

Theoretically dipole moment can be calculated as-

Dipole moment \(=\) charge on electron \(\times\) bond length of the molecule

Dipole moment \(=1.6 \times 10^{-19} \times 2.29 \times 10^{-10}\)

Dipole moment \(=3.664 \times 10^{-20}\)

As we have calculated the dipole moment. So, now let us calculate

Percentage ionic character as Percentage ionic character \(=\frac{6.226 \times 10^{-19}}{3.664 \times 10^{20}} \times 100\)

Percentage ionic character \(=16.9 \%\)

This value is approximately equal to \(17 \%\).

The numerical value of gas constant R,in lit atm mol⁻¹ K⁻¹ is 0.0821.

The letter R is the universal has constant which has many applications in equations like the ideal gas equation, Nernst equation, etc. It relates the energy or pressure-volume statistic to the temperature scale. It is known to be equivalent to the Boltzmann constant. The universal gas constant was noted as a product of the Boltzmann constant and the Avogadro’s number and is represented as: R=N_Ak_b

Here, N_A is Avogadro's number and k_b is the Boltzmann constant.

The universal gas constant has different values according to the different units that it can take. The value for the SI units of the universal gas constant is 8.314JK⁻¹mol⁻¹.

For a reversible isothermal expansion process, the work done is given by\(\mathrm {W=-n R T \ln \frac{V_2}{V_1}}\)
Given that \(\mathrm {n=0.5 \mathrm{~mol}, V_1=2 \mathrm{~L}, V_2=5 \mathrm{~L}, T=273+27=}\) \(300 \mathrm{~K}\) and \(\mathrm R=1.987 \mathrm{cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).
Substituting these values in the equation, we get\(\mathrm {W_{\max } =0.5 \times 1.987 \times 300 \times \ln \left(\frac{5}{2}\right)} \)\(=0.5 \times 1.987 \times 300 \times 0.916=303.4 \mathrm{cal}\)
According to first law of thermodynamics, for isothermal expansion\(\mathrm {q=\Delta U+W}\)
Given that \(\mathrm q=200 \mathrm{cal}\). Substituting values of \(\mathrm q\) and \(\mathrm W\) in the above equation, we get\(\Delta \mathrm U=200-303.4\)\(=-103.4 \mathrm{cal}\)

Hydrogen gas combines with nitrogen and forms ammonia. The general chemical equation for this reaction is:

\(\mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{NH}_{3}(g)\)

To balance the above equation, the number of atoms should be made equal on each side. The reactant side contains two atoms of nitrogen, hence two molecules of ammonia should be produced. Multiply ammonia with two. If two molecules of ammonia are being produced, then six hydrogen atoms should be consumed, multiple hydrogen molecules with three. On doing these steps we observe that the number of atoms of both nitrogen and hydrogen are the same on the reactant as well as the product side. Hence, the balanced chemical equation is:

\(3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \)

The process of removing impurities from impure metal is called refining.

Electrolytic refining is a widely used method of refining as it is a cost-effective and efficient method.

During the process, the impure metals are taken as anode and, pure metals are taken as the cathode. As electrolysis proceeds, the pure metal keeps depositing on the cathode and, the impurities settle at the bottom.

The major air pollutant is carbon monoxide (CO). 

The pollutants which destroy the quality of air and make it unsuitable for living beings and plants are known as air pollutants. The major air pollutant is carbon monoxide (CO). It is one of the most harmful pollutants. Other gaseous air pollutants are carbon dioxide, oxides of sulfur, nitrogen, hydrocarbons, hydrogen sulfide, ozone and other oxidants. Particulate pollutants include mist, dust, smog, fumes, etc.

Depression in freezing point is a colligative property which depends upon the amount of the solute.

\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}\)

Thus, for a given solvent and given concentration, \(\Delta T_{f}\) is directly porportional to \(i\) (Van't Hoff factor) i.e. maximum \(T_{f}\) (and hence lowest freezing point) will correspond to maximum value of \(i \).

(A) \(A l_{2}\left(S O_{4}\right)_{3} \stackrel{H_{2} O}{\longrightarrow} 2 A l^{3+}+3 S O_{4}^{2-}\)

Here, Van't Hoff factor, i = 5

(B) \(C_{5} H_{10} O_{8} \stackrel{H_{2} O}{\longrightarrow}\) No ionization

Here, Van't Hoff factor, i = 1

(C) \(K I \stackrel{H_{2} O}{\longrightarrow}K^{+}+I^{-}\)

Here, Van't Hoff factor, i = 2

(D) \(C_{12} H_{22} O_{11} \stackrel{H_{2} O}{\longrightarrow}\) No ionization

Here, Van't Hoff factor, i = 2

Therefore, \(0.10 M(\approx 0.10 \mathrm{~m})\) aqueous solution will have the lowest feezing point.

In a galvanic cell, the reactions taking place in the anodic half-cell and the cathodic half-cell will beoxidation and reduction respectively.

In a galvanic cell in an anode half-cell, oxidation of \(Zn\) takes place, and in a cathodic half-cell\(Cu ^{2 +}\) gets reduced.

Due to the presence of intermolecular forces, the gases show deviation from ideal behaviour. Also, the presence of intermolecular forces makes the liquefaction of gas possible. Inert gases could not have been liquefied if these intermolecular van der Waal's forces were not present within a molecule.