Chemistry Test - 33

As given, the temperature is increased from \(\mathrm{T}_{1}\) to \(\mathrm{T}_{2}\), then new constants become \(\mathrm{k}_{3}\) and \(\mathrm{k}_{4}\) respectively.

Rate constant increases with increase in temperature.

So, \(k_{1}<\mathrm{k}_{3}\) and \(\mathrm{k}_{2}<\mathrm{k}_{4}\)

On electrolysis of dilute sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will beOxygen gas.

When hydroxide ions are discharged, oxygen gas is liberated at anode. However, if sulphate ions are discharged, sulphur dioxide gas is liberated. Since the value of the electrode potential for the discharge of hydroxide ions is less than that of sulphate ions, the hydroxide ions will get discharged preferentially at anode. Due to this, oxygen gas will liberate in preference to sulphur dioxide gas at anode.During the electrolysis of dilute aqueous sulphuric acid, using platinum electrodes, oxygen gas is liberated at anode.

The group of peptidase, amylase, rennincontains biocatalysts. In humans salivary amylase enzyme breaks down starch. Renin is a proteolytic enzyme related to pepsin that synthesized by chief cells in the stomach of some animals. Peptidases are enzymes that cleave peptide bonds, yielding proteins and peptides.Biocatalyst is the substance which is used in the biochemical reaction and this increases the speed of the biochemical reaction. The enzyme accelerates the chemical reaction and that enzyme will not change the state of the equilibrium. This type of enzyme is known as biocatalyst.

\(2,2,4\)-Trimethylpentane has all the four types \(\left(1^{\circ}, 2^{\circ}, 3^{\circ}, 4^{\circ}\right)\) of carbon atoms.

\(\mathrm{C}_{1}\) and \(\mathrm{C}_{5}\) are \(1^{\circ}, \mathrm{C}_{3}\) is \(2^{\circ}, \mathrm{C}_{4}\) is \(3^{\circ}\) and \(\mathrm{C}_{2}\) is \(4^{\circ}\)

\(2,3,4\)-Trimethylpentane has \(1^{\circ}, 3^{\circ} \mathrm{C}\) atoms.

Neo-Pentane has \(1^{\circ}, 4^{\circ}\) C atoms.

Separating the oxidation and reduction reaction from the redox reaction:

\(3 \mathrm{~N}_{2} \mathrm{H}_{4}+2 \mathrm{BrO}_{3}^{-} \rightarrow 3 \mathrm{~N}_{2}+2 \mathrm{Br}^{-}+6 \mathrm{H}_{2} \mathrm{O}\)

Assigning the oxidation number on central atom (N and Br) in each molecules by considering oxidation number of \(\mathrm{H}=+1, \mathrm{O}=-2\),we get oxidation state as:

\(3\overset {-2}{N}_2H_4 + 2[\overset{+5}{Br}O_3]^- \rightarrow 3\overset{0}{N_2} + 2\overset{-1}{Br^-} +6H_2O\)

As the oxidation number of \(N\) changes from \(-2\) to 0 as:

\(\mathrm{N}_{2} \mathrm{H}_{4} \rightarrow N_{2}\)

Its a oxidation reaction where \(N_{2} H_{4}\) gets oxidized.

Similarly, as the oxidation number of \(B r\) changes from \(+5\) to \(-1\) as:

\(\mathrm{BrO}_{3}^{-} \rightarrow B r^{-}\)

Its a reduction reaction where \(\mathrm{BrO}_{3}^{-}\) gets reduced.

\(+3\) oxidation state is common for all lanthanides.

• \([\mathrm{Xe}] 4 \mathrm{f}^{1-14} 5 \mathrm{~d}^{0-1} 6 \mathrm{~s}^{2}\) is the general configuration of Lanthanides.
• It may seem \(+2\) should be the common oxidation state but relative stability depends on other factors as well.
• Lanthanides most commonly shows \(+3\) oxidation state due to high hydration enthalpy than of \(+2\) oxidation state.

Given:

Mole fraction of solute in first solution \(= 0.2\)

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of solute, i.e.,

\(\frac{p^{\circ}-p}{p^{\circ}}=\frac{n}{n+N}\)

or, \(\frac{\Delta p}{p^{\circ}}=\frac{n}{n+N} \)

The decrease in vapour pressure is \(10\) mm Hg:

\(\frac{10}{p^{\circ}}=0.2\)

\(\therefore p^{\circ}=50 \mathrm{~mm}\) ....(1)

For other solution of same solvent, the decrease in vapor pressure is \(20\) mm Hg:

\(\frac{20}{p^{\circ}}=\frac{n}{n+N}\)

or, \(\frac{20}{50}=\frac{n}{n+N}\) (from (1))

\(0.4=\frac{n}{n+N}\) (mole fraction of solute)

\(\because\) Mole fraction of solvent \(+\) mole fraction of solute \(=1\)

So, mole fraction of solvent \(=1-0.4=0.6\)

Group\(13\) elements group’s elements have smaller atomic radii.

Group \(13\) elements have smaller atomic radii and ionic radii than those of alkaline earth metals and alkali metals due to the greater effective nuclear charge, atomic radii increases on going down the group with an abnormality at gallium.

Among the following mixtures, dipole-dipole as the major interaction is present inAcetonitrile and acetone.

cis-[Co(en) \(\left.{ }_{2} \mathrm{Cl}_{2}\right]\) compounds is expected to exhibit optical isomerism.

Compound 'a' does not have an element of symmetry and thus exhibits optical isomerism.