Chemistry Test - 12

The electrode reaction is:

\(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}\)

The quantity of electricity passed \(=\) Current \(\times\) Time

\(=0.5\) (amp) \(\times 2 \times 60 \times 60\) (sec)

\(=3600 \mathrm{C}\)

From the electrode reaction, it is clear that \(96500 \mathrm{C}\) of electricity deposit \(\mathrm{Ag}=108 \mathrm{~g}\)

\(3600 \mathrm{C}\) of electricity will deposit \(\mathrm{Ag}\)

\(=\frac{108}{96500} \times 3600\)

\(=4.03 \mathrm{~g}\)

Calculation of thickness:

Let the thickness of silver deposited be \(x~ \mathrm{cm}\).

Mass \(=\) Volume \(\times\) Density

\(=\) Area \(\times\) Thickness \(\times\) Density

(\(\because\)Volume = Area \(\times\) thickness)

\(4.03 \mathrm{~g}=900\left(\mathrm{ cm}^{2}\right) \times x(\mathrm{ cm}) \times 10.5\left(\mathrm{ g} \mathrm{~cm}^{-3}\right)\)

\(x=\frac{4.03}{900 \times 10.5}\)

\(x=4.26 \times 10^{-4} \mathrm{~cm}\)

(D) \(\mathrm{H}_{2} \mathrm{S}\)

Let oxidation number of sulphur in \(H_{2} S\) be \(x\).

\(\therefore 2+x=0\)

\(x=-2\)

(A) \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Let oxidation number of sulphur in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) be \(x\).

\(\therefore 2+x-4(2)=0\)

\(x=+6\)

(B) \(S O_{2}\)

Let oxidation number of sulphur in \(S O_{2}\) be \(x\).

\(x+(2 \times-2)=0\)

\(x-4=0\)

\(x=+4\)

(C)\(\mathrm{H}_{2} \mathrm{SO}_{3}\)

Let oxidation number of sulphur in \(\mathrm{H}_{2} \mathrm{SO}_{3}\) be \(x\).

\(\therefore 2+x+(3)(-2)=0\)

\(x=4\)

The substance that decreases the activity of catalysts is poisoner.

Promoters/Activators: Substance which themselves are not catalyst but its presence can increase the catalytic activity of catalyst. A promoters increase the number of active sites on the surface.

Catalytic Poisons/ Anti catalysts/ Catalyst Inhibitor are those substances which themselves are not catalyst but whose presence decreases the activity of the catalyst. Poisoning is due to preferential adsorption of poison on the surface of the catalyst.

The name of the compound \(\left[ Co \left( NH _{3}\right)_{5} NO _{2}\right] Cl _{2}\) will be Pentaamminenitrocobalt(III) chloride. 

Ligands are named in alphabetical order. The oxidation number of cobalt is \(+3\) as the oxidation numbers of the nitro group and chlorine are \(- 1\) each. Option (A) is ruled out as cobalt has \(+2\) oxidation state. Option (B) and (D) are ruled out as Co is written as cobaltate which indicates that Co is part of a complex anion. Also, in option (D), nitroso should be replaced with nitro.

The correct order of basicity of the given compounds is 2 < 1 < 3 < 4.

According to the structure given in the question:

(1) ; one lone pair

(2) ; one lone pair

(3) ; This donation helps the sp²- hybridized nitrogen atoms to donate lone pair of its own with greater effort.

(4)  Much more donation here than the previous case

We know that, 

(i) Donation capacity of nitrogen increases if electron donating group(s) (+I or +R-effect containing) is/are attached with it.

(ii) More nitrogen atoms (or donor atoms) means more basicity (in general).

(iii) sp²-hybridized nitrogen is less capable of donating lone pair than sp³-hybridized.

From the second point, we know that 4>3 and both will be greater than 1 and 2. Out of 1 and 2, 1 has an sp³ hybridized N which is bonded to electron releasing methyl group and therefore is more basic than 2.

Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of electrolyte.

Given,

Specific Conductance, \((\kappa)=1.01 \times 10^{-2} {ohm}^{-1} {~cm}^{-1}\)

Molarity, \((M)=0.1 M\)

We know the relation,

\(\Lambda_{m}=\frac{\kappa \times 1000}{M}\)

\(=\frac{1.01 \times 10^{-2} \times 1000}{0.1}\)

\(=1.01 \times 10^{-2} \mathrm{{ohm}^{-1} {~cm}^{2} {~mol}^{-1}}\)

Positive deviation from Raoult's law: When the partial vapour pressure of each component ( A and B) consequently the total vapour pressure is greater than the pressure expected on the basis of Raoult's law then the deviation is termed as positive deviation.

Cause of positive deviation: This type of deviation is observed by the solution in which the forces of attraction between A-A molecules and between B-B molecules is greater than the forces of attraction between A-B molecules.

\(\gamma_{A}-B<\gamma_{A}-A\) or \(\quad \gamma_{B}-B\)

Negative deviation from Raoult's law: When the partial vapour pressure of each component of the solution is less than the vapour pressure expected on the basis of Raoult's law then the deviation is called as negative deviation.

Causes of negative deviation: This type of deviation is shown by solutions in which the forces of attraction between \(\mathrm{A}-\mathrm{A}\) and \(\mathrm{B}-\mathrm{B}\) molecules is less than the forces of attraction between \(\mathrm{A}\) and B molecules.

\(\gamma A-B>\gamma A-A\) Or \(\gamma B-B\)

Tishchenko reaction is a disproportionation reaction that allows the preparation of esters from two equivalents of an aldehyde.

Hence the correct option is (C).

The reaction: \(n-\mathrm{BuBr}+\mathrm{KCN} \stackrel{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}-\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} n-\mathrm{BuCN}\) is an example of Nucleophile substitution..

This is an example of bimolecular nucleophilic substitution reaction \(\mathrm{SN}^{2}\). Cyanide ion acts as a nucleophile and bromide ion acts as a leaving group. It is one step reaction and the mechanism is as shown above.

When Benzoyl chloride reacts with hydrogen in presence of palladium catalyst and barium sulphate, benzaldehyde is formed. This is known as Rosenmund reaction.

It is a reduction reaction where hydrogen is added to benzoyl chloride in presence of a catalyst. The catalyst is poisoned with sulfur or quinoline to prevent over reduction of the compound to alcohol.

The alcohol formed then may react with the benzoyl chloride to form esters, thus the presence of catalyst poison is very important.

The reaction can be given as follows:

\(\mathrm{A}: \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}, \quad \mathrm{~B}: \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\)

\(\begin{aligned}&\mathrm{n}_{\mathrm{A}}=\frac{10}{188}=0.053, \quad \mathrm{n}_{\mathrm{B}}=\frac{80}{202}=0.396 \\&\mathrm{x}_{\mathrm{A}}=0.118, \quad \mathrm{x}_{\mathrm{B}}=0.882\end{aligned}\)

\(\mathrm{P}_{\mathrm{A}}=\mathrm{x}_{\mathrm{A}} \mathrm{P}_{\mathrm{A}}^{0}=183 \times 0.118=21.594 \mathrm{mmHg}\)

\(\mathrm{P}_{\mathrm{B}}=\mathrm{x}_{\mathrm{B}} \mathrm{P}_{\mathrm{B}}^{0}=0.882 \times 127=112.014 \mathrm{~mm} \mathrm{Hg}\)

Total pressure of solution \(=21.594+112.014=133.608 \mathrm{~mm} \mathrm{Hg}\) \(\mathrm{x}_{\mathrm{A}}^{\prime}=\) Mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}\) in vapour \(=\frac{21.594}{133.608}=0.161\)

\(\mathrm{x}_{\mathrm{B}}^{\prime}=\) Mole fraction of \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2}\) in vapour \(=1-0.161=0.839\)

The rate of a chemical reaction depends on the nature of reactants because the number of bonds broken in the reactant molecules and the number of bonds formed in product molecules changes.

In other words, the ease with which bonds are formed and the ease with which bonds are broken determines the rate of the chemical reaction.

Lipidsare not enzymes. Lipids are group of macromolecules (large molecules) found in the body, consisting wide group of substances like fat and oils, certain vitamins (A,D,E,K), etc.

Despite the fact that lipids are not enzymes, certain enzymes require lipids as coenzymes -sort of vital accessories- to carry out their biological function, the absence of which makes the enzyme inactive. Vitamin K dependent carboxylase is example of an enzyme that requires a lipid coenzyme (reduced Vitamin K).

\(50 \%\) reaction completes in 2 hours. That is, half life of reaction is 2 hours

\(t _{\left(\frac{1}{2}\right)}=2\) hours. Also given that, \(t _{\left(\frac{3}{4}\right)}=4\) hours

We can observe that,

\(t _{\left(\frac{3}{4}\right)}=2 \times t _{\left(\frac{1}{2}\right)}\)

Thus, relation holds true only in first order