Chemistry Test - 6

The mass number of atoms is the total number of protons + the total number of neutrons.

So mass number \(=3+4=7\)

Now valency of atom \(=8\) - mass number of that atom.

Valency of atom \(=8-7=1\)

So, the compound needs \(1\) electron to achieve stability.

Also, As the compound has \(3\) protons so its electronic configuration will be \(1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{1}\)

So it achieves a stable electronic configuration the molecule will lose \(1\) electron. Its valency is \(1\).

Moles of \({N}_{2} {O}_{4}=0.1\)

Moles of \({NO}_{2}({~g})=\) ?

\({N}_{2} {O}_{4}{(~g)} \rightleftarrows 2 {NO}_{2}({~g})\)

\({k}_{{p}}=0.14\)

\({N}_{2} {O}_{4} \rightleftarrows 2 {NO}_{2}\)

Let \(\alpha\) be degree of dissociation

At \({t}=0, \quad 0.1\)

\(0.1(1-\alpha)\)

At \({t}={t}_{{eq}} \quad 0.1 \times 2 \alpha\)

Total number of moles of gas \(=0.1(1-\alpha)+0.1 \times 2 \alpha=0.1(1+\alpha)\)

Pressure in the vessel is directly proportional to the number of Moles of gas:

\({PV}+{nRT}\)

At \(0.1\) mole , pressure was 1 atom

At \(0.1(1+\alpha)\) mole , pressure will become \(1+\alpha\)

\({P}\) artial pressure of \({N}_{2} {O}_{4}=\frac{0.1(1-\alpha)}{0.1(1+\alpha)} \times 1+\alpha\)

P artial pressure of \({NO}_{2}=\frac{0.1 \times 2 \alpha}{0.1(1+\alpha)} \times 1+\alpha\)

\({K}_{{p}}=\frac{{pNO}_{2}}{{pN}_{2} {O}_{4}}\)

\({n}_{\mathrm{NO}_{2}}=0.1 \times 2 \alpha\)

\(=0.1 \times 0.37 \quad(\alpha=\) very small \(=0.19)\)

\(=0.037\) mole

We know that,

Equivalent weight of hydrogen \(=1\)

Equivalent weight of magnesium \(=12\)

Substitute values in the following equation

\(\frac{\text { Mass of } \mathrm{H}_{2} \text { deposited }}{\text { Mass of } \mathrm{Mg} \text { deposited }}=\frac{\text { Eq.Wt. of } \mathrm{H}_{2}}{\text { Eq.Wt. of } \mathrm{Mg}}\)

\(=\frac{1}{12}\)

The Friedel Crafts acylation reagent is normally composed of an acyl halide mixed with a Lewis acid catalyst such as aluminium chloride. This produces an acylium cation R − C ≡ O⁺. Such electrophiles are not exceptionally reactive.

If Aniline and AlCl₃ along with Acid Chloride are taken for Friedel Crafts Acylation, Aniline reacts with electrophile AlCl₃ to give a coordination complex, which precipitates out and the reaction does not proceed. This complex is electron-withdrawing and it deactivates the ring for acylation reactions.

Order of reactivity of different alcohols towards esterification is \(1^{\circ}>2^{\circ}>3^{\circ}\).

Thus, as the steric hinderance (or bulkiness) increases from primary to secondary to tertiary alcohol, the order of esterification decreases.

\(\mathrm{A}+\mathrm{T} / \mathrm{C}+\mathrm{G}\) is not constant, is inconsistent regarding double helical structure of DNA.  Each molecule of DNA is a double helix formed from two complementary strands of nucleotides held together by hydrogen bonds between \(\mathbf{G}-\mathrm{C}\) and \(A-T\) base pairs. Duplication of the genetic information occurs by the use of one DNA strand as a template for formation of a complementary strand.The density of DNA decreases on heating as hydrogen bonds breakdown. The amount of adenine is always equal to that of thymine, and the amount of guanine is always equal to that of cytosine i.e \(A=T\) and \(G=C\). The base ratio \(\mathrm{A}+\mathrm{T} / \mathrm{C}+\mathrm{G}\) may vary form species to species, but is constant for a given species.

Graphite is one of the allotropes of carbon. Unlike diamond, it is an electrical conductor and a good lubricant. Graphite is a good conductor of electricity due to the presence of free valence electrons.

This property of graphite is due to the carbon atoms arranged in different layers and each atom is covalently bonded on three of its neighbouring atoms in the same layer. The fourth valence electrons of each atom are present between different layers and are free to move about. These free electrons in graphite make it a good conductor of electricity.

In presence of sunlight, the chlorine molecule undergoes homolytic fission to form chlorine free radical. It reacts with \(\left({CH}_{3}\right)_{3} {CH}\) to form \(\left({CH}_{3}\right)_{3} {C}\). free radical. Thus the reaction proceeds by free radical mechanism.

Thus the carbocation is not involved in the mechanism for the reaction \(\left({CH}_{3}\right)_{3} {CH}+\) \({Cl}_{2}+{hv} \rightarrow\)

The normal elements were located in Groups I to VII, transition elements were present in Group VIII. Groups I to VII have been classified to two sub-groups, whereas three elements were meant to be Group VIII.

Periods from 4th to 7th were divided into first series and second series. Elements showing similar properties were located in the same group. Example: Lithium, rubidium, potassium were present in the first group.

The reason why groups in the Mendeleev periodic table were parted into two subgroups having elements possessing the same physical property and chemical properties and they are given the labels as A and B. -The element’s properties of subgroups look alike to each other more markedly than the properties of those elements of the two-sub group groups.