Applying a coating of zinc methods is suitable for preventing an iron frying pan from rusting.

• Applying paint and grease are not suitable solutions to prevent an iron frying pan from rusting as the paint can melt by flame and can be destroyed in repeated use.
• Applying a coating of zinc is a better option as it forms a coating of corrosion-resistant zinc which prevents corrosive substances from reaching the more delicate part of the metal.

The basic unit of nucleic acid is nucleotide. Nucleotides are organic molecules that serve as the monomers, or subunits, of nucleic acids like DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). The building blocks of nucleic acids, nucleotides are composed of a nitrogenous base, a five-carbon sugar (ribose or deoxyribose), and at least one phosphate group.

Given:

Weight of solute $(w)=1g$

Weight of solvent $(W)=75g$

Boiling point of solution $={100.114}^{\circ }C$

Boiling point of solvent $={100}^{\circ }C$

Therefore, 

$\mathrm{△}T=100.114-100={0.114}^{\circ }C$

Molecular weight of solute $(m)=60.1$

Boiling point elevation constant $(K)=?$

We know that:

$m=\frac{1000\times K\times w}{\mathrm{\Delta }T\times W}$

or, $K=\frac{m\times \mathrm{\Delta }T\times W}{1000\times w}$

$=\frac{60.1\times 0.114\times 75}{1000\times 1}$

$=\frac{513.8}{1000}$

$K=0.513$

${\mathrm{Cu}}^{2+}+{\mathrm{NH}}_3\stackrel{{\mathrm{K}}_1}{\rightleftharpoons }{\left[\mathrm{Cu}\left({\mathrm{NH}}_3\right)\right]}^{2+}$

${\left[\mathrm{Cu}\left({\mathrm{NH}}_3\right)\right]}^{2+}+{\mathrm{NH}}_3\stackrel{{\mathrm{K}}_2}{\rightleftharpoons }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_2\right]}^{2+}$

${\mathrm{K}}_3$

${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_2\right]}^{2+}+{\mathrm{NH}}_3\stackrel{{\mathrm{K}}_3}{\rightleftharpoons }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_3\right]}^{2+}$

${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_3\right]}^{2+}+{\mathrm{NH}}_3\stackrel{{\mathrm{K}}_4}{\rightleftharpoons }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$

${\mathrm{Cu}}^{2+}+4{\mathrm{NH}}_3\stackrel{\mathrm{K}}{\rightleftharpoons }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}v$

$\text{ So, }\mathrm{K}={\mathrm{K}}_1\times {\mathrm{K}}_2\times {\mathrm{K}}_3\times {\mathrm{K}}_4$

$={10}^4\times 1.58\times {10}^3\times 5\times {10}^2\times {10}^2$

$\therefore \mathrm{K}=7.9\times {10}^{11}$

Where $\mathrm{K}=$ Equilibrium constant for formation of ${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$ So, equilibrium constant ( ${\mathrm{K}}^{\prime }$ ) for dissociation

$\text{ of }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}\text{ is }\frac{1}{\mathrm{K}}$

${\mathrm{K}}^{\prime }=\frac{1}{\mathrm{K}}$

${\mathrm{K}}^{\prime }=\frac{1}{7.9\times {10}^{11}}=1.26\times {10}^{-12}=(\mathrm{x}\times {10}^{-12})$

So, $x=1.26\approx 1.0$.

Oxidation is the gain of oxygen and reduction is the loss of oxygen.

Lead is getting reduced due to gain of electrons and decrease in oxidation no.

Similarly, $C$ is accepting $O$, therefore, $C$ is oxidising to $C{O}_2$.

Promethium ($Pm$) is the element which prepared only by synthetic methods. It is not present in nature. It is the only synthetic radioactive lanthanoid.

Most promethium is used only in research. A little promethium is used in specialised atomic batteries. These are roughly the size of a drawing pin and are used for pacemakers, guided missiles and radios. The radioactive decay of promethium is used to make a phosphor give off light and this light is converted into electricity by a solar cell. Promethium can also be used as a source of $x$-rays and radioactivity in measuring instruments.

Given:

Number of moles $=0.25$ moles

Mass of solvent $=500\mathrm{g}$

Depression in freezing temperature, $\mathrm{\Delta }{T}_f={2.79}^{\circ }C$

Freezing point depression constant, $K_f=1.86$

Van't hoff factor $=i$

We know that,

$\mathrm{\Delta }{T}_f=i\times K_t\times m$

So,

Molality, $m=\frac{\text{ Number of moles }}{\text{ Solvent in kg}}=\frac{0.25\times 1000}{500}=0.5\mathrm{m}$

$2.79=i\times 1.86\times 0.5$

$i=3$

Therefore, the number of ions furnished $=3$

The increasing order of the first ionisation enthalpies of the elements B, P, S and $\mathrm{F}$ (lowest first) is\(P

In general as we move from left to right in a period, the ionisation enthalpy increases with increasing atomic number. The ionisation enthalpy decreases as we move down a group. $P\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^3\right)$ has a stable half filled electronic configuration than $\mathrm{S}\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^4\right)$. For this reason, ionisation enthalpy of $P$ is higher than S.

The complex ${\left[Co{\left(N{H}_3\right)}_5\left(N{O}_2\right)\right]}^{2+}$ and ${[Co{\left(N{H}_3\right)}_5(ONO)]}^{2+}$ are called linkage isomers due to presence of $N{O}_2$ ligand. It is an ambidentate ligands are capable of coordinating in more than one way i.e., $N{O}_2^{-}/ON{O}^{-}$.

Ionic species are stabilized by the dispersal of charge.

• The carboxylate ion $\left({\mathrm{F}}_2{\mathrm{CHCOO}}^{-}\right)$is the most stable.
• The negative charge on $\mathrm{O}$ is dispersed through resonance in the carboxylate group.
• The negative charge is further dispersed due to negative inductive effect (-I effect) of two $\mathrm{F}$ atoms present on alpha carbon atoms.