Chemistry Test - 24

Dextrosedoes not represent a disaccharide.

Disaccharides are the smallest and commonest oligosaccharides found in nature. They are formed by the condensation of two monosaccharide molecules. The most important disaccharides are sucrose, lactose, and maltose.

A deuterium nucleus consists of one proton and one neutron.

Benzene is a polymer of Ethyne.

\(\mathrm{Be}^{2+}\) ion has small size and high charge. Hence, it has high polarizing power and can attract several water molecules. Thus it has highest hydration energy among the given ions.

The expression ofMolality of concentration of a solution is independent of temperature.

We know that:

Molality \(=\frac{\text {Number of moles of solute }}{\text {Mass of solvent in kilograms }}\)

Since mass is unaffected by the change in temperature, therefore, molality is independent of temperature.

Butylated hydroxy toluene (BHT) and Butylatedhydroxy anisole (BHA) are antioxidantsin food.

Atmospheric pollution is generally studied as tropospheric and stratospheric pollution. The presence of ozone in the stratosphere prevents about 99.5 per cent of the sun’s harmful ultraviolet (UV) radiations from reaching the earth’s surface and thereby protecting humans and other animals from its effect.

Thymine is a pyrimidine nucleobase that is uracil in which the hydrogen at position 5 is replaced by a methyl group. It has a role as a human metabolite, an Escherichia coli metabolite and a mouse metabolite. It is a pyrimidine nucleobase and a pyrimidone.

Thymine is a natural product found in Synechocystis, Tectitethya, and other organisms with data available. 

Thymine is most often represented as a ring-like structure called a pyrimidine. Each respective base is in the shape of a ring-like form and there are two types. A purine is a base that contains a double ring form.

Given reaction is:

\(2 \mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{HCl}+\mathrm{O}_2\)

Also given,

Bond energies for:

\(C l-C l=242.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \)

\(\mathrm{H}-\mathrm{Cl}=431.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \)

\( O-H=464 \mathrm{~kJ} \mathrm{~mol}^{-1} \)

\( O=O=442 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

We know that,

From Hess' Law,

\(\Delta H=\text { B.E. of } \)

\((2 \times C l-C l)+(2 \times 2 \times O-H)-(4 \times H-C l)+(O=O) \)

\(=2 \times 242.8+4 \times 464-4 \times 431.8-442 \)

\( =172.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

First, we have to count the number of electrons in all the given species.

In \((He)\), we know that the number of electrons in helium is 2 but the Bohr'stheory is applicable only for one-electron species. That's why Bohr's theory isnot applicable for helium \((He)\) atoms.

In \(\left(H e^{2+}\right)\), we know that the number of the electron in helium is 2 but \(+2\)indicates that loss of two electrons to become helium ion \(\left(H e^{2+}\right)\), Thusthere is no election and we know that Bohr's theory is applicable one-electronspecies. That's why Bohr's theory is not applicable for helium ion \(\left(H e^{2+}\right)\).

In \(\left(L i^{2+}\right)\), we know that the number of the electron in lithium is 3 but \(+2\)indicates that loss of two electrons to become helium ion \(\left(L i^{2+}\right)\), Thusthere is one election left and we know that Bohr's theory is applicable oneelectron species.

That's why Bohr's theory is applicable for lithium-ion\(\left(L i^{2+}\right) .\)