Chemistry Test - 46

Given: Weight of Urea \(=12 {~g}\)

And, weight of sucrose \(=68.4 {~g}\)

We know that: Number of moles \(=\frac{\text{Weight}}{\text{Molar mass}}\)

Therefore, Moles of urea \(=\frac{12}{60}=0.2\)

(Molar mass of Urea is \(60 {gm}\). )

Moles of sucrose \(=\frac{68.4}{342}=0.2\)

(Molar mass of sucrose is \(342 {gm}\))

Since, there are equal moles of solute in equal volumes of water, the mole fraction is the same. As per Raoult's law, the relative lowering of pressure will also be the same.

C₂H₆ compound has 7 covalent bonds. The structure of the Ethane molecule is given below.

When plaster of Paris is heated above \(393 K\) its water of crystallization is lost and the leftover anhydrous calcium sulphate \(CaSO _{4}\) is called the dead burnt plaster.

Ethane is an example of a saturated carbon compound.Saturated hydrocarbons are carbon compound that contains only single bonds between carbon atoms.Ethane is a colourless, odourless gas.Ethane is isolated on an industrial scale from natural gas and as a petrochemical by-product of petroleum refining.The chemical formula of Ethane is \(\mathrm{C}_{2} \mathrm{H}_{6}\).Ethane can be found underground in liquid form in rock formations.

Zns has zinc blende type structure (i.e., ccp structure). The S^2- ions are present at the corners of the cube and at the centre of each face. Zinc ions occupy half of the tetrahedral sites. Each zinc ion is surrounded by four sulphide ions which are disposed towards the corner of the regular tetrahedron. Similarly, S^2- ion is surrounded by four Zn²⁺ ions.

Thus, the correct statement is coordination number (4 : 4); ccp; Zn²⁺ ion in the alternate tetrahedral voids.

We know that:

\(\mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O} ; \quad E^{\circ}=1.59 \mathrm{~V}\)

\(2 \mathrm{NO}+2 \mathrm{H}^{+} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} ; E^{\circ}=1.59\)

In basic medium,

\(2 \mathrm{NO}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{N}_{2} \mathrm{O}+2 \mathrm{OH}^{-}\)

Let \(E^{\circ}=x\)

Now, in alkaline medium,

\(\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}\)

\(\Rightarrow K=10^{-14}\)

or, \(2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{H}^{+}+2 \mathrm{OH}^{-} ; K=10^{-28}\)

We know that, 

\(E^{\circ}=\frac{0.059}{n} \log K\)

\(=\frac{0.059}{2} \log 10^{-28}\)

\(=0.826\)

For standard potential,

\(E_{\text {acidic }}^{\circ}+E_{\text {basic }}^{\circ}=E_{\text {cell }}^{\circ}\)

\(\Rightarrow 1.59+x=0.826\)

\(\Rightarrow x=0.764\)

Lyases are the'enzymes which cause cleavage, removal of groups without hydrolysis, addition of groups to double bonds or removal of a group producing double bond, e.g histidine decarboxylase break histidine to histamine and \(\mathrm{CO}_{2}\).