Chemistry Test-26

The partial pressure of toluene in the vapour phase is:

\(\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{T}}^0 \mathrm{X}_{\mathrm{T}}=37\) torr \(\times 0.5=18.5\) torr

Here, \(\mathrm{P}_{\mathrm{T}}^0\) is the vapour pressure of pure toluene and \(\mathrm{X}_{\mathrm{T}}\) is the mole fraction of toluene in the liquid phase.

The partial pressure of benzene in the vapour phase is:

\(\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^0 \mathrm{X}_{\mathrm{B}}=119\) torr \(\times 0.5=59.5\) torr

Here, \(\mathrm{P}_{\mathrm{B}}^0\) is the vapour pressure of pure benzene and \(\mathrm{X}_{\mathrm{B}}\) is the mole fraction of benzene in the liquid phase.

The total pressure is \(\mathrm{P}_{\mathrm{T}}+\mathrm{P}_{\mathrm{B}}=18.5\) torr \(+59.5\) torr \(=78\) torr

The mole fraction of toluene in the vapour phase is \(\mathrm{Y}_{\mathrm{T}}\)

It is given by the expression \(\mathrm{Y}_{\mathrm{T}}=\frac{\mathrm{P}_{\mathrm{T}}}{\text { Total pressure }}=\frac{18.5}{78}=0.237\)

So, the mole fraction of toluene in the vapour phase is 0.237.

When an electric current is passed through the copper sulfate solution, free copper ions get drawn to the electrode connected to the negative terminal of the battery and get deposited on it.

The postulates of Dalton’s atomic theory:

• The matter is composed of exceedingly small particles called atoms.
• An atom is the smallest unit of an element that can participate in a chemical change.
• An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of that element.
• Atoms of one element differ in properties from atoms of all other elements.
• Atoms are neither created nor destroyed during a chemical change.

Dalton used three laws of chemical reactions as a basis for his theory:

• The Law of Conservation of Mass.
• The Law of Definite Proportions.
• The Law of Multiple Proportions.

The oxidation number of barium in barium peroxide and barium oxide remains the same, which is \(+2\). The oxidation number of oxygen changes. In barium peroxide, it is \(-1\), whereas in barium oxide, it is \(-2\) and in oxygen, it is 0.

For balancing any chemical reaction, the elements that are present in reactant should also be present in the product. The correct initial reaction of the above reaction without coefficients is:

\(\mathrm{Na}_{2} \mathrm{CO}_{3}+x \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

In the above reaction, the total number of atoms in \(\mathrm{Na}\) is \(2\) in the reactant side, while the number of atoms is \(1\) in the product side. So, for balancing \(\mathrm{Na}\) element, we should add \(2\) coefficients in product side \(\mathrm{Na}\) element. Similarly, we should add \(2\) coefficients in the \(\mathrm{HCl}\) compound also.

Now, we compare the number of atoms per element, this process starts with every single element. After balancing, the corrected reaction would be:

\(\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Now, the above reaction is accurately balanced. So, the coefficient of \(\mathrm{HCl}\), instead of \(x\) is \(2\).

Chalk dipped in ink is an example of adsorption and absorption simultaneously. The ink particles get absorbed in the chalk while the solution's water i.e, the water present in ink gets absorbed. The molecules undergoing absorption are taken up by the volume and in adsorption, the molecules are taken up by the surface. So, adsorption is a surface phenomenon and absorption is a bulk phenomenon. If we have a vessel in which we have a solid bulky substance that is dissolved in a liquid that has solid particles dissolved in it, we know after some time these solid particles tend to accumulate on the surface of the bulky substance. Now, if we say this accumulation of the bulky substance is because of the unbalanced or residual forces present on the solid body itself and this phenomenon is exactly what we term as adsorption. The only thing we make sure is that these particles do not penetrate the bulk itself.

The disappearance of one substance into another so that the absorbed losses its identifying characteristics while the absorbing substance retains most of its original physical aspects used in refining to selectively remove specific components from process streams.

Depending on the forces that are responsible for binding the molecules the adsorption is mainly divided into two types:

Sorption is a process in which both adsorption (It is a process of accumulation of a substance at the surface of the solid or liquid) and absorption (Accumulation of the substance within the bulk of a solid or liquid) take place.

Hybridization = a rounded atoms to central atom + \(\frac{1}2\) (valence e- of central atom -bonded e-)

So, hybridization of \({XeF_6}=6+\frac{1}2(8-6)= 7\)

i.e. \(7=\) (bonded electron)+ (Lone pair)

So, hybridization \(=7=\operatorname{sp^3} {d^3} \)

Its hybridization is \(sp^3d^3\) however its geometry is capped or distorted octahedral(square bipyramidal) one and not pentagonal bipyramidal. The fluorine atoms occupy the vertices of the octahedron and lone pairs keep on moving in space to minimize repulsion, and hence distorting the octahedral geometry.

Defect of Mass during fission of 1 uranium atom \((\Delta M)=U+n-(M o+X e+ 2n)\)

Application:

We have,

\(n=1.0087\)

\(\mathrm{U}=235.0439\)

Mo \(=97.9054\)

\(X e=135.9170\)

Therefore,

\(\Delta \mathrm{M}=235.0439+1.0087-(97.9054+135.9170+2(1.0087))=0.2128 \mathrm{amu}\)

Now, Mass in \(\mathrm{kg}=0.2128 \times 1.67 \times 10^{-27} \mathrm{~kg}\)

Energy \(=\Delta \mathrm{M} \times \mathrm{C}^{2}\) (Here, \(\mathrm{C}\) is speed of light)

Thus,

Energy \(=0.2128 \times 1.67 \times 10^{-27} \mathrm{~kg} \times\left(3 \times 10^{8}\right)=3.198 \times 10^{-11} \mathrm{~J}\)

Number of atoms in \(1 \mathrm{gm}\) is Avagadro Number =

\(\frac{6.02 \times 10^{23}}{235.0439}=0.0256 \times 10^{23}\)

Thus, Energy Relased by \(1 \mathrm{gm}=3.198 \times 10-11 \times 0.0256 \times 1023\) \(=8.1 \times 10^{10} \mathrm{~J}\)

\({R}-{C} \equiv\) Alkyl Cyanide

\(\therefore\) In alkyl cyanide, alkyl group is attached to C of CN.