Chemistry Test-9

According to the thermodynamics first law, the energy can neither be created nor destroyed but can be transformed from one form to the other form, change in internal energy of a system is related to its heat, and work is done as follows:

ΔE = Q + W

Where,

ΔE is a change in the internal energy of the system.

Q is a heat absorbed by the system = + 300 cal

W is the work done by the system on the surroundings = −200 cal

Calculating the change in internal energy of the system from done using the equation,

ΔE = Q + W

⇒ ΔE = 300 cal + (−200 cal)

⇒ ΔE = 100 cal

Therefore, the change in internal energy of the system is 100 cal.

Lecithin and cephalin both are phospholipids. Lecithin is probably the most common phospholipid which contains the amino alcohol, choline. It is found in egg yolks, wheat germ, and soybeans. Lecithin is extracted from soybeans for use as an emulsifying agent in foods. Lecithin is an emulsifier because it has both polar and non-polar properties, which enable it to cause the mixing of other fats and oils with water components. Cephalins are phosphoglycerides that contain ehtanolamine or the amino acid serine attached to the phosphate group through phosphate ester bonds. These are found in most cell membranes, particularly in brain tissues. They also important in the blood clotting process as they are found in blood platelets.

\(\mathrm{XeF}_{6}\) dissolves in anhydrous HF to give a good conducting yellow solution which has conductivity due to formation of \(\mathrm{HF}_{2}^{-}\)and \(\mathrm{XeF}_{5}^{+}\)ions. Conducting solution is that solution which allows the electricity to pass through it. It is shown by the below given reaction,

\(\mathrm{HF}+\mathrm{XeF}_{6} \rightarrow \mathrm{XeF}_{5}^{+}+\mathrm{HF}_{2}^{-}\)

Option (C): In Lanthanide series ionic radii of trivalent ions decreases as atomic number increases. This is known as lanthanide contraction. But the decrease is too small in comparison with atomic number. The decrease in size can be attributed to poor shielding of electrons present in \(4f\) orbital.

Thus, \(\mathrm{La}^{+3}>\mathrm{Ce}^{+3}>\mathrm{Lu}^{+3}\)

So, option (C) is in incorrect order.

Option (A): Among actinides also there is a decrease in ionic radii due to poor shielding effect of electrons present in \((\mathrm{n}-2) \mathrm{f}\) orbital. This is known as actinoid contraction.

Thus, \(\mathrm{Ac}^{3+}>\mathrm{Th}^{3+}>\mathrm{Lr}^{3+}\)

Option (B): More the oxidation number, more quickly it will get reduced, which means it is a strong oxidizing agent. The order of oxidation number in the given ions is:

\(\mathrm{V}^{+5}<\mathrm{Cr}^{+6}<\mathrm{Mn}^{+7}\)

Option (D): In solution, hydrates are formed and the stability of hydrates directly depends on oxidation number. More the oxidation number more is stability.

\(\mathrm{V}^{+3}>\mathrm{V}^{+2}\)

The \(\mathrm{H}^{+}\)concentration at which the colloidal particles have no charge is known as the isoelectric point. At this point, the stability of colloidal particles becomes very less, and they do not move under the influence of the electric field. The sol particles at the isoelectric point do not show the electrophoresis process. Isoelectric point (pl or IEP) is the \(\mathrm{pH}\) of a solution or dispersion at which the net charge on the molecules or colloidal particles is zero. In electrophoresis, there is no motion of the particles in an electric field at the isoelectric point. The net charge (the algebraic sum of all the charged groups present) of any amino acid, peptide or protein, will depend upon the \(\mathrm{pH}\) of the surrounding aqueous environment. For example, alanine can have a charge of \(+1,0\), or \(-1\), depending on the \(\mathrm{pH}\) of the solution in which it is dissolved.

The van der Waals constant b is four times the molecular volume of 1 mole of a gas.

Volume of one molecule of a gas of radius \(r=\frac{4}{3} \pi r^{3}\)

Co-volume \(=\frac{4}{3} \pi(2 \mathrm{r})^{3}\)

\(=8 \times \frac{4}{3} \pi \mathrm{r}^{3}\)

Above co-volume is occupied by two molecules.

So, co-volume per molecule is \(=4 \times \frac{4}{3} \pi r^{3}\)

Since co-volume of 1 mole of gas = Number of molecule in 1 mole gas \(\times\) co-volume of 1 molecule

\(=\mathrm{N}_{\mathrm{A}} \times \frac{4}{3} \pi \mathrm{r}^{3}\)

Therefore,

\(b=4 \times\) co-volume of 1 mole of gas

\(\Rightarrow \mathrm{b}=4 \times \mathrm{N}_{\mathrm{A}} \times \frac{4}{3} \pi \mathrm{r}^{3}\)

\(\Rightarrow \mathrm{b}=4 \times\left(\frac{4}{3} \pi \mathrm{r}^{3}\right) \mathrm{N}_{\mathrm{A}}\)

The \(-\mathrm{NO}_{2}\) group is electron withdrawing group, so it deactivates the benzene ring largely, \(-\mathrm{Cl}\) atom is also deactivate the benzene ring but this deactivation is lower than \(-\mathrm{NO}_{2}\) group; while \(-\mathrm{CH}_{3}\) group activates the benzene ring. Hence, on the basis of activation and deactivation they have the following order of electrophilic attraction (because more activated group attracts electrophile easily).

The balanced chemical equation for the reaction of zinc with sulfuric acid is as shown below:

\(\mathrm{Zn}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{H}_{2}(\mathrm{~g}) \uparrow\)

1 mole of zinc reacts with 1 mole of sulfuric acid to produce 1 mole of hydrogen. The balanced chemical equation for the reaction of zinc with sodium hydroxide is as shown below:

\(\mathrm{Zn}+2 \mathrm{NaOH}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]+\mathrm{H}_{2}(\mathrm{~g}) \uparrow\)

1 mole of zinc reacts with 2 moles of sodium hydroxide to produce 1 mole of hydrogen. Thus, the ratio of volumes of hydrogen evolved is \(1:1\).

The coordination number of a central metal atom in the complex is determined bythe number of ligands around the metal ion bonded by sigma bonds.It is also defined as the number of atoms, ions, or molecules that a central atom or ion holds as its nearest neighbours in a complex.

For example, in the complex ions \(\left[\mathrm{PtCl}_{6}\right]^{2-}\) and \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\), the coordination number of \(\mathrm{Pt}\) and \(\mathrm{Ni}\) are 6 and 4 respectively.

Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion, incoordination or paralysis or some other disturbances of nervous system. These are classified as follows:

(i) Non-narcotic (non-addictive) analgesics

(ii) Narcotic drugs

Aspirin,Ibuprofen and paracetamol belong to the class of non-narcotic analgesics. Morphine,Heroin and codeine belong to the class of narcotic analgesics.

(I) As temperature rises \(\mathrm{K}_{\mathrm{a}}\) will increase \(\mathrm{K}_{\mathrm{b}}\) will also increase since \(E_{\mathrm{a}}<\mathrm{E}_{\mathrm{b}}\) and \(\mathrm{K}_{\mathrm{a}}>\mathrm{K}_{\mathrm{b}}\).

(II) At lower temperature \(\mathrm{K}_{\mathrm{a}}\) will fall and \(\mathrm{K}_{\mathrm{b}}\) will fall but as \(\mathrm{E}_{\mathrm{a}}<\mathrm{E}_{\mathrm{b}}\) and \(\mathrm{K}_{\mathrm{a}}^{\prime}>\mathrm{K}_{\mathrm{b}}^{\prime}\).

(II) As temperature rises very high \(\mathrm{K}_{\mathrm{a}}\) and \(\mathrm{K}_{\mathrm{b}}\) will try to close each other as they are exponential functions of temperature.

Leaching of \(A g_{2} S\) is carried out by heating it with a dilute solution of \(\mathrm{NaCN}\) in presence of \(\mathrm{O}_{2}\). This forms a soluble complex \(\left[A g(C N)_{2}\right]^{-}\).

\(\mathrm{Ag}_{2} \mathrm{~S}+4 \mathrm{NaCN} \rightleftharpoons 4 \mathrm{Na}^{+}+2\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{2-}+S^{2}\)

Here sulphide ion is oxidised to sulphate ion by oxygen. Thus, soluble material can be separated from insoluble material. Treatment of the cynaide compound with zinc gives displacement reaction in which silver is obtained.

\(2 N a\left[A g(C N)_{2}\right]+Z n \rightarrow\left[Z n(C N)_{4}\right]^{2-}+2 A g \downarrow\)

Given,

\(M_{1}=0.2 \mathrm{M}\)

\(M_{2}=0.1 \mathrm{M}\)

\(V_{1}=20 \mathrm{~mL}\)

\(V_{2}=35 \mathrm{~mL}\)

\(V_{f}=100 \mathrm{~mL}\)

\(\mathrm{M}_{\mathrm{NaOH}}(\) resultant \()=\frac{M_{1} \times V_{1}+M_{2} \times V_{2}}{V_{f}}\)

\(=\frac{20 \times 0.2+35 \times 0.1}{100}\)

\(=0.075 \mathrm{M}\)

Milli-equivalent of \(\mathrm{NaOH}=\) Milli-equivalent of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

Let the weight of impure sample be \(x\).

\(\Rightarrow 40 \times 0.075=\frac{x \times 0.90}{90} \times 2 \times 1000\)

\(\Rightarrow x=0.15 \mathrm{~g}\)

The electronegativity difference between \(\mathrm{M}_{1}\) and \(O\) is \(0.1\), which indicates \(M_{1}-O\) bond will be covalent, since \(O-\mathrm{H}\) bond having more ionic character thus bond will break and \(\mathrm{H}^{+}\)ions will release and acidic solution is formed. Whereas difference between electronegativity of \(\mathrm{M}_{2}-\mathrm{O}\) bond is \(2.3\), thus, \(\mathrm{M}_{2}-\mathrm{OH}\) bond will break. Thus, solution will be basic in nature.

Cyanide ion is a strong nucleophile. It attacks carbonyl carbon. Thus when 2,4-pentadione is treated with excess KCN and dil acetic acid, two molecules of HCN are added across two carbonyl groups. This is followed by the hydrolysis of cyanide group in presence of dil sulfuric acid to form 2,4-dihydroxy-2,4-dimethyl pentane dioic acid. Crystallization, followed by heating gives the final product lactone.