Chemistry Test - 2

Minerals associated with redox reactions are Fe, Cu.

• Fe is an important constituent of proteins like ferredoxin and cytochromes which are involved in the transfer of electrons. It is reversibly oxidized from ${\mathrm{Fe}}^{2+}$ to ${\mathrm{Fe}}^{3+}$ during electron transfer. It activates the catalyze enzyme and is essential for the formation of chlorophyll.
• Copper helps in the formation of starch. It is required for the overall metabolism in plants. It is associated with certain enzymes involved in redox reactions and is reversibly oxidized from $C{u}^{+}$ to $C{u}^{2+}$.

The acylation of benzene is called Friedel and craft reaction.

Acylation is the substitution of an acyl group into an organic compound. In case of benzene, the acyl group is substituted into the benzene ring. This reaction is also known as Friedel-Crafts acylation of Benzene.

${\mathrm{C}}_6{\mathrm{H}}_6+{\mathrm{CH}}_3\mathrm{COCl}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CO}{\mathrm{CH}}_3+\mathrm{HCl}$

Tritium is a beta-emitting radioactive isotope of hydrogen. Its nucleus consists of one proton and two neutrons, making it three times as heavy as a hydrogen nucleus (with its one proton) and one-and-a-half times as heavy as deuterium (which contains one proton and only one neutron).

For the reaction,

${\mathrm{m}}_1\mathrm{A}+{\mathrm{m}}_2\mathrm{B}\to {\mathrm{n}}_1\mathrm{C}+{\mathrm{n}}_2\mathrm{D}$

At equilibrium,

$-\frac{1}{{\mathrm{m}}_1}\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=+\frac{1}{{\mathrm{n}}_1}\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}$

$\Rightarrow -\frac{\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}}{\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}}=\frac{{\mathrm{m}}_1}{{\mathrm{n}}_1}$

Sucrose (cane sugar), lactose (milk sugar) and maltose are disaccharides in which two monosaccharides are hold together by a glycosidic linkage.

(A) In sucrose, C1 of α−D− glucose and C2 of β−D− fructose are linked together (ii).

(B) In lactose, C1 of β - D-galactose and C4 of β-D-glucose are linked together (i).

(C) In maltose, C1 of α− D-glucose is linked to C4 of another α−D− glucose unit (iii).

So, option (c) is the correct answer.

The product formed by the reaction of an aldehyde with a primary amine is Schiff base. It is a substituted imine.

${\mathrm{CH}}_3-\mathrm{C}(\mathrm{O})-{\mathrm{CH}}_3+{\mathrm{H}}_2\mathrm{N}-\mathrm{R}\stackrel{{\mathrm{H}}^{+}}{⟶}{\mathrm{CH}}_3-\mathrm{C}\left({\mathrm{CH}}_3\right)=\mathrm{N}-\mathrm{R}+{\mathrm{H}}_2\mathrm{O}$

$$Schiff's base

In the mixture of nitrite and nitrate ions, the nitrite ions can be removed by heating with urea and supluric acid.

The reaction takes place as follows:

$2{\mathrm{NaNO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{HNO}}_2$

$2{\mathrm{HNO}}_2+\mathrm{CO}{\left({\mathrm{NH}}_2\right)}_2⟶{\mathrm{CO}}_2+{\mathrm{N}}_2+3{\mathrm{H}}_2\mathrm{O}$

In the above reaction, we can see that the products, nitrogen and carbon dioxide are in gaseous form and thus, do not stay in the solution. This reaction is exothermic and thus, will occur spontaneously without the supply of any extra energy.

Inert gases condense into the solid state at very low temperature as there is strong attractive force between the atoms.

In solid state, Van der Waals attractive forces are predominant between the atoms. The attractive force increases with the size of the atom as a result of the increase in polarizability and the decrease in ionization potential.

As we know,

The angular momentum is given as, 

$\mathrm{L}=\mathrm{m}\mathrm{vr}$

The angular momentum is on integer multiple of $\frac{\mathrm{h}}{2\pi }$.

Then we can write,

$\mathrm{mvr}=\frac{\mathrm{nh}}{2\pi }$

For, $\mathrm{n}=1$

$\mathrm{mvr}=\frac{\mathrm{h}}{2\pi }$

From ideal gas equation we know,

$\mathrm{PV}=\mathrm{nRT}$

$\Rightarrow \mathrm{PV}=\frac{\mathrm{W}}{\mathrm{M}}\mathrm{RT}$

$\Rightarrow \mathrm{P}=\frac{\mathrm{W}}{\mathrm{V}}\cdot \frac{\mathrm{RT}}{\mathrm{M}}$

$\Rightarrow \mathrm{P}=\mathrm{d}\cdot \frac{\mathrm{RT}}{\mathrm{M}}[\because \mathrm{d}=\frac{\mathrm{W}}{\mathrm{V}}]$

For a fixed amount of gas at a fixed temperature, $\mathrm{M}$ and $\mathrm{T}$ is constant.

$\therefore \mathrm{P}\propto \mathrm{d}$

So graph between pressure (P) and density (d) is a straight line.

Also, $\frac{\mathrm{P}}{\mathrm{d}}=\frac{\mathrm{RT}}{\mathrm{M}}$

$\Rightarrow \frac{\mathrm{P}}{\mathrm{d}}\propto \mathrm{T}[\text{ as }\mathrm{R},\mathrm{M}=\text{ constant }]$

$\therefore$ When temperature $\mathrm{T}$ increases then slope of graph between $\mathrm{P}$ and $\mathrm{d}$ increases.

As ${\mathrm{T}}_3>{\mathrm{T}}_2>{\mathrm{T}}_1$ then slope of ${\mathrm{T}}_3$ will be highest then ${\mathrm{T}}_2$ and then ${\mathrm{T}}_1$.

So, graph (B) is the right graph.