Chemistry Test - 8

Antibodies are immune system-related proteins called immunoglobulins. Each antibody consists of four polypeptides– two heavy chains and two light chains joined to form a "Y" shaped molecule. The amino acid sequence in the tips of the "Y" varies greatly among different antibodies.

As it is the heat required for a unit mass of the body so specific heat doesn’t depend on the mass of that body.Specific heat capacity of anybody/material is a property of that body/material. It is the same for each and every molecule of that body. So it doesn’t depend on the shape of the body. Thus it is independent of the mass and shape of the body.

Photochemical smog can be suppressed by radical traps. When the compounds are sprayed to the atmosphere, they generate free radicals which readily combine with free radical precursors of photochemical smog. Diethyl hydroxylamine has been found to possess smog inhibiting characteristics.

Equivalent conductance is defined as conducting power of all the ions produced by one gram equivalent of electrolyte.

The equivalent weight of:

\({BaCl}_{2}=\frac{\text { molecular weight }}{2}\)

\(\lambda_{{m}}^{\infty}\) for \({BaCl}_{2}=\lambda_{{m}\left({Ba}^{2-}\right)}^{\infty}+2 \lambda_{{m}\left({Cl}^{-}\right)}^{\infty}\)

\(\lambda_{{eq}}^{\infty}\) for \({BaCl}_{2}=\frac{1}{2} \lambda_{{m}\left({Ba}^{2-}\right)}^{\infty}+\lambda_{{m}\left({Cl}^{-}\right)}^{\infty}\)

\(=\frac{127}{2}+76\)

\(=139.5 {ohm}^{-1} {~cm}^{2} {eq}^{-1}\)

Given,

\([A]_{0}=2.0 m,[~A]=0.15 ~m, t=200 ~min\)

For first order reaction

\(A_{0}=\) Initial concentration

\(A=\) Final concentration

Rate constant, \(k=\frac{2.303}{t} \log \frac{[A]_{0}}{[~A]}\)

\(=\frac{2.303}{200} \log \frac{2.0}{0.15} \)

\(=\frac{2.303}{200}(\log 200-\log 15) \)

\(=\frac{2.303}{200} \times(2.3010-1.1761) \)

\(=\frac{2.303 \times 1.1249}{200} \)

\(=0.01295 \min ^{-1}\)

Now, half life, \(t_{\frac{1}{2}}=\frac{0.6932}{k}\)

\(=\frac{0.6932}{0.01295}\)

\(=53.52 ~min\)

In \(\mathrm{NH}_{3}\), charge from three bonds was moving towards nitrogen in the direction of lone pairs. In \(\mathrm{NF}_{3}\) and \(\mathrm{BF}_{3}\), due to structure, pyramidal and trigonal planar, \(\mathrm{NF}_{3}\) has more dipole moment. The resultant dipole for \(\mathrm{BF}_{3}\) is zero.

Let the volume of vessel \(={V} {cm}^{3}\)

\(\therefore\) Volume of air in the flask at \(27^{\circ} {C}={V} {cm}^{3}\)

According to Charle's law,

\(\frac{{V}_{1}}{{~T}_{1}}=\frac{{V}_{2}}{{~T}_{2}}\)

\({V}_{2}=\frac{{V}_{1} \times {T}_{2}}{{~T}_{1}}\)

Where,

\(V_{1}\) = Volume of the vessel

\(V_{2}\) = Volume expelled

\(T_{1}\) = False temperature

\(T_{2}\) = True temperature

Here, \({V}_{1}={V} {cm}^{3}\),

\({T}_{1}=273+27=300 {~K}\)

\({T}_{2}=273+477=750 {~K}\)

Substituting the values, we have

\(V_{2}=\frac{V c m^{3} \times 750 K}{300 K}=2.5 {Vcm}^{3}\)

\(\therefore\) Volume expelled \(=2.5 {~V} {~cm}^{3}-{V} {cm}^{3}\)

\(=1.5 {~V} {~cm}^{3}\)

\(\therefore\) Fraction of the air expelled

\(=\frac{1.5 {Vcm}^{3}}{2.5 {Vcm}^{3}}=\frac{3}{5}\)

The absolute zero pressure can be attained at a temperature of -273°C or 0 Kelvin. Absolute zero is the lowest limit of the thermodynamic temperature scale.

C₃H₉N can have 4 structural isomers. And its isomers are

Metre to centimetre conversion:

\(1 \mathrm{~m}=100 \mathrm{~cm}=10^{2} \mathrm{~cm}\)

We know that,

The radius of atomic nucleus (r):

\(r=10^{-15} \mathrm{~m}\)

\(r=10^{-15} \times 10^{2} \mathrm{~cm}\)

\(r=10^{(-15+2)} \mathrm{cm}\)

\(r=10^{-13} \mathrm{~cm}\)

The radius of atomic nucleus is of the order \(10^{-13} \mathrm{~cm}\).