Chemistry Test - 47

\(\mathrm{pH}\) of \(75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}\) will be equal to \(1\)

Total volume \(=75 \mathrm{~mL}+25 \mathrm{~mL}=100 \mathrm{~mL}\)

Number of \(\mathrm{mmol}\) of \(\mathrm{HCl}=75 \mathrm{~mL} \times \frac{\mathrm{M}}{5}=15 \mathrm{mmol}\)

Number of \(\mathrm{mmol}\) of \(\mathrm{NaOH}=25 \mathrm{~mL} \times \frac{\mathrm{M}}{5}=5 \mathrm{mmol}\)

\(5 \mathrm{mmol}\) of \(\mathrm{NaOH}\) will neutralise \(5 \mathrm{mmol}\) of \(\mathrm{HCl}\).

\(15-5=10 \mathrm{mmol}\) of \(\mathrm{HCl}\) will remain.

\(\left[\mathrm{H}^{+}\right]=[\mathrm{HCl}]=\frac{10 \mathrm{mmol}}{100 \mathrm{~mL}}=0.1 \mathrm{M}\)

\(\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]\)

\(\mathrm{pH}=-\log _{10} 0.1 \mathrm{M}=1\)

Since at \(n=1\) the population of electrons is maximum i.e., at the ground state. So, maximum excitation will take place from \(n=1\) to \(n=2\).

Thus, \(\mathrm{n}=2\) is the possible excited state.

Now, we have the formula for the energy of H-atom. \((\mathrm{E_n}) _{H}=-13.6 \frac{\mathrm{z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}\), where \(Z\) = atomic number

\(\mathrm{Z}\) for \(\mathrm{H}\) -atom \(=1\) \(\therefore\left(\mathrm{E}_{n}\right)_{H}=13.6 \times \frac{1}{2^{2}} \mathrm{eV}\) \(=\frac{136}{4} \mathrm{eV}=-3.4 \mathrm{eV}\)

Given, \(\left(t_{1 / 2}\right)_1=54 \mathrm{~min}\)

\(\underset{\text{A}}{\left(\mathrm{t}_{1 / 2}\right)_2}=\underset{\text{B}}{18 \mathrm{~min}}\)

\(t=0,{ }^{\prime} x^{\prime} \mathrm{M} \quad t=0,{ }^{\prime} x^{\prime} \mathrm{M}\)

To calculate, \(\left[\mathrm{A}_{\mathrm{t}}\right]=16 \times\left[\mathrm{B}_{\mathrm{t}}\right] \ldots\) (i) 

time \(=\) ?

For Ist order reaction, \(\left[A_t\right]=\frac{A_0}{(2)^n}\)

\(\mathrm{n}=\) number of half-lives

From Eq. (i), \(\left[\mathrm{A}_{\mathrm{t}}\right]=16 \times\left[\mathrm{B}_{\mathrm{t}}\right]\)

\(\frac{\mathrm{x}}{(2)^{\mathrm{n}_1}}=\frac{\mathrm{x}}{(2)^{\mathrm{n}_2}} \times 16\)

\((2)^{\mathrm{n}_2}=(2)^{\mathrm{n}_1} \times(2)^4 \)

\(\Rightarrow \mathrm{n}_2=\mathrm{n}_1+4 \)

\(\frac{\mathrm{t}}{\left(\mathrm{t}_{1 / 2}\right)_2}=\frac{\mathrm{t}}{\left(\mathrm{t}_{1 / 2}\right)_1}+4 \Rightarrow \mathrm{t}\left(\frac{1}{18}-\frac{1}{54}\right)=4 \)

\(\Rightarrow \mathrm{t}=\frac{4 \times 18 \times 54}{36}=108 \mathrm{~min}\)

\(CH_3 - \underset{OH}{\underset{|}{C}H} - CH = \underset{CH_3}{\underset{|}{C} -} CHO\)

Numbering of parent chain will occur from right hand side because functional group like aldehyde is given lowest numbering.

\(\overset{5}{C}H_3 - \underset{OH}{\underset{|}{\overset{4}{C}}H} - \overset{3}{C}H = \underset{CH_3}{\underset{|}{\overset{2}{C}} -} \overset{1}{C}HO\)

4-hydroxy-2-methylpent-2-en-1-al

Essential amino acids are those which cannot be synthesized by organisms in the body and are obtained from plants.The \(9\) essential amino acids are: histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.

Non-essential amino adds can be synthesized by the organism and may not be the requisite components of the diet, e.g., serine, cysteine, proline, glycine, alanine, asparagine, glutamine and tyrosine.

All the mentioned compounds are examples of molecular solids:

• \(I_{2}\) is an example of non polar molecular solid,
• Solid \(\mathrm{NH}_{3}\) is an example of polar molecular solid and
• Ice is an example of hydrogen bonded molecular solid.

The \(\mathrm{C}-\mathrm{X}\) bond in haloalkanes is more polar than the \(\mathrm{C}-\mathrm{X}\) in haloarenes.

In haloarenes, due to resonance, charge distribution or rotation of charge on benzene nucleus takes place therefore, the \(\mathrm{C}-\mathrm{X}\) bond in haloalkanes ( which are not aromatic) is more polar than the \(\mathrm{C}-\mathrm{X}\) bond in haloarenes (aromatic).

Given: Vapour pressure of liquid A, \(P_{A} =120~ mm\)

Vapour pressure of liquid B, \(P_{B} =180~ mm\)

Number of moles of liquid A,  \({n}_{{A}}=2\)

Number of moles of liquid B, \({n}_{{B}}=3\)

Therefore,

Mole fraction of liquid A,  \({X}_{{A}}=\frac{n_A}{n_A+n_B}\)\(=\frac{2}{5}\)

Mole fraction of liquid B,  \({X}_{{B}}=\frac{n_B}{n_A+n_B}\)\(= \frac{3}{5}\)

We know that: Vapor Pressure of solution, \({P}={X}_{{A}} {P}_{{A}}+{X}_{{B}} {P}_{{B}}\)

\(=\frac{2}{5} \times 120+\frac{3}{5} \times 180\)\(=48+108\)

\(P=156 {~mm} {~Hg}\)

If the accumulation of gas on the surface of a solid occurs on account of weak Van der Waals’ forces, the adsorption is termed as physical adsorption or physisorption.

Characteristics of Physisorption are:

• It is not Specific in nature
• It is a Reversible process
• It depends on thenature of gas. More easily liquefiable gases are adsorbed readily.

Enthalpy of adsorption is low.Low temperature is favourable for adsorption.There is no necessity of activation energy and it depends on the surface area. Thus, porous surfaces are good adsorbents.

In the presence of a catalyst, the heat evolved or absorbed during the reaction remains unchanged.The enthalpy change value of the reaction will not be affected by a catalyst. A catalyst will only lower the required activation energy for the reactions. Since it will lower the activation energy for both the forward and reverse reactions to the same extent, the net change in enthalpy is zero.