Chemistry Test - 24

Given reaction is:

\(\mathrm{Cl}_{2}+\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}+\mathrm{H}_{2} O\)

The balanced reaction will be:

\(3 \mathrm{Cl}_{2}+6 \mathrm{OH}^{-} \longrightarrow 5 \mathrm{Cl}^{-}+\mathrm{ClO}_{3}+3 \mathrm{H}_{2} \mathrm{O}\)

Oxidation number of \(C l_{2}=0\)

Oxidation number \(C l\) in \(C l^{-}=-1\)

Oxidation number of\(\mathrm{Cl}\) in \(\mathrm{ClO}_{3}:\)

\(\Rightarrow x+3 \times(-2) =-1\)

\(\Rightarrow x=+5\)

So, \(\mathrm{Cl}_{2}\) is oxidised as well as reduced.

Molar mass of \({KI}=39+127=166 {~g} {~mol}^{-1}\)

\(20 \%\) aqueous solution of KI means \(20 {~g}\) of \({KI}\) is present in \(100 {~g}\) of solution.

Therefore,

Mass of \(KI=20 {~g}\)

That is,

\(20 {~g}\) of \({KI}\) is present in \((100-20) {g}\) of water \(=80 {~g}\) of water

We know that:

Molality of the solution \(=\frac{\text { Moles of } K I}{\text { Mass of water in } {kg}}\)

\(=\frac{\frac{\text {Mass of KI}}{\text {Molar Mass of KI}} }{\text { Mass of water in } {kg}}\)

\(=\frac{\frac{20}{166}}{0.08} {~m}\)

\(=1.506 {~m}\)

\(=1.51 {~m}\)

The potential (E) of the pH electrode may be written by means of the Nernst equation as:

\(\mathrm{E}=\mathrm{E}_{0}-\frac{2.3036 \mathrm{RT}}{\mathrm{F}}(\Delta \mathrm{pH})\)

We have to find change in open-circuit voltage across the electrode (i.e.) \(\left(E_{0}-E\right)\)

\(\mathrm{E}_{0}-\mathrm{E}=\frac{2.3036 \mathrm{RT}}{\mathrm{F}}(\Delta \mathrm{pH})\)

Given,

\(\mathrm{pH}_{1}=6, \mathrm{pH}_{2}=8\)

\(\mathrm{R}=\) gas constant \(=8.31 \frac{\text { volt }-\text { columb }}{\mathrm{mol}-\mathrm{k}}\)

\(\mathrm{F}=96500\) coulombs \(\mathrm{mol}^{-1}\)

\(\mathrm{E}_{0}-\mathrm{E}=\frac{2.3036 \times 8.31 \times 298}{96500}(8-6)\)

\(=118 \mathrm{mV}\)

Dissociation of Potassium Sulphate \(\left(\mathrm{K}_2 \mathrm{SO}_4\right)\),

\(\mathrm{K}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_4^{2-}\)

i (Van't Hoff factor) \(=3\)

We know that, \(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \mathrm{m}\)

where, \(\mathrm{K}_{\mathrm{f}}\) is molal depression constant and \(\mathrm{m}\) is molality.

=3×4×0.03=0.36K

Given,

[A]₀ = 2 g

[A]_t = 0.2 g

The rate constant for a first order reaction, K= \(4.606 \times 10^{-3} \mathrm{~s}^{-1}\)

For a first order reaction,

\(K=\frac{2.303}{t} \log \frac{[A]_0}{[A]_t}\)

Where,

K = First order rate constant

[A]₀ = Initial concentration

[A]_t = Concentration at time 't'

\(\Rightarrow t=\frac{2.303}{4.606 \times 10^{-3}} \log \frac{2}{0.2}\)

\(\Rightarrow t=\frac{1000}{2} \log {10}\)

As we know,

\(\log {10}=1\)

\(\Rightarrow t=500 ~s\)

According to the question,

(i)\(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

(ii) \(\mathrm{C}\) (s) \(+2 \mathrm{O}_{2}\) (g) \(\rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

(iii) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}\) (g) \(\rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Thus, the desired equation is the one that represents the formation of \(\mathrm{CH}_{4}(\mathrm{~g})\) that is as follows:

\(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g}) ; \Delta_{\mathrm{f}} \mathrm{H}_{\mathrm{CH}_{4}}=\Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{c}}+2 \Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{H}_{2}}-\Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{CO}_{2}}\)

Substituting the values in the above formula :

Enthalpy of formation \(\mathrm{CH}_{4}(\mathrm{g})=(-393.5)+2 \times(-285.8)-(-890.3)=-74.8 \mathrm{kJmol}^{-1}\)

The arrangement of elements in the Modem Periodic Table is based on increasing atomic number in the horizontal rows.

In 1869, Russian chemist Dmitri Mendeleev created the framework that became the modern periodic table, leaving gaps for elements that were yet to be discovered. While arranging the elements according to their atomic weight, if he found that they did not fit into the group he would rearrange them.

Given,

Wavelength \(=5800 Å=5800 \times 10^{-10} \mathrm{~m}\)

As we know,

The speed of light,

\(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

(a) Wavenumber is given as,

\(\bar{\nu}=\frac{1}{\lambda}\)

\(=\frac{1}{5800 \times 10^{-10} }\)

\(=1.724 \times 10^{6} \mathrm{~m}^{-1}\)

\(=1.724 \times 10^{4} \mathrm{~cm}^{-1}\)

(b) The frequency is given as,

\(\nu=\frac{c}{\lambda}\)

\(=\frac{3 \times 10^{8} }{5800 \times 10^{-10}}\)

\(=5.172 \times 10^{14}\) Hz

Coal is the chief source ofAromatic hydrocarbons.

Simple aromatic hydrocarbons come from two main sources: Coal and petroleum. Coal is a complex mixture of a large number of compounds, most of which are long-chain compounds. If coal is heated to about 1000°C in the absence of air (oxygen), volatile components, the so-called tar oil, are stripped out.

In propanone((CH₃)₂CO), the functional group present is a ketone (- C=O) group.

Acetone also is known as Propanone is an organic compound with the formula (CH₃)₂CO.

In propanol (C₃H₇OH), the functional group present is an alcohol (- OH) group.

In propanoic acid (C₂H₅COOH), the functional group present is a carboxylic acid (- COOH) group. 

In ethanal (CH₃CHO), the functional group present is an aldehyde(- CHO) group.

In propanal (C₂H₅CHO), the functional group present is an aldehyde(- CHO) group.