Chemistry Test - 25

\(\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}\)

\(\mathrm{i}=1+\alpha\)

\(\Rightarrow \alpha=(\mathrm{i}-1)\)

Now,

\(\mathrm{K}_{\mathbf{a}}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}\)

\(\Rightarrow \mathrm{K}_{\mathrm{a}}=\frac{(\mathrm{i}-1)^{2}}{1-(\mathrm{i}-1)}\)

\(\Rightarrow \mathrm{K}_{\mathbf{a}}=\frac{(\mathrm{i}-1)^{2}}{2-\mathrm{i}}\)

Oxygen of iron oxide is removed and added to carbon monoxide to form carbon dioxide.

The reaction takes place as follows:

\(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Fe}(l)+3 \mathrm{CO}_{2}(g)\)

Carbon mono oxide : oxidation

Iron oxide : reduction

The heats of adsorption in physisorption or physical adsorption lies in the range of 10−40 kJ/mol.

It is less as compared to chemisorption as there is no chemical reaction involved in it. The heat of chemisorption lie in the range of 80 to 240kJmol^−1.

Mercurous chloride is \(H g C l_{2}\)

\(H g C l_{2} \rightleftharpoons H g^{2+}+2 C l^{-}\)

\(\quad\)\(\quad\)\(\quad\)\(\quad\)\(~S\)\(\quad\)\(\quad\)\(\quad\)\(2S\)

\([\frac{K_{s p}(Hg_{2} Cl_{2})}{4}]^{\frac{1}{3}}\)

Mercuric chloride is soluble in ethanol, methanol, acetone, ethyl acetate, and diethyl ester. It is slightly soluble in acetic acid, pyridine, and carbon disulfide. Its solubility in water at 20°C is 69 g/l.

Mole ratio of \(\mathrm{H}, \mathrm{C}\) and \(\mathrm{N}\)

\(=\frac{8.7}{1}: \frac{74}{12}: \frac{17.3}{14} \)

\( =8.7: 6.167: 1.23 \)

\(=\frac{8.7}{1.23}: \frac{6.167}{1.23}: \frac{1.23}{1.23} \)

\(=7: 5: 1\)

\(\therefore\) Emperical formula \(=\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}\)

\(\therefore\) Molecular formula \(=\left(\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}\right)_n\)

Given molecular mass \(=162\)

Molecular mass of \(\left(\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}\right)_{\mathrm{n}}\)

\(=(5 \times 12+7 \times 1+14) \times \mathrm{n} \)

\(=(81) \times \mathrm{n} \)

\(\therefore 81 \times \mathrm{n}=162 \)

\(\Rightarrow \mathrm{n}=2\)

\(\therefore\) Molecular formula \(=\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_2\)

Hint: Lithium, nitrogen and oxygen combine to form a compound which can be called as lithium nitrate. Lithium nitrate on heating forms multiple products which includes the formation of alkali oxide, oxygen and nitrogen dioxide. The alkali oxide formed in this reaction is lithium oxide.

The atoms in the periodic table combine with other atoms and form molecules attached through different bonds. Lithium is an electropositive element and also a metal, nitrogen and oxygen are non-metallic. These three atoms combine to form a chemical compound called lithium nitrate.

The molecular formula of lithium nitrate is \(\mathrm{LiN}_3\), as it has elements other than oxygen and hydrogen. It can be considered as an inorganic compound.

Inorganic compounds upon heating give multiple products. The products formed on heating lithium nitrate are alkali oxide, oxygen and nitrogen dioxide. The alkali can be a metal that belongs to the alkali earth metals. Here, the alkali metal is lithium.

The chemical reaction involved can be written as:

\(4 \mathrm{LiNO}_3 \rightarrow 2 \mathrm{Li}_2 \mathrm{O}+4 \mathrm{~N} \mathrm{O}_2+\mathrm{O}_2\)

Thus, the products are lithium oxide \(\left(\mathrm{Li}_2 \mathrm{O}\right)\), oxygen \(\left(\mathrm{O}_2\right)\) and nitrogen dioxide \(\left(\mathrm{NO}_2\right)\).

The first sample has a Mo to S mass ratio is: \(\frac{1.50 \mathrm{~g} \mathrm{~of} \mathrm{~Mo}}{1.00 \mathrm{~g} \mathrm{~of} \mathrm{~S}}\) 

In the second sample, we know the mass of \(\mathrm{S}(2.50 \mathrm{~g})\) and the mass of Mo is unknown. The mass ratio of Mo to \(\mathrm{S}\) in the second sample is, therefore, \(\frac{\text { Mass of Mo }}{2.50 \mathrm{~g} \text { of S }}\)

Now we equate them, because the two ratios must be equal according to law of definite proportions.

\(\frac{\text { Mass of Mo }}{2.50 \mathrm{~g} \text { of } \mathrm{S}}=\frac{1.50 \mathrm{~g} \text { of Mo }}{1.00 \mathrm{~g} \text { of } \mathrm{S}}\)

On Solving, the mass of Mo is: \(=2.50 \mathrm{~g}\) of S \(\times \frac{1.50 \mathrm{~g} \text { of Mo }}{1.00 \mathrm{~g} \text { of } \mathrm{S}}\) \(=3.75 \mathrm{~g}\) of Mo 

In this case, \({ }_{35}^{80} \mathrm{Br}\),

\(\mathrm{Z}=35 \)

\(\mathrm{~A}=80\)

Number of protons \(=\) number of electrons \(= Z = 35\)

Number of neutrons \(= 80 – 35 = 45\)

\({CH}_{3} {COONa}\) is salt of strong base and weak acid. 

\({NaOH}+{CH}_{3} {COOH} \rightarrow {CH}_{3} {COONa}+{H}_{2} {O}\)

\(100 {~mL}\) of \(0.2 {M} {CH}_{3} {COOH}=0.2 \times 100=20 {~m} {~mol}\) of \({CH}_{3} {COOH}\)

\(100 {~mL}\) of \(0.2 {M} {NaOH}=0.2 \times 100=20 {~m}\) mol of \({NaOH}\)

\(20 {~m}\) mol of \({NaOH}\) reacts with \(20 {~m}\) mol of \({CH}_{3} {COOH}\) to form \(20 {~m}\) mol of \({CH}_{3} {COONa}\).

Concentration of \({CH}_{3} {COONa}=20 {mmoles} / 200 {ml}=0.1 {M}\)

\({pH}=\frac{p K w}{2}+\frac{p K a}{2}+\frac{\log C}{2}\)

\({pH}=7+2.37+\frac{1}{2}[\log (0.1)]\)

\({pH}=9.37+\frac{1}{2}[-1]\)

\(=9.37-0.5\)

\(=8.87\)