Chemistry Test - 5

Copper dust is not used as a reducing agent in the manufacture of dyestuffs and paints.  

Copper powders are used in very many applications, markets and technologies by virtue of the diverse range of physico-chemical properties.

According to molecular orbital theory \(\mathrm{O}_{2}\) molecule is more paramagnetic because it has two unpaired electrons in the antibonding molecular orbital.

The neutral oxygen is paramagnetic according to MO theory because it ends up with two unpaired electrons in two degenerate pi antibonding molecular orbitals.

The other two are paramagnetic because they have an odd number of electrons so it doesn’t matter what kind of bonding they are involved in, the electrons cannot be all paired up.

\(O_{2}^{2-}\) and \( O_{2}^{2+}\) these would also be diamagnetic as the double negative would have filled up the \(\mathrm{pi}^{*}\) orbitals and the double-positive version would have left both of them empty.

Be exhibits the diagonal relationship with Al.

The elements of the 2nd period, show resemblance in properties with elements of the 3rd period, placed diagonally. This is called the diagonal relationship. Beryllium (Be) shows a diagonal relationship with aluminium (Al).

The representation of the Galvanic cell is done as below:

Oxidation (left part) || Reduction (right part).

Aqueous elements have to be near the salt bridge in the representation. One has to remember that, in an electrochemical cell, reduction occurs at cathode and oxidation occurs at the anode. 

Here in this question, option (A) satisfies the representation rules and thus, is the correct answer.

\({Zn}+2 {Ag}^{+} \longrightarrow {Zn}^{2+}+2 {Ag}\) can be represented as:

\(Z n_{(s)}|Z n_{(a q)}^{2+} \| A g_{(a q)}^{+}| A g_{(s)}\)

Phenol has activating (electron releasing) \(-\mathrm{OH}\) group and bromine water supplies \(\mathrm{Br}^{+}\)ion easily, hence under such conditions reaction does not stop at monobromo or dibromo stage but a fully brominated (2, 4, 6,-tribromophenol compound is the final product.

A bicycloalkene cannot have the formula of C₂H₆.

C₂H₆ has two degree of unsaturation (two H₂ less than saturated hydrocarbons), therefore it can be a diene, a cycloalkene or a bicycloalkane but it cannot be a bicycloalkene because it has three degrees of unsaturation.

The reaction of combustion of methane takes place as follows:

\(\underset{\mathrm{1 ~mole}}{\mathrm{CH}_{4}}+\underset{\mathrm{2 ~mole}}{2 \mathrm{O}_{2}} \longrightarrow \underset{\mathrm{1 ~mole}}{\mathrm{CO}_{2}}+\underset{\mathrm{2 ~mole}}{2 \mathrm{H}_{2} \mathrm{O}}\)

Here, we know that:

Mass of 1 mole methane is \(16 ~g~\)i.e.,\((12 + 4 \times 1)\).

Since 1 mole of methane on combustion produces 2 moles of \(\mathrm{H}_{2} \mathrm{O}\).

Therefore, \(16 g\) of methane on combustion, produces \(36 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\).

Both gas A and Hydrogen \(\left(\mathrm{H}_2\right)\) gas have same volume at same pressure. Let both 's volume is \(\mathrm{V}\) and pressure \(\mathrm{P}\).

For gas A :

Pressure \(=\mathrm{P}\)

Temperature \((T)=300 \mathrm{~K}\)

Volume \(=\mathrm{V}\)

Mass \(=3 \mathrm{~g}\)

Molar mass \(=\mathrm{Mgm} / \mathrm{mol}\)

using ideal gas equation,

\(\mathrm{PV}=\mathrm{nRT} \)

\(\Rightarrow \mathrm{PV}=\frac{3}{\mathrm{M}} \times \mathrm{R} \times 300 \ldots .(1)\)

For Hydrogen,

Pressure \(=\mathrm{P}\)

Temperature \((T)=200 \mathrm{~K}\)

Volume \(=\mathrm{V}\)

Mass \(=0.2 \mathrm{~g}\)

Molar mass \(=2 \mathrm{gm} / \mathrm{mol}\)

Using ideal gas equation,

\(\mathrm{PV}=\frac{0.2}{2} \times \mathrm{R} \times 200 \text {. ....(2)}\)

From (1) and (2), we get

\(\frac{3}{\mathrm{M}} \times \mathrm{R} \times 300=\frac{0.2}{2} \times 2 \times 200 \)

\(\Rightarrow \mathrm{M}=45\)

The rate for the reaction between ionic compounds cannot be determined because they are generally instantaneous reactions.

Ionic compounds readily dissociate into ions, which react with each other instantaneously to form products. Therefore the rate of these reactions cannot be determined.

At cathode : 2Na⁺ + 2e^– \(\Rightarrow\) 2Na

At anode : 2Cl^– \(\Rightarrow\) Cl₂ + 2e^–

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Net reaction: 2Na⁺ + 2Cl^– \(\Rightarrow\) 2Na + Cl₂

From Faraday’s first law of electrolysis,

\(\mathrm{W}=\mathrm{Z} \times \mathrm{I} \times \mathrm{t}\)

=\(\frac{E}{96500} \times \mathrm{I} \times \mathrm{t}\)

No. of moles of \(\mathrm{Cl}_2\) gas \(\times \mathrm{Mol}\). wt. of \(\mathrm{Cl}_2\) gas

=\(\frac{E q . w t . \text { of } \mathrm{Cl}_2 \text { gas } \times I \times t}{96500}\)

\(\Rightarrow 0.10 \times 71=\frac{35.5 \times 3 \times t}{96500}\)

\(\Rightarrow t=\frac{0.10 \times 71 \times 96500}{35.5 \times 3}\)

=6433.33 sec

=107.22 \(\mathrm{~min} \simeq 110 \mathrm{~min}\)