Chemistry Test - 19

We know that:

The molar mass of oxygen \(=16\)

The molar mass of iron \(=55.85\)

Given:

Percentage of iron by mass \(=69.9 \%\)

Percentage of oxygen by mass \(=30.1 \%\)

Now,

Relative mass of iron in iron oxide \(=\frac{\text { Percentage of iron by mass }}{\text { At. mass of iron }}=\frac{69.9}{55.85}=1.25\)

Relative mass of oxygen in iron oxide \(=\frac{\text { Percentage of oxygen by mass }}{\text { At. mass of oxygen }}=\frac{30.1}{16}=1.88\)

Now,

Simplest molar ratio of iron to oxygen \(=1.25: 1.88=2: 3\)

Therefore the empirical formula of iron oxide is \({Fe}_{2} {O}_{3}\).

\(\mathrm{C}_{2} \mathrm{H}_{6}\) compound has \(7\) covalent bonds.

The structure of the Ethane molecule is given below.

From the above structure, we can see that the Ethane molecule has \(6~ \mathrm{C}-\mathrm{H}\) covalent bonds and one \(\mathrm{C-C}\) covalent bond. Therefore, on adding the covalent bonds of \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{H}\) covalent bonds, we can say that the Ethane molecule has \(7\) covalent bonds.

The compound contains 3 carbons. The alkane with 3 carbons is "propane". There is a -CHO group (aldehyde) in this compound. Hence the name would contain the " suffix".

The IUPAC name would then be:

Propane- 'e' +'al' \(\rightarrow\) Propanal

\(\begin{aligned} & \% \text { of } \mathrm{H}=\frac{2}{18} \times \frac{\text { wt. of } \mathrm{H}_2 \mathrm{O}}{\text { wt. of organic compound }} \times 100 \\ & =\frac{2}{18} \times \frac{0.4428}{0.492} \times 100 \\ & =0.11 \times 0.9 \times 100 \\ & =0.099 \times 100=9.9 \\ & \% \text { of } \mathrm{C}=\frac{12}{44} \times \frac{0.7938}{0.492} \times 100 \\ & =0.27 \times 1.61 \times 100 \\ & =43.47 \\ & \% \text { Oxygen }=100-(43.47+9.9) \\ & =100-53.37 \simeq 46\end{aligned}\)

Given, \(\left[\mathrm{K}_{\mathrm{sp}}\right]_{\mathrm{Pbl}_2}=8 \times 10^{-9}\)

To calculate solubility of \(\mathrm{PbI}_2\) in \(0.1 \mathrm{M}\) solution of \(\mathrm{Pb}\left(\mathrm{NO}_3\right)_2\).

\(\text { (II) } \mathrm{Pbl}_2(\mathrm{~s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+21 \)

\(\mathrm{~S} 2 \mathrm{~S} \)

\( \therefore\left[\mathrm{Pb}^{2+}\right]=\mathrm{S}+0.1 \approx 0.1 \)

\(\because \mathrm{S}<<0.1\)

\(\text { Now, } \mathrm{K}_{\mathrm{sp}}=8 \times 10^{-9} \)

\({\left[\mathrm{~Pb}^2\right]\left[\mathrm{I}^2\right]^2=8 \times 10^{-9}} \)

\( 0.1 \times(2 \mathrm{~S})^2=8 \times 10^{-9} \)

\( 4 \mathrm{~S}^2=8 \times 10^{-8} \Rightarrow \mathrm{S}=141 \times 10^{-6} \mathrm{M} \)

\( \mathrm{x}=141\)

Transition metal compounds are usually coloured. This is due to the electronic transition within the d-orbitals.

Transition elements (also known as transition metals) are elements that have partially filed d orbitals.

Many ionic and covalent compounds of transition elements are coloured and this colouring property is due to \(d-d\) electronic excitation. The colour of the compound is complementary to the colour absorbed.

\(\begin{aligned} & \text { Moles at equilibrium }=1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2} \\ & \mathrm{i}=\frac{\text { Total molesat equilibrium }}{\text { Initial moles }}=\frac{1-\frac{\alpha}{2}}{1} \\ & \mathrm{i}=\frac{1-\frac{0.8}{2}}{1}=0.6 \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \mathrm{m} \\ & 2=0.6 \times 5 \times\left(\frac{\mathrm{w}}{\frac{122}{30}}\right) \times 1000 \\ & \mathrm{w}=2.4 \mathrm{~g}\end{aligned}\)

A reaction, \(\mathrm{A+B \rightarrow C+D+q}\) is found to have a positive entropy change. The reaction will bepossible at any temperature.

For a reaction to be spontaneous, \(\Delta \mathrm{G}\) should be negative \(\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}\)

According to the question, for the given reaction,

\(\Delta \mathrm{S}=\) positive

\(\Delta \mathrm{H}=\) negative (since heat is evolved)

That results in \(\Delta \mathrm{G}=\) negative

Therefore, the reaction is spontaneous at any temperature.

Among the following mixtures, dipole-dipole as the major interaction is present inAcetonitrile and acetone.

"Acidic hydrogen means it has a tendency to be released as \(H^{+}\) ion. So, if any H-atom is attached to another atom or group of atoms with higher electronegativity, that \(\mathrm{H}\)-atom can be released very easily as \(H^{+}\) ion."

(A) 

The hydrogen marked with red color is an acidic hydrogen, due to the presence of hydrogens in between electron withdrawing groups.

(B) 

The hydrogens marked with red color are acidic hydrogens, due to the presence of hydrogens in between electron withdrawing groups.

The hydrogen marked with red color is acidic hydrogen, due to the presence of hydrogens in between electron withdrawing groups.

The hydrogen marked with red color is acidic hydrogen, due to the attachment of hydrogen directly to the oxygen.

Oxygen atoms have higher electronegativity than carbon.

So, the hydrogen which is attached to oxygen is more acidic than the hydrogen atoms that are attached to carbon atoms (carbon atoms later attached to electron withdrawing groups).