Chemistry Test - 8

It deals with energy changes of microscopic systems thisstatement is not true regarding the laws of thermodynamics.

The laws ofthermodynamics deal with energy changes of macroscopic systems involving alarge number of molecules rather than microscopic systems containing a fewmolecules. Laws of thermodynamics apply only when a system is in equilibrium ormoves from one equilibrium state to another equilibrium state.

\({C H}_{3} {COOH}+{HF} \rightleftharpoons {CH}_{3} {COOH}_{2}^{+}+{F}^{-}\)

\(HF\) gives \({H}^{+}\) to the \({CH}_{3} {COOH},\)

So, it is acid its conjugate base is \({F}^{-}\)

The total number of electrons that take part in forming bonds in \(\mathrm{N}_{2}\) is \(6 .\)

Structure of \(N_{2}\) is \(N \equiv N\).

We know that the atomic number of nitrogen is 7 and the electronic configuration of nitrogen is 2,5 which means two electrons are present in s-orbital and five electrons in \(\mathrm{p}-\) orbital.

When two nitrogen atoms will react, three electrons (from five valence electrons) of each nitrogen atom are shared between them so that both will obtain a stable configuration.

In \(N_{2}\) there are 3 bonds between two nitrogen atoms and each bond is formed by 2 electrons.

The full outer shells with the shared electrons are now stable forming a covalent bond and this can be shown through the following diagram:

So, 3 bonds make it \(3 \times 2=6\) electrons in total.

Let the volume of the round bottomed flask be \(\mathrm{V}\).

Then, the volume of air inside the flask at \(27^{\circ} \mathrm{C}\) is \(\mathrm{V}\).

Now,

\(V_{1}=V\)

\(T_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}, V_{2}=?\)

\(T_{2}=477^{\circ} \mathrm{C}=750 \mathrm{~K}\)

According to Charles's law,

\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

\(\Rightarrow V_{2}=\frac{V_{1} T_{2}}{T_{2}}\)

\(=\frac{750 \mathrm{~V}}{300}\)

\(=2.5 \mathrm{~V}\)

Therefore, volume of air expelled out \(=2.5 \mathrm{~V}-\mathrm{V}=1.5 \mathrm{~V}\)

So, fraction of air expelled out \(=\frac{1.5 \mathrm{~V}}{2.5 \mathrm{~V}}=\frac{3}{5}\)

\(\mathrm{FCH}_{2} \mathrm{CHO}\) is most reactive towards nucleophilic addition since presence of most electronegative \(F\) withdraws electron from carbon of carbonyl group making it more polar.

Feedback inhibition is a cellular control mechanism in which an enzyme’s activity is inhibited by the enzyme’s end product. This mechanism allows cells to regulate how much of an enzyme’s end product is produced. Inhibition of enzyme activity at an active site by an inhibitor, which is structurally similar to a substrate

Since \({A}\) is present at the 8 corners, number of \({A}\) ions per unit cell \(=\frac{1}{8} \times 8=1\)

Since \(B\) is present at the centre of 6 faces, number of \(B\) per unit cell \(=\frac{1}{2} \times 6=3\)

\(\therefore\) The ratio of \(A\) and \(B\) is \(1: 3\), so the formula of the compound is \(A B_{3}\).

The chemical formula of halothane is \(\mathrm{CF}_{3} \mathrm{CHClBr}\).

• Halothane, the brandname Fluothane, is a general anaesthetic.  
• It can be used to induce or maintain anaesthesia.

The molecular structure of halothane is as follows:

The ions \(\mathrm{Mg}^{+2}\) and \(\mathrm{PO}_{4}^{-3}\) combine to form compound \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Total charge on cation \(=3(2+)=+6\)

Total charge on anion \(=2(-3)=-6\)

Therefore,

Total charge on cation \(=\) Total charge on anion