Chemistry Test-19

1-phenylethanone

We know

\(\frac{0.693}{t_{1 / 2}}=\frac{2.303}{t} \log \left(\frac{N_{0}}{N}\right)\)

Given \(t_{1 / 2}=1386 sec\) , \(t=80 sec , N_{0}=1\)

\(N =\) remaining fraction after \(80 sec\)

\(\frac{0.693}{1386}=\frac{2.303}{80} \log \left(\frac{1}{N}\right)\)

On solving

\(N =0.96\)

Fraction decayed \(=1-0.96=0.04\)

percentage decay \(=0.04 \times 100=4\)

\(\mathrm{NH}_{4} \mathrm{NO}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2}\)

\(2\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}\)

\(\mathrm{Ba}\left(\mathrm{N}_{3}\right)_{2} \rightarrow \mathrm{Ba}+3 \mathrm{~N}_{2}\)

\(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{N}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

The compound that does not produce nitrogen has by the thermal decomposition is\(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

The heat change associated with reactions at constant volume is due to the difference in the internal energy of the reactants and the products.

First Law of Thermodynamics dictates that \(\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}\)

At constant \(\mathrm{V}, \Delta \mathrm{W}=0\)

So, \(\Delta \mathrm{Q}=\Delta \mathrm{U}\)

So at constant volume, heat change takes place due to a change in the internal energy of the system.

Let us consider

\(\lambda_A=\lambda\)

Therefore, \(\lambda_B=2 \lambda\)

If \(N _0\) is the number of atoms in \(A\) and \(B\) at \(t =0\), then initial rate of disintegration of \(A=\lambda N _0\) and initial rate of disintegration of \(B=2 \lambda N_0\)

Since \(\lambda_{ B }=2 \lambda_{ A }\)

So, \(\left(T_{\frac{1}{ 2}}\right)_B=\frac{1}{ 2}(T_{\frac{ 1}{ 2}})_A\)

After one half life of \(A\)

\((\frac{- dN}{dt} )_{ A }=\frac{\lambda N _0}{2} \ldots \ldots . .(1)\)

Similarly, after two half lives of \(B\)

\((\frac{- dN}{dt} )_{ B }=\frac{2 \lambda N _0}{4}=\frac{\lambda N _0}{2} \ldots \ldots \ldots(2)\)

From equations (1) and (2), we get,

\((\frac{-d N}{d t})_A=(\frac{-d N}{d t})_B\)

Therefore, the value of \(n=1\) i.e after one half life of \(A\), the rate of disintegration of both will be equal.

Uracil is pyrimidine base. Pyrimidine is an aromatic heterocyclic organic compound similar to pyridine. Three types of nucleobases are pyrimidine derivatives: cytosine(C), thymine(T), and uracil(U). Uracil is present in RNA and thymine is present in DNA.

A heterogeneous mixture has components whose proportions vary throughout the sample or is simply and mixture that is not uniform in composition.

Suspension is a heterogeneous mixture of larger particles whereas solution is a homogenous mixture.

Mixture of \(\mathrm{\left[ Conc~ HNO _3\right.}\) conc \(\mathrm{\left.H _2 SO _4\right]}\) gives \(\mathrm{{ }^{+} NO _2}\) which acts as electrophile and in nitrobenzene \(\mathrm{- NO_2}\) group is a meta directing group so nitronium ion attacks at meta position.

John Dalton Postulates about atoms.

All matter is made up of tiny, indivisible particles called atoms.

All atoms of a specific element are identical in mass, size, and other properties. However, atoms of different element exhibit different properties and vary in mass and size.

Atoms can neither be created nor destroyed. Furthermore, atoms cannot be divided into smaller particles.

Atoms of different elements can combine with each other in fixed whole-number ratios in order to form compounds.

Atoms can be rearranged, combined, or separated in chemical reactions.