Chemistry Test-30

The reason for the drug's classification is as follows:

(i) The pharmacological impact is based on: To physicians this definition is helpful. It provides a whole array of drugs to classify drugs to different diseases.

(ii) Based on action on the drugs: This is based on a drug's action on a specific biochemical process.

(iii) Building on chemical structure: The range of drugs that share common structural characteristics and have similar pharmacological activity.

(iv) Building on molecular structure: Many medications have the same action function on targets. In such cases this distinction is useful.

Oxidation happens at the anode and reduction at the cathode due to a potential difference between the two electrodes. As the reaction proceeds, these potentials come close to each other until they become equal. When this happens, no more electrons flow - as current exists only when there is a potential difference. This is when the cell stops working.

Given,

The nuclide of the element plutonium, \(= {}_{94}Pu^{242}\)

To Find: the number of neutrons, \(n\).

To calculate \(n\), the formula used:

Number of neutrons = Atomic mass - Number of protons

\(n=A-Z\)

A nuclide of an element denotes its symbol atomic mass(\(A\)), and number of protons(\(Z\)) of that element.

Number of protons, \(\mathrm{Z}\) is written as a subscript of the atomic symbol, and the atomic mass(\(A\)) is written as a superscript.

Applying the above formula:

\(n=A-Z\)

here, the number of protons \(\mathrm{Pu}, \mathrm{Z}=94\)

The atomic number of \(\mathrm{Pu}, \mathrm{A}=242\)

\(n =242-94 \)

\(=148\)

\(n=148\)

Therefore, the number of neutrons is \(148\).

The product of the reaction is \(\mathrm{RNC}\).

The general form of the reaction is:

When it is cold, the moisture in the air condenses on the surface of dust particles, forming tiny droplets. These droplets are colloidal in nature and form a blanket of fog or mist in the atmosphere. Clouds are colloidal systems as they are aerosols consisting of water droplets suspended in the air.

\(\mathrm{Fe}(\mathrm{s})+\mathrm{CuSO}_{4}(\mathrm{aq}) \rightarrow \mathrm{FeSO}_{4}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\)

The above reaction is an example of a displacement reaction. In the above reaction, iron has displaced or removed another element,copper, from copper sulphate solution.

The reaction in which one atom or a group ofatoms of a compound is replaced by anotheratom, is called a displacement reaction. Generally,a more reactive metal displaces a less reactivemetal from its salt solution in displacementreaction.

From the data, it is clear that the difference between the radii of ground state and excited state is

\(r_{n}-r_{1}=0.79 \times 10^{-9} \mathrm{m}\)

\(\mathrm{r}_{1}=0.0529 \times 10^{-9} \mathrm{m}\)

\(r_{n}-0.0529 \times 10^{-9} \mathrm{m}=0.79 \times 10^{-9} \mathrm{m}\)

\(r_{n}=0.84 \times 10^{-9} \mathrm{m}\)

\(\mathrm{Also}, \mathrm{r}_{\mathrm{n}}=0.0529 \times 10^{-9} \times \mathrm{n}^{2}\)

\(\frac{0.84 \times 10^{-9}}{0.0529 \times 10^{-9}}\)

\(n^{2}=15.87 \approx 16\)

\(n=4\)

Thus, electrons excited from first orbit to fourth orbit and number of spectral lines \(=\frac{\left(n_{2}-n_{1}\right)\left(n_{2}-n_{1}+1\right)}{2}\)

\(=6\)