Chemistry Test - 25

Some medicines are more effective in the colloidal form because of the large surface area and easy assimilation.

A colloid is a heterogeneous system in which one substance is dispersed (dispersed phase) as very fine particles in another substance called dispersion medium.

Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

Various applications of collids are:

• Purification of drinking water
• Cleansing action of soaps and detergents as Medicines
• Photographic plates and films

Given reaction,

\(2 C u_{(a q)}^{+} \rightarrow C u_{(s)}+C u_{(a q)}^{2+}\)

And, \(E_{c e l l}^0 = 0.36 ~V\)

It is evident from the cell reaction that it involves the transfer of one electron so that \(\mathrm{n}=1\)

We know a relation:

\(\log K=\frac{E_{c e l l}^{0} \times n}{0.0591}\)

\(\log K=\frac{0.36 \times 1}{0.0591}=6.09\)

\(K =\operatorname{antilog} 6.09=1.2 \times 10^{6}\)

The process of conversion of ammonia into nitrate nitrogen is called nitrification. It is done in two steps- Nitrite formation, Nitrate formation. Nitrite formation: Ammonia are oxidized to nitrites by Nitrococcus, Nitrosomonas.

The nitrogen cycle has four steps:

• Nitrogen Fixation: Conversion of molecular nitrogen into the inorganic nitrogenous compound.
• Ammonification: Conversion of the dead organic nitrogenous compound into ammonia.
• Nitrification: Oxidation of ammonia into nitrates.
• Denitrification: Nitrites or nitrates converts back into molecular nitrogen.

An ideal solution is a solution in which the enthalpy of solution is zero or close to zero.

\(C _{2} H _{5} Br\ \&\ C _{2} H _{5} I\) has same enthalpy even Benzene \& toluene both has same enthalpy.

\(\Delta H =0\).

Rate of reaction depends on temperature, concentration of the reactants and presence of catalyst. A catalyst speeds up a chemical reaction with being chemically changed upon completion of reaction.

Example:

1. At higher temperature, particles collide more frequently and with greater similarity.
2. \(\mathrm{CH_{4}(g)+2 O_{2}(g) \longrightarrow CO_{2}(g)+2 H_{2} O}\) Faster rate
3. \(\mathrm{C}_{25} \mathrm{H}_{52}(\mathrm{~g})+38 \mathrm{O}_{2} \longrightarrow 25 \mathrm{CO}_{2}(\mathrm g)+26 \mathrm{H}_{2} \mathrm{O}\) Slow rate

So, reaction depends on nature of reactants.

A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.

For first order reaction,

\([\mathrm{A}]=\left[\mathrm{A}_{0}\right] \mathrm{e}^{-\mathrm{kt}}\)

Where,

\(\mathrm{k}=\) First order rate constant

\([\mathrm{A}]_{0}=\) Initial concentration

\([\mathrm{A}]=\) Concentration at time \(\mathrm{t}\)

From the above equation we can draw:

Thus we can say that the concentration of reactants will exponentially decrease with time.

The reaction of decomposition of phosphine can be written as:

\(\mathrm{4 P H_{3} \rightarrow P_{4}+6 H_{2}}\)

The balanced chemical reaction indicates that four moles of phosphine generates one mole of phosphorus and six moles of hydrogen.

The phosphorus exists as a triatomic molecule in elemental form. It is a solid and therefore, its volume change is negligible according to gas laws.

The volume of phosphine given is \(100 \mathrm{~mL}\). The amount of hydrogen produced can be calculated as:

Amount of hydrogen \(=\frac{6}{4} \times\) amount of phosphine

\(=\frac{6}{4} \times 100 \mathrm{~mL}\)

\(=150 \mathrm{~mL}\)

Thus, \(100 \mathrm{~mL}\) of phosphine gas generates \(150 \mathrm{~mL}\) of hydrogen gas.

The change in volume \(=150-100\)

\(=50 \mathrm{~mL}\)

Interstitial compounds have high melting points in comparison to pure metals. They are very hard and retain metallic conductivity. They are chemically inert i.e., not reactive.

• Interstitial compounds are those which are formed when small atoms like H, C, N, B etc. are trapped inside the crystal lattice of metals.
• They are generally non-stoichiometric and neither typically ionic or covalent in nature.

It is a subsociative compound. In which \(\mathrm{K_{2}}\) is the cation part and \((\mathrm{Zn(OH)_{4}})\) is the negative part.

\((\mathrm{Zn(OH)_{4}})\) is a complex ion and here ligand \(\mathrm{OH}\)is anionic ligand.

As we know,Anionic ligands end in -o rather than -ide andIndicate numbers of ligands with the appropriate Greek prefix: di-, tri-, tetra-, penta-, hexa-, etc. Here numbers of ligand are \(4\) i.e., tetra.So it will be named as tetrahydroxo.

If the metal is in an anionic complex ion, use the ending -ate. After this we will write oxidation state of metal in roman.Here metal is Zinc \((Zn)\) and its oxidation state is \(2\).So it will be named asPotassium tetrahydroxozincate(II).

Chlorine atom in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is attached to \(\mathrm{sp}^{3}\) carbon.

The structure of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is as follows:

As from the structure, it can be seen that chlorine atom in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is attached to a carbon with \(4 \sigma\) bond. 

Thus, it is attached to the \(\mathrm{sp}^{3}\) carbon atom.