Chemistry Test - 12

The correct name of \(\left[ Pt \left( NH _{3}\right)_{4} Cl _{2}\right]\left[ P tCl _{4}\right]\) is: Tetraaminedichloro platinum (IV) tetrachloro platinate(II)

First, we have to write the cationic part, of the coordinate compound

Here, the ligands are Amine and Chloride. Since there are 4 amine groups we'll name it as tetramine and 2 chloride groups will be named as dichloro. Also, we have platinum metal as the central metal atom in a cation, and it's oxidation state is +4 .

Now for the other (anionic part):

Here the ligand is chloride. Since there are 4 chlorides so we'll call it tetrachloride. We have the central metal atom as platinum again, but since it is in the anionic coordination sphere.

Given,

The rate of reaction becomes 2 times for every 10° rise in temperature.

The temperature is increased from 30°C to 80°C, thus the temperature is raised by 50°C.

Thus, the temperature coefficient of the reaction = 2 when the temperature is increased by 50°C.

As we know,

\(\frac{r_2}{r_1} = 2^n\).....(1)

Where,

\(r_2\) = Rate at 80°C

\(r_1\) = Rate at 30°C

And n \(= \frac{ΔT}{10}\)

When temperature increases from 30°C to 80°C, change in temperature,

n \(= \frac{80-30}{10} = 5\)

From equation (1), we get

\(\frac{r_2}{r_1} = 2^5\)

= 32 times

By this relation:

\(\mathrm{I.E.+E . A}={275+86=361}\) \(\mathrm{kcal ~mol}^{-1}\)

\(=361 \times 4.184=1510.42 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Where \(\mathrm{I.E.}\) = Ionization enthalpy

\(\mathrm{E.A.}\) = Electron affinity (electron gain enthalpy)

Electronegativity \(=\frac{\mathrm{I.E.}+\mathrm{E} . \mathrm{A} .}{540}\)

So, electronegativity \(=\frac{1510.42}{540}\)

\(\Rightarrow 2.797=2.8\)

The combustion of diborane includes following reaction:

\({B}_{2} {H}_{6({~g})}+3 {O}_{2({~g})} \longrightarrow {B}_{2} {O}_{3({~s})}+3 {H}_{2} {O}_{({g})}\)

According to Hess Law of heat summation :

\(\Delta {H}_{\text {combustion }}=\left[\Delta {H}_{{formB}_{2} {O}_{3}}({~s})+3 \Delta {H}_{{formH}_{2} {O}}({g})\right]-\Delta {H}_{{formB}_{2} {H}_{6}}\)

\(\Delta {H}_{\text {combustion }}=\text { Heat of combustion of diborane }=-1941 {~kJ}\)

\(\Delta {H}_{{formB}_{2} {O}_{3}({~g})}=\text { Heat of formation of borane oxide }=-2368 {~kJ}\)

\(\Delta {H}_{{formH}_{2} {O}({g})}=\text { Heat of formation of water }=-241.8 {~kJ}\)

\(\Delta \mathrm{H}_{\text {formB }_{2}} {H}_{6}({~g})=\) Heat of formation fo diborane Putting the values in the equation above,

We get :

\(-1941=[-2368+3(-241.8)]-\Delta {H}_{{formB}_{2} {H}_{6}} \)

\(\Delta {H}_{{formB}_{2} {H}_{6}}=[-2368-725.4]+1941 \)

\(=-3093.4+1941 \)

\(=-1152.4 {~kJ} / {mol} .\)

The heat of the formation of diborane is \(=-1152.4 {~kJ} / {mol} .\)

The conversion of atomic hydrogen into ordinary hydrogen is an exothermic reaction. Since atomic hydrogen is very unstable it readily combines with another hydrogen atom to form a hydrogen molecule which is highly stable therefore energy is released.

Initial At eqn. \(\quad N_{2} O_{4} \quad \rightarrow N O_{2}\)

\(\quad\quad\quad\quad\quad\quad\quad {1mol}\) \(\quad\quad\quad{0}\)

At eqn. \(\quad\quad\quad\left(\begin{array}{ll}1-0.2) m o l & 0.4 m o l\end{array}\right.\)

\(\quad\quad\quad\quad\quad\quad=0.8 \mathrm{~mol}\)

Total moles after dissociation, \(n_{2}=0.8+0.4=1.2\)

Initial temperature \(=300 K P_{1} V=n_{1} R T_{1}\) or \(1 \times V=1 \times R \times 300\)

Temperature after dissociation \(=600 K\) No. of moles after dissociation, \(n_{2}-1.2 P_{2} \times V=1.2 \times R \times 600\)

Dividing eqn (ii) by eqn (i) \(P_{2}=1.2 \times \frac{600}{300}=2.4\) \(atm\)

The compounds in solid, liquid or gaseous state which carry carbon in its molecule are known as organic compounds.

It is a large class of chemical compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen.

Friedrich Wohler discovered in the early 1800s, organic compounds can be synthesized from minerals and other non-organic materials in the laboratory.

Lipids, Cellulose are the most common organic compounds.

The few carbon-containing compounds not categorised as organic include carbides, carbonates and cyanides.

Organic compounds can be broadly classified as acyclic (open chain) or cyclic (closed chain).

In the periodic table, the Actinoid element or the actinide element, any of a series of 15 consecutive chemical elements from actinium to lawrencium (atomic numbers 89–103).

As a group, Actinoid elements are significant largely because of their property of being radioactive.

Curium (Cm), a synthetic chemical element having atomic number 96 comes under the Actinoid series of the periodic table.

Actinoid elements are actinium, thorium, protactinium, uranium, neptunium, plutonium, americium, curium, berkelium, californium, einsteinium, fermium, mendelevium, nobelium and lawrencium.

Given that,

Bond length, \(\mathrm{d=2.00}\) Å \(\mathrm{=2 \times 10^{-10}~m}\)

Theoretical Dipole Moment, \(\mathrm{ =5.12 \times10^{-30}~Cm}\)

We know that,

Charge, \(\mathrm{Q=1.6 \times10^{-19}}\)

Therefore, from:

Dipole moment \(=\) Charge \(\times\) Bond Length

\(=1.6 \times10^{-19} \times 2 \times 10^{-10}~m\)

\(=32 \times 10^{-30}\mathrm{~Cm}\)

We also know that:

\(\%\)Ionic Character \(=\frac{\text{Observed Dipole Moment}}{\text{Theoritical Dipole Moment}}\times100\)

\(=\frac{5.12 \times10^{-30}}{32 \times 10^{-30}}\times 100\)

\(=0.16 \times 100\)

\(=16\%\)

Sucrose isnot an artificial sweetener. It is a naturally occurring sugar.

• Sucrose is produced naturally in plants from which table sugar is refined.
• It is disaccharide, a molecule composed of two monosaccharides: glucose and fructose.
• Sucralose,Alitame andSaccharin areartificial sweeteners.