Chemistry Test - 9

The substance that decreases the activity of catalysts is poisoner.

Promoters/Activators: Substance which themselves are not catalyst but its presence can increase the catalytic activity of catalyst. A promoters increase the number of active sites on the surface.

Catalytic Poisons/ Anti catalysts/ Catalyst Inhibitor are those substances which themselves are not catalyst but whose presence decreases the activity of the catalyst. Poisoning is due to preferential adsorption of poison on the surface of the catalyst.

Increasing the temperature of the substance, increases the fraction of molecules which collide with energies greater than Ea. For every \(10^{\circ} \mathrm{C}\) rise in temperature, the fraction of molecules having energy equal to or greater than Ea gets doubled leading to doubling the rate of reaction.

So, the number of fruitful collisions gets doubled.

Molar mass of \({KI}=39+127=166 {~g} {~mol}^{-1}\)

\(20 \%\) aqueous solution of KI means \(20 {~g}\) of \({KI}\) is present in \(100 {~g}\) of solution.

Therefore,

Mass of \(KI=20 {~g}\)

That is,

\(20 {~g}\) of \({KI}\) is present in \((100-20) {g}\) of water \(=80 {~g}\) of water

We know that:

Molality of the solution \(=\frac{\text { Moles of } K I}{\text { Mass of water in } {kg}}\)

\(=\frac{\frac{\text {Mass of KI}}{\text {Molar Mass of KI}} }{\text { Mass of water in } {kg}}\)

\(=\frac{\frac{20}{166}}{0.08} {~m}\)

\(=1.506 {~m}\)

\(=1.51 {~m}\)

Matters are generally categorized into 3 states:

• Solid: Particles are held very close to each other and can't move.
• Liquid: Particles are close to each other but they can move around.
• Gas: Particles are far from each other as compared to those present in solid or liquid states and their movement is easy and fast. 

Two more states are discovered under extreme circumstances:

• Plasma: A superheated matter, so hot that the electrons are ripped away from the atoms forming an ionized gas. 
• Bose-Einstein Condensate: A group of atoms cooled to absolute zero temperature (-273.15 K).

Given,

Molecular formula =Be_nAl₂Si₆O₁₈

The oxidation states of each element of given molecular formula:

Be = +2

Al = +3

Si = +4

O = -2

(2n) + (3 × 2) + (4 + 6) + (−2 × 18) = 0

⇒2n + 30 − 36 = 0

⇒2n = 6

⇒n = 3

Given:

Mass of acetic acid, \(\mathrm{w_1}=75 \mathrm{~g}\) 

Molar mass of ascorbic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\),

\(\mathrm{M_2}=6 \times 12+8 \times 1+6 \times 16\)

\(=176 \mathrm{~g} \mathrm{~mol}^{-1}\)

Lowering of melting point, 

\(\Delta \mathrm{T}_{\mathrm{f}}=(1.5+273)-(0+273)=1.5 \mathrm{~K}\)

We know that:

\(\Delta \mathrm{T}_{\mathrm{f}}=\frac{\mathrm{K}_{\mathrm{f}} \times 1000 \times \mathrm{w}_{2}}{\mathrm{M}_{2} \times \mathrm{w}_{1}}\)

\(\mathrm{w}_{2}=\frac{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{M}_{2} \times \mathrm{w}_{1}}{\mathrm{~K}_{\mathrm{f}} \times 1000}\)

\(\mathrm{w}_{2}=\frac{1.5 \times 176 \times 75}{3.9 \times 1000}\)

\(\mathrm{w}_{2}=5.08 \mathrm{~g}\)

Therefore, \(5.08 \mathrm{~g}\) of ascorbic acid is needed to be dissolved.

Coal is the chief source ofAromatic hydrocarbons.

Simple aromatic hydrocarbons come from two main sources: Coal and petroleum. Coal is a complex mixture of a large number of compounds, most of which are long-chain compounds. If coal is heated to about 1000°C in the absence of air (oxygen), volatile components, the so-called tar oil, are stripped out.

Given that:

Vapour density \( =39\)

Molecular formula \(= (C H)_{n}\)

Then,Molar mass \(=(12+1) \times n=13 n\)

We know that:Molar mass \(=2 \times\) vapour density

\(13 n=2 \times 39\)

\(n=6\)

Therefore, the compound is \((C H)_{6}\) or \(C_{6} H_{6}\)

Using ideal gas equation,

\(P V=n R T\)

\(P V=\frac{w}{M} R T\)

\(P=\frac{w}{M} \frac{R T}{V}\)

As per question the formula will be,

\(P=\left[\frac{w_{C}+w_{S_{2}}+w_{C S_{2}}}{M_{C}+M_{S_{2}}+M_{C S_{2}}}\right] \times \frac{R T}{V}\)

where,

\(P=\) total pressure \(=?\)

\(\mathrm{R}=\) gas constant

\(\mathrm{T}=\) temperature \(=1027^{\circ} \mathrm{C}=273+1027=1300 \mathrm{~K}\)

\(V=\) volume \(=13 \mathrm{~L}\)

\(w_{c}=\) mass of \(C=12 \mathrm{~g}\)

\(w_{S_{2}}=\) mass of \(S_{2}=64 \mathrm{~g}\)

\(w_{C S_{2}}=\) mass of \(C S_{2}=76 \mathrm{~g}\)

\(M_{c}=\) molar mass of \(\mathrm{C}=12 \mathrm{~g} / \mathrm{mole}\)

\(M_{S_{2}}=\) molar mass of \(S_{2}=64 \mathrm{~g} / \mathrm{mole}\)

\(M_{C S_{2}}=\) molar mass of \(C S_{2}=76 \mathrm{~g} / \mathrm{mole}\)

Now put all the given values in the above formula, we get:

\(P=\frac{12 g+64 g+76 g}{12 g / \text { mole }+64 g / \text { mole }+76 g / \text { mole }} \times \frac{R \times 1300 K}{13 L}\)

\(P=100 R\)

Therefore, the total pressure is \(100 R\).

For an elementary reaction, the molecularity is equal to the order of the reaction. Thus, the rate law for the given reaction will be:

\(\mathrm{r}=\mathrm{k}[\mathrm{A}]^{2}[\mathrm{B}]\)

Now, the concentration here is basically moles per unit volume.

So, decreasing the volume by half will double the concentration of both the reactants.

Thus, the rates can be compared as:

\(r_{1}=k[A]^{2}[\mathrm{~B}]\)

\(\mathrm{r}_{2}=\mathrm{k}[2 \mathrm{~A}]^{2}[2 \mathrm{~B}]\)

\(=8 \mathrm{k}[A]^{2}[B]\)

\(=8 \mathrm{r}_{1}\)

Therefore, rate increases by eight times.