What role does adsorption play in heterogeneous catalysis?
Question: What role does adsorption play in heterogeneous catalysis? Solution: Heterogeneous catalysis: A catalytic process in which the catalyst and the reactants are present in different phases is known as a heterogeneous catalysis. This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps: (i)Adsorption of reactant molecules on the catalyst surface. (ii)Occurrence of a chemical reaction through the formation...
Read More →The area of the largest triangle that
Question: The area of the largest triangle that can be inscribed in a semi-circle of radius r is (a) $2 r$ (b) $r^{2}$ (c) $r$ (d) $\sqrt{r}$ Solution: The triangle with the largest area will be symmetrical as shown in the figure. Let the radius of the circle ber. Hence, $\operatorname{ar}(\triangle \mathrm{ABC})=\frac{1}{2}(r)(2 r)$ $=r^{2}$ sq. unit Therefore the answer is (b)....
Read More →What do you understand by activation of adsorbent?
Question: What do you understand by activation of adsorbent? How is it achieved? Solution: By activating an adsorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are: (i)By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it. (ii)Some specific treatments can also lead to the activation of the adsorbent. For example, wood charcoal is activated by heating it between 650 K and 1330 K in v...
Read More →In the given figure, ∠ABC = 90° and BD ⊥ AC.
Question: In the given figure, ABC= 90 andBDAC. IfBD= 8 cm,AD= 4 cm, findCD. Solution: It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse. In $\triangle D B A$ and $\triangle D C B$, we have : $\angle B D A=\angle C D B$ $\angle D B A=\angle D C B=90^{\circ}$ Therefore, by AA similarity theorem, we get: $\triangle D B A \sim \triangle D C B$ $\Rightarrow \frac{B D}{C D}=\frac{A D}{B D}$ $\Rightarrow C D=\frac{B D^{2}}{A D}$ $C D=\f...
Read More →What is an adsorption isotherm?
Question: What is an adsorption isotherm? Describe Freundlich adsorption isotherm Solution: The plot between the extent of adsorption $\left(\frac{x}{m}\right)$ against the pressure of gas $(P)$ at constant temperature $(T)$ is called the adsorption isotherm. Freundlich adsorption isotherm: Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature. From the given plot it is clear ...
Read More →If the area of a square is same as the area of a circle,
Question: If the area of a square is same as the area of a circle, then the ratio of their perimeters, in terms of , is (a) $\pi: \sqrt{3}$ (b) $2: \sqrt{\pi}$ (c) $3: \pi$ (d) $\pi: \sqrt{2}$ Solution: We have given that area of a circle of radiusris equal to the area of a square of sidea. $\therefore \pi r^{2}=a^{2}$ $\therefore a=\sqrt{\pi} r$ We have to find the ratio of the perimeters of circle and square. $\therefore \frac{\text { Perimeter of circle }}{\text { Perimeter of square }}=\frac...
Read More →In the given figure, ∠ABC = 90° and BD ⊥ AC.
Question: In the given figure, ABC= 90 andBDAC. IfAB= 5.7 cm,BD= 3.8 cm andCD= 5.4 cm, findBC. Solution: It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse. In $\triangle B D C$ and $\triangle A B C$, we have : $\angle A B C=\angle B D C=90^{\circ}$ (given) $\angle C=\angle C$ (common) By AA similarity theorem, we get: $\triangle B D C \sim \triangle A B C$ $\frac{A B}{B D}=\frac{B C}{D C}$ $\Rightarrow \frac{5.7}{3.8}=\frac{B C}{5...
Read More →In the given figure, ∠CAB = 90° and AD ⊥ BC. Show that ∆ BDA ∼ ∆ BAC. If AC = 75 cm,
Question: In the given figure, CAB= 90 andADBC. Show that ∆BDA ∆BAC. IfAC= 75 cm,AB= 1 m andBC= 1.25 m, findAD. Solution: In $\triangle B D A$ and $\triangle B A C$, we have : $\angle B D A=\angle B A C=90^{\circ}$ $\angle D B A=\angle C B A \quad($ Common $)$ Therefore, by AA similarity theorem, $\triangle B D A \sim \triangle B A C$ $\Rightarrow \frac{A D}{A C}=\frac{A B}{B C}$ $\Rightarrow \frac{A D}{0.75}=\frac{1}{1.25}$ $\Rightarrow A D=\frac{0.75}{1.25}$ $=0.6 \mathrm{~m}$ or $60 \mathrm{~...
Read More →What are the factors which influence the adsorption of a gas on a solid?
Question: What are the factors which influence the adsorption of a gas on a solid? Solution: There are various factors that affect the rate of adsorption of a gas on a solid surface. (1) Nature of the gas: Easily liquefiable gases such as NH3, HCl etc. are adsorbed to a great extent in comparison to gases such as H2, O2etc. This is because Van der Waals forces are stronger in easily liquefiable gases. (2) Surface area of the solid The greater the surface area of the adsorbent, the greater is the...
Read More →Give reason why a finely divided substance is more effective as an adsorbent.
Question: Give reason why a finely divided substance is more effective as an adsorbent. Solution: Adsorption is a surface phenomenon. Therefore, adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent....
Read More →What is the difference between physisorption and chemisorption?
Question: What is the difference between physisorption and chemisorption? Solution:...
Read More →The corresponding sides of two similar triangles ABC and DEF are BC = 9.1 cm and EF = 6.5 cm.
Question: The corresponding sides of two similar trianglesABCandDEFareBC= 9.1 cm andEF= 6.5 cm. If the perimeter of ∆DEFis 25 cm, find the perimeter of ∆ABC. Solution: It is given that $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$. Therefore, their corresponding sides will be proportional.Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides. $\Rightarrow \frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle D E F...
Read More →Distinguish between the meaning of the terms adsorption and absorption.
Question: Distinguish between the meaning of the terms adsorption and absorption. Give one example of each. Solution: Adsorptionis a surface phenomenon of accumulation of molecules of a substance at the surface rather than in the bulk of a solid or liquid. The substance that gets adsorbed is called the adsorbate and the substance on whose surface the adsorption takes place is called the adsorbent. Here, the concentration of the adsorbate on the surface of the adsorbent increases. In adsorption, ...
Read More →Why is it essential to wash the precipitate with water before estimating it quantitatively?
Question: Why is it essential to wash the precipitate with water before estimating it quantitatively? Solution: When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities....
Read More →What modification can you suggest in the Hardy-Schulze law?
Question: What modification can you suggest in the Hardy-Schulze law? Solution: Hardy-Schulze law states that the greater the valence of the flocculating ion added, the greater is its power to cause precipitation. This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze l...
Read More →What is the role of desorption in the process of catalysis?
Question: What is the role of desorption in the process of catalysis? Solution: The role of desorption in the process of catalysis is to make the surface of the solid catalyst free for the fresh adsorption of the reactants on the surface....
Read More →Why is the ester hydrolysis slow in the beginning
Question: Why is the ester hydrolysis slow in the beginning and becomes faster after sometime? Solution: Ester hydrolysis can be represented as: Ester $+$ Water $\longrightarrow$ Acid $+$ Alcohol The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts....
Read More →The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm respectively.
Question: The perimeters of two similar trianglesABCandPQRare 32 cm and 24 cm respectively. IfPQ= 12 cm. findAB. Solution: It is given that triangles ABC and PQR are similar.Therefore, $\frac{\text { Perimeter }(\triangle A B C)}{\text { Perimeter }(\triangle P Q R)}=\frac{A B}{P Q}$ $\Rightarrow \frac{32}{24}=\frac{A B}{12}$ $\Rightarrow A B=\frac{32 \times 12}{24}=16 \mathrm{~cm}$...
Read More →Why is it necessary to remove CO when ammonia is obtained by
Question: Why is it necessary to remove CO when ammonia is obtained by Habers process? Solution: It is important to remove CO in the synthesis of ammonia as CO adversely affects the activity of the iron catalyst, used in Habers process....
Read More →Why are powdered substances more effective adsorbents
Question: Why are powdered substances more effective adsorbents than their crystalline forms? Solution: Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increases and physisorption is directly proportional to the surface area of the adsorbent....
Read More →In the given figure, if ∠ADE = ∠B, show that ∆ADE ∼ ∆ABC.
Question: In the given figure, if ADE= B, show that ∆ADE ∆ABC. IfAD= 3.8 cm,AE= 3.6 cm,BE= 2.1 cm andBC= 4.2 cm, findDE. Solution: Given: $\angle A D E=\angle A B C$ and $\angle A=\angle A$ LetDEbexcm Therefore, by AA similarity theorem, $\triangle A D E \sim \triangle A B C$ $\Rightarrow \frac{A D}{A B}=\frac{D E}{B C}$ $\Rightarrow \frac{3.8}{3.6+2.1}=\frac{x}{4.2}$ $\Rightarrow x=\frac{3.8 \times 4.2}{5.7}=2.8$ Hence,DE= 2.8 cm...
Read More →Why does physisorption decrease with the increase of temperature?
Question: Why does physisorption decrease with the increase of temperature? Solution: Physisorption is exothermic in nature. Therefore, in accordance with Le-Chatelieres principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature....
Read More →Write any two characteristics of Chemisorption.
Question: Write any two characteristics of Chemisorption. Solution: 1.Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate. 2.Like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent....
Read More →In the given figure
Question: In the given figure $\triangle O A B \sim \triangle O C D$. If $A B=8 \mathrm{~cm}, B O=6.4 \mathrm{~cm}, O C=3.5$ and $C D=5 \mathrm{~cm}$, Find (i)OA(ii)DO. Solution: (i) LetOAbexcm. $\because \triangle O A B \sim \triangle O C D$ $\therefore \frac{O A}{O C}=\frac{A B}{C D}$ $\Rightarrow \frac{x}{3.5}=\frac{8}{5}$ $\Rightarrow x=\frac{8 \times 3.5}{5}=5.6$ Hence,OA= 5.6 cm(ii) LetODbeycm $\because \triangle O A B \sim \triangle O C D$ $\therefore \frac{A B}{C D}=\frac{O B}{O D}$ $\Ri...
Read More →If the radius of a circle is diminished by 10%,
Question: If the radius of a circle is diminished by 10%, then its area is diminished by(a) 10%(b) 19%(c) 20%(d) 36% Solution: Letxbe the initial radius of the circle. Therefore, its area is $\pi x^{2}$ .(1) It is given that the radius is diminished by 10%, therefore, its new radius is calculated as shown below, new radius $=x-0.10 x$ $\therefore$ new radius $=0.90 x$ $\therefore$ new area $=(0.90 x)^{2}$ $\therefore$ new area $=0.81 x^{2}$ Now we will find the percentage decreased in the area. ...
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