Describe the manufacture of
Question: Describe the manufacture of H2SO4by contact process? Solution: Sulphuric acid is manufactured by the contact process. It involves the following steps: Step (i): Sulphur or sulphide ores are burnt in air to form SO2. Step (ii): By a reaction with oxygen, SO2is converted into SO3in the presence of V2O5as a catalyst. $2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \stackrel{\mathrm{v}_{3} \mathrm{O}_{3}}{\longrightarrow} 2 \mathrm{SO}_{3(g)}$ Step (iii): SO3produced is absorbed on H2SO4to give H2...
Read More →Describe the manufacture of
Question: Describe the manufacture of H2SO4by contact process? Solution: Sulphuric acid is manufactured by the contact process. It involves the following steps: Step (i): Sulphur or sulphide ores are burnt in air to form SO2. Step (ii): By a reaction with oxygen, SO2is converted into SO3in the presence of V2O5as a catalyst. $2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \stackrel{\mathrm{v}_{3} \mathrm{O}_{3}}{\longrightarrow} 2 \mathrm{SO}_{3(g)}$ Step (iii): SO3produced is absorbed on H2SO4to give H2...
Read More →∆ABC is right-angled at A and AD ⊥ BC.
Question: ∆ABCis right-angled atAandADBC. IfBC= 13 cm andAC= 5 cm, find the ratio of the areas of ∆ABCand ∆ADC. Solution: In $\triangle A B C$ and $\triangle A D C$, we have: $\angle B A C=\angle A D C=90^{\circ}$ $\angle A C B=\angle A C D \quad$ (common) By AA similarity, we can conclude that $\triangle B A C \sim \triangle A D C$. Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides. $\therefore \frac{\operatorname{ar}(\Delta B A C)}{\...
Read More →Which aerosols deplete ozone?
Question: Which aerosols deplete ozone? Solution: Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen....
Read More →Knowing the electron gain enthalpy values for
Question: Knowing the electron gain enthalpy values for O Oand O O2as 141 and 702 kJ mol1respectively, how can you account for the formation of a large number of oxides having O2species and not O? (Hint: Consider lattice energy factor in the formation of compounds). Solution: Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be. Lattice energy is directly proportional to the charge carried by an ion. When a metal combines wit...
Read More →Why is dioxygen a gas but sulphur a solid?
Question: Why is dioxygen a gas but sulphur a solid? Solution: Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively formppbonds and form O2(O==O) molecule. Also, the intermolecular forces in oxygen are weak van der Walls, which cause it to exist as gas. On the other hand, sulphur does not form M2molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid....
Read More →In the given figure, DE ∥ BC. If DE = 3 cm, BC = 6 cm and ar(∆ADE) = 15 cm2,
Question: In the given figure,DE∥BC. IfDE= 3 cm, BC = 6 cm and ar(∆ADE) = 15 cm2, find the area of ∆ABC. Solution: It is given thatDE∥ BC $\therefore \angle A D E=\angle A B C$ (Corresponding angles) $\angle A E D=\angle A C B$ (Corresponding angles) By AA similarity, we can conclude that $\triangle A D E \sim \triangle A B C$. $\therefore \frac{\operatorname{ar}(\Delta A D E)}{\operatorname{ar}(\Delta A B C)}=\frac{D E^{2}}{B C^{2}}$ $\Rightarrow \frac{15}{\operatorname{ar}(\Delta A B C)}=\frac...
Read More →The sum of the radii of two circles is 140 cm
Question: The sum of the radii of two circles is 140 cm and the difference of their circumferences is 88 cm. Find the diameters of the circles. Solution: Let the radius of two circles be $r_{1} \mathrm{~cm}$ and $r_{2} \mathrm{~cm}$ respectively. Then their circumferences are $C_{1}=2 \pi r_{1} \mathrm{~cm}$ and $C_{2}=2 \pi r_{2} \mathrm{~cm}$ respectively and their areas are $A_{1}=\pi r_{1}^{2} \mathrm{~cm}^{2}$ and $A_{2}=\pi r_{2}^{2} \mathrm{~cm}^{2}$ respectively. It is given that the sum...
Read More →Justify the placement of O, S, Se,
Question: Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation. Solution: The elements of group 16 are collectively called chalcogens. (i)Elements of group 16 have six valence electrons each. The general electronic configuration of these elements isns2np4, wherenvaries from 2 to 6. (ii)Oxidation state: Asthese elements have six valence electrons (ns2np4), they should display an oxidation s...
Read More →Solve the following
Question: Can PCl5act as an oxidising as well as a reducing agent? Justify. Solution: PCl5can only act as an oxidizing agent. The highest oxidation state that P can show is +5. In PCl5, phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation state and act as an oxidizing agent....
Read More →The side of a square is 10 cm.
Question: The side of a square is 10 cm. Find the area of circumscribed and inscribed circles. Solution: It is given that the side of square is 10 cm. So, the diameter of circle inscribed the square is 10 cm. We know that the areaAof circle inscribed the square is $A=\pi r^{2}$ Substituting the value of radius of inscribed circle $r=5 \mathrm{~cm}$, $A=3.14 \times 5 \times 5$ $=78.5 \mathrm{~cm}^{2}$ Hence the area of circle inscribed the square is $78.5 \mathrm{~cm}^{2}$ Now we will find the di...
Read More →Give the disproportionation reaction of
Question: Give the disproportionation reaction of H3PO3. Solution: On heating, orthophosphorus acid (H3PO3) disproportionates to give orthophosphoric acid (H3PO4) and phosphine (PH3). The oxidation states of P in various species involved in the reaction are mentioned below....
Read More →In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm,
Question: In the given figure, $A B C$ is a triangle and $P Q$ is a straight line meeting $A B$ in $P$ and $A C$ in $Q$. If $A P=1 \mathrm{~cm}, P B=3 \mathrm{~cm}, A Q=1.5 \mathrm{~cm}, Q C=4.5 \mathrm{~cm}$, prove that are of $\triangle A P Q$ is $\frac{1}{16}$ of the area of $\triangle A B C$. Solution: We have: $\frac{A P}{A B}=\frac{1}{1+3}=\frac{1}{4}$ and $\frac{A Q}{A C}=\frac{1.5}{1.5+4.5}=\frac{1.5}{6}=\frac{1}{4}$ $\Rightarrow \frac{A P}{A B}=\frac{A Q}{A C}$ Also, $\angle A=\angle A$...
Read More →Why does nitrogen show catenation properties less than phosphorus?
Question: Why does nitrogen show catenation properties less than phosphorus? Solution: Catenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness ofthe NN single bond as compared to the PP single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the NN single bond....
Read More →Write main differences between the properties of
Question: Write main differences between the properties of white phosphorus and red phosphorus. Solution:...
Read More →The circumference of two circles are in
Question: The circumference of two circles are in the ratio 2 : 3. Find the ratio of their areas. Solution: Let the radius of two circles beandrespectively. Then their circumferences areandrespectively and their areas areandrespectively. It is given that, $\frac{C_{1}}{C_{2}}=\frac{2}{3}$ $\frac{2 \pi r_{1}}{2 \pi r_{2}}=\frac{2}{3}$ $\frac{r_{1}}{r_{2}}=\frac{2}{3}$ Now we will calculate the ratio of their areas, $\frac{A_{1}}{A_{2}}=\frac{\pi r_{1}^{2}}{\pi r_{2}^{2}}$ $=\frac{r_{1}^{2}}{r_{2}...
Read More →Nitrogen exists as diatomic molecule and phosphorus
Question: Nitrogen exists as diatomic molecule and phosphorus as P4. Why? Solution: Nitrogen owing to its small size has a tendency to formppmultiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule, N2. On moving down a group, the tendency to formppbonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the P4state....
Read More →Solve the following
Question: Explain why NH3is basic while BiH3is only feebly basic. Solution: NH3is distinctly basic while BiH3is feebly basic. Nitrogen has a small size due to which the lone pair of electronsis concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides d...
Read More →The areas of two similar triangles are 100 cm2 and 64 cm2 respectively.
Question: The areas of two similar triangles are 100 cm2and 64 cm2respectively. If a median of the similar triangle is 5.6 cm, find the corresponding median of the other. Solution: Let the two triangles beABCandPQRwith mediansAMandPN,respectively. Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding medians. $\therefore \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{A M^{2}}{P N^{2}}$ $\Rightarrow \frac...
Read More →Explain why
Question: Explain why NH3is basic while BiH3is only feebly basic. Solution: NH3is distinctly basic while BiH3is feebly basic. Nitrogen has a small size due to which the lone pair of electronsis concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides d...
Read More →Solve the following
Question: Why does R3P=O exist but R3N=O does not (R = alkyl group)? Solution: N(unlike P) lacks thed-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence,R3N=O does not exist....
Read More →The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?
Question: The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint:Can be explained on the basis ofsp3hybridisation in NH3and onlyspbonding between hydrogen and other elements of the group]. Solution: Hydride NH3PH3AsH3SbH3 HMH angle 107 92 91 90 The above trend in the HMH bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion bet...
Read More →A sheet of paper is in the form of a rectangle ABCD
Question: A sheet of paper is in the form of a rectangle ABCD in which AB = 40 cm and AD = 28 cm. A semi-circular portion with BC as diameter is cut off. Find the area of the remaining paper. Solution: The length and width of rectangleABCDis given byandrespectively. Now, we will find the area of rectangle. Area of rectangle $=l \times w$ $=40 \times 28$ $=1120 \mathrm{~cm}^{2}$ It is given that a semicircular portion with BC as diameter is cutoff from rectangle. So, radius of semicircle $=\frac{...
Read More →The areas of two similar triangles are 81 cm2 and 49 cm2, respectively. If the altitude of one triangle is 6.3 cm,
Question: The areas of two similar triangles are 81 cm2and 49 cm2, respectively. If the altitude of one triangle is 6.3 cm, find the corresponding altitude of the other triangle. Solution: It is given that the triangles are similar.Therefore, theratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.Let the two triangles be ABC and ...
Read More →Give the resonating structures of
Question: Give the resonating structures ofNO2andN2O5. Solution: (1) (2)...
Read More →