What happens when
Question: What happens when H3PO3is heated? Solution: H3PO3,on heating, undergoes disproportionation reaction to form PH3and H3PO4. The oxidation numbers of P in H3PO3,PH3, and H3PO4are +3, 3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction....
Read More →If a chord of a circle of radius 28 cm
Question: If a chord of a circle of radius 28 cm makes an angle of 90 at the centre, then the area of the major segment is (a) $392 \mathrm{~cm}^{2}$ (b) $1456 \mathrm{~cm}^{2}$ (c) $1848 \mathrm{~cm}^{2}$ (d) $2240 \mathrm{~cm}^{2}$ Solution: Area of major segment, $=$ Area of circle $-\left[\frac{\pi \theta}{360}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right](r)^{2}$ $=\pi(28)^{2}-\left(\frac{\pi}{4}-\frac{1}{2}\right)(28)^{2}$ $=784 \pi-196(\pi-2)$ $=2240 \mathrm{~cm}^{2}$ So the answer i...
Read More →What is the basicity of
Question: What is the basicity of H3PO4? Solution: H3PO4 Since there are three OH groups present in H3PO4,its basicity is three i.e., it is a tribasic acid....
Read More →Write a balanced equation for the hydrolytic reactionz of
Question: Write a balanced equation for the hydrolytic reaction of PCl5in heavy water. Solution:...
Read More →The area of a sector whose perimeter
Question: The area of a sector whose perimeter is four times its radius r units, is (a) $\frac{r^{2}}{4}$ sq. units (b) $2 r^{2}$ sq. units (c) $r^{2}$ sq.units (d) $\frac{r^{2}}{2}$ sq. units Solution: We know that perimeter of the sector $=2 r+\frac{\theta}{360} \times 2 \pi r$. We have given that perimeter of the sector is four times the radius. $2 r+\frac{\theta}{360} \times 2 \pi r=4 r$ Subtracting 2rfrom both sides of the equation, $\therefore \frac{\theta}{360} \times 2 \pi r=4 r-2 r$ $\t...
Read More →What happens when
Question: What happens when PCl5is heated? Solution: All the bonds that are present in PCl5are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger than the axial ones. Therefore, when PCl5is heated strongly, it decomposes to form PCl3....
Read More →What happens when white phosphorus is heated with
Question: What happens when white phosphorus is heated with concentrated NaOHsolution in an inert atmosphere of CO2? Solution: White phosphorous dissolves in boiling NaOH solution (in a CO2atmosphere) to give phosphine, PH3....
Read More →Bond angle in
Question: Bond angle inis higher than that in PH3. Why? Solution: In PH3, P issp3hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated withsp3bonding is changed to pyramidal. PH3combines with a proton to formin which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. H...
Read More →What is the covalence of nitrogen in
Question: What is the covalence of nitrogen in N2O5? Solution: From the structure of N2O5, it is evident that the covalence of nitrogen is 4....
Read More →The radius of a circle is 20 cm.
Question: The radius of a circle is 20 cm. It is divided into four parts of equal area by drawing three concentric circles inside it. Then, the radius of the largest of three concentric circles drawn is (a) $10 \sqrt{5} \mathrm{~cm}$ (b) $10 \sqrt{3} \mathrm{~cm}$ (c) $10 \sqrt{5} \mathrm{~cm}$ (d) $10 \sqrt{2} \mathrm{~cm}$ Solution: The circle can be divided into four parts of equal area by drawing three concentric circles inside it as, It is given that OB = 20 cm. Let OA =x. Since the circle ...
Read More →How does ammonia react with a solution of
Question: How does ammonia react with a solution of Cu2+? Solution: NH3acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion....
Read More →Mention the conditions required to maximise the yield of ammonia.
Question: Mention the conditions required to maximise the yield of ammonia. Solution: Ammonia is prepared using the Habers process. The yield of ammonia can be maximized under the following conditions: (i)High pressure (200 atm) (ii)A temperature of700 K (iii)Use of a catalyst such as iron oxide mixed with small amounts of K2O and Al2O3...
Read More →Solve the following
Question: Why is N2less reactive at room temperature? Solution: The two N atoms in N2are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, N2is less reactive at room temperature....
Read More →Solve the following
Question: Why is BiH3the strongest reducing agent amongst all the hydrides of Group 15 elements? Solution: As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since the stability of hydrides decreases on moving from NH3to BiH3, the reducing character of the hydrides increases on moving from NH3to BiH3....
Read More →Why are pentahalides more covalent than trihalides?
Question: Why are pentahalides more covalent than trihalides? Solution: In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides....
Read More →If the perimeter of a circle is equal to that of a square,
Question: If the perimeter of a circle is equal to that of a square, then the ratio of their areas is(a) 13 : 22(b) 14 : 11(c) 22 : 13(d) 11 : 14 Solution: We have given that perimeter of circle of radiusris equal to square of sidea. $\therefore$ perimeter of the circle $=$ perimeter of the square $\therefore 2 \pi r=4 a$ $\therefore r=\frac{4 a}{2 \pi}$ $\therefore r=\frac{2 a}{\pi}$ Now we will substitute value ofrin the following equation $\frac{\text { Area of circle }}{\text { Area of squar...
Read More →If the perimeter of a circle is equal to that of a square,
Question: If the perimeter of a circle is equal to that of a square, then the ratio of their areas is(a) 13 : 22(b) 14 : 11(c) 22 : 13(d) 11 : 14 Solution: We have given that perimeter of circle of radiusris equal to square of sidea. $\therefore$ perimeter of the circle $=$ perimeter of the square $\therefore 2 \pi r=4 a$ $\therefore r=\frac{4 a}{2 \pi}$ $\therefore r=\frac{2 a}{\pi}$ Now we will substitute value ofrin the following equation $\frac{\text { Area of circle }}{\text { Area of squar...
Read More →Predict conditions under which Al might be expected to reduce MgO.
Question: Predict conditions under which Al might be expected to reduce MgO. Solution: Above 1350C, the standard Gibbs free energy of formation of Al2O3from Al is less than that of MgO from Mg. Therefore, above 1350C, Al can reduce MgO....
Read More →Two chords AB and CD of a circle intersect at a point outside the circle. Prove that
Question: Two chords AB and CD of a circle intersect at a point outside the circle. Prove that (a) $\triangle \mathrm{PAC} \sim \triangle \mathrm{PDB}$ (b) PA. PB = PC. PD Solution: Given: AB and CD are two chordsTo Prove: (a) $\triangle \mathrm{PAC} \sim \triangle \mathrm{PDB}$ (b) PA. PB $=$ PC. PD Proof: $\angle \mathrm{ABD}+\angle \mathrm{ACD}=180^{\circ}$ ........(1) (Opposite angles of a cyclic quadrilateral are supplementary) $\angle \mathrm{PCA}+\angle \mathrm{ACD}=180^{\circ}$ ............
Read More →Outline the principles of refining of metals by the following methods:
Question: Outline the principles of refining of metals by the following methods: (i)Zone refining (ii)Electrolytic refining (iii)Vapour phase refining Solution: (i) Zone refining: This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves along with it. As a r...
Read More →What is the role of graphite rod in the electrometallurgy of aluminium?
Question: What is the role of graphite rod in the electrometallurgy of aluminium? Solution: In the electrometallurgy of aluminium, a fused mixture of purified alumina (Al2O3), cryolite (Na3AlF6) and fluorspar (CaF2) is electrolysed. In this electrolysis, graphite is used as the anode and graphite-lined iron is used as the cathode. During the electrolysis, Al is liberated at the cathode, while CO and CO2are liberated at the anode, according to the following equation. If a metal is used instead of...
Read More →Name the processes from which chlorine is obtained as a by-product.
Question: Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis? Solution: In the electrolysis of molten NaCl, Cl2is obtained at the anode as a by product. $\mathrm{NaCl}_{\text {(melt) }} \longrightarrow \mathrm{Na}_{\text {(melt) }}^{+}+\mathrm{Cl}^{-}{ }_{\text {(melt) }}$ At cathode: $\mathrm{Na}^{+}{ }_{(\text {melt })}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}_{(s)}$ At anode: $\mathrm{Cl}^{-}{ }_{(...
Read More →In the following figure, the area of the shaded region is
Question: In the following figure, the area of the shaded region is (a) $3 \pi \mathrm{cm}^{2}$ (b) $6 \pi \mathrm{cm}^{2}$ (c) $9 \pi \mathrm{cm}^{2}$ (d) $7 \pi \mathrm{cm}^{2}$ Solution: In the figure, $\angle C=\angle B=90^{\circ}$ and $\angle D=60^{\circ}$, $\therefore \angle A+\angle B+\angle C+\angle D=360^{\circ}$ $\angle A+90^{\circ}+90^{\circ}+60^{\circ}=360^{\circ}$ $\therefore \angle A=120^{\circ}$ Area of shaded region $=\frac{\theta}{360} \times \pi r^{2}$ $\therefore$ Area of shad...
Read More →In a circle, two chords AB and CD intersect at a point inside the circle. Prove that
Question: In a circle, two chords AB and CD intersect at a point inside the circle. Prove that (a) $\triangle \mathrm{PAC} \sim \triangle \mathrm{PDB}$ (b) PA.PB = PC. PD Solution: Given: AB and CD are two chordsTo Prove: (a) $\triangle \mathrm{PAC} \sim \triangle \mathrm{PDB}$ (b) $\mathrm{PA} . \mathrm{PB}=\mathrm{PC} . \mathrm{PD}$ Proof: In $\triangle \mathrm{PAC}$ and $\triangle \mathrm{PDB}$ $\angle \mathrm{APC}=\angle \mathrm{DPB}$ (Vertically Opposite angles) $\angle \mathrm{CAP}=\angle ...
Read More →In the following figure, If ABC is an equilateral triangle,
Question: In the following figure, If ABC is an equilateral triangle, then shaded area is equal to (a) $\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right) r^{2}$ (b) $\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) r^{2}$ (c) $\left(\frac{\pi}{3}+\frac{\sqrt{3}}{4}\right) r^{2}$ (d) $\left(\frac{\pi}{3}+\sqrt{3}\right) r^{2}$ Solution: We have given that ABC is an equilateral triangle. $\therefore \angle A=60^{\circ}$ As we know that, $\angle B C A=\frac{1}{2} m(\angle B O C)$. $\therefore 60^{\circ}=\...
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