The choice of a reducing agent in a particular case depends onm thermodynamic factor.
Question: The choice of a reducing agent in a particular case depends onm thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples. Solution: The above figure is a plot of Gibbs energyvs. T for formation of some oxides. It can be observed from the above graph that a metal can reduce the oxide of other metals, if the standard free energy of formationof the oxide of the former is more negative than the latter. For example, sinceis more negative than, A...
Read More →Out of C and CO,
Question: Out of C and CO, which is a better reducing agent for ZnO ? Solution: Reduction of ZnO to Zn is usually carried out at 1673 K. From the above figure, it can be observed that above 1073 K, the Gibbs free energy of formation of CO from C and above 1273 K, the Gibbs free energy of formation of CO2from C is lesser than the Gibbs free energy of formation of ZnO. Therefore, C can easily reduce ZnO to Zn. On the other hand, the Gibbs free energy of formation of CO2from CO is always higher tha...
Read More →ABCD is a quadrilateral in which AD = BC. If P, Q, R , S be the mid points of AB,
Question: ABCD is a quadrilateral in which AD = BC. If P, Q, R , S be the mid points of AB, AC, CD and BD respectively, show that PQRS is a rhombus. Solution: In △ABC, P and Q are mid points of AB and AC respectively. So, $\mathrm{PQ} \| \mathrm{BC}$, and $\mathrm{PQ}=\frac{1}{2} \mathrm{BC}$ ..........(1) Similarly, In $\triangle \mathrm{ADC}, \mathrm{QR}=\frac{1}{2} \mathrm{AD}=\frac{1}{2} \mathrm{BC}$ ............(2) Now, In $\triangle \mathrm{BCD}, \mathrm{SR}=\frac{1}{2} \mathrm{BC}$ .........
Read More →The value of
Question: The value of ${ }_{\mathrm{l}} \mathrm{G}^{\theta}$ for formation of $\mathrm{Cr}_{2} \mathrm{O}_{3}$ is $-540 \mathrm{kJmol}^{-1}$ and that of $\mathrm{Al}_{2} \mathrm{O}_{3}$ is $-827 \mathrm{kJmol}^{-1} .$ Is the reduction of $\mathrm{Cr}_{2} \mathrm{O}_{3}$ possible with Al? Solution: The value of $\Delta_{1} \mathrm{G}^{\theta}$ for the formation of $\mathrm{Cr}_{2} \mathrm{O}_{3}$ from $\mathrm{Cr}\left(-540 \mathrm{kJmol}^{-1}\right)$ is higher than that of $\mathrm{Al}_{2} \mat...
Read More →If he area of a sector of a circle is
Question: If he area of a sector of a circle is $\frac{7}{20}$ of the area of the circle, then the sector angle is equal to (a) $110^{\circ}$ (b) $130^{\circ}$ (c) $100^{\circ}$ (d) $126^{\circ}$ Solution: We have given that area of the sector is $\frac{7}{20}$ of the area of the circle. Therefore, area of the sector $=\frac{7}{20} \times$ area of the circle $\therefore \frac{\theta}{360} \times \pi r^{2}=\frac{7}{20} \times \pi r^{2}$ Now we will simplify the equation as below, $\frac{\theta}{3...
Read More →Why is zinc not extracted from
Question: Why is zinc not extracted from zinc oxide through reduction using CO? Solution: The standard Gibbs free energy of formation of ZnO from Zn is lower than that of CO2from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from ZnO through reduction using CO....
Read More →If the area of a sector of a circle is
Question: If the area of a sector of a circle is $\frac{5}{18}$ of the area of the circle, then the sector angle is equal to (a) $60^{\circ}$ (b) $90^{\circ}$ (c) $100^{\circ}$ (d) $120^{\circ}$ Solution: We have given that area of the sector is $\frac{5}{18}$ of the area of the circle. Therefore, area of the sector $=\frac{5}{18} \times$ area of the circle $\therefore \frac{\theta}{360} \times \pi r^{2}=\frac{5}{18} \times \pi r^{2}$ Now we will simplify the equation as below, $\frac{\theta}{36...
Read More →How is leaching carried out in case of
Question: How is leaching carried out in case of low grade copper ores? Solution: In case of low grade copper ores, leaching is carried out using acid or bacteria in the presence of air. In this process, copper goes into the solution as Cu2+ions. The resulting solution is treated with scrap iron or H2to get metallic copper....
Read More →In the following figure, the ratio of the areas of two sectors S1 and S2 is
Question: In the following figure, the ratio of the areas of two sectors S1and S2is (a) 5 : 2(b) 3 : 5(c) 5 : 3(d) 4 : 5 Solution: Area of the sector, $S_{1}=\frac{\theta_{1}}{360} \times \pi r^{2}$ Area of the sector, $S_{2}=\frac{\theta_{2}}{360} \times \pi r^{2}$ Now we will take the ratio, $\frac{S_{1}}{S_{2}}=\frac{\frac{\theta_{1}}{360} \times \pi r^{2}}{\frac{\theta_{2}}{360} \times \pi r^{2}}$ Now we will simplify the ratio as below, $\frac{S_{1}}{S_{2}}=\frac{\theta_{1}}{\theta_{2}}$ Su...
Read More →What is the role of cryolite in the metallurgy of aluminium?
Question: What is the role of cryolite in the metallurgy of aluminium? Solution: Cryolite (Na3AlF6) has two roles in the metallurgy of aluminium: 1.To decrease the melting point of the mixture from 2323 K to 1140 K. 2.To increase the electrical conductivity of Al2O3....
Read More →Why copper matte is put in silica lined converter?
Question: Why coppermatteis put in silica lined converter? Solution: Coppermattecontains Cu2S and FeS. Coppermatteis put in a silica-lined converter to remove the remaining FeO and FeS present in thematteas slag (FeSiO3). Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining FeS and FeO are converted to iron silicate (FeSiO3) and Cu2S is converted into metallic copper....
Read More →If the area of a sector of a circle bounded by an arc
Question: If the area of a sector of a circle bounded by an arc of length 5 cm is equal to 20 cm2, then the radius of the circle(a) 12 cm(b) 16 cm(c) 8 cm(d) 10 cm Solution: We have given length of the arc and area of the sector bounded by that arc and we are asked to find the radius of the circle. Iflis the length of the arc, A is the area of the arc andris the radius of the circle, then we know the expression of the area of the sector in terms of the length of the arc and radius of the circle....
Read More →Differentiate between “minerals” and “ores”.
Question: Differentiate between minerals and ores. Solution: Minerals are naturally occurring chemical substances containing metals. They are found in the Earths crust and are obtained by mining. Ores are rocks and minerals viable to be used as a source of metal. For example, there are many minerals containing zinc, but zinc cannot be extracted profitably (conveniently and economically) from all these minerals. Zinc can be obtained from zinc blende (ZnS), calamine (ZnCO3), Zincite (ZnO) etc. Thu...
Read More →How is ‘cast iron’ different from ‘pig iron”?
Question: How is cast iron different from pig iron? Solution: The iron obtained from blast furnaces is known as pig iron. It contains around 4% carbon and many impurities such as S, P, Si, Mn in smaller amounts. Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3%) than pig iron. Unlike pig iron, cast iron is extremely hard and brittle....
Read More →In the following figure, the area of segment ACB is
Question: In the following figure, the area of segmentACBis (a) $\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) r^{2}$ (b) $\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right) r^{2}$ (c) $\left(\frac{\pi}{3}-\frac{\sqrt{2}}{3}\right) r^{2}$ (d) None of these Solution: We have to find area of segment ACB. Area of the $\mathrm{ACB}$ segment $=\left(\frac{\pi \theta}{360}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right) r^{2}$ We know that $\theta=120^{\circ}$. Substituting the values we get, Area of th...
Read More →Giving examples,
Question: Giving examples, differentiate between roasting and calcination. Solution: Roasting is the process of converting sulphide ores to oxides by heating the ores in a regular supply of air at a temperature below the melting point of the metal. For example, sulphide ores of Zn, Pb, and Cu are converted to their respective oxides by this process. On the other hand, calcination is the process of converting hydroxide and carbonate ores to oxides by heating the ores either in the absence or in a...
Read More →How can you separate alumina from silica in bauxite ore associated with silica?
Question: How can you separate alumina from silica in bauxite ore associated with silica? Give equations, if any. Solution: To separate alumina from silica in bauxite ore associated with silica, first the powdered ore is digested with a concentrated NaOH solution at 473 523 K and 35 36 bar pressure. This results in the leaching out of alumina (Al2O3) as sodium aluminate and silica (SiO2) as sodium silicate leaving the impurities behind. Then, CO2gas is passed through the resulting solution to ne...
Read More →In the following figure, the area of the segment PAQ is
Question: In the following figure, the area of the segmentPAQ is (a) $\frac{a^{2}}{4}(\pi+2)$ (b) $\frac{a^{2}}{4}(\pi-2)$ (c) $\frac{a^{2}}{4}(\pi-1)$ (d) $\frac{a^{2}}{4}(\pi+1)$ Solution: We have to find area of segment PAQ. Area of the $\mathrm{PAQ}$ segment $=\left(\frac{\pi \theta}{360}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right) r^{2}$ We know that $\theta=90^{\circ}$. Substituting the values we get, Area of the PAQ segment $=\left(\frac{\pi \times 90}{360}-\sin 45 \cos 45\right) a...
Read More →Describe a method for refining nickel.
Question: Describe a method for refining nickel. Solution: Nickel is refined by Monds process. In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex. Then, the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature (450 470 K) to obtain pure nickel metal....
Read More →What criterion is followed for the selection of the stationary phase in chromatography?
Question: What criterion is followed for the selection of the stationary phase in chromatography? Solution: The stationary phase is selected in such a way that the components of the sample have different solubilitys in the phase. Hence, different components have different rates of movement through the stationary phase and as a result, can be separated from each other....
Read More →What is meant by the term “chromatography”?
Question: What is meant by the term chromatography? Solution: Chromatography is a collective term used for a family of laboratory techniques for the separation of mixtures. The term is derived from Greek words chroma meaning colour and graphein meaning to write. Chromatographic techniques are based on the principle that different components are absorbed differently on an absorbent. There are several chromatographic techniques such as paper chromatography, column chromatography, gas chromatograph...
Read More →In the following figure, the shaded area is
Question: In the following figure, the shaded area is (a) $50(\pi-2) \mathrm{cm}^{2}$ (b) $25(\pi-2) \mathrm{cm}^{2}$ (c) $25(\pi+2) \mathrm{cm}^{2}$ (d) $5(\pi-2) \mathrm{cm}^{2}$ Solution: Area of the shaded region is- $=\left[\frac{\pi \theta}{360}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right](r)^{2}$ $=\left(\frac{\pi}{4}-\frac{1}{2}\right)(10)^{2}$ $=25(\pi-2) \mathrm{cm}^{2}$ So the answer is (b)....
Read More →State the role of silica in the metallurgy of copper.
Question: State the role of silica in the metallurgy of copper. Solution: During the roasting of pyrite ore, a mixture of FeO and Cu2O is obtained. The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as slag. If the sulphide ore of copper contains iron, then silica (SiO2) is added as flux before roasting. Then, FeO combines with silica to form iron silicate, FeSiO3(slag)....
Read More →If diameter of a circle is increased by 40%,
Question: If diameter of a circle is increased by 40%, then its area increase by(a) 96%(b) 40%(c) 80%(d) 48% Solution: If $d$ is the original diameter of the circle, then the original radius is $\frac{d}{2}$. $\therefore$ area of the circle $=\pi\left(\frac{d}{2}\right)^{2}$ $\therefore$ area of the circle $=\pi \times \frac{d^{2}}{4}$ If diameter of the circle increases by 40%, then new diameter of the circle is calculated as shown below, That is new diameter $=d+0.4 d$ $=1.4 d$ $\therefore$ ne...
Read More →State the role of silica in the metallurgy of copper.
Question: State the role of silica in the metallurgy of copper. Solution: During the roasting of pyrite ore, a mixture of FeO and Cu2O is obtained. The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as slag. If the sulphide ore of copper contains iron, then silica (SiO2) is added as flux before roasting. Then, FeO combines with silica to form iron silicate, FeSiO3(slag)....
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