Write chemical reactions taking place
Question: Write chemical reactions taking place in the extraction of zinc from zinc blende. Solution: The different steps involved in the extraction of zinc from zinc blende (ZnS) are given below: (i) Concentration of ore First, the gangue from zinc blende is removed by the froth floatation method. (ii) Conversiontooxide (Roasting) Sulphide ore is converted into oxide by the process of roasting. In this process, ZnS is heated in a regular supply of air in a furnace at a temperature, which is bel...
Read More →The area of a circle whose area and circumference
Question: The area of a circle whose area and circumference are numerically equal, is (a) $2 \pi$ sq. units (b) $4 \pi$ sq. units (c) $6 \pi$ sq. units (d) $8 \pi$ sq. units Solution: We have given that circumference and area of a circle are numerically equal. Let it bex. Let r be the radius of the circle, therefore, circumference of the circle isand area of the circle will be. Therefore, from the given condition we have, $2 \pi r=x \ldots \ldots \ldots(1)$ $\pi r^{2}=x$.........(2) Therefore, f...
Read More →Write down the reactions taking place in different zones in the blast furnace
Question: Write down the reactions taking place in different zones in the blast furnace during the extraction of iron. Solution: During the extraction of iron, the reduction of iron oxides takes place in the blast furnace. In this process, hot air is blown from the bottom of the furnace and coke is burnt to raise the temperature up to 2200 K in the lower portion itself. The temperature is lower in the upper part. Thus, it is the lower part where the reduction of iron oxides (Fe2O3and Fe3O4) take...
Read More →Name the common elements present in the anode mud in electrolytic refining of copper.
Question: Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present ? Solution: In electrolytic refining of copper, the common elements present in anode mud are selenium, tellurium, silver, gold, platinum, and antimony. These elements are very less reactive and are not affected during the purification process. Hence, they settle down below the anode as anode mud....
Read More →Out of C and CO,
Question: Out of C and CO, which is a better reducing agent at 673 K? Solution: At $673 \mathrm{~K}$, the value of $\Delta \mathrm{G}_{\left(\mathrm{co}, \mathrm{CO}_{2}\right)}$ is less than that of $\Delta \mathrm{G}_{(\mathrm{c}, \mathrm{co})}$. Therefore, $\mathrm{CO}$ can be oxidised more easily to $\mathrm{CO}_{2}$ than $\mathrm{C}$ to $\mathrm{CO}$. Hence, $\mathrm{CO}$ is a better reducing agent than $\mathrm{C}$ at $673 \mathrm{~K}$....
Read More →If AB is a chord of length
Question: If $A B$ is a chord of length $5 \sqrt{3} \mathrm{~cm}$ of a circle with centre $O$ and radius $5 \mathrm{~cm}$, then area of sector $O A B$ is (a) $\frac{3 \pi}{8} \mathrm{~cm}^{2}$ (b) $\frac{8 \pi}{3} \mathrm{~cm}^{2}$ (c) $25 \pi \mathrm{cm}^{2}$ (d) $\frac{25 \pi}{3} \mathrm{~cm}^{2}$ Solution: We have to find the area of the sector OAB. We have, $\mathrm{AM}=\frac{5 \sqrt{3}}{2}$ So, $\sin \angle \mathrm{AOM}=\frac{5 \sqrt{3}}{2(5)}$ Hence, $\angle \mathrm{AOM}=60^{\circ}$ Theref...
Read More →Explain:
Question: Explain:(i)Zone refining(ii)Column chromatography. 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 with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten ...
Read More →Why is the extraction of copper from pyrites more difficult than that from
Question: Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction? Solution: The Gibbs free energy of formation (ΔfG) of Cu2S is less than that ofand. Therefore, H2and C cannot reduce Cu2S to Cu. On the other hand, the Gibbs free energy of formation ofis greater than that of. Hence, C can reduce Cu2O to Cu. $\mathrm{C}_{(s)}+\mathrm{Cu}_{2} \mathrm{O}_{(s)} \longrightarrow 2 \mathrm{Cu}_{(s)}+\mathrm{CO}_{(g)}$ Hence, the extraction of copper fr...
Read More →What is the role of depressant in froth floatation process?
Question: What is the role of depressant in froth floatation process? Solution: In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores (ZnS and Pbs), NaCN is used as a depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form Na2[Zn(CN)4]. $4 \mathrm{NaCN}+\mathrm{ZnS} \lo...
Read More →Copper can be extracted by hydrometallurgy but not zinc.
Question: Copper can be extracted by hydrometallurgy but not zinc. Explain. Solution: The reduction potentials of zinc and iron are lower than that of copper. In hydrometallurgy, zinc and iron can be used to displace copper from their solution. $\mathrm{Fe}_{(s)}+\mathrm{Cu}_{(a q)}^{2+} \longrightarrow \mathrm{Fe}_{(a q)}^{2+}+\mathrm{Cu}_{(s)}$ But to displace zinc, more reactive metals i.e., metals having lower reduction potentials than zinc such as Mg, Ca, K, etc. are required. But all these...
Read More →The area of a circular path of uniform width h surrounding a circular region of radius r is
Question: The area of a circular path of uniform width h surrounding a circular region of radius r is (a) $\pi(2 \mathrm{r}+\mathrm{h}) \mathrm{r}$ (b) $\pi(2 \mathrm{r}+\mathrm{h}) h$ (c) $\pi(\mathrm{h}+\mathrm{r}) \mathrm{r}$ (d) $\pi(\mathrm{h}+\mathrm{r}) h$ Solution: We have $O A=r$ $A B=h$ Therefore, radius of the outer circle will be $r+h$. Now we will find the area between the two circles. Area of the circular path = area of the outer circle - area of the inner circle $\therefore$ Area ...
Read More →Is it true that under certain conditions,
Question: Is it true that under certain conditions, Mg can reduce SiO2and Si can reduce MgO? What are those conditions? Solution: $\mathrm{Mg}_{(s)}+\frac{1}{2} \mathrm{O}_{2(g)} \longrightarrow \mathrm{MgO}_{(s)}\left[\Delta G_{\left(\mathrm{Mg}, \mathrm{M}_{8} \mathrm{O}\right)}\right]$ $\mathrm{Si}_{(s)}+\mathrm{O}_{2(g)} \longrightarrow \mathrm{SiO}_{2(s)}\left[\Delta G_{\left(\mathrm{Si}, \mathrm{SiO}_{2}\right)}\right]$ The temperature range in which $\Delta G_{(\mathrm{Mg}, \mathrm{Mg} \m...
Read More →The reaction,
Question: The reaction, $\mathrm{Cr}_{2} \mathrm{O}_{3}+2 \mathrm{Al} \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+2 \mathrm{Cr} \quad\left(\Delta G_{\mathrm{o}}=-421 \mathrm{~kJ}\right)$ is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature? Solution: The change in Gibbs energy is related to the equilibrium constant,Kas $\Delta G=-\mathrm{R} T \ln K$ At room temperature, all reactants and products of the given reaction are in ...
Read More →If the difference between the circumference and radius of a circle is 37 cm,
Question: If the difference between the circumference and radius of a circle is 37 cm, then its area is (a) $154 \mathrm{~cm}^{2}$ (b) $160 \mathrm{~cm}^{2}$ (c) $200 \mathrm{~cm}^{2}$ (d) $150 \mathrm{~cm}^{2}$ Solution: We have given the difference between circumference and radius of the circle. Let C be the circumference,rbe the radius and A be the area of the circle. Therefore, from the given condition we have $C-r=2 \pi r-r$ $\therefore 37=2 \pi r-r$ $\therefore 37=r(2 \pi-1)$ $\therefore r...
Read More →What is the significance of leaching in the extraction of aluminium?
Question: What is the significance of leaching in the extraction of aluminium? Solution: In the extraction of aluminium, the significance of leaching is to concentrate pure alumina (Al2O3) from bauxite ore. Bauxite usually contains silica, iron oxide, and titanium oxide as impurities. In the process of leaching, alumina is concentrated by digesting the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar. Under these conditions, alumina (Al2O3) dissolves as sodium meta-al...
Read More →Which of the ores mentioned in Table 6.1 can be concentrated by
Question: Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method? Solution: If the ore or the gangue can be attracted by the magnetic field, then the ore can be concentrated by the process of magnetic separation. Among the ores mentioned in table 6.1, the ores of iron such as haematite (Fe2O3), magnetite (Fe3O4), siderite (FeCO3), and iron pyrites (FeS2) can be separated by the process of magnetic separation....
Read More →The area of the circle that can be inscribed
Question: The area of the circle that can be inscribed in a square of side 10 cm is (a) $40 \pi \mathrm{cm}^{2}$ (b) $30 \pi \mathrm{cm}^{2}$ (c) $100 \pi \mathrm{cm}^{2}$ (d) $25 \pi \mathrm{cm}^{2}$ Solution: We know that ABCD is a square of length 10 cm. A circle is inscribed in the square therefore, all the sides of the square are become tangents of the circle. By, the tangent property, we have $A P=P D=5$ $A Q=Q B=5$ $B R=R C=5$ $C S=D S=5$ If we join PR then it will be the diameter of the ...
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 its radius is(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. We know that area of the sector $=\frac{\theta}{360} \times \pi r^{2}$. Length of the arc $=\frac{\theta}{360} \times 2 \pi r$ Now we will substitute the values. Area of the sector $=\frac{\theta}{360} \times \pi...
Read More →Comment on the statement that
Question: Comment on the statement that colloid is not a substance but a state of substance. Solution: Common salt (a typical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 1 nm and 1000 nm, it behaves as a colloid. Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size...
Read More →If the perimeter of a sector of a circle
Question: If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area is (a) $58 \mathrm{~cm}^{2}$ (b) $52 \mathrm{~cm}^{2}$ (c) $25 \mathrm{~cm}^{2}$ (d) $56 \mathrm{~cm}^{2}$ Solution: We know that perimeter of a sector of radius $t=2 r+\frac{\theta}{360} \times 2 \pi r$ We have given perimeter of the sector and radius of the sector and we are asked to find the area of the sector. For that we have to find the sector angle. Therefore, substituting the corresponding values ...
Read More →What are micelles?
Question: What are micelles? Give an example of a micellers system. Solution: Micelle formation is done by substances such as soaps and detergents when dissolved in water. The molecules of such substances contain a hydrophobic and a hydrophilic part. When present in water, these substances arrange themselves in spherical structures in such a manner that their hydrophobic parts are present towards the centre, while the hydrophilic parts are pointing towards the outside (as shown in the given figu...
Read More →Give four uses of emulsions.
Question: Give four uses of emulsions. Solution: Four uses of emulsions: (i)Cleansing action of soaps is based on the formation of emulsions. (ii)Digestion of fats in intestines takes place by the process of emulsification. (iii)Antiseptics and disinfectants when added to water form emulsions. (iv)The process of emulsification is used to make medicines....
Read More →Explain the following terms:
Question: Explain the following terms: (i)Electrophoresis(ii)Coagulation (iii)Dialysis(iv)Tyndall effect. Solution: (i) Electrophoresis: The movement of colloidal particles under the influence of an applied electric field is known as electrophoresis. Positively charged particles move to the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated. (ii) Coagulation: The process of settling down...
Read More →What is shape selective catalysis?
Question: What is shape selective catalysis? Solution: A catalytic reaction which depends upon the pore structure of the catalyst and on the size of the reactant and the product molecules is called shape-selective catalysis. For example, catalysis by zeolites is a shape-selective catalysis. The pore size present in the zeolites ranges from 260-740 pm. Thus, molecules having a pore size more than this cannot enter the zeolite and undergo the reaction....
Read More →Describe some features of catalysis by zeolites.
Question: Describe some features of catalysis by zeolites. Solution: Zeolites are alumino-silicates that are micro-porous in nature. Zeolites have a honeycomb-like structure, which makes them shape-selective catalysts. They have an extended 3D-network of silicates in which some silicon atoms are replaced by aluminium atoms, giving them an AlOSi framework. The reactions taking place in zeolites are very sensitive to the pores and cavity size of the zeolites. Zeolites are commonly used in the petr...
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