Maharashtra Board Class 10 Science Part – 1 Solution Chapter 8 – Metallurgy
Balbharati Maharashtra Board Class 10 Science Part – 1 Solution Chapter 8: Metallurgy. Marathi or English Medium Students of Class 10 get here Metallurgy full Exercise Solution.
|
Std |
Maharashtra Class 10 |
|
Subject |
Science Part – 1 Solution |
| Chapter |
Metallurgy |
Chapter: -8
(Metallurgy)
Exercise: –
1) Write names.
a) Alloy of sodium with mercury.
b.) Molecular formula of the common ore of aluminium.
c.) The oxide that forms salt and water by reacting with both acid and base.
d.) The device used for grinding an ore.
e.) The non-metal having electrical conductivity.
f.) The reagent that dissolves noble metals.
Ans: – a)amalgam or Na(Hg) is the alloy of sodium with mercury.
b) Bauxite or Al2O3.2H2O is the ore of common ore of aluminum.
c) Metal oxide after reaction with acid and base forms salt. Example: – Al2O3 is are acting with acid and base forms salt and water.
d) Ball mill is used for grinding an ore.
e) The non metal used to conduct electric for example graphite which is an non metal used for electrical conductivity.
f) The reagent which dissolves noble metal is aqua Regia.
2) Make pairs of substances and their properties.
Substance: – (a. Potassium bromide b. Gold c. Sulphur d. Neon)
Property: – (1. Combustible2. Soluble in water 3. No chemical reaction4.High ductility).
Ans: –
| Substance | Property |
| a. Potassium bromide | 2.soluble in water |
| b. Gold | 4.Highly ductility |
| c. Sulphur | 1.Combustible |
| d. Neon | 3.No chemical reaction |
3) Identify the pairs of metals and their ores from the following.
Group A :-(a. Bauxite b. Cassiterite c. Cinnabar)
Group B :-(i. Mercury ii. Aluminiumiii. Tin).
Ans: –
| Group A | Group B |
| a. Bauxite | ii. Aluminium |
| b. Cassiterite | ii. Tin |
| c. Cinnabar | i. Mercury |
4) Explain the terms.
a) Metallurgy b.) Ores c.) Minerals d.) Gangue.
Ans: – a) Metallurgy: -In which process a metal can be extracted and refining from it ore is known as metallurgy. There is various process of extracting a metal.
b) Ores: – In which substance any metal can be extracted or purified from it is known as ores economically of that metal.
c) Minerals: – Minerals which is a homogeneous solid with having different composition of any metal present in it is known as minerals.
d) Gangue: – Different types of impurities like limestone, mica, dust, soil, sand. Which is present in an ore is called gangue.
5.) Write scientific reasons.
a.) Lemon or tamarind is used for cleaning copper vessels turned greenish.
Ans: – As in present of air the copper vessels upper surface coated with copper carbonate by using lemon ut removed. The citric acid present in lemon react with the layer.
b.) Generally the ionic compounds have high melting points.
Ans: – There exists electrostatic force of attraction in ionic compound. The different cation and anion present in the ionic compounds are attached by this force which needs more energy to melt.
c.) Sodium is always kept in kerosene.
Ans: – In the presence of oxygen and water molecules in air the sodium molecules react with it and form NaOH or sodium hydroxide which is an exothermic reaction so we don’t kept sodium in air.
d) Pine oil is used in froth flotation.
Ans: –The hydrophobic nature of pine oil, it floats on water with foam this property of pine oil makes it usable in forth flotation. As a result, we see by forth floatation many ores are purified.
e) Anodes need to be replaced from time to time during the electrolysis of alumina.
Ans: – Alumina are used as anode in an electrolysis process because of reduction of alumina oxygen evolve in anode. As a result, cathode fill with aluminium.
6.) When a copper coin is dipped in silver nitrate solution, a glitter appears on the coin after some time. Why does this happen? Write the chemical equation.
Ans: – If we dipped a copper coin in silver nitrate solution a glitter appears on the coin after some time after reaction with silver nitrate cupric nitrate is made and silver is deposit. This silver appears on coin make it blue.
2AgNO3 + Cu —-> Cu(NO3)2 + 2Ag.
7.) The electronic configuration of metal ‘A’ is 2,8,1 and that of metal ‘B’ is 2,8,2. Which of the two metals is more reactive? Write their reaction with dilute hydrochloric acid.
Ans: – The meta A with configuration 2,8,1 will be sodium. And the metal B with configuration of 2,8,2 will be magnesium. From these two metal configuration we can easily say that metal A sodium is more reactive as only one electron is present in its valence shell.
Reaction of A is , 2Na + 2HCl —-> 2NaCl + H2.
Reaction of B , Mg + 2HCl —–> MgCl2 + H2.
8.) Draw a neat labelled diagram.
a.) Magnetic separation method.
Ans: –
b.) Froth floatation method.
Ans: –
c) Electrolytic reduction of alumina.
Ans: –
d) Hydraulic separation method.
Ans: –
9.) Write chemical equation for the following events.
a.) Aluminium came in contact with air.
Ans:- 4Al + 3O2 —-> 2Al2O3.
b.) Iron filings are dropped in aqueous solution of copper sulphate.
Ans: – Fe + CuSO4 —-> FeSO4.
c.) A reaction was brought about between ferric oxide and aluminium.
Ans: – 2Al + Fe2O3 —-> Al2O3 + 2Fe.
d.) Electrolysis of alumina is done.
Ans:- 2Al2O3 —-> 4Al(+3) + 6O2(-2) ; cathode :- 4Al(+3) 12e —->4Al ; anode :- 6O2(-2) —-> 3O2 + 12e , C +O2 —-> CO2.
e.) Zinc oxide is dissolved in dilute hydrochloric acid.
Ans: – ZnO + 2HCl —-> ZnCl2 + H2O.
10.) Complete the following statement using every given options. During the extraction of aluminum………….
a.) Ingredients and gangue in bauxite.
Ans: – We all know that the main ingredients of fluorspar(CaF2), Na3Al6. And the gangue in bauxite is SiO2, TIO2 etc.
b.) Use of leaching during the concentration of ore.
Ans: – As there are present different impurities in bauxite like SiO2, TiO2 we use the leaching to make it pure, this process is known as Hall’s process.
c.) Chemical reaction of transformation of bauxite into alumina by Hall’s process.
Ans: –There are different chemical reaction happened which are
2Al2O3.2H2O + Na2CO3 —-> 2NaAlO2 + CO2 + H2O ; 2NaAlO2 + 2H2O + CO2 —-> 2Al(OH)3 + Na2CO3 ; 2Al(OH)3 —-> Al2O3 + 3H2O.
d.) Heating the aluminium ore with concentrated caustic soda.
Ans:- Al2O3 + 2NaOH —-> 2NaAlO2 + H2O.
After aluminium oxide heating at 1400 to 1500℃ for 2 to 8hr with NaOH solution sodium aluminate is formed.
11.) Divide the metals Cu, Zn, Ca, Mg, Fe, Na, Li into three groups, namely reactive metals, moderately reactive metals and less reactive metals.
Ans: – The reactive metals: -Ca, Na.
Moderately reactive metal: – Mg, Li, Zn.
Less reactive metal: -Fe, Cu.
Here is your solution of Maharashtra Board Class 10 Science Part – 1 Chapter 8 Metallurgy
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