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SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Science Metals and Non-metals

65 questions · 65 still being checked

Short Answer Questions 37–46 (part 5 of 7)

  1. Exercise 37

    Iqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in reaction mixture. He made the same observations when this element was treated with hydrochloric acid. Suggest how can he identify the produced gas. Write chemical equations for both the reactions.

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    NCERT’s answer
    The produced gas can be identified by bringing a burning match stick near the reaction vessel, a pop sound is produced \[\begin{aligned} & \mathrm{M}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{MO}_2+\mathrm{H}_2 \\ & \mathrm{M}+2 \mathrm{HCl} \rightarrow \mathrm{MCl}_2+\mathrm{H}_2 \end{aligned} \] The element is a metal
    M gives bubbles with both a base and an acid — that amphoteric behaviour identifies M as zinc.To identify the gas: bring a burning splinter to the mouth of the tube; the gas burns with a pop, confirming hydrogen.\[\mathrm{Zn(s)} + 2\mathrm{NaOH(aq)} \rightarrow \mathrm{Na_2ZnO_2(aq)} + \mathrm{H_2(g)} \] \[\mathrm{Zn(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{ZnCl_2(aq)} + \mathrm{H_2(g)} \]Answer: The gas is hydrogen; M is zinc, an amphoteric metal.

  2. Exercise 38

    During extraction of metals, electolytic refining is used to obtain pure metals. (a) Which material will be used as anode and cathode for refining of silver metal by this process? (b) Suggest a suitable electrolyte also. (c) In this electrolytic cell, where do we get pure silver after passing electric current?

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    NCERT’s answer
    (a)
    Anode : Impure silver
    Cathode : Pure silver
    (b)
    Electrolyte: Silver salt, such as \(\displaystyle \mathrm{AgNO}_3\)
    (c)
    We get pure silver at cathode
    (a) Anode: a bar of impure silver. Cathode: a strip of pure silver. (b) Electrolyte: acidified aqueous silver nitrate, \(\displaystyle \mathrm{AgNO_3(aq)}\). (c) Pure silver deposits on the cathode; insoluble impurities fall below the anode as anode mud.At anode: \[\mathrm{Ag(s)} \rightarrow \mathrm{Ag^+(aq)} + e^- \] At cathode: \[\mathrm{Ag^+(aq)} + e^- \rightarrow \mathrm{Ag(s)} \]Answer: Pure silver collects at the cathode.

  3. Exercise 39

    Why should the metal sulphides and carbonates be converted to metal oxides in the process of extraction of metal from them?

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    NCERT’s answer
    It is easier to obtain metal from its oxide, as compared from its sulphides and carbonates.
    Sulphide and carbonate ores are chemically stable; reducing them to the metal directly is difficult. Converting them first to the oxide gives a compound that a reducing agent such as carbon reduces far more readily.Calcination (carbonate): \[\mathrm{MCO_3(s)} \xrightarrow{\Delta} \mathrm{MO(s)} + \mathrm{CO_2(g)} \] Roasting (sulphide): \[2\mathrm{MS(s)} + 3\mathrm{O_2(g)} \xrightarrow{\Delta} 2\mathrm{MO(s)} + 2\mathrm{SO_2(g)} \]Answer: Because the oxide of a metal is much easier to reduce than its sulphide or carbonate.

  4. Exercise 40

    Generally, when metals are treated with mineral acids, hydrogen gas is liberated but when metals (except Mn and Mg), treated with HNO3\displaystyle \mathrm{HNO}_3, hydrogen is not liberated, why?

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    NCERT’s answer
    It is because \(\displaystyle \mathrm{HNO}_3\) is a strong oxidising agent. It oxidises the \(\displaystyle \mathrm{H}_2\) produced to \(\displaystyle \mathrm{H}_2 \mathrm{O}\).
    A metal above hydrogen normally displaces \(\displaystyle \mathrm{H_2}\) from an acid. \(\displaystyle \mathrm{HNO_3}\) is a strong oxidising agent: it oxidises the \(\displaystyle \mathrm{H_2}\) to water as it forms and is itself reduced to nitrogen oxides. Only Mn and Mg, with very dilute \(\displaystyle \mathrm{HNO_3}\), still give \(\displaystyle \mathrm{H_2}\).\[\mathrm{Zn(s)} + 2\mathrm{HCl(dil.)} \rightarrow \mathrm{ZnCl_2(aq)} + \mathrm{H_2(g)} \] \[\mathrm{Zn(s)} + 4\mathrm{HNO_3(conc.)} \rightarrow \mathrm{Zn(NO_3)_2(aq)} + 2\mathrm{NO_2(g)} + 2\mathrm{H_2O(l)} \] \[\mathrm{Mg(s)} + 2\mathrm{HNO_3(very\ dil.)} \rightarrow \mathrm{Mg(NO_3)_2(aq)} + \mathrm{H_2(g)} \]Answer: \(\displaystyle \mathrm{HNO_3}\) oxidises the \(\displaystyle \mathrm{H_2}\) as it forms; only Mn and Mg release it, with very dilute acid.

  5. Exercise 41

    Compound X and aluminium are used to join railway tracks. (a) Identify the compound X (b) Name the reaction (c) Write down its reaction.

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    NCERT’s answer
    (a)
    X — \(\displaystyle \mathrm{Fe}_2 \mathrm{O}_3\) (b) Thermite reaction
    (c)
    \(\displaystyle \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+2 \mathrm{Al}(\mathrm{s}) \rightarrow 2 \mathrm{Fe}(\mathrm{l})+\mathrm{Al}_2 \mathrm{O}_3(\mathrm{~s})+\mathrm{Heat}\)
    (a) X is iron(III) oxide, \(\displaystyle \mathrm{Fe_2O_3}\). (b) The reaction is the thermit reaction. (c) \[\mathrm{Fe_2O_3(s)} + 2\mathrm{Al(s)} \rightarrow 2\mathrm{Fe(l)} + \mathrm{Al_2O_3(s)} + \text{Heat} \]The reaction is highly exothermic, so it is set off well clear of anything flammable; the molten iron produced fills the gap between the rails and fuses them on cooling.Answer: X is \(\displaystyle \mathrm{Fe_2O_3}\); the thermit reaction welds the rails with molten iron.

  6. Exercise 42

    When a metal X is treated with cold water, it gives a basic salt Y with molecular formula XOH (Molecular mass = 40\displaystyle 40) and liberates a gas Z which easily catches fire. Identify X, Y and Z and also write the reaction involved.

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    NCERT’s answer
    X — Na, Y — NaOH, Z — \(\displaystyle \mathrm{H}_2\) \(\displaystyle 2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2+\) Heat energy
    \[M(\mathrm{XOH}) = M(\mathrm{X}) + 16 + 1 = 40\ \mathrm{g\,mol^{-1}} \Rightarrow M(\mathrm{X}) = 23\ \mathrm{g\,mol^{-1}} \Rightarrow \mathrm{X} = \mathrm{Na} \]So X is sodium, Y is \(\displaystyle \mathrm{NaOH}\), and Z is hydrogen, which catches fire readily in air.\[2\mathrm{Na(s)} + 2\mathrm{H_2O(l)} \rightarrow 2\mathrm{NaOH(aq)} + \mathrm{H_2(g)} + \text{Heat} \]Answer: X = Na, Y = NaOH, Z = \(\displaystyle \mathrm{H_2}\).

  7. Exercise 43

    A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.

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    NCERT’s answer
    X — Carbon; Y — Diamond and Z — Graphite
    Diamond: rigid three-dimensional network, no free electrons. Graphite: flat layers with delocalised electrons.Answer: X = carbon; Y = diamond; Z = graphite.

  8. Exercise 44

    The following reaction takes place when aluminium powder is heated with MnO2\displaystyle \mathrm{MnO}_2
    3MnO2(s)+4Al(s)→3Mn(l)+2Al2O3(l)+\displaystyle 3 \mathrm{MnO}_2(\mathrm{s})+4 \mathrm{Al}(\mathrm{s}) \rightarrow 3 \mathrm{Mn}(\mathrm{l})+2 \mathrm{Al}_2 \mathrm{O}_3(\mathrm{l})+ Heat
    (a)
    Is aluminium getting reduced? (b) Is MnO2\displaystyle \mathrm{MnO}_2 getting oxidised?

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    NCERT’s answer
    (a)
    No, because oxygen is added to aluminium therefore, it is getting oxidised
    (b)
    No, since manganese has lost oxygen therefore, it is getting reduced.
    (a) No — aluminium is oxidised, not reduced: its oxidation number rises from $\displaystyle 0$ to +$\displaystyle 3$ as it loses electrons. (b) No — \(\displaystyle \mathrm{MnO_2}\) is reduced, not oxidised: manganese's oxidation number falls from +$\displaystyle 4$ to $\displaystyle 0$ as it gains electrons.\[\mathrm{Al} \rightarrow \mathrm{Al^{3+}} + 3e^- \quad \text{(oxidation)} \] \[\mathrm{Mn^{4+}} + 4e^- \rightarrow \mathrm{Mn} \quad \text{(reduction)} \]Answer: (a) No, Al is oxidised. (b) No, \(\displaystyle \mathrm{MnO_2}\) is reduced.

  9. Exercise 45

    What are the constituents of solder alloy? Which property of solder makes it suitable for welding electrical wires?

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    NCERT’s answer
    Solder is an alloy of lead and tin. Low melting point of solder makes it suitable for welding electrical wires.
    Solder is an alloy of lead (\(\displaystyle \mathrm{Pb}\)) and tin (\(\displaystyle \mathrm{Sn}\)). It melts far below the copper wire, so it flows over the joint while the wire stays solid. \[T_{mp}(\text{solder}) \approx 456\ \mathrm{K} \ll T_{mp}(\mathrm{Cu}) = 1358\ \mathrm{K} \]Answer: \(\displaystyle \mathrm{Pb}\) and \(\displaystyle \mathrm{Sn}\); its low melting point makes it suitable for welding electrical wires.

  10. Exercise 46

    A metal A, which is used in thermite process, when heated with oxygen gives an oxide B, which is amphoteric in nature. Identify A and B. Write down the reactions of oxide B with HCl and NaOH.

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    NCERT’s answer
    A — Al; B — \(\displaystyle \mathrm{Al}_2 \mathrm{O}_3\) \(\displaystyle \mathrm{Al}_2 \mathrm{O}_3+6 \mathrm{HCl} \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2 \mathrm{O}\) \(\displaystyle \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{NaOH} \rightarrow 2 \mathrm{NaAlO}_2+\mathrm{H}_2 \mathrm{O}\)
    A is aluminium, the reducing metal of the thermite reaction; heating it in oxygen gives B. \[4\mathrm{Al(s)} + 3\mathrm{O_2(g)} \xrightarrow{\Delta} 2\mathrm{Al_2O_3(s)} \] B (\(\displaystyle \mathrm{Al_2O_3}\)) is amphoteric — it reacts with both an acid and a base. \[\mathrm{Al_2O_3(s)} + 6\mathrm{HCl(aq)} \rightarrow 2\mathrm{AlCl_3(aq)} + 3\mathrm{H_2O(l)} \] \[\mathrm{Al_2O_3(s)} + 2\mathrm{NaOH(aq)} \rightarrow 2\mathrm{NaAlO_2(aq)} + \mathrm{H_2O(l)} \]Answer: A = Al, B = \(\displaystyle \mathrm{Al_2O_3}\); reacts with HCl to give \(\displaystyle \mathrm{AlCl_3}\) + \(\displaystyle \mathrm{H_2O}\), and with NaOH to give \(\displaystyle \mathrm{NaAlO_2}\) + \(\displaystyle \mathrm{H_2O}\).