Q.1: Why should magnesium ribbon be cleaned before burning in air ?
Ans: Magnesium ribbon is a very reactive metal. When stored it reacts with oxygen to form a layer of magnesium oxide on its surface. This layer of magnesium oxide being a stable compound prevents further reaction of magnesium with oxygen. The magnesium is cleaned before burning in air to remove this layer so that the metal can be exposed to air properly.
Q.2: Write the balanced equation for the following chemical reactions:
(i) Hydrogen + Chlorine --> Hydrogen chloride
(ii) Sodium + Water --> Sodium hydroxide + Hydrogen
Ans:
(i) H2 (g) + Cl2 (g) --> 2HCl (g)
(ii) 2Na (s) + 2H2O (liq) --> 2NaOH (aq) + H2 (g).
Q.3: Write a balanced chemical equation with state symbols for the following chemical reactions:
(i) Solution of barium chloride and sodium sulphate in water to give insoluble barium sulphate and the solution of sodium chloride.
(ii) Sodium hydroxide solution (in water) react with hydrochloric acid solution (in water) to produce sodium chloride and water.
Ans:
(i) BaCl2 (aq) + Na2SO4 (aq) --> BaSO4 (s) + 2NaCl (aq)
(ii) NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (liq)
(Page 10)
Q.1: A reaction of a substance X is used for white washing.
(i) Name the substance X and write its formula.
(ii) Write the reaction of the substance X named in (i) above with water.
Ans:
(i) The substance X is lime which is used for white washing. Its formula is CaO.
(ii) CaO + H2O --> Ca(OH)2 + Heat
Q.2: Why is the amount of gas collected in one of the test tubes in Activity 1.7, double of the amount collected in the other ? Name this gas.
Ans: water contains two parts of hydrogen and one part oxygen. Therefore, during the electrolysis of water the amount of hydrogen gas collected in one of the test tubes is double than that of the oxygen produced and collected in the other test tube.
(Page 13)
Q.1: Why does the colour of copper sulphate solution change when an iron nail is dipped in it ?
Ans: Iron is more reactive than copper. So, when an iron nail is dipped in a copper sulphate solution, iron displaces copper from its solution to form iron sulphate, which is green in color.
Fe (s) + CuSO4 (aq) --> FeSO4 (aq) + Cu (s)
Hence, the blue color of copper sulphate solution changes into green color because of this displacement reaction.
Q.2: Give an example of the double displacement reaction other than the one given in Activity 1.10.
Ans: When lead (II) nitrate is mixed with potassium iodide, potassium nitrate and lead iodide are formed.
Pb(NO3)2 + 2KI --> 2KNO3 + PbI2
Na2CO3 (aq) + CaCl2 (aq) --> CaCO3 (s) + 2NaCl (aq)
Here sodium carbonate and calcium chloride exchange ions to form two new compounds.
Q.3: Identify the substances that are oxidized and the substances that are reduced in the following reactions:
(i) 4Na(s) + O2(g) --> 2Na2O(s)
(ii) CuO(s) + H2(g) --> Cu(s) + H2O(liq)
Ans:
(i) Sodium (Na) is oxidized to Na2O.
(ii) CuO (Copper oxide) is reduced to Cu, while H2 gas is oxidized to H2O
Ans: Magnesium ribbon is a very reactive metal. When stored it reacts with oxygen to form a layer of magnesium oxide on its surface. This layer of magnesium oxide being a stable compound prevents further reaction of magnesium with oxygen. The magnesium is cleaned before burning in air to remove this layer so that the metal can be exposed to air properly.
Q.2: Write the balanced equation for the following chemical reactions:
(i) Hydrogen + Chlorine --> Hydrogen chloride
(ii) Sodium + Water --> Sodium hydroxide + Hydrogen
Ans:
(i) H2 (g) + Cl2 (g) --> 2HCl (g)
(ii) 2Na (s) + 2H2O (liq) --> 2NaOH (aq) + H2 (g).
Q.3: Write a balanced chemical equation with state symbols for the following chemical reactions:
(i) Solution of barium chloride and sodium sulphate in water to give insoluble barium sulphate and the solution of sodium chloride.
(ii) Sodium hydroxide solution (in water) react with hydrochloric acid solution (in water) to produce sodium chloride and water.
Ans:
(i) BaCl2 (aq) + Na2SO4 (aq) --> BaSO4 (s) + 2NaCl (aq)
(ii) NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (liq)
(Page 10)
Q.1: A reaction of a substance X is used for white washing.
(i) Name the substance X and write its formula.
(ii) Write the reaction of the substance X named in (i) above with water.
Ans:
(i) The substance X is lime which is used for white washing. Its formula is CaO.
(ii) CaO + H2O --> Ca(OH)2 + Heat
Q.2: Why is the amount of gas collected in one of the test tubes in Activity 1.7, double of the amount collected in the other ? Name this gas.
Ans: water contains two parts of hydrogen and one part oxygen. Therefore, during the electrolysis of water the amount of hydrogen gas collected in one of the test tubes is double than that of the oxygen produced and collected in the other test tube.
(Page 13)
Q.1: Why does the colour of copper sulphate solution change when an iron nail is dipped in it ?
Ans: Iron is more reactive than copper. So, when an iron nail is dipped in a copper sulphate solution, iron displaces copper from its solution to form iron sulphate, which is green in color.
Fe (s) + CuSO4 (aq) --> FeSO4 (aq) + Cu (s)
Hence, the blue color of copper sulphate solution changes into green color because of this displacement reaction.
Q.2: Give an example of the double displacement reaction other than the one given in Activity 1.10.
Ans: When lead (II) nitrate is mixed with potassium iodide, potassium nitrate and lead iodide are formed.
Pb(NO3)2 + 2KI --> 2KNO3 + PbI2
Na2CO3 (aq) + CaCl2 (aq) --> CaCO3 (s) + 2NaCl (aq)
Here sodium carbonate and calcium chloride exchange ions to form two new compounds.
Q.3: Identify the substances that are oxidized and the substances that are reduced in the following reactions:
(i) 4Na(s) + O2(g) --> 2Na2O(s)
(ii) CuO(s) + H2(g) --> Cu(s) + H2O(liq)
Ans:
(i) Sodium (Na) is oxidized to Na2O.
(ii) CuO (Copper oxide) is reduced to Cu, while H2 gas is oxidized to H2O
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