PAGE 6
1. Why should a magnesium ribbon be cleaned before burning in air?
Answer:
It is cleaned to remove Magnesium oxide coating and
impurities from the ribbon. Then only it can burn in air.
2. Write the balanced equation for the following chemical
reactions.
(i)
Hydrogen + Chlorine → Hydrogen chloride
(ii)
Barium chloride +
Aluminium sulphate → Barium sulphate + Aluminium chloride
(iii)
Sodium + Water → Sodium hydroxide +
Hydrogen
Answer:
(i)
H2
+ Cl2 → 2HCl
(ii)
3BaCl2
+ Al2(SO4)3 → 3BaSO4
+ 2AlCl3
(iii)
2Na + 2H2O
→ 2NaOH + H2
3. Write a balanced chemical equation with state symbols for
the following reactions.
(i) Solutions of barium chloride and sodium sulphate in water
react to give insoluble barium sulphate and the solution of sodium chloride.
(ii) Sodium hydroxide solution (in water) reacts with
hydrochloric acid solution (in water) to produce sodium chloride solution and
water.
Answer:
(i)
BaCl2 (aq)
+ Na2SO4 (aq) → BaSO4 (s)
+ 2NaCl (aq)
(ii)
NaOH (aq)
+ HCl (aq) → NaCl (aq) + H2O (l)
PAGE 10
1. A solution 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.
Answer:
(i)
X= Calcium
oxide (Quick Lime). Chemical formula = CaO.
(ii)
Calcium
oxide reacts vigorously with water to form calcium hydroxide (slaked lime).
CaO (s) + H2O (l) → Ca(OH)2 (aq)
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.
Answer:
The gas is hydrogen. At cathode, double amount of hydrogen gas is collected
in test tube because during the break down of water, 2H molecule is released
with 1 oxygen molecule.
PAGE 13
1. Why does the colour of
copper sulphate solution change when an iron nail is dipped in it?
Answer:
When an iron nail is dipped in CuSO4 solution,
iron displaces copper from the CuSO4 because iron is more reactive
than copper. So the colour of the CuSO4 solution changes.
Fe (s) + CuSO4 (aq) → FeSO4 (aq)
+ Cu (s)
2. Give an example of a
double displacement reaction other than the one given in Activity 1.10.
Answer:
2KBr (aq) + BaI2 (aq) → 2KI (aq)
+ BaBr2 (aq)
3. Identify the substances
that are oxidised 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
(l)
Answer:
(i)
Sodium
(Na) is oxidised and oxygen is reduced.
(ii) Copper oxide (CuO) is reduced and H2 is oxidised.