Class 10 Ncert intext Question Answers of chapter Chemical reaction and equations

Q1. Why should a magnesium ribbon be cleaned before burning in air

Ans:- magnesium ribbon should be cleaned with burning in air in order to remove the white layer of magnesium oxide (mgo) from its surface which hinders the burning of magnesium
Q2. 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

Ans:- (I) to balance a chemical equation
         Hydrogen+ chlorine ----> hydrogen chloride

STEP1: writing the skeletel chemical equation :
      H2 + c1 2 ----->HC1
STEP2: balancing the equation

(ii) to balance a chemical equation :
   Barium chloride + Aluminium sulphate ------> barium sulphate + Aluminium chloride
Step 1---> writing the skeletal chemical equation :
Bac12+Al2(so4) 3 ------> BaSo4  + A1c13
Step2-----> balancing the equation :
       3BACI2 + AL2(SO4)3------>3BaSO4 + 2AICI3
(III) to balance a chemical equation :
Sodium + water----->sodium hydroxide + hydrogen

Step1----> writing the skeletal chemical equation :
2na + 2h2o ------>2Naoh+ h2
Q3--- write a balanced chemical equation with state chemical symbols for the following reactions:
(I) solution of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride .

(II) sodium hydroxide solution reacts with hydrochloric acid solution to produce sodium chloride solution and water

Ans (I) to balance a chemical equation
Barium chloride + sodium sulphate ----->
                        Barium sulphate+ sodium chloride

Step 1----->  writing the skeletal chemical equation
BaCI2 (AQ) + Na2 so4(AQ)------> baso4(s) + 2NaCI(AQ)
(II) to balance a chemical equation
            Sodium hydroxide + hydrochloric acid -----> sodium chloride + water
Step1----> write the skeletal chemical equation NaOh (AQ) + hci (AQ)------> NaCi
(AQ) + h2o(l)
Step2 balancing the equation
   NaOh (AQ) + HCI (AQ) ------> NaCi (AQ) + h2o (l)
Q1-- A solution of a substance 'x' is used for whitewashing
(I) name the substance 'x' is write its formula .
(II) write the reaction of substance 'X' named in (I) above the water .
Ans---> (I)for whitewashing a solution quicklime and its formula is CaO [calcium oxide]
(II) when quick lime react with water the following reaction takes place :
CaO(s) + H2O(l) -----> ca(Oh)2 (aq)
Q2 why is the amount of gas collected in one of the test tubes on your book activity 1.7 double of the amount collected in the other ? Name this gas
Ans----> during electrolysis, water decomposes to form hydrogen and oxygen gases
2H2O(l)-------> in the above reaction the balanced chemical equation shows that on electrolysis the water decomposes to form hydrogen and oxygen in the ratio of 2:1 by volume .
Q1---> Why does the colour of copper sulphate solution change when an iron nail dipped in it ??
Ans--->iron is more reactive element than copper when an iron nail is dipped in copper sulphate solution which is blue in colour , iron being more reactive than copper displaces copper from copper sulphate and forms iron (II) sulphate which is green in  colour .
Cuso4 (aq) + Fe (s) -------> feso4 (AQ) + cu(s)
Q2----> give an example of a double displacement reaction other than the one given in activity 1.10
Ans---> AgNo3(aq) +. NaCI(aq)-------> AgCi (s) + NaNo3(aq)
Q3----identify the substances that are oxidised and the substances that are reduced in the following reactions
(I) 4Na(s) + O2(g) -----> 2Na2 O(s) (II) CuO(S) + H2(g)----> Cu(s) + h2O(l)

Ans----> (I) 4Na(s) + O2(g) -----> 2Na2O (s)
Na has gained oxygen and forms Na2O
So, Na is Oxidised and O2 is reduced
(II) CuO(s) + H2(g) -----> Cu(s) + H2O(l)
CuO has lost oxygen and forms Cu so CuO is reduced and H2  which has gained oxygen and oxidised .


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