NCERT Solutions Class 10 Science Chapter 1 Chemical Reactions And Equations – Here are all the NCERT solutions for Class 10 Science Chapter 1. This solution contains questions, answers, images, explanations of the complete Chapter 1 titled Chemical Reactions And Equations of Science taught in class 10. If you are a student of class 10 who is using NCERT Textbook to study Science, then you must come across Chapter 1 Chemical Reactions And Equations. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Science Chapter 1 Chemical Reactions And Equations in one place.
NCERT Solutions Class 10 Science Chapter 1 Chemical Reactions And Equations
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| Class | 10 |
| Subject | Science |
| Book | Science |
| Chapter Number | 1 |
| Chapter Name |
Chemical Reactions And Equations |
NCERT Solutions Class 10 Science chapter 1 Chemical Reactions And Equations
Class 10, Science chapter 1, Chemical Reactions And Equations solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: Why should a magnesium ribbon be cleaned before burning in air?
Ans : Magnesium is very reactive metal. When stored it reacts with oxygen to form a layer magnesium oxide on its surface. This layer of magnesium oxide is quite stable and prevents further reaction of magnesium with oxygen. The magnesium ribbon is cleaned by sand paper to remove this layer so that the underlying metal can be exposed into air.
Q.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
Ans : (i) \( \mathrm { H } _ { 2 ( g ) } + \mathrm { Cl } _ { 2 ( g ) } \longrightarrow 2 \mathrm { HCl } _ { ( \mathrm { g } ) }\) (ii) \(\ 3 \mathrm { BaCl } _ { 2 ( \mathrm { s } ) }\)+\( \mathrm { Al } _ { 2 } \left( \mathrm { SO } _ { 4 } \right) _ { 3 ( \mathrm { s } ) }\) ->\( 3 \mathrm { BaSO } _ { 4 ( s ) }\)+\( 2 \mathrm { A } \left| \mathrm { Cl } _ { 3 ( \mathrm { s } ) }\right.\) (iii) \( 2 \mathrm { Na } _ { ( s ) } + 2 \mathrm { H } _ { 2 } \mathrm { O } _ { ( l ) } \longrightarrow 2 \mathrm { NaOH } _ { ( a q ) } + \mathrm { H } _ { 2 ( \mathrm { g } ) }\)
Q.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.
Ans : (i) \( \mathrm { BaCl } _ { 2 ( \mathrm { aq } ) } + \mathrm { Na } _ { 2 } \mathrm { SO } _ { 4 ( \mathrm { aq } ) }\) ->\( \mathrm { BaSO } _ { 4 ( s ) } + 2 \mathrm { NaCl } _ { ( \mathrm { aq } ) }\) (ii) \( \mathrm { NaOH } _ { ( a q ) } + \mathrm { HCl } _ { ( a q ) }\) ->\( \mathrm { NaCl } _ { ( a \mathrm { q } ) } + \mathrm { H } _ { 2 } \mathrm { O } _ { ( j ) }\)
Q.4: A solution of a substance ‘X’ is used for whitewashing.
(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 calcium oxide. Its chemical formula is CaO. (ii) Calcium oxide reacts vigorously with water to form calcium hydroxide (slaked lime). \( \mathrm { CaO } _ { ( s ) } \quad + \quad \mathrm { H } _ { 2 } \mathrm { O } _ { ( l ) } \longrightarrow \mathrm { Ca } ( \mathrm { OH } ) _ { 2 ( \mathrm { oH } ) }\)
Q.5: 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\( \mathrm { H } _ { 2 } \mathrm { O }\) contains two parts hydrogen and one part oxygen. Therefore, the amount of hydrogen and oxygen produced during electrolysis of water is in a 2:1 ratio. During electrolysis, since hydrogen goes to one test tube and oxygen goes to another, the amount of gas collected in one of the test tubes is double of the amount collected in the other.
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