##### Class 10^{th} Physics & Chemistry AP Board Solution

**Improve Your Learning**- What is a balanced chemical equation? Why should chemical equations be balanced? (AS1)…
- NaOH + H2SO4 Na2SO4H2O Balance the following chemical equations. (AS1)…
- Hg(NO3)2 + KI HgI2 + KNO3 Balance the following chemical equations. (AS1)…
- H2 + O2 H2O Balance the following chemical equations. (AS1)
- KClO3KCl + O2 Balance the following chemical equations. (AS1)
- C3H8 + O2 CO2 + H2O Balance the following chemical equations. (AS1)…
- Zinc +Silver nitrateZinc nitrate+ Silver. Write the balanced chemical equations for the…
- Aluminum + copper chloride Aluminum chloride+ Copper. Write the balanced chemical…
- Hydrogen + Chlorine. Hydrogen chloride Write the balanced chemical equations for the…
- Ammonium nitrate Nitrous Oxide + water. Write the balanced chemical equations for the…
- Calcium hydroxide(aq) + Nitric acid(aq) Water(1)+ Calcium nitrate(aq) Write the balanced…
- Magnesium(s)+ Iodine(g)→ Magnesium iodide(g) Write the balanced chemical equation for the…
- Magnesium(s)+ Hydrochloric acid(aq)→ Magnesium chloride(aq)+ Hydrogen(g) Write the…
- Zinc(s) + Calcium chloride(aq)→ Zinc chloride(aq) + calcium(s) Write the balanced chemical…
- Write an equation for decomposition reaction where energy is supplied in the form of teat/…
- What do you mean by precipitation reaction? (AS1)
- How chemical displacement reactions differ from chemical decomposition reaction? Explain…
- Name the reactions taking place in the presence of sunlight? (AS1)…
- Why does respiration considered as an exothermic reaction? Explain. (AS1)…
- What is the difference between displacement and double displacement reaction? Write…
- MnO2 + 4HCl MnCl2 + 2H2O + Cl2 In the above equation, name the compound which is oxidized…
- Give two examples for oxidation-reduction reaction. (AS1)
- In the refining of silver the recovery of silver from silver nitrate solution involved…
- What do you mean by corrosion? How can you prevent it? (AS1)
- Explain rancidity. (AS1)
- C6H12O6 C2H5OH+CO2 Balance the following chemical equations including the physical states.…
- Fe + O2 Fe2O2 Balance the following chemical equations including the physical states.…
- NH3 +Cl2 N2 +NH4Cl Balance the following chemical equations including the physical states.…
- Na+ H2O NaOH + H2 Balance the following chemical equations including the physical states.…
- Barium chloride and sodium sulphate aqueous solutions react to give insoluble Barium…
- Sodium hydroxide reacts with hydrochloric acid to produce sodium chloride and water.…
- Zinc pieces react with dilute hydrochloric acid to liberate hydrogen gas and forms zinc…
- A shiny brown colored element 'X' on heating in air becomes black in colour. Can you…
- Why do we apply paint on iron articles? (AS7)
- What is the use of keeping food in air tight containers? (AS7)

**Improve Your Learning**

- What is a balanced chemical equation? Why should chemical equations be balanced? (AS1)…
- NaOH + H2SO4 Na2SO4H2O Balance the following chemical equations. (AS1)…
- Hg(NO3)2 + KI HgI2 + KNO3 Balance the following chemical equations. (AS1)…
- H2 + O2 H2O Balance the following chemical equations. (AS1)
- KClO3KCl + O2 Balance the following chemical equations. (AS1)
- C3H8 + O2 CO2 + H2O Balance the following chemical equations. (AS1)…
- Zinc +Silver nitrateZinc nitrate+ Silver. Write the balanced chemical equations for the…
- Aluminum + copper chloride Aluminum chloride+ Copper. Write the balanced chemical…
- Hydrogen + Chlorine. Hydrogen chloride Write the balanced chemical equations for the…
- Ammonium nitrate Nitrous Oxide + water. Write the balanced chemical equations for the…
- Calcium hydroxide(aq) + Nitric acid(aq) Water(1)+ Calcium nitrate(aq) Write the balanced…
- Magnesium(s)+ Iodine(g)→ Magnesium iodide(g) Write the balanced chemical equation for the…
- Magnesium(s)+ Hydrochloric acid(aq)→ Magnesium chloride(aq)+ Hydrogen(g) Write the…
- Zinc(s) + Calcium chloride(aq)→ Zinc chloride(aq) + calcium(s) Write the balanced chemical…
- Write an equation for decomposition reaction where energy is supplied in the form of teat/…
- What do you mean by precipitation reaction? (AS1)
- How chemical displacement reactions differ from chemical decomposition reaction? Explain…
- Name the reactions taking place in the presence of sunlight? (AS1)…
- Why does respiration considered as an exothermic reaction? Explain. (AS1)…
- What is the difference between displacement and double displacement reaction? Write…
- MnO2 + 4HCl MnCl2 + 2H2O + Cl2 In the above equation, name the compound which is oxidized…
- Give two examples for oxidation-reduction reaction. (AS1)
- In the refining of silver the recovery of silver from silver nitrate solution involved…
- What do you mean by corrosion? How can you prevent it? (AS1)
- Explain rancidity. (AS1)
- C6H12O6 C2H5OH+CO2 Balance the following chemical equations including the physical states.…
- Fe + O2 Fe2O2 Balance the following chemical equations including the physical states.…
- NH3 +Cl2 N2 +NH4Cl Balance the following chemical equations including the physical states.…
- Na+ H2O NaOH + H2 Balance the following chemical equations including the physical states.…
- Barium chloride and sodium sulphate aqueous solutions react to give insoluble Barium…
- Sodium hydroxide reacts with hydrochloric acid to produce sodium chloride and water.…
- Zinc pieces react with dilute hydrochloric acid to liberate hydrogen gas and forms zinc…
- A shiny brown colored element 'X' on heating in air becomes black in colour. Can you…
- Why do we apply paint on iron articles? (AS7)
- What is the use of keeping food in air tight containers? (AS7)

###### Improve Your Learning

**Question 1.**What is a balanced chemical equation? Why should chemical equations be balanced? (AS1)

**Answer:**Balanced chemical equation: A chemical equation in which the number of atoms of reactants and number of atoms of products is called a balanced equation.

Every chemical equation should be balanced because:

⇒According to the law of conservation of mass, atoms are neither created not destroyed in chemical reactions.

⇒It means the total mass of the products formed in chemical reaction must be equal to the mass of reactants consumed.

**Question 2.**Balance the following chemical equations. (AS1)

NaOH + H_{2}SO_{4} Na_{2}SO_{4}H_{2}O

**Answer:**NaOH + H_{2}SO_{4} Na_{2}SO_{4}H_{2}O

Balanced equation: 2NaOH + H_{2}SO_{4}Na_{2}SO_{4} + 2H_{2}O

Explanation:

⇒Step 1: Write the given unbalanced equation

NaOH + H_{2}SO_{4}Na_{2}SO_{4} + H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider sodium atom. If we multiply 2 in the reactant (in NaOH), we will get the equal number of atoms as in product (Na_{2}SO_{4})

⇒Step 4: Write the resulting equation:

2NaOH + H_{2}SO_{4}Na_{2}SO_{4} + H_{2}O

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of oxygen, hydrogen and Sulphur atoms are unequal on the two sides. First balance the hydrogen number.

⇒Step 6: Now, let us consider hydrogen atom. If we multiply 2 in the product (in H_{2}O), we will get the equal number of atoms as in reactants (in 2NaOH and H_{2}SO_{4})

⇒Step 7: Write the resulting equation:

2NaOH + H_{2}SO_{4}Na_{2}SO_{4} + 2H_{2}O

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

2NaOH + H_{2}SO_{4}Na_{2}SO_{4} + 2H_{2}O

**Question 3.**Balance the following chemical equations. (AS1)

Hg(NO_{3})_{2} + KI HgI_{2} + KNO_{3}

**Answer:**Hg(NO_{3})_{2} + KI HgI_{2} + KNO_{3}

Balanced equation: Hg(NO_{3})_{2} + 2KI HgI_{2} + 2KNO_{3}

Explanation:

⇒Step 1: Write the given unbalanced equation

Hg(NO_{3})_{2} + KI HgI_{2} + KNO_{3}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. First, let us consider iodine atom. If we multiply 2 in the reactant (in KI), we will get the equal number of atoms as in product (HgI_{2})

⇒Step 4: Write the resulting equation:

Hg(NO_{3})_{2} + 2KI HgI_{2} + KNO_{3}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of oxygen, nitrogen and potassium atoms are unequal on the two sides.

First balance the potassium number.

⇒Step 6: Now, let us consider potassium atom. If we multiply 2 in the product (KNO_{3}), we will get the equal number of atoms as in reactant (in KI)

⇒Step 7: Write the resulting equation:

Hg(NO_{3})_{2} + 2KI HgI_{2} + 2KNO_{3}

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

Hg(NO_{3})_{2} + 2KI HgI_{2} + 2KNO_{3}

**Question 4.**Balance the following chemical equations. (AS1)

H_{2} + O_{2} H_{2}O

**Answer:**H_{2} + O_{2} H_{2}O

Balanced equation: 2H_{2} + O_{2} 2H_{2}O

Explanation:

⇒Step 1: Write the given unbalanced equation

H_{2} + O_{2} H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. First, let us consider oxygen atom. If we multiply 2 in the product (in H_{2}O), we will get the equal number of atoms as in reactant (O_{2})

⇒Step 4: Write the resulting equation:

H_{2} + O_{2} 2H_{2}O

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of hydrogen atoms are unequal on the two sides.

⇒Step 6: Now, let us consider hydrogen atom. If we multiply 2 in the reactant (H_{2}), we will get the equal number of atoms as in product (in 2H_{2}O)

⇒Step 7: Write the resulting equation:

2H_{2} + O_{2} 2H_{2}O

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

2H_{2} + O_{2} 2H_{2}O

**Question 5.**Balance the following chemical equations. (AS1)

KClO_{3}KCl + O_{2}

**Answer:**KClO_{3}KCl + O_{2}

Balanced equation: 2KClO_{3}2KCl + 3O_{2}

Explanation:

⇒Step 1: Write the given unbalanced equation

KClO_{3}KCl + O_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider oxygen atom. If we multiply 2 in the product (in KClO_{3}) and 3 in the reactant (in O_{2}) we will get the equal number of atoms in both sides.

⇒Step 4: Write the resulting equation:

2KClO_{3}KCl + 3O_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of potassium and chlorine atoms are unequal on the two sides. First balance the potassium atom.

⇒Step 6: If we multiply 2 in the product (in KCl), we will get the equal number of atoms as in reactant (in 2KClO_{3})

⇒Step 7: Write the resulting equation:

2KClO_{3}2KCl + 3O_{2}

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

2KClO_{3}2KCl + 3O_{2}

**Question 6.**Balance the following chemical equations. (AS1)

C_{3}H_{8} + O_{2} CO_{2} + H_{2}O

**Answer:**C_{3}H_{8} + O_{2} CO_{2} + H_{2}O

Balanced equation: C_{3}H_{8} + 5O_{2} 3CO_{2} + 4H_{2}O

Explanation:

⇒Step 1: Write the given unbalanced equation

C_{3}H_{8} + O_{2} CO_{2} + H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 4 in the product (in H_{2}O), we will get the equal number of atoms as in reactant (in C_{3}H_{8})

⇒Step 4: Write the resulting equation:

C_{3}H_{8} + O_{2} CO_{2} + 4H_{2}O

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of carbon and oxygen atoms are unequal on the two sides.

First balance the carbon atom.

⇒Step 6: If we multiply 3 in the product (in CO_{2}), we will get the equal number of atoms as in reactant (in C_{3}H_{8})

⇒Step 7: Write the resulting equation:

C_{3}H_{8} + O_{2} 3CO_{2} + 4H_{2}O

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is not balanced yet. As the number of oxygen atoms are unequal on the two sides.

⇒Step 9: Now, we consider oxygen atoms. If we multiply 5 in the reactant (in O_{2}), we will get the equal number of atoms as in products (in 3CO_{2} and 4H_{2}O)

⇒Step 10: Write the resulting equation:

C_{3}H_{8} + 5O_{2} 3CO_{2} + 4H_{2}O

We find that the equation is balanced now.

⇒Step 11: Write down the final balanced equation:

C_{3}H_{8} + 5O_{2} 3CO_{2} + 4H_{2}O

**Question 7.**Write the balanced chemical equations for the following reactions. (AS1)

Zinc +Silver nitrateZinc nitrate+ Silver.

**Answer:**Zn + AgNO_{3}→ Zn(NO_{3})_{2} + Ag

Balanced equation: Zn + 2AgNO_{3}→ Zn(NO_{3})_{2} + 2Ag

Explanation:

⇒Step 1: Write the given unbalanced equation

Zn + AgNO_{3}→ Zn(NO_{3})_{2} + Ag

Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider nitrogen atom first. If we multiply 2 in the reactant (in AgNO_{3}), we will get the equal number of atoms as in product (Zn(NO_{3})_{2})

⇒Step 4: Write the resulting equation:

Zn + 2AgNO_{3}→ Zn(NO_{3})_{2} + Ag

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of silver atoms are unequal on the two sides.

⇒Step 6: Now, let us consider silver atom. If we multiply 2 in the product (in Ag), we will get the equal number of atoms as in reactant (in 2AgNO_{3})

⇒Step 7: Write the resulting equation:

Zn + 2AgNO_{3}→ Zn(NO_{3})_{2} + 2Ag

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

Zn + 2AgNO_{3}→ Zn(NO_{3})_{2} + 2Ag

**Question 8.**Write the balanced chemical equations for the following reactions. (AS1)

Aluminum + copper chloride Aluminum chloride+ Copper.

**Answer:**Al + CuCl_{2} → AlCl_{3} + Cu

Balanced equation: 2Al + 3CuCl_{2} → 2AlCl_{3} + 3Cu

Explanation:

⇒Step 1: Write the given unbalanced equation

Al + CuCl_{2} → AlCl_{3} + Cu

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider chlorine atom first. If we multiply 2 in the product (in AlCl_{3}) and 3 in the reactant (in CuCl_{2}) we will get the equal number of atoms in both sides.

⇒Step 4: Write the resulting equation:

Al + 3CuCl_{2} → 2AlCl_{3} + Cu

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of aluminum and copper atoms are unequal on the two sides.

First balance the aluminum atom.

⇒Step 6: If we multiply 2 in the reactant (in Al), we will get the equal number of atoms as in product (in 2AlCl_{3})

⇒Step 7: Write the resulting equation:

2Al + 3CuCl_{2} → 2AlCl_{3} + Cu

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is not balanced yet. As the number of copper atoms are unequal on the two sides.

⇒Step 9: If we multiply 3 in the product (in Cu), we will get the equal number of atoms as in product (in 3CuCl_{2})

⇒Step 10: Write the resulting equation:

2Al + 3CuCl_{2} → 2AlCl_{3} + 3Cu

⇒Step 11: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is now balanced.

⇒Step 12: Write down the final balanced equation:

2Al + 3CuCl_{2} → 2AlCl_{3} + 3Cu

**Question 9.**Write the balanced chemical equations for the following reactions. (AS1)

Hydrogen + Chlorine. Hydrogen chloride

**Answer:**H_{2} + Cl_{2}→ HCl

Balanced equation: H_{2} + Cl_{2}→ 2HCl

Explanation:

⇒Step 1: Write the given unbalanced equation

H_{2} + Cl_{2}→ HCl

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. First, let us consider hydrogen atom. If we multiply 2 in the product (in HCl), we will get the equal number of atoms as in reactant (H_{2})

⇒Step 4: Write the resulting equation:

H_{2} + O_{2} 2HCl

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is now balanced yet.

⇒Step 6: Write down the final balanced equation:

2H_{2} + Cl_{2} 2HCl

**Question 10.**Write the balanced chemical equations for the following reactions. (AS1)

Ammonium nitrate Nitrous Oxide + water.

**Answer:**NH_{4}NO_{3} → N_{2}O + H_{2}O

Balanced equation: NH_{4}NO_{3} → N_{2}O + 2H_{2}O

Explanation:

⇒Step 1: Write the given unbalanced equation

NH_{4}NO_{3} → N_{2}O + H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider hydrogen atom first. If we multiply 2 in the product (in H_{2}O), we will get the equal number of atoms as in the reactant (in NH_{4}NO_{3})

⇒Step 4: Write the resulting equation:

NH_{4}NO_{3} → N_{2}O + 2H_{2}O

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is now balanced yet.

⇒Step 6: Write down the final balanced equation:

NH_{4}NO_{3} → N_{2}O + 2H_{2}O

**Question 11.**Write the balanced chemical equation for the following and identity the type of reaction in each case. (AS1)

Calcium hydroxide_{(aq)} + Nitric acid_{(aq)} Water_{(1)}+ Calcium nitrate_{(aq)}

**Answer:**Ca(OH)_{2} + HNO_{3}→ H_{2}O + Ca(NO_{3})_{2}

Balanced equation: Ca(OH)_{2} + 2HNO_{3}→ 2H_{2}O + Ca(NO_{3})_{2}

Type of reaction: Double displacement reaction

**Question 12.**Write the balanced chemical equation for the following and identity the type of reaction in each case. (AS1)

Magnesium_{(s)}+ Iodine_{(g)}→ Magnesium iodide_{(g)}

**Answer:**Mg + I_{2}→ MgI_{2}

Balanced equation: Mg + I_{2}→ MgI_{2}

Type of reaction: Decomposition reaction

**Question 13.**Write the balanced chemical equation for the following and identity the type of reaction in each case. (AS1)

Magnesium_{(s)}+ Hydrochloric acid_{(aq)}→ Magnesium chloride_{(aq)}+ Hydrogen_{(g)}

**Answer:**Mg + HCl → MgCl_{2} + H_{2}

Balanced equation: Mg + 2HCl → MgCl_{2} + H_{2}

Type of reaction: Displacement reaction

**Question 14.**Write the balanced chemical equation for the following and identity the type of reaction in each case. (AS1)

Zinc_{(s)} + Calcium chloride_{(aq)}→ Zinc chloride_{(aq)} + calcium_{(s)}

**Answer:**Zn + CaCl_{2}→ ZnCl_{2} + Ca

Balanced equation: Zn + CaCl_{2}→ ZnCl_{2} + Ca

Type of reaction: Displacement reaction

**Question 15.**Write an equation for decomposition reaction where energy is supplied in the form of teat/ light/ electricity. (AS1)

**Answer:**On heating calcium carbonate, it decomposes to calcium oxide and carbon dioxide. The reaction is:

CaCO_{3}→ CaO + CO_{2}

The above reaction is an example of thermal decomposition reaction in which decomposition take places in the presence of heat.

Note: Decomposition reaction is a reaction in which only one reactant decomposes into two or more products.

**Question 16.**What do you mean by precipitation reaction? (AS1)

**Answer:**Precipitation reaction – When two reactants exchange their constituents chemically and form products in which one of them product is insoluble in water, the reaction takes place is called precipitation reaction.

For example: when aqueous lead nitrate and potassium iodide are mixed together, they form lead iodide which is insoluble in water. Lead iodide is called precipitate. The reaction is:

Pb(NO_{3})_{2} + KI → PbI_{2} + 2KNO_{3}

precipitate

**Question 17.**How chemical displacement reactions differ from chemical decomposition reaction? Explain with an example for each. (AS1)

**Answer:**Difference between displacement and double displacement reaction:

Note:

Displacement reaction –

Decomposition reaction –

**Question 18.**Name the reactions taking place in the presence of sunlight? (AS1)

**Answer:**The decomposition reaction which occurs in the presence of sunlight is called photochemical reaction.

For example: 2AgBr → 2Ag + Br_{2}

⇒In the above reaction, silver bromide (AgBr) decomposes to silver and bromine in sunlight.

⇒Light colour of AgBr changes to gray due to sunlight.

**Question 19.**Why does respiration considered as an exothermic reaction? Explain. (AS1)

**Answer:**Exothermic reaction: A reaction in which heat is released when reactants changes into products.

Respiration is considered as an exothermic reaction because:

⇒In respiration, a large amount of heat energy is released when oxidation of glucose takes place.

**Question 20.**What is the difference between displacement and double displacement reaction? Write equations for these reactions? (AS1)

**Answer:**Difference between displacement and double displacement reaction:

Note: Double displacement reaction –

**Question 21.**MnO_{2} + 4HCl MnCl_{2} + 2H_{2}O + Cl_{2}

In the above equation, name the compound which is oxidized and which is reduced? (A S1)

**Answer:**In the given reaction, MnO_{2} + 4HCl MnCl_{2} + 2H_{2}O + Cl_{2}

HCl is oxidized to Cl_{2} and MnO_{2} is reduced to MnCl_{2}

Explanation: As we know that oxidation is reaction that involves the removal of hydrogen and reduction is a reaction that involves the removal of oxygen.

Thus, HCl is oxidized and MnO_{2} is reduced.

**Question 22.**Give two examples for oxidation-reduction reaction. (AS1)

**Answer:**Oxidation – reduction reaction: When oxidation and reduction takes place at the same time, then the reaction is called oxidation-reduction reaction.

Oxidation: losing of electrons Reduction: gaining of electrons

For example:

CuO + H_{2}→ Cu + H_{2}O

⇒In the reaction, CuO loses oxygen atom which means that reduction of CuO (copper oxide) takes place.

⇒H_{2} (hydrogen) takes up oxygen atom.

⇒As a result, formation of water takes place. This means hydrogen undergoes oxidation.

Another example:

**Question 23.**In the refining of silver the recovery of silver from silver nitrate solution involved displacement by copper metal. Write the reaction involved. (AS1)

**Answer:**The reaction involved is:

2AgNO_{3} + Cu → Cu(NO_{3})_{2} + 2Ag

In the above reaction, when copper is mixed with silver nitrate, copper displaces the silver from silver nitrate and form its own ions as Cu(NO_{3})_{2}

**Question 24.**What do you mean by corrosion? How can you prevent it? (AS1)

**Answer:**Corrosion – When some metals are exposed to moisture, air, acids etc. they tarnish due to the formation of metals oxide on their surface. This process is called corrosion.

Corrosion can be prevented by:

⇒Shielding the surface, the metal surface from oxygen and moisture.

⇒By painting, oiling, greasing, galvanizing, chrome plating or making alloys.

⇒Galvanizing is a method of protecting iron from rusting by coating them a thin layer of zinc.

**Question 25.**Explain rancidity. (AS1)

**Answer:**When we use old, left over cooking oil for making foodstuff, it is found to have foul odour called rancidity.

⇒Rancidity is an oxidation reaction.

⇒If food is cooked in such oil, its taste also changes.

⇒When oils or fats are left aside for a long time, they undergo air oxidation and become rancid.

⇒Rancidity in the food stuff cooked in oil or ghee is prevented by using antioxidants.

**Question 26.**Balance the following chemical equations including the physical states. (AS1)

C_{6}H_{12}O_{6} C_{2}H_{5}OH+CO_{2}

**Answer:**Balanced equation: C_{6}H_{12}O_{6(s)} 2C_{2}H_{5}OH_{(s)} +2CO_{2(g)}

Explanation:

⇒Step 1: Write the given unbalanced equation

C_{6}H_{12}O_{6} C_{2}H_{5}OH+CO_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 2 in the product (in C_{2}H_{5}OH), we will get the equal number of atoms as in reactant (in C_{6}H_{12}O_{6})

⇒Step 4: Write the resulting equation:

C_{6}H_{12}O_{6} 2C_{2}H_{5}OH+CO_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of carbon and oxygen atoms are unequal on the two sides.

First balance the carbon atom.

⇒Step 6: If we multiply 2 in the product (in CO_{2}), we will get the equal number of atoms as in reactant (in C_{6}H_{12}O_{6})

⇒Step 7: Write the resulting equation:

C_{6}H_{12}O_{6} 2C_{2}H_{5}OH+2CO_{2}

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

C_{6}H_{12}O_{6} 2C_{2}H_{5}OH+2CO_{2}

**Question 27.**Balance the following chemical equations including the physical states. (AS1)

Fe + O_{2} Fe_{2}O_{2}

**Answer:**Fe + O_{2} Fe_{2}O_{2}

Balanced equation: 2Fe_{(s)} + O_{2(g)} Fe_{2}O_{2(s)}

Explanation:

⇒Step 1: Write the given unbalanced equation

Fe + O_{2} Fe_{2}O_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider Fe atom. If we multiply 2 in the reactant (in Fe), we will get the equal number of atoms as in product (in Fe_{2}O_{2})

⇒Step 4: Write the resulting equation:

2Fe + O_{2} Fe_{2}O_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is balanced now.

⇒Step 6: Write down the final balanced equation:

2Fe + O_{2} Fe_{2}O_{2}

**Question 28.**Balance the following chemical equations including the physical states. (AS1)

NH_{3} +Cl_{2} N_{2} +NH_{4}Cl

**Answer:**NH_{3} +Cl_{2} N_{2} +NH_{4}Cl

Balanced equation: 8NH_{3(g)} + 3Cl_{2(g)} N_{2(g)} + 6NH_{4}Cl_{(s)}

Explanation:

⇒Step 1: Write the given unbalanced equation

NH_{3} +Cl_{2} N_{2} +NH_{4}Cl

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 4 in the reactant (in NH_{3}) and 3 in product (in NH_{4}Cl), we will get the equal number of atoms as in both the sides.

⇒Step 4: Write the resulting equation:

4NH_{3} +Cl_{2} N_{2} + 3NH_{4}Cl

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of chlorine and nitrogen atoms are unequal on the two sides.

First balance the chlorine atom.

⇒Step 6: If we multiply 3 in the reactant (in Cl_{2}) and 2 in the product (in 3NH_{4}Cl), we will get the equal number of atoms as in both the sides.

⇒Step 7: Write the resulting equation:

4NH_{3} + 3Cl_{2} N_{2} + 6NH_{4}Cl

_{⇒}Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is not balanced yet. As the number of nitrogen atoms are unequal on the two sides.

Balance the nitrogen atom.

⇒Step 9: If we multiply 2 in the reactant (in 4NH_{3}) we will get the equal number of atoms as in products (in N_{2} and 6NH_{4}Cl)

⇒Step 10: Write the resulting equation:

4NH_{3} + 3Cl_{2} N_{2} + 6NH_{4}Cl

_{⇒}Step 11: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 12: Write down the final balanced equation:

4NH_{3} + 3Cl_{2} N_{2} + 6NH_{4}Cl

**Question 29.**Balance the following chemical equations including the physical states. (AS1)

Na+ H_{2}O NaOH + H_{2}

**Answer:**Na+ H_{2}O NaOH + H_{2}

Balanced equation: 2Na_{(s)} + 2H_{2}O_{(aq)}2NaOH_{(aq)} + H_{2(g)}↑

Explanation:

⇒Step 1: Write the given unbalanced equation.

Na+ H_{2}O NaOH + H_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 2 in the reactant (in H_{2}O) and 2 in product (in NaOH), we will get the equal number of atoms as in both the sides.

⇒Step 4: Write the resulting equation:

Na+ 2H_{2}O 2NaOH + H_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of sodium atoms are unequal on the two sides.

Balance the sodium atom.

⇒Step 6: If we multiply 2 in the reactant (in Na), we will get the equal number of atoms as in product (in 2NaOH).

⇒Step 7: Write the resulting equation:

2Na+ 2H_{2}O 2NaOH + H_{2}

_{⇒}Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

2Na+ 2H_{2}O 2NaOH + H_{2}

**Question 30.**Balance the chemical equation by including the physical states of the substances for the following reactions. (AS1)

Barium chloride and sodium sulphate aqueous solutions react to give insoluble Barium sulphate and aqueous solution of sodium chloride.

**Answer:**BaCl_{2} + Na_{2}SO_{4}→ BaSO_{4} + NaCl

Balanced equation:

Explanation: BaCl_{2(aq)} + Na_{2}SO_{4(aq)}→ BaSO_{4(s)}↓+ 2NaCl_{(aq)}

⇒Step 1: Write the given unbalanced equation

BaCl_{2} + Na_{2}SO_{4}→ BaSO_{4} + NaCl

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. First, let us consider sodium atom. If we multiply 2 in the product (in NaCl), we will get the equal number of atoms as in product (Na_{2}SO_{4})

⇒Step 4: Write the resulting equation:

BaCl_{2} + Na_{2}SO_{4}→ BaSO_{4} + 2NaCl

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is balanced now.

⇒Step 6: Write down the final balanced equation:

BaCl_{2} + Na_{2}SO_{4}→ BaSO_{4} + 2NaCl

**Question 31.**Balance the chemical equation by including the physical states of the substances for the following reactions. (AS1)

Sodium hydroxide reacts with hydrochloric acid to produce sodium chloride and water.

**Answer:**NaOH + HCl → NaCl + H_{2}O

Balanced equation: NaOH_{(aq)} + HCl_{(aq)}→ NaCl_{(aq)} + H_{2}O_{(l)}

Explanation:

⇒Step 1: Write the given unbalanced equation

NaOH + HCl → NaCl + H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

We find that the equation is already balanced.

⇒Step 3: Write down the balanced equation:

NaOH + HCl → NaCl + H_{2}O

**Question 32.**Balance the chemical equation by including the physical states of the substances for the following reactions. (AS1)

Zinc pieces react with dilute hydrochloric acid to liberate hydrogen gas and forms zinc chloride.

**Answer:**Zn + HCl → ZnCl_{2} + H_{2}

Balanced equation: Zn_{(s)} + 2HCl_{(aq)}→ ZnCl_{2(aq)} + H_{2(g)}↑

Explanation:

⇒Step 1: Write the given unbalanced equation

Zn + HCl → ZnCl_{2} + H_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having uSnequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 2 in the reactant (in HCl), we will get the equal number of atoms as in product (in H_{2})

⇒Step 4: Write the resulting equation:

Zn + 2HCl → ZnCl_{2} + H_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is balanced now.

⇒Step 6: Write down the final balanced equation:

Zn + 2HCl → ZnCl_{2} + H_{2}

**Question 33.**A shiny brown colored element 'X' on heating in air becomes black in colour. Can you predict the element ‘X’ and black colored substance filmed? How do you support your predictions? (AS2)

**Answer:**⇒It is given that ‘X’ is a shiny brown colored element. This means ‘X’ is a copper metal.

⇒When copper (X) is heated in air, it becomes black in color due to the deposit of copper oxide on its surface.

⇒The reaction takes place:

2Cu + O_{2}→ 2CuO

Brown Black

**Question 34.**Why do we apply paint on iron articles? (AS7)

**Answer:**We apply paint on iron articles to prevent the corrosion and rusting of iron. It decreases the rate of the process of rusting of iron.

**Question 35.**What is the use of keeping food in air tight containers? (AS7)

**Answer:**The use of keeping food in air tight containers is to prevent oxidation and to slow down the oxidation process, otherwise the food undergoes oxidation and becomes rancid.

**Question 1.**

What is a balanced chemical equation? Why should chemical equations be balanced? (AS1)

**Answer:**

Balanced chemical equation: A chemical equation in which the number of atoms of reactants and number of atoms of products is called a balanced equation.

Every chemical equation should be balanced because:

⇒According to the law of conservation of mass, atoms are neither created not destroyed in chemical reactions.

⇒It means the total mass of the products formed in chemical reaction must be equal to the mass of reactants consumed.

**Question 2.**

Balance the following chemical equations. (AS1)

NaOH + H_{2}SO_{4} Na_{2}SO_{4}H_{2}O

**Answer:**

NaOH + H_{2}SO_{4} Na_{2}SO_{4}H_{2}O

Balanced equation: 2NaOH + H_{2}SO_{4}Na_{2}SO_{4} + 2H_{2}O

Explanation:

⇒Step 1: Write the given unbalanced equation

NaOH + H_{2}SO_{4}Na_{2}SO_{4} + H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider sodium atom. If we multiply 2 in the reactant (in NaOH), we will get the equal number of atoms as in product (Na_{2}SO_{4})

⇒Step 4: Write the resulting equation:

2NaOH + H_{2}SO_{4}Na_{2}SO_{4} + H_{2}O

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of oxygen, hydrogen and Sulphur atoms are unequal on the two sides. First balance the hydrogen number.

⇒Step 6: Now, let us consider hydrogen atom. If we multiply 2 in the product (in H_{2}O), we will get the equal number of atoms as in reactants (in 2NaOH and H_{2}SO_{4})

⇒Step 7: Write the resulting equation:

2NaOH + H_{2}SO_{4}Na_{2}SO_{4} + 2H_{2}O

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

2NaOH + H_{2}SO_{4}Na_{2}SO_{4} + 2H_{2}O

**Question 3.**

Balance the following chemical equations. (AS1)

Hg(NO_{3})_{2} + KI HgI_{2} + KNO_{3}

**Answer:**

Hg(NO_{3})_{2} + KI HgI_{2} + KNO_{3}

Balanced equation: Hg(NO_{3})_{2} + 2KI HgI_{2} + 2KNO_{3}

Explanation:

⇒Step 1: Write the given unbalanced equation

Hg(NO_{3})_{2} + KI HgI_{2} + KNO_{3}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. First, let us consider iodine atom. If we multiply 2 in the reactant (in KI), we will get the equal number of atoms as in product (HgI_{2})

⇒Step 4: Write the resulting equation:

Hg(NO_{3})_{2} + 2KI HgI_{2} + KNO_{3}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of oxygen, nitrogen and potassium atoms are unequal on the two sides.

First balance the potassium number.

⇒Step 6: Now, let us consider potassium atom. If we multiply 2 in the product (KNO_{3}), we will get the equal number of atoms as in reactant (in KI)

⇒Step 7: Write the resulting equation:

Hg(NO_{3})_{2} + 2KI HgI_{2} + 2KNO_{3}

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

Hg(NO_{3})_{2} + 2KI HgI_{2} + 2KNO_{3}

**Question 4.**

Balance the following chemical equations. (AS1)

H_{2} + O_{2} H_{2}O

**Answer:**

H_{2} + O_{2} H_{2}O

Balanced equation: 2H_{2} + O_{2} 2H_{2}O

Explanation:

⇒Step 1: Write the given unbalanced equation

H_{2} + O_{2} H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. First, let us consider oxygen atom. If we multiply 2 in the product (in H_{2}O), we will get the equal number of atoms as in reactant (O_{2})

⇒Step 4: Write the resulting equation:

H_{2} + O_{2} 2H_{2}O

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of hydrogen atoms are unequal on the two sides.

⇒Step 6: Now, let us consider hydrogen atom. If we multiply 2 in the reactant (H_{2}), we will get the equal number of atoms as in product (in 2H_{2}O)

⇒Step 7: Write the resulting equation:

2H_{2} + O_{2} 2H_{2}O

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

2H_{2} + O_{2} 2H_{2}O

**Question 5.**

Balance the following chemical equations. (AS1)

KClO_{3}KCl + O_{2}

**Answer:**

KClO_{3}KCl + O_{2}

Balanced equation: 2KClO_{3}2KCl + 3O_{2}

Explanation:

⇒Step 1: Write the given unbalanced equation

KClO_{3}KCl + O_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider oxygen atom. If we multiply 2 in the product (in KClO_{3}) and 3 in the reactant (in O_{2}) we will get the equal number of atoms in both sides.

⇒Step 4: Write the resulting equation:

2KClO_{3}KCl + 3O_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of potassium and chlorine atoms are unequal on the two sides. First balance the potassium atom.

⇒Step 6: If we multiply 2 in the product (in KCl), we will get the equal number of atoms as in reactant (in 2KClO_{3})

⇒Step 7: Write the resulting equation:

2KClO_{3}2KCl + 3O_{2}

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

2KClO_{3}2KCl + 3O_{2}

**Question 6.**

Balance the following chemical equations. (AS1)

C_{3}H_{8} + O_{2} CO_{2} + H_{2}O

**Answer:**

C_{3}H_{8} + O_{2} CO_{2} + H_{2}O

Balanced equation: C_{3}H_{8} + 5O_{2} 3CO_{2} + 4H_{2}O

Explanation:

⇒Step 1: Write the given unbalanced equation

C_{3}H_{8} + O_{2} CO_{2} + H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 4 in the product (in H_{2}O), we will get the equal number of atoms as in reactant (in C_{3}H_{8})

⇒Step 4: Write the resulting equation:

C_{3}H_{8} + O_{2} CO_{2} + 4H_{2}O

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of carbon and oxygen atoms are unequal on the two sides.

First balance the carbon atom.

⇒Step 6: If we multiply 3 in the product (in CO_{2}), we will get the equal number of atoms as in reactant (in C_{3}H_{8})

⇒Step 7: Write the resulting equation:

C_{3}H_{8} + O_{2} 3CO_{2} + 4H_{2}O

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is not balanced yet. As the number of oxygen atoms are unequal on the two sides.

⇒Step 9: Now, we consider oxygen atoms. If we multiply 5 in the reactant (in O_{2}), we will get the equal number of atoms as in products (in 3CO_{2} and 4H_{2}O)

⇒Step 10: Write the resulting equation:

C_{3}H_{8} + 5O_{2} 3CO_{2} + 4H_{2}O

We find that the equation is balanced now.

⇒Step 11: Write down the final balanced equation:

C_{3}H_{8} + 5O_{2} 3CO_{2} + 4H_{2}O

**Question 7.**

Write the balanced chemical equations for the following reactions. (AS1)

Zinc +Silver nitrateZinc nitrate+ Silver.

**Answer:**

Zn + AgNO_{3}→ Zn(NO_{3})_{2} + Ag

Balanced equation: Zn + 2AgNO_{3}→ Zn(NO_{3})_{2} + 2Ag

Explanation:

⇒Step 1: Write the given unbalanced equation

Zn + AgNO_{3}→ Zn(NO_{3})_{2} + Ag

Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider nitrogen atom first. If we multiply 2 in the reactant (in AgNO_{3}), we will get the equal number of atoms as in product (Zn(NO_{3})_{2})

⇒Step 4: Write the resulting equation:

Zn + 2AgNO_{3}→ Zn(NO_{3})_{2} + Ag

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of silver atoms are unequal on the two sides.

⇒Step 6: Now, let us consider silver atom. If we multiply 2 in the product (in Ag), we will get the equal number of atoms as in reactant (in 2AgNO_{3})

⇒Step 7: Write the resulting equation:

Zn + 2AgNO_{3}→ Zn(NO_{3})_{2} + 2Ag

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

Zn + 2AgNO_{3}→ Zn(NO_{3})_{2} + 2Ag

**Question 8.**

Write the balanced chemical equations for the following reactions. (AS1)

Aluminum + copper chloride Aluminum chloride+ Copper.

**Answer:**

Al + CuCl_{2} → AlCl_{3} + Cu

Balanced equation: 2Al + 3CuCl_{2} → 2AlCl_{3} + 3Cu

Explanation:

⇒Step 1: Write the given unbalanced equation

Al + CuCl_{2} → AlCl_{3} + Cu

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider chlorine atom first. If we multiply 2 in the product (in AlCl_{3}) and 3 in the reactant (in CuCl_{2}) we will get the equal number of atoms in both sides.

⇒Step 4: Write the resulting equation:

Al + 3CuCl_{2} → 2AlCl_{3} + Cu

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of aluminum and copper atoms are unequal on the two sides.

First balance the aluminum atom.

⇒Step 6: If we multiply 2 in the reactant (in Al), we will get the equal number of atoms as in product (in 2AlCl_{3})

⇒Step 7: Write the resulting equation:

2Al + 3CuCl_{2} → 2AlCl_{3} + Cu

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is not balanced yet. As the number of copper atoms are unequal on the two sides.

⇒Step 9: If we multiply 3 in the product (in Cu), we will get the equal number of atoms as in product (in 3CuCl_{2})

⇒Step 10: Write the resulting equation:

2Al + 3CuCl_{2} → 2AlCl_{3} + 3Cu

⇒Step 11: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is now balanced.

⇒Step 12: Write down the final balanced equation:

2Al + 3CuCl_{2} → 2AlCl_{3} + 3Cu

**Question 9.**

Write the balanced chemical equations for the following reactions. (AS1)

Hydrogen + Chlorine. Hydrogen chloride

**Answer:**

H_{2} + Cl_{2}→ HCl

Balanced equation: H_{2} + Cl_{2}→ 2HCl

Explanation:

⇒Step 1: Write the given unbalanced equation

H_{2} + Cl_{2}→ HCl

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. First, let us consider hydrogen atom. If we multiply 2 in the product (in HCl), we will get the equal number of atoms as in reactant (H_{2})

⇒Step 4: Write the resulting equation:

H_{2} + O_{2} 2HCl

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is now balanced yet.

⇒Step 6: Write down the final balanced equation:

2H_{2} + Cl_{2} 2HCl

**Question 10.**

Write the balanced chemical equations for the following reactions. (AS1)

Ammonium nitrate Nitrous Oxide + water.

**Answer:**

NH_{4}NO_{3} → N_{2}O + H_{2}O

Balanced equation: NH_{4}NO_{3} → N_{2}O + 2H_{2}O

Explanation:

⇒Step 1: Write the given unbalanced equation

NH_{4}NO_{3} → N_{2}O + H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider hydrogen atom first. If we multiply 2 in the product (in H_{2}O), we will get the equal number of atoms as in the reactant (in NH_{4}NO_{3})

⇒Step 4: Write the resulting equation:

NH_{4}NO_{3} → N_{2}O + 2H_{2}O

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is now balanced yet.

⇒Step 6: Write down the final balanced equation:

NH_{4}NO_{3} → N_{2}O + 2H_{2}O

**Question 11.**

Write the balanced chemical equation for the following and identity the type of reaction in each case. (AS1)

Calcium hydroxide_{(aq)} + Nitric acid_{(aq)} Water_{(1)}+ Calcium nitrate_{(aq)}

**Answer:**

Ca(OH)_{2} + HNO_{3}→ H_{2}O + Ca(NO_{3})_{2}

Balanced equation: Ca(OH)_{2} + 2HNO_{3}→ 2H_{2}O + Ca(NO_{3})_{2}

Type of reaction: Double displacement reaction

**Question 12.**

Write the balanced chemical equation for the following and identity the type of reaction in each case. (AS1)

Magnesium_{(s)}+ Iodine_{(g)}→ Magnesium iodide_{(g)}

**Answer:**

Mg + I_{2}→ MgI_{2}

Balanced equation: Mg + I_{2}→ MgI_{2}

Type of reaction: Decomposition reaction

**Question 13.**

Write the balanced chemical equation for the following and identity the type of reaction in each case. (AS1)

Magnesium_{(s)}+ Hydrochloric acid_{(aq)}→ Magnesium chloride_{(aq)}+ Hydrogen_{(g)}

**Answer:**

Mg + HCl → MgCl_{2} + H_{2}

Balanced equation: Mg + 2HCl → MgCl_{2} + H_{2}

Type of reaction: Displacement reaction

**Question 14.**

Write the balanced chemical equation for the following and identity the type of reaction in each case. (AS1)

Zinc_{(s)} + Calcium chloride_{(aq)}→ Zinc chloride_{(aq)} + calcium_{(s)}

**Answer:**

Zn + CaCl_{2}→ ZnCl_{2} + Ca

Balanced equation: Zn + CaCl_{2}→ ZnCl_{2} + Ca

Type of reaction: Displacement reaction

**Question 15.**

Write an equation for decomposition reaction where energy is supplied in the form of teat/ light/ electricity. (AS1)

**Answer:**

On heating calcium carbonate, it decomposes to calcium oxide and carbon dioxide. The reaction is:

CaCO_{3}→ CaO + CO_{2}

The above reaction is an example of thermal decomposition reaction in which decomposition take places in the presence of heat.

Note: Decomposition reaction is a reaction in which only one reactant decomposes into two or more products.

**Question 16.**

What do you mean by precipitation reaction? (AS1)

**Answer:**

Precipitation reaction – When two reactants exchange their constituents chemically and form products in which one of them product is insoluble in water, the reaction takes place is called precipitation reaction.

For example: when aqueous lead nitrate and potassium iodide are mixed together, they form lead iodide which is insoluble in water. Lead iodide is called precipitate. The reaction is:

Pb(NO_{3})_{2} + KI → PbI_{2} + 2KNO_{3}

precipitate

**Question 17.**

How chemical displacement reactions differ from chemical decomposition reaction? Explain with an example for each. (AS1)

**Answer:**

Difference between displacement and double displacement reaction:

Note:

Displacement reaction –

Decomposition reaction –

**Question 18.**

Name the reactions taking place in the presence of sunlight? (AS1)

**Answer:**

The decomposition reaction which occurs in the presence of sunlight is called photochemical reaction.

For example: 2AgBr → 2Ag + Br_{2}

⇒In the above reaction, silver bromide (AgBr) decomposes to silver and bromine in sunlight.

⇒Light colour of AgBr changes to gray due to sunlight.

**Question 19.**

Why does respiration considered as an exothermic reaction? Explain. (AS1)

**Answer:**

Exothermic reaction: A reaction in which heat is released when reactants changes into products.

Respiration is considered as an exothermic reaction because:

⇒In respiration, a large amount of heat energy is released when oxidation of glucose takes place.

**Question 20.**

What is the difference between displacement and double displacement reaction? Write equations for these reactions? (AS1)

**Answer:**

Difference between displacement and double displacement reaction:

Note: Double displacement reaction –

**Question 21.**

MnO_{2} + 4HCl MnCl_{2} + 2H_{2}O + Cl_{2}

In the above equation, name the compound which is oxidized and which is reduced? (A S1)

**Answer:**

In the given reaction, MnO_{2} + 4HCl MnCl_{2} + 2H_{2}O + Cl_{2}

HCl is oxidized to Cl_{2} and MnO_{2} is reduced to MnCl_{2}

Explanation: As we know that oxidation is reaction that involves the removal of hydrogen and reduction is a reaction that involves the removal of oxygen.

Thus, HCl is oxidized and MnO_{2} is reduced.

**Question 22.**

Give two examples for oxidation-reduction reaction. (AS1)

**Answer:**

Oxidation – reduction reaction: When oxidation and reduction takes place at the same time, then the reaction is called oxidation-reduction reaction.

Oxidation: losing of electrons Reduction: gaining of electrons

For example:

CuO + H_{2}→ Cu + H_{2}O

⇒In the reaction, CuO loses oxygen atom which means that reduction of CuO (copper oxide) takes place.

⇒H_{2} (hydrogen) takes up oxygen atom.

⇒As a result, formation of water takes place. This means hydrogen undergoes oxidation.

Another example:

**Question 23.**

In the refining of silver the recovery of silver from silver nitrate solution involved displacement by copper metal. Write the reaction involved. (AS1)

**Answer:**

The reaction involved is:

2AgNO_{3} + Cu → Cu(NO_{3})_{2} + 2Ag

In the above reaction, when copper is mixed with silver nitrate, copper displaces the silver from silver nitrate and form its own ions as Cu(NO_{3})_{2}

**Question 24.**

What do you mean by corrosion? How can you prevent it? (AS1)

**Answer:**

Corrosion – When some metals are exposed to moisture, air, acids etc. they tarnish due to the formation of metals oxide on their surface. This process is called corrosion.

Corrosion can be prevented by:

⇒Shielding the surface, the metal surface from oxygen and moisture.

⇒By painting, oiling, greasing, galvanizing, chrome plating or making alloys.

⇒Galvanizing is a method of protecting iron from rusting by coating them a thin layer of zinc.

**Question 25.**

Explain rancidity. (AS1)

**Answer:**

When we use old, left over cooking oil for making foodstuff, it is found to have foul odour called rancidity.

⇒Rancidity is an oxidation reaction.

⇒If food is cooked in such oil, its taste also changes.

⇒When oils or fats are left aside for a long time, they undergo air oxidation and become rancid.

⇒Rancidity in the food stuff cooked in oil or ghee is prevented by using antioxidants.

**Question 26.**

Balance the following chemical equations including the physical states. (AS1)

C_{6}H_{12}O_{6} C_{2}H_{5}OH+CO_{2}

**Answer:**

Balanced equation: C_{6}H_{12}O_{6(s)} 2C_{2}H_{5}OH_{(s)} +2CO_{2(g)}

Explanation:

⇒Step 1: Write the given unbalanced equation

C_{6}H_{12}O_{6} C_{2}H_{5}OH+CO_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 2 in the product (in C_{2}H_{5}OH), we will get the equal number of atoms as in reactant (in C_{6}H_{12}O_{6})

⇒Step 4: Write the resulting equation:

C_{6}H_{12}O_{6} 2C_{2}H_{5}OH+CO_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of carbon and oxygen atoms are unequal on the two sides.

First balance the carbon atom.

⇒Step 6: If we multiply 2 in the product (in CO_{2}), we will get the equal number of atoms as in reactant (in C_{6}H_{12}O_{6})

⇒Step 7: Write the resulting equation:

C_{6}H_{12}O_{6} 2C_{2}H_{5}OH+2CO_{2}

⇒Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

C_{6}H_{12}O_{6} 2C_{2}H_{5}OH+2CO_{2}

**Question 27.**

Balance the following chemical equations including the physical states. (AS1)

Fe + O_{2} Fe_{2}O_{2}

**Answer:**

Fe + O_{2} Fe_{2}O_{2}

Balanced equation: 2Fe_{(s)} + O_{2(g)} Fe_{2}O_{2(s)}

Explanation:

⇒Step 1: Write the given unbalanced equation

Fe + O_{2} Fe_{2}O_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, let us consider Fe atom. If we multiply 2 in the reactant (in Fe), we will get the equal number of atoms as in product (in Fe_{2}O_{2})

⇒Step 4: Write the resulting equation:

2Fe + O_{2} Fe_{2}O_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is balanced now.

⇒Step 6: Write down the final balanced equation:

2Fe + O_{2} Fe_{2}O_{2}

**Question 28.**

Balance the following chemical equations including the physical states. (AS1)

NH_{3} +Cl_{2} N_{2} +NH_{4}Cl

**Answer:**

NH_{3} +Cl_{2} N_{2} +NH_{4}Cl

Balanced equation: 8NH_{3(g)} + 3Cl_{2(g)} N_{2(g)} + 6NH_{4}Cl_{(s)}

Explanation:

⇒Step 1: Write the given unbalanced equation

NH_{3} +Cl_{2} N_{2} +NH_{4}Cl

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 4 in the reactant (in NH_{3}) and 3 in product (in NH_{4}Cl), we will get the equal number of atoms as in both the sides.

⇒Step 4: Write the resulting equation:

4NH_{3} +Cl_{2} N_{2} + 3NH_{4}Cl

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of chlorine and nitrogen atoms are unequal on the two sides.

First balance the chlorine atom.

⇒Step 6: If we multiply 3 in the reactant (in Cl_{2}) and 2 in the product (in 3NH_{4}Cl), we will get the equal number of atoms as in both the sides.

⇒Step 7: Write the resulting equation:

4NH_{3} + 3Cl_{2} N_{2} + 6NH_{4}Cl

_{⇒}Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is not balanced yet. As the number of nitrogen atoms are unequal on the two sides.

Balance the nitrogen atom.

⇒Step 9: If we multiply 2 in the reactant (in 4NH_{3}) we will get the equal number of atoms as in products (in N_{2} and 6NH_{4}Cl)

⇒Step 10: Write the resulting equation:

4NH_{3} + 3Cl_{2} N_{2} + 6NH_{4}Cl

_{⇒}Step 11: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 12: Write down the final balanced equation:

4NH_{3} + 3Cl_{2} N_{2} + 6NH_{4}Cl

**Question 29.**

Balance the following chemical equations including the physical states. (AS1)

Na+ H_{2}O NaOH + H_{2}

**Answer:**

Na+ H_{2}O NaOH + H_{2}

Balanced equation: 2Na_{(s)} + 2H_{2}O_{(aq)}2NaOH_{(aq)} + H_{2(g)}↑

Explanation:

⇒Step 1: Write the given unbalanced equation.

Na+ H_{2}O NaOH + H_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 2 in the reactant (in H_{2}O) and 2 in product (in NaOH), we will get the equal number of atoms as in both the sides.

⇒Step 4: Write the resulting equation:

Na+ 2H_{2}O 2NaOH + H_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is not balanced yet. As the number of sodium atoms are unequal on the two sides.

Balance the sodium atom.

⇒Step 6: If we multiply 2 in the reactant (in Na), we will get the equal number of atoms as in product (in 2NaOH).

⇒Step 7: Write the resulting equation:

2Na+ 2H_{2}O 2NaOH + H_{2}

_{⇒}Step 8: Now check whether the equation is balanced or not by comparing the atoms.

We find that the equation is balanced now.

⇒Step 9: Write down the final balanced equation:

2Na+ 2H_{2}O 2NaOH + H_{2}

**Question 30.**

Balance the chemical equation by including the physical states of the substances for the following reactions. (AS1)

Barium chloride and sodium sulphate aqueous solutions react to give insoluble Barium sulphate and aqueous solution of sodium chloride.

**Answer:**

BaCl_{2} + Na_{2}SO_{4}→ BaSO_{4} + NaCl

Balanced equation:

Explanation: BaCl_{2(aq)} + Na_{2}SO_{4(aq)}→ BaSO_{4(s)}↓+ 2NaCl_{(aq)}

⇒Step 1: Write the given unbalanced equation

BaCl_{2} + Na_{2}SO_{4}→ BaSO_{4} + NaCl

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having unequal no. of atoms on both sides. First, let us consider sodium atom. If we multiply 2 in the product (in NaCl), we will get the equal number of atoms as in product (Na_{2}SO_{4})

⇒Step 4: Write the resulting equation:

BaCl_{2} + Na_{2}SO_{4}→ BaSO_{4} + 2NaCl

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is balanced now.

⇒Step 6: Write down the final balanced equation:

BaCl_{2} + Na_{2}SO_{4}→ BaSO_{4} + 2NaCl

**Question 31.**

Balance the chemical equation by including the physical states of the substances for the following reactions. (AS1)

Sodium hydroxide reacts with hydrochloric acid to produce sodium chloride and water.

**Answer:**

NaOH + HCl → NaCl + H_{2}O

Balanced equation: NaOH_{(aq)} + HCl_{(aq)}→ NaCl_{(aq)} + H_{2}O_{(l)}

Explanation:

⇒Step 1: Write the given unbalanced equation

NaOH + HCl → NaCl + H_{2}O

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

We find that the equation is already balanced.

⇒Step 3: Write down the balanced equation:

NaOH + HCl → NaCl + H_{2}O

**Question 32.**

Balance the chemical equation by including the physical states of the substances for the following reactions. (AS1)

Zinc pieces react with dilute hydrochloric acid to liberate hydrogen gas and forms zinc chloride.

**Answer:**

Zn + HCl → ZnCl_{2} + H_{2}

Balanced equation: Zn_{(s)} + 2HCl_{(aq)}→ ZnCl_{2(aq)} + H_{2(g)}↑

Explanation:

⇒Step 1: Write the given unbalanced equation

Zn + HCl → ZnCl_{2} + H_{2}

⇒Step 2: Compare the number of atoms of reactants with the number of atoms of products.

⇒Step 3: Now, first we consider the element having uSnequal no. of atoms on both sides. Thus, first let us consider hydrogen atom. If we multiply 2 in the reactant (in HCl), we will get the equal number of atoms as in product (in H_{2})

⇒Step 4: Write the resulting equation:

Zn + 2HCl → ZnCl_{2} + H_{2}

⇒Step 5: Now check whether the equation is balanced or not by comparing the atoms

We find that the equation is balanced now.

⇒Step 6: Write down the final balanced equation:

Zn + 2HCl → ZnCl_{2} + H_{2}

**Question 33.**

A shiny brown colored element 'X' on heating in air becomes black in colour. Can you predict the element ‘X’ and black colored substance filmed? How do you support your predictions? (AS2)

**Answer:**

⇒It is given that ‘X’ is a shiny brown colored element. This means ‘X’ is a copper metal.

⇒When copper (X) is heated in air, it becomes black in color due to the deposit of copper oxide on its surface.

⇒The reaction takes place:

2Cu + O_{2}→ 2CuO

Brown Black

**Question 34.**

Why do we apply paint on iron articles? (AS7)

**Answer:**

We apply paint on iron articles to prevent the corrosion and rusting of iron. It decreases the rate of the process of rusting of iron.

**Question 35.**

What is the use of keeping food in air tight containers? (AS7)

**Answer:**

The use of keeping food in air tight containers is to prevent oxidation and to slow down the oxidation process, otherwise the food undergoes oxidation and becomes rancid.