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Online Education NCERT Solutions For Class 9 Science Chapter 3 Atoms and Molecules

These Solutions are part of Online Education NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Chemistry) Chapter 3 – Atoms and Molecules solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 3 – Atoms and Molecules Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

More Resources

• NCERT Solutions for Class 9 Science
• HOTS Questions for Class 9 Science
• Value Based Questions in Science for Class 9
• NCERT Exemplar Solutions for Class 9 Science
• Previous Year Question Papers for CBSE Class 9 Science

NCERT IN TEXT PROBLEMS

IN TEXT QUESTIONS

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6.0 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. (CBSE 2012)
Answer:
The chemical reaction leading to products is :
sodium carbonate + ethanoic acid ———–> sodium ethanoate + carbon dioxide + water.
Mass of reactants = (5.3 + 6.0) = 1.3 g
Mass of products = (8.2 + 2.2 + 0.9) = 11.3 g.
The reactants and products have same mass. This means that there was no loss of mass during the reaction. Therefore, the data is in agreement with law of conservation of mass.

Question 2.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react with 3 g of hydrogen gas ?
Answer:
According to available data,
Mass of oxygen combining with 1 g of hydrogen = 8 g.

Question 3:
Which postulate of Dalton’s Atomic theory is the basis of law of conservation of mass ? (CBSE 2012)
Answer:
The law of conservation mass is based on the following postulate of Dalton’s Atomic theory.
“Atoms can neither be created nor destroyed during a physical change or a chemical reaction.”

Question 4.
Which postulate of Dalton’s Atomic theory can explain the law of definite proportions ?
Answer:
The law of definite proportions is based on the following postulate of Dalton’s Atomic theory.
“All atoms of a particular element are identical in every respect. This means that they have same mass, same size and also same chemical properties.”

Question 5.
Define atomic mass unit.
Answer:
Atomic mass unit may be defined as :
The mass of one-twelfth (1/12) of the mass of one atom of carbon taken as 12 u. It is represented as 1 u (unified mass).

Question 6.
Why is not possible to see an atom with naked eye ?
Answer:
It is not possible to see an atom with naked eye because of its extremely small size. For example, the radius of an atom of hydrogen is of the order of 10^-10 m. Actually an atom is regarded as a microscopic particle. These microscopic particles cannot be seen with naked eye.

Question 7.
Write down the formulae of :
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide.
Answer:

Question 8.
Write the names of the compounds represented by the following formulae :
(i) Al₂(SO₄)₃
(ii) CaCl₂
(Hi) K₂SO₄
(iv) KNO₃
(v) CaCO₃.
Answer:
(i) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate

Question 9.
What is meant by the term chemical formula ?
Answer:
Molecule represents a group of two or more atoms (same or different) chemically bonded to each other and held tightly by strong attractive forces. Molecules are represented in terms of symbols of constituting atoms and it is known as chemical formula.

Question 10.
How many atoms are present in
(i) H₂S molecule
(ii) PO₄^3- ion ?
Answer:
(i) Three
(ii) Five.

Question 11.
Calculate the molecular masses of :
(i) H₂
(ii) O₂
(iii) Cl₂
(iv) CO₂
(v) CH₄
(vi) C₂H₆
(vii) C₂H₄
(viii) NH₃
(ix) CH₃OH.
Answer:
(i) Hydrogen (H₂)
Molecular mass of H₂ = 2 x Atomic mass of H = (2 x 1 u) = 2 u.
(ii) Oxygen (O₂)
Molecular mass of O₂ = 2 x Atomic mass of O = (2 x 16 u) = 32 u.
(iii) Chlorine (CI₂)
Molecular mass of Cl₂ = 2 x Atomic mass of Cl = (2 x 35’5 u) = 71 u.
(iv) Carbon dioxide (CO₂)
Molecular mass of CO₂ = (1 x Atomic mass of C) + (2 x Atomic mass of O)
= (1 x 12 u) + (2 x 16 u) = 12 u + 32 u = 44 u.
(v) Methane (CH₄)
Molecular mass of CH₄ = ( 1 x Atomic mass of C) + (4 x Atomic mass of H)
= ( 1 x 12 u) + (4 x 1 u) = 16 u.
(vi) Ethane (C₂H₆)
Molecular mass of C₂H₆ = (2 x Atomic mass of C) + (6 x Atomic mass of H)
= (2 x 12 u) + (6 x 1 u) = 30 u.
(vii) Ethylene (C₂H₄)
Molecular mass of C₂H₄ =(2 x Atomic mass of C) + (4 x Atomic mass of H)
= (2 x 12 u) + (4 x 1 u) = 28 u.
(viii) Ammonia (NH₃)
Molecular mass of NH₃ = (1 x Atomic mass ofN) + (3 x Atomic mass of H)
= (1 x 14 u) + ( 3 x 1 u) = 17 u.
(ix) Methyl alcohol (CH₃OH)
Molecular mass of CH₃OH = (1 x Atomic mass of C) + (4 x Atomic mass of H)
+ (1 x Atomic mass of O)
= (1 x 12 u) + (4 x 1 u) + (1 x 16 u) = 32 u.

Question 12.
Calculate the formula unit mass of :
(i) ZnO
(ii) Na₂O
(iii) K₂CO₃.
Given : Atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and O = 16 u.
Answer:
(i) Formula unit mass of ZnO (Zinc oxide)
= (1 x Atomic mass of Zn) +(1 x Atomic mass of O)
= (1 x 65 u) + (1 x 16 u) = 81 u.
(ii) Formula unit mass of Na₂O (Sodium oxide)
= (2 x Atomic mass of Na) + (1 x Atomic mass of O)
= (2 x 23 u) + (1 x 16 u) = 62 u.
(iii) Formula unit mass of K₂CO₃ (Potassium carbonate).
= (2 x Atomic mass of K) + (1 x Atomic mass of C) + (3 x Atomic mass of O)
= (2 x 39 u) + (1 x 12 u) + (3 x 16 u) = 138 u

Question 13.
Calculate the number of moles of the following :
(i) 52 g of He
(ii) 12.044 x 10²³ atoms of He.
Answer:

Question 14.
Calculate the number of particles in each of the following :
(i) 46 g of sodium atoms
(ii) 8 g of oxygen (O₂)
(iii) 0.1 mole of carbon atoms.
Answer:

Question 15.
If one mole of carbon weighs 12 grams, what is the mass (in gram) of one atom of carbon ?
Answer:

Question 16.
Which has more number of atoms ?
(a) 100 grams of sodium
(b) 100 grams of iron
(Given : atomic mass of Na = 23 u ; Fe = 56 u)
Answer:

NCERT END EXERCISE

Question 1.
0.24 g of sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron
and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:

Question 2.
When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is formed. What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.0 g of oxygen ? Which law of chemical combination will govern your answer ? (CBSE 2011, 2012)
Answer:
Carbon and oxygen react to form carbon dioxide according to the equation Carbon (C) + Oxygen (O₂) > Carbon dioxide (CO₂)

In the first case :
3.0 g of carbon are burnt in 8.0 g of oxygen to form 11.0 g of CO₂ In the second case :
3.0 g of carbon must also combine with 8.0 g of oxygen only. This means that (50 – 8) = 42 g of oxygen will remain unreacted.
The mass of CO₂ in this case must be also 11 g.
The answer is based on Law of constant proportions.
In the second case, only 8.0 g of oxygen react although 50.0 g are available. This shows that the mass of carbon dioxide (11.0 g) formed depends upon the mass of carbon (3.0 g) or the substance present in smaller amount. In general, the substance (element or compound) present in smaller amount in a reaction limits the participation of the other reactants. It is quite often called limiting reactant. Carbon is the limiting reactant in this case. It limits the participation of oxygen and also the formation of carbon dioxide.

Question 3.
What are polyatomic ions ? Give examples. (CBSE 2015)
Answer:
Polyatomic ions are the group of atoms which carry either positive charge (cations) or negative charge (anions). For example,

1. Carbonate ion (CO₃)^2-
2. Nitrate ion (NO₃)^–
3. Ammonium ion (NH₄)⁺
4. Phosphate ion (PO₄)^3-.

Question 4.
Write the chemical formulae of the following :
Answer:

Question 5.
Give the names of the elements present in the following compounds :
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer:
The names of elements present can be given only if the chemical formula of the compound is known. For example,
(a) Quick lime : It is the commercial name of the compound. Its chemical name is calcium oxide and the
chemical formula is CaO.
Elements present : calcium (Ca) : oxygen (O).
(b) Hydrogen bromide : The chemical formula of the compound is HBr
Elements present : hydrogen (H) ; bromine (Br).
(c) Baking powder : It is the commercial name of the compound. Its chemical name is sodium hydrogen carbonate and the chemical formula is NaHCO₃
Elements present : sodium (Na), hydrogen (H), carbon (C), oxygen (O).
(d) Potassium sulphate : The chemical formula of the compound is K₂SO₄
Elements present : potassium (K), sulphur (S), oxygen (O).

Question 6.
Calculate the molar mass of the following substances :
(a) Ethyne, C₂H₂
(b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
(d) Nitric acid, HNO₃
(e) Hydrochloric acid, HCl.
Answer:
(a) Ethyne, C₂H₂
Molar mass of C₂H₂ = (2 x Atomic mass of C) + (2 x Atomic mass of H)
= (2 x 12 u) + (2 x 1 u) = 26 u.
(b) Sulphur molecule, S₈
Molar mass of S₈ = 8 x Atomic mass of S = (8 x 32 u) = 256 u
(c) Phosphorus molecule, P₄
Molar mass of P₄ =4 x Atomic mass of P = (4 x 31 u) = 124 u
(d) Nitric acid, HNO₃
Molar mass of HNO₃ = (1 x Atomic mass of H) + (1 x Atomic mass of N) + ( 3 x Atomic mass of O)
= (1 x 1 u) + (1 x 14 u) + (3 x 16 u) = 63 u.
(e) Hydrochloric acid, HCl.
Molar mass of HCl = (1 x Atomic mass of H) + (1 x Atomic mass of Cl)
= (1 x 1 u) + (1 x 35.5 u) = 36.5 u.

Question 7.
What is the mass of :
(a) 1 mole of nitrogen atoms ?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27) ?
(c) 10 moles of sodium sulphite (Na₂SO₃) ?
Answer:
(a) 1 mole of nitrogen atoms
Mass of 1 mole of nitrogen (N) atoms = 14 u
(b) 4 moles of aluminium atoms
Mass of 1 mole of aluminium (Al) atoms = 27 u
Mass of 4 moles of aluminium (Al) atoms = 4 x 27 = 108 u
(c) 10 moles of sodium sulphite (Na₂SO₃)
Molar mass of Na₂SO₃ = 2 x Atomic mass of Na + Atomic mass of S + 3 x Atomic mass of O
= 2 x 23 + 32 + 3 x 16 = 126 u
1 mole of sodium sulphite has mass = 126 u
10 moles of sodium sulphite have mass = 10 x 126 = 1260 u.

Question 8.
Convert into moles
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
Answer:

Question 9.
What is the mass of :
(a) 0.2 mole of oxygen atoms ?
(b) 0.5 mole of water molecules ?
Answer:
(a) 0.2 mole of oxygen atoms
Mass of 1 mole of oxygen (O) atoms = 16 u
Mass of 0.2 mole of oxygen (O) atoms = 0.2 x 16 = 3.2 u
(b) 0.5 mole of water molecules
Mass of 1 mole of water (H₂O) molecules = 2 x 1 + 16 = 18 u
Mass of 0.5 mole of water (H2O) molecules = 0.5 x 18 = 9 u.

Question 10.
Calculate the number of molecules of sulphur (Sg) present in 16 g of solid sulphur.
Answer:

Question 11.
Calculate the number of aluminium ions in 0.051 g of aluminium oxide (Al₂O₃). (CBSE 2012)
Answer:

VERY SHORT ANSWER QUESTIONS

Question 1.
Out of atoms and molecules, which can exist independently ?
Answer:
Molecules can exist independently. However, the atoms of noble gases (He, Ne, Ar, Kr, Xe) can also exist independently.

Question 2.
What does the symbol ‘u’ represent ?
Answer:
The symbol V represents unified mass.

Question 3.
Write the chemical symbols and Latin names of
(i) gold
(ii) mercury ?
Answer:
(i) Au (Aurum)
(ii) Hg (Hydrargyrum).

Question 4.
Are the mass of a molecule of a substance and its molar mass same ?
Answer:
No, they are different.

Question 5.
How are mass, molar mass and number of moles of a substance related to each other ?
Answer:

Question 6.
Avogardo’s number represents how many particles ?
Answer:
Avogadro’s number (Ng) represents 6.022 x 10²³ particles.

Question 7.
Give an example of
(i) a divalent anion
(ii) a trivalent cation
(iii) a monovalent anion.
Answer:
(i) (SO₄)^2-
(ii) Al₃⁺
(iii) Cl^–

Question 8.
Calculate the molar mass of ethyl alcohol (C₂H₅OH).
Answer:
Molar mass of C₂H₅OH= (2 x gram atomic mass of C) + (6 x gram atomic mass of H)
+ (1 x Atomic mass of O)
= (2 x 12 g) + (6 x 1 g) + (1 x 16 g) = 46 g.

Question 9.
If one mole of oxygen atoms weigh 16 grams, calculate the mass of one atom of oxygen (in grams).
(CBSE 2014)
Answer:

Question 10.
What is the valency of calcium in CaCO₃ ?
Answer:
The valency of Ca in CaCO₃ is 2+ (i.e. Ca²⁺).

Question 11.
What happens to an element ‘A’ if its atom gains two electrons ?
Answer:
It changes to a divalent anion (A^2-).

Question 12.
Why is a cation so named ?
Answer:
When electric current is passed through the solution of a salt like sodium chloride (NaCl), the positive ion (Na⁺) migrates towards cathode (negative electrode). It is therefore, called cation. Please remember that
• Positive ion migrating towards cathode on passing electric current is known as cation.
• Negative ion migrating towards anode on passing electric current is known as anion.

Question 13.
An element Z forms an oxide with formula Z₂O₃. What is its valency ?
Answer:
The valency of the element Z in Z₂O₃ is 3+.

Question 14.
The valency of an element A is 4. Write the formula of its oxide.
Answer:
The formula of the oxide is A₂O₄ or AO₂.

Question 15.
An element X has valency 3 while the element Y has valency 2. Write the formula of the compound between X and Y.
Answer:
The formula of the compound between X and Y is X₂Y₃.

Question 16.
Formula of the carbonate of a metal M is M₂CO₃. Write the fomula of its chloride.
Answer:
The valency of the metal (M) in M₂CO₃ is (1+) i.e. metal exists as M+ ion. Therefore, the formula of metal chloride is MCl.

Question 17.
What do you understand from the statement “relative atomic mass of sulphur is 32”. (CBSE 2012)
Answer:
This means that an atom of sulphur is 32 times heavier as compared to 1/12 of the mass of 1 atom of C — 12(1 u).

Question 18.
Calculate the formula unit mass of CaCl₂.
Answer:
Formula unit mass of CaCl₂ (Calcium chloride)
= (1 x Atomic mass of Ca) + (2 x Atomic mass of Cl)
= (1 x 40 u) + (2 x 35.5 u) = 111 u.

Question 19.
Which of the following represents the correct chemical formula ?
(a) NaSO₄
(b) CaPO₄
(c) ZnS
(d) AlSO₄.
Answer:
The formula (c) represents the correct formula. Both the ions are divalent i.e. Zn²⁺ and S^2-. The name of the compound is zinc sulphide.

Question 20.
Sample A contains one gram molecules of oxygen molecules and sample B contains one mole of oxygen molecules. What is the ratio of the number of molecules in both the sample ?
Answer:
One gram molecules and one gram mole contain the same number of molecules (6.022 x 10²³). Therefore, the ratio is 1 : 1.

Question 21.
Gram molecular mass of ammonia (NH₃) is 17 g. Is it correct to regard it as formula unit mass also ?
Answer:
No, it is not correct. Ammonia exists in molecular form and is not an ionic compound made up of cation and anion. Therefore, it cannot have formula unit mass. It has only molecular mass.

Question 22.
Give one example each of polyatomic element and polyatomic ion.
Answer:
Polyatomic element (P₄) ; Polyatomic ion (SO₄^2-).

Question 23.
Name the element which is used as a reference for the atomic masses of the elements.
Answer:
1/12 of the mass of carbon atom taken as 12 u is used as a reference for the atomic masses of the elements.

Question 24.
Four samples of water [H₂O] are collected from different sources. Each sample on analysis was found to contain same percentage of oxygen. Which law of chemical combination is demonstrated by the above observation ?
(CBSE 2014)
Answer:
The law of constant combination is demonstrated by this observation.

SHORT ANSWER QUESTIONS

Question 25.
List the elements present in
(i) quick lime
(ii) sodium hydrogen carbonate.
Answer:
(i) The chemical name of quick lime is calcium oxide. Its chemical formula is CaO. Elements present are Ca and O.
(ii) The chemical formula of sodium hydrogen carbonate is NaHCO₃. The elements present are Na, H, C and O.

Question 26.
Cenvert into moles :
(i) 20 g of water
(ii) 20 g of carbon dioxide
Answer:

Question 27.
(a) How many moles are present in 11.5 g of sodium ?
(b) The mass of an atom of element (X) is 2.0 x 10^-23 g. Calculate its atomic mass.
Answer:

Question 28.
(a) How many particles are represented by 0.25 mole of an element ?
(b) Out of 4 g of methane (CH₄) and llg of CO₂, which has more molecules ?
Answer:

Question 29.
What is the number of molecules present in 1.5 mole of ammonia (NH₃) ?
Answer:

Question 30.
Write the chemical symbols of two elements
(a) Which are formed from the first letter of the elements name.
(b) Whose name has been taken from the names of the elements in Latin.
Answer:
(a) Boron (B), carbon (C)
(b) Argentum (Ag), Kalium (K).

Question 31.
(a) Four samples of carbon dioxide (CO₂) were prepared by using different methods. Each sample on analysis was found to contain 27.27% carbon by mass. Name the law which is in agreement with this observation.
(b) Explain why the number of atoms in one mole of hydrogen gas is double the number of atoms in one mole of helium gas.
Answer:
(a) Carbon dioxide consists of elements carbon and oxygen. Since the percentage of carbon in each sample is fixed, that of oxygen must be also fixed. This is according to law of constant proportions.
(b) Hydrogen gas is diatomic in nature (H₂) while helium gas is monoatomic (He). As a result, the number of atoms in one mole of hydrogen. (2 x N_A) are expected to be double as compared to number of atoms in one mole of helium (N_A).

Question 32.
10²² atoms of an element ‘X’ are found to have a mass of 930 mg. Calculate the molar mass of the element ‘X’?
Answer:

Question 33.
(a) If the valency of carbon is 4 and that of sulphur is 2, write the formula of the compound formed between carbon and sulphur atoms. Also name the compound.
(b) What is wrong with the statement T mole of hydrogen ?
Answer:
(a) The formula of compound can be written by exchanging the valencies (cross-over). Therefore, the expected formula is C₂S₄ or CS₂. The compound is called carbon disulphide.
(b) The statement is not correct. We must always write whether hydrogen is in atomic form or molecular form. The correct statement is :

Question 34.
T mole of hydrogen atoms or one mole of hydrogen molecules.
Identify the cations and anions in the following compounds :
(a) CH₃COONa
(b) NH₃
(c) NH₄Cl
(d) SrCl₂.
Answer:
(a) CH₃COO^–, Na⁺
(b) It is a molecular compound
(c) NH₄⁺ ,Cl^– (d) Sr²⁺, 2Cl^–

Question 35.
Classify the following based on atomicity
(a) O₃
(b) P₄
(c) S₈. (CBSE 2012)
Answer:
Atomicity is the number of atoms present in one molecule of an element. Based upon this, the atomicity of different molecules may be expressed as :
(a) 3
(b) 4
(c) 8.

Question 36.
Give the formulae of the compounds that will be formed from the following sets of elements.
(a) Calcium and fluorine
(b) Magnesium and oxygen
(c) Sodium and sulphur
(d) Carbon and chlorine
(e) Carbon and sulphur
(f) Nitrogen and hydrogen.
Answer:
(a) CaF₂
(b) MgO
(c) Na₂S
(d) CCl₄
(e) CS₂
(f) NH₃.

Question 37.
Which of the following symbols of elements are incorrect. Write correct symbols
(a) Iron (Ir)
(b) Gold (Au)
(c) Manganese (M)
(d) Potassium (Po)
(e) Zinc (ZN)
(f) Calcium (Ca).
Answer:
(a) Iron (Fe)
(c) Manganese (Mn)
(d) Potassium (K)
(e) Zinc (Zn).

Question 38.
Verify by calculation that :
4 moles of CO₂ and 6 moles of H₂O do not have same mass in grams.
Answer:
Molar mass of CO₂ = 12 + 2 x 16 = 44 g
1 mole of CO₂ has mass = 44 g
4 moles of CO₂
have mass = 44 x 4 = 176 g
Molar mass of H₂O = 2 x 1 + 16 =18 g
1 mole of H₂O has mass = 18 g
6 moles of H₂O have mass = 6 x 18 = 108 g.
This shows that these have different masses in grams.

Question 39.
A sample of vitamin C contains 2.48 x 10²⁵ oxygen atoms. How many moles of oxygen atoms are present in the sample ?
Answer:

Question 40.
Calculate the total number of ions in 0.585 g of sodium chloride.
Answer:

Question 41.
A flask contains 4.4 g of CO₂ gas. Calculate
(a) How many moles of CO₂ gas does it contain ?
(b) How many molecules of CO₂ gas are present in the sample.
(c) How many atoms of oxygen are present in the given sample.
(Atomic mass of C = 12 u. O = 16 u)
Answer:

Question 42.
Determine the molecular mass of :
(i) NH₄OH
(ii) K₂CO₃
(iii) CH₃COOH Given atomic masses :
H = 1.0 u ; O = 16.0 u ; C = 12.0 ; K = 39.0 u ; N = 14.0 u
Answer:
(i) Molecular mass of NH₄OH
= Atomic mass of N + 5 x (Atomic mass of H) + Atomic mass of O
= 14 + 5 x 1 + 16 = 14 + 5 + 16 = 35 u
(ii) Molecular mass of K₂CO₃
= 2 x (Atomic mass of K) + Atomic mass of C + 3 x (Atomic mass of O)
= 2 x 39 + 12 + 3 x 16 = 138 u
(iii) Molecular mass of CH₃COOH
= 2 x (Atomic mass of C) + 4 x (Atomic mass of H) + 2 x (Atomic mass of O)
= 2 x 12 + 4 x 1+2 x 16 = 60 u

Question 43.
(a) What is Avogadro Constant ?
(b) Calculate the number of particles present in 56 g of N₂ molecule.
Answer:
(a) Avogadro’s number is the number of particles (atoms, ions, molecules etc.) present in one mole of any substance. It is denoted either as N_A or as N_Q. The number is also called Avogadro’s constant because its value is fixed (6.022 x 10²³) irrespective of the nature of the particles.
(b) Molar mass of N₂ = 14 x 2 = 28 g
28 g of N₂ have molecules or particles = N_A = 6.022 x 10²³
56 g of N₂ have molecules or particles = N_A = 6.022 x 10²³ x 2 = 1.204 x 10²⁴

Question 44.
(a) An element ‘X’ exhibits variable valencies 3 and 5. Write the formulae of the chlorides of the element.
(b) What is the ratio by mass of the elements present in the chemical formula of magnesium oxide.
Answer:
(a) The formulae of the chlorides of the element ‘X’ = XCl₅ and XCl₃.
(b) The chemical formula of magnesium oxide is MgO. The elements are present in the ratio of 24 : 16 or 3 : 2.

Question 45.
Give the example of trivalent cation and monovalent anion. Write the formula of the compound formed by their combination. .
Answer:
Trivalent cation : Al³⁺ ; Monovalent anion : Cl^–
Formula of the compound : AlCl₃.

Question 46.
(a) State the law of conservation of mass
(b) What mass of silver nitrate will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrate if the law of conservation of mass is true ?
Answer:
(a) The total mass of the products in a physical change or a chemical reaction is equal to the total mass of the reactants that have combined. The law may also be stated in another form. The mass can neither he created nor destroyed in a physical change or a chemical reaction.
(b) The chemical reaction leading to products is :
Mass of reactants = x g + 5.85 g Mass of products = 14.35 g + 8.5 g According to law of conservation of mass
Mass of reactants = Mass of products . (x + 5.85) g = (14.35 + 8.5) g
x = (22.85 – 5.85) g = 17.0 g

LONG ANSWER QUESTIONS

Question 47.
(a) The mass of one molecule of a substance is 4.65 x 10^-23 g. What is its molecular mass ? What could this substance be ?
(b) Which have more molecules? 10 g of sulphur dioxide (SO₂) or 10 g of oxygen (O₂) ?
Answer:

Question 48.
Which has more atoms ?
(a) 10 g of nitrogen (N₂)
(b) 10 g of ammonia (NH₃)
Answer:

Question 49.
(a) Explain with the help of a labelled diagram an activity for the verification of law of conservation of mass.
(b) Find the number of atoms present in 100 g of sodium and 100 g of iron. (Given that Na = 23 u ; Fe = 56 u ;
Answer:
(a) The total mass of the products in a physical change or a chemical reaction is equal to the total mass of the reactants that have combined. The law may also be stated in another form. The mass can neither he created nor destroyed in a physical change or a chemical reaction.
In other words, the mass remains unchanged or conserved in a chemical reaction. The law is also known as the Law of Indestructibility of Matter.
Explanation : Let us try to analyse as to what happens in a chemical reaction. To understand the same, let us consider a chemical reaction between barium chloride and sodium sulphate. When the solutions of these reactants prepared separately in water are mixed, the following chemical reaction takes place :
Barium chloride + Sodium sulphate———– > Barium sulphate + Sodium chloride (white precipitate)
If we look at this reaction, we find that the exchange of constituents has taken place between the reactants, i.e. chloride part of barium chloride has been exchanged by sulphate part of sodium sulphate. As such, no loss or gain in mass is expected. In other words, the mass remains conserved or does not change.

Question 50.
(a) Give one point of difference between an atom and an ion.
(b) Give one example each of a polyatomic cation and a polyatomic anion.
(c) Identify the correct chemical name of FeSO₄ from the given names Ferrous sulphate, Ferrous sulphide, Ferrous sulphite.
(d) Write the chemical formula for the chloride of magnesium.
Answer:
(a) An atom is always neutral in the sense that it does not carry any charge. An ion carries either positive charge (cation) or negative charge (anion).
(b) Ammonium (NH₄)⁺ ion is a polyatomic cation while sulphate (SO₄)^2- ion is a polyatomic anion.
(c) Correct chemical name of FeSO₄ is ferrous sulphite.
(d) Chemical formula for the chloride of magnesium is MgCl₂.

Question 51.
(a) What do the following observations stand for ?
(i) 2O
(ii) 3O₂
(b) Which amongst the following has more number of atoms and how much ?
(i) 11.5 g of sodium
(ii) 15.0 g of calcium
Answer:
(a) (i) 2O represent two atoms of oxygen (or the gram atoms of oxygen
(ii) 3O₂ represent three molecules of oxygen or three gram moles of oxygen.

NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Hope given NCERT Solutions for Class 9 Science Chapter 3 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

Though speech is essentially an essay, it is significantly different. Whereas an essay or an article is read by people at a distance in time and place, a speech is delivered ‘live’ to an audience. This fact determines, to a great extent, the language and style of the composition.

A speech consists of the words spoken by a person to a gathering. The speaker could be a leader, a learned man, a man in a position of power and authority or an ordinary person, a student. The ‘address’, the language and style are determined by the occasion and the audience.

This grammar section explains Online Education English Grammar in a clear and simple way. There are example sentences to show how the language is used. NCERT Solutions for Class 11 English will help you to write better answers in your Class 11 exams. Because the Solutions are solved by subject matter experts. https://ncertmcq.com/speech-writing-format-cbse-class-11/

Online Education for Speech Writing Format CBSE Class 11 Examples, Samples, Topics

Speech Writing Class 11

♦ Format:

(a) Address the gathering, e.g., Respected Principal, teachers, students, Good Morning.
(b) Speech in three or four paragraphs.
(c) A speech ends with Thank you. The speaker thanks the listeners for their patient hearing. Content:

♦ Content:

According to the occasion, you may include quotations. Organise the speech into introduction, main points and conclusion.

Language and style should be a mingling of the serious and conversational modes.

Speech Topics For Class 11 Question 1.
You are Mukesh/Mukta of Saket. You have been reading the news items on Nithari killings. Write a speech in 150-200 words to be given in the morning assembly, stressing the need of keeping a close vigil on the anti-social elements of the locality and taking care of young children of the area so that they don’t fall prey to bad characters.
Answer:

Crimes Against Children
by Mukta (XIA)

Respected Principal, teachers and dear friends. Good Morning. I am Mukta of Class XI. Today I would like to share my views with you on crimes against children. It is a horrifying reality with which we have to contend.

Some recent tragic events have opened our eyes to the magnitude of crimes against children. Mostly, children are kidnapped for ransom if they are from rich families, or for begging and stealing if they are from ordinary families. Cases of child abuse are growing in alarming proportions.

The most disgusting and horrifying of these cases was the one discovered at Noida, where the owner- servant team carried out dastardly crimes against children behind closed doors. Investigations are unveiling horrible acts.

In such a scenario, it is imperative that parents take good care of their children. Children should be trained not to.trust strangers.

The security of a neighbourhood depends on the alertness of the citizens. We have a tremendous responsibility towards our youngsters.

I hope my words will make you think about the matter.
Thank you

Format Of Speech Writing For Class 11 Question 2.
Children usually come to school without taking breakfast in the morning and eat junk food from the school canteen. This habit affects adversely the performance of students in academics and sports. Write a speech in 150-200 words to be given in your school assembly about ‘How Health is Affected by Lifestyle’.
Answer:

How Health is Affected by Lifestyle

Respected Principal, teachers and dear friends,

Good morning, I am Mamta of XI A, and I wish to speak to you on a very important matter. I wish to speak to you on how our health is affected by our lifestyle.

Yesterday, a class VI student fainted in the assembly. The reason was that she had not taken her breakfast of porridge, fruit and milk. She wanted noodles only with coffee!

Friends, the breakfast her mother had prepared was nutritious, it would have given her energy for the whole day. If we fill our stomach, meal after meal, with junk food, how shall we grow tall and strong? How will we get the energy for mental work?

Like food, rest is an important part of our lifestyle. My friend Roshan was unable to answer his English paper in the unit test; he was feeling sleepy. He had studied till 3 a.m. because he thought English could be studied in a few hours. Had he been regular in work and study, he would have done well, for he is very intelligent.

My last word is for my obese friends who form almost 30% of our student body. Exercise, exercise and regular exercise is the answer to your problems.

I sincerely hope you will improve your lifestyle to be healthier and stronger.

Thank You.

♦ Analysis:

1. Notice that the speech opens with an address to the listeners. They are addressed in order of importance. Do not forget to address the student audience.
2. Notice the conclusion. The speech ends with thanking the audience.
3. Notice that the language and style are more conversational than that of an article.
4. The structure of the content of a speech is very similar to that of an article/essay, i.e., having an introduction, main body and conclusion.

Speech Format Class 11 Question 3.
A recent survey by the Ministry of Health shows that the growing tendency of obesity among children today is caused by addiction to junk food and lack of physical exercise, and it adversely affects their learning power. Write a speech in 150-200 words to be delivered in the morning assembly of the school highlighting the importance of good health and motivating the children to eat healthy, balanced diet.
Answer:

Side Effects of Junk Food

Respected Principal, Vice Principal, teachers, and dear friends. Good morning.

I am grateful to the Principal, Dr N. Ravi, for giving me this opportunity to speak on a matter that is close to the heart of all teenagers.

As I stand here, I can see about eight to ten overweight students in each class line. Surely, they are embarrassed and have become the butt of jokes. To look fit and trim is vital for a good personality.

Friends, I want you to look at your daily routine and see what you are doing to get this bulky figure. Are you excessively fond of cold drinks, pizzas and hamburgers? Do you love to sit before TV and munch wafers and packaged chips of all brands ? Is your favorite snack during the long break everyday a spicy oily ‘samosa’ or ‘bread pakora’?

Later in the evening do you just sit indoors and work on your computer, complete your homework or watch TV ?

If the answer to these questions is ‘Yes’, you know why you have this ungainly figure. In order to be slim and athletic, you need to eat right, exercise briskly in fresh air and live a disciplined life. The food that your mother prepares is what you should consume rather than junk food.

Thank you

Format Of Speech Writing Class 11 Question 4.
With the visual media technology overtaking the print-media, you have observed a decline in the people’s interest in reading books. You are disturbed by this excessive dependence of the young on television and computer games as the source of entertainment. Write a speech in 150-200 words for your school magazine on the “Pleasures of Reading” or “The Company of Books. ”
Answer:

The Pleasures of Reading
by ABC

Reading affords pleasure like no other activity. It is deeply fulfilling and satisfying. While reading, the reader is in direct communication with the writer, however distant he may be in time or space. The reader is fully focussed and enjoys the content as well as the beauty and power of language. Reading offers an infinite variety according to one’s taste. Knowledge, imagination and the reader’s own appreciation of the written word are some of the rewards of reading. If one is adventurous, one can venture into unknown cultures through books. For example, a reading of Tolstoy or Chekhov gives us an insight into the life of the Russians as a reading of Dickens gives us of the British, particularly of London dwellers.

Reading makes us more sensitive to human greatness as well as human suffering. A reading of Premchand’s short stories acquaints us with the strengths and weaknesses of human character. The classics of literature are eternal in their appeal.

Unfortunately, today most of us visit the celluloid version of Sharat Chandra’s ‘Devdas’ three four times, and the original printed one not even once.

Speech Writing Format Class 11 Question 5.
The world that we are living in is under a great threat of extinction. Large scale pollution, deforestation and industrialization are suffocating the environment. You, as a concerned citizen, want to share the causes and effects of global warming. You would also like to give a few suggestions on how to contribute to the making of a better world. As Vivek/Vineeta Arora, Head Boy/Girl of the school, prepare a speech on this issue for the morning assembly under the heading “The Hazards of Global Warming”. (150-200 words)
Answer:

The Hazards of Global Warming

Respected Principal, teachers and dear friends,

This morning, I wish to draw your attention to the catastrophe that is slowly and surely coming towards us. I am talking about the hazards of global warming.

Some of my friends may not be aware of ‘global warming’. This term means the rise of the temperature of the atmosphere surrounding our planet. This is happening because of over industrialisation, large-scale pollution and deforestation. Hot emissions from industry and release of chemicals in the air are making the atmosphere of the earth warmer.

What are the hazards of global warming? Already, the polar ice caps have started melting. This can bring disaster to the world. Huge glaciers in the Himalayas and other mountains are shrinking. This will cause a rise in the sea level and will eventually lead to the flooding of coastal areas. Global warming will also bring about changes in the world climate. There will be excess rains, floods, unbearable heat and cold. These changes will surely bring disaster to the flora and fauna of our planet.

What can we do about this problem? The countries need to exercise discipline ajpout putting up industries. Effective measures to control pollution are needed.

Friends, I have tried to acquaint you with the problem the whole world is facing.

Thank you.

Speech Writing Class 11 Format Question 6.
You are the Class Representative, Class XI of Gandhi Memorial School, Pushp Vihar. You have been asked to prepare a speech on the need to imbibe principles of truthfulness, gentleness and cooperation to counter the growing violence in today’s world. Write this speech in 150-200 words.
Answer:

Respected Principal, teachers and dear friends,

Today, I wish to speak to you on the very familiar subject of values like truthfulness, gentleness and cooperation. If we wish to counter the growing violence around us, these values are the only recourse open to us. (Develop these points).

Truthfulness – the basis of social stability – will rid society of deceit and fraud – example of Gandhiji! who could defeat the mighty British with the weapon of truth.

Gentleness – tolerance of the other colour, religion and caste, patience and compassion advocated by Buddha and Mahavir.

Cooperation – inevitable in the era of globalization.

Thank you

Speech Class 11 Question 7.
You are Rajini/Rajat Gaur. After seeing some obese children in your school, you are worried about
the craze for junk food and electronic gadgets in students. Write a speech to be read out in the morning assembly highlighting the need to have good eating habits and the benefits of outdoor activities. (Word limit: 150-200 words)
Answer:

Respected principal, teachers and dear friends. Today I am standing before you to express my views on the topic “obesity among school children”. Obesity has become a cause of concern for one and all. With the entry of companies like McDonalds, Kentucky chicken, wimpeys, Pizza Huts, etc. the concept of fast food has spread like a wild fire. More and more school children have taken to chocolates and soft drinks without worrying for colories. The result is obesity increasing by leaps and bounds among school children.

‘Junk food’ appeals to our taste buds but our digestive system repels it. Oil and extra fat harm our liver. We put on flesh and increase our weight. Fresh home cooked meals are hygenic, better and full of vitamins. Minerals can be easily digested. Our food should consist of coarse cereals, green and leafy vegetables, fruits, and pulses, etc. To keep ourselves agile, energetic and robust, we must do physical exercises, play games like football, badminton, kabbaddi, volleyball, etc.

By eating healthy food, doing regular physical activities not only our body will be healthy and toned our minds would remain active. I hope you all will give some consideration to my suggestion and try to instil it in your daily routine.

Thank you all for giving me a patient hearing.

Speech Writing Examples For Class 11 Question 8.
Yoga is beneficial. Speak about the benefits of yoga in the morning assembly. (Word limit: 150-200)
Answer:

Respected Principal, teachers and friends, a very good morning to one and all. Today I am standing before you to express my views on the benefits of practising yoga.

Many of us perform yoga but do we know what this entails and why is it performed? Basically yoga is a means to attain balance in our body, i.e., gaining strength, increasing flexibility and attaining spirituality. Postures and asanas are an indispensable part of yoga. These are performed keeping in mind all the body parts, combining breathing practices, and include guided mediation, Yoga is a union between body and mind. In other words, we can say it is a balance between mind and helps us keep ourselves fit and fine.

Yoga has the potential to cure many diseases like respiratory problem, abdominal problem and diseases related to the nervous system. Yoga helps in releasing toxins, reduces stress and increases concentration. In short, yoga helps us keep disease-free, happy, wealthy, wise and toned up. Yoga is a never-ending process. The deeper you delve into it the more fruitfull will be the results.

I would like to end my speech on this note.

Here we are providing Online Education for Class 11 Geography Important Extra Questions and Answers Chapter 11 Water in the Atmosphere. Important Questions for Class 11 Geography are the best resource for students which helps in class 11 board exams.

Online Education Important Questions for Class 11 Geography Chapter 11 Water in the Atmosphere

Water in the Atmosphere Important Extra Questions Very Short Answer Type

Question 1.
What is the greatest source of atmospheric moisture?
Answer:
The surface of the ocean is the greatest source of atmospheric moisture.

Question 2.
What is meant by humidity?
Answer:
The state of the atmosphere with respect to its content of moisture or water vapour is called humidity.

Question 3.
What is meant by sensible temperature?
Answer:
It is the cold or heat felt by the human body, depending not only on actual temperature but also on relative humidity and wind.

Question 4.
Name three states in which water exists.
Answer:
Water exists in three states or forms:

1. Solid (ice),
2. Liquid (water),
3. Gas (water vapour).

Question 5.
What is the measure of absolute humidity?
Answer:
Absolute humidity is measured in grams per cubic metre (g/m³).

Question 6.
What is the dew point?
Answer:
It is the temperature at which air, on cooling, becomes saturated with water vapour, and below which condensation begins and dew forms.

Question 7.
Why dew and frost form on grass quite readily?
Answer:
Because grass is a good radiator and, therefore, cools quickly.

Question 8.
Identify various forms of precipitation.
Answer:
Various forms of precipitation include rainfall, snowfall, hail, sleet and drizzle.

Question 9.
Name three types of precipitation.
Answer:
The three types of precipitation are convectional, orographic and cyclonic precipitation.

Question 10.
How does precipitation take place?
Answer:
Precipitation takes place as a result of the continued condensation and growth of the moisture particles until they become too large to remain suspended in the air.

Question 11.
What is the source of atmospheric moisture?
Answer:
The oceanic surface is the main source of atmospheric moisture.

Question 12.
Which factors control evaporation?
Answer:
There are three main factors which control evaporation:

1. aridity,
2. temperature,
3. movement of air.

Question 13.
Name the various types of clouds.
Answer:
Cirrus clouds, stratus clouds, stratocumulus and cumulus clouds.

Question 14.
What are the different types of fog?
Answer:
Fogs are of three types: radiating, advection and frontal fog.

Question 15.
What is condensation?
Answer:
Cooling of saturated air is called condensation.

Question 16.
What are the different types of rainfall?
Answer:
Rainfall is of three types: convectional, orographic and cyclonic rainfall.

Question 17.
What is sublimation?
Answer:
The process of conversion of ice into liquid or vapour is known as sublimation.

Question 18.
Name the temperature at which a given sample of air becomes saturated.
Answer:
Dew point.

Question 19.
What do you mean by saturated air?
Answer:
The air that contains moisture to full capacity.

Question 20.
What is relative humidity?
Answer:
The ratio of the amount of water vapour actually presents in the air at a particular temperature to the amount of water vapour required to saturate the same air at the same temperature.

Question 21.
What is sleet?
Answer:
The falling of frozen raindrops and frozen melted snow on the earth’s surface is known as sleet.

Question 22.
Name the various types of humidity.
Answer:
Absolute humidity, specific humidity and relative humidity.

Water in the Atmosphere Important Extra Questions Short Answer Type

Question 1.
Describe various processes involved in the change of state (form) of water.
Answer:
Water exists in three forms, solid (ice), liquid (water) and gas-(vapour). All three states are interchangeable. The change of solid to ‘ liquid is called melting; from liquid to solid is freezing; from liquid to gas is evaporation and from gas to liquid is condensation. But sometimes ice directly converts into vapour or vice-versa. This is called sublimation.

Question 2.
Describe various sources of atmospheric moisture on the earth.
Answer:
The surface of the ocean is the greatest source of atmospheric moisture. Besides, smaller water bodies, vegetation and damp surfaces also contribute significant amounts of moisture.

Question 3.
What is relative humidity and how is it measured?
Answer:
The amount of water vapour in the air compared with the amount that would be present if the air was saturated at that temperature. It is expressed in percentage determined by dividing the absolute humidity by water holding capacity of air. If the air is saturated, its relative humidity is 100%; if only half saturated, the relative humidity. is 50%. Since the capacity of air for absorbing and retaining moisture varies with temperature, the relative humidity of air mass can be altered by merely lowering its temperature without changing the actual amount of moisture present in it.

Question 4.
Discuss the reasons which impede the formation of dew or frost.
Answer:
On cool nights in early winter, when Radiation from the ground is relatively rapid, the air in contact with the ground may be chilled to the dew point and moisture condenses on the leaves and grass in the form of dew. If the dew point is below the freezing point, the condensation takes place in the form of ice crystals or frost. The formation of dew or frost is impeded by

1. Dry air,
2. Wind mixes the air and prevents its lower portion from reaching the dew point, and
3. Clouds which slow down terrestrial radiation.

Question 5.
What are the middle clouds?
Answer:
The clouds which form at height of2000-7000 km. above the ground are called middle clouds. Altostratus and altocumulus are examples.

Altostratus – It is a uniform sheet of cloud, grey or bluish in colour and usually has a fibrous structure.

Altocumulus – These are flattened globular masses of clouds arranged in lines or waves. They have large globules with shadows.

Question 6.
Why does the amount of water vapour decrease rapidly with altitude?
Answer:
Water vapour in the atmosphere is the result of the evaporation of water from the earth’s surface, a continual process dependent on air. Since the air becomes thinner with the altitude, so also does the water vapour.

Question 7.
What is condensation and how does it take place?
Answer:
Condensation is the process of water vapour changing to a liquid state. If the air is cooled below its dew point, some of the air’s water vapour becomes liquid. Thus, any amount of cooling of saturated air starts the process of condensation. Condensation may start with the addition of any further water vapour to the saturated air, or with the reduction of its temperature.

Question 8.
What is precipitation? What are the conditions to determine the form of precipitation?
Answer:
Precipitation is a process of condensation of water vapour in ‘ the atmosphere which falls to the earth in the form of rain, snow, hail or sleet. These are the various forms of precipitation.
Conditions for precipitation.

1. There should be evaporation so that the relative humidity is „ high in the air.
2. There should be adiabatic cooling of the moist air.
3. There should be dust particles in the air.

Question 9.
What are the effects of humidity?
Answer:
The main effects of humidity are:

1. The skin of the human body starts to crack due to low relative humidity.
2. It causes joint pain in the human body.
3. The high relative humidity is harmful to the human body.
4. The joints of furniture get loosened.

Question 10.
Why does the amount of water vapour decrease rapidly with altitude?
Answer:
The amount of water in the atmosphere is variable from place to place and from time to time. It decreases rapidly with altitude. The temperature decreases as the altitude increases.

Question 11.
How is rainfall caused?
Answer:
The moist air is cooled, saturated and condensation takes place. Water vapour condenses on a large number of dust particles to form clouds.

The cloud droplets merge to form raindrops. When these raindrops become so heavy that air cannot hold them, these fall on the earth in the form of rainfall. The release of moisture in the form of drops of water is called rainfall.

Question 12.
Describe the conditions for the formation of dew.
Answer:
Following are the favourable conditions for the formation of dew:

1. Clear sky
2. Calmness, i.e., the wind should not blow
3. High relative humidity
4. Rich vegetation cover
5. Long nights

Water in the Atmosphere Important Extra Questions Long Answer Type

Question 1.
Distinguish between:
1. Relative humidity and Absolute humidity.
2. Evaporation and Condensation.
3. Dew and Frost.
Answer:
1. Relative humidity and Absolute humidity:
Relative humidity:
The amount of water vapour in the air compared with the amount that would be present if the air was saturated at that temperature. It is expressed in percentage determined by dividing the absolute humidity by water holding capacity of air. If the air is saturated, its relative humidity is 100%; if only half saturated, the relative humidity. is 50%. Since the capacity of air for absorbing and retaining moisture varies with temperature, the relative humidity of air mass can be altered by merely lowering its temperature without changing the actual amount of moisture present in it.

Absolute humidity:
It is the actual amount of water vapour present per unit volume of air and it is measured in grams per cubic meter.

2. Evaporation and Condensation
Evaporation is the process by which a solid or liquid is converted to a gaseous state or vapour. It occurs whenever energy is transported to an evaporated surface and temperature rises. The molecules become more mobile and overcome the forces binding them to the water surface and break away resulting in evaporation.

Condensation:
Condensation is the process of water vapour changing to a liquid state. If the air is cooled below its dew point, some of the air’s water vapour becomes liquid. Thus, any amount of cooling of saturated air starts the process of condensation. Condensation may start with the addition of any further water vapour to the saturated air, or with the reduction of its temperature.

3. Dew and Frost:
Dew: When the moisture is deposited in the form of water droplets on cooler surfaces of solid objects, such as stones, grass blades and plant leaves, it is known as dew. Clear sky, calm air, high relative humidity, cold and long nights are the ideal conditions for the formation of dew.

Frost: Frost forms on cold surfaces at a dew point which is at or below the freezing point when an excess of moisture is deposited in the form of minute ice crystals instead of droplets of water.

Question 2.
Write short notes on:
(a) Convectional precipitation
(b) Cyclonic precipitation
(c) Orographic precipitation
Answer:
(a) Convectional precipitation:
It occurs when moist air over the heated ground becomes warmer than the surrounding air and is forced to rise, expand, cool and yield some of its moisture.

Convectional rain is common in low latitudes and usually comes in the form of short heavy showers, often during the hottest part of the day. Thunder and lightning generally accompany the showers. If the convection currents are especially strong and turbulent, hail is formed.

(b) Cyclonic precipitation:
It occurs in low-pressure areas, where cyclonic winds coming from various directions converge and force the large volumes of light air to rise and cause rain.

Cyclonic precipitation is generally associated with the passage of depression in middle or high latitudes, as the warm moist air mass of the depression rises and over-rides colder, heavier air.

(c) Orographic precipitation:
It is caused due to relief. It occurs when the moist wind is forced to rise over a mountain or any other height. Precipitation occurs due to cooling of moisture-laden air as it rises over a high relief barrier. It happens, particularly on the windward slopes! On the other side of the barrier or on the leeward side of the slopes, there develops rain shadow area with no or very little rainfall because as the moisture-laden wind descends down the slope, after precipitating on the windward side, it becomes warmer and drier. Such a situation occurs along the western coasts of Qudra, North America and South America. The moist air from the Arabian Sea is forced by Sahyadri hills of the Western Ghats to rise up resulting in expansion, cooling and rainfall.

Amount of precipitation depends on slope, the height of the hills, temperature and moisture content of the air mass. On the other side of the mountain, the descending wind is devoid of moisture and, hence, does not give rain. The eastern slope of Sahyadri mountains in India falls in the rain shadow area.

Question 3.
Describe the factors controlling the rate of evaporation and evapotranspiration.
Answer:
vaporation is the process by which liquid water or ice changes into water vapour. It occurs whenever energy is transported to an evaporated surface and temperature rises. The molecules become more mobile and overcome the forces binding them to the water surface and break away resulting in evaporation.

Evapotranspiration, on the other hand, is the amount of moisture transferred to the atmosphere by evaporation of liquid and solid water plus transpiration from living tissues principally from the plants.

Factors like the temperature of the air, source of moisture, vegetation cover, latitude, permeability and water retention effect evaporation and evapotranspiration.

Evaporation depends upon air temperature. Greater the temperature, higher is the rate of evaporation and so on. Therefore, the highest rate of evaporation occurs in deserts in conditions Of great heat and lack of surface cone.

The surface of oceans is the greatest source of atmospheric moisture.

Vegetation is an important factor, particularly in evapotranspiration. Because, under transpiration, the loss of water takes place from the green plants resulting in a stream of water with dissolved mineral salts surfing up through the plants.

Question 4.
Describe the various forms of precipitation and the major precipitation regimes of the world.
Answer:
Precipitation results from the continued condensation and growth of the moisture particles until they become too large to remain suspended in the air.

If condensation takes place at a temperature above 0°C, the resulting precipitation is in the form of rain.

If the process through a layer of colder air on the way down, it may freeze and fall as sleet.

In a strong turbulent current of a thunderstorm, water drops may be carried upward into freezing temperatures and eventually fall as hail. In fact, violent air currents may keep hailstones shuttling up and down until they grow to enormous sizes.

When the more. moisture crystallises into ice particles from vapour at a temperature below freezing point, it called snow. If snowflakes grow without interference, they form beautiful symmetrical, six-sided Crystals.

Ice-storms result when rain, already near the freezing point, falls on colder ground and vegetation and freezes upon contact.

Major precipitation regimes:
(a) Convectional precipitation:
It occurs when moist air over the heated ground becomes warmer than the surrounding air and is forced to rise, expand, cool and yield some of its moisture.

Convectional rain is common in low latitudes and usually comes in the form of short heavy showers, often during the hottest part of the day. Thunder and lightning generally accompany the showers. If the convection currents are especially strong and turbulent, hail is formed.

(b) Cyclonic precipitation:
It occurs in low-pressure areas, where cyclonic winds coming from various directions converge and force the large volumes of light air to rise and cause rain.

Cyclonic precipitation is generally associated with the passage of depression in middle or high latitudes, as the warm moist air mass of the depression rises and over-rides colder, heavier air.

(c) Orographic precipitation:
It is caused due to relief. It occurs when the moist wind is forced to rise over a mountain or any other height. Precipitation occurs due to cooling of moisture-laden air as it rises over a high relief barrier. It happens, particularly on the windward slopes! On the other side of the barrier or on the leeward side of the slopes, there develops rain shadow area with no or very little rainfall because as the moisture-laden wind descends down the slope, after precipitating on the windward side, it becomes warmer and drier. Such a situation occurs along the western coasts of Qudra, North America and South America. The moist air from the Arabian Sea is forced by Sahyadri hills of the Western Ghats to rise up resulting in expansion, cooling and rainfall.

Amount of precipitation depends on slope, the height of the hills, temperature and moisture content of the air mass. On the other side of the mountain, the descending wind is devoid of moisture and, hence, does not give rain. The eastern slope of Sahyadri mountains in India falls in the rain shadow area.

Question 5.
Why dew is not formed on cloudy, windy nights and in shady places?
Answer:
Dew is formed when objects radiate heat thoroughly so that the moist air coming into contact with them may be sufficiently cooled down and the water vapour condenses into water droplets. The clouds act as a blanket for the earth’s surface. They check the radiation of heat so it does not radiate heat thoroughly. Hence, dew is not formed on cloudy nights and in shady places.

Similarly, dew is not formed on windy nights, because the winds blow fast and the moist air does not come long enough into contact with cold objects to get cooled. Hence, condensation does not take place and no dew can be formed.