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Selina Concise Chemistry Class 8 ICSE Solutions – Atomic Structure

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 4 Atomic Structure. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Exercise

1. Fill in the blanks.

(a) Dalton said that atoms could not be divided
(b) An ion which has a positive charge is called a cation.
(c) The outermost shell of an atom is known as valence shell.
(d) The nucleus of an atom is very hard and dense.
(e) Neutrons are neutral particles having mass equal to that of protons.
(f) Isotopes are the atoms of an element having the same atomic number but a different mass number.

2. Write ‘true’ or ‘false’ for the following statements:

(a) An atom on the whole has a positive charge.
false
(b) The maximum number of electrons in the first shell can be 8.
false
(c) The central pad of the atom is called nucleus.
True.

3. Give the following a suitable word/phrase.

(a) The sub-atomic particle with negative charge and negligible mass.
(b) Protons and neutrons present in the nucleus.
(c) The electrons present in the outermost shell.
(d) Arrangement of electrons in the shells of an atom.
(e) The number of protons present in the nucleus of an atom.
(f) The sum of the number of protons and neutrons of an atom.
(g) Atoms of same element with same atomic number but a different mass number.
(h) The smallest unit of an element which takes part in a chemical reaction.

Answer:

(a) Neutron
(b) Mass number
(c) Valency
(d) Orbits or Valence shells
(e) Atomic number
(f) Mass number
(g) Isotopes
(h) Atom

4. Multiple Choice Questions

(a) The outermost shell of an atom is known as

1. valency
2. valence electrons
3. nucleus
4. valence shell

(b) The number of valence electrons present in magnesium is

1. two
2. three
3. four
4. five

(c) The sub atomic particle with negative charge is

1. proton
2. neutron
3. electron
4. nucleon

(d) If the atomic number of an atom is 17 and mass number is 35 then number of neutron will be

1. 35
2. 17
3. 18
4. 52

(e) The number of electrons in an atom is equal to number of

1. protons in a neutral atom
2. neutrons in a neutral atom
3. nucleons in a neutral atom
4. none of the above

(f) The sum of number of protons and number of neutrons present in the nucleus of an atom is called its

1. mass number
2. atomic number
3. number of electrons
4. all of the above

Question 5.
Name three fundamental particles of the atom. Give the symbol with charge, on each particle.
Answer:
The fundamental particles of the atom are: electrons, protons and neutrons.

Particle	Symbol	Charge
electron	e–	-1 or 1.602 x 10-19 C. Where -1 represent its one unit negative electrical charge
proton	p+	+ 1 or 1.602 x 10-19 C. Where +1 represents one unit +ve electrical charge.
neutron	no	0

Question 6.
Define the following terms:
(a) Atomic number
(b) Mass number
(c) Nucleons
(d) Valence shell
Answer:
(a) Atomic number: Atomic number refers to the number of protons present in an atom. It is denoted by Z. Example: An atom of oxygen contains 8 proton Therefore its atomic number is 8.
(b) Mass number: Mass number refers to the sum of the number of protons and neutrons present in the nucleus of an atom and denoted by A Mass number = Number of protons + Number of neutrons.
(c) Nucleons: The protons and neutrons collectively are known as nucleons.
(d) Valence Shell: The outermost shell of an atom is known as its valence shell.

Question 7.
Mention briefly the salient features of Dalton’s atomic theory (five points).
Answer:
Salient features of Dalton’s atomic theory:

1. Matter consists of very small and indivisible particles called atoms, which can neither be created nor can be destroyed.
2. The atoms of an element are alike in all respects i.e. size, mass, density, chemical properties but they differ from the atoms of other elements.
3. Atoms of an element combine in small numbers to form molecules of the element.
4. Atoms of one element combine with atoms of another element in simple whole number ratio to form molecules of compounds.
5. Atoms are the smallest units of matter that can take part in a chemical reaction during which only rearrangement of atoms takes place.

Question 8.
(a) What are the two main features of Rutherford’s atomic model?
(b) State its one drawback.
Answer:
(a) According to Rutherford’s model an atom consists of:

1. The centrally located nucleus: The nucleus is a centrally located positively charged mass. The entire mass of the atom is concentrated in it. It is the densest part of the atom. Its size is very small as compare to the atom as a whole.
2. The outer circular orbits: Electrons revolve in circular orbits (shell) in the space available around the nucleus. An atom is electrically neutral i.e., number of protons and electrons present in an atom are equal.

(b) Rutherford’s atomic model could not explain the stability of the atom as it is like a solar system, the sun is at the centre and the planets revolve around it, in an atom the electrons revolve around the centrally located nucleus containing protons.

Question 9.
What are the observations of the experiment done by Rutherford in order to determine the structure of an atom?
Answer:
Following were the observations made by Rutherford:

1. Most of the alpha particle passed straight through the foil without any deflection from their path.
2. A small fraction of them were deflected from their original path by small angles.
3. Only a few particles bounced back.

Question 10.
State the mass number, the atomic number, number of neutrons and electronic configuration of the following atoms.

Also, draw atomic diagrams for them.
Answer:

Question 11.
What is variable valency? Name two elements having variable valency and state their valencies.
Answer:
Variable valency: Some elements exhibit more than one valency. They are said to have variable valency, e.g. Iron, copper, tin, lead.
Iron         Fe       Fe²⁺ or Fe³⁺
Copper   cu        cu⁺ or cu²⁺

Question 12.
The atomic number and the mass number of sodium are 11 and 23 respectively. What information is conveyed by this statement?
Answer:
Atomic number = 11; No of protons = 11
Mass number = 23 = Number of protons + Number of neutrons.
No of neutrons = 23-11 = 12.

Question 13.
Draw the diagrams representing the atomic structures of the following:
(a) Nitrogen (b) Neon
Answer:

Question 14.
Explain the rule with example according to which electrons are filled in various energy levels,
Answer:
The maximum number of electrons that can be present in any shell or orbit of an atom is given by the formula 2n², where n is the serial number of the shell.
Therefore:
K shell, n = 1, no. of electrons = 2 x 1² = 2
L shell, n = 2, no. of electrons = 2 x 2² = 8
M shell, n = 3, no. of electrons = 2 x 3² = 18
N shell, n = 4, no. of electrons = 2 x 4²= 32
Electrons are not accommodated in a given shell, unless the inner shells are filled.
That is, the shells are filled in a stepwise manner.
Example:

Question 15.
The atom of an element is made up of 4 protons, 5 neutrons and 4 electrons. What is its atomic number and mass number?
Answer:
Protons = 4, neutrons = 5, electrons = 4
Atomic number = 4,
Mass number = 4 + 5 = 9

Question 16.
(a) What are the two main parts of which an atom is made of?
(b) Where is the nucleus of an atom situated ?
(c) What are orbits or shells of an atom ?
Answer:
(a)

1. The centrally located nucleus
2. The outer circular orbits.

(b) The nucleus is a centrally located positively charged mass.
(c) The circular orbits (shell present) in the space available around the nucleus on which electrons revolve are called orbits or shells of an atom.

Question 17.
What are isotopes? How does the existence of isotopes contradict Dalton’s atomic theory?
Answer:
Atoms of an element must have the same atomic number, but their mass number can be different due to the presence of a different number of neutrons. These atoms of an element having a different number of neutrons are called groups.
According to Dalton’s theory, all atoms of an element are similar to all respects, for example, they have the same shape, size etc. and have similar physical and chemical properties like mass, density and reactivity. Whereas isotopes of an element have atoms that are similar as they have same number of protons and electrons but differ in the number of neutrons. So, the isotopes have atoms that are not similar in all aspects.

Question 18.
Complete the table below by identifying A, B, C, D, E and F.
Answer:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 22 Data Handling. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Data Handling Exercise 22A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Arrange the following data as an array (in ascending order):
(i) 7, 5, 15, 12, 10, 11, 16
(ii) 6.3, 5.9, 9.8, 12.3, 5.6, 4.7
Solution:
(i) Ascending order = 5, 7, 10, 11, 12, 15, 16
(ii) Ascending order = 4.7, 5.6, 5.9, 6.3, 9.8, 12.3

Question 2.
Arrange the following data as an array (descending order):
(i) 0 2, 0, 3, 4, 1, 2, 3, 5
(ii) 9.1, 3.7, 5.6, 8.3, 11.5, 10.6
Solution:
(i) Descending order = 5, 4, 3, 3, 2, 2, 1, 0
(ii) Descending order = 11.5, 10.6, 9.1, 8.3, 5.6, 3.7

Question 3.
Construct a frequency table for the following data:
(i) 6, 7, 5, 6, 8, 9, 5, 5, 6, 7, 8, 9, 8, 10, 10, 9, 8, 10, 5, 7, 6, 8.
(ii) 3, 2, 1, 5, 4, 3, 2, 5, 5, 4, 2, 2, 2, 1, 4, 1, 5, 4.
Solution:

Question 4.
Following are the marks obtained by 30 students in an examinations.

Taking class intervals 0-10, 10-20, ……… 40-50 ; construct a frequency table.
Solution:

Question 5.
Construct a frequency distribution table for the following data ; taking class-intervals 4-6, 6-8, ……… 14-16.

Solution:

Question 6.
Fill in the blanks:
(i) Lower class limit of 15-18 is ………
(ii) Upper class limit of 24-30 is ……..
(iii) Upper limit of 5-12.5 is ………
(iv) If the upper and the lower limits of a class interval are 16 and 10 ; the class-interval is ……..
(v) If the lower and the upper limits of a class interval are 7.5 and 12.5 ; the class interval is ……..
Solution:
(i) Lower class limit of 15 – 18 is 15.
(ii) Upper class limit of 24 – 30 is 30.
(iii) Upper limit of 5 – 12.5 is 12.5
(iv) If the upper and lower limits of a class interval are 16 and 10 ; the class interval is 10 – 16
(v) If the lower and upper limits of a class interval are 7.5 and 12.5 ; the class interval is 7.5 – 12.5

Data Handling Exercise 22B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Hundred students from a certain locality use different modes of travelling to school as given below. Draw a bar graph.

Solution:

Question 2.
Mr. Mirza’s monthly income is Rs. 7,200. He spends Rs. 1,800 on rent, Rs. 2,700 on food, Rs. 900 on education of his children ; Rs. 1,200 on Other things and saves the rest.
Draw a pie-chart to represent it.
Solution:

Question 3.
The percentage of marks obtained, in different subjects by Ashok Sharma (in an examination) are given below. Draw a bar graph to represent it.

Solution:

Question 4.
The following table shows the market position of different brand of tea-leaves.

Draw it-pie-chart to represent the above information.
Solution:

Question 5.
Students of a small school use different modes of travel to school as shown below:

Draw a suitable bar graph.
Solution:

Question 6.
For the following table, draw a bar-graph

Solution:

Question 7.
Manoj appeared for ICSE examination 2018 and secured percentage of marks as shown in the following table:

Represent the above data by drawing a suitable bar graph.
Solution:

Question 8.
For the data given above in question number 7, draw a suitable pie-graph.
Solution:
∵ 60 + 45 + 42 + 48 + 75 = 270

Question 9.
Mr. Kapoor compares the prices (in Rs.) of different items at two different shops A and B. Examine the following table carefully and represent the data by a double bar graph.

Solution:

Question 10.
The following tables shows the mode of transport used by boys and girls for going to the same school.

Draw a double bar graph representing the above data.
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 23 Probability. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Probability Exercise 23 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A die is thrown, find the probability of getting:
(i) a prime number
(ii) a number greater than 4
(iii) a number not greater than 4.
Solution:
A die has six numbers : 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6

Question 2.
A coin is tossed. What is the probability of getting:
(i) a tail? (ii) ahead?
Solution:
On tossing a coin once,
Number of possible outcome = 2

Question 3.
A coin is tossed twice. Find the probability of getting:
(i) exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Solution:

Question 4.
A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?
Solution:
Total no. of letters in the word ‘PENCIL = 6
Total Number of Consonant = ‘PNCL’ i.e. 4

Question 5.
A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:
(i) a red ball
(ii) not a red ball
(iii) a white ball.
Solution:
Total number of possible outcomes = 3

Question 6.
6. In a single throw of a die, find the probability of getting a number
(i) greater than 2
(ii) less than or equal to 2
(iii) not greater than 2.
Solution:
A die has six numbers = 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6

Question 7.
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size.
A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:

Question 8.
In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will be an odd number.
(iii) will not be an even number.
Solution:

Question 9.
In a single throw of a die, find the probability of getting :
(i) 8
(ii) a number greater than 8
(iii) a number less than 8
Solution:
On a die the numbers are 1, 2, 3, 4, 5, 6 i.e. six.
∴ Number of possible outcome = 6

Question 10.
Which of the following can not be the probability of an event?

Solution:
The probability of an event cannot be
(ii) 3.8 i.e. the probability of an even cannot exceed 1.
(iv) i.e. -0.8 and
(vi) -2/5, This is because probability of an even can never be less than 1.

Question 11.
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball,
(ii) a black ball
Solution:

Question 12.
Three identical coins are tossed together. What is the probability of obtaining:
all heads?
exactly two heads?
exactly one head?
no head?
Solution:

Question 13.
A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?
Solution:

Question 14.
Two coins are tossed together. What is the probability of getting:
(i) at least one head
(ii) both heads or both tails.
Solution:

Question 15.
From 10 identical cards, numbered 1, 2, 3, …… , 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 2 (ii) 3
(iii) 2 and 3 (iv) 2 or 3
Solution:

Question 16.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12
Solution:

Question 17.
A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number greater than 3
(iii) a number other than 3 and 5
(iv) a number less than 6
(v) a number greater than 6.
Solution:

Question 18.
Two coins are tossed together. Find the probability of getting:
(i) exactly one tail
(ii) at least one head
(iii) no head
(iv) at most one head
Solution:

Selina Concise Chemistry Class 8 ICSE Solutions – Physical and Chemical Changes

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 2 Physical and Chemical Changes. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 2 Physical and Chemical Changes

Exercise

Question 1.
Define:
(a) a physical change, (b) a chemical change.
Answer:
(a) Physical Change: A physical change is a temporary change in which no new substance is formed and the chemical composition of the original substance remains the same, even though its physical properties like colour, state, shape, size etc. might change.
(b) Chemical Change: A chemical change is permanent change in which new substances are formed whose chemical composition and physical and chemical properties are different from those of in original substance.

Question 2.
Classify the following as a physical or a chemical change.
(a) Drying of wet clothes
(b) Manufacture of salt from sea water
(c) Butter getting rancid
(d) Boiling of water
(e) Burning of paper
(f) Melting of wax
(g) Burning of coal
(h) Formation of clouds
(i) Making of a sugar solution
(j) Glowing of an electric bulb
(k) Curdling of milk
Answer:
Physical change
(a) Drying of wet clothes
(b) Manufacture of salt from sea water
(d) Boiling of water
(f) Melting of wax
(h) Formation of clouds
(i) Making of a sugar solution
(j) Glowing of an electric bulb.
Chemical change
(c) Butter getting rancid
(e) Burning of paper
(g) Burning of coal
(k) Curdling of milk

Question 3.
Fill in the blanks.
Answer:
(a) The process of a liquid changing into a solid is called freezing.
(b) A change, which alters the composition of a substances, is known as a chemical change.
(c) There is no change in the composition of the substance during a physical change.
(d) The reaction in which energy is evolved is called exothermic reaction.

Question 4.
Given reason:
(a) Freezing of water to ice and evaporation of water are physical changes.
(b) Burning of a candle is both a physical and chemical change.
(e) Burning of paper is a chemical change.
(d) Cutting of a cloth piece is a physical change, though it cannot be reversed.
Answer:
(a) Freezing of water to ice and evaporation of water are physical change because water can be brought back to its original (liquid) form by

1. We can heat the ice to bring it back to water.
2. We can cool down the vapours to bring it back to water.

(b) When a candle is lighted, some of the solid wax first melts and turns into liquid, then it turns into vapours to produce a flame. New substances CO₂ and H₂O vapours are formed alongwith the evolution of light and heat energy. This shows a chemical change. When some of the molten wax drops to the floor, it again solidifies. Which shows a physical change. Thus the melting of candle wax is a physical change and the production of CO₂ and H₂O represents chemical change.

(c) When a piece of paper is burnt a new substance ash is produced. Even when the burning is stopped, the ash cannot be changed back into paper. This shows that the formation of the ash from paper is a permanent and irreversible change.

(d) Because it does not change chemical composition of cloth and the change is only in the state, size, shape, colour, texture or the smell of some or all of the substances that undergo physical change.

Question 5.
Give four difference between physical and chemical changes.
Answer:
The differences are Physical and Chemical Changes:
Physical change

1. In a physical change no new substance is formed and the chemical composition of substance remains same. There are changes only in physical properties and state.
2. Temporaiy change which can be reversed by simple physical methods.
3. Weight of original substance doesn’t change
4. Energy like heat, light etc. may or may not be absorbed or released

Chemical change

1. In a chemical change new substance with entirely different chemical composition and properties is formed.
2. Permanent change and irreversible
3. Weight of original substances may increase or decrease
4. Energy like heat, light etc. are given out or absorbed.

Selina Concise Chemistry Class 8 ICSE Solutions – Chemical Reactions

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 6 Chemical Reactions. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 6 Chemical Reactions

Exercise – I

Question 1.
(a) Define a chemical reaction.
(b) What happens during a chemical reaction?
(c) What do you understand by a chemical bond?
Answer:
(a) Any chemical change in a matter which involves transformation into one or more substances with entirely different properties is called a chemical reaction.
(b) A chemical reaction involves breaking of chemical bonds between the atoms or groups of atoms of reacting substances and rearrangement of atoms making new bonds to form new substances.

(c) A chemical bond is the attractive force that holds the atoms of a molecule together, in a compound.

Question 2.
Give one example each of which illustrates the following characteristics of a chemical reaction:
(a) evolution of a gas
(b) change of colour
(c) change in state
Answer:
(a) When Zinc reacts with dil. sulphuric acid. Hydrogen gas is evolved, with an effervescence

(b) When blue coloured copper sulphate reacts with hydrogen sulphide gas, a black coloured substance copper sulphide is formed.

(c) The reaction between hydrogen sulphide and chlorine (both gases) produces sulphur (solid) and hydrogen chloride (gas).

Question 3.
How do the following help in bringing about a chemical change?
(a) pressure (b) light
(c) catalyst (d) heat.
Answer:
(a) Some chemical reactions take place when reactants are subjected to high pressure.
e.g: Nitrogen and hydrogen when subjected to high pressure produce ammonia gas.

(b) Some chemical reactions can take place in the presence of light. Ex. Photosynthesis.

(c) A catalyst can either increases or decreases the rate of a chemical reaction and some chemical reactions need a catalyst to change the rate of the reaction, in case it is too slow or too fast.

1. Positive catalyst: When a catalyst increases the rate of reaction finely divided iron is used as a positive catalyst in the manufacturing of ammonia from hydrogen and nitrogen.
   
2. Negative Catalyst: When a catalyst decreases the rate of reaction.
   Ex. Phosphoric acid act as a negative catalyst to decrease the rate of the decomposition of hydrogen peroxide.

(d) Some chemical reactions take place only in the presence of heat.
e.g. When lead nitrate is heated, it breaks into lead monoxide, nitrogen dioxide, and oxygen.

Question 4.
(a) Define catalyst.
(b) What are (i) positive catalysts and (ii) negative catalysts? Support your answer with one example for each of them.
(c) Name three biochemical catalysts found in the human body.
Answer:
(a) Catalyst: A catalyst is a substance that either increases or decreases the rate of a chemical reaction without itself undergoing any chemical change during the reaction.

(b) (i) Positive catalyst: When a catalyst increases the rate of chemical reaction, it is called a positive catalyst.
e.g. when potassium chlorate heated to 700°C decomposes to evolve oxygen gas, when MnO₂ has added the decomposition takes place at 300°C

(ii) Negative catalyst: When a catalyst decreases the rate of chemical reaction it is called a negative catalyst.
Example. Phosphoric acid acts as a negative catalyst to decrease the rate of the decomposition of hydrogen peroxide. Alcohol too acts as a negative catalyst in certain chemical reactions.

(c) Biochemical catalysts found in the human body:

1. Pepsin
2. Trypsin
3. lipase.

Question 5.
What do you observe when
(a) dilute sulphuric acid is added to granulated zinc?
(b) a few pieces of iron are dropped in a blue solution of copper sulphate?
(c) silver nitrate is added to a solution of sodium chloride?
(d) ferrous sulphate solution is added to an aqueous solution of sodium hydroxide.
(e) solid lead nitrate is heated?
(f) when dilute sulphuric acid is added to barium chloride solution?
Answer:
(a) When Zinc reacts with dilute sulphuric acid, hydrogen gas is evolved with effervescence.
Zn + dil. H₂SO₄ → Zn SO₄ + H₂.

(b) When a few pieces of iron are dropped into a blue coloured copper sulphate solution, the blue colour of the solution fades and eventually turns green.

(c) When a solution of silver nitrate is added to a solution of sodium chloride, white insoluble ppt. of silver chloride is formed.
AgNO₃ (aq) + NaCl (aq) → AgCl (ppt) + NaNO₃ (aq)

(d) When ferrous sulphate solution is added to sodium hydroxide solution, a dirty green ppt. of ferrous hydroxide is formed.
FeSO₄ (aq) + 2NaOH (aq) → Fe(OH)₂ ↓ + Na₂SO₄(aq)

(e) When solid lead nitrate is heated, it decomposes to produce light yellow solid lead monoxide, reddish-brown nitrogen dioxide gas and colourless oxygen gas.

(f) When few drops of dilute sulphuric acid is added to barium chloride solution, a white precipitate of barium sulphate is formed.

Question 6.
Complete and balance the following chemical equations:

Answer:

Exercise – II

Question 1.
1. Fill in the blanks.
(a) A reaction in which two or more substances combine to form a single substance is called a combination reaction.
(b) A catalyst is a substance which changes the rate of a chemical reaction without undergoing a chemical change.
(c) The formation of gas bubbles in a liquid during a reaction is called effervescence
(d) The reaction between an acid and a base is called a neutralization reaction.
(e) Soluble bases are called alkalis.
(f) The chemical change involving iron and hydrochloric acid illustrates a displacement reaction.
(g) In the type of reaction called double decomposition reaction, ions two compounds exchange their positive and negative radicals ions respectively.
(h) A catalyst either increases or decreases the rate of a chemical change but itself remains unchanged at the end of the reaction.
(i) The chemical reaction between hydrogen and chlorine is a combination reaction
(j) When a piece of copper is added to silver nitrate solution, it turns blue in colour.

Question 2.
Classify the following reactions as a combination, decomposition, displacement, precipitation, and neutralization. Also, balance the equations.


Answer:

Question 3.
Define:
(a) precipitation (b) neutralization (c) catalyst
Answer:
(a) Precipitation: A chemical reaction in which two compounds in their aqueous state react to form an insoluble salt as one of the product.
Acid + Base → Salt + Water
Example.

(b) Neutralization: A chemical reaction in which a base or an alkali reacts, with an acid to produce a salt and water only.

(c) Catalyst: A catalyst is a substance that either increases or decreases the rate of a chemical reaction without itself undergoing any chemical change.

here iron act as a catalyst and increases the rate of a chemical reaction.

Question 4.
Explain the following types of chemical reactions giving two examples for each of them.
(a) combination reaction
(b) decomposition reaction
(c) displacement reaction
(d) double decomposition reaction
Answer:
(a) Combination reaction: A reaction in which two or more substances combine to form a single substance is called a combination reaction.
A + B → AB
e.g (i) When iron and sulphur are heated together, they combine to form iron sulphide.

(ii) When carbon bums in oxygen to form a gaseous compound called carbon dioxide.

(b) Decomposition reaction: A reaction in which a compound breaks up due to the application of heat into two or more simple substances is called a decomposition reaction.

e.g. (i) Mercuric oxide when heated, decomposes to form two elements mercury and oxygen

(ii) CaCO₃ when heated decomposes to calcium oxide and carbon dioxide.

(c) Displacement reaction: A reaction in which a more active element displaces a less active element from a compound is called a displacement reaction.
AB + C → CB + A
e.g. (i) Zinc, displaces copper from copper sulphate solution.
Zn + CuSO₄ (aq) → ZnSO₄ (aq) + Cu
(ii) Iron piece when added to copper sulphate solution, copper is displaced.
Fe + CuSO₄ → FeSO₄ + Cu.

(d) Double decomposition reaction: A chemical reaction in which two compounds in their aqueous state exchange their ions to form new compounds is called a double decomposition reaction.
AB + CD → CB + AD
e.g. (i) AgNO₃ + HCl  AgCl + HNO₃(aq)
(ii) NaOH (aq) + HCl (aq)  NaCl (aq) + H₂O.

Question 5.
Write the missing reactants and products and balance the equations.

Answer:

Question 6.
How will you obtain it?
(a) Magnesium oxide from magnesium.
(b) Silver chloride from silver nitrate.
(c) Nitrogen dioxide from lead nitrate.
(d) Zinc chloride from zinc.
(e) Ammonia from nitrogen.
Also, give balanced equations for the reactions
Answer:
(a) Magnesium when burnt in air (oxygen) Magnesium oxide is formed

(b) When silver nitrate solution reacts with sodium chloride, silver chloride is formed.

(c) Lead nitrate when heated nitrogen oxide is obtained

(d) Zinc when reacts with hydrochloric acid zinc chloride and hydrogen (g) is formed.
Zn + 2HCl → ZnCl₂ + H₂

(e) Nitrogen when reacts with hydrogen at 450°C and under 200 atm, ammonia is formed.

Question 7.
What do you observe when
(a) Iron nail is kept in copper sulphate solution for some time.
(b) Phenolphthalein is added to sodium hydroxide solution.
(c) Blue litmus paper is dipped in dilute hydrochloric acid.
(d) Lead nitrate is heated.
(e) Magnesium ribbon is burnt in oxygen.
(f) Ammonia is brought in contact with hydrogen chloride. gas.
Answer:
(a) A brown layer of copper gets deposited on an iron nail. This is due to a chemical reaction.
Fe (s) + CuSO₄ (aq) → FeSO₄ (aq) + Cu (s)
(b) Solution turns pink.
(c) Blue litmus turns red in an acid solution.
(d) The pale yellow solid is lead monoxide, the reddish-brown gas is nitrogen dioxide and the colourless gas is oxygen.

(e) Magnesium ribbon bums with a dazzling white light and produces a white powder which is magnesium oxide.
The reaction can be represented as
2Mg + O₂ → 2MgO (white powder)
(f) Ammonia and hydrogen chloride, both compounds, combine to form a compound, ammonium chloride.

Question 8.
Give reason:
(a) A person suffering from acidity is advised to take an antacid.
(b) Acidic soil is treated with quick lime.
(c) Wasp sting is treated with vinegar.
Answer:
(a) An antacid neutralizes stomach acidity.
(b) If the soil is acidic it can be treated with a base like quick lime, to make it neutral.
(c) Wasp stings are alkaline and can be neutralized by vinegar which is a weak acid.

Question 9.
What is meant by the metal reactivity series? State its importance, (any two points).
Answer:
A list in which the metals are arranged in the decreasing order of their chemical reactivity is called the metal reactivity series.

Special features of the activity series:

1. The ease with which a metal in solution loses an electron(s) and forms a positive ion decreases down the series, i.e. from potassium to gold.
2. Hydrogen is included in the activity series because, as metals do, it too loses an electron and becomes positively charged (H⁺) in most chemical reactions.
3. The series facilitates the comparative study of metals in terms of the degree of their reactivity.
4. The compounds of the metals (oxides, carbonates, nitrates and hydroxides) too can be easily compared.

Question 10.
What are oxides? Give two examples of each of the following oxides.
(a) Basic oxide (b) Acidic oxide
(c) Amphoteric oxide (d) Neutral oxide
Answer:
An oxide is a compound that essentially contains oxygen in its molecule, chemically combined with a metal or a non-metal.

Question 11.
Define exothermic and endothermic reactions. Give two examples of each.
Answer:
Exothermic reactions: The chemical reaction in which heat is given out is called exothermic reactions. It causes rise in temperature. .
e.g. (i) When carbon bums in oxygen to form carbon dioxide, a lot of heat is produced.
C + O₂ → CO₂ + heat.
When water is added to quicklime a lot of heat is produced which boils the water.
CaO + H₂O → Ca (OH)₂ + Heat.

Endothermic reaction: A chemical reaction in which heat is absorbed is called endothermic reaction. It causes fall in temperature.
e.g. (i) When nitrogen and oxygen together are heated to a temperature of about 3000°C, nitric oxide gas is formed.
N₂ + O₂ + heat → 2NO (g)
(ii) Decomposition of calcium carbonate into carbon dioxide and calcium oxide when heated to a 1000°C.
CaCO₃ + Heat → CaO (s) + CO₂ (g)

Question 12.
State the effect of:
(a) an endothermic reaction
(b) an exothermic reaction on the surroundings.
Answer:
(a) Carbon dioxide present in the atmosphere is trapped by infrared radiations, gives rise to temperature which is exothermic reaction.
(b) The melting of glaciers by global warming.

Question 13.
What do you observe when
(a) an acid is added to a basic solution.
(b) ammonium chloride is dissolved in water.
Answer:
(a) A chemical reaction in which a base or an alkali reacts with an acid to produce a salt and water.
Acid + Base → Salt + Water
(b) Dissolution of ammonium chloride in water is an endothemic reaction in which heat energy is absorbed.

Selina Concise Chemistry Class 8 ICSE Solutions – Matter

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 1 Matter. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 1 Matter

Exercise

Question 1.
Define:
(a) matter
(b) intermolecular force of attraction.
Answer:
(a) Matter is anything which has mass, occupies space and can be percieved by our senses.
Example: Air, Book.
(b) The molecules of matter are always in motion and attract each other with a force called intermolecular force of attraction due to which they are held together.

Question 2.
What are the three states of matter ? Define each of them with two examples.
Answer:
The three states of matter are:
solids, liquids and gases

• Solids — A solid has a definite shape and definite volume.
  Example – wood, stone, iron, ice etc.
• Liquid — A liquid has a definite volume but not definite shape.
  Example — water, juice, milk, oil, etc.
• Gases — A gas neither has definite shape nor a definite volume.
  Example – air, hydrogen, oxygen, watervapour etc.

Question 3.
Define interconversion of states of matter. What are the two factors responsible for the change of states of matter?
Answer:
The process by which matter changes from one state to another and back to original state, without any change in its chemical composition is called interconversion state of matter.
Two factors responsible for change of state of matter are: change in
(i) Temperature (ii) Pressure

Question 4.
State the main postulates of kinetic theory of matter.
Answer:
The main postulates of the theory are:

1. Matter is composed of very small particles called atoms and molecules.
2. The constituent particles of a kind of matter are identical in all respects.
3. These particles have space or gaps between them which is known as interparticular or intermolecular space.
4. There exists a force of attraction between the particles of matter which holds them together. This force of attraction is known as interparticular or intermolecular force of attraction.
5. Particles of matter are always in a state of random motion and possess kinetic energy, which increases with increase in temperature and vice-versa.

Question 5.
What happens to water if
(a) it is kept in a deep freezer
(b) it is heated
Explain the phenomenon of change of state of water.
Answer:
(a) When water is kept in a deep freezer, it gets cooled and change into ice at 0°C ice.

(b) Water on heating changes into steam at 100°C

Phenomenon of change of state of water:
Water is a liquid under ordinary conditions but, when it is kept in a deep freezer, it changes into ice at 0°C and when ice is kept at room temperature again changes back into liquid water.
Similarly, water on heating change into steam at 100°C, which on cooling changes back into liquid water. But there is no change in the chemical composition of water. When its state changes from liquid to solid or liquid to gaseous state.

Question 6.
(a) State the law of conversation of mass.
(b) What do you observe when barium chloride solution is mixed with sodium sulphate solution?
Answer:
(a) “Matter can neither be created nor be destroyed in a chemical reaction”. However, it may change from one form to another in the process.
It can also be stated as, “In a chemical reaction, the total mass of the reactants is equal to the total mass of the products”.

(b)

We will observe that a white insoluble solid (precipitate) of barium sulphate is formed along with a solution of sodium chloride. Wait for ten minutes to complete the reaction and the solid formed to settle down.
Weigh the content again and note the reading.
We will observe that,
total mass of the apparatus + reactants = total mass of apparatus + products
Hence the law of conservation of mass is verified.

Question 7.
Give reasons:
(a) A gas can fill the whole vessel in which it is enclosed.
(b) Solids cannot be compressed.
(c) Liquids can flow.
(d) When magnesium is burnt in air, there is an increase in mass after the reaction.
Answer:
(a) Because, in gases, the molecules are free to move.
They are not stuck to each other and the intermolecular force of attraction is least in the gases. So the gas almost filled the whole vessel in which it is enclosed.

(b) In solids, particles are closely packed. There is a strong force of attraction and the intermolecular space is almost zero. Therefore the molecules are not free to move, which makes them hard and rigid. So solids can not be compressed.

(c) In liquids intermolecular force is weaker because the particles are not closely packed and hence there is large intermolecular space. So molecules in a liquids can move randomely and hence liquids can flow easily.

(d) When magnesium ribbon is burnt in air, a white solid, magnesium oxide is formed. The mass of magnesium oxide is more than the mass of magnesium. This is because mass of oxygen used is not taken. If that is considered, the total mass of the reactants and the products is found to be almost equal.

Question 8.
Fill in the blanks:

(a) The change of a solid into a liquid is called melting or fusion.
(b) The process in which a solid directly changes into a gas is called sublimation.
(c) The change of water vapour into water is called condensation.
(d) The temperature at which a liquid starts changing into its vapour state is evaporation or vaporisation.

Question 9.
Give two examples for each of the following:
(a) The substances which sublime.
(b) The substances which do not change their state on heating.
Answer:
(a) Camphor, iodine, naphthalene, ammonium chloride, dry ice (solid carbon dioxide), etc.
(b) Gases do not change their state on heating.
Example: O₂.

Question 10.
Define:
(a) Diffusion.
(b) Brownian motion.
Answer:
(a) Diffusion: The intermixing of two or more substances due to the motion of their particles in order to get a uniform mixture is called ‘diffusion’.
(b) Brownian motion: The haphazard, random motion of suspended particles on the surface of a liquid or in air is called ‘Brownian motion’.

Question 11.
When sodium chloride is added to a definite volume of water and stirred well, a solution is formed, but there is no increase in the level of water. Why?
Answer:
This is because there is some space between the particles of water in which the salt particles get accomodated when dissolved.

Question 12.
What do you observe when a gas jar which appears empty is inverted over a gas jar containing Bromine vapours? Name the phenomenon.
Answer:
When a gas jar full of bromine vapours (reddish brown) is inverted over a gas jar containing air over it. It is observed that after sometime, the reddish brown vapours of bromine also spread out into the upper jar. This mixing is called diffusion. The rate of diffusion is the fastest in gases and the slowest in solids. It increases with an increase in temperature.

Question 13.
Why can a piece of chalk be broken easily into smaller pieces while a coal piece cannot be broken easily?
Answer:
The particles of matter have force acting between them. This force keeps the particles together. The strength of this force of attraction is lesser in chalk, hence it could be broken easily into smaller pieces.
But the strength of inter-molecular force of attraction is very strong in coal, therefore it is not possible to break them into small pieces.

Selina Concise Chemistry Class 8 ICSE Solutions – Elements, Compounds and Mixtures

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 3 Elements, Compounds and Mixtures. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 3 Elements, Compounds and Mixtures

Exercise 3(A)

Question 1.
Define: (a) Elements (b) Compounds
Answer:
(a) Elements: Element is a substance which cannot be broken further into simpler substances and has a definite set of properties. Elements are made up of only one kind of atoms.
(b) Compounds: Compounds are pure substances composed of two or more elements in definite proportion by mass and has properties, entirely different from those of its constituents elements.
Compound, are made up of different types of atoms combined chemically.

Question 2.
Give two examples for each of the following:
(a) Metals (b) Non-metals
(c) Metalloids (d) Inert gases
Answer:
(a) Metals: Iron, silver, gold.
(b) Non-metals: Carbon, sulphur, oxygen.
(c) Metalloids: Antimony, silicon, boron.
(d) Inert gases: Helium, argon, neon.

Question 3.
Differentiate between:
(a) Pure and impure substances
(b) Homogenous and heterogenous substances
Answer:
(a) Pure substances —

1. Pure substances have definite composition and definite physical and chemical properties.
2. They are all homogeneous i.e. their composition is uniform throughout the bulk.
3. Examples: Elements and compounds.

Impure substances —

1. Impure substances are made up of two or more pure substances mixed together in any proportion.
2. They may be homogeneous or hetergeneous i.e. their composition is not uniform throughout the bulk.
3. They are all mixtures.
   Examples: air, sea water, petroleum, a solution of sugar in water are all impure substances.

(b) Homogeneous mixture — is a mixture where the components that make up the mixture are uniformly distributed throughout the mixture.
Example — air, sugar water, rain water.
Heterogeneous mixture — is a mixture, where the components of the mixture are not uniform or have localized regios with different properties.
Example—Cereal in milk, vegetable soup.

Question 4.
Write the chemical name of the following and also give their molecular formulae:
(a) Baking soda (b) Vinegar
(c) Marble (d) Sand
Answer:
(a) Sodium bicarbonate (Baking soda) — NaHCO₃
(b) Acetic acid (Vinegar) — CH₃COOH
(c) Calcium carbonate (Marble) — CaCO₃
(d) Silicon dioxide (Sand) — SiO₂

Question 5.
Name:
(a) a soft metal
(b) a metal which is brittle
(c) a non-metal which is lustrous
(d) a liquid metal
(e) a metal which is a poor conductor of electricity.
(f) a non-metal which is a good conductor of electricity.
(g) a liquid non-metal
(h) the hardest naturally occurring substance
(i) an inert gas
Answer:
(a) Gold
(b) Zinc
(c) Iodine
(d) Mercury
(e) Tungsten
(f) Graphite
(g) Bromine
(h) Diamond
(i) Neon, helium

Question 6.
How is sodium chloride different from its constituent elements ?
Answer:
The properties of sodium chloride are completely different from those of sodium and chlorine. Sodium is a soft, highly reactive metal. Chlorine is a poisonous non-metallic gas while sodium chloride is a very useful non poisonous compound which is added to our food to get minerals and also to add taste to it.

Question 7.
Why is iron sulphide a compound ?
Answer:
Iron sulphide is a compound which can be broken into the elements iron and sulphur they both have different properties. The properties of compound are entirely different from there of its constituents elements.

Exercise 3(B)

Question 1.
Classify the following substances into compounds and mixtures:
Answer:
Carbon dioxide, air, water, milk, common, salt, blood, fruit juice, iron sulphide.
Carbon dioxide — (Compound)
air — (Mixture)
water — (Compound)
milk — (Mixture)
common salt — (Compound)
blood — (Mixture)
fruit juice — (Mixture)
iron sulphide — (Compound)

Question 2.
Give one example for each of the following types of mixtures
(a) solid-solid homogenous mixture
(b) solid-liquid heterogenous mixture
(c) misicible liquids
(d) liquid-gas homogenous mixture
Answer:
(a) Solid-solid homogenous mixture — Alloys of metals e.g. brass, bronze stainless steel etc.
(b) Solid-liquid heterogenous mixture — Sand and water, mud and water, sugar and oil.
(c) Misicible liquids — water and ethanol.
(d) Liquid-gas homogenous mixture — Air

Question 3.
Suggest a suitable technique to separate the constituents of the following mixtures. Also give the reason for selecting the particular method.
(a) Salt from sea water
(b) Ammonium chloride from sand
(c) Chalk powder from water
(d) Iron from sulphur
(e) Water and alcohol
(f) Sodium chloride and potassium nitrate
(g) Calcium carbonate and sodium chloride
Answer:

(a) The technique used to separate the salt from seawater is Evaporation.
Reason – Because this method is used to separate the components of the homogeneous solid-liquid mixture. In this method, sea water is collected in a shallow bed and allowed to evaporate in the sun. When all the water is evaporated, salt is left behind. By this method, we only get solid and liquid is evaporated in its vapour form.

(b) Technique used to separate Ammonium chloride from sand is sublimation.
Because this method is used for solid mixtures in which one of the components can sublime on heating. In this method, Ammonium chloride changes into vapours on heating and salt is left behind.

(c) Technique used to separate chalk powder from water is filtration.
Reason – Because this process is used to separate the components of a heterogeneous solid-liquid mixture in which solids are lights and insoluble in liquids. Substances used as filters are sand filter paper at C. These filters allows the liquid to pass through them, but not solids.

(d) Technique to separate iron from sulpher is magnetic separation.
Because, this method is used when one of the component of mixture is Iron. Iron gets attracted towards the magnet and hence get separated.

(e) Technique used to separate water and Alcohol is Fractional Distillation.
Because in this method, the vapours of water is left behind in the original vessel as the alcohol boils at lower temperature than water. Thus these two liquids can be separated.

(f) Technique used is Fractional-crystallisation.
Because: This method is used when solubility of solid components of mixture and different in the same solvent. Here, sodium chloride and potassium nitrate. Both are soluble in water but solubility of potassium nitrate is more.

(g) Technique used is Solvent Extraction Method: Because, by this method, salts get dissolve in water while calcium carbonate being insoluble in water settles down in the container. And hence get separated about.

Question 4.
(a) Define mixture.
(b) Why is it necessary to separate the constituents of a mixture.
(c) State four differences between compounds and mixtures.
Answer:
(a) “Mixtures can be defined as. a kind of matter which is formed by mixing two or more pure substances (elements and compounds) in any proportion, such that they do not undergo any chemical change and retain their individual properties. Therefore they are impure substances.

(b) Because: The mixtures contain unwanted substances which may be harmful and may degrade the properties of mixtures. So we, need to separated them and extract useful substances.
This is necessary because
(i) It removes unwanted and harmful substances
(ii) to obtain pure and useful substances them.
Example: Sea water is rich in common salt which is an important ingredient of our food to add taste and nutrients. But sea water, cannot be directly used to get the salt.
Hence, it is necessary to separate both.

(c) Compound

1. A compound is formed from its constituent elements as a result of chemical reaction.
2. A compound is always homogeneous in nature.
3. In a compound the elements are present in a fixed ratio by weight.
4. The components of a compound can’t be separated by physical methods but can be separated by chemical methods only.
5. The properties of a compound are different from those of its elements.
6. The formation of a compound from its elements is accompanied by energy changes.

Mixture

1. A mixture is obtained form its (elements, compounds) components as a result of physical change.
2. The mixtures can be homogeneous or heterogeneous.
3. In a mixture the components can be present in any ratio.
4. The components of a mixture can be separated by physical methods only.
5. The properties of a mixture lie between those of-its components.
6. The formation of a mixture from its constituents is not accompanied by energy changes.

Question 5.
(a) What is chromatography ? For which type of mixture is it used ?
(b) What are the advantages of chromatography.
Answer:
(a) This is one of the latest techniques to separate the coloured components of a mixture when all the components are very similar in their properties. Example: Components of ink are separated by this method. Ink is a mixture of different dyes, which are separated by chromatography because some of the dyes are less soluble and some are more soluble in a solvent.

(b)

1. A very small quantity of the substance can be separated.
2. Components with very similar physical and chemical properties can be separated.
3. It identifies the different constitutes of a mixture.
4. It also helps in quantitive estimation of components of a mixture.

6. Choose the most appropriate answer from the options given below:

(a) a mixture of sand and ammonium chloride can be separated by

1. filtration
2. distillation
3. sublimation
4. crystallisation

(b) A pair of metalloids are

1. Na and Mg
2. B and Si
3. C and P
4. HeandAr

(c) Which of the following property is not shown by compounds?

1. They are heterogeneous.
2. They are homogeneous.
3. They have definite molecular formulae.
4. They have fixed melting and boiling points.

(d) A solvent of Iodine is

1. Water
2. Kerosene oil
3. Alcohol
4. Petrol

(e) Which of the gas is highly soluble in water ?

1. Ammonia
2. Nitrogen
3. Carbon monoxide
4. Oxygen

Selina Concise Chemistry Class 8 ICSE Solutions – Hydrogen

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 7 Hydrogen. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 7 Hydrogen

Points to Remember:

1. Hydrogen is the most abundant element found in the universe.
2. Hydrogen is much more common in the form of compounds. The most important compound of hydrogen is water.
3. The chief sources of hydrogen are water, acids and alkalies.
4. Hydrogen is prepared by the action of water, acids or alkalies on active metals.
5. Electrolysis of water results in the formation of hydrogen and oxygen.
6. Hydrogen is lighter than air.
7. Hydrogen bums in air with pop sound.
8. Hydrogen acts as a reducing agent.
9. Hydrogen is used to produce oxyhydrogen flame and in weather forecast balloons.

Exercise

1. Fill in the blanks:

(a) Hydrogen is lighter than air.
(b) Hydrogen is sparingly soluble in water.
(c) Hydrogen bums with a oxyhydrogen pale blue flame and pop sound is heard.
(d) A metal sodium hydrogen in the reactivity series gives hydrogen with water.
(e) Hydrogen reacts with metal oxides to form metal and water.
(f) Oxidation is the removal of hydrogen and addition of oxygen.
(g) In redox reaction oxidation and reduction occur simultaneously.

2. Indicate which of the following statements are true and which are false:

(a) Hydrogen molecule is monovalent.
(b) The removal of hydrogen from a substance is called reduction.
(c) Nitric acid can not be used to prepare hydrogen by its action on active metals ?
(d) The reaction between hydrogen and nitrogen to form ammonia is reversible.
(e) Zinc can liberate hydrogen from water, acid and alkali solution.
(f) Hydrogen is combustible as well as a supporter of combustion.
(g) Hydrogen gas is easily liquefiable.
Answer:
(a) False
(b) True
(c) False. Hydrogen cannot be prepared by the action of nitric acid on metals because it also releases nitrous oxide and nitric oxide.
(d) True
(e) True
(f) False
(g) False

3. Complete and balance the following equations:


Answer:

4. Give reasons for the following:

(a) Hydrogen be used as a fuel?
(b) Though hydrogen is lighter than air it cannot be collected by downward displacement of air.
(c) A pop sound produced when hydrogen is burnt?
(d) Helium replaced hydrogen in weather observation balloons?
(e) Nitric acid not used for the preparation of hydrogen gas?
(a) Because of its high heat of combustion, it is used as a fuel.
Answer:
(a) Coal gas, water gas and liquid hydrogen are some significant fuel.
(b) Since hydrogen is lighter than air. it is possible to collect the gas by downward displacement of air. But it is not safe to do so since a mixture of hydrogen and air can lead to an explosion.
(c) Impure hydrogen gas bums in air with a pop sound. This is because of the presence of impurities in it.
(d) If there is small leakage of hydrogen in a balloon, it forms a mixture with air that can explode. So helium has replaced hydrogen.
(e) Hydrogen cannot be prepared by the action of nitric acid on metals because it also releases nitrous oxide and nitric oxide and oxides the hydrogen to form water.

5. Name the following:
(a) Two metals which give hydrogen with cold water.
(b) A metal which liberates hydrogen only when steam is passed over red hot metal.
(c) The process in which oxygen is added or hydrogen is removed.
(d) A metallic oxide which can be reduced into metal by hydrogen.
Answer:
(a) Sodium (Na) and Potassium (K) give hydrogen with cold water.
(b) Iron
(c) Oxidation
(d) Copper oxide (CuO)

6. (a) Name the chemicals required to prepare hydrogen
gas in the laboratory.
(b) Give a balanced chemical equation for the reaction.
(c) Draw a neat and well-labelled diagram for the laboratory preparation ofhydrogen.
(d) How is hydrogen gas collected?
Answer:
(a) Granulated Zinc and dil. Hydrochloric acid.
(b) Zn + 2 HCl → ZnCl₂ + H₂ (g)
(c)

(d) Hydrogen gas is collected by the down-ward displacement of water.

7. How would you show that hydrogen:
(a) is a non-supporter of combustion?
(b) is lighter than air?
Answer:
(a) Hold a hydrogen gas-filled jar with its mouth downwards.
Place a lighted candle inside the jar. The candle gets extinguished but the gas bums with a pop sound. This shows that hydrogen is a non-supporter of combustion.

(b) Take a delivery tube and place one of its ends in a soap solution kept in a trough and the other one in a flat bottom jar as shown in the figure. The soap bubbles containing hydrogen rise upward the air. The rising soap bubbles prove that hydrogen is lighter than air.

Hydrogen-filled soap bubbles rising upward in the soap solution and into the air shows that hydrogen is lighter than air.

8. Hydrogen is a good reducing agent: What do you understand by the above statement? Explain with the help of copper oxide as an example.
Answer:
Hydrogen acts as a good reducing agent means, when hydrogen gas is passed over hot metallic oxides of copper, lead, iron, etc. it removes oxygen from them and thus reduces them to their corresponding metal.
Let us consider the following example, in each of which metallic oxide react with hydrogen. Metallic oxide act as oxidising agents and hydrogen acts as a reducing agent.

9. (a) Name a process by which hydrogen gas is manufactured.
(b) Give equations for the reactions.
(c) How is hydrogen separated from carbon dioxide and carbon monoxide?
Answer:
(a) Commercially, hydrogen is prepared by Bosch process.

(b) (i) Steam is passed over hot coke at 1000°C in a furnace called converters. As a result water gas is produced which is a mixture of carbon monoxide and hydrogen gases.

This reaction is endothermic in nature.
(ii) Water gas is mixed with excess of steam and passed over a catalyst ferric oxide (Fe₂O₃) and a promotor chromium trioxide (Cr₂O₃).

This reaction is exothermic in nature

(c) (i) The products are hydrogen, carbon dioxide and some unreacted carbon monoxide. Hydrogen is separated from carbon dioxide by passing the mixture through water under pressure, in which carbon dioxide gets dissolved leaving behind hydrogen. Carbon dioxide can also be separated by passing it through caustic potash (KOH) solution.
2KOH + CO₂ → K₂CO₃ + H₂O
(ii) To separate carbon monoxide the gaseous mixture is passed through ammoniacal cuprous chloride in which carbon monoxide dissolves leaving behind hydrogen.

Thus hydrogen gas is obtained.

10. Match the statements in Column A with those in Column B.

Answer:

11. State four uses of hydrogen:
Answer:

1. Hydrogen with oxygen produces an oxy-hydrogen flame which is used for cutting and welding.
2. Hydrogen gas is used as a fuel.
3. Hydrogen is used for the hydrogenation of vegetable oil.
4. Hydrogen gas is used extensively in the manufacture of ammonia gas, which is used to produce fertilizers.

12. Define:
(a) catalytic hydrogenation (b) oxidation
(c) reduction (d) redox reaction
Answer:
(a) Catalytic hydrogenation: catalytic hydrogenation is a process by which hydrogen gas is passed through vegetable oils in the presence of catalysts like Ni, Pt or Pd to convert them into solid vanaspati ghee.

(b) Oxidation: A reaction in which a substance combines with oxygen or in which hydrogen is removed is called an oxidation reaction.
Example: H₂S + Cl → 2HCl + S

(c) Reduction: Those reactions in which hydrogen combines with a substance or oxygen is removed from a substance, are known as reduction reactions.

(d) Redox reaction: Redox reactions are those in which reduction and oxidation both takes place simultaneously i.e. one substance is reduced while the other gets oxidised.

13. Multiple Choice Questions

(a) Equal volumes of hydrogen and chlorine are exposed to diffused sunlight to prepare

1. hydrogen chloride
2. water
3. sodium hydroxide
4. hydrochloric acid

(b) The metal which reacts with cold water to produce hydrogen is

1. magnesium
2. aluminium
3. calcium
4. iron

(c) In metal activity series the more reactive metals are at

1. top
2. bottom
3. middle
4. none

(d) Hydrogen is responsible for producing

1. heat and light
2. hydrogenated oil
3. fertilizers
4. all of the above

(e) Hydrogen is

1. combustible
2. non-combustible
3. supporter of combustion
4. neither supporter nor combustible

(f) Water gas is a mixture of

1. carbon monoxide and oxygen
2. carbon monoxide and hydrogen
3. hydrogen and oxygen
4. hydrogen and nitrogen.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

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Representing 3-D in 2-D Exercise 19 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If a polyhedron has 8 faces and 8 vertices, find the number of edges in it.
Solution:
Faces = 8
Vertices = 8
using Eulers formula,
F + V – E = 2
8 + 8 – E = 2
-E = 2 – 16
E= 14

Question 2.
If a polyhedron has 10 vertices and 7 faces, find the number of edges in it.
Solution:
Vertices = 10
Faces = 7
Using Eulers formula,
F + V – E = 2
7 + 10 – E = 2
-E = -15
E = 15

Question 3.
State, the number of faces, number of vertices and number of edges of:
(i) a pentagonal pyramid
(ii) a hexagonal prism
Solution:
(i) A pentagonal pyramid
Number of faces = 6
Number of vertices = 6
Number of edges = 10

(ii) A hexagonal prism
Number of faces = 8
Number of vertices = 12
Number of edges = 18

Question 4.
Verily Euler’s formula for the following three dimensional figures:

Solution:
(i) Number of vertices = 6
Number of faces = 8
Number of edges = 12
Using Euler formula,
F + V – E = 2
8 + 6 – 12 = 2
2 = 2 Hence proved.

(ii) Number of vertices = 9
Number of faces = 8
Number of edges = 15
Using, Euler’s formula,
F + V – E = 2
9 + 8 – 15 = 2
2 = 2 Hence proved.

(iii) Number of vertices = 9
Number of faces = 5
Number of edges = 12
Using, Euler’s formula,
F + V – E = 2
9 + 5 – 12 = 2
2 = 2 Hence proved.

Question 5.
Can a polyhedron have 8 faces, 26 edges and 16 vertices?
Solution:
Number of faces = 8
Number of vertices = 16
Number of edges = 26
Using Euler’s formula
F + V – E
⇒ 8 + 16 – 26 ≠ -2
⇒ -2 ≠ 2
No, a polyhedron cannot have 8 faces, 26 edges and 16 vertices.

Question 6.
Can a polyhedron have:
(i) 3 triangles only ?
(ii) 4 triangles only ?
(iii) a square and four triangles ?
Solution:
(i) No.
(ii) Yes.
(iii) Yes.

Question 7.
Using Euler’s formula, find the values of x, y, z.

Solution:

Question 8.
What is the least number of planes that can enclose a solid? What is the name of the solid.
Solution:
The least number of planes that can enclose a solid is 4.
The name of the solid is Tetrahedron.

Question 9.
Is a square prism same as a cube?
Solution:
Yes, a square prism is same as a cube.

Question 10.
A cubical box is 6 cm x 4 cm x 2 cm. Draw two different nets of it.
Solution:

Question 11.
Dice are cubes where the sum of the numbers on the opposite faces is 7. Find the missing numbers a, b and c.

Solution:

Question 12.
Name the polyhedron that can be made by folding each of the following nets:

Solution:
(i) Triangular prism. It has 3 rectangles and 2 triangles.
(ii) Triangular prism. It has 3 rectangles and 2 triangles.
(iii) Hexagonal pyramid as it has a hexagonal base and 6 triangles.

Question 13.
Draw nets for the following polyhedrons:

Solution:
Net of hexagonal prism:

Net of pentagonal pyramid:

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles)

Constructions Circles Exercise 19 – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Draw a circle of radius 3 cm. Mark a point P at a distance of 5 cm from the centre of the circle drawn. Draw two tangents PA and PB to the given circle and measure the length of each tangent.
Solution:
Steps of Construction:

1. Draw a circle with centre O and radius 3 cm.
2. From O, take a point P such that OP = 5 cm
3. Draw a bisector of OP which intersects OP at M.
4. With centre M, and radius OM, draw a circle which intersects the given circle at A and B.
5. Join AP and BP.
   AP and BP are the required tangents.
   On measuring AP = BP = 4 cm

Question 2.
Draw a circle of diameter of 9 cm. Mark a point at a distance of 7.5 cm from the centre of the circle. Draw tangents to the given circle from this exterior point. Measure the length of each tangent.
Solution:

1. Draw a circle of diameter 9 cm, taking O as the centre.
2. Mark a point P outside the circle, such that PO = 7.5 cm.
3. Taking OP as the diameter, draw a circle such that it cuts the earlier circle at A and B.
4. Join PA and PB.
   Thus, PA and PB are required tangents. PA = PB = 6 cm

Question 3.
Draw a circle of radius 5 cm. Draw two tangents to this circle so that the angle between the tangents is 45º.
Solution:
Steps of Construction:

1. Draw a circle with centre O and radius BC = 5 cm
2. Draw arcs making an angle of 180º- 45º = 135º at O such that ∠AOB = 135º
3. AT A and B, draw two rays making an angle of 90º at each point which meet each other at point P, outside the circle.
4. AP and BP are the required tangents which make an angle of 45º with each other at P.

Question 4.
Draw a circle of radius 4.5 cm. Draw two tangents to this circle so that the angle between the tangents is 60º.
Solution:
Steps of Construction:

1. Draw a circle with centre O and radius BC = 4.5 cm
2. Draw arcs making an angle of 180º – 60º = 120º at O such that ∠AOB = 120º
3. AT A and B, draw two rays making an angle of 90º at each point which meet each other at point P, outside the circle.
4. AP and BP are the required tangents which make an angle of 60º with each other at P.

Question 5.
Using ruler and compasses only, draw an equilateral triangle of side 4.5 cm and draw its circumscribed circle. Measure the radius of the circle.
Solution:
Steps of construction:

1. Draw a line segment BC = 4.5 cm
2. With centers B and C, draw two arcs of radius 4.5 cm which intersect each other at A.
3. Join AC and AB.
4. Draw perpendicular bisectors of AC and BC intersecting each other at O.
5. With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.
   This is the required circumcircle of triangle ABC.
   On measuring the radius OA = 2.6 cm

Question 6.
Using ruler and compasses only.
(i) Construct triangle ABC, having given BC = 7 cm, AB – AC = 1 cm and ∠ABC = 45°.
(ii) Inscribe a circle in the ∆ABC constructed in (i) above. Measure its radius.
Solution:
Steps of Construction:

i) Construction of triangle:

• Draw a line segment BC = 7 cm
• At B, draw a ray BX making an angle of 45º and cut off BE = AB – AC = 1 cm
• Join EC and draw the perpendicular bisector of EC intersecting BX at A.
• Join AC.
  ∆ABC is the required triangle.

ii) Construction of incircle:

• Draw angle bisectors of ∠ABC and ∠ACB intersecting each other at O.
• From O, draw perpendiculars OL to BC.
• O as centre and OL as radius draw circle which touches the sides of the ∆ABC. This is the required in-circle of ∆ABC.
  On measuring, radius OL = 1.8 cm

Question 7.
Using ruler and compasses only, draw an equilateral triangle of side 5 cm. Draw its inscribed circle. Measure the radius of the circle.
Solution:
Steps of Construction:

1. Draw a line segment BC = 5 cm
2. With centers B and C, draw two arcs of 5 cm radius each which intersect each other at A.
3. Join AB and AC.
4. Draw angle bisectors of ∠B and ∠C intersecting each other at O.
5. From O, draw OL ⊥ BC.
6. Now with centre O and radius OL, draw a circle which will touch the sides of ∆ABC
   On measuring, OL = 1.4 cm

Question 8.
Using ruler and compasses only,
(i) Construct a triangle ABC with the following data:
Base AB = 6 cm, BC = 6.2 cm and ∠CAB – 60°
(ii) In the same diagram, draw a circle which passes through the points A, B and C and mark its centre as O.
(iii) Draw a perpendicular from O to AB which meets AB in D.
(iv) Prove that AD = BD
Solution:
Steps of Construction:

1. Draw a line segment AB = 6 cm
2. At A, draw a ray making an angle of 60º with BC.
3. With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C.
4. Join BC.
   ∆ABC is the required triangle.
5. Draw the perpendicular bisectors of AB and AC intersecting each other at O.
6. With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.
7. From O, draw OD ⊥ AB.
   Proof: In right ∆OAD and ∆OBD
   OA = OB (radii of same circle)
   Side OD = OD (common)
   ∴ ∆OAD ≅ ∆OBD (RHS)
   ⇒ AD = BD (CPCT)

Question 9.
Using ruler and compasses only construct a triangle ABC in which BC = 4 cm, ∠ACB = 45° and perpendicular from A on BC is 2.5 cm. Draw a circle circumscribing the triangle ABC.
Solution:
Steps of Construction:

1. Draw a line segment BC = 4 cm.
2. At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm.
3. From E, draw another perpendicular line EY.
4. From C, draw a ray making an angle of 45º with CB, which intersects EY at A.
5. Join AB.
   ∆ABC is the required triangle.
6. Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
7. With centre O, and radius OB, draw a circle which will pass through A, B and C.
   Measuring the radius OB = OC = OA = 2 cm

Question 10.
Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O.
(i) What do you call the point O?
(ii) What is the relation between the distances OA, OB and OC?
(iii) Does the perpendicular bisector of BC pass through O?
Solution:
Steps of Construction:

1. O is called the circumcentre of circumcircle of ∆ABC.
2. OA, OB and OC are the radii of the circumcircle.
3. Yes, the perpendicular bisector of BC will pass through O.

Question 11.
The bisectors of angles A and B of a scalene triangle ABC meet at O.
i) What is the point O called?
ii) OR and OQ are drawn perpendiculars to AB and CA respectively. What is the relation between OR and OQ?
iii) What is the relation between angle ACO and angle BCO?
Solution:
Steps of Construction:

1. O is called the incentre of the incircle of ∆ABC.
2. OR and OQ are the radii of the incircle and OR = OQ.
3. OC is the bisector of angle C
   ∴ ∠ACO = ∠BCO

Question 12.
i) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm.
ii) Find its incentre and mark it I.
iii) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle.
Solution:
Steps of Construction:

1. Draw a line segment BC = 6 cm.
2. With centre B and radius 8 cm draw an arc.
3. With centre C and radius 5 cm draw another arc which intersects the first arc at A.
4. Join AB and AC.
   ∆ABC is the required triangle.
5. Draw the angle bisectors of ∠B and ∠A intersecting each other at I. Then I is the incentre of the triangle ABC
6. Through I, draw ID ⊥ AB
7. Now from D, cut off \(D P=D Q=\frac{2}{2}=1 \mathrm{cm}\)
8. With centre I, and radius IP or IQ, draw a circle which will intersect each side of triangle ABC cutting chords of 2 cm each.

Question 13.
Construct an equilateral triangle ABC with side 6 cm. Draw a circle circumscribing the triangle ABC.
Solution:
Steps of Construction:

1. Draw a line segment BC = 6 cm
2. With centers B and C, draw two arcs of radius 6 cm which intersect each other at A.
3. Join AC and AB.
4. Draw perpendicular bisectors of AC, AB and BC intersecting each other at O.
5. With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.
   This is the required circumcircle of triangle ABC.

Question 14.
Construct a circle, inscribing an equilateral triangle with side 5.6 cm.
Solution:
Steps of Construction:

1. Draw a line segment BC = 5.6 cm
2. With centers B and C, draw two arcs of 5.6 cm radius each which intersect each other at A.
3. Join AB and AC.
4. Draw angle bisectors of ∠B and ∠Cintersecting each other at O.
5. From O, draw OL ⊥ BC.
6. Now with centre O and radius OL, draw a circle which will touch the sides of ∆ABC
   This is the required circle.

Question 15.
Draw a circle circumscribing a regular hexagon of side 5 cm.
Solution:
Steps of Construction:

1. Draw a regular hexagon ABCDEF with each side equal to 5 cm and each interior angle 120º.
2. Join its diagonals AD, BE and CF intersecting each other at O.
3. With centre as O and radius OA, draw a circle which will pass through the vertices A, B, C, D, E and F.
   This is the required circumcircle.

Question 16.
Draw an inscribing circle of a regular hexagon of side 5.8 cm.
Solution:
Steps of Construction:

1. Draw a line segment AB = 5.8 cm
2. At A and B, draw rays making an angle of 120° each and cut off AF = BC = 5.8 cm
3. Again F and C, draw rays making an angle of 120° each and cut off FE = CD = 5.8 cm.
4. Join DE. Then ABCDEF is the regular hexagon.
5. Draw the bisectors of ∠A and ∠B intersecting each other at O.
6. From O, draw OL ⊥ AB
7. With centre O and radius OL, draw a circle which touches the sides of the hexagon.
   This is the required in circle of the hexagon.

Question 17.
Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.
Solution:
Steps of Construction:

1. Draw a circle of radius 4 cm with centre O
2. Since the interior angle of regular hexagon is 60o, draw radii OA and OB such that ∠AOB = 60°
3. Cut off arcs BC, CD, EF and each equal to arc AB on given circle
4. Join AB, BC, CD, DE, EF, FA to get required regular hexagon ABCDEF in a given circle.
   The circle is the required circum circle, circumscribing the hexagon.

Question 18.
Draw a circle of radius 3.5 cm. Mark a point P outside the circle at a distance of 6 cm from the centre. Construct two tangents from P to the given circle. Measure and write down the length of one tangent.
Solution:
Steps of Construction:

1. Draw a line segment OP = 6 cm
2. With centre O and radius 3.5 cm, draw a circle
3. Draw the midpoint of OP
4. With centre M and diameter OP, draw a circle which intersect the circle at T and S
5. Join PT and PS.
   PT and PS are the required tangents. On measuring the length of PT = PS = 4.8 cm

Question 19.
Construct a triangle ABC in which base BC = 5.5 cm, AB = 6 cm and m∠ABC =120˚.
i. Construct a circle circumscribing the triangle ABC.
ii. Draw a cyclic quadrilateral ABCD so that D is equidistant from B and C.
Solution:
Steps of Construction:
i.

a. Draw a line BC = 5.4 cm.
b. Draw AB = 6 cm, such that m ∠ABC = 120°.
c. Construct the perpendicular bisectors of AB and BC, such that they intersect at O.
d. Draw a circle with O as the radius.
ii.
(e) Extend the perpendicular bisector of BC, such that
it intersects the circle at D.
(f) Join BD and CD.
(g) Here BD = DC.

Question 20.
Using a ruler and compasses only :
(i) Construct a triangle ABC with the following data: AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°.
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure ∠BCP.
Solution:
Steps of constructions:

1. Draw a line segment BC = 6 cm.
At B, draw a ray BX making an angle of 120° with BC.
With B as centre and radius 3.5 cm, cut-off AB = 3.5 cm.
Join AC
Thus, ABC is the required triangle.

2. Draw perpendicular bisector MN of BC which cuts BC at point o.
With O as centre and radius = OB, draw a circle.
Draw angle bisector of ∠ABC which meets the cirde at point P.
Thus, point P is equidistant from AB and BC

3. On measuring, ∠BCP = 30°

Question 21.
Construct a ∆ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle.
Solution:
Steps of construction:

1. Draw BC = 6.5 cm.
2. With B as centre, draw an arc of radius 5.5 cm.
3. With C as oentre, draw an arc of radius 5 cm.
   Let this arc meets the previous arc at A.
4. Join AB and AC to get ∆ABC.
5. Draw the bi sectors of ∠ABC and ∠ACB.
   Let these bisectors meet each other at O.
6. Draw ON ⊥ BC.
7. With O as centre and radius ON, draw a inarcle that touches all the sides of ∆ABC
8. By measurement, radius ON = 1.5 cm

Question 22.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence :
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Solution:
Steps of construction:

1. Draw AB = 5.5 cm
2. Construct ∠BAR = 1050
3. With centre A and radius 6 cm, aut off arc on AR at C.
4. Join BC. ABC is the required triangle.
5. Draw angle bisector BD of ∠ABC, which is the loss of points equidistant from BA and BC.
6. Draw perpendicular bisector EF of BC, which is the locus of points equidistant from B and C.
7. BD and EF intersect each other at point P.
   Thus, P satisfies the above two lod.
   By measurement, PC = 4.8 cm

Question 23.
Construct a regular hexagon of side 5 cm. Hence construct all its lines of symmetry and name them. (2016)
Solution:
Steps of construction:

1. Draw AF measuring 5 cm using a ruler.
2. With A as the centre and radius equal to AF, draw an arc above AF.
3. With F as the centre, and same radius cut the previous arc at Z
4. With Z as the centre, and same radius draw a circle passing through A and F.
5. With A as the centre and same radius, draw an arc to cut the circle above AF at B.
6. With B as the centre and same radius, draw an arc to cut the circle at C.
7. Repeat this process to get remaining vertices of the hexagon at D and E.
8. Join consecutive arcs on the circle to form the hexagon.
9. Draw the perpendicular bisectors of AF, FE and DE.
10. Extend the bisectors of AF, FE and DE to meet CD, BC and AB at X, L and O respectively.
11. Join AD, CF and EB.

These are the 6 lines of symmetry of the regular hexagon.

Question 24.
Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct:
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents.   (2016)
Solution:
Steps for construction:

1. Draw AB = 5 cm using a ruler.
2. With A as the centre cut an arc of 3 cm on AB to obtain C.
3. With A as the centre and radius 2.5 cm, draw an arc above AB.
4. With same radius, and C as the centre draw an arc to cut the previous arc and mark the intersection as O.
5. With O as the centre and radius 2.5 cm, draw a circle so that points A and C lie on the circle formed.
6. Join OB.
7. Draw the perpendicular bisector of OB to obtain the mid-point of OB, M.
8. With the M as the centre and radius equal to OM, draw a circle to cut the previous circle at points P and Q.
9. Join PB and QB. PB and QB are the required tangents to the given circle from exterior point B.

QB = PB = 3 cm
That is, length of each tangent is 3 cm.

Solution 25.
Steps of construction :

1. Draw a line AB = 7 cm
2. Taking P as centre and same radius, draw an arc of a circle which intersects AB at M.
3. Taking M as centre and with the same radius as before drawn an arc intersecting previously drawn arc, at point N.
4. Draw the ray AX passing through N, then ∠XAB = 60°
5. Taking A as centre and radius equal to 5 cm, draw an arc cutting AX at C.
6. Join BC
7. The required triangle ABC is obtained.
8. Draw angle bisector of ∠CAB and ∠ABC
9. Mark their intersection as O
10. With O as center, draw a circle with radius OD

Solution 26.
Steps for construction :

1. Draw BC = 6.8 cm.
2. Mark point D where BD = DC = 3.4 cm which is mid-point of BC.
3. Mark a point A which is intersection of arcs AD = 4.4 cm and AB = 5 cm from a point D and B respectively.
4. Join AB, AD and AC. ABC is the required triangle.
5. Draw bisectors of angle B and angle C which are ray BX and CY where I is the incentre of a circle.
6. Draw incircle of a triangle ABC.

Solution 27.
Steps for construction :

1. Draw concentric circles of radius 4 cm and 6 cm with centre of O.
2. Take point P on the outer circle.
3. Join OP.
4. Draw perpendicular bisectors of OP where M is the midpoint of OP.
5. Take a distance of a point O from the point M and mark arcs from M on the inner circle it cuts at point A and B respectively.
6. Join PA and PB.

We observe that PA and PB are tangents from outer circle to inner circle are equal of a length 4.5 cm each.

Solution 28.
Steps for construction :

1. Draw BC = 7.2 cm.
2. Draw an angle ABC = 90°using compass.
3. Draw BD perpendicular to AC using compass.
4. Join BD.
5. Draw perpendicular bisectors of AB and BC which intersect at I, where I is the circumcentre of a circle.
6. Draw circumcircle using circumcentre I. we get radius of a circle is 4.7 cm.

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