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You can Download Periodic Table and Electronic Configuration Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration

Periodic Table and Electronic Configuration Text Book Questions and Answers

Kerala Syllabus 10th Standard Chemistry Chapter 1 Text Book Page No: 7

→ What is the basis of classification of elements in the periodic table?
Answer:
Atomic Number

Sslc Chemistry Chapter 1 Questions And Answers Text Book Page No: 8

→ Atomic number of sodium is 11 Electronic configuration – 2,8,1
GroupNumber — …………..
Period number — …………
Answer:
Group Number — 1
Period number — 3

→ Is the group 1 element a metal or a nonmetal?
Answer:
Metal

→ Write the electronic configuration of sodium and argon and complete the Table.

Answer:

This article mainly deals with the chemical formula of lithium oxide, the structural lithium oxide formula with its properties and uses.

Periodic Table And Electronic Configuration Class 10 Notes Text Book Page No: 9

→ How many electrons are present in the M shell, the outermost shell of argon?
Answer:
8

→ What is the maximum number of electrons that can be accommodated in the M Shell?
Answer:
18

→ The ‘K’ shell, which is the first shell, has 1 subshell. The next ‘L’ shell has 2, and so on. What will be the number of subshells in the ‘M’ shell and ‘N’
M = ……………… , N = ……………….
Answer:
M = 3, N = 4

→ Which subshell is common to all shells?
Answer:
S

Sslc Chemistry Chapter 1 Notes Kerala Syllabus Text Book Page No: 10

→ Complete the Table 1.3

Answer:

→ Complete the Table 1.4

Answer:

→ What is the maximum number of electronics that can be accommodated in the ‘s’?
Answer:
2

→ What may be the maximum number of electrons to be filled in the ‘p’ subshell?
Answer:
6

Periodic Table And Electronic Configuration Class 10 Text Book Page No: 11.

The atomic number of hydrogen is 1(₁H)

→ How many electrons are present?
Answer:
1

→ In which shell is the electron filled?
Answer:
‘K’ shell

→ In which subshell?
Answer:
S

→ How many electrons are present in helium (₂He)?
Answer:
2

Follow along with the alkene reactions cheat sheet.

Chemistry Class 10 Chapter 1 Kerala Syllabus Text Book Page No: 12

→ Complete the subshell electronic configuration?
Answer:
1s²

→ Write the electronic configuration of Lithium (3Li)
Answer:
1s² 2s¹

→ Complete the electronic configuration of beryllium?
Answer:
Be[Z=4] -1s² 2s²

→ Write the electronic configuration of Boron
Answer:
B[Z=5] -1s¹ 2s² 2p¹

→ Write the electronic configuration of Carbon
Answer:
C[Z=6] – 1s² 2s² 2p²

→ Complete the Table 1.6

Answer:

Class 10 Chemistry Chapter 1 Periodic Table And Electronic Configuration Text Book Page No: 13

→ How was the shell wise electronic configuration of potassium written?
Answer:
2, 8, 8, 1

→ Compare the energies of Is and 2s subshells. Which one has lower energy?
Answer:
1s < 2s

Sslc Chemistry Notes Chapter 1 Kerala Syllabus Question 14.
Among the 3s & 3p subshells which has higher energy?
Answer:
3s < 3p

→Among the 3d & 4s subshells which has higher energy?
Answer:
4s < 3d

→ Write down the subshells in the increasing order of their energies.
Answer:
1s <2s <2p <3s <3p <4s <3d <4p

→ Write the subshell wise electronic configu-ration of potassium.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

→The electronic configuration of scandium (2lSc) is
Answer:
s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²

Periodic Table And Electronic Configuration Kerala Syllabus Text Book Page No: 14

→ Write the electronic configu ration of 22Ti, 23V, the two elements after Sc.
Answer:
₂₂Ti — 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
₂₃V — 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²

→Which is the noble gas preceding sodium (11Na)?
Answer:
Neon(Ne)

→ Write its subshell electronic configuration.
Answer:
₁₀Ne – 1s² 2s² 2p⁶

→ Subshell electronic configuration of sodium?
Answer:
₁₁Na – 1s² 2s² 2p⁶ 3s¹

Periodic Table And Electronic Configuration Class 10 Important Questions Text Book Page No: 15

→ Using the symbol of neon, write the subshell electronic configuration of sodium?
Answer:
[Ne] 3s¹

→ Complete the Table 1.7

Answer:

→ Write the subshell electronic configuration of 24Cr
Answer:
₂₄Cr – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

→ On the basis of this, identify the correct electronic configuration of 29Cu from those given below:
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s² – False
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ – True

Periodic Table And Electronic Configuration Class 10 Question Paper Text Book Page No: 16

If the subshell wise electronic configuration of an atom is 1s² 2s² 2p⁶ 3s², find answers to the following:

→ How many shells are present in this atom?
Answer:
3

→Which are the subshells of each shell?
Answer:
K — Is, L — 2s, 2p, M — 3s

→Which is the subshell to which the last electron was added?
Answer:
3s

→ What is the total number of electrons in the atom?
Answer:
12

→ What is its atomic number?
Answer:
12

→ How can the subshell electronic configuration be written in a short form?
Answer:
[Ne]3s²

Periodic Table And Electronic Configuration Class 10 Questions And Answers Text Book Page No: 17

→ Complete the Table 1.8

Answer:

→ Which is the subshell of lithium to which the last electron was added?
Answer:
S

→ What about the subshell to which the last electron of nitrogen was added
Answer:
p

→ What is the relation between the subshell to which the last electron was added and the block to which the element belongs?
Answer:
The subshell in which the last electron enters represent the block in which the element belongs.

→ Write the subshell electronic configuration of the following elements and find the blocks to which they belong.
a. 4Be: ………………..
b. 26Fe……………..
c. 18Ar: ……………
Answer:
a. 4Be : 1s² 2s² — s block
b. 26Fe : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² — d block
c. 18Ar : 1s² 2s² 2p⁶ 3s² 3p⁶ — p block

Text Book Page No: 18

→ Complete the Table 1.9

Answer:

→ Complete the Table 1.10

Answer:

→ What is the relation between number of electrons present in the last ‘s’ subshell and their group number?
Answer:
The number of electrons in the outermost ‘s’
subshell = The group number

Text Book Page No: 20

→ When the s block elements react, do they donate or accept electrons?
Answer:
They donate electrons.

→ Which type of chemical bond is usually formed?
Answer:
Ionic bonds

→ How many electrons are donated by the first group elements in chemical reactions ?-
Answer:
One

→ How many electrons are donated by the second group elements in chemical reaction?
Answer:
Two

→ Complete the table 1.11

Answer:

→ ‘s’ block elements are present at the extreme left side of the periodic table. Relating to their position, what other characteristics can be listed out?
Answer:

• More metallic character s
• Less ionization energy
• Less electronegativity
• Lose of electrons in chemical reaction
• Compounds are mostly ionic
• Oxides and hydroxides are basic in nature

Text Book Page No: 21

→ Which are the group included in the p block
Answer:
13, 14, 15, 16, 17, 18

→ In which subshell did the filling of the last electron take place?
Answer:
p subshell

→ Complete the table 1.13.

Answer:

Text Book Page No: 22

The outermost subshell wise electronic configuration of an element Y (Symbol is hot real) is 3s2 3p4.

→ To which period and group does this element belongs to?
Answer:
Period = 3, Group = 16

→ Write down the outermost subshell electronic configuration of the element coming just below it in the same group?
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴

→ Find out examples of elements in such different states with the help of the periodic tables?
Answer:
Solid – Li, Be, B, C, Na, Mg, Al, Si
Liquid – Br
Gas – H, He, N, O, F, Ne

→ Which element has the highest ionization energy in each period?
Answer:
Group 18 elements.

Text Book Page No: 23

→ The elements having the highest electronegativity is in the p block. Find its name and position?
Answer:
Fluorine F, Period – 2, p block, Group 17

→ Analyze the general characteristics of the p block elements and prepare a note on this?
Answer:

• The outermost p subshell of the p block elements contains 1 to 6 electrons.
• Elements showing positive oxidation state and negative oxidation state are members of this block.
• There are metals and nonmetals in these blocks.
• Elements in the solid, liquid and gaseous states are present in p block.

→ Complete the table 1.14

Answer:

→ Which element has a valency 1?
Answer:
Y

→Which element shows metallic character?
Answer:
X

→ Which element has the highest ionization energy?
Answer:
Y

→ Write the chemical formula of the compound formed by the combination of X and Y and label the oxidation states?
Answer:
Compound: X Y₂
Oxidation state: X²⁺, Y^1-

→ Where is the position of d block elements in the periodic table?
Answer:
3rd Group to 12th Group

→ From which period onwards does the d block begin?
Answer:
4

Text Book Page No: 24

→ Put a tick mark ✓’ against the statements below, which are applicable to d block elements.
Answer:
1. ‘✓’ These are metals.
2. ‘✓’ The last electron is filled in the penultimate shell.
3. ‘✗’ In the case of these elements in the 4th period, the last electron is filled in 4s.
4. ‘✓’ These are found in groups 3 to 12 of the periodic table.

Text Book Page No: 25

→ Complete the table 1.16

Answer:

→ How does Fe change to Fe²⁺?
Answer:
By losing 2 electrons from 4s valence subshell.

Text Book Page No: 26

→ Write down the subshell electronic configuration of Fe21.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶

There is only a small difference of energy between the outermost s subshell and the penultimate d subshell of transition elements.

→ If so, which will be the subshell from which iron loses the third electron?
Answer:
From 3d sub-shell

→ Write the electronic configuration of Fe3+ on the basis of this.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

→ Write the subshell electronic configuration of Manganese (Mn).
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²

→ Complete the table 1.17

Answer:

Text Book Page No: 27

→ Examine these compounds available. Find more colored compounds and extend the list.
Answer:

• Copper sulfate CuSO₄.5H₂O – blue,
• Copper nitrate Cu(NO₃)₂.6H₂O – pink.
• Potassium permanganate KMnO₄ – violet.
• Ferrous sulfate FeSO₄.7H₂O – Green,
• Ferrous nitrate (Fe(NO₃)₂.6H₂O) – light green

Text Book Page No: 28

→ List out the dements of the s block?
Answer:
A, B

→ Which elements shows+2 oxidation state?
Answer:
B, C, D

→ Which elements contains 5 electrons in the outermost shell?
Answer:
E

→ Which is the element that has 5 p electrons in the outermost shell?
Answer:
G

→ Which are the elements in which the last electron enters the d subshell?
Answer:
C, D

→ Which element has the highest ionization energy?
Answer:
H

→ Which is the highly reactive nonmetal?
Answer:
G

→ Which elements show -2 oxidation state?
Answer:
F

Text Book Page No: 29

The outermost electron configuration of an element in this is 2s² 2p⁶

→ Which is the element?
Answer:
H

→ Write down the complete subshell electronic configuration?
Answer:
Is² 2s² 2p⁶

→ Write any two characteristics of this element?
Answer:

• Noble element / gases.
• The outermost shell is completely filled

→ Write the chemical number of questions, the answer of which is an element in the table
Answer:
A G

Periodic Table and Electronic Configuration Let Us Assess

Question 1.
Based on the hints given, find out the atomic number and write down the subshell electronic configuration of elements (Symbols used are not real).
i. A – period 3 group 17
ii. B – period 4 group 6
Answer:
A₁₇ — 1s² 2s² 2p⁶ 3s² 3p⁵
B₂₄ — 1S² 2s² 2p⁶ 3S² 3p⁶ 3d⁵ 4S¹

Question 2.
When the last electron of an atom was filled in the 3d subshell, the subshell electronic configuration was recorded as 3d8 Answer the questions related to this atom.
1. Complete subshell electronic configuration
2. Atomic number
3. Block
4. Period number
5. Group number
Answer:
1. Complete subshell electronic configuration:
1 s², 2s², 2p⁶, 3s², 3p⁶, 3d⁸, 4s²
2. Atomic number: 28
3. Block : d
4. Period number: 4
5. Groupnumber : 8 + 2 = 10

Question 3.
Pick out the wrong ones from the subshell electronic configuration given below.
a. 1s² 2s² 2p⁷
b. 1s² 2s² 2p²
c. 1s² 2s² 2p⁵ 3s²
d. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
Answer:
Wrong electronic configuration
a. 1s², 2s², 2p⁷
(2p maximum 6 electrons only)

c. 1s², 2s², 2p⁵, 3s¹ (electrons are filled in 3s only after filling 6 electrons in 2p)

d. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d², 4s¹ (electrons are filled in 3d only after filling 2 electrons in 4s)

Question 4.
The element X in group 17 has 3 shells. If so,
a. Write the subshell electronic configuration of the element.
b. Write the period number,
c. What will be the chemical formula of the compound formed if the element X reacts with element Y of the third period which contains one electron in the p subshell?
Answer:
a. Three shells are K, L, M. The subshells are 1s, 2s, 2p, 3s, 3p, 3d
Group number: 17
Electrons in last shell: 7
Shell electronic configuration: 2,8,7
Sub-shell electronic configuration 1s², 2s², 2p⁶, 3s², 3p⁵

b. Period-3

c. Y – Third period
∴ shells – 3
1 electron in p – subshell
Total electrons in valence shell 2+1=3 (2 electrons in s + 1 electron in p)
Valency of x – 1(1 electron is recieved – electro negative atom)
Valency of y – 3 (3 electrons are lost – electro positive atom)
Therefore they combine to form compounds with chemical formula YX3
(Symbol of electropositive element first followed by electro negative element).

Question 5.
The element Cu with atomic number 29 undergoes chemical reaction to formation with oxidation number +2.
a. Write down the subshell electronic configuration of this ion.
b. Can this element show variable valency? Why?
c. Write down the chemical formula of one compound formed when this element reacts with chlorine (17CI).
Answer:
a. ₂₉Cu — 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s¹
Cu²⁺ — 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁹

b. Yes. One electron can be lost from 4s subshell and can’exist as Cu⁺ ion, It is a d-block element.

c. Copper react with chlorine to form two compounds Cu⁺, Cu²⁺ ions react with chlorine to form CuCl and CuCl2 respectively.

Question 6.
Certain subshells of an atom are given below. 2s, 2d, 3f, 3d, 5s, 3p
a. Which are the subshells that are not possible?
b. Give the reason.
Answer:
a. Not possible sub-shells are 2d, 3f

b. d – subshell is not possible in 2^nd shell
f – subshell is not possible in 3^rd shell

Periodic Table and Electronic Configuration Extended Activities

Question 1.
Prepare the comprehensive table which indicates the name, symbol, electron configuration, subshell configuration of elements having atomic number 1 to 36?
Answer:

Question 2.
Some information related to the elements of the p bllock in the 17th group of the periodic table are given in the table below. Complete the table and analyze the following questions?
Answer:

a. What is the family names of elements belonging to the 17^th group?
Answer:
Halogen

b. What is their common valency?
Answer:
1

c. Which element has the highest electro negativity ?
Answer:
F

d. Which element has the highest ionization energy?
Answer:
F

e. List out the name and chemical formula of the compounds formed by these elements with block elements?
Answer:

• sodium chloride – NaCl
• potassium chloride – KCl
• magnesium chloride- MgCl₂
• calcium chloride- CaCl₂
• magnesium fluoride- MgF₂
• calcium fluoride – Ca F₂
• sodium iodide – Nal
• potassium iodide – KI
• potassium bromide – KBr
• potassium fluoride – KF

Periodic Table and Electronic Configuration Orukkam Questions and Answers

Question 1.
Complete the table of details about shells and subshells.

a. No of electrons in KLMN shell.
b. No of electrons in each shell.

c. Which subshell is common to all sub-shells?
d. Write names of subshells in accordance with increasing energy level,
e. Identify the incorrect subshell electronic configuration.
– 1s²
– ls² 2p⁶
– ls² 2s² 2p⁶
– 1s² 2s² 2p⁶ 3s² 3p²
Answer:

a, K – 2 ; L – 8 ; M – 18 ; N – 32

c. s- Subshell

d. 1s <2s <2p <3s <3p <4s <3d <4p <5s <4d <5p <6s <4f <5s

e. 1s² 2p⁶

Question 2.
Atomic number of iron is 26. It exhibits Fe²⁺, Fe³⁺ oxidation state. Write the subshell electronic configuration.

Answer:

Question 3.
Manganese, a d-block element exhibits I different oxidation state. Why?
a. Include chemical formulae of more compounds of manganese in the table, write their ; oxidation state and subshell electronic configuration.

b. Write the oxidation number and subshell electronic configuration K, Cl and O.
Answer:
Manganese shows different oxidation states because in manganese 4s and 3d subshell electrons take part in chemical reactions.

Question 4.
Find out atomic number, group, block period using subshell electronic configuration and then complete the table.

Answer:

Question 5.
Write down the characteristics of s,d,p, f block elements
Answer:
s-block elements:
Elements in which last electron enters into s-subshell are called s-block elements. It contains group I elements (Alkali metals) and group II elements (Alkaline earth metals).

1st group elements lose one electron during chemical combination. Therefore its oxidation state is +1.

2nd group elements lose two electrons from valence shell during chemical combination and their oxidation state is +2.

The highest shell number in a sub-shell electronic configuration is the period number of that element.

1. Group number characteristics = no.of electrons in valence sub-shell.
2. s block ionization energy & electro negativity decreases downwards.
3. Metallic character & reactivity increases downwards.
4. Lose electrons during chemical combination j and they form ionic compounds.
Their oxides and hydroxides are basic.
Their atomic radii are high in a period.

p-block elements:

• Last electron enters into p-subshell.
• Group 13 -18 elements.
• Highly reactive elements are non-metals – group 17,
• These are elements with positive and negative oxidation state.

Group number of p-block elements = electrons in last p-subshell + 12

d-block elements:

• Last electron enters into penultimate d-subshell
• Known as transition elements.
• Metals
• Shows similarity in group and period.
• Variable oxidation states.
• Form coloured compounds.

Group = electrons in ‘d’-subshdl + electrons in s-subshell.

f-block elements:

• Last electron enters into antepenultimate f sub-shell.
• Contains Lanthanoids and Actinoids.
• Variable oxidation state.
• Most of the Actinoids are radioactive.
• Most of the elements are artificial.
• U, Th, Pu are used in nuclear reactors.
• Some elements are used as catalyst in pet-roleum industry.

Periodic Table and Electronic Configuration Evaluation Questions

Question 1.
Write down subshell electronic configuration of Cu¹⁺ and Cu²⁺
Answer:
Cu¹⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
Cu²⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹

Question 2.
How many ‘s’ subshell electrons are in 1s², 2s², 2p⁶, 3s², 3p²
Answer:
6 Electrons

Question 3.
11,17,10 are the atomic number of elements X, Y, and Z.
a. Write down their subshell electronic configuration, group, block, period,
b. Write the molecular formulae of the compound formed when any two of the above elements are combined.
c. Write down the oxidation numbers of the elements in those compounds. Write the subshell electronic configuration of both ions.
Answer:

b. X Y

Question 4.
Element ‘X’ is having atomic number 28, it gives two electrons to element ‘Y’.
a Write down the electronic configuration of ‘X’ and its ion
b. In which block ‘X’ belongs?
c. Write down the characteristics of that block
Answer:
a. X₂₈ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
X²⁺– 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸

b. d block Compound

c. 1. It exhibits variable oxidation states
2. Forms colored compounds
3. Last electron enters d subshell

Question 5.

a. Write down the group and period of each element.
b. What are the use of writing electronic configuration this fashion?
Answer:

b. Group and period of the element can be identified easily. In the Same way long electron configuration can be avoided.

Question 6.
₂₄Cr – [Ar] 3d⁵ 4s¹
₂₉Cu – [Ar] 3d¹⁰ 4s¹
Why chromium and copper exhibits such electronic configuration ?
Answer:
Half filled and completely filled subshells are most stable. Change in the electronic configuration of ₂₄Cr &, ₂₉Cu is due to .this. The electrons in these elements are arranged in such away to give these elements stability.

Periodic Table and Electronic Configuration SCERT Questions and Answers

Question 1.
The electronic configuration of the elements A, B, C, Dare given below.
A – 1s² 2s² 2p⁶ 3s² 3p⁴
B – 1s² 2s² 2p⁶ 3s²
C – 1s² 2s² 2p⁶ 3s² 3p⁵
D – 1s² 2s² 2p⁶ 3s¹
a. Which of these elements show +2 oxidation state?
b. Which metal belongs to 17^th group?
c. Which is the period number of the element A ? What is the basis of your findings?
d. Which of these elements can form basic Oxides?
Answer:
a. B

b. C

c. Period number: 3, Period number = No.of shells

d. B, D

Question 2.
Two compounds of iron are jpven below.
FeSO₄ Fe₂(SO₄)₃
(The oxidation state of sulfate radical is-2)
a. Which ofthese compounds show +2 oxidation state for Fe?
b. Which compounds has Fe³⁺ ion?
c. Write the subshell electronic configuration of Fe³⁺ ion.
d. Why do transition elements show variable oxidation states?
Answer:
a., FeSO₄
b. Fe₂(SO₄)₃
c. Fe³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

d. The energy difference between the outer most ‘s’ subshell and penultinate ‘d’ subshell is very small. Hence under suitable conditions, the electrons in ‘d’ subshell also take part in chemical reaction.

Question 3.
Identify the incorrect electronic configurations and correct them.
i) 1s² 2s² 2p³
ii) 1s² 2s² 2p⁶ 3s¹
iii) 1s² 2s² 2p⁶ 2d⁷
iv) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴
Answer:
iii). 1s² 2s² 2p⁶ 3s² 3p⁵ .
iv). 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

Question 4.
Complete the table.

Answer:
a. – 2
b. 1
c. 17,
d. – 1
e. 12
f. +12

Question 5.
a. Two compounds XY₂, XZ₄ are given. The oxidation state of Z is 1. What will be the oxidation state of Y ?
b. Write the molecular formula of the compound formed by Y when it combines with aluminum (Al) having oxidation state +3.
Answer:
a. Y= – 2 (oxidation state of X is +4)
b. Al₂ Y₃

Question 6.
Pick out the statements which suit to f-block elements.
a. All of them are naturally occurring elements.
b. Uranium and Thorium are f block elements.
c. Last electrons is filled in the shell pre-ceding the outermost shell.
d. last electrons are filled up in the antepenultimate shell.
e. Includes some radioactive elements.
f. Some of them are used as catalyst in petroleum industry.
Answer:
b, d, e, f

Question 7.
The atomic number of four elements are given below. (The symbols ore not real)
A – 8
B – 10
C – 12
D – 18
a. Write the sub-shell electronic configuration of the elements,
b. Which of them are inert gases?
c. Write the chemical formula of the compound formed by two elements other than inert gases.
Answer:
a. A – 1s² 2s² 2p⁴
B – 1s² 2s² 2p⁶
C – 1s² 2s² 2p⁶ 3s²
D – 1s² 2s² 2p⁶ 3s² 3p⁶

b. B, D

c. CA, (C₂ A₂ is simplified and written as CA)

Question 8.
The subshell electronic configuration of two elements ends as follows. (Symbols are not real)
P – 3s² Q – 3p⁴
a. Write the complete subshell electronic configuration.
b. Find out the oxidation state of each element.
c. The chemical formula of the compound formed by these elements is PQ. Is this statement correct? Justify your answer.
Answer:
a. P – 1s² 2s² 2p⁶ 3s²
Q – 1s² 2s² 2p⁶ 3s² 3p⁴

b. P = +2, Q = – 2 :

c. Right, valency of both P and Q is ‘2’

Question 9.
Match the following.

Answer:

Question 10.
The atomic number of two elements are given below.
Si – 14 Ni – 28
a. Write the subshell electronic configu-ration of these elements.
b. Find out the group and period of each element.
Answer:
a. Si – 1s² 2s² 2p⁶ 3s² 3p²
Ni – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
b. Si – Period Number – 3, Group number – 14 Ni – Period Number – 4, Group number – 10

Question 11.
The element ‘X’ has 4 shells and its 3d subshell has 6 electrons. (Symbol is not real)
a. Write the complete electronic configu-ration of the element.
b. What is its group number? Which is the block?
c. Write any two characteristics of the block to which element X belongs to.
d. From which subshell the electrons are lost when the element X shows +2 oxidation state.
Answer:
a. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
b. Group number – 8, Block – d
c. All of them are metals
d – block elements are placed in group 3 to group 12
d. s – Sub shell

Question 12.
The outermost electronic configuration of the element A is 2s² 2p². (Symbol is not real)
a. Find out the group number and block of the element.
b. Write the chemical formula of the compound formed by A when it combines with chlorine.
c. Write the complete electronic configuretion of the element just below ‘A’ in the j periodic table.
Answer:
a. Group number – 14, Block – P
b. ACl₄
c. 1s² 2s² 2p⁶ 3s² 3p²

Question 13.
The figure of an incomplete periodic table is given below.

a. Which one of these elements shows -2 oxidation state?
b. Which of these elements have 3 electrons in their outermost p subshell?
c. Which element has the highest atomic radius? Which one has the least?
d. Which of these elements show variable oxidation state?
e. Which of these elements has the highest ionization energy?
Answer:
a. G

b.F

c. The element having highest atomic radius – A
The element having lowest Atomic radius – H

d. D, C

e. H

Question 14.
Examine the given electronic configurations.
A – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
B – 1s² 2s² 2p⁶ 3s¹
C – 1s² 2s² 2p¹ 3s² 3p⁶
D – 1s² 2s² 2p⁶ 3s²
E – 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
a. Which of these elements belongs to 4th period?
b. Which elements belongs to the same group ?
c. Which element doesn’t participate in chemical reactions generally ?
d. Which element has highest metallic character ?
Answer:
a. A, E
b. B, E
c. C
d. E

Question 15.
The atomic number of the elements X and Y are 20, 26 respectively. When these elements combine with chlorine, three compounds XCl₂, YCl₂, YCl₃ are formed.
a. What is the specialty of the oxidation number of Y, compared to that of X?
b. Explain the reason for this, on the basis of the subshell based electronic configuration.
Answer:
a. Element X has constant oxidation state. Y shows variable oxidation states.
b. X₂₀ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s²
X₂₆ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁶
Y is a transitional element. In chemical reactions only two elections in ‘ s’ subshell or besides ‘s’ subshell electrons ‘d’ sub shell electrons also take part.

Periodic Table and Electronic Configuration Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
Arrange the following sub-shells in the in-creasing order of energy 5p, 2s, 4f, 3s, 4s, 3d, 6s
Answer:
2s < 3s < 4s < 3d < 5p < 6s < 4f

Question 2.
Last electron in f-block elements goes to
a. Which shell? Outer shell/Penultimate shell /Antepenultimate shell
b. Which sub-shell? Outer f-subshell Penultimate f-subshell/Antepenulti mate f-subshell.
Answer:
a. Antepenultimate shell
b. Antepenultimate f-sub-shell

Question 3.
Sub-shell electronic configuration of X is given below.
1s², 2s², 2p⁵
a. The element Y is coming just below the element in same group. Then write the sub-shell electronic configuration of Y.
b. Write the sub-shell electronic configuration of the element next to X in same period.
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p²
b. Is², 2s², 2p⁶

Question 4.
A compound of vanadium pentoxide (V20;) is used as catalyst.
a. What is the oxidation state of vanadium in this compound?
b. How vanadium ion is represented?
c. Write the sub-shell electronic configuration of this ion (V – 23)
Answer:
a. +5
b. V⁵⁺
c. 1s², 2s², 2p⁶, 3s², 3p⁶

Short Answer Type Questions (Score 2)

Question 5.
Find the wrong electronic configurations from the following. What is wrong in these?
a. 1s²,2s²,2p⁶,3s²,3p⁶,3d⁹,4s²
b. 1s²
c. 1s², 2s¹, 2p⁶
d. 1s², 2s², 2p⁶, 3s², 3p²
Answer:
(a) and (c) are wrong electronic con figurations.

In (a) one electron from 4s is to be trans-ferred to 3d since completely filled configu-rations are more stable. So the correct electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s¹

In (c) electrons are filled in 2p only after filling electrons in 2s.

Question 6.
Group and period number of two elements are given.
P – group 17, period – 3
Q – group 2, period – 3
a. Write the sub-shell electronic configuration of each.
b Write the chemical formula of the compound formed by their combination.
Answer:
Answer:
a. P – 1s², 2s², 2p⁶, 3s², 3p⁵
Q – 1s², 2s², 2p⁶, 3s²

b. Q is electropositive. P is electro negative;
∴Chemical formula QP2

Question 7.
Write the reason for the statement given
below.
a. d-block elements in the same period show similarity.
b. Transition elements show variable oxidation state.
Answer:
a. Valence shell electrons of d-block elements in same periods are almost same. Valence shell electrons are entering in chemical reaction. Therefore they shows similarity.

b. Energy of electrons in s-subshell and inner d- subshells are almost same. Therefore s- electrons or s and d electrons take part in chemical reaction and show variable oxidation state.

Short Answer Type Questions (Score 3)

Question 8.
Write the sub-shell electronic configuration of following elements. Predict the block, group and period. (Symbols are not real)
a. M – 27 b. N – 19 c. P – 15
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁷, 4s²
block – d; group – 9; period – 4.
b. 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹
block – s; group – 1; period – 4
c. 1s², 2s², 2p⁶, 3s², 3p³
block – p; group – 15; period – 3

Question 9.
Observe the model of periodic table.

a. Which element is having S electrons in valence shell?
b. Which elements are having 2 electrons in valence sub-shell?
c. Which element is having last electron in3p?
d. Which element ends with electronic configuration 4d⁵, 5s¹ ?
Answer:
a. B;
b.A, C;
c. C, D;
d. E

Question 10.
Calculate oxidation state of transition elements in the following compounds.
Answer:
a. KMnO4 – Mn – 7⁺
b. Cr₂ O₃ – Cr – 3⁺
c. K₂Cr₂O₇ – Cr – 6⁺

Question 11.
Atomic number of some elements are given. A – 15, B – 8, C – 11, D – 18, E – 20, F – 34, G – 10
a. Which are the elements in same period?
b. Which are the elements in same group?
Answer:
A -1s², 2s², 2p⁶, 3s², 3p³
(group -15 period – 3)
B – 1s², 2s², 2p⁴ (group –16 period – 2)
C – 1s², 2s², 2p⁶, 31 (group – 1 period – 3)
D – 1s², 2s², 2p⁶, 3s², 3p⁶ (group – 18 period – 3)
E – 1s², 2s², 2p⁶, 3s², 3p⁶, 4s².(group – 2 period – 4)
F – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁴ (group – 16 period – 4)
G – 1s², 2s², 2p⁶ (group – 18 period – 2)
a. A, D same period ;
B, G same period 1
b. B, F same group ;
D, G same group

Long Answer Type Questions (Score 4)

Question 12.
Electronic configuration of some elements are given. Write answers to the following questions.
i. [Ne] 3s²
ii. [Ar ] 3d²,4s²
iii. [Xe] 6s²
iv. [Ne]3s²
v [Ne] 3s²,3p⁵
a. Which metal is having high reactivity?
b. Which is having possibility of formation of colored compounds?
c. Which is the non-metal?
d Which element shows the possibility of +2 oxidation state?
Answer:
a. [Xe] 6s¹
b. [Ar ] 3d², 4s²
c. [Ne] 3s², 3p⁵
d. [Ne] 3s²,[Ar] 3d², 4s²

Question 13.
Pick the wrong statement from the following.
a. Elements with atomic number 5 belong to group 15.
b. Electronic configuration of scandium (Atomic number 21) is 2,8,8,3.
c. d-block elements are known as transition elements.
d. All s-block elements are metals.
Answer:
a. Wrong. It belongs to group 13.
b. Wrong. Electronic configuration
2,8,9,2 (1s², 2s², 2p⁶, 3s², 3p⁶,3d¹, 4s²)
c. Correct .
d. Correct

Question 14.
Look at the Bohr model of X-atom.

a. Write the sub-shell electronic configuration of this atom.
b. Mention the compounds in which d-subshell electrons are taking part in chemical reaction during their formation.
XCl₂, XO₂, X₂O₇
c. Write the electronic configuration of X ions in the above three compounds.
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵, 4s²

b. XCl₂ – ion X²⁺ (electrons in 4s only)
XO₂ – ion X+⁴ (2 electrons in 4s and 2 electrons in 3d)
X₂O₇ – ion X⁷⁺ (2 electrons in 4s and 5 electrons in 3d)
XO₂, X₂O₇ d – subshell electrons are taking part in chemical reaction during the formation of X207

c. X²⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵
X⁴⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d³
X⁷⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶

Question 15.
Select the suitable one from the following columns A, B, C.

Answer:

You can Download Periodic Table Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 1 Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 4 Periodic Table

Periodic Table Textual Questions and Answers

Earlier Attempts for Classification of Elements

Kerala Syllabus 9th Standard Chemistry Notes Chapter 4 Question 1.
Explain the earlier attempt of classification by Lavoiser?
Answer:
Antoine Lavoisier classified the known elements into metals and nonmetals. But he was not able to duly classify metalloids.

9th Class Chemistry Periodic Table Kerala Syllabus Question 2.
Explain Newland’s law of octaves?
Answer:
Newlands arranged elements in the increasing order of atomic mass. He noticed that every eighth element has properties similar to those of the first elements. But this peculiarity could be noticed in elements upto calcium only.

Octaves of Newlands

9th Class Chemistry 4th Chapter Kerala Syllabus Question 3.
Define Mendeleev’s periodic law?
Answer:
In 1869 Mendeleev arranged the known 63 elements in horizontal and vertical columns and gave shape to the periodic table. He found that the chemical and physical properties of elements repeat at a regular intervals when they were arranged in the increasing order of atomic masses. Based on this Mendeleev proposed the periodic law of elements. The law states that physical and chemical properties of elements are periodic function of their atomic masses.

Chemistry Notes For Class 9 Periodic Table Kerala Syllabus Question 4.
What is meant by groups and periods in the periodic table?
Answer:
The vertical columns in the periodic table are known as groups and the horizontal rows are called periods.

Kerala Syllabus 9th Standard Chemistry Notes Question 5.
Evaluate the Mendeleev’s periodic table and find the following.
a) Total number of periods
b) Total number of Groups
c) Are the same elements showing similar properties arranged in the same group or same period?
Answer:
a) 6
b) 8
c) Same group

Class 9 Chemistry Notes Kerala Syllabus Question 6.
Advantages of Mendeleev’s periodic table?
Answer:
1. For the first time elements were comprehensively classified in such a way that elements of similar properties were placed in the same group. This has made the study of chemistry easy.
2. When the classification was made in such a way that the elements of similar properties came in the same group. It was noticed that certain their proper group. The reason for this was wrongly determined atomic masses and consequently, those wrong atomic masses were corrected.
Eg. The atomic mass of beryllium was known to be 14. Mendeleev reassessed it as a and assigned beryllium a proper place.
3. Columns were left vacant for elements which were not known at the time and their properties were predicted also. This gives an impetus to experiments in chemistry.

Ex Mendeleev give names Eka aluminum and Eka silicon to those elements which were to come below aluminum and silicon respectively in the periodic table and predicted their properties. Later when these elements gallium and germanium were discovered the prediction of Mendeleev turned out to be true.

Hsslive Guru 9th Chemistry Kerala Syllabus Question 7.
Explain the limitation of Mendeleev’s Periodic table?
Answer:
1. Elements with large difference in properties were included in the same group.
eg. Hard metal like copper [Cu], Silver [Ag] were included along with soft metals like sodium [Na] potassium [K],
2. No proper position could be given to element hy¬drogen. Non-metallic hydrogen was placed along with metals like sodium [Na] and potassium [K]
3. The increasing order of atomic mass was not strictly followed throughout.
eg. Co and Ni, Te and I
4. As isotopes are atoms of same element having different atomic masses, they should have been given different position while arranging them in the order of atomic mass. But this was not done.

Counting Atomic Calculator is a free online tool that displays the atomic mass for the given chemical formula.

Modern Period Table

Periodic Table 9th Class Pdf Kerala Syllabus Question 8.
State and explain modern periodic table and mod-ern periodic law?
Answer:
In 1913 Mosely through his x-ray diffraction experiments proved that the properties of elements depended on the atomic number not on the atomic mass.

According to this the periodic law of Mendeleev and the periodic table were modified consequently the modern periodic table was prepared by arranging elements in the increasing order of atomic number. The modern periodic law states that the physical and chemical properties of elements are periodic function of their atomic number.

Periodic Table Chapter Class 9 Kerala Syllabus Question 9.
How many periods in the modern periodic table?
Answer:
7

Labour India Class 9 Chemistry Kerala Syllabus Question 10.
Which is the shortest period?
Answer:
I period

9th Class Chemistry Chapter 4 Kerala Syllabus Question 11.
Number of elements in the third period?
Answer:
8

Class 9 Chemistry Periodic Table Kerala Syllabus Question 12.
Total number of groups?
Answer:
18

Periodic Table In 9th Class Kerala Syllabus Question 13.
Explain representative elements?
Answer:
Elements of group 1 and 2 also those in groups of 13 – 18 are called representative elements it belongs to metals, nonmetals, and metalloids.

Periodic Table Chemistry Class 9 Kerala Syllabus Question 14.
Do in representative elements do they include metalloids [eg. Si, Ge, As, Sb…) exhibiting the characteristics of metals and non-metals?
Answer:
Yes

Periodic Table Notes Pdf Class 9 Kerala Syllabus Question 15.
As there elements existing in solid, liquid and gaseous state find examples?
Answer:

• In solid-state- sodium, aluminum, carbon
• In liquid state – Bromine
• In gaseous state – oxygen, neon, argon

9th Class Chemistry Chapter 4 Notes Kerala Syllabus Question 16.
Write the electronic configuration of elements with atomic number 1-10
Answer:

Element	Atomic number	Electronic configuration
Hydrogen	1	1
Helium	2	2
Lithium	3	2, 1
Beryllium	4	2, 2
Boron	5	2, 3
Carbon	6	2, 4
Nitrogen	7	2, 5
Oxygen	8	2, 6
Fluorine	9	2, 7
Neon	10	2, 8

The atom of the elements of these group show the periodically in electron filling they contain 1 -8 electron in their outermost shell. The elements of these groups are called representative elements.

Noble Gases

Kerala Syllabus 9th Standard Notes Chemistry Question 17.
List the elements in group 18
Answer:
Helium, Neon, Argon, Krypton, Xenon, Radon

9 Class Chemistry Chapter 4 Kerala Syllabus Question 18.
Now try to write their electronic configuration
Answer:
₂He – 2
₁₀Ne – 2, 8
₁₈Ar – 2, 8, 8
₃₆Kr – 2, 8, 18, 8
₅₄Xe – 2, 8, 18, 18, 8
₈₆Rn – 2, 8, 18, 32, 18, 8

Question 19.
How many electrons are there in the outermost shell of each element?
Answer:
8

Question 20.
The elements do not normally take part in chemical reactions. Find the reason?
Answer:
They have a stable configuration in the outermost shell.

Transition Elements

Question 21.
Which group of elements belong to transition elements?
Answer:
Elements of group 3-12 in the periodic table are transition elements.

Question 22.
Find out whether elements familiar to you are present in these groups?
Answer:
Copper, silver, gold, iron

Question 23.
Aren’t transition elements metals?
Answer:
Yes

Question 24.
What are the characteristics of transition elements?
1. They from coloured compound,
2. They show similarity in properties as well as in a period.
3. In compounds, they exhibit different oxidation state
eg. Fe²⁺ and Fe³⁺

Lanthanides and Actinoids

Question 25.
Which element is next to lanthanum with atomic number 57 of group 6 in the periodic table?
Answer:
Cerium with atomic No. 58

Question 26.
Find out the position allotted to the elements with atomic number 58-71?
Answer;
Separate position at the bottom of the periodic table.

Question 27.
Is the same way aren’t the elements with atomic number 90 to 103 of period 7 give separate positions at the bottom of the periodic table?
Answer:
Yes. These elements are called inner transition elements.

Question 28.
What is meant by inner transition elements?
Answer:
Inner transition elements from Cerium [Ce] to Lutecium [Lu] of period 6 are called lanthanides. Inner transition elements from Thorium (Th] to Lewrencium [Lr] of period 7 are called actinoids. Lanthanoids are also called rare earth. Actinoids are man-made artificial elements (except thorium and uranium).

Periodic trends in the periodic table

Question 29.
Electronic configuration of group I elements of the periodic table are given
Answer:

Element	Atomic number	Electron configuration	Group	Period
H	1	1	1	1
Li	3	2, 1	1	2
Na	11	2, 8, 1	1	3
K	19	2, 8, 8, 1	1	4
Rb	37	2, 8, 18, 8, 1	1	5
Cs	55	2, 8, 18, 18, 8, 1	1	6
Fr	87	2, 8, 18, 32, 18, 8, 1	1	7

Question 30.
What is the peculiarity seen in the electronic configuration of the outer most shell of these elements?
Answer:
All these elements we can see one electron in the outermost shell.
Hence elements of group I exhibit similarity in chemical properties.

Question 31.
Which are the electrons shows the chemical properties of elements?
Answer:
Outermost electrons.

Question 32.
Is there any relationship between the group number and the number of electrons present in the outermost shell? What is it?
Answer:
Same, group number equal tot he number of election in the outermost shell for the elements in groups 1 and 2.

Question 33.
Observe figure the electronic configuration of the second-period elements of the group from 13-18 given below.

(i) Won’t we get the group number of these elements by adding 10 to the number of elements by adding 10 to the number of electrons in the outermost shell?
Answer:
Yes

(ii) Analyze table 3.1 and find whether there is any relation between the number of shells in an atom and the number of periods?
Answer:
Number of shells in an atom and the period number is same.

Size of an Atom in Group

Question 34.
Are you familiar with the Bohr model of an atom? See the Bohr model of atoms of certain elements, in group I.

(i) Which among them is the biggest?
Answer:
Potassium (K)

(ii) Which one is the smallest?
Answer:
Hydrogen [H]

(iii) What happens to the size of an atom when we move down the group?
Answer:
Increases

(iv) What is the reason for this?
Answer:
Number of shells increases.
As we move from top to bottom of a group in the periodic table the size of the atom increases as there is an increase in the number of shells.

Atomic Size in Period

See (Fig 3.3) the representation of Bohr model of elements with atomic numbers 3 to 9 in the second period of the periodic table.

Question 35.
Is there are in the number of shells with the increase in atomic number?
Answer:
No.

Question 36.
What happens to the nuclear charge with increase in atomic number?
Answer:
On moving from left to right in a period, as nuclear charge increases, the force of attraction on the outer-most electrons increases and consequently the size of atom decreases.

Ionisation Energy

Question 37.
You have understood how sodium chloride is formed by combining sodium and chlorine atoms. The Bohr model of sodium and chlorine are given below
Answer:

(i) Which among these atoms lose electrons?
Answer:
Sodium atom

(ii) Which one gains electrons
Answer:
Chlorine atom

Question 38.
How the ions are formed?
Answer:
Atom becomes charged when there is transfer of electrons [Lose or gain electrons] they are called ions.

Question 39.
Define ionization energy?
Answer:
The amount of energy required to liberate the most loosely bound electrons from the outermost shell of an isolated gaseous atom of an element is called ionization energy.

Question 40.
What are the factors affecting the ionization energy? Nuclear charge
Answer:
Size of the atom

Question 41.
When the size of an atom increases, does the attraction of the nucleus on the outermost electron increase or decrease?
Answer:
Decrease

Question 42.
Then what is the change in ionization energy?
Answer:
As the size of atom increases ionization energy decreases.

Question 43.
Can you find out how ionization energy changes as we move from top to bottom in a group?
Answer:
Ionization energy decreases.

Question 44.
What is the general trend in the variation of ionization energy on moving across a period from left to right?
Answer:
Ionization energy increases

Question 45.
Find how ionization energy changes with increase in nuclear charge?
Answer:
On moving from left to right in a period, as nuclear charge increases, the size of the atom decrease hence ionization energy increase.

Question 46.
Define electronegativity?
Answer:
In the case of two atoms joined by a covalent bond, electronegativity is the ability of each atom to attract the bonded electrons.

Question 47.
How size of an atom influence the electronegativity?
Answer:
As the size of an atom increases the distance between the nucleus and the outermost electron increases, hence the electronegativity decreases. As we move in the same period form left to right size of atom decrease hence electronegativity increases.

Question 48.
What is the basis for the chemical properties of metals and non-metals?
Answer:
Metals are the elements which give away the electrons and those that accept electrons are generally non-metals. Metals are electropositive elements because they lose electrons to form positive ions. Non-metals are called electronegative elements because they gain electrons in chemical reactions to form negative ions.

Question 49.
What is relationship between metallic character and the size of an atom?
Answer:
As the size of the atom increases metallic character also increases.

Question 50.
How do the metallic character and nonmetallic character vary while moving from left to right in a period? Arriving at a conclusion by assessing the size of atom?
Answer:
In the periodic table, while moving from to top to bottom in groups metallic character generally in-creases while non-metallic character decreases.
In a period as we move form left to right metallic character generally decreases while non-metallic character increases.

Question 51.
Don’t you think that there is a relationship between ionization energy and metallic -non-metallic character? Is the element with highest ionization energy metallic or non-metallic?
Answer;
Non-metallic

Question 52.
Then what about those having the low ionization energy?
Answer:
Metals

Question 53.
Isn’t there a relationship between electronegativity and metallic, non-metallic character? Explain the relationship?
Answer:
Non-metals are more electronegative.

Metalloids

Question 54.
Explain metalloids?
Answer:
Elements exhibiting the properties of both metal as well as nonmetal are called metalloids, eg. Silicon [Si], germanium [Ge] Arsenic [As], Antimony [Sb] and Tellurium [Te] belongs to this category.

Question 55.
You must understand certain periodic trends in the periodic table? Based on these (✓)the correct option given below in table 3.7.

Answer:

Trends	In a group from  top to bottom	In period from  left to right
Size of atom	✓ Increases/  decreases	Increases/  decreases ✓
Metallic character	✓ Increases/ decreases	Increases/ decreases ✓
Non-metallic character	Increases/ decreases ✓	✓ Increases/ decreases
Ionization energy	Increases/ decreases ✓	✓ Increases decreases
Electronegativity	Increases/ decreases ✓	✓ Increases/ decreases

Let Us Assess

Question 1.
The table given below lists the contributions and names of scientists who made earlier attempts in the classification of elements. Fill in the blanks.

Answer:

Contribution/Findings	Name of Scientist
Triads	Dobereiner
Law of octaves	Newlands
Classification of elements into metals and non-metals	Antonie Lavoisier
Modern periodic law	Henry Moseley

Question 2.
Complete the table
Answer:

Element	Atomic number	Electronic  configuration	Group  number	Period  number
Lithium	3	2,1	1	2
Oxygen	8	2J3	16	2
Argon	18	2,8,8	18	3
Calcium	20	2,8,8,2	2	4

Question 3.
Symbols of certain elements are given. Write their electronic configuration and find the period and group in which they are included.

Answer:
a) \(_{6}^{12} C\)
Electronic configuration 2, 4
Period – 2
Group – 14
b) \(_{12}^{24} \mathrm{Mg}\)
Electronic configuration 2,8,2
Period-3
Group – 2
c) \(_{17}^{35} \mathrm{Cl}\)
Electronic configuration 2,8,3
Period – 3
Group – 17
d) \(_{13}^{27} \mathrm{Al}\)
Electronic configuration 2,8,3
Period – 3
Group – 13
e) \(\begin{array}{l}{20} \\ {10}\end{array} \mathrm{Ne}\)
Electronic configuration 2,8
Period – 2
Group -18

Question 4.
There are three shells in the atom of element ‘X’, 6 electrons are present in its outermost shell.
a) Write the electronic configuration of the element.
b) What is its atomic number?
c) In which period does this element belong?
d) In which group is this element included?
e) Write the name and symbol of this element.
f) To which family of element does is this element belong to?
g) Draw and illustrate the Bohr atom model of this element.
Answer:
a) 2, 8, 6
b) 16
c) 3
d) 16
e) Sulphur, ‘S’
f) Oxygen family

Question 5.
Electronic configurations of elements P, Q, R, and S are given below. (These are not actual symbols).
P – 2, 2
Q -2, 8, 2
R – 2, 8, 5
S – 2, 8
a) Which among these elements are included in the same period?
b) Which are those included in the same group?
c) Which among them is a noble gas?
d) To which group and period does the element R belong?
Answer;
a) P and S, Q and R – belongs to same period
b) P and Q, belongs to same group
c) S
d) R belongs to 3rd period and 15th group

Question 6.
An incomplete form of the periodic table is given below. Write answers to the questions connecting the position of elements in it.

a) Which is the element with the biggest atom in group 1?
b) Which is the element having very lowest ionization energy in group 1?
c) Which element has the smallest atom in period 2?
d) Which among them are transition elements?
e) Which of the elements L and M has the lowest electronegativity?
f) Among B and I which has higher metallic character?
g) Which among these are included in the halogen family?
h) Which is the element that resembles E the most in its properties?
Answer:
a) D
b) D
c) M
d) G, H
e) L
f) B
g) M, N
h) F

Periodic Table Model Questions and Answers

Question 1.
Symbols of certain elements are given write down the electronic configuration and find the period and group in which they are included.

Answer:
a) \(\begin{array}{l}{23} \\ {11}\end{array} \mathrm{Na}\)
Electron configuration 2, 8, 1
Period – 3
Group – 1
b) \(_{17}^{35} \mathrm{Cl}\)
Electron configuration -2, 8, 7
Period – 3
Group-17
c) \(_{9}^{19} \mathrm{F}\)
Electron configuration -2, 7
Period – 3
Group – 17

Question 2.
A, B, C, D are four elements. The electronic configuration is given below and find the answers in the following (Hint. The symbols are not real)
A – 2, 2
B – 2, 8, 5
C – 2, 7
D – 2, 8, 2
a) Find the elements belongs to same period?
b) Find the elements belongs to same group.
c) ‘C’ belongs to which period and group?
Answer:
a) B, D, and A, C because the number of shells are same.
b) A, D because the number of electrons in the outermost shell is same.
c) ‘C’ belongs to second period and 17th group

Question 3.
Table given below lists the contributions and names of scientists who made earlier attempts in the classification of elements. Make them in the correct order.

Contribution/Findings	Name of scientist
Octet Rule	John Dalton
Triads	New Lands
Modern periodic table	Lavoisier
Classify into metals	Henry Moseley
Non-metals
Atomic theory	Dobereiner

Answer:

Contribution/Findings	Name of Scientist
Triads	Dobereiner
Law of octaves	Newlands
Classification of elements into metals and non-metals	Antonie Lavoisier
Modern periodic law	Henry Moseley

Question 4.
An incomplete form of the periodic table is given below.
write answers in the question connecting the position of elements in it.

1. Which element has the largest atomic size in group I?
2. Write the transition elements?
3. Which element has the lowest ionization energy in the 2nd period?
4. Which element belongs to Noble gas elements?
5. Compare L, M which element has the lowest electronegativity?
6. Write the element belongs to Halogen family?
Answer:
1. D
2. F, G, H
3. C
4. N
5. L
6. M

You can Download Gas Laws Mole Concept Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept

Gas Laws Mole Concept Text Book Questions and Answers

Gas Laws And Mole Concept Questions And Answers Kerala Syllabus 10th Text Book Page No: 33

→ Complete the table 2.1

Answer:

→ If a gas which is kept in a cylinder having a volume of 1 liter, is completely transferred to another 5-liter cylinder then what will be the volume of the gas?
Answer:
5 liter

→ Press the piston after closing the nozzle of the syringe. What will happen to the volume of air inside the syringe?
Answer:
Volume decreases

→ Explain it on the basis of the distance between the molecules of gas and their freedom of movement?
Answer:
Gases molecules are separated from each other by a large distance. As a result, there will be a lot of vacant spaces. So as the piston is pressed, the molecUles come closer and this volume decreases

Use the Molarity Calculator Chemistry to calculate the mass, volume or concentration required to prepare a solution of compound of known molecular weight.

Sslc Chemistry Chapter 2 Kerala Syllabus  Text Book Page No: 34

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

→ What assumption can be made regarding the possibility of collision between gas molecules?
Answer:
The molecules collide each other.

→ Which energy gained due to the movement of molecules? Potential energy/Kinetic, energy.
Answer:
Kinetic energy

→ When a gas is heated, temperature is increased. What happens to the movement of molecules if the temperature of the gas is increased?
Answer:
Speed of motion increased

→ Asa result, what happens to the energy of the molecules?
Answer:
Energy of molecules increase

→ Volume
Answer:
The space needed for a substance to occupy is its volume. The volume of solids and liquids are definite. But the volume of a gas is the volume of its container in which it is present.

→ Pressure
Answer:
The force exerted at unit area is pressure, Therefore, force at unit area/ pressure
\(=\frac{\text {Force exerted at the surface}}{\text {Area of the surface}}\)

→ Temperature
Answer:
The average Kinetic energy of all the molecules in a substance is its temperature

Sslc Chemistry Chapter 2 Questions And Answers Text Book Page No: 35

→ Is there any change in the number of molecules?
Answer:
No

→ What happens to the pressure when the volume is decreased?
Answer:
Pressure increases

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

Mass to Moles Calculator — The quantity of substance n in moles is equal to the mass m in grams divided by the molar mass M in g/mol.

→ What changes can you observe in the volume of the gas inside the syringe?
Answer:
Volume is formed to be decreasing.

→ What about decreasing the pressure?
Answer:
Volume increasing

→ What relation do you arrive at between pressure and volume of the gas?
Answer:
As pressure increases volume decreases when pressure is reduced volume increased.

Gas Laws And Mole Concept Kerala Syllabus 10th Text Book Page No: 36

→ The size of the air bubbles rising from the bottom of an aquarium increases. Can you explain the reason?
Answer:
As the bubbles move upward, the pressure on them decreases. This causes increase in volume. So as the bubbles move upward, their size increases.

→ What do you observe?
Answer:
Ink rises through the tube

→ What is the reason for the rising of the ink upwards?
Answer:
When the bottle is placed in hot water, air inside the bottle becomes hot. This causes expansion of air. This pushes ink in the tube. So ink rises through the tube.

→ What did you observe on cooling the bottle after taking it out? Why?
Answer:
Ink comes down. Because as air becomes cool, Its volume decreases.

→ What can you infer about the relation between the volume and temperature of a gas?
Answer:
When temperature is increased volume of gas increases. Similarly, when temperature is decreased, The volume decreases.

→ Complete the table 2.2

Answer:

Sslc Chemistry Chapter 2 Notes Kerala Syllabus Text Book Page No: 37

→ In which unit is the temperature stated?
Answer:
Kelvin (K)

→ What happens to the volume when the temperature is increased?
Answer:
Volume increases.

→ If an inflated ballon is kept in sunlight, it will burst. What may be the reason for this?
Answer:
When the ballon placed in sun light, temperature increases. so volume of air inside the ballon increases. Thus ballon expands and finally bursts.

→ What happens to the volume of the gas when its pressure is decreased or temperature is increased. volume increased/decreased.
Answer:
volume increased

→ If the temperature and pressure are kept constant how can we increase the volume?
Answer:
Fill some more gas.

→ Fill the cylinder with a little more gas. Does the number of molecules increase or decrease now?
Answer:
Then number of molecules increases

→ What is the relation between the volume and number of molecules?
Answer:
When the number of molecules increases, volume increases.

→ According to Avagadro’s law when the temperature and pressure remain constant on which factor does the volume of gas depend?
Answer:
Depends on the number of molecules.

Kerala Syllabus 10th Standard Chemistry Chapter 2 Text Book Page No: 38

→ If the mass of a coin 5g, then what will be the mass of thousand coins?
Answer:
5 x 1000 = 5000g

→ If the mass of coins in a bag is 50,000 g, then how many coins will be there?
Answer:
\(\frac { 50000 }{ 5 }\) = 10000

→ Like this we can calculate the number of coins on the basis of mass. Can‘t we?
Answer:
Yes, it becomes easy.

→ Is their any relation between the mass and the number, if the particles are of the same mass.
Answer:
Yes

→ What may be the method of stating the mass of atoms?
Answer:
The mass of an atom is compared to the mass, of another atom and expressed as a number which shows how many times it is heavier than the other atom. The atomic mass of elements are expressed by considering 1/12 mass of an atom of carbon-12 as one unit.

→ What do you understand from the statement that the atomic mass of Helium is 4?
Answer:
Atomic mass of Helium is 4. That is mass of one atom of Helium is 4 times of 1/12^th mass of carbon atom.

Enter the formula and press “calculate” to work out the molar mass calculator with steps, the number of moles in 1 g and the percentage by mass of each element.

Chemistry Class 10 Chapter 2 Kerala Syllabus Text Book Page No: 39

→ How many oxygen atoms combine with one carbon atom?
Answer:
2 Oxygen atoms

→ How many oxygen atoms combine with 1000. carbon atoms?
Answer:
2000 oxygen atoms.

→ How many atoms are present in 12g carbon?
Answer:
6.022 x 10B carb Answer:
2 × 6.022 × 10²³ oxygen atoms

→ What will be the mass of these tabs ?
Answer:
2 × 16 = 32g

Gas Laws And Mole Concept Extra Questions 10th Text Book Page No: 40

→ Complete the table 2.5

Answer:

→ 1GAM sodium means 23g sodium. This contains 6.022 × 1023 atoms. If so, how many GAM is present in 69 g sodium? How many atoms are present in it?
Answer:
\(\frac { 69 }{ 23 }\) = 3 GAM,
3 × 6.22 × 10²³ sodium atoms.

Gas Laws And Mole Concept Notes Pdf Kerala Syllabus 10th Text Book Page No: 41

→ How many GAMs are present in each the samples given below? Calculate the, number of atoms present in each of die sample ? (Atomic mass N= 14, O= 16)
1. 42g Nitrogen,
2. 80g Oxygen
Answer:
1. 42g Nitrogen:
No.of GAM = \(\frac { 42 }{ 14 }\) = 3GAM
No.of atoms = 3 × 6.022 × 10²³

2. 80g Oxygen:
No.of GAM = \(\frac { 80 }{ 16 }\) = 5GAM
No.of atoms = 5 × 6.022 × 10²³

→ Complete the table 2.6

Answer:

→ Calculate the molecular mass of glucose (C₆ H₁₂ O₆) and sulphuric acid (H₂ SO₄)
Answer:
Molecular mass of glucose
= 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 g
Molecular mass of sulphuric acid
2 × 1+ 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 g

Gas Laws And Mole Concept Questions And Answers Pdf 10th Text Book Page No: 42

→ Complete the table 2.7

Answer:

→ What is the molecular mass of oxygen?
Answer:
32 g

→ How many GMM is present in 32g oxygen?
Answer:
1 GMM

→ How many molecules are present in it?
Answer:
6.022 × 10²³ oxygen molecules

→ How many GMM is present in 28 gm nitrogen?
Answer:
1 GMM

→ How many molecules are present in N₂?
Answer:
6.022 × 10²³ N₂, molecules

→ How many GMM is present in 18 gm water?
Answer:
1 GMM

→ How many H₂O molecules arf present in it?
Answer:
6.022 × 10²³ H₂O Molecules

→ Calculate the number of GMM present in 96g oxygen?
Answer:
\(\frac { 96 }{ 32 }\) = 3GMM

Sslc Chemistry Chapter 2 Notes Pdf Kerala Syllabus Text Book Page No: 43

How many GMM are present in each of the given samples? Calculate the number of molecules present in each sample ?

→ 360 g glucose (Molecular mass = 180)
Answer:
No.of GMM = \(\frac { 360 }{ 180 }\) = 2 GMM
No. of molecules = 2 × 6.022 × 10²³

→ 90g water (Molecular mass = 18)
Answer:
No.of GMM = \(\frac { 90}{ 18 }\) = 5 GMM
No. of molecules = 5 × 6.022 × 10²³

→ How many molecules of water are present . in one mole of water ?
Answer:
6.022 × 10²³ water molecules

→ What is its mass?
Answer:
18 g

→How many GMM is present in it?
Answer:
1 GMM

Sslc Chemistry 2nd Chapter Notes Kerala Syllabus Text Book Page No: 44

22.4 L of a gas at STP = 1 mole
44.8 Lofa gas at STP = = 2 mole
224 L of a gas at STP = \(\frac { 224 }{ 22.4 }\) = 10 mole

Gas Law And Mole Concept Kerala Syllabus 10th Text Book Page No: 45

Complete the flow chart given below, related to one mole of substance.

Answer:

Gas Laws Mole Concept Let Us Assess

Gas Laws And Mole Concept Notes Kerala Syllabus 10th Question 1.
Examine the date given in the table (Temperature and number of molecules of the gas are kept constant).
a. Calculate P × V
b. Which is the gas law related to this?
Answer:
a. 8L atm
b. Boyle’s law

Gas Laws And Mole Concept Pdf Kerala Syllabus 10th Question 2.
Analyse the situations given below and explain the gas law associated with it.
a. When an inflated balloon is immersed in water, its size decreases.
b. A balloon is being inflated
Answer:
a. Avogadro’s law
b. Boyle’s law

Gas Laws And Mole Concept Class 10 Kerala Syllabus Question 3.
Certain data regarding various gases kept under the same conditions of temperature and pressure are given below.

a. Complete the table?
b. Which gas law is applicable here?
Answer:
a.

b. Avogadro’s law

Hss Live Guru 10th Chemistry Kerala Syllabus Question 4.
a. Calculate the mass of 112 L CO₂ gas kept at STP (molecular mass = 44)
b. How many molecules of CO₂ are present in it?
Answer:

Sslc Chemistry Chapter 2 Gas Laws And Mole Concept Question 5.
Calculate the volume of 170g of ammonia at STP ? (Molecular mass 17)
Answer:
Number of moles = \(\frac { Given mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10 moles
Volume at STP = mole × 22.4 L = 10 × 22.4 L = 224 L

Hsslive Chemistry 10th Kerala Syllabus Question 6.
Find out the number of moles of molecules present in the samples given below (GMM-N₂=28g, H₂O= 18g)
a. 56g N₂
b. 90g H₂O
Answer:
a. Number of mol molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 56 }{ 28 }\) = 2

b. Number of mol molecules = \(\frac { 90 }{ 18 }\) = 5

10th Chemistry 2nd Chapter Kerala Syllabus Question 7.
The molecular mass of ammonia is 17.
a. How much is the GMM of ammonia?
b. Find out the number of moles of molecules present in 170g of ammonia.
c. Calculate the number of ammonia molecules present in the above sample of ammonia?
Answer:
a. GMM of ammonia (NH3) = 14 + 3 × 1 = 17g = 1GMM

b. Number of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10

c. Number Of molecules = Mole × 6.022 × 10²³
= 10 × 6.022 × 10²³

Class 10 Chemistry Chapter 2 Kerala Syllabus Question 8.
The molecule’s mass of oxygen is 32.
a. What is the GMM of O₂
b. How many moles of molecules are there in 64g of oxygen? How many molecules are there in it?
c. Calculate the number of oxygen atoms present in 64g of oxygen?
Answer:
a. GMM of O₂ =2 × 16 = 32 g

b. No. of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 64 }{ 32 }\) = 2
Number of molecules = mole × 6.022 × 10²³
= 2 × 6.022 × 10²³
c. c. Number of atoms = Number of molecules × number of atoms in one molecules Number of atoms in one molecules of oxygen (O₂) =2
∴ total number of atoms = 2 × 6.022 × 10²³ × 2
=4 × 6.022 × 10²³

Gas Laws Mole Concept Extended activities

Hss Live Guru Chemistry 10 Kerala Syllabus Question 1.
How many grams of carbon and oxygen are required to get the same number of atoms as in one gram of Helium?
Answer:
GAM of Helium = 4 g
Number of mole atoms in 4 g of Helium = 6.022 × 10²³
Number of atoms in 1 gofHelium = \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
GAM ofCarbon = 12 g
∴ Number of atoms in 12gofCarbon=6.022 x 10²³
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
Carbon atoms = \(3 \mathrm{g}\left(12 \times \frac{1}{4}\right)\)
GAM of Oxygen = 16 g.
∴ Number of atoms in 16 g of Oxygen = 6.022 × 10²³
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 10²³
Oxygen atoms = \(4 \mathrm{g}\left(16 \times \frac{1}{4}\right)\)

Kerala Syllabus 10th Standard Chemistry Guide Question 2.
Examine the samples given:
a. 20 g of He
b. 44.8 L of NH₃ at STP
c. 67.2 L of N₂ at STP
d. 1 mol of H₂SO₄
e. 180 g of water.
i. Arrange the samples in increasing order of the number of molecules in each.
ii. What will be the ascending order of the total number of atoms?
iii. What will be the masses of samples b, c, and d?
Answer:

Mole Concept Class 11 Question 3.
In 90 grams of water.
a. How many molecules are present?
b.What will be the total number of atoms?
c. What will be the total number of electrons in this sample?
Answer:
a. GMM= 18 g
∴ Number of molecules
\(\frac { 90 }{ 18 }\) × 6.022 × 10²³ = 5 × 6.022 × 10²³ 18

b. ∴ Number of atoms = 3 × 5 × 6.022 × 10²³ = 15 × 6.022 × 10²³

c. Electrons in 1 H atom = 1
Electrons in 1 O atom = 8
Total electrons in H₂O molecule? 10
Total electrons in 90 g H₂O = 10 × 5 × 6.022 × 10²³ = 50 × 6.022 × 10²³

Gas Laws Mole Concept Orukkam Questions and Answers

Question 1.
a. Complete the table based on the data given in the box.

b. Express atomic weight and molecular weight in grams. How many moles is this? Find out the number of Atoms or molecules in it?

Answer:
a.

b.

Question 2.
Complete the table based on the molecules given in the first column and then answer the question given below.

a. 10 Mole of water = …………. g …………… Molecules
5 mole of CaO = ……….. g ………… Molecules
2 Mole of H₂SO₄ = …………. g ………… Molecules
1/2 Mole of Al₂O₃= …………. g ………….. Molecules
Answer:

a. 10 Mole of water = 10 × 18 = 180g, 10 × 6.022 × 10²³ Molecules
5 mole of CaO = 280g, 5 × 6.022 × 10²³ Mol-ecules
2 Mole of H₂SO₄ = 196g, 2 × 6.022 × 10²³ Mol-ecules
2- Mole of AlO = 100g, 1/2 × 6.022 × 10²³

Question 3.
Complete the table.

Answer:

Question 4.
Complete the data.

Answer:

Question 5.
Based on the reaction given below, write the answers for the questions.
N₂ + 3H₂ → 2NH₃ ;
a. Write the ratio of reactant molecules and product molecules.
b. How many moles of Ammonia forms when we take 2 moles of Nitrogen and six moles of Hydrogen?
c. Two moles of Nitrogen and three moles of hydrogen are taken in jar? Will they react together?
d. How many moles of Nitrogen and Hydrogen is needed for rearing 20 moles of Ammonia?
Answer:
a. 1:3:2
b. 2 Mole
c. No
d. 10 Mole Nitrogen 30 Mole Hydrogen

Question 6.
Balance the given equation and then write down the answers for the questions given below.
CH₄+O₂ → CO₂ + H₂O
a. How many moles of C02 formed when 20 moles of Methane burns in air?
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
b. Based on the equation above, How many moles of CO₂ is formed when 10 moles of Ethane is burned in air ?
Answer:
a. When 1 mole of methane bum in air 1 mole of CO₂ gas is formed.
When 20 moles methane burn 20
moles of CO₂ are formed.
The molecular weight of 20 moles = 20 × 44 = 880 g

b. When 2 moles of ethane is burned 4 moles of CO2 is formed.
The number of moles of CO₂ when 1 mole of ethane is burned = 4/2 = 2 mole.
The number of moles when 10 moles of ethane bums = 2 x 10 = 20 mole.
Weight of 20 moles = 20 × 44 = 880 g

Question 7.
Based on the given equation write down the answers.
2H₂+ O₂ → H₂O
a. How much Oxygen and Hydrogen is needed for making 1800g of water vapor?
b. How many moles of Oxygen is needed for the reaction with one mole of Hydrogen?
Answer:
a. 36 g of water vapour can be made using 4 g hydrogen.
The amount of hydrogen required to make 1 g water vapour = 4/36
The amount of hydrogen required to produce 1800 g watervapour= 4/26 × 1800 = 200 g
Mass of oxygen = 1800 – 200 = 1600 g

b. 0.5 Mole

Gas Laws Mole Concept Evaluation Questions

Question 1.
Find out the number of moles of hydrogen and Oxygen atoms present in 10 moles of HCI.
Answer:
One mole HC₁ contains 1 mole of hydrogen and 1 mole of chlorine.
Hydrogen contained in 10 moles of HC₁ = 1 × 10 =10 mol,
Clatom = 10 × 1 = 10 mol

Question 2.
Find out the mass of Hydrogen atom and chlorine atom in 10 moles of HCI.
Answer:
Mass of 10 mole hydrogen atom = 10 × 1 = 10g
Mass of 10 mole chlorine atom =10 × 35.5 = 355 g.

Question 3.
a Find out the mass of one mole of CaCO3. How many moles of calcium present in 1000g CaCO₃?
b. How many moles of Oxygen present in 1000gms of CaCO₃?
Answer:
a. Mass of 1 mole of CaCO₃ = 40 + 12 + 48 = 100 gram.
No of moles ofCa in 1000 g CaCO₃ = \(\frac { 1000 }{ 100 }\) = 10 mol

b. No of moles of Ca in 10 moles CaCO₃ =10 × 1 = 10 mol
No of moles of oxygen in 10 moles of CaCO₃ = 10 × 3 = 30mol.

This Combined Gas Law Calculator can help you estimate either the pressure, temperature or the volume of gas.

Question 4.
Find out number of moles of water formed when 4gms of Hydrogen and 32 gms of Oxygen combined together. What is the result when 5 gms of Hydrogen and 32 gms of Oxygen combined together?
Answer:
Ans. When 4 g of hydrogen and 32g of oxygen are combined 36 g of water \(\frac { 37 }{ 18 }\) = 2 mol
5g H + 32 gO → 37g H₂O
No of moles in 3 7 g of water = \(\frac { 37 }{ 18 }\) = 2.055 mol

Question 5.
a How much grams of NaCl is needed for making 2 molar solution (NaCl – 58.5). Wh-at is the amount of water needed for this?
b. How will you change a two molar solution of Sodium Chloride into 5 major?
Answer:
a. Mass of 2 moles of NaCl = 2 × 58.5 = 117 g
1 liter water is required for this.

b. When 2 moles of NaCl is dissolved in 4 liter of water 5 molar solution is obtained.

Finally, you encounter how to find molar concentration step-by-step manually, and if your preference indulges with instant calculations.

Question 6.
How many moles of Cl₂ present in 11.2 L of same in STP? Find out the mass of this?
Answer:
No.of moles present in 11.2 litre of chlorine = \(\frac { 11.2 }{ 22.4 }\) = 0.5 mol
Mass of 0.5 moles of chlorine = 0.5 × 35.5 = 17.759 g.

Question 7.
Find out the mass of Oxygen atom in 44.8L of CO₂ in STP.
Answer:
No of moles in 44.8 litre of CO₂ \(\frac { 44.8 }{ 22.4 }\) 2 mol,
1 Mole of CO₂ contains 1 mole of C and 1 mole of O₂
∴ 2 Mole of CO₂ contains 2 mole of O₂ or 4 mole of oxygen atom.
Man of oxygen atom 4 × 16 = 64 g

Question 8.
Find out the amount of CO₂ formed when the burning of one mole of Ethane.
Answer:
2 moles of CO₂ is formed when 1 mole of ethane burns.
Mass of 2 moles of CO₂ = 2 × 44 = 88g

Question 9.
Why are atomic mass of some elements are infractions ?
Answer:
The atomic masses of some elements are infractions because they exist as a mixture of isotopes of different masses. The fractional atomic masses arise because of this mixture.
Average mass = \(\frac { Total mass of all atoms }{ numbers of atoms }\)

Gas Laws Mole Concept SCERT Questions and Answers

Question 1.
One GAM substance contains Avogadro number of particles in it.
a. How many particles are there in Avogadro number ?
b. Write the number of atoms present in each of the following.
i. 32g Sulphur
ii. 32g Oxygen
iii. 32g Carbon
(Atomic mass S = 32, O = 16, C = 12)
Answer:
a. 6.022 × 10²³
b. i.6.022 × 10²³
ii. 2 x 6.022 × 10²³
iii \(\frac { 32 }{ 12 }\) × 6.022 × 10²³

Question 2.
a. Group the following into pairs having same number of atoms.
A. 2g Hydrogen
B. 16g Oxygen
C. 14g Nitrogen
D. 8g Helium (Atomic mass H=1, O= 16, N =14, He=4)
b. How many atoms are present in each pair?
Answer:
a. A, D2g Hydrogen, 8g Helium
B, C 16g Oxygen, 14g Nitrogen

b. A, D – 2 × 6.022 × 10²³
B,C – 6.022 × 10²³

Question 3.
N₂ + 3H₂ → 2NH₃
a. What is the ratio between the reactant molecules in the above reaction?
b. Complete the following table.

Answer:
a. 1:3
b. a – 2 NH₃,
b – l H₂,
c – 12H₂,
d – 4NH₃

Question 4.
2H₂ + O₂ → 2H₂O
a. What is the ratio between the reactant molecules in the above reaction?
b How many O₂ molecules are required to react 100 H₂ molecules completely?
c. How many water molecules are formed when 1000 H₂ molecules are reacted completely
Answer:
a. 2:1
b. 50 O₂ molecules
c. 1000 H₂O molecules

Question 5.
Complete the following table. (All the elements given are diatomic. Atomic mass O=16, N=14, CI=35.5)

Answer:
a. 6.022 × 10²³
b. 6.022 × 10²³
c. 71 g
d. 14 g

Question 6.
A sample of substances are given.

Hint: Molecular mass NH3 = 17, N2 = 28, H2SO4 = 98, O2=32
a. Which of these samples have same number of molecules?
b. Which of these samples has least number of molecules?
Answer:
a. 68g NH₃, 128gO₃
b. 49 g H₂SO₄

Question7.
Pick out the correct statements from the following. Also, correct the incorrect statements.
a The number of molecules present in 1 mol hydrogen and 1 mol oxygen are same,
b. 2 mol chlorine contains 4 x 6.022 x 10²³ chlorine molecules.
c. The mass of 1/2 mol nitrogen gas is 14 g.
d. 0.5 mol water has the mass 9g. There are 6.022 × 10²³ H20 molecules in it. (Atomic mass H = 1, O = 16, CI= 35.5, N= 14)
Answer:
Correct statements – a, c
No. of molecules in 2 mol chlorine is 2 × 6.022 × 10²³
Mass of 0.5 mol water is 9g. So it contains 0.5 × 6.022 × 10²³ H₂O molecules.

Question 8.
Complete the following.

Answer:
a. 2 × 6.022 × 10²³
b. 1GMM
c. 6.022 × 10²³

Question 9.
67.2 L of Carbon dioxide gas is filled in a cylinder at STP.
a. Calculate the mass of CO₂ present in it. (Atomic mass- C = 12, O = 16)
b. Calculate the number of molecules present in the cylinder.
Answer:
a. Molecular mass of CO₂=12 × 1+16 × 2 = 12 + 32 = 44
No. of moles in 67.2L CO₂ at STP = \(\frac { 67.2L }{ 22.4L }\) = 3
b. 3 × 6.022 × 10²³

Question 10.

Answer:
a. 2
b. 2 × 6.022 × 10²³
c. 17 g
d. 51 g
e. 3
f. 3 × 22.4 L

Question 11.
CH₄ + 2O₂ → CO₂ + 2H₂O
The equation describes the combustion of methane in air.
a. How many moles of oxygen is required for the complete combustion of 16g CH₄?
b. Calculate the amount of CO₂ formed when 100g of CH₄ is completely burnt?
Answer:
a. 2 mol
b. Amount of CO₂ produced by the combustion of 16g CH4 = 44g
Amount of CO₂ produced by the combustion of 1 gm CH4 = \(\frac { 44 }{ 16 }\) g
Amount of CO₂ produced by the combustion of 100g CH4 = \(\frac { 44 }{ 16 }\) × 100g

Question 12.
45 g glucose is taken in a beaker and made into 1 L (MM = 180).
a. Calculate the molarity of the solution,
b. Above solution is made up to 2 L by adding more water. What will be the molarity of the resultant solution?
c. How will you prepare IM solution of glucose with the same quantity (45 g) of glucose?
Answer:
a. 0.25
b. M = \(\frac { n }{ v }\) = \(\frac { 0.25 }{ 2 }\)
c. Add 250 ml water in 45 g glucose

Question 13.
Two gases occupy equal volume at STP are shown below.
(Atomic mass S = 32, O =16, N = 14

a. Find the mass of the gas in B.
b. Calculate the number of molecules present in B.
Answer:
a. No. of moles in 320g SO₂ = \(\frac { 320 }{ 64 }\) = 5
Mass of 5 mol NO₂ = 5 × 46 = 230 g
b. No.of molecules present in B = 5 × 6.022 × 10²³

Question 14.
The balanced chemical equation of a reaction (at STP) is given below.
2H₂(g) + O₂(g) → 2H₂O(g)
a. Calculate the volume of oxygen required to combine completely with 224 L of the hydrogen at STP.
b. Calculate the mass of water formed as a result of the reaction (a).
Answer:
a. 112 L
b. Volume of water obtained when 224 L hydrogen completely reacts with oxygen = 224 L
No. of moles in 224 L water = \(\frac { 224 }{ 22.4 }\) = 10
Massof 10 mol water = 10 × 18 = 180 g

Question 15.
Complete the table.

Hint: (MM – CO₂ = 44, CH₄ = 16, SO₂ = 64)
Answer:
a. 67.2 l
b. 132 g
c. 1/4.
d. 4g
e. 11.2 L
f. 1/2.

Question 16.
Analyse the following equation
2NO(g) + O₂(g) → 2NO₂(g)
a. Calculate the number of the moles of NO required to combine completely with 112 L of Oxygen at STP.
b. Calculate the mass of NO₂ formed when 112L of oxygen reacts completely?
Answer:
a. 10 mol
b. 2NO(g) + O₂(g) → 2NO₂(g) (2 : 1: 2)
No. of moles in 112L O₂ = 5 mol
According to equation no. of moles of NO₂ obtained by reacting oxygen completely with nitric oxide = 2
No. of moles of NO₂ obtained by reacting 5 mol oxygen completely = 10
Massof 10 mol NO₂ = 10 × 46 = 460g

Question 17.
The chemical equation of the decomposition of calcium carbonate is given below.
CaCO₃→ CaO + CO₂
(HintMM: CaCO₃ – 100, CaO – 56, CO₂ – 44)
a. Calculate the mass of CaCO3 required to get 224 g of CaO?
b. Calculate the number of CO2 molecules fronted when 224g of CaO is obtained?
Answer:
a. CaCO₃ → CaO + CO₂
100g 56g 44g
I I : I
Amount of CaCO₃ required to get 56g of CaO = 100g
Amount of CaCO₃ required to get 1 g of CaO = \(\frac { 100 }{ 56 }\)
Amount of CaCO₃ required to get 224 g of CaO = \(\frac { 100 }{ 56 }\) × 224 = 400g
b. 4 × 6.022 × 10²³

Question 18.
You are requested to make 20 moles of NaCl into packets of 100g each. (Hint^Molecular mass of NaCl is 58.5)
a. How many packets of NaCl can be prepared?
b. Is there any NaCl remaining? If so, how much?
Answer:
a. Mass of 20 mol NaCl = 20 × 58.5 = 1170g
1170 g NaCl can be made into 11 packets with 100g each.
b. Remaining NaCl = 1170 – 1100 = 70g

Gas Laws Mole Concept Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
GAM of Hydrogen is 1 g.
a. How many number of atoms are there in 1 g of Hydrogen?
b. Find the mass of 1 atom of hydrogen.
Answer:
a. 1g hydrogen = 6.022 × 10²³ atoms
b. Mass of 1 atom of hydrogen
= \(\frac{1 g}{6.022 \times 10^{23}}\) = 1.66 × 10-24 g

Question 2.
Number of molecules of substance is 3.011 × 10²⁴.
a. What is the number of molecules of 1 mole of any substance?
b. Find the number of moles of 3.011 × 10²⁴ molecules.
Answer:
a. 6.022 ×10²³
b. Number of moles of molecules = \(\frac { Number of molecules }{ N_A }\)
= \(\frac{3.011 \times 10^{24}}{6.022 \times 10^{23}}=5\)

Short Answer Type Questions (Score 2)

Question 3.
Fill the patterns.

Answer:
a. GMM
b. 22.4
c. Number of molecules
d. Number of molecules
e. Volume in litres
f. Mass

Question 4.
Identify the incorrect statements from those given with respect to the arrangements of molecules in gases.
a. The minute molecules are present without any freedom of movement
b. Collision take place between the molecules,
c. Increasing the number of molecules at constant volume causes the decrease in number of collisions.
d. The energy of molecules are comparatively high.
Answer:
a. The statements (a) and (c) are not correct

Question 5.
The pressure of 20L of a gas kept at 300 K is found to be 2 atoms. If the pressure is increased to 3 atom at the same temperature, what will be the new volume?
Answer:
According to Boyles law, P_V = a constant
Therefore, P₁ V₁ = P₂ V₂
Here, P₁ = 2atm V₁ = 20L P₂ = 3atm V₂=?
∴ 2 × 20 = 3 × V₂
Thus, V₂ = \(\frac { 2 × 20 }{ 3 }\) = 13.3 L

Charles Law Calculator is a free online tool that displays the volume of gas that tends to expand when heated.

Question 6.
If the temperature of 5L of a gas at atmospheric pressure is changed from 200K to 50 K, what will be the volume?
Answer:
According to Charles law, \(\frac { V }{ T }\) = a constant

Question 7.
What will be mass of 89.6 L of ammonia (NH3) gas at STP?
Answer:
89.6 L of NH3 gas at STP = \(\frac { 89.6 }{ 22.4 }\) = 4 mol
GMM of NH3 = 14 + 3 = 17g
Mass = mole × GMM = 4 × 17 = 68g

Question 8.
8. a. What is molar volume?
b. What is the molar volueofa gas at STP?
Answer:
a. The volume of one mole of a gas is called molar volume,
b. 22.4 L

Question 9.
Look at the balanced equation given.
2NaOH + CO₂ → Na₂ CO₃ + H₂0
a. Find out the mass of NaOH needed for 264 g CO₂ to react completely.
b. Find out the total number of moles of water molecules when CO₂ reacts.
Answer:
a. GMM of CO₂ = 44 g
∴Number of moles in the molecule of 264 g CO₂ = \(\frac { 264 }{ 44 }\) = 6
According to the equation NaOH needed for the reaction of 1 mole CO₂= 2 moles
∴ NaOH needed for the reaction of 6 moles CO₂ = 2 × 6 = 12 moles
GMM of NaOH = 23 + 16 + 1 = 40 g
Total mass of NaOH = 12 × 40 = 480 g

b. H₂O formed when lmole CO₂ reacts = 1 mole
∴ Total number of moles of water molecules when 6 mole CO₂ reacts = 6 moles

Question 10.
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
This is the equation of ignition of cooking gas butane.
Calculate the volume of CO₂ in STP during the complete ignition of 14 kg of cooking gas.
Answer:
Mass of Butane (C₄H₁₀) = 14 kg = 1400g
GMM of C₄H₁₀ = 4 × 12 + 10 × 1 = 58 g
∴ Number of moles in molecules = \(\frac { 1400 }{ 58 }\) = 241.38
Amount of CO₂when 2 moles of C₄ H₁₀ ignites = 8 moles of C₄H₁₀ ignites = \(\frac { 8 }{ 2 }\) × 965.52 moles
∴ Volume of CO₂ formed in STP
= 965.52 × 22.4 L = 21627.65 L

Question 11.
Write down the preparation of 100 ml NaOH solution of 0.1 M.
Answer:
GMM of NaOH = 40 g
Molarity = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}=\frac{\mathbf{n}}{\mathbf{v}}\)
M = 0.1 V=100ml = 0.1 L
0.1 = n/0.1
∴ n = 0.1 × 0.1 = 0.01
Mass needed to prepare 100 ml NaOH in 0.1M = 0.01 × 40 = 0.4g
Take 0.4 g NaOH in a beaker. Dissolve it Hilly by adding a little amount of water. Then, again add water to make it 100 ml.

Question 12.
The molarity of250 ml of Na₂CO₃ solution is 0.5 M. Find the mass of Na₂CO₃.
Answer:

Question 13.
63 g HNO₃ is in the dilute solution of 200 ml HNO₃ (Nitric acid). Find the molarity.
Answer:
a.

Short Answer Type Questions (Score 3)

Question 14.
Some equations related to gas laws are given below.
i. V α P
ii. \(\frac { V }{ T }\) = a constant
iii. V α n
iv. P_v = a constant
a. Which of these are correct?
b. Write the gas law to which it is related for the correct equations.
Answer:
a. Equations (ii), (iii) and (iv) are correct

b. (ii) Charles law
(iii) Avogadro’s law
(iv) Boyles law

Question 15.
In 100 g of CaCO₃
a. Find out the number of moles of each element and atom.
b. Find out the total number of atoms of each element.
Answer:

Question 16.
Fill the blanks in the given table.

Answer:

Long Answer Type Questions (Spore 4)

Question 17.
See CO₂ gas is taken in a cylinder provided with a piston. The cylinder is dipped in hot water.
a. What happens to the movement of CO₂ molecules?
b. What change do you expect in the position of the piston?
c. What is the relation between temperature and the volume of a gas?
d State this gas law.
Answer:
a. As the temperature increase, the energy of molecules increases. This increases the speed of the motion of molecules.
b. Piston is pulled in the upward direction. So piston moves upward.
c. As temperature increases, volume increases.
d. At constant pressure, the volume of a definite mass of gas is directly proportional to the temperature in Kelvin scale. (Charles law)

Question 18.
The molecular formula of ammonium sulfate is (NH₄)₂SO₄.
a. Find the gram molecular mass (GMM) of ammonium sulfate.
b. Calculate the number of molecules and atoms in 1.32g of ammonium sulfate.
Answer:
a. GMM of (NH₄)₂SO₄
= (14+4) × 2 + 32 + 4 × 16 = 36 + 32 + 64
= 132 g

Question 19.
Fill in the blanks of the table given below.

Answer:

Question 20.
See the diagram given below.

Answer:
a. 196g
b. 2 × 6.022 × 10²³
c. 2 GMM
d. 2 × 6.022 × 10²³

Question 21.
Write in pairs, equal number of atoms from those given below,
a. 2g Hydrogen
b. 16 g Oxygen
c. 14 g Nitrogen
d. 8 g Helium
Answer:

Question 22.
Certain compounds and its masses are given,
i) 68 g NH₃
ii) 28 g N₂
iii) 9 g H₂O
iv) 128 g O₂
a. Which of these compounds have equal number of molecules?
b. How many molecules are there?
c. How many atoms are there in 9 g of water?
Answer:

Question 23.
368 g NO ₂gas is given. Find the answers of each one given below,
a. GMM of NO₂
b. Number of moles of molecules of 368 g NO₂
c. Number of molecules
d. Number of atoms
e. Volume in STP
Answer:
a. GMM of NO₂
= 14 + 2 × 16 = 46 g

b. Number of mole = \(\frac { 368 }{ 46 }\) = 8

c. Number of molecules = Number of moles × NA
= 8 × 6.022 × 10²³

d. Number of atoms = Number of atoms in one molecule × Number of molecules
= 3 × 8 × 6.022 × 10²³ = 24 × 6.022 × 10²³

e. Volume in STP = Number of moles × 22.4
L = 8 × 22.4 L = 179.2 L

Question 24.
Find out the GMMofthe following. Also find out the total number of atoms.
a. 20 g Nitrogen (H₂)
b. 88.75 g Chlorine (Cl₂)
c. 4 g Calcium (Ca₂) ,
d. 7.75 g phosphorus (p₄)
(H = 1, Cl =35.5, Ca =40, P=31)
Answer:
a. 20 g of hydrogen (H2): GMM = \(\frac { 20 }{ 2 }\) = 10
Number of molecules = 10 × 6.022 × 10²³
Total number of atoms = 2 × 10 × 6.022 × 10²³
= 20 × 6.022 × 10²³

b. 88.75g of chlorine
Number of GMM = \(\frac { 88.75 }{ 71 }\) = 1.25
Number of molecules = 1.25 × 6.022 × 10²³
Total number of atoms = 2 × 1.25 × 6.022 × 10²³
= 2.5 × 6.022 × 10²³

c. 4 g Calcium (Ca)
Number of GMM = \(\frac { 4 }{ 40 }\) = \(\frac { 1 }{ 10 }\) = 0.1
Number of molecules = 0.1 × 6.022 × 10²³
Total number of atoms = 1 × 0.1 × 6.022 × 10²³
= 0.1 × 6.022 × 10²³

d. 7.75 g Phosphorus Number of GMM = \(\frac { 7.75 }{ 31 }\) = \(\frac { 1 }{ 4 }\) = 0.25
Number of molecules = 0.25 × 6.022 × 10²³

You can Download The Last Leaf (Story) Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Physics Solutions Part 2 Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State State Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power

A The Trio (Story) Textual Questions and Answers

Hss Live Guru 9th Physics Kerala Syllabus Chapter 5 Question 1.
Observe figure try to write down the activities shown in them.
Answer:

1. A man pushes a trolly.
2. Batting of a cricket ball.
3. Pushing a wall.

Kerala Syllabus 9th Standard Physics Notes Chapter 5 Question 2.
Write down more activities familiar to you
Answer:

1. A man carrying a load
2. Throwing a ball
3. Lifting the bag on to the shoulder
4. Pushing a car into motion

Hsslive Guru 9th Physics Kerala Syllabus Chapter 5 Question 3.
You have understood that a force is to be applied on a body to do an activity. Find out the source of applied force for every activity and note them down in the table.


Answer:

Activity	Source of applied force
Falling of a mango	The earth
A trolley being pushed	The person pushing
Batting of a cricket ball	Batsman
Pushing a wall	The person pushing
Lifting a load	The person lifting

Objects undergo displacement only when the force is applied on it them.

Displacement takes place in the direction of force applied	No displacement for the body in the direction of force applied
1. A cricket ball when hit by a bat	1. A wall is pushed
2. A trolley is being pulled.	2. A car is pushed by sitting inside the car
3. Climbing a ladder with a load on head.
4. Falling of a mango from mango tree.

Work :

Work is said to be done only when a body under¬goes displacement in the direction of the applied force.

work energy and power Question 4. Observe figure and write down the situation where work is said to be done
Answer:
The person loaded on head and goes upward.

Kerala Syllabus 9th Physics Notes Chapter 5 Question 5.
A boy pushed an object of mass 30Kg horizontally across the floor through 50m. Another boy pushed an object of mass 50kg across the same floor through 50m. Both of them gave the same speed for moving the objects.
a) Who applied greater force here?
b) In which case was the work greater?
c) Write down a factor influencing work.
Answer:
a) Second boy applied greater force
b) More work is done in the second situation
c) The Factor influencing work is force.

Learn Mass Transfer MCQ questions & answers are available for a Chemical Engineering students.

Hss Live Guru Physics 9th Kerala Syllabus Chapter 5 Question 6.
A boy pushed an object of mass 30kg across a horizontal floor through 20m. Another boy pushed the same body through 30m on the same floor with the same speed.
a) Who pushed a greater distance here?
b) What about the force applied
c) Who did the greater work?
d) Which is the factor influencing work here?
Answer:
a) Second boy
b) Force is same
c) Second boy done greater work
d) The Factor influencing work is displacement

The factor affecting work done are force (F) and displacement (S)
Equation for calculating work done Work W = Fs
Unit of work is Joule (J), Kilojoule (KJ)
1KJ = 1000 J

If a force of F newton is applied continuously on a body and the body undergoes a displacement of s metre in the direction of the force, then the work done by the applied force is W = Fs

Get the free “Nuclear Equation Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

Physics Notes For Class 9 Kerala Syllabus Chapter 5 Question 7.
When a force of 10N is applied continuously on a body it undergoes a displacement of 2m find out the magnitude of the work done?
Answer:
F = 10N, s = 2m
Work W= Fs = 10 × 2 = 20 Nm
Look at the Figure

Hsslive Guru Physics Class 9 Kerala Syllabus Chapter 5 Question 8.
A body of mass m kg is placed on a table. What are the forces experienced by this body?
Answer:
Weight of the body applies downwards and the table applies an equal force upwards.

Hss Live 9th Physics Kerala Syllabus Chapter 5 Question 9
In which directions do these forces act?
Answer:
Forces acting on both directions, upward direction and downward direction.

9th Physics Notes Kerala Syllabus Chapter 5 Question 10.
A book of mass 100g is raised to the top of a table of height 1 m. Find the magnitude of the work done by the force applied against the gravitational force(g = 10m/s²)
Answer:
m = 100g = 0.1kg
g = 10m/s²
h = 1m
W = mgh
= 0.1 × 10 × 1 = 1J
1 J is the amount of work done to raise a body of mass 100 g through a height of 1m.

Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Question 11.
If a force of 50N is applied on a body and it under¬goes a displacement of 2m in the direction of the force, calculate the amount of work done.
Answer:
F = 50N
s = 2m
W = Fs = 50 x 2 = 100J

Hss Live Guru 9 Physics Kerala Syllabus Chapter 5 Question 12.
a) If a force of 200N is applied on a table of mass 50kg, it undergoes a displacement of 0.5m in the direction of force. Calculate the amount of work done.
b) If the same table is raised by 3 m, what would be the work done against the gravitational force?
Answer:
a) F= 200N, s = 0.5m
W = Fs = 200 x 0.5 = 100J
b) m = 50 kg, g = 10m/s²
h = 3 m
W = mgh = 50 x 10 x 3 = 1500J
Observe Figure

Let a body mass m be pulled by a force F. If the body has a displacement s in the direction of the force, then the work done by the force F, Wf = Fs Here the displacement produced is in the direction of the force itself.

Kerala Syllabus Class 9 Physics Solutions Chapter 5 Question 13.
Write whether this work is negative or positive.
Answer:
Work is positive

9th Class Physics Notes Kerala Syllabus Chapter 5 Question 14.
The displacement is opposite to the frictional force, is the work done by the frictional force positive or negative
Answer:
If the displacement is in opposite direction work done by frictional force is negative.

Class 9 Physics Kerala Syllabus Chapter 5 Question 15.
In which direction is the force of gravity on the body?
Answer:
Direction of the gravitational force will be in the downward direction

9th Standard Physics Notes Kerala Syllabus Chapter 5 Question 16.
Is there a displacement for the body in the direction of the gravitational force?
Answer:
No displacement occurs in the direction of the gravitational force.
When a body on a floor is pulled and if it is displaced in the direction of the applied force, the work done by the applied force will be positive and the work done by the frictional force exerted by the floor will be negative.

Energy

Question 17.
What is the work to be done to raise a body of mass m kg through h meter?
Physics chapter 5 work and energy Answer:
w = mgh
Energy is the capacity to do work. Unit is Joule (J)

work energy book Question 18. In daily life, we use different forms of energy for various activities. List the forms of energy familiar to you
Answer:

1. Mechanical energy
2. Heat energy
3. Electrical energy
4. Chemical energy
5. Light energy
6. Nuclear energy

There are two type of Mechanical energy Kinetic energy and potential energy

Kinetic Energy

Pulling the toy car backwards a little and allow it to hit the plastic ball

Question 19.
What happened to the ball when the moving car hit it?
Answer:
Ball moves forward

10 to the 9th power Question 20.
How did the car get the energy to move the ball forward?
Answer:
The energy is obtained from the motion of toy car
Conclusion:

• Moving objects possess energy
• The energy possessed by a body by virtue of its motion is the Kinetic energy

Experiment:
Let’s do another activity. Allow a powder tin to slide down a polished, inclined plane as shown in the figure and let it hit a toy car. Try to measure the displacement of the toy car. Repeat the experiment by increasing the height of the inclined plane and filling the tin with sand.

Observation:

• The displacement of the toy car is greater When it is hit by the powder filled with sand
• Also the displacement is greater when it is dropped from a greater height

Conclusion: Kinetic energy depends on mass (m) and velocity (v)
Derivation of the equation for calculating K.E
Work W = Fs
As per II law of motion
F = ma
∴ W = mas
According to third equation of motion
v² = u² + 2as = 0 + 2as (u = 0) = 2as
∴ as = \(\frac{v^{2}}{2}\)
W = mas , if we put \(\frac{v^{2}}{2}\) instead of as

Work is equal to the magnitude of kinetic energy
i.e, Kinetic energy K = \(\frac { 1 }{ 2 }\) mv2,
M – mass, V – Velocity
When a body of mass m moves with a velocity v, its kinetic energy will be K = \(\frac { 1 }{ 2 }\) mv2

Question 21.
A man having a mass of 70kg is riding a scooter of mass 80kg. What is the total kinetic energy if the velocity of the scooter is 10m/s?
Answer:
m = 70kg + 80kg = 150kg
v = 10m/s

Question 22.
A car of mass 1500 kg is moving at a velocity of 20m/s. What is its kinetic energy?
Answer:
m = 1500 kg
v = 20m/s
K = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 1500 × 20²
= 300000J = 300kJ

Question 23.
A boy of mass 50kg is riding a bicycle with a speed 2m/s. The bicycle has a mass of 10kg. Calculate the total kinetic energy?
Answer:
m = 50kg + 10kg = 60kg
v = 2m/s
K = \(\frac { 1 }{ 2 }\) mv²
1/2 × 60 × 2² = 120J

Potential Energy

When we lift a body in perpendicular direction, the work is said to be done against gravitational force. According to figure, maximum work is done when it reaches at a height fb from the group, That means as the height increases work done also increases.

• The energy received by the body is equal to the work done on it.
• The body attains more energy when the height from the ground level increase

The energy possessed by a body due to its position is the potential energy.
ie, Potential energy
U = mgh
m – mass, g – acceleration due to gravity
h = height from the ground

Question 24.
Identify more situation in which potential energy is acquired by virtue of position
Answer:

• Coconut in a coconut tree
• Water stores in huge reservoirs
• Objects placed above the buildings,

Inference:
Height increases, potential energy increases

Question 25.
Write down situations in which potential energy varies.
Answer:

• Falling of a coconut from a coconut tree
• Pumping water to a tank at a height
• Climbing a ladder

Question 26.
Calculate the potential energy of a body of mass 1 kg at a height of 6m from the ground?
Answer:
Mass m = 1 kg, Acceleration due to gravity
g = 10m/s²
h = 6 m,
U = mgh = 1 × 10 × 6 = 60J

Question 27.
A bird of mass 0.5 kg is flying at the same speed at the same height of 5m. In this state, if its Kinetic energy and potential energy are equal.
a) What is the potential energy of the bird?
b) What is the velocity of the bird?
Answer:
a) m = 0.5kg
g = 10m/s²
h = 5m
u = mgh
= 0.5 × 10 × 5 = 25J
b) Kinetic energy K = 25J (∴U = K)

Question 28.
Write other examples for getting potential energy due to strain
Answer:

• A Stretched bow
• An elongated rubber band
• Compressed spring
• Compressed spring in a toy car

Conclusion: The factor which gives potential energy are position and strain

Law Of Conservation Of Energy

Question 29.
What form of energy does the flower pot have when it is on the sunshade of a building?
Answer:
Potential energy

Question 30.
While the flower pot is falling down, what forms of energy does it possess?
Answer:
potential energy and kinetic energy

Question 31.
Answer:
Does its potential energy increase/decrease when the pot falls down.
Answer:
Potential energy decreases

Question 31.
Will the kinetic energy increase/decrease at that time?
Answer:
Kinetic energy increases

Question 33.
What energy transformation takes place just before the flower pot reaches the ground?
Answer:
Potential energy is converted completely into kinetic energy

Question 34.
Let the mass of the flower pot 15kg and the height of the sunshade 4m.
a) When the flower pot is on the sunshade, what is its potential energy? (g = 10m/s2).
b) When it is on the sunshade, what is its kinetic energy?
c) If so, what is its total energy?
Answer:
U = mgh
= 15 × 10 × 4
= 600J
b) Kinetic energy will be zero
c) Total energy = PE + KE
= 600J + 0 = 600J

Question 35.
While falling, when the flower pot is at a height of 2m from the ground, what will be its kinetic energy?
Answer:
K = \(\frac { 1 }{ 2 }\) mv²
u = 0, g = 10m/s² ,
s = 4 – 2 = 2m
v² = u2² + 2as
=0 + 2 × 10 × 2 = 40
K = 1/2 × 15 × 40
= 300J

Question 36.
What is the potential energy when it is at a height of 2m from the ground? What is the total energy now?
Answer:
Potential energy at 2m height
U = mgh = 15 × 10 × 2 = 300J
∴ Total energy = 300 + 300 = 600J

Question 37.
What is the kinetic energy of the flower pot just before it touches the ground?
Answer:
K = \(\frac { 1 }{ 2 }\) mv²
v² = u² + 2as
= 0 + 2 x 10 x 4 = 80
Kinetic energy K = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 15 × 80 = 600J

Question 38.
The potential energy
U = mgh
= 15 x 10 x 10 x = 0.
What will be the total energy?
Answer:
Total energy = 600 + 0 = 600J

Question 39.
To sum up the amount of energy at each situation:
a) When on the sunshade
b) When at a height of 2m from the ground
c) Just before hitting the ground
Answer:
a) When on the sunshade = 600J
b) At a height of 2m from the ground = 600J
c) Just before hitting the ground = 600J
Energy can neither be created nor be destroyed. It can be transformed from one from to another

Sun is the major source of energy. We utilize solar energy in different ways. It is renewable source of energy Plants prepare food by using sunlight. From these fossil fuels, firewood are formed. Windmills works by utilizing kinetic energy of wind energy from infra¬red rays. Tidal energy is obtained due to gravitation. When we consider any energy sources, they were originated from solar energy.

Power

Question 40.
Given below is the information regarding the working of pumps in three neighboring houses. Complete the table (g = 10 m/s²).

Answer:
b) 150000J
c) 150000J

Question 41.
Is the amount of work done by the pump to fill water in the three tanks equal?
Answer:
Amount of work done is equal

Question 42.
Find the amount of work done per second by each pump.

Answer:

Pump	Work done	Time (s) (J)	Work done per second (J/S)
A	150000	100	1500
B	150000	200	750
C	150000	400	375

Amount of work done per second is referred as power of the pump.
Work done per unit time or rate of doing work is power
Power = \(\frac{\text { work }}{\text { time }}, P=\frac{w}{t}\)
Units of power
watt (W), kilowatt (KW), Horse power (H.P)
1 kW= 1000W
1 HP = 746W

Question 43.
If a man of mass 70kg climbs up a mountain of height 30m in 5 minutes, what is his power?
Answer:
m = 70kg, g = 10m/s²
h = 30m
Work W = mgh = 70 × 10 × 30 = 21000J
Time t = 5mt = 5 × 60 = 300s
Power p = \(\frac { w }{ t }\) = \(\frac { 21000 }{ 300 }\) = 70W

Question 44.
If a man of mass 50kg takes 60s to climb up 20 steps, each 15cm high, calculate his power.
Answer:
m = 50kg, g = 10m/s²
h = 15cm × 20 = 300cm = 8m
t = 60s
work W = mgh = 50 × 10 × 3 = 1500J
Power p = \(\frac { w }{ t }\) = \(\frac { 1500 }{ 60 }\) = 25W

Let Us Assess

Question 1.
A boy is trying to push the concrete pillar of the building using a force of 300N. Calculate the amount of work done by the boy.
Answer:
F= 300N, s = 0
w = Fs = 300 × 0 = 0

Question 2.
From what you have learnt of potential energy and kinetic energy write down the from of energy possessed by the bodies given below.
a) water in a dam
b) Stretched rubber band
c) Mango falling from a tree
Answer:
a) Potential energy
b) Potential energy
c) Potential energy lesser. Kinetic energy greater

Question 3.
Calculate the kinetic energy of an athlete of mass 60kg running with a velocity 10 m/s
m = 60kg
v= 10m/s

Question 4.
A stone of mass 2kg is thrown upwards from the ground with a velocity of 3m/s. When it reaches maximum height, calculate its potential energy.
Answer:
u = 3m/s v = 0
a = g = -10m/s² (Velocity of a body moving upwards is decreasing so acceleration is negative)
v² = u² + 2as
0 = 32 + 2 × -10 × s
0 = 9 – 20s
20s = 9, s = \(\frac { 9 }{ 20 }\)
U = mgh = 20 × 10 × \(\frac { 9 }{ 20 }\) = 90J

Question 5.
The heart of a healthy person beats 72 times per minute and each beat uses up about 1J of energy. Calculate the power of the heart.
Answer:
W= 1 x 72 = 72J
t = 1mt = 60s
P = \(\frac { W }{ t }\) = \(\frac { 72 }{ 60 }\) = 1.2w

Question 6.
Which among the following is a vector quantity?
a) work
b) momentum
c) power
d) energy
Answer:
momentum

Question 7.
If the velocity of an object is doubled, its kinetic energy becomes
a) 2 times
b ) 1/2
c) 4 times
e) 1/4
Answer:
4 times

Question 8.
An object of mass 1 kg is falling from a height 10 m. What be the work done while falling?
a) 10 J
b) 1J
c) 100 J
d)1000 J
Answer:
w = mgh = 1 × 10 × 10 = 100J

Question 9.
Which one among the following is correct?

Answer:
W = p × t

Question 10.
A roller weighing 1 tonne is being dragged along a road. What is the work done against gravity? Why?
Answer:
Zero, No displacement in the direction of force.

Question 11.
Is it possible for an object to possess energy without momentum? Give one example for such a situation.
Answer:
Yes, possible. A coconut on a coconut tree has potential energy, but it has no momentum.

Question 12.
Say whether the following are positive work or negative work.
1) The work done by a person drawing water from a well using a rope without a pulley.
2) Work done by gravitational force in this situation
3) Work done by the frictional force while and object is sliding down along an inclined plane.
4) Work done by the force is moving along a plane surface.
Answer:

1. Positive work
2. Negative work
3. Negative work
4. Positive work

Question 13.
How much joule is 1 kWh?
Answer:
1 kWh = 1000 × 60 × 60 = 3600000J

Question 14.
Find out the work done against the gravitational force in the situations given below.
1. A child is standing still with a bundle of books of mass 5 kg
2. With the same bundle of books, she travels 1m along a plane surface with a speed 5 m/s
3. The bundle of books is lifted onto the top of a cupboard having 1m height (g=10m/s2).
Answer:
1. 0
2. 0
3. W = mg = 5 × 10 × 1 = 50 J

Question 15.
A ball of mass 0.4 kg is thrown vertically upward with a velocity 14 m/s. Calculate its kinetic energy and potential energy after 1 s.
(Hint: v + at, s = ut + 1/2at²)
Answer:
v = u + at
= 14 + 10 × 1
= 24 m/s
KE = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 0.4 × 24
= 115.2 J
S = ut + 1/2 at²
= 14 × 1 + 1/2 × 10 × 1²
= 14 + 5 = 19 m
E = mgh = 0.4 × 10 × 19
= 76 J

Question 16.
An object of mass 1000 kg is travelling with a velocity 72 km/h. Calculate the work done to bring it rest.
Answer:
m = 1000 kg
v = 72 km/h = 20 m/s
Work = difference of kinetic energy
= \(\frac { 1 }{ 2 }\) mv²
= \(\frac { 1 }{ 2 }\) × 1000 × 20 × 20 = 400000J

Question 17.
Estimate the work done on a object of mass 80 kg to change its velocity from 5 m/s to 10 m/s
Answer:
m = 80 kg
Work = difference of kinetic energy

Work, Energy and Power More Questions

Question 1.
Fill in the blanks
a) Work done when a body of mass 100g is lifted up to a height of 1 meter is ……….
b) The two factors related to potential energy to body are………….
c) 1HP = ……… Watt
Answer:
1 Joule
b) Position, Strain
c) 746W

Question 2.
Classify the following into work done and work is not done
1. A mango falling from a mango tree
2. Pushing a table while sitting on it
3. Pushing a wall
4. Kicking a football
5. A trolley is moving forward
6. Standing with a load above the head
Answer:
Work done:
1. A mango falling from a mango tree
2. Kicking a football
3. A trolling is moving forward
Work is not done:
1. Pushing a table while sitting on it
2. Pushing a wall
3. Standing with a load on the head

Question 3.
When Lekshmi is applied a force of 50N on an object it undergoes a displacement of 2m. When Vinitha applied the same force on itundergoesadisplacementof3m if so
a) Which person done more work?
b) Give reason
c) Calculate the work done by each person?
Answer:
a) Vinitha
b) Displacement is greater
c) Work done by Lekshmi
W = Fs = 50N × 2m = 100J
Work done by Vinitha
W = Fs = 50N × 3m = 150J

Question 4.
Calculate work done when a boy of mass 40kg climb¬ing a staircase of height 50cm, g = 10m/s2
Answer:
W= Fs
F = mg = 40 × 10 = 400N
S = h = 0.5m
W= 400 × 0.5 = 200J

Question 5.
a) What is mean by kinetic energy?
b) How mass and velocity affect kinetic energy?
c) A body of mass 20kg is moving with a velocity of 5m/s. Calculate the kinetic energy?
Answer:
a) The energy possessed by a body by virtue of its motion is called kinetic energy
b) Mass increases, kinetic energy increases mass . is doubled kinetic energy also doubled. Velocity increases, kinetic energy increases, Velocity is doubled kinetic energy becomes four times.
c) Kinetic energy KE = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 20 × 5² = 1/2 × 20 × 25 = 250J

Question 6.
a) Write the equation for finding potential energy?
b) indicate the representation of each letter
Answer:
a) U = mgh
b) M – mass,
g – acceleration
h – height from the ground

Question 7.
State law of conservation of energy?
Answer:
Energy can neither be created nor be destroyed. It can be transformed from one form to another.

Question 8.
A store of mass 5kg was raised from the ground to the second floor of height 7m and from there to the third floor of height 3m from the second floor. Calcu¬late the potential energy of the store with respect to the ground floor and the second floor?
Answer:
Mass m = 5kg
Height from ground to third floor
h = 7m + 3m
g = 10m/s²
Potential energy with respect to the ground
U = mgh = 5kg × 10m/s² × 10 = 500J
Height from second floor to third floor h = 3m
Potential energy with respect to the second Floor
U = mgh = 5kg × 10m/s × 3 = 150J

Question 9.
a) What is power?
b) The time takes to move an object at a distance of 5m is 10s is the force applied is 30N, what is the power?
Answer:
a) Rate of doing work is power
b) Power P = E
W = Fs = 30 × 5 = 150J
t = 10sec.
Power P = \(\frac { 150 }{ 10s }\) = 15W

You can Download Moving Images Questions and Answers, Kerala SSLC 10th IT Theory Questions and Answers Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus SSLC IT Theory Questions and Answers Chapter 9 Moving Images

Moving Images Questions and Answers

Section -1

Question 1.
The animation software in IT @ School GNU/Linux is
a) Synfig studio
b) Tupi
c) Gimp
d) Paint
Answer:
a) Synfig studio

Question 2.
Which type of animation software is synfig studio.
a) Proprietary
b) not able to take copy
c) pay and use
d) free software
Answer:
d) free software

Question 3.
To prepare a film, How many times images appear one after the other continuously in front of our eyes in one second?
a) 48
b) 12
c) 24
d) 36
Answer:
c) 24

Question 4.
Say the important stage of an animation film?
a) character designing
b) background
c) characters
d) action
Answer:
a) character designing

Question 5.
Construction of a storyboard is a preparation of
a) Drawing a picture
b) finding time zones
c) producing an animation
d) character designing
Answer:
c) Producing an animation

Question 6.
Synfig Studio is a free-animation software
a) Drawing pictures
b) Studio
c) Three dimensional
d) Two dimensional
Answer:
d) Two dimensional

Question 7.
Synfig studio software is designed by
a) Stall men
b) Robert B Quattlebaum
c) Leslie Lamport
d) Donald Knuth
Answer:
b) Robert B Quattlebaum

Question 8.
What versions of animation software can run in GNU/Linux and Microsoft Windows?
a) Grass
b) Notepad
c) Synfig studio
d) Pencil
Answer:
c) Synfig studio

Question 9.
Which tool is used to fill colors to the objects in synfig studio?

Answer:

Question 10.
Which tool is used to draw rectangular objects in synfig studio?

Answer:

Question 11.
Which tool is used to move the objects in synfig studio?

Answer:

Question 12.
Expand FPS
a) Frames per system
b) Frames per second
c) File properties settings
d) Frames per hour
Answer:
b) Frames per second

Question 13.
Animation is done by fast and continuous movements of images in a two-dimensional canvas. The images are known as
a) characters
b) FPS
c) Frames
d) Scene
Answer:
c) Frames

Question 14
Which menu is used to studio
a) File
b) view
c) window canvas
d)canvas
Answer:
d)canvas

Question 15.
To change the FPS in Synfig studio, by clicking
a) canvas → properties → time
b) view → pause
c) window → toolbox
d) canvas properties → image
Answer:
a) canvas → properties → time

Question 16.
We can import Images into synfig studio and use them
a) odf
b) vector
c) Pdf
d) Bitmap
Answer:
d) Bitmap

Question 17.
FPS = 24, Time =5, for an animation. Find the total number of frames in that animation
a) 24,
b) 48
c) 120
d) 920
Answer:
c) 120

Question 18.
The frames that represent important positions are known as
a) Tweening
b) current time
c) key
d)keyframe
Answer:
d) keyframe

Question 19.
What happens when you press

play button
a) Animation works
b) Animation stops
c) no change
d) to save
Answer:
a) Animation works

Question 20.
The software fills up the frames in between two keyframes is called.
a) Interpolation
b) keyframe
c) Tweening
d) Edit
Answer:
c) Tweening

Question 21.
Rey frame utility is set in one of the panels in synfig audio. Give the names of that panel
a) Time track panel
b) layers panel
c) parameters panel
d) panel
Answer:
c) parameters panel

Question 22.
Name the extension of project file in synfig studio?
a). svg
b) .pdf
c) .sifz
d) .ods
Answer:
c) .sifz

Question 23.
In which panel to give the number of frames in current time?
a) Time track panel
b) layers panel
c) meters panel
d) panel
Answer:
a) Time track panel

Question 24.
What is the current time to animation from first frame
a) 0 f
b) 60 f
c) 120f
d) 121 f
Answer:
a) 0 f

Question 25.
What is the name for the frame with current time is ‘o’ f
a) first keyframe
b) last keyframe
c) middle keyframe
d) keyframe
Answer:
a) first keyframe

Question 26.
Give the order of activity to export a project file on synfig studio
a) File → Render
b) File → Save
c) File → Export
d) File → Save as
Answer:
a) File → Render

Question 25.
Which is not a member related to synfig studio?
a) parameters panel
b) time track panel
c) panel
d) layers panel
Answer:
c) panel

Question 28.
The utility to animate the flap its wings of a bird is
a) joining wings
b) adding time loop layer
c) copy of the first wing
d) adding the layer
Answer:
b) adding time loop layer

Question 29.
Write the activity to include an image to synfig studio
a) file → open
b) file → new
c) file → import
d) file → save
Answer:
c) file → import

Question 29.
in which menu import facility is available”?
a) Edit
b) file
c) canvas
d) windows
Answer:
b) file

Question 30.
Select ofie not group
a) . dv
b) .flv
c) . mpeg
d). svg
Answer:
d) .svg

Question 32.
What is the maximum value of current time?
a) 20 f
b) 60 f
c) 120 f
d) 1 f ,
Answer:
c) 120 f

Question 33.
What is the least value of current time?
a) 60 f
b) 120 f
c) 40 f
d) Of
Answer:
d)0f ‘

Question 34.
Keyframe are constructed in different type or same type
a) same type
b) different type
c) not constructing a keyframe
d) Equal type
Answer:
b) different type

Question 35.
Select any two free animation software’s
a) Anim studio
b) pencil
c) synfig studio
d) adobe flash
Answer:
b) pencil
and
c) synfig studio

Question 36.
An important stage in preparing animation is character designing. What is the meaning of character designing?
a) characters
b) Draw the characters
c) bringing character to life with humanity & personality
d) Giving life to the story
Answer:
b) Draw the characters
and
c) bringing character to life with humanity & personality

Question 37.
What are the uses of multi-colored buttons on the handle of a image?
a) To switch on the Animate editing mode
b) To adjust size
c) To take copy
d) To rotate if needed.
Answer:
b) To adjust size
and
d) To rotate if needed

Question 38.
Give the two activities done before starting animation from first keyframe
a) Current time is 60 f
b) Current time is 0 f
c) Before editing the motion, animate the edit mode button is active
d) press the play button to see the animation
Answer:
b) Current time is 0 f
and
c) Before editing the motion, animate the edit mode button is active

Question 39.
Sky and one star are drawn in synfig studio canvas. In the layers panel we can see a rectangle 001, and star 001. What are the tools use dot draw them
a) Circle tool
b) Fill tool
c) Star tool
d) Rectangle tool
Answer:
c) Star tool
and
d) Rectangle tool

Question 40.
Animation became much easier in film industry, with the arrival of new technologies. What are they?
a) computers
b) Radio
c) Animation software’s
d) office software.
Answer:
a) computers
and
c) Animation software’s

Question 41.
Create animations is one of the stages in animation film. There are two other stages before creating an animation. What are they?
a) To save animation project
b) Character designing
c) Playing an animation
d) Preparation of storyboards
Answer:
b) Character designing
and
d) Preparation of storyboards

Question 42.
Utilities of synfig studio windows are given below. Select one of each set and make a list.

Set-I

a) Database panel
b) Work area
c) Layers Panel
d) Task area
Answer:
c) Layers Panel

Set -II

a) Audio track
b) Video track
c) Time track panel
d) Fill
Answer:
c) Time track panel

Set-III

a) Parameters panel
b) Table
c) Storyboard
d) Path
Answer:
a) Parameters panel

Set -IV

a) Database window
b) tooIX
c) Editor window
d) Window
Answer:
b) tooIX

pH Calculator is a free online tool that displays the pH value for the given chemical solution.

Question 43
Select one video formats from each set

Set I

a) PNG
b) gif
c) xcf
d) sifz
Answer:
b) gif

Set II

a) dv
b) ods
c) odf
d) odp
Answer:
a) dv

Set III

a) jpg
b) flv
c) jpg
d) py
Answer:
b) flv

Set -IV

a) Html
b) ph
c) htm
d) mpeg
Answer:
d) mpeg

Question 44.
There are a lots of job opportunities related to animation in Government/public sectors of India and abroad. A list of job opportunities are given below. Select one from each set.

Set-I

a) Cinema production
b) Still photograph
c) Newspaper
d) Shops
Answer:
a) Cinema production

Set-II

a) Schools
b) advertising agencies
c) building construction
b) advertising agencies
Answer:
d) Library

Set-III

a) Radio
b) FM radio
c) TV
d) Tape recorder
c) TV

Set-IV

a) mobile
b) computer
c) calculator
d) computer games
Answer:
d) computer games

Question 45.
Images are included in synfig studio through import menu what are the uses of handles on the image? Select one form each set.

Set-1

a) To join the parts of an image
b) To arrange its position
c) To animation
d) Frame
Answer:
b) To arrange its position

Set – II

a) To arrange its size
b) To reduce its size
c) To increase its size
d) Can’t change its size
Answer:
a) To arrange its size

Set-II

a) To rotate
b) Can’t rotate
c) To remove the image
d) To include and image
Answer:
a) To rotate

Set-IV

a) To add a background image
b) Creating a scene
c) To adjust the views of the images
d) To change its background-color
Answer:
c) To adjust the views of the images

Question 46.
The layers are seen in the layers panel of a synfig studio some activities related to the layers are given in sets. Select correct activities from each set.

Set-1

a) Can’t change its order
b) can change its order
c) Arrange the order before including the layer
d) Layers are not visible
Answer;
b) can change its order

Set -II

a) Can’t group together
b) Objects
c) Can group together
d) Can’t change its color
Answer:
c) Can group together

Set-III

a) To take a copy of a layer
b) Layer is fixed
c) Can edit animation
d) Can’t delete the layer
Answer:
a) To take a copy of a layer

Set-IV

a) Removing the layer may effect he other
b) Can delete a layer
c) Can edit animation
d) Can’t delete the layers
Answer:
b) Can delete a layer

You can Download Publishing Questions and Answers, Kerala SSLC 10th IT Theory Questions and Answers Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus SSLC IT Theory Questions and Answers Chapter 2 Publishing

Publishing Questions & Answers

Question 1.
Clickthrough Insert → fields → more fields in writer software. We get a field window contains utility.
a) File
b) Edit
c) Database
d) None
Answer:
c) Database

Question 2.
Formattings of a text is copped to another text is using tool
a) Clone formatting
b) style
c) Paragraph formatting
d) styles and formatting
Answer:
a) Clone formatting

Question 3.
Which is the Clone formatting tool?

Answer:

Question 4.
A word can define headings, subheadings and para-graph separately is called
a) Style
b) Insert
c) Formatting
d) Clone
Answer:
a) Style

Question 5.
In which menu lies the Index and Tables
a) File
b) Insert
c) Edit
d) Data
Answer:
b) Insert

Question 6.
Where does contain the table of contents in your text?
a) Middle of the text
b) Not entered
c) At the beginning
d) At the end
Answer:
c) At the beginning

Question 7
The Software developed by Leslie Lamport is
a) Linux
b) Windows
c) Android
d) LaTex
Answer:
d) LaTex

Question 8.
The program for type setting technical documents is developed by Donald Knuth is
a) LaTex
b) Tex.
c) Linux
d) Windows
Answer:
b) Tex

Question 9.
If you click at the word preface on the table of contents in writer. What message appears?
a) Ctrl-click to follow link
b) Click to follow link
c) Follow link
d) Click to follow link to close the window
Answer:
a) Ctrl-click to follow link

Question 10.
Which menu lies the field’s utility?
a) Edit
b) Insert
c) File
d) formatting
Answer:
b) Insert

Question 11.
Which software having the mail merge facility?
a) Libre office writer
b) Python
c) Firefox
d) Calculator
Answer:
a) Libre office writer

The Full form of DSLR is Digital Single Lens Reflex Cameras.

Question 12.
To include the names and address one by one in a table in the letters using a feature called
a) Style
b) Mail merge
c) Database
d) Indexable
Answer:
b) Mail merge

Question 13.
Which style is used to the headings more attractive?
a) Character style
b) Page style
c) Paragraph style
d) List style
Answer:
c) Paragraph style

Question 14.
Anu was prepared a style named main head. Then he selects the all headings in a report and clicks on the style named main head. What happens for the headings?
a) All headings are changed to same type
b) No change to the headings
c) Headings are changed to different sizes
d) Headings are seen in different colors
Answer:
a) AI1 headings are changed to same type

Question 15.
The use of organizer tab in styles and formattings window for creating a new style is?
a) No use at all
b) To applying the style
c) Giving the name for the style
d) Don’t give a name for the style
Answer:
c) Giving the name for the style

Questions 16.
Which software includes the styles and formattings utility
a) Cale
b) Python
c) Writer
d) Sun clock
Answer:
c) Writer

Question 17.
Pick out the correct statement related to a frame
a) Can’t able to type inside the frame
b) Frame can move at any place on the page
c) Frame can’t move on the page
d) Can’t able to insert an image in a frame
Answer:
b) Frame can move at any place on the page

Question 18.
Where did you get the fields window?
a) Insert →Fields → More fields
b) Insert → Media → Fields
c) Format → More fields
d) Tools → More fields
Answer:
a) Insert → Fields → More fields

Question 19.
A correct statement related to clone formatting is
a) Use clone formatting for a large report
b) Formattings of headings can be changed by changing the formatting of each heading separately
c) Formattings of headings can be changed by changing the formatting of one heading
d) Formattings of heading cannot be copped to another
b) Formattings of headings can be changed by changing the formatting of each heading separately

Section – II

Question 20.
Right-click on the heading (styles name) on a paragraph style, we get the utilities
a) Modify
b) Colour
c) New
d) Pattern
Answer:
a) Modify

Question 21.
Libre office writer file prepared in Malayalam fonts and English fonts, on which fonts are changed to change the styles of the files?
a) If the file is in Malayalam fonts, then change the CTL font
b) If the file is in English fonts, then change the CTL font
c) If the file is in Malayalam fonts, then change western text font
d) If the file is in English fonts, then change the Western text font
Answer:
a and b

Question 22.
Select the correct statements
a) The styles in a writer are changed and use them
b) The styles in a writer cannot be changed
c) New styles cannot be created
d) New styles can prepared in a writer.
Answer:
a and d

Question 23.
Select two correct statements related to a frame
a) Can’t move from one part of a page
b) To place text or images within a document separated from the main contents
c) To place anywhere in the page
d) Not separated from the main contents
Answer:
b and c

Question 24.
Participant’s cards are prepared by mail merge utility and select print from file menu. Then you get a window to select two utilities to save the output file what are they?
a) Save as single document
b) Save as image
c) Save as individual documents
d) Save picture as
Answer:
a and c

Question 25.
Using mail merge to preparing the letters for the parents about the Youth Festival. What are the preparations we make for this?
a) Making a list of parents address in Libre Office writer
b) Making a list of parents addresses in Libre Office Calc
c) Preparing the letters to the parents in Libre office writer
d) Preparing the letters to the parents in Libre office calc
Answer:
b and c

Question 26.
Why did scientific articles and research reports are prepared by LaTex Software?
a) Facility to use English fonts
b) Facility to use Malayalam fonts
c) Several features for typesetting Symbols
d) Several features for typesetting equations.
Answer:
c and d

Question 27.
What are the difficulties to prepare a text /document using clone formatting?
a) Separate formattings the contents of a text is difficult
b) Formattings of heading can be changed by changing the formattings of each heading separately
c) Formattings of headings/ paragraph can be changed by changing one
d) One by one formatting is very easy
Answer:
a and b

Question 28.
The order of preparing a table of contents to the school report are following. Select one from each set.

Set -1

a) Close the report
b) Type the school report
c) Correct the report
d) Open the prepared school report
Answer:
d) Open the prepared school report

Set – II

a) Click on the last page of the report
b) Click on the place where the table of contents to be inserted
c) Insert the table of contents on the cover page
d) Click on the blank space of the report
Answer:
b) Click on the place where the table of contents to be inserted

Set – III

a) Select Index and Tables from Insert Menu
b) Select fields from Insert menu
c) Select modify
d) Select styles and formatting from insert menu
Answer:
a) Select Index and Tables from Insert Menu

Set – IV

a) Close the window
b) A new window appears, give a heading and background color for the table of contents and click OK.
c) Click cancel on the new window
d) Leave the window without any change
Answer:
b) A new window appears, give a heading and background color for the table of contents and click OK.

Question 29.
What are the styles included in styles and formatting window?

Set -1

a) Paragraph style
b) Western-style
c) Formatting style
d) More style
Answer;
a) Paragraph style

Set – II

a) Apply style
b) Character styles
c) Style Box
d) Clone formatting
Answer:
b) Character styles

Set – III

a) Heading style
b) Report style
c) Frame style
d) Life style
Answer:
c) Frame style

Set-IV

a) Page style
b) Heading 1
c) Head
d) Address
Answer:
a) Page style

Question 30.
The uses of mail merge utility in LibreOffice writer.

Set-I

a) Preparing Notice
b) Preparing the letter for parents about youth festival
c) Writing the mark list
d) To prepare address
Answer:
b) Preparing the letter for parents about youth festival

Set – II

a) Creating a style
b) Taking a print
c) To taking a copy
d) To prepare participant’s card
Answer:
d)To prepare participant’s card

Set – III

a) To prepare certificate
b) To prepare a letter to a friend
c) To prepare a document
d) To prepare a study report
Answer:
a) To prepare certificate

Set – IV

a) To calculate the achievement
b) To prepare electricity bill
c) To apply style
d) To prepare the table of contents
Answer:
b) To prepare electricity bill

Question 31.
Which are the styles we are using from paragraph style? Make a list using from each set

Set – I

a) Table
b) Report
c) Heading
d) Contents
Answer:
c) Heading

Set – II

a) Caption
b) Font
c) Footer
d) Color
Answer:
a) Caption

Set-III

a) Find
b) Link
c) Object
d) Index
Answer:
d) Index

Set-IV

a) Body
b) Media
c) Text body
d) Frame
Answer:
c) Text body