CBSE Class 10

On this page, you will find Chemical Reactions and Equations Class 10 Notes Science Chapter 1 Pdf free download. CBSE NCERT Class 10 Science Notes Chapter 1 Chemical Reactions and Equations will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Science Chapter 1 Notes Chemical Reactions and Equations

Chemical Reactions and Equations Class 10 Notes Understanding the Lesson

1. Chemical reaction: The reaction in which the original state of the particles changes and it cannot be reversed by simple physical means, is known as a chemical reaction.
Examples: fermentation of grapes, burning of wood, etc. Burning of wood produces charcoal and we cannot get back wood from charcoal on reversing the conditions.

• Chemical reaction is accompanied by change in state, colour, evolution of gas or change in temperature. The chemical reaction is represented as
  Reactants → Products
• Example of a chemical reaction is burning of magnesium ribbon with a dazzling white flame to form a white powder (magnesium oxide).
  2Mg + O₂ → 2MgO

2. Chemical equation: Representation of a chemical reaction in terms of chemical symbols and formulae of the reactants and products is known as chemical equation. A chemical equation represents the reactants, products and their physical states symbolically.
For example,
Magnesium + Oxygen → Magnesium oxide
2Mg(s) + O₂(g) → 2MgO(s)
The substances that undergo chemical change in the reaction, i.e., magnesium and oxygen, are the reactants. The new substance, magnesium oxide, formed during the reaction is the product.

Writing a chemical reaction in terms of chemical equation: A chemical equation represents a chemical reaction. While writing a chemical equation the reactants are written on the left hand side of the equation while products on the right hand side.
Examples:

3. Balanced chemical equation: The chemical equation in which the number of atoms of different elements is same on both sides of the arrow is called a balanced chemical equation.
This is in accordance to the law of conservation of mass.
Let us try to balance the following chemical equation:
Fe + H₂O → Fe₃O₄ + H₂
Number of atoms of different elements present in the unbalanced equation

Now select the element which has the maximum number of atoms Fe304. There are four oxygen atoms on the RHS and only one on the LHS.

To balance the oxygen atoms

So, multiplying H₂O molecules by four, we get
Fe + 4H₂O → Fe₃O₄ + H₂

To balance H atoms, make the number of molecules of hydrogen as four on the RHS.
Fe + 4H₂O → Fe₃O₄ + 4H₂

To equalise Fe, we multiply Fe atoms by three on the LHS.
Hence,
3Fe + 4H₂O → Fe₃O₄ + 4H₂ (Balanced equation)

To make chemical equation more informative gaseous, liquid, aqueous and solid states of reactants and products are represented by the notations (g), (l), (aq) and (s) respectively.
Hence,  3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

4.Types of chemical reactions
(a) Combination reaction: The reactions in which two or more substances combine to form a new substance is called combination reaction.
For example,

• 2Mg(s) + O₂ (fe)→ 2MgO(s)
• CaO(s) + H₂O(l) → Ca(OH)₂(aq)

(b) Decomposition reaction: The reaction in which a single compound breaks up into two or more simpler substances is called decomposition reaction. For example,

2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g)+ O₂(g)

The decomposition of a substance by passing electric current through it is known as electrolysis.

The decomposition of a substance on heating is known as thermal decomposition.

The decomposition of a substance by absorbing light energy is called photochemical decomposition.
For example,

The above two reactions are used in black and white photography.

Decomposition reactions are opposite of combination reactions.

(c) Displacement reaction: The chemical reaction in which a more reactive element displaces a less reactive element from its salt solution is known as displacement reaction. For example,

• Zn(s) + CuSO₄(aq)→  ZnSO₄(aq) + Cu(s)
• Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂ (aq) + 2Ag(s)

(d) Double displacement reaction: In this reaction two different atoms or group of atoms are mutually exchanged.

A white, insoluble substance, i.e., BaS04 is formed which is called precipitate.
Precipitation Reaction-Any reaction that produces a precipitate is called a precipitation reaction.

(e) Oxidation: Oxidation is the gain of oxygen or loss of hydrogen.

(f) Reduction: Reduction is the loss of oxygen or gain of hydrogen.

Redox reaction: The reaction in which one reactant gets oxidised while other gets reduced.

Exothermic reactions: Reaction in which heat is released along with the formation of products.
For example, \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(g)\)
Respiration and decomposition of vegetable matter into compost are exothermic reactions.

Endothermic reactions: The reactions which require energy in form of heat, light or electricity are called endothermic reactions.
\(2 \mathrm{Ba}(\mathrm{OH})_{2}+2 \mathrm{NH}_{4} \mathrm{Cl} \longrightarrow \text { Heat } \mathrm{2BaCl}_{2}+2 \mathrm{NH}_{4} \mathrm{OH}\)

5. Corrosion: The process of slow deterioration of some metals like iron, copper and silver into their compounds due to their reaction with oxygen, water, acids, gases, etc. present in the atmosphere is called corrosion.
Rusting The process in which iron reacts with oxygen and moisture present in the air to form a reddish brown coating called rust on its surface.

6. Rancidity: The taste and odour of food materials containing fat and oil changes when they are left exposed to air for a long time. This is called rancidity. It is caused due to oxidation of fat and oil present in food material. It can be prevented by using various methods such as by adding antioxidants to the food materials, storing food in air tight containers and by flushing out air with nitrogen.

Class 10 Science Chapter 1 Notes Important Terms

Chemical reaction is a process in which old bond breaks up and new bonds are formed.

Chemical equation is the representation of a chemical reaction in terms of chemical symbols and formulae.

Combination reaction is a reaction in which two or more substances combine to form a new substance.

Decomposition reaction is a reaction in which a single compound breaks up into two or more simpler substances.

Displacement reaction is a reaction in which a more reactive element displaces a less reactive element from its salt solution.

Redox reaction is the reaction in which oxidation and reduction takes place simultaneously.

On this page, you will find Probability Class 10 Notes Maths Chapter 15 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 15 Probability will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 15 Notes Probability

Probability Class 10 Notes Understanding the Lesson

Theoretical probability: The theoretical (or classical) probability of an event E [denoted by P(E)] is given by
\(\mathrm{P}(\mathrm{E})=\frac{\text { Number of outcomes favourable to } \mathrm{E}}{\text { Number of all possible outcomes of the experiment }} \text { i.e., } \frac{n(\mathrm{A})}{n(\mathrm{S})}\)

Number of all possible outcomes of the experiment when the outcomes of the experiment are equally likely.

Equally likely outcomes: All the outcomes of an experiment are said to be equally likely when the chances of there occurrence are equal.
e.g. When a coin is tossed, the two possible outcomes are head and tail, which are equally likely.

Elementary event: An outcome of a random experiment is called an elementary event. e.g. In tossing a coin, possible outcomes are head and tail.
⇒ H and T are elementary events.

• The sum of the probabilities of all the elementary events of an experiment is 1.
• For an events E, P(E) + P(\(\overrightarrow{\mathrm{E}}\)) = 1, where \(\overrightarrow{\mathrm{E}}\) is the event ‘Not E’. E and \(\overrightarrow{\mathrm{E}}\) are called complementary events.
• If P(E) = 1, then E is called ‘sure or certain event’.
• If P(E) = 0, then E is impossible event.
• For any event E,
  0 < P(E) <1

On this page, you will find Statistics Class 10 Notes Maths Chapter 14 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 14 Statistics will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 14 Notes Statistics

Statistics Class 10 Notes Understanding the Lesson

The measures of central tendency are:

• Arithmetic mean or mean
• Median
• Mode

Mean of Raw Data

Mean of n observations x_1; x₂, x₃, ..x_n is given by
\(\bar{x}=\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}=\frac{\Sigma x_{i}}{n}\)
where ∑ (sigma) means “summation of’.

1. Mean of Grouped Data

(i) Direct method:
\(\bar{x} \doteq \frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)
For each class interval, Class mark = \(\frac{\text { Lower limit }+\text { Upper limit }}{2}\)

(ii) Short-cut method or assumed mean method:

Where a = assumed mean

(iii) Step-deviation method:

h=Class-Size

2. Mode of Groped Data
Class with the maximum frequency is called the modal class.
\(\text { Mode }=l+\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right] \times h\)

Where l = lower limit of the modal class
h = size of the class-interval
f₁ = frequency of the modal class
f₀ – frequency of the class preceeding the modal class
f₂ = frequency of the class succeeding the modal class

3. Median of Ungrouped Data
To find the median of ungrouped data, first arrange the data values of the observations in the ascending or descending order. Then,

4. Median of Grouped Data
\(\text { Median }=l+\left[\frac{\frac{n}{2}-c \cdot f \cdot}{f}\right] \times h\)
l – lower limit of median class.
n = number of observation
c.f. = cumulative frequency of the class preceeding the median class
f = frequency of the median class
h = class size

Median class: Class whose cumulative frequency is greater than (and nearest to) \(\frac{n}{2}\)

On this page, you will find Surface Areas and Volumes Class 10 Notes Maths Chapter 13 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 13 Surface Areas and Volumes will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 13 Notes Surface Areas and Volumes

Surface Areas and Volumes Class 10 Notes Understanding the Lesson

Surface area: Surface area of an object is the measure of the total area that the surface of an object occupies.

Volume: Volume of an object is the measure of space occupied by the object.

Basic Solids: In standard X, we have studied the surface area and volume of solids. Here we will study more about them.

1. Cuboid
(i) Surface area of cuboid = 2(lb + bh + lh) sq.unit
where l is the length
b is the breadth
h is the height

(ii) Area of four walls of cuboid
= 2(l + b) x h
= [Perimeter of floor x Height] sq. unit

(iii) Surface area of cuboid without roof or lid
= lb + 2 [bh + Ih] sq. unit

(iv) Volume of cuboid = l x b x h unit

(v) Diagonal of cuboid or length of longest rod kept =\(\sqrt{l^{2}+b^{2}+h^{2}}\)unit

2. Cube
Let each edge of a cube be of length a unit. Then
(i) Surface area of cube = 6 side² = 6a² unit

(ii) Surface area of four walls of cube = 4 side²
= 4a² sq. unit

(iii) Surface area of a cube without lid (or rod) of a cube
= 5a² sq. unit.

(iv) Length of longest diagonal (or rod) of a cube
\(=\sqrt{a^{2}+a^{2}+a^{2}}=\sqrt{3}\) aunit

(v) Volume of cube = a³ unit

3. Cylinder

(i) Curved surface area of cylinder = 2πr x h
= Perimeter of base x height sq. unit

(ii) Total surface area of cylinder
= CSA + Area of 2 circular ends of cylinder
= 2πrh + 2πr² = 2πr (r + h)

(iii) Volume of cylinder =πr²h

(iv) Volume of material in hollow pipe = Exterior volume – Interior volume
= πR²h – πr²h = πh [R² – r²]

(v) Total surface area of hollow cylinder
= CSA of outer and inner cylinder + 2(area of base ring)
= 2πRh + 2πrh + 2(πR² – πr²)
= 2π(R + r)h + 2π(R² – r²) = 2π(R + r) (h + R – r)

Note:

• Two ends of cylinder are circles having each area = πr²
• Mass of cylinder = Volume of cylinder x Density
  M = V x ρ

4. Cone

h – OA = height of cone
r = OB = radius of cone
l = AB = slant height of cone

(i) \(l=\sqrt{r^{2}+h^{2}}\) units

(ii) Curved surface area of cone or lateral
surface area of cone = πrl sq. unit

(iii) Total surface area of cone = CSA + Area of circular base
= πrl + πr² – πr(r + l) sq. unit

(iv) Volume of cone =\( \frac{1}{3}\) πr²h cu.unit

5. Sphere

• Surface area of sphere = 4πr² unit
• Volume of sphere = \( \frac{4}{3}\)cu.unit

6. Hemisphere

• Curved surface area of hemisphere = 4πr² sq unit
• Volume of hemisphere = \( \frac{2}{3}\) πr² cu unit
• Total surface area of hemisphere = 2πr² + πr² = 3πr² sq. unit

7. Spherical shell

(i) Total surface area of spherical shell = 4πR² + 4πr²
= 4πr(R² + r²) sq. unit

(ii) Volume of spherical shell = \( \frac{4}{3}\)π(R³– r³) cu . unit

Shapes of Frustum

(i) Slant height of frustum = \(\sqrt{(\mathrm{R}-r)^{2}+h^{2}}\) unit
(ii) Curved surface area of frustum = π(R + r)l sq. unit
(iii) Total surface area of frustum of cone
= πl (R + r) + πR² + πr² sq. unit

(iv) Volume of frustum of cone = \(\frac{1}{3}\)πh (R² + r² + Rr) sq. unit

Volume of Combination Solids

The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the two basic solids.

Conversion of Solid from One Shape to Another

If we melt the candle in the shape of cylinder and pour it into a conical vessel, then it changes into the conical shape. Thus, volume of cylindrical candle = Volume of conical solid.

On this page, you will find Areas related to Circles Class 10 Notes Maths Chapter 12 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 12 Areas related to Circles will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 12 Notes Areas related to Circles

Areas related to Circles Class 10 Notes Understanding the Lesson

1. Circle: Circle is set of all points in a plane which are at the fixed distance from a fixed point i.e., centre. Centre: Mid-point of a circle is called centre of a circle.

2. Radius: The distance between the centre of a circle to the circumference of the circle.
It is denoted by r or R.

3. Chord: A line segment which joins the two points in the circumference of the circle.

4. Diameter: It is the longest chord which passes through the centre of a circle. It is denoted by d or D.
Diameter = 2 x radius
\(\text { Radius }=\frac{\text { Diameter }}{2}\)

5. Circumference of a Circle (or Perimeter)
A perimeter is a path that surrounds a two dimensional shape.
The perimeter of a circle is called its circumference. Circumference
\(\frac{\text { Circumference }}{\text { Diameter }}=\pi\)
Circumference = n x diameter = π x 2r = 2πr
or
The value of π is \(\frac{22}{7}\) or 3.14 (approximately)

6. Arc: An arc is the part of the circumference of a circle.
An arc AB is denoted as \(\widehat{\mathrm{AB}}\)

Length of arc AB is l \(\widehat{\mathrm{AB}}\) or l.
Length of an arc of a sector of angle
\(\theta=\frac{2 \pi r \theta}{360^{\circ}}\)

7. Sector: The portion (or part) of circular region enclosed by two radii and the corresponding arc is called a sector of the circle.

8. Minor and Major sector of the circle: Shaded region OAPB is called sector or minor sector of a circle with centre O.
∠AOB is called angle of the sector.

And OAQB is called major sector.
Angle of major sector = 360° – ∠AOB.

9. Area of Circle

Let r be the radius of circle. If we cut the circle in sectors and arrange then we see this figure like a rectangle whose length is \(\frac{1}{2}\) 2πr = πr and breadth is r.
Hence,
Area of circle = Area of rectangle
= l x b = \(\frac{1}{2}\) 2πr = r
Area of circle = πr²

10. Segment: The portion or part of a circular region enclosed between a chord and the corresponding arc is called a segment of the circle.
The shaded region APB is the minor segment.
And the region AQB is the major segment.
Area of circle = πr²
Area of the sector of an angle \(\theta=\frac{\pi r^{2} \theta}{360^{\circ}}\)

Area of major sector = Area of circle – Area of minor sector
Area of minor segment = Area of sector – Area of ΔOAB\(\frac{\pi r^{2} \theta}{360^{\circ}}\)– area of ΔOAB
Area of major segment = Area of circle – Area of minor segment

11. Area of Combinations of Plane Figures
In our daily life we have observed various plane figures which are combinations of two or more figures and i also in the form of various interesting designs like flower beds, curtains, drain covers, window designs, ] designs on table covers.  To calculate areas of such figures, we field the area of shapes used and then and/subtract as per need.