Questions Related to NCERT

In butanone, the suffix “-one” shows the presence of a ketone group (C=O) attached to a middle carbon atom. Its structure is CH3–CO–C2H5.
So, the functional group in butanone is ketone (–CO–).

Ethane (C2H6) has 7 covalent bonds. Each carbon atom joins with three hydrogen atoms and one other carbon atom. This makes 6 C–H bonds and 1 C–C bond. All the bonds are single bonds. So, ethane has a total of 7 covalent bonds.

(a) Ethyne (C₂H₂)
= (2 × 12) + (2 × 1)
= 24 + 2
= 26 g/mol

(b) Sulphur molecule (S₈)
= 8 × 32
= 256 g/mol

(c) Phosphorus molecule (P₄)
= 4 × 31
= 124 g/mol

(d) Hydrochloric acid (HCl)
= (1 × 1) + (1 × 35.5)
= 1 + 35.5
= 36.5 g/mol

(e) Nitric acid (HNO₃)
= (1 × 1) + (1 × 14) + (3 × 16)
= 1 + 14 + 48
= 63 g/mol

(a) Quick lime (CaO)   ---->   Calcium and Oxygen
(b) Hydrogen bromide (HBr)    ---->   Hydrogen and Bromine
(c) Baking powder (NaHCO₃)    ----->   Sodium, Hydrogen, Carbon, and Oxygen
(d) Potassium sulphate (K₂SO₄)   ------>   Potassium, Sulphur, and Oxygen

(a) Magnesium chloride  ---->  MgCl₂
(b) Calcium oxide  ---->   CaO
(c) Copper nitrate   ----->  Cu(NO₃)₂
(d) Aluminium chloride   ---->  AlCl₃
(e) Calcium carbonate   ------>   CaCO₃

Polyatomic ions are ions that contain two or more atoms joined together by covalent bonds and carry a net positive or negative charge. These atoms act as a single charged unit during chemical reactions.

Examples:

• Ammonium ion (NH₄⁺)
• Hydroxide ion (OH⁻)
• Nitrate ion (NO₃⁻)
• Sulphate ion (SO₄²⁻)
• Carbonate ion (CO₃²⁻)

Thus, polyatomic ions are groups of atoms that behave as one charged particle.

From the data:
3.0 g of carbon + 8.0 g of oxygen   ---->   11.0 g of carbon dioxide

This means 3 g of carbon requires 8 g of oxygen to form 11 g of carbon dioxide.

Now, if 3 g of carbon is burnt in 50 g of oxygen, carbon is the limiting reactant because it can combine only with 8 g of oxygen. The remaining oxygen will be left unused.

So, the amount of carbon dioxide formed will still be 11 g.

Mass of carbon dioxide formed = 11.0 g

Law governing the reaction:
This is governed by the Law of Constant Proportion, which states that a chemical compound always contains the same elements in the same fixed ratio by mass.

Total mass of the compound = 0.24 g
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g

Percentage of boron = (0.096 / 0.24) × 100 = 40%
Percentage of oxygen = (0.144 / 0.24) × 100 = 60%

Therefore, the percentage composition of the compound is:
Boron = 40%
Oxygen = 60%

Density of the packet = Mass / Volume = 500 g / 350 cm³ = 1.43 g/cm³

Since the density of the packet (1.43 g/cm³) is greater than the density of water (1 g/cm³), the packet will sink in water.

Mass of water displaced = Volume of packet × Density of water
= 350 cm³ × 1 g/cm³
= 350 g

Final Answer:
The packet will sink in water, and the mass of water displaced is 350 g.

Density of the substance = Mass / Volume = 50 g / 20 cm³ = 2.5 g/cm³

Since the density of the substance (2.5 g/cm³) is greater than the density of water (1 g/cm³), the substance will sink in water.

A block of plastic comes up to the surface of water because the buoyant force acting on it is greater than its weight. Since the density of plastic is less than the density of water, the upward buoyant force pushes it up until it reaches the surface.

The buoyant force on an object immersed in a liquid acts vertically upward, that is, opposite to the direction of gravity.

(a) Time to rise to top = half the total time = 6/2 = 3 s.
Initial velocity u = g × time = 10 × 3 = 30 m/s.

(b) Maximum height H = u² / (2g) = (30)² / (2 × 10) = 900 / 20 = 45 m.

(c) Position after t = 4 s (take upward as positive, origin at thrower):
y = ut − ½ g t² = 30×4 − ½×10×4² = 120 − 80 = 40 m above the thrower.

Final answers:
(a) 30 m/s upward
(b) 45 m
(c) 40 m above the thrower (after 4 s)

Height of tower = 100 m
Acceleration due to gravity, g = 10 m/s²

Let the stones meet after t seconds.

For the first stone (falling from the top):
Initial velocity, u₁ = 0
Distance fallen, s₁ = ½ g t² = 5t²

For the second stone (projected upwards):
Initial velocity, u₂ = 25 m/s
Distance covered, s₂ = u₂t – ½ g t² = 25t – 5t²

They meet when the total distance covered by both is equal to the height of the tower.
s₁ + s₂ = 100
5t² + (25t – 5t²) = 100
25t = 100
t = 4 seconds

Distance fallen by the first stone, s₁ = 5t² = 5 × 4² = 80 m
Therefore, they meet 20 m above the ground (100 – 80 = 20).

Final Answer:
Time when they meet = 4 seconds
Height above the ground where they meet = 20 meters

Mass of Earth, M₁ = 6 × 10²⁴ kg
Mass of Sun, M₂ = 2 × 10³⁰ kg
Distance between Earth and Sun, r = 1.5 × 10¹¹ m
Gravitational constant, G = 6.67 × 10⁻¹¹ N·m²/kg²

Formula:
F = G × M₁ × M₂ / r²

Substituting values:
F = (6.67 × 10⁻¹¹ × 6 × 10²⁴ × 2 × 10³⁰) / (1.5 × 10¹¹)²
F = (8.004 × 10⁴⁴) / (2.25 × 10²²)
F = 3.56 × 10²² N

Therefore, the gravitational force between the Earth and the Sun is 3.56 × 10²² N.

Initial velocity (u) = 40 m/s
Acceleration due to gravity (g) = –10 m/s² (negative because gravity acts downward)
At maximum height, final velocity (v) = 0

Using the equation of motion:
v² = u² + 2as

0 = (40)² + 2(–10)s
0 = 1600 – 20s
20s = 1600
s = 80 m

Maximum height = 80 m

When the stone returns to the ground:

Net displacement = 0 (since it comes back to the starting point)
Total distance covered = Distance up + Distance down = 80 + 80 = 160 m

Final Answers:
Maximum height = 80 m
Net displacement = 0
Total distance covered = 160 m

Height (h) = 19.6 m
Initial velocity (u) = 0 (since the stone is released from rest)
Acceleration due to gravity (g) = 9.8 m/s²

Using the equation of motion:
v² = u² + 2gh

Substitute the values:
v² = 0 + 2 × 9.8 × 19.6
v² = 384.16
v = √384.16
v = 19.6 m/s

Therefore, the final velocity of the stone just before touching the ground is 19.6 m/s.

We know that Weight = Mass × Acceleration due to gravity (W = m × g)

On Earth:
m = 10 kg
g = 9.8 m/s²
Wₑ = 10 × 9.8 = 98 N

On Moon:
Gravity on Moon = (1/6) × gravity on Earth
gₘ = 9.8 / 6 = 1.63 m/s²
Wₘ = 10 × 1.63 = 16.3 N

Therefore,
Weight on Earth = 98 N
Weight on Moon = 16.3 N

A sheet of paper falls slower than a crumpled paper ball because of air resistance.
The flat sheet has a larger surface area, so it experiences more air resistance while falling.
When the paper is crumpled into a ball, its surface area facing the air becomes smaller, reducing air resistance.

Therefore, the crumpled paper ball falls faster than the flat sheet.

No, the friend will not agree with the weight of the gold.

This is because the value of acceleration due to gravity (g) is greater at the poles and smaller at the equator.

Since weight = mass × gravity (W = m × g), the weight of the same gold will be less at the equator than at the poles.

Therefore, the gold will weigh slightly less at the equator, even though its mass remains the same.