Questions Related to Atom & Molecules
Updated on November 6, 2025 | By Learnzy Academy
Q21. Give the names of the elements present in the following compounds: (a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate
(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
Q22. Write the chemical formulae of the following: (a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate
(a) Magnesium chloride ----> MgCl₂
(b) Calcium oxide ----> CaO
(c) Copper nitrate -----> Cu(NO₃)₂
(d) Aluminium chloride ----> AlCl₃
(e) Calcium carbonate ------> CaCO₃
Q23. What are polyatomic ions? Give examples.
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.
Q24. When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
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.
Q25. A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
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%