Chemistry — Worked Solutions (Year 10 Science)
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Worked examples for Year 10 Science chemistry. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Year 10 Science — Chemistry. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
For equations, balance atoms one element at a time and don't forget the state symbols.
Example 1 — Balancing a combustion equation
Question
Methane ($\text{CH}_4$) burns completely in oxygen to produce carbon dioxide and water. Write a balanced chemical equation, including state symbols, and name the type of reaction.
Solution
Start with the unbalanced equation: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$.
Balance carbon (already 1 each), then hydrogen: 4 H on the left, so put a 2 in front of water → $2\text{H}_2\text{O}$.
Now oxygen: the right has $2 + 2 = 4$ O atoms, so put a 2 in front of $\text{O}_2$.
$\text{CH}_4{}_{(g)} + 2\text{O}_2{}_{(g)} \rightarrow \text{CO}_2{}_{(g)} + 2\text{H}_2\text{O}_{(l)}$.
This is combustion — a fuel reacting with oxygen. Always balance oxygen last because it appears in two products.
Let's write what we know first: methane plus oxygen makes carbon dioxide and water, so $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$.
Balancing means having the same number of each atom on both sides, because atoms can't be created or destroyed. Carbon is fine (1 each). Hydrogen has 4 on the left but only 2 on the right, so we double the water to $2\text{H}_2\text{O}$.
Now count oxygen on the right: 2 from $\text{CO}_2$ plus 2 from $2\text{H}_2\text{O}$ = 4. To match, we need $2\text{O}_2$ on the left.
$\text{CH}_4{}_{(g)} + 2\text{O}_2{}_{(g)} \rightarrow \text{CO}_2{}_{(g)} + 2\text{H}_2\text{O}_{(l)}$.
A fuel burning in oxygen like this is called combustion — that's why it releases heat and light.
Unbalanced: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$.
- C: 1 = 1 ✓
- H: 4 left → $2\text{H}_2\text{O}$ on right
- O: right now has 4 → $2\text{O}_2$ on left
Balanced: $\text{CH}_4{}_{(g)} + 2\text{O}_2{}_{(g)} \rightarrow \text{CO}_2{}_{(g)} + 2\text{H}_2\text{O}_{(l)}$.
Type: combustion.
Where the marks go
- 1 mark: Correct reactants and products in formula form
- 1 mark: Correctly balanced coefficients ($1, 2, 1, 2$)
- 1 mark: Correct state symbols included
- 1 mark: Identifies the reaction as combustion
Key idea
Balance an equation so each element has equal atoms on both sides; a fuel reacting with oxygen to release energy is combustion.
Example 2 — Atomic structure and the periodic table
Question
A neutral atom of sodium ($\text{Na}$) has an atomic number of 11 and a mass number of 23. State the number of protons, neutrons and electrons it contains, and explain why sodium is placed in Group 1 of the periodic table.
Solution
Atomic number = number of protons = 11. In a neutral atom, electrons equal protons, so 11 electrons.
Neutrons = mass number − atomic number = $23 - 11 = 12$.
So: 11 protons, 12 neutrons, 11 electrons.
Sodium sits in Group 1 because it has 1 electron in its outer shell (electron arrangement 2, 8, 1). The number of outer-shell electrons sets the group — get that arrangement right and the group follows.
The atomic number tells us the number of protons, which is 11 for sodium. Because the atom is neutral, the positive protons must be balanced by an equal number of negative electrons — so 11 electrons too.
For neutrons, we use mass number − atomic number, since the mass number counts protons plus neutrons: $23 - 11 = 12$ neutrons.
So sodium has 11 protons, 12 neutrons and 11 electrons.
Now for the group: if we arrange those 11 electrons in shells we get 2, 8, 1 — one electron in the outermost shell. The periodic table groups elements by their number of outer-shell electrons, and one outer electron places sodium in Group 1, which is why it reacts the way it does.
- Protons = atomic number = 11
- Electrons = protons (neutral) = 11
- Neutrons = mass − atomic number = $23 - 11 = 12$
Electron arrangement: 2, 8, 1 → 1 outer-shell electron.
Group is set by outer-shell electrons → Group 1.
Where the marks go
- 1 mark: Correct number of protons (11) and electrons (11)
- 1 mark: Correct number of neutrons ($23 - 11 = 12$)
- 1 mark: Explains Group 1 placement using 1 outer-shell electron (2, 8, 1)
Key idea
Protons = atomic number, neutrons = mass number − atomic number, and the number of outer-shell electrons sets the periodic table group.
Frequently asked questions
Step-by-step solutions to Chemistry questions in Year 10 Science, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Year 10 Science, using the methods and notation expected in exams and assessments.