Chemistry - Myanmar Matriculation Core Concepts
The Chemistry domain of the Myanmar Matriculation examination rewards candidates who can move fluently between three representations of the same idea: the symbol equation, the particle picture, and the measured quantity. Questions test definitions and periodic trends, but the marks that separate grades usually come from calculations and from writing correct formulae and balanced equations.
Rules worth memorising exactly
- Amount of substance: n = m/M; for gases n = V/22.4 dm³ at STP; n = c × V with V in dm³; n = N/N_A.
- Volume conversion: 1 dm³ = 1000 cm³, so 25.0 cm³ = 0.0250 dm³. Convert before substituting.
- Trends across a period: atomic radius falls, first ionisation energy rises, electronegativity rises. Down a group all three reverse.
- In ionic equations, cancel spectator ions but keep state symbols; a precipitate, gas or weak electrolyte must be written as a molecule or solid, not split into ions.
Worked example
4.00 g of NaOH is dissolved and made up to 250 cm³. Find the concentration and the mass of HCl that neutralises 25.0 cm³ of it. M(NaOH) = 40.0, so n = 4.00/40.0 = 0.100 mol; c = 0.100/0.250 = 0.400 mol/dm³. In 25.0 cm³: n(NaOH) = 0.400 × 0.0250 = 1.00 × 10⁻² mol. NaOH + HCl → NaCl + H₂O is 1:1, so n(HCl) = 1.00 × 10⁻² mol and mass = 1.00 × 10⁻² × 36.5 = 0.365 g.
Traps that cost marks
- Substituting cm³ straight into c = n/V, giving answers 1000 times too large.
- Using 22.4 dm³ for gas volumes measured at room conditions rather than STP.
- Confusing mass number with relative atomic mass, or ignoring isotope abundance in an average.
- Losing the final mark by omitting state symbols, or by writing half-equations whose charges do not balance.