Interactive Study Guide

Number Bases SS1 · Mathematics

A number base is the number of distinct digits a place-value system uses (base 10 uses 0-9, base 2 uses 0-1). It is tested every year in conversions and in arithmetic done directly in a given base.

  • Base n uses digits 0 to n−1; each place is a power of n. Example: 2013₅ = 2×5³ + 0×5² + 1×5 + 3 = 258₁₀.
  • Base n to base 10: multiply each digit by its place value nᵏ and add. Fractions use negative powers: 0.101₂ = 1/2 + 1/8 = 0.625.
  • Base 10 to base n (whole numbers): divide repeatedly by n and read the remainders from the last to the first (bottom to top).
  • Decimal fraction to base n: multiply the fractional part repeatedly by n; the whole-number parts, read top to bottom, are the digits after the point.
  • Base a to base b (neither is 10): convert to base 10 first, then to base b. Binary to octal: group digits in threes from the point; to base 4: in twos.
  • Arithmetic in base n: add or multiply as usual, but carry when the total reaches n. In base 2, 1 + 1 = 10, so 1 + 1 + 1 = 11.
  • Subtraction in base n: borrowing gives n (not 10) to the digit. To find an unknown base, write the equation in base 10 and solve, e.g. 24ₓ = 14 gives 2x + 4 = 14, so x = 5.
  • Always write the base as a subscript on the answer, and check that every digit is smaller than the base (e.g. 5 cannot appear in base 4).
Worked example: Evaluate 11011₂ + 1011₂ in base 2. Convert: 11011₂ = 27 and 1011₂ = 11, sum = 38. Divide 38 by 2 repeatedly: remainders from bottom to top give 100110₂. Direct binary addition gives the same answer.
Exam trap: Reading the remainders from top to bottom instead of bottom to top, and carrying 10 instead of the base value when adding or multiplying. Forgetting the base subscript also loses marks.
Modular Arithmetic SS1 · Mathematics

Modular arithmetic works with remainders after division by a fixed number (the modulus), like the 12-hour clock or days of the week. It simplifies problems involving cycles and is a standard WAEC/JAMB topic.

  • a ≡ b (mod n) means n divides (a − b), i.e. a and b leave the same remainder on division by n. Example: 23 ≡ 2 (mod 7).
  • To reduce a number mod n, divide by n and keep the remainder, which lies from 0 to n−1. Example: 63 mod 5 = 3, since 63 = 12×5 + 3.
  • Addition and multiplication: (a + b) mod n = ((a mod n) + (b mod n)) mod n, and similarly for ×. Reduce first to keep numbers small.
  • Subtraction: if the result is negative, add multiples of n until it lies from 0 to n−1. Example: 3 − 5 ≡ −2 ≡ 5 (mod 7).
  • Solving linear congruences such as 3x ≡ 4 (mod 7): try x = 0, 1, …, n−1 (or add multiples of 7 to 4 until divisible by 3): 4 + 14 = 18, so x = 6.
  • Real-life use: clock arithmetic (mod 12 or 24) and days of the week (mod 7). Example: 100 days after Monday is 100 mod 7 = 2, so Wednesday.
  • Working mod n, the remainders form the set {0, 1, …, n−1}; mod 5 gives {0,1,2,3,4}. Tables of addition and multiplication mod n are built from this set.
  • Powers: reduce the base first, then look for a repeating pattern. Example: 2ⁿ mod 5 repeats 2, 4, 3, 1, so 2⁶ ≡ 4 (mod 5).
Worked example: Solve 3x ≡ 4 (mod 7). Add multiples of 7 to 4 until the result is divisible by 3: 4, 11, 18. So 3x = 18 and x = 6. Check: 3 × 6 = 18 = 2×7 + 4, so the remainder is 4.
Exam trap: Leaving a negative or too-large answer instead of reducing it into 0 to n−1, and using ordinary division to solve congruences. Also confusing the modulus with the quotient.
Indices and Standard Form SS1 · Mathematics

Indices (powers) show repeated multiplication, and standard form writes very large or small numbers compactly. The laws of indices are the basis of many algebra and logarithm questions.

  • Laws: aᵐ × aⁿ = aᵐ⁺ⁿ; aᵐ ÷ aⁿ = aᵐ⁻ⁿ; (aᵐ)ⁿ = aᵐⁿ; (ab)ⁿ = aⁿbⁿ.
  • Special indices: a⁰ = 1 (a ≠ 0); a⁻ⁿ = 1/aⁿ; a^(1/n) = ⁿ√a; a^(m/n) = (ⁿ√a)ᵐ. Example: 8^(2/3) = 2² = 4.
  • Indicial equations: write both sides with the same base, then equate the powers. Example: 8ˣ = 4^(x+1) gives 2^(3x) = 2^(2x+2), so x = 2.
  • If the unknown is the base, equate the powers instead: xⁿ = c gives x = c^(1/n). For equations like 2ˣ + 2ˣ⁺¹ = 12, factorise: 2ˣ(1 + 2) = 12, so 2ˣ = 4, x = 2.
  • Standard form is A × 10ⁿ with 1 ≤ A < 10 and n an integer. Example: 0.000456 = 4.56 × 10⁻⁴ and 3 200 000 = 3.2 × 10⁶.
  • Multiplying in standard form: multiply the A parts, add the powers of 10, then readjust. Example: (3 × 10⁴)(5 × 10³) = 15 × 10⁷ = 1.5 × 10⁸.
  • Adding or subtracting in standard form: first write both numbers with the same power of 10, then combine the A parts.
  • Significant figures and decimal places are often combined with standard form; leading zeros are not significant.
Worked example: Solve 8ˣ = 4^(x+1). Write 8 = 2³ and 4 = 2², so 2^(3x) = 2^(2x+2). Equate powers: 3x = 2x + 2, so x = 2. Check: 8² = 64 and 4³ = 64.
Exam trap: Writing 15 × 10⁷ as the final standard-form answer (A must be below 10), and adding powers when the bases differ. Treating a⁻ⁿ as negative instead of as a reciprocal is also common.
Logarithms SS1 · Mathematics

A logarithm is the power to which a base must be raised to give a number: logₐN = x means aˣ = N. Logarithms turn multiplication into addition and are used with tables or calculators in WAEC.

  • Definition: logₐN = x ⇔ aˣ = N. Example: log₂ 8 = 3 because 2³ = 8; log₁₀ 1000 = 3.
  • Laws: log(MN) = log M + log N; log(M/N) = log M − log N; log Mᵖ = p log M; logₐa = 1; logₐ1 = 0.
  • The log of a negative number or of zero is undefined, so reject any solution that makes a log argument ≤ 0.
  • Change of base: logₐb = log b / log a. Also logₐb = 1/log_b a. Useful for log₂ values on a base-10 calculator.
  • Common logs have base 10. A common log = characteristic (whole part) + mantissa (decimal part, from the tables). log 450 = 2.6532, characteristic 2; for 0.045 the characteristic is 2̄ (−2).
  • Reading 4-figure tables: find the mantissa from the row and column, add the 'difference' column for the fourth figure, then attach the characteristic. Antilog reverses this.
  • Calculation by logs: add logs to multiply and subtract to divide, halve for square roots, then take the antilog to give the answer.
  • Graph of y = 10ˣ: passes through (0, 1), always positive, rises steeply, and has the x-axis as asymptote. Its reflection in y = x is y = log x.
Worked example: Solve log₂x + log₂(x − 2) = 3. Combine: log₂[x(x − 2)] = 3, so x(x − 2) = 2³ = 8. Then x² − 2x − 8 = 0, giving (x − 4)(x + 2) = 0. x = −2 is rejected because log₂(−2) is undefined, so x = 4.
Exam trap: Writing log(M + N) as log M + log N, and forgetting to reject solutions that give a negative log argument. In table work, errors in the characteristic of numbers less than 1 are common.
Sets and Venn Diagrams SS1 · Mathematics

A set is a well-defined collection of objects called elements. Set notation and Venn diagrams are used to solve counting problems with up to three sets.

  • Notation: x ∈ A (x belongs to A); {1, 2, 3} lists elements; {x : x is a prime less than 10} is set-builder form. n(A) is the number of elements.
  • Types: empty set ∅ or { } (n = 0); finite and infinite sets; universal set 𝒰 or ξ; equal sets (same elements); equivalent sets (same n); disjoint sets (no common element).
  • Subset: A ⊂ B means every element of A is in B. A set with n elements has 2ⁿ subsets, including ∅ and itself, and 2ⁿ − 1 proper subsets.
  • Union A ∪ B: elements in A or B or both. Intersection A ∩ B: elements in both. Complement A′: elements of 𝒰 not in A. Difference A \ B: in A but not in B.
  • Two-set formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Always n(A) + n(A′) = n(𝒰).
  • Three-set formula: n(A ∪ B ∪ C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C).
  • Venn diagram method: begin with the innermost region (all three sets, or both sets), then subtract it from each overlap and work outwards. The total of all regions equals n(𝒰).
  • Shading: draw 𝒰 as a rectangle and sets as overlapping circles; shade the stated region such as (A ∪ B)′ or A ∩ B′. De Morgan: (A ∪ B)′ = A′ ∩ B′.
Worked example: In a class of 40, 25 offer Mathematics, 20 offer Physics and 5 offer neither. Students offering at least one = 40 − 5 = 35. Both = 25 + 20 − 35 = 10. So 15 offer Mathematics only and 10 offer Physics only.
Exam trap: Putting the overlap total into the circle intersection without subtracting it from the 'only' regions, and forgetting the students outside all circles. Mixing up ⊂ with ∈ is also common.
Variation SS1 · Mathematics

Variation describes how one quantity changes in proportion to others, using a constant of proportionality k. It is used for formula-based problems in science and commerce.

  • Direct variation: y ∝ x means y = kx; the ratio y/x is constant and the graph of y against x is a straight line through the origin.
  • Inverse variation: y ∝ 1/x means y = k/x, so xy = k. When x doubles, y halves.
  • Joint variation: z ∝ xy means z = kxy (z varies directly as both). If z ∝ x/y, then z = kx/y (z varies directly as x and inversely as y).
  • Partial variation: y is partly constant and partly varies with x, i.e. y = a + bx. Two sets of data give two equations to find a and b.
  • Method: (1) write the proportion, (2) insert k, (3) substitute the given values to find k, (4) rewrite the formula with k, (5) answer the question.
  • Powers and roots: y ∝ x² gives y = kx²; y ∝ √x gives y = k√x; y ∝ 1/x² is the inverse square law. Area ∝ r² and volume ∝ r³ are common cases.
  • Without finding k: if y ∝ x², then y₁/y₂ = x₁²/x₂². Percentage change questions: x up 10% under y ∝ x² makes y 1.1² = 1.21 times, i.e. 21% up.
  • Always state the final formula and use units where given.
Worked example: y varies directly as x², and y = 18 when x = 3. Then y = kx², so 18 = 9k and k = 2. The formula is y = 2x², so when x = 5, y = 2 × 25 = 50.
Exam trap: Forgetting to find k first, and using direct variation when the question says 'inversely'. In partial variation, students wrongly write y = kx and omit the constant term.
Simple Equations and Change of Subject SS1 · Mathematics

Changing the subject means rearranging a formula so that a chosen letter stands alone on one side. It also covers substituting values and solving simple equations, and it is vital throughout science and mathematics.

  • The subject is the letter alone on one side, with coefficient 1. To change it, undo operations in reverse order, doing the same to both sides.
  • Linear equations with brackets or fractions: clear fractions by multiplying every term by the LCM, expand brackets, collect like terms, then solve.
  • Powers and roots: to free a squared letter, take the square root of both sides; to free a root, square both sides. Example: V = πr²h gives r = √(V/πh).
  • If the subject appears twice, collect those terms on one side and factorise. Example: y = (x + 1)/(x − 1) gives y(x − 1) = x + 1, so x(y − 1) = y + 1, hence x = (y + 1)/(y − 1).
  • Formulas with roots: square both sides first. Example: T = 2π√(l/g) gives T² = 4π²l/g, so l = gT²/4π².
  • Substitution: replace letters with given numbers, keeping brackets for negative values. Evaluate powers before multiplication and addition (BODMAS).
  • Simultaneous equations by elimination or substitution may appear with formula problems. Always check the answer in the original equation.
  • Fractions: if a/b = c/d then ad = bc (cross-multiply).
Worked example: Make l the subject of T = 2π√(l/g). Divide by 2π: T/2π = √(l/g). Square both sides: T²/4π² = l/g. Multiply by g: l = gT²/4π². For T = 2 and g = 10 with π = 3.14, l = 10 × 4/(4 × 9.8596) ≈ 1.01.
Exam trap: Moving a term across without changing its sign, and squaring only part of one side, e.g. squaring a + b as a² + b². Taking the root of only one side is also common.
Quadratic Equations SS1 · Mathematics

A quadratic equation has the form ax² + bx + c = 0 (a ≠ 0) and has at most two roots. It is solved by factorisation, completing the square or the formula, and the graph is a parabola.

  • Factorisation: find two numbers whose product is ac and sum is b, split the middle term and factorise. Then set each bracket to zero. Example: 2x² − 5x − 3 = (2x + 1)(x − 3), so x = −1/2 or 3.
  • Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a. Use it when factorising is not obvious or when answers are to be given to decimal places.
  • Discriminant D = b² − 4ac: D > 0 gives two distinct real roots, D = 0 gives equal roots, D < 0 gives no real roots.
  • Completing the square: divide by a, move the constant, add (b/2a)² to both sides, write as a perfect square and take roots. x² + 6x − 7 = 0 gives (x + 3)² = 16, so x = 1 or −7.
  • Sum of roots = −b/a and product = c/a. To form an equation from roots α and β: x² − (α + β)x + αβ = 0.
  • Graph of y = ax² + bx + c: a parabola, U-shaped if a > 0 and ∩-shaped if a < 0. The turning point has x = −b/2a and the roots are where it meets the x-axis.
  • Drawing the graph: make a table of values, plot with a smooth curve, and read off roots, the minimum or maximum, and the line of symmetry.
  • Word problems: define a variable, form the equation (areas, consecutive numbers, ages) and reject any root that is impossible, such as a negative length.
Worked example: Solve 2x² − 5x − 3 = 0. ac = −6, so we need numbers with product −6 and sum −5: −6 and 1. Then 2x² − 6x + x − 3 = 2x(x − 3) + 1(x − 3) = (2x + 1)(x − 3). So x = −1/2 or x = 3.
Exam trap: Not rearranging to = 0 before factorising, and sign errors in −b or in b² − 4ac when b is negative. Leaving out the ± in the formula or giving only one root loses marks.
Logical Reasoning and Truth Tables SS1 · Mathematics

Logical reasoning studies statements that are either true or false and how they combine using connectives such as 'and', 'or', 'not' and 'if … then'. Truth tables list every possible truth value to test arguments.

  • A statement (proposition) is a sentence that is either true or false, not both. Questions, commands and opinions are not statements. Simple statements are written p, q, r.
  • Negation ~p (not p): has the opposite truth value to p. Double negation: ~(~p) = p.
  • Conjunction p ∧ q (p and q): true only when both p and q are true. Disjunction p ∨ q (p or q): false only when both are false.
  • Implication p ⇒ q (if p then q): false only when p is true and q is false. Biconditional p ⇔ q (p if and only if q): true when p and q have the same truth value.
  • Truth table for n statements has 2ⁿ rows (4 for p, q; 8 for p, q, r). Build columns for each sub-expression and finish with the whole compound statement.
  • A tautology is always true, and a contradiction is always false. Example: p ∨ ~p is a tautology; p ∧ ~p is a contradiction.
  • Related forms: converse of p ⇒ q is q ⇒ p; inverse is ~p ⇒ ~q; contrapositive is ~q ⇒ ~p, which is equivalent to the original.
  • Valid arguments: a conclusion is valid if it is true whenever all premises are true. Example: p ⇒ q and p lead to q.
Worked example: Let p: '2 is even' (true) and q: '5 is even' (false). Then p ∧ q is false, p ∨ q is true, ~q is true, and p ⇒ q is false because a true statement leads to a false one.
Exam trap: Treating p ⇒ q as false when p is false (it is true then), and mixing up the converse with the contrapositive. Missing rows in the truth table is another common error.
Geometric Constructions and Loci SS1 · Mathematics

Constructions draw exact angles and lines using only a ruler, pair of compasses and protractor where allowed. A locus is the path of a point that moves under a given rule, and it is used to describe regions.

  • Perpendicular bisector of a line AB: draw arcs of equal radius (more than half of AB) from A and B; join the two intersection points. Every point on it is equidistant from A and B.
  • Angle bisector: draw an arc cutting both arms, then equal arcs from those points; the line to the vertex divides the angle in two. Its points are equidistant from the two arms.
  • Construct 60° with an equilateral-triangle arc, 90° with a perpendicular, and 30° and 45° by bisecting 60° and 90°. Also 120°, 75°, 105° and 135° by combining or bisecting.
  • Perpendicular from a point to a line, and a line parallel to another through a point, use arcs of equal radius; keep all construction arcs visible.
  • Triangle construction: from three sides (SSS) use arcs; from two sides and the included angle (SAS) or two angles and a side (ASA) use a protractor or constructed angles.
  • Locus rules: a point at a fixed distance from a point is a circle; equidistant from two points is their perpendicular bisector; equidistant from two lines is the angle bisector; at a fixed distance from a line is a pair of parallel lines.
  • Loci problems combine several conditions; the required region is where all are met, using inequalities like 'less than 3 cm from P' (inside the circle).
  • Label the points, give the scale if asked, and shade or mark the final region clearly.
Worked example: Locate the points equidistant from A and B and exactly 3 cm from A, where AB = 5 cm. Draw the perpendicular bisector of AB and a circle of radius 3 cm centred at A. They meet at two points, which are the answer.
Exam trap: Erasing construction arcs, and using a protractor when the question says 'using ruler and compasses only'. Mixing up the locus 'equidistant from two points' with 'equidistant from two lines' is common.
Euclidean Geometry: Angle Theorems SS1 · Mathematics

Euclidean theorems give reliable rules about angles in straight lines, parallel lines and triangles. They are used to find missing angles and to give reasons in proofs.

  • Angles on a straight line sum to 180°; angles at a point sum to 360°; vertically opposite angles are equal.
  • Parallel lines: corresponding angles are equal, alternate angles are equal, and co-interior (allied) angles sum to 180°.
  • Angle sum of a triangle is 180°. For a quadrilateral it is 360°; for an n-sided polygon it is (n − 2) × 180°.
  • Exterior angle theorem: an exterior angle of a triangle equals the sum of the two interior opposite angles. Example: if the exterior angle is 110° and one interior opposite angle is 50°, the other is 60°.
  • Isosceles triangle: base angles are equal; equilateral triangle: all angles are 60°. In a right-angled triangle the two acute angles add to 90°.
  • Exterior angles of any convex polygon sum to 360°; each exterior angle of a regular n-gon is 360°/n and the interior angle is 180° − 360°/n.
  • Pythagoras: in a right-angled triangle, a² + b² = c², where c is the hypotenuse. Triples such as 3, 4, 5 and 5, 12, 13 are common.
  • Geometry reasons: state the theorem (e.g. 'alternate angles', 'angle sum of triangle') beside each step.
Worked example: In triangle ABC, side BC is extended to D, angle ACD = 110° and angle A = 50°. By the exterior angle theorem 110° = 50° + angle B, so angle B = 60°. Check: 50° + 60° + 70° = 180°, with angle ACB = 180° − 110° = 70°.
Exam trap: Calling corresponding angles 'alternate' (or the reverse), and using the wrong triangle sides for the exterior angle. Forgetting to give reasons when asked to prove or show also loses marks.
Trigonometric Ratios SS1 · Mathematics

Trigonometric ratios link the angles of a right-angled triangle to its side lengths. They are used for heights, distances and bearings, and are extended to any angle using the unit circle.

  • SOH-CAH-TOA: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. Also tan θ = sin θ / cos θ.
  • Special angles: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3; sin 45° = cos 45° = √2/2, tan 45° = 1; sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.
  • Complementary angles: sin θ = cos(90° − θ). Identity: sin²θ + cos²θ = 1.
  • Unit circle: for a point (x, y) at angle θ, cos θ = x, sin θ = y. Signs by quadrant (ASTC): 1st all positive; 2nd sin only; 3rd tan only; 4th cos only.
  • Angles 90° to 360°: sin(180° − θ) = sin θ, cos(180° − θ) = −cos θ, tan(180° − θ) = −tan θ. Also sin(360° − θ) = −sin θ. Example: sin 150° = 1/2.
  • Graphs: y = sin x and y = cos x have amplitude 1 and period 360°, ranging from −1 to 1. y = tan x has period 180° and vertical asymptotes at 90° and 270°.
  • Applications: angle of elevation (looking up from the horizontal) and depression (looking down). Draw a clear right-angled triangle before using a ratio.
  • Using tables or calculator: make sure the calculator is in degree mode, and give the answer to the stated accuracy.
Worked example: A ladder 10 m long leans against a wall at 60° to the ground. The height reached is opposite the angle and the ladder is the hypotenuse, so height = 10 × sin 60° = 10 × 0.8660 = 8.66 m.
Exam trap: Choosing the wrong side as opposite or adjacent, and leaving the calculator in radian mode. Ignoring the sign of the ratio in the 2nd, 3rd and 4th quadrants is also common.
Mensuration: Arcs, Sectors, Areas and Volumes SS1 · Mathematics

Mensuration is the measurement of lengths, areas and volumes of plane and solid shapes. It covers circle parts such as arcs, sectors and segments, and solids such as cylinders, cones and spheres.

  • Circle: circumference = 2πr and area = πr². Use π = 22/7 or 3.142 as stated in the question.
  • Arc length = (θ/360) × 2πr. Sector area = (θ/360) × πr². Perimeter of a sector = arc + 2r.
  • Segment area = area of sector − area of triangle = (θ/360)πr² − ½r² sin θ. For a minor segment subtract; for a major segment add the triangle.
  • Cone: a sector of radius l and arc length 2πr forms a cone of base radius r. Slant height l = √(r² + h²); the sector angle is (r/l) × 360°.
  • Cuboid: V = lbh; surface area = 2(lb + lh + bh). Cylinder: V = πr²h; curved surface area = 2πrh; total surface area = 2πr(r + h).
  • Cone: V = ⅓πr²h; curved surface area = πrl; total = πr(r + l). Sphere: V = ⁴⁄₃πr³; surface area = 4πr². Pyramid: V = ⅓ × base area × h.
  • Prism: V = cross-sectional area × length. Frustum: volume of the large cone minus the small cone removed.
  • Keep units consistent: cm², m² for area and cm³, m³ for volume (1 m³ = 1 000 000 cm³; 1 litre = 1000 cm³).
Worked example: A sector has radius 7 cm and angle 90°. Arc = (90/360) × 2 × 22/7 × 7 = 11 cm. Area = (90/360) × 22/7 × 7² = 38.5 cm². A cylinder of radius 7 cm and height 10 cm has V = 22/7 × 49 × 10 = 1540 cm³.
Exam trap: Using diameter instead of radius, and mixing perimeter and area units. Using vertical height instead of slant height in curved surface area of a cone is also frequent.
Data Presentation SS1 · Mathematics

Data presentation organises raw data into tables and diagrams so patterns are easy to see. WAEC tests frequency tables, bar charts, histograms and pie charts, including reading and drawing them.

  • Raw data is tallied into a frequency table: variable, tally and frequency (f). The sum of frequencies Σf is the total number of observations.
  • Grouped data uses class intervals such as 10–19 or 10–<20. Class boundaries (e.g. 9.5–19.5), class width and class mid-point (lower + upper)/2 are used.
  • Bar chart: separate bars of equal width with gaps, bar height equal to frequency; used for discrete or categorical data. Label both axes and give a title.
  • Histogram: for continuous grouped data, bars touch, with class boundaries on the x-axis. Area is proportional to frequency; with equal widths, height is frequency.
  • Pie chart: sector angle = (frequency/total) × 360°. The angles must add up to 360°, and each sector is labelled with its item or percentage.
  • Frequency polygon: join the mid-points of the tops of the histogram bars with straight lines; extend to the axis at both ends.
  • Reading diagrams: total from the sum of heights; fractions or percentages as (part/total) × 100%. The modal class is the one with the highest frequency.
  • Choose the diagram to suit the data: pie chart for proportions, bar chart for categories, histogram for continuous data, line graph for trends over time.
Worked example: In a survey of 180 students, 45 chose football. Pie chart angle = (45/180) × 360° = 90°. If 60 chose athletics, its angle = (60/180) × 360° = 120°.
Exam trap: Drawing histogram bars with gaps or using class limits instead of boundaries. In pie charts, angles that do not total 360° and forgetting to find the total first are common errors.
Logarithm Calculations and Approximations SS2 · Mathematics

Logarithms turn multiplication, division, powers and roots into addition, subtraction and multiplication, and four-figure tables make hard calculations quick. WAEC and JAMB still test log-table work, especially with numbers less than one.

  • Laws: log(ab) = log a + log b; log(a/b) = log a − log b; log aⁿ = n log a; log ⁿ√a = (log a)/n. Common logarithms use base 10, so log 10 = 1 and log 1 = 0.
  • A logarithm has two parts: the characteristic (whole number part) and the mantissa (positive decimal part read from the tables). For 345 = 3.45 × 10², log = 2.5378.
  • For numbers greater than 1, the characteristic is (number of digits before the decimal point) − 1. So log 62.8 has characteristic 1 and equals 1.7980.
  • For numbers less than 1, the characteristic is negative, written with a bar: 0.0345 = 3.45 × 10⁻², so log 0.0345 = 2̄.5378, meaning −2 + 0.5378. The mantissa stays positive.
  • Reading four-figure tables: find the first two digits in the row, the third digit in the column, then add the 'difference' column value for the fourth digit. Antilog tables reverse this process.
  • Adding and subtracting bar numbers: 2̄.5378 + 1.7980 = 0.3358 (the carry +1 combines with −2 to give −1, then +1 more). Dividing a bar number: make the negative part divisible first, e.g. 3̄.4 ÷ 2 = (4̄ + 1.4) ÷ 2 = 2̄.7.
  • Approximation: give answers to the stated number of significant figures or decimal places. Log work is accurate to 3 or 4 s.f., so round the final answer only, never intermediate steps.
  • Calculation layout: write a table of number | logarithm, add or subtract the logs, then take the antilog of the result. Antilog of a bar characteristic fixes the decimal point position, e.g. antilog 2̄.1234 = 0.01327.
Worked example: Evaluate 0.0345 × 62.8 using logarithm tables. log 0.0345 = 2̄.5378; log 62.8 = 1.7980. Add: 2̄.5378 + 1.7980 = 0.3358. Antilog of 0.3358 = 2.167 (characteristic 0 so one digit before the point). Answer ≈ 2.167 (3 d.p.).
Exam trap: Treating the mantissa of a bar number as negative (writing 2̄.5378 as −2.5378) and wrong carrying when adding or halving bar logarithms. Also misplacing the decimal point after taking the antilog.
Sequences and Series (A.P. and G.P.) SS2 · Mathematics

A sequence is an ordered list of numbers following a rule, and a series is the sum of its terms. Arithmetic (A.P.) and geometric (G.P.) progressions are the standard cases in WAEC and JAMB.

  • A sequence is an ordered set of numbers u₁, u₂, u₃, …; a series is the sum of the terms, e.g. 2 + 5 + 8 + 11. The nth term is written Tₙ or uₙ.
  • A.P.: each term differs from the previous one by a constant common difference d = T₂ − T₁. The nth term is Tₙ = a + (n − 1)d, where a is the first term.
  • Sum of the first n terms of an A.P.: Sₙ = (n/2)[2a + (n − 1)d] = (n/2)(a + l), where l is the last term.
  • G.P.: each term is the previous one multiplied by a constant common ratio r = T₂/T₁. The nth term is Tₙ = arⁿ⁻¹.
  • Sum of the first n terms of a G.P.: Sₙ = a(rⁿ − 1)/(r − 1) when r > 1, and Sₙ = a(1 − rⁿ)/(1 − r) when r < 1.
  • Sum to infinity of a G.P. exists only when |r| < 1: S∞ = a/(1 − r). Example: 8 + 4 + 2 + … has S∞ = 8/(1 − ½) = 16.
  • Useful link: Tₙ = Sₙ − Sₙ₋₁. Three numbers a, b, c are in A.P. if 2b = a + c, and in G.P. if b² = ac (b is the arithmetic or geometric mean).
  • To find the number of terms, set Tₙ equal to the last term and solve for n; n must be a positive whole number. Word problems on savings, salaries and depreciation reduce to A.P. or G.P.
Worked example: The first term of an A.P. is 5 and the common difference is 3. Find the sum of the first 20 terms. S₂₀ = (20/2)[2(5) + 19(3)] = 10(10 + 57) = 10 × 67 = 670.
Exam trap: Mixing up the formulas (using a + nd instead of a + (n − 1)d) and using the G.P. sum formula with the wrong sign for r < 1. Using S∞ when |r| ≥ 1 is also wrong.
Simultaneous Linear and Quadratic Equations SS2 · Mathematics

Simultaneous equations are solved together to find values of the unknowns that satisfy all of them. A linear and a quadratic equation together give two pairs of solutions, which are the points where a line meets a curve.

  • Linear pair (two unknowns): solve by elimination (make coefficients equal, then add or subtract) or by substitution. Example: x + y = 5 and x − y = 1 give x = 3, y = 2.
  • Linear and quadratic pair: make one variable the subject of the linear equation, substitute into the quadratic equation, and solve the resulting quadratic equation.
  • Solve the quadratic by factorisation or the formula x = [−b ± √(b² − 4ac)]/2a. Each root of x is then put back into the LINEAR equation to find its matching y.
  • There are usually two solution pairs, written as (x₁, y₁) and (x₂, y₂). They are the coordinates where the line cuts the curve.
  • Discriminant b² − 4ac of the substituted quadratic: > 0 gives two points of intersection, = 0 gives one (line is a tangent), < 0 gives none (no real solutions).
  • Always check each pair in the original equations. Reject nothing unless the problem context (length, age, number of items) forbids a negative or non-integer value.
  • Word problems: define unknowns, form one linear equation and one quadratic (for example from perimeter and area of a rectangle), then solve as above.
Worked example: Solve y = x + 1 and x² + y² = 25. Substitute: x² + (x + 1)² = 25, so 2x² + 2x − 24 = 0, so x² + x − 12 = 0, so (x + 4)(x − 3) = 0. x = 3 gives y = 4; x = −4 gives y = −3. Solutions: (3, 4) and (−4, −3).
Exam trap: Substituting the x values into the quadratic to find y, which produces extra false pairs. Giving only the x values, or squaring (x + 1) wrongly as x² + 1.
Circle Theorems SS2 · Mathematics

Circle theorems relate angles and lengths formed by chords, arcs, tangents and cyclic quadrilaterals. They are used to find unknown angles and to give reasons in geometry questions.

  • Angle at the centre is twice the angle at the circumference subtended by the same arc: ∠AOB = 2∠ACB.
  • Angles in the same segment (subtended by the same arc) are equal. Angle in a semicircle is 90°.
  • Cyclic quadrilateral: opposite angles are supplementary (sum to 180°). An exterior angle equals the interior opposite angle.
  • A tangent is perpendicular to the radius at the point of contact. Two tangents from an external point are equal in length.
  • Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.
  • Angles subtended by equal arcs (or equal chords) are equal; the angle at the circumference is proportional to the arc length, and equal chords are equidistant from the centre.
  • Perpendicular from the centre to a chord bisects the chord. Intersecting chords: AP × PB = CP × PD. Tangent-secant: (tangent length)² = external part × whole secant.
  • In answers, state the reason with each step, e.g. 'angles in the same segment' or 'opposite angles of a cyclic quadrilateral', since marks are given for reasons.
Worked example: ABCD is a cyclic quadrilateral with ∠ABC = 105°. Find ∠ADC. Opposite angles of a cyclic quadrilateral add up to 180°, so ∠ADC = 180° − 105° = 75°.
Exam trap: Using the theorem on a quadrilateral that is not cyclic, mixing up the angle at the centre with the angle at the circumference, and picking the wrong segment in the alternate segment theorem. Leaving out reasons also loses marks.
Sine and Cosine Rules, Bearings and Distances SS2 · Mathematics

The sine and cosine rules solve any triangle, not just right-angled ones, and are used with bearings to find distances and directions in navigation problems.

  • Sine rule: a/sin A = b/sin B = c/sin C. Use it when given two angles and a side, or two sides and an angle opposite one of them.
  • Cosine rule: a² = b² + c² − 2bc cos A, and cos A = (b² + c² − a²)/2bc. Use it when given two sides and the included angle, or all three sides.
  • Area of a triangle = ½ab sin C (two sides and the included angle). Angles in a triangle sum to 180°.
  • Ambiguous case of the sine rule: given two sides and a non-included angle, there may be two triangles because sin θ = sin (180° − θ). Check which angle fits.
  • Bearings are measured clockwise from North, written with three figures, e.g. 045°, 270°. East is 090°, South 180°, West 270°.
  • Back bearing: add or subtract 180°. If the bearing of B from A is 040°, the bearing of A from B is 220°. Use parallel North lines and alternate or co-interior angles.
  • Distance problems (ships, aircraft, towers): draw a clear diagram with North lines, mark the angles from the bearings, split into triangles, and apply the sine or cosine rule.
  • Angles of elevation and depression are measured from the horizontal; the angle of elevation of A from B equals the angle of depression of B from A.
Worked example: A ship sails 12 km on a bearing of 040°, then 16 km on a bearing of 130°. Find its distance from the start and bearing from the start. The turn is 130° − 40° = 90°, so d² = 12² + 16² = 400 and d = 20 km. The angle at the start from the first leg is tan⁻¹(16/12) = 53.1°, so the bearing = 40° + 53.1° = 093.1°.
Exam trap: Measuring bearings anticlockwise or from the wrong North line, writing bearings with fewer than three figures, and using the cosine rule with an angle that is not between the two given sides. Calculators must be in degree mode.
Measures of Location and Dispersion SS2 · Mathematics

Measures of location (mean, median, mode) describe the centre of data, while measures of dispersion (range, variance, standard deviation) describe how spread out it is. They are used on both raw and grouped data in WAEC and JAMB.

  • Mean of raw data = Σx/n. For a frequency table, mean = Σfx/Σf. Median is the middle value of ordered data; mode is the most frequent value.
  • Grouped mean: use class mid-points x, then mean = Σfx/Σf. Alternatively use an assumed mean A: mean = A + Σfd/Σf with d = x − A.
  • Grouped median = L + [(N/2 − F)/f] × c, where L is the lower class boundary of the median class, F the cumulative frequency before it, f its frequency and c the class width.
  • Grouped mode = L + [Δ₁/(Δ₁ + Δ₂)] × c, where L is the lower boundary of the modal class, Δ₁ = f − f₀ (before) and Δ₂ = f − f₁ (after). It can also be read from a histogram.
  • Range = highest value − lowest value. For grouped data the median can be read from the cumulative frequency curve (ogive) at N/2; quartiles at N/4 and 3N/4; inter-quartile range = Q₃ − Q₁.
  • Variance = Σf(x − x̄)²/Σf = Σfx²/Σf − x̄². Standard deviation = √variance. For raw data, variance = Σ(x − x̄)²/n.
  • Standard deviation is in the same units as the data; variance is in squared units. A small standard deviation means the values cluster closely around the mean.
  • Class boundaries for class intervals like 6–10 are 5.5–10.5. Always use boundaries, not limits, for the median and mode formulas.
Worked example: Classes 1–5, 6–10, 11–15, 16–20 have frequencies 3, 7, 6, 4. Mid-points are 3, 8, 13, 18, so Σfx = 9 + 56 + 78 + 72 = 215 and Σf = 20, mean = 215/20 = 10.75. Median class is 6–10: 5.5 + [(10 − 3)/7] × 5 = 10.5. Modal class 6–10: Δ₁ = 4, Δ₂ = 1, mode = 5.5 + (4/5) × 5 = 9.5.
Exam trap: Using class limits instead of boundaries, dividing by the number of classes instead of Σf, and forgetting to square-root the variance. Also forgetting the cumulative frequency F is the total BEFORE the median class.
Differentiation SS3 · Mathematics

Differentiation finds the rate of change (gradient) of a function. It is used for tangent gradients, rates of change and maximum or minimum problems.

  • Limit concept: dy/dx = lim (h→0) [f(x+h) − f(x)]/h, the gradient of the tangent to the curve at x.
  • Power rule: if y = axⁿ then dy/dx = naxⁿ⁻¹; the derivative of a constant is 0, and each term is differentiated separately.
  • Product rule: y = uv gives dy/dx = u(dv/dx) + v(du/dx). Quotient rule: y = u/v gives dy/dx = [v(du/dx) − u(dv/dx)]/v².
  • Function of a function (chain rule): dy/dx = (dy/du) × (du/dx), e.g. y = (3x + 2)⁵ gives dy/dx = 15(3x + 2)⁴.
  • Gradient of a curve at a point = dy/dx at that x. Tangent: y − y₁ = m(x − x₁); normal gradient = −1/m.
  • Stationary (turning) points occur where dy/dx = 0. Solve for x, then substitute into y to get the point.
  • Nature test: d²y/dx² < 0 gives a maximum, d²y/dx² > 0 gives a minimum. Check also that the sign of dy/dx changes.
  • Rates and optimisation: write the quantity as a function of one variable, differentiate, set to 0, e.g. maximum area for a fixed perimeter. Trig: d/dx(sin x) = cos x, d/dx(cos x) = −sin x.
Worked example: For y = 3x³ − 6x² + 2: dy/dx = 9x² − 12x = 3x(3x − 4) = 0, so x = 0 or x = 4/3. d²y/dx² = 18x − 12: at x = 0 it is −12 (maximum, y = 2); at x = 4/3 it is 12 (minimum, y = −14/9).
Exam trap: Forgetting to substitute x back into y for the turning point's y-coordinate, and mixing up the sign test for maximum and minimum. Also dropping the minus sign in the quotient rule numerator order (v·u′ − u·v′).
Integration SS3 · Mathematics

Integration is the reverse of differentiation and is used to find areas under curves. It also recovers a function from its rate of change.

  • Integration is the reverse of differentiation: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C, for n ≠ −1.
  • The constant of integration C must be included in an indefinite integral; it is found using a given point, e.g. the curve passes through (1, 3).
  • Constant multiples and sums: ∫(ax² + bx + c) dx = ax³/3 + bx²/2 + cx + C; integrate each term separately.
  • Definite integral: ∫ₐᵇ f(x) dx = F(b) − F(a). No constant C is needed, because it cancels.
  • Area under a curve y = f(x) between x = a and x = b (above the x-axis) is ∫ₐᵇ y dx, in square units.
  • If the curve is below the x-axis the integral is negative; take the positive (numerical) value for area, or split the region at the roots.
  • Area between a curve and a line = ∫(upper − lower) dx between the intersection points, found by solving the two equations together.
  • Useful trig results: ∫cos x dx = sin x + C and ∫sin x dx = −cos x + C. Differentiation then integration recovers the original function.
Worked example: Find the area under y = x² + 1 from x = 0 to x = 3. Area = ∫₀³ (x² + 1) dx = [x³/3 + x]₀³ = (9 + 3) − 0 = 12 square units.
Exam trap: Omitting + C in indefinite integrals, and wrongly raising the power without dividing by the new power (writing x³ instead of x³/3). Also subtracting limits in the wrong order: it is F(upper) − F(lower).
Matrices and Determinants SS3 · Mathematics

A matrix is a rectangular array of numbers used to organise data and solve simultaneous equations. Determinants and inverses of 2 × 2 matrices are examined every year.

  • Order of a matrix = rows × columns. Addition and subtraction need matrices of the same order and are done element by element.
  • Scalar multiplication: k[a b; c d] = [ka kb; kc kd], every element is multiplied by k.
  • Multiplication: (m × n)(n × p) gives m × p; row of the first times column of the second. In general AB ≠ BA.
  • Determinant of A = [a b; c d] is |A| = ad − bc. If |A| = 0 the matrix is singular and has no inverse.
  • Inverse: A⁻¹ = (1/(ad − bc)) [d −b; −c a]: swap the leading-diagonal elements, change the signs of the other two, divide by the determinant.
  • Identity matrix I = [1 0; 0 1], and AA⁻¹ = A⁻¹A = I.
  • Solving ax + by = p, cx + dy = q: write as A[x; y] = [p; q] so [x; y] = A⁻¹[p; q]. Cramer's method gives the same answer.
  • Transpose swaps rows and columns. Matrices can model data, e.g. prices × quantities to give total cost.
Worked example: Solve 2x + y = 7 and x + 3y = 11. |A| = 2×3 − 1×1 = 5. A⁻¹ = (1/5)[3 −1; −1 2]. [x; y] = (1/5)[3×7 − 11; −7 + 2×11] = (1/5)[10; 15] = [2; 3]. So x = 2, y = 3.
Exam trap: Forgetting to divide by the determinant, or changing the wrong signs when finding the inverse. Also multiplying matrices element by element instead of row by column.
Vectors in a Plane SS3 · Mathematics

A vector has both magnitude and direction, unlike a scalar. Vectors in a plane describe displacement, force and velocity and are handled using components.

  • A vector can be written as a column [x; y], in i, j form (xi + yj), or as a magnitude with a direction (bearing or angle).
  • Magnitude of v = xi + yj is |v| = √(x² + y²). Direction: θ = tan⁻¹(y/x) measured from the x-axis.
  • Position vector OA of point A(x, y) is [x; y]. Vector AB = OB − OA, that is, 'end minus start'.
  • Addition and subtraction are done component by component: (a₁i + b₁j) ± (a₂i + b₂j) = (a₁ ± a₂)i + (b₁ ± b₂)j. Geometrically, use the triangle or parallelogram law.
  • Scalar multiple kv has magnitude |k||v|, is parallel to v, and is reversed in direction if k < 0. Equal vectors have equal components.
  • Dot (scalar) product: a·b = a₁a₂ + b₁b₂ = |a||b|cos θ. The result is a scalar, not a vector.
  • Angle between vectors: cos θ = (a·b)/(|a||b|). If a·b = 0 the vectors are perpendicular; i·i = j·j = 1 and i·j = 0.
  • Unit vector in the direction of v = v/|v|. Resultant of forces or displacements is the vector sum.
Worked example: For a = 3i + 4j and b = 2i − j: a·b = 6 − 4 = 2; |a| = 5; |b| = √5. cos θ = 2/(5√5) = 0.1789, so θ = 79.7°.
Exam trap: Using start minus end instead of end minus start for AB, and treating the dot product as a vector. Also adding magnitudes instead of components when finding a resultant.
Probability SS3 · Mathematics

Probability measures how likely an event is, as a number from 0 to 1. It is used in games of chance, selection and sampling problems.

  • P(E) = number of favourable outcomes / total number of outcomes in the sample space, and 0 ≤ P(E) ≤ 1.
  • Sample space S is the set of all possible outcomes, e.g. two coins: {HH, HT, TH, TT}; two dice give 36 outcomes.
  • P(E) = 0 for an impossible event, P(E) = 1 for a certain event, and P(not E) = 1 − P(E).
  • Mutually exclusive events cannot happen together: P(A or B) = P(A) + P(B). In general P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
  • Independent events (one does not affect the other, e.g. with replacement): P(A and B) = P(A) × P(B).
  • Dependent events (without replacement): P(A and B) = P(A) × P(B given A), with the second probability using the reduced total.
  • Tree diagrams: multiply along the branches, add the results of the different routes; all branches from a point sum to 1.
  • Sum of probabilities of all outcomes = 1. Expected frequency = probability × number of trials.
Worked example: A bag has 4 red and 6 blue balls. Two are drawn without replacement. P(both red) = 4/10 × 3/9 = 12/90 = 2/15.
Exam trap: Using the original total on the second draw when there is no replacement, and adding instead of multiplying for 'and' events. Also giving probabilities greater than 1 or forgetting to simplify fractions.
Concepts and States of Matter SS1 · Physics

Matter is anything that has mass and occupies space, and it exists mainly as solid, liquid or gas. Understanding the molecular structure explains properties such as shape, volume, compressibility and changes of state.

  • Matter is anything that has mass and occupies space. The three common states are solid, liquid and gas; plasma is sometimes listed as a fourth state.
  • Solids: molecules are closely packed in fixed positions, held by strong forces, and only vibrate. They have definite shape and volume and are almost incompressible.
  • Liquids: molecules are close but free to slide past each other, with weaker forces than solids. They have definite volume but take the shape of the container.
  • Gases: molecules are far apart, move randomly at high speed, with very weak forces. They have no fixed shape or volume, fill the container and are easily compressed.
  • Molecular (kinetic) theory: matter is made of tiny particles in constant motion; heating increases their kinetic energy and can change the state (melting, boiling, evaporation, condensation, freezing, sublimation).
  • Evidence for molecular motion: Brownian motion (random zig-zag motion of smoke or pollen particles seen under a microscope) and diffusion (a gas or dye spreading through another substance).
  • Intermolecular forces: cohesion is attraction between molecules of the same substance; adhesion is attraction between molecules of different substances (e.g. water wetting glass).
  • Evaporation occurs at any temperature from the surface only and causes cooling; boiling occurs at a fixed temperature throughout the liquid. Density = mass/volume (kg/m³).
Worked example: Why can a gas be compressed but a solid cannot? In a gas the molecules are far apart with large empty spaces, so they can be pushed closer; in a solid they are already tightly packed with almost no spaces.
Exam trap: Confusing Brownian motion (evidence of molecular motion) with diffusion, and saying evaporation and boiling are the same. Remember that liquids have definite volume but no definite shape.
Fundamental and Derived Quantities, Units, Instruments SS1 · Physics

Physical quantities are measurable properties, classified as fundamental or derived, each with an SI unit. Accurate measurement with instruments such as the vernier calipers and micrometer screw gauge is basic to all practical physics.

  • Fundamental quantities: length (metre, m), mass (kilogram, kg), time (second, s), electric current (ampere, A), temperature (kelvin, K), amount of substance (mole, mol), luminous intensity (candela, cd).
  • Derived quantities are combinations of fundamental ones: area (m²), volume (m³), density (kg/m³), speed (m/s), acceleration (m/s²), force (N = kg m/s²), work (J = N m), power (W = J/s), pressure (Pa = N/m²).
  • Dimensions express a quantity in terms of M, L, T: speed = LT⁻¹, acceleration = LT⁻², force = MLT⁻², energy = ML²T⁻², density = ML⁻³, pressure = ML⁻¹T⁻².
  • Dimensional analysis checks whether an equation is homogeneous: both sides must have the same dimensions. Quantities added or subtracted must have the same dimensions and units.
  • Vernier calipers: reading = main scale reading + (coinciding vernier division × least count). Least count = 0.01 cm (0.1 mm) for a 10-division vernier on a 1 mm main scale; check and subtract zero error.
  • Micrometer screw gauge: reading = sleeve (main scale) reading + thimble reading × 0.01 mm. Pitch is usually 0.5 mm with 50 thimble divisions, giving least count 0.01 mm. Use the ratchet to avoid over-tightening.
  • Zero error: if the instrument does not read zero when closed, the correction is subtracted from the observed reading (a positive zero error is subtracted; a negative one is added). Other tools: metre rule, stopwatch, beam balance, measuring cylinder.
  • Unit prefixes: kilo (10³), mega (10⁶), giga (10⁹), centi (10⁻²), milli (10⁻³), micro (10⁻⁶), nano (10⁻⁹). Convert, e.g., 1 g/cm³ = 1000 kg/m³.
Worked example: A vernier calipers has main scale reading 2.3 cm and the 6th vernier division coincides with a main scale division; least count = 0.01 cm. Reading = 2.3 + 6 × 0.01 = 2.36 cm. If zero error is +0.02 cm, the true length = 2.36 − 0.02 = 2.34 cm.
Exam trap: Treating a derived unit such as the newton as fundamental, and forgetting to correct for zero error or to convert units (cm³ to m³). Mass (kg) is fundamental but weight (N) is a derived force.
Position, Distance and Displacement SS1 · Physics

Position, distance and displacement describe where an object is and how far it has moved. They introduce the key difference between scalar and vector quantities used throughout mechanics.

  • Position is the location of a point relative to a reference point (origin), described using coordinates or a distance and direction from the origin.
  • Distance is the total length of the path travelled; it is a scalar (magnitude only), measured in metres (m), and is never negative.
  • Displacement is the shortest straight-line distance from the starting point to the final point in a stated direction; it is a vector, measured in metres (m), and can be positive, negative or zero.
  • Scalars have magnitude only: distance, speed, mass, time, energy, temperature. Vectors have magnitude and direction: displacement, velocity, acceleration, force, momentum.
  • If an object returns to its starting point, its displacement is zero although the distance travelled is not. Distance ≥ magnitude of displacement always.
  • Vectors are added by the triangle or parallelogram law; for perpendicular vectors the resultant R = √(a² + b²) at angle θ = tan⁻¹(b/a). Vectors in the same line add or subtract.
  • Position-time graph: the slope gives velocity. Displacement on a straight line is positive in the chosen positive direction and negative in the opposite direction.
  • A vector can be resolved into components: horizontal = R cos θ, vertical = R sin θ.
Worked example: A boy walks 6 m east then 8 m north. Distance = 6 + 8 = 14 m. Displacement = √(6² + 8²) = 10 m at tan⁻¹(8/6) = 53.1° north of east.
Exam trap: Quoting displacement as the total path length, or giving a displacement without direction. Going round a complete circuit or track gives zero displacement.
Speed, Velocity, Acceleration and Equations of Motion SS1 · Physics

This topic covers how fast and in what direction objects move and how their motion changes. The equations of motion and motion graphs are among the most examined calculation areas in WAEC and JAMB.

  • Speed = distance/time (m/s, scalar). Velocity = displacement/time (m/s, vector). Average speed = total distance ÷ total time. Convert: 1 km/h = 5/18 m/s.
  • Acceleration = change in velocity/time = (v − u)/t, in m/s². Negative acceleration (retardation or deceleration) means the speed is decreasing.
  • Equations of motion (constant acceleration): v = u + at; s = ut + ½at²; v² = u² + 2as; s = ½(u + v)t. Here u is initial velocity, v final velocity, s displacement, t time.
  • Free fall: use a = g ≈ 10 m/s² (or 9.8 m/s²) downward; for an object thrown upward, a = −g and velocity is zero at maximum height, with time up equal to time down.
  • Velocity-time graph: slope = acceleration; area under the graph = displacement (distance). A horizontal line means constant velocity; a sloping straight line means uniform acceleration.
  • Distance-time graph: slope = speed. A horizontal line means the body is at rest; a straight sloping line means uniform speed; a curve means changing speed.
  • Uniform velocity means constant speed in a straight line. Uniform circular motion has constant speed but changing velocity, so it is accelerated.
  • Area under a v-t graph for a trapezium = ½(sum of parallel sides) × width; use this to find the distance travelled in a given interval.
Worked example: A car starts with u = 10 m/s and accelerates at 2 m/s² for 5 s. v = u + at = 10 + 2(5) = 20 m/s. s = ut + ½at² = 10(5) + ½(2)(25) = 50 + 25 = 75 m.
Exam trap: Using the equations when acceleration is not constant, forgetting to convert km/h to m/s, and mixing up the slope and area of a graph. Take care with signs for upward and downward motion.
Friction SS1 · Physics

Friction is the force that opposes relative motion between surfaces in contact. It is useful (walking, braking) and harmful (wear, heat), so it must be understood and controlled.

  • Friction is a force that opposes the motion, or tendency of motion, between two surfaces in contact. It acts parallel to the surfaces, opposite to the direction of motion.
  • Static friction acts when there is no motion and adjusts up to a maximum called limiting friction. Dynamic (kinetic or sliding) friction acts when surfaces slide and is slightly less than limiting friction.
  • Coefficient of friction μ = F/R, where F is the frictional force and R the normal reaction (no unit). On a horizontal surface R = mg, so F = μmg.
  • Laws of friction: F is proportional to R; F is independent of the area of contact (for the same normal reaction); kinetic friction is nearly independent of speed; F depends on the nature of the surfaces.
  • Advantages: walking, gripping, braking, writing, nails and screws holding, lighting a match. Disadvantages: wear and tear, heat loss, reduced efficiency of machines, energy wastage.
  • Methods of reducing friction: lubrication with oil or grease, using ball or roller bearings, polishing surfaces, streamlining (against fluid friction), using wheels, air cushions.
  • Friction on an inclined plane: normal reaction R = mg cos θ, and the body just slides when tan θ = μ (θ is the angle of friction).
  • Fluid friction (viscous drag) acts on objects moving through liquids and gases; it increases with speed and gives rise to terminal velocity.
Worked example: A 20 kg block on a horizontal floor just begins to slide when a horizontal force of 80 N is applied (g = 10 m/s²). R = mg = 200 N, so μ = F/R = 80/200 = 0.4.
Exam trap: Writing that friction depends on the area of contact, or using R = mg on an incline (it is mg cos θ). Also, μ has no unit, and friction is not always a hindrance.
Newton's Laws of Motion SS1 · Physics

Newton's three laws explain how forces change the motion of objects. They are the foundation of dynamics and are used to solve many force and acceleration problems.

  • First law (inertia): a body remains at rest or in uniform motion in a straight line unless acted on by a resultant external force. Inertia is the reluctance of a body to change its state; it depends on mass.
  • Examples of inertia: passengers jerk forward when a bus brakes suddenly; a coin stays when the card under it is flicked away; seat belts are used for safety.
  • Second law: the rate of change of momentum is proportional to the applied resultant force and takes place in the direction of the force. For constant mass, F = ma, with F in newtons (N), m in kg, a in m/s².
  • One newton is the force that gives a mass of 1 kg an acceleration of 1 m/s². Weight W = mg (a force in N); mass is constant but weight varies with location.
  • Third law: to every action there is an equal and opposite reaction. The two forces act on different bodies, so they do not cancel each other.
  • Examples of the third law: recoil of a gun, rocket and jet propulsion, swimming (pushing water backwards), walking, a book resting on a table (weight and normal reaction).
  • Resultant force: when forces balance, the resultant is zero and the body is in equilibrium or moves with constant velocity. Net force = ma for unbalanced forces.
  • Linear momentum p = mv (kg m/s); the second law can be written F = (mv − mu)/t. In a lift, apparent weight = m(g + a) going up with acceleration and m(g − a) going down with acceleration a.
Worked example: A resultant force of 6 N acts on a 3 kg trolley. a = F/m = 6/3 = 2 m/s². After 4 s from rest, v = at = 2 × 4 = 8 m/s.
Exam trap: Thinking action and reaction cancel (they act on different bodies), confusing mass with weight, and forgetting to use the resultant force (e.g. applied force minus friction) in F = ma.
Work, Energy, Power and Simple Machines SS1 · Physics

Work, energy and power describe how forces transfer energy and how quickly it is done. Simple machines show how a small effort can move a large load.

  • Work done = force × distance moved in the direction of the force, W = Fs (joule, J). If the force is at angle θ to the motion, W = Fs cos θ. No movement means no work.
  • Energy is the capacity to do work (J). Kinetic energy KE = ½mv²; gravitational potential energy PE = mgh. Other forms: chemical, heat, electrical, nuclear, sound, light.
  • Law of conservation of energy: energy cannot be created or destroyed, only converted from one form to another. For a falling body, loss in PE = gain in KE, so v = √(2gh).
  • Power = work done/time = energy/time (watt, W). P = Fv for a body moving at constant velocity. 1 kW = 1000 W; 1 horsepower ≈ 746 W.
  • Simple machines: lever, pulley, inclined plane, wheel and axle, screw, gear, wedge. They make work easier by changing the size or direction of a force.
  • Mechanical advantage MA = load/effort (no unit). Velocity ratio VR = distance moved by effort/distance moved by load. Efficiency = (MA/VR) × 100% = (work output/work input) × 100%.
  • For an inclined plane VR = 1/sin θ = length/height; for a wheel and axle VR = R/r; for a pulley system VR = number of pulleys supporting the load (number of rope segments).
  • No machine is 100% efficient because of friction and the weight of moving parts; efficiency is always less than 100%, and a machine cannot give out more work than is put in.
Worked example: A boy of mass 50 kg climbs 4 m in 20 s (g = 10 m/s²). Work = mgh = 50 × 10 × 4 = 2000 J; power = 2000/20 = 100 W. A machine with load 400 N, effort 100 N and VR = 8 has MA = 4 and efficiency = 4/8 × 100 = 50%.
Exam trap: Using mass instead of weight for force, forgetting that efficiency is MA/VR (not VR/MA), and mixing up the units of energy (J) and power (W). Also, carrying a load horizontally does no work against gravity.
Heat and Temperature SS1 · Physics

Heat is energy in transit while temperature measures how hot a body is. Temperature scales and thermometers allow us to measure and compare thermal states accurately.

  • Heat is a form of energy transferred because of a temperature difference; its unit is the joule (J). Temperature is the degree of hotness or coldness of a body, measured in kelvin (K) or °C.
  • Temperature measures the average kinetic energy of molecules; heat depends on mass, temperature change and material. Heat flows from a hot body to a cold body.
  • Fixed points: the lower fixed point (ice point) is 0 °C, 32 °F, 273 K; the upper fixed point (steam point) is 100 °C, 212 °F, 373 K (at standard pressure). Fundamental interval = 100 °C.
  • Conversions: T(K) = θ(°C) + 273; °F = (9/5)°C + 32. For any thermometer, unknown temperature θ = [(X_θ − X₀)/(X₁₀₀ − X₀)] × 100 °C.
  • Liquid-in-glass thermometers: mercury (good conductor, expands uniformly, opaque, freezes at −39 °C) and alcohol (low freezing point, needs dye). A narrow bore and thin bulb give sensitivity.
  • Clinical thermometer: mercury type with a constriction that keeps the mercury level; range about 35–42 °C. Six's maximum and minimum thermometer records extremes of temperature.
  • Other thermometers: constant-volume gas thermometer (accurate), resistance thermometer (platinum), thermocouple (wide range, fast), pyrometer (very high temperatures, no contact).
  • Sensitivity, range, linearity and accuracy depend on the thermometric property used (length of a liquid column, resistance, e.m.f., pressure of a gas).
Worked example: Convert 50 °C to kelvin and °F. K = 50 + 273 = 323 K; °F = (9/5)(50) + 32 = 90 + 32 = 122 °F.
Exam trap: Confusing heat with temperature, adding 273 incorrectly (use 273 K or 273.15 K as instructed), and forgetting the 32 offset in Fahrenheit conversions. A temperature change of 1 K equals 1 °C change.
Light Propagation SS1 · Physics

Light is a form of energy that travels in straight lines and enables us to see. Its rectilinear propagation explains shadows, eclipses and the pinhole camera.

  • Light is an electromagnetic wave that travels in vacuum at 3.0 × 10⁸ m/s. Luminous bodies (sun, lamp) produce their own light; non-luminous bodies (moon) are seen by reflected light.
  • A ray is a single line showing the direction of travel of light; a beam is a collection of rays and may be parallel, convergent or divergent.
  • Media: transparent (glass, clear water) transmit light fully; translucent (frosted glass, oiled paper) transmit partly; opaque (wood, metal) block light.
  • Rectilinear propagation: light travels in straight lines in a uniform medium. Evidence: sharp shadows, pinhole camera images, and inability to see round corners.
  • Shadows: a point source gives only an umbra (total shadow); an extended source gives an umbra and a penumbra (partial shadow).
  • Solar eclipse: the moon lies between the sun and the earth, so the moon's shadow falls on the earth. Lunar eclipse: the earth lies between the sun and the moon, so the moon enters the earth's shadow.
  • Pinhole camera: forms an inverted real image on the screen; image size/object size = image distance/object distance. Moving the screen away enlarges the image but makes it dimmer.
  • Speed of light is greater than sound; lightning is seen before thunder is heard. Distance = speed × time can be used to estimate distance of a storm.
Worked example: A 2 m tall object is 10 m from a pinhole and the screen is 0.5 m behind the pinhole. Image height = 2 × (0.5/10) = 0.1 m = 10 cm, inverted.
Exam trap: Saying the image in a pinhole camera is upright or virtual, and mixing up solar and lunar eclipses. Remember an umbra is total shadow and a penumbra partial shadow.
Reflection of Light and Mirrors SS1 · Physics

Reflection is the bouncing of light from a surface, and mirrors use it to form images. Plane and curved mirrors have many everyday uses and the mirror formula is a common calculation.

  • Laws of reflection: the incident ray, reflected ray and normal at the point of incidence lie in the same plane; the angle of incidence i equals the angle of reflection r.
  • Plane mirror image: virtual, upright, laterally inverted, same size as the object, and as far behind the mirror as the object is in front. A rotating mirror by θ turns the reflected ray by 2θ.
  • Two inclined plane mirrors at angle θ form n = (360°/θ) − 1 images; parallel mirrors form an infinite number. A periscope uses two parallel plane mirrors at 45°.
  • Curved mirror terms: pole P, centre of curvature C, principal focus F, principal axis, radius of curvature r = 2f. A concave mirror converges light (real focus); a convex mirror diverges light (virtual focus).
  • Mirror formula: 1/f = 1/u + 1/v; magnification m = v/u = image height/object height. Use the real-is-positive convention: concave f positive, convex f negative; virtual image distances negative.
  • Concave mirror images: object beyond C gives real, inverted, diminished image; at C real, inverted, same size; between C and F real, inverted, magnified; at F image at infinity; inside F virtual, upright, magnified.
  • Convex mirror always forms a virtual, upright, diminished image. Uses: concave in shaving mirrors, car headlamps, dentist mirrors; convex as driving mirrors for a wider field of view.
  • Ray diagram rules: a ray parallel to the axis reflects through F; a ray through F reflects parallel to the axis; a ray through C returns along itself.
Worked example: An object is 30 cm from a concave mirror of focal length 10 cm. 1/v = 1/10 − 1/30 = 2/30, so v = 15 cm (real). m = v/u = 15/30 = 0.5, so the image is real, inverted and diminished.
Exam trap: Using wrong sign conventions in the mirror formula, and saying plane mirror images are real. Also, r = 2f for spherical mirrors; do not confuse C with F.
Projectiles SS2 · Physics

A projectile is a body thrown into the air that moves under gravity alone. The topic lets you predict how long it stays up, how high it goes and how far it lands.

  • A projectile moves under gravity only (air resistance ignored): horizontal velocity u cosθ stays constant, while vertical motion has constant acceleration g = 10 m/s² downward.
  • Launch at speed u and angle θ: horizontal component ux = u cosθ, vertical component uy = u sinθ. Resolve first, then apply equations of motion separately to each direction.
  • Time of flight (level ground): T = 2u sinθ/g. The time to reach the top is T/2 = u sinθ/g because vertical velocity is zero at maximum height.
  • Maximum height: H = u² sin²θ/(2g). At the top the vertical velocity is zero but the horizontal velocity u cosθ remains, so the speed there is u cosθ.
  • Range on level ground: R = u² sin2θ/g = u cosθ × T. Maximum range occurs at θ = 45°; angles θ and (90° − θ) give the same range.
  • For a body projected horizontally from height h: time of fall t = √(2h/g), independent of its horizontal speed; range = u t. Path is a parabola.
  • Compare: at 45° range is largest; at 90° range is zero and height is largest (H = u²/2g). Doubling u multiplies range and height by 4 and time by 2.
  • Speed on landing (level ground) equals launch speed u, and the landing angle equals the launch angle, directed below the horizontal.
Worked example: A ball is kicked at 40 m/s at 30° to the horizontal (g = 10 m/s²). T = 2 × 40 × sin30°/10 = 2 × 40 × 0.5/10 = 4 s. H = (40 × 0.5)²/(2 × 10) = 400/20 = 20 m. R = 40² × sin60°/10 = 1600 × 0.866/10 ≈ 138.6 m.
Exam trap: Using u instead of its components, and forgetting that horizontal velocity is constant. Students also write sin²θ as sin θ², use 2θ wrongly in H, or forget that 45° gives maximum range only on level ground.
Linear Momentum and Impulse SS2 · Physics

Linear momentum is the product of mass and velocity, and it is conserved in collisions when no external force acts. It explains collisions, recoil of guns and rocket propulsion.

  • Linear momentum p = mv, a vector measured in kg m/s (or N s). Its direction is that of the velocity, so give opposite directions opposite signs.
  • Newton's second law in momentum form: F = (mv − mu)/t = change in momentum/time. Force is the rate of change of momentum.
  • Impulse = Ft = change in momentum (mv − mu), measured in N s. Area under a force-time graph equals impulse; a longer contact time reduces the force (airbags, cricket catching).
  • Principle of conservation of linear momentum: in an isolated system the total momentum before collision equals the total momentum after: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
  • Elastic collision: both momentum and kinetic energy are conserved, and bodies separate. Inelastic collision: momentum is conserved but some kinetic energy becomes heat and sound.
  • Perfectly inelastic collision: bodies stick together and move with common velocity v = (m₁u₁ + m₂u₂)/(m₁ + m₂); the greatest kinetic energy loss occurs here.
  • Recoil: gun of mass M and bullet of mass m start at rest, so 0 = mv + MV, giving recoil velocity V = −mv/M. The heavier gun recoils slowly.
  • Rockets and jets work by expelling gas backwards at high speed; the gas gains backward momentum, so the rocket gains equal forward momentum.
Worked example: A 2 kg trolley moving at 6 m/s hits a stationary 4 kg trolley and they stick. Momentum before = 2 × 6 = 12 kg m/s; common velocity v = 12/(2 + 4) = 2 m/s. KE before = ½ × 2 × 6² = 36 J; after = ½ × 6 × 2² = 12 J; loss = 24 J.
Exam trap: Forgetting direction signs when bodies move towards each other, and claiming kinetic energy is conserved in every collision. Also confusing impulse (N s) with force (N), and mixing grams with kilograms.
Equilibrium of Forces, Moments and Couples SS2 · Physics

A body is in equilibrium when the forces and turning effects on it balance. The topic covers parallel forces, centre of gravity, moments and couples used in beams, levers and balances.

  • Moment of a force = force × perpendicular distance from the pivot to the line of action of the force, unit N m. It may be clockwise or anticlockwise.
  • Conditions for equilibrium of a rigid body: (1) the vector sum of forces is zero; (2) the sum of moments about any point is zero. Principle of moments: sum of clockwise moments = sum of anticlockwise moments.
  • Parallel forces: the resultant of like parallel forces is their sum, acting between them; for unlike parallel forces it is their difference. Reactions at supports balance weight and loads.
  • Centre of gravity is the point where the whole weight of a body appears to act. For a uniform regular body it is at the geometric centre (midpoint of a uniform rule at the 50 cm mark).
  • Types of equilibrium: stable (returns after small tilt, c.g. rises), unstable (topples, c.g. falls) and neutral (stays in new position, c.g. height unchanged, e.g. a ball on a table).
  • Stability increases with a low centre of gravity and a wide base; a vertical line from the c.g. must fall within the base, otherwise the body topples.
  • A couple is two equal, opposite, parallel forces with different lines of action. Moment of a couple = one force × perpendicular distance between the forces; it causes rotation only, no translation.
  • Examples of couples: turning a steering wheel, tap or key. Resultant force of a couple is zero, so no single force can balance it; only an equal opposite couple can.
Worked example: A uniform metre rule of weight 4 N is pivoted at the 40 cm mark. What weight W at the 0 cm mark balances it? The rule's weight acts at 50 cm, 10 cm from the pivot (anticlockwise side opposite W): 4 × 10 = W × 40, so W = 40/40 = 1 N.
Exam trap: Measuring distance along the rule rather than perpendicular to the force, and forgetting the weight of a uniform rule acting at its centre. Learners also say a couple has a resultant force.
Elastic Properties of Solids SS2 · Physics

Elasticity is the ability of a solid to regain its shape after a deforming force is removed. It helps engineers choose materials for springs, cables and bridges.

  • Elasticity: property of a material to return to its original size and shape when the deforming force is removed. Plastic materials stay deformed.
  • Hooke's law: within the elastic limit, extension e is directly proportional to the applied force F, so F = ke. k is the force constant (N/m); the graph of F against e is a straight line through the origin.
  • Elastic limit: the maximum force beyond which a material does not return to its original length. Beyond it the material is permanently stretched; yield point and breaking point follow.
  • Stress = force/cross-sectional area (N/m² or Pa). Strain = extension/original length (no unit, a ratio).
  • Young's modulus E = stress/strain = FL/(Ae), unit N/m² (Pa). It is a property of the material (steel ≈ 2 × 10¹¹ Pa), valid within the limit of proportionality.
  • Energy stored in a stretched wire or spring = ½Fe = ½ke² (area under the F–e graph), in joules.
  • Springs in series: extension adds, so 1/k = 1/k₁ + 1/k₂. Springs in parallel: force constants add, k = k₁ + k₂.
  • A thicker, shorter wire of the same material extends less under the same load; E does not depend on the wire's dimensions.
Worked example: A steel wire 2 m long, cross-sectional area 1 × 10⁻⁶ m², extends by 1 mm under a 100 N load. Stress = 100/10⁻⁶ = 1 × 10⁸ Pa; strain = 0.001/2 = 5 × 10⁻⁴; E = 10⁸/(5 × 10⁻⁴) = 2 × 10¹¹ Pa.
Exam trap: Forgetting to convert mm or cm² to m and m², and treating strain as having a unit. Hooke's law is also wrongly applied beyond the elastic limit.
Thermal Expansion SS2 · Physics

Thermal expansion is the increase in size of a substance when its temperature rises. It matters in railway tracks, bridges, bimetallic strips and thermometers.

  • Solids, liquids and gases expand when heated because particles vibrate or move more vigorously and, on average, move further apart. Gases expand most, solids least.
  • Linear expansivity α = ΔL/(L₁ΔT), unit per kelvin (K⁻¹) or °C⁻¹. So ΔL = αL₁Δθ and the new length L₂ = L₁(1 + αΔθ).
  • Area expansivity β = ΔA/(A₁Δθ) and cubic (volume) expansivity γ = ΔV/(V₁Δθ). For the same solid, β ≈ 2α and γ ≈ 3α.
  • Expansion of a hole: a hole in a metal plate expands as if it were made of the same metal, so the hole gets larger when heated.
  • Applications: gaps between rail tracks and bridge sections, rollers under bridges, bimetallic strips in thermostats and fire alarms (brass expands more than iron), riveting, fitting metal tyres.
  • Disadvantages and precautions: sagging power cables in hot weather, cracking of thick glass when heated suddenly (Pyrex has low α), loosening of joints.
  • Liquids have apparent and real expansion: real expansivity = apparent expansivity + expansivity of container. Water is anomalous: it contracts from 0°C to 4°C, has maximum density at 4°C.
  • Bimetallic strip bends towards the metal with the smaller expansivity, because the other metal is longer when heated.
Worked example: A steel rail 5 m long has α = 1.2 × 10⁻⁵ K⁻¹. Its temperature rises from 20°C to 60°C. ΔL = αL₁Δθ = 1.2 × 10⁻⁵ × 5 × 40 = 2.4 × 10⁻³ m = 2.4 mm.
Exam trap: Using final length instead of original length, or mixing β, γ with α (β = 2α, γ = 3α). Candidates also forget that water is densest at 4°C, not at 0°C.
Gas Laws SS2 · Physics

Gas laws relate the pressure, volume and temperature of a fixed mass of gas. They are used in tyre pressure, balloons, pumps and weather problems.

  • Boyle's law: for a fixed mass of gas at constant temperature, pressure is inversely proportional to volume: PV = constant, so P₁V₁ = P₂V₂. The P–V graph is a hyperbola; P against 1/V is a straight line.
  • Charles' law: at constant pressure, volume is directly proportional to absolute temperature: V/T = constant, V₁/T₁ = V₂/T₂. The V–T graph extrapolates to 0 K (−273°C).
  • Pressure law: at constant volume, pressure is directly proportional to absolute temperature: P/T = constant, P₁/T₁ = P₂/T₂.
  • Absolute (Kelvin) temperature: T(K) = θ(°C) + 273. Always convert to kelvin before using Charles', pressure or general gas laws.
  • General (ideal) gas law: P₁V₁/T₁ = P₂V₂/T₂. Ideal gas equation PV = nRT, with R = 8.31 J mol⁻¹ K⁻¹; n is the number of moles.
  • Standard temperature and pressure (s.t.p.): 273 K (0°C) and 760 mmHg (1.01 × 10⁵ Pa). Pressure units: 1 atm = 760 mmHg = 101 325 Pa.
  • Kinetic theory explanation: pressure comes from molecular collisions with the walls; higher temperature gives greater average KE, so more frequent and forceful collisions.
  • Real gases obey the laws closely at low pressure and high temperature; they deviate near liquefaction because of intermolecular forces.
Worked example: A gas of volume 600 cm³ at 27°C and 100 kPa is heated to 87°C at 200 kPa. T₁ = 300 K, T₂ = 360 K. V₂ = P₁V₁T₂/(P₂T₁) = (100 × 600 × 360)/(200 × 300) = 360 cm³.
Exam trap: Using Celsius temperatures in the gas law, which gives wrong answers. Students also apply Boyle's law when temperature changes, or invert the ratio of pressures.
Heat Transfer, Specific Heat and Latent Heat SS2 · Physics

Heat is energy that flows from a hot body to a cold one, and different materials store and change it differently. This topic explains cooking, insulation, cooling and changes of state.

  • Conduction: heat transfer through a solid by vibrating particles and free electrons, without movement of the material. Metals are good conductors; air, wool and wood are poor conductors (insulators).
  • Convection: heat transfer in fluids (liquids and gases) by movement of warmer, less dense fluid upward and cooler fluid downward. Examples: sea breeze, boiling water, ventilation.
  • Radiation: transfer by electromagnetic (infrared) waves; needs no medium and travels through a vacuum at 3 × 10⁸ m/s. Dull black surfaces absorb and emit best; shiny surfaces reflect.
  • Vacuum flask: vacuum stops conduction and convection, silvered walls reduce radiation, and the cork or plastic stopper reduces conduction.
  • Heat capacity C = Q/Δθ (J/K). Specific heat capacity c = Q/(mΔθ), unit J kg⁻¹ K⁻¹; so Q = mcΔθ. Water has a high c (4200 J kg⁻¹ K⁻¹).
  • Method of mixtures: heat lost by hot body = heat gained by cold body (ignoring losses). Electrical method: IVt = mcΔθ.
  • Specific latent heat L = Q/m (J/kg): energy to change state without temperature change. Fusion for ice ≈ 3.34 × 10⁵ J/kg; vaporization of water ≈ 2.26 × 10⁶ J/kg.
  • Cooling by evaporation: the fastest molecules leave the liquid, so the remaining liquid cools. Evaporation occurs at any temperature; boiling happens at the boiling point throughout the liquid.
Worked example: How much heat changes 0.5 kg of ice at 0°C to water at 20°C? (L = 3.34 × 10⁵ J/kg, c = 4200 J kg⁻¹ K⁻¹). Melting: 0.5 × 3.34 × 10⁵ = 167 000 J. Warming: 0.5 × 4200 × 20 = 42 000 J. Total = 209 000 J.
Exam trap: Forgetting the latent heat term when a change of state occurs, or using Q = mcΔθ during melting. Candidates also mix up heat capacity (J/K) with specific heat capacity (J kg⁻¹ K⁻¹).
Wave Properties and the Wave Equation SS2 · Physics

A wave transfers energy from one place to another without transferring matter. The wave equation links speed, frequency and wavelength for all waves.

  • A wave is a disturbance that transfers energy without net movement of the medium. Mechanical waves need a medium; electromagnetic waves do not.
  • Transverse waves: particles vibrate perpendicular to the direction of travel (light, water, radio waves, string). Longitudinal waves: vibration parallel to travel, with compressions and rarefactions (sound).
  • Terms: amplitude (maximum displacement, m), wavelength λ (distance between successive crests, m), period T (s), frequency f (Hz = cycles per second), with f = 1/T.
  • Wave equation: v = fλ, where v is speed in m/s. For a given medium v is constant, so increasing f decreases λ. Frequency is set by the source and does not change on entering a new medium.
  • Two points are in phase if they are a whole number of wavelengths apart; a phase difference of 180° (π rad) corresponds to half a wavelength.
  • Wave behaviours: reflection (angle i = angle r), refraction, diffraction, interference. Superposition of two waves gives constructive (in phase) or destructive (out of phase) interference.
  • Stationary waves form when two identical waves travel in opposite directions: nodes (zero amplitude) and antinodes (maximum amplitude) are separated by λ/2.
  • Ripple tank is used to study water waves; a stroboscope or slow-motion makes the pattern appear still. Speed of waves in water depends on depth.
Worked example: A source vibrates at 50 Hz and produces waves of wavelength 0.8 m. v = fλ = 50 × 0.8 = 40 m/s; period T = 1/f = 1/50 = 0.02 s.
Exam trap: Believing that frequency changes when a wave enters a new medium (it does not; speed and wavelength change). Learners also confuse transverse with longitudinal waves and read wavelength from a graph of displacement against time (that gives period).
Refraction and Diffraction SS2 · Physics

Refraction is the bending of a wave as it changes speed on entering another medium, and diffraction is its spreading around obstacles or through gaps. They explain lenses, optical fibres and why we hear around corners.

  • Refraction: change in direction of a wave when its speed changes on passing from one medium to another. Frequency stays constant; speed and wavelength change.
  • Light going from air into glass bends towards the normal; from glass into air it bends away from the normal. Along the normal (i = 0) there is no bending.
  • Snell's law: sin i/sin r = constant = refractive index n of the second medium relative to the first. Absolute n = c/v = (speed in vacuum)/(speed in medium), no unit; n of glass ≈ 1.5, water ≈ 1.33.
  • Real and apparent depth: n = real depth/apparent depth. An object under water looks shallower than it really is; a pool appears less deep.
  • Total internal reflection needs light going from a denser to a less dense medium with the angle of incidence greater than the critical angle C, where sin C = 1/n.
  • Applications of total internal reflection: optical fibres (communications and endoscopes), prism periscopes and binoculars, reflectors on roads, and mirages.
  • Diffraction: spreading of waves around obstacles or through gaps. It is greatest when the gap size is about the same as the wavelength; long wavelengths (radio, sound) diffract more than light.
  • Dispersion: splitting of white light into colours by a prism, because different colours have different speeds and refractive indices (violet bends most, red least).
Worked example: Light enters glass of refractive index 1.5 from air at an incidence of 45°. sin r = sin45°/1.5 = 0.7071/1.5 = 0.4714, so r ≈ 28.1°. The critical angle for this glass is C = sin⁻¹(1/1.5) ≈ 41.8°.
Exam trap: Measuring angles from the surface instead of from the normal, and using sin C = n instead of 1/n. Total internal reflection does not occur when light goes from a less dense to a denser medium.
Sound Waves and Resonance SS2 · Physics

Sound is a longitudinal mechanical wave produced by vibrating objects. Understanding its speed, pitch and resonance explains musical instruments and echoes.

  • Sound is produced by vibrating bodies and travels as longitudinal waves of compressions and rarefactions. It needs a material medium and cannot travel through a vacuum.
  • Speed of sound: about 330–340 m/s in air at room temperature, faster in liquids (~1500 m/s in water) and fastest in solids (~5000 m/s in steel). It rises with temperature.
  • Echo: reflection of sound; speed v = 2d/t, where d is the distance to the reflecting surface. The echo must arrive at least 0.1 s after the original sound to be heard separately.
  • Pitch depends on frequency (higher f means higher pitch); loudness depends on amplitude; quality (timbre) depends on the number and strength of harmonics (overtones) present.
  • Resonance: a system vibrates with maximum amplitude when driven at its natural frequency. Examples: sonometer, swing, tuning fork over an air column, shattering of a glass.
  • Closed tube: fundamental L = λ/4, with only odd harmonics (f, 3f, 5f). Open tube: fundamental L = λ/2, with all harmonics (f, 2f, 3f). The open tube has a richer sound.
  • Resonance tube experiment: successive resonant lengths in a closed tube differ by λ/2, so v = 2f(L₂ − L₁), which also eliminates end correction.
  • Vibrating string: fundamental frequency f = (1/2L)√(T/μ), where T is tension and μ mass per unit length. Overtones are at 2f, 3f, ... Ultrasound (f > 20 kHz) is used in sonar and scanning.
Worked example: A closed tube 0.17 m long resonates at its fundamental with air at 340 m/s. λ = 4L = 4 × 0.17 = 0.68 m, so f = v/λ = 340/0.68 = 500 Hz. The next resonance occurs at 3λ/4 = 0.51 m.
Exam trap: Using L = λ/2 for a closed tube (it is λ/4), and calling loudness 'pitch'. Students also forget that a sound echo travels to the wall and back, so the distance is d = vt/2.
Electrostatics and Coulomb's Law SS2 · Physics

Electrostatics studies electric charges at rest and the forces and fields they produce. It explains lightning, charging by friction and the operation of capacitors and lightning conductors.

  • Two kinds of charge: positive (deficiency of electrons) and negative (excess of electrons). Like charges repel, unlike charges attract. Charge is measured in coulomb (C); the electron has e = 1.6 × 10⁻¹⁹ C.
  • Charging methods: friction (rubbing transfers electrons), contact, and induction (earthing in the presence of a charged body gives an opposite charge). Charge is conserved.
  • Conductors allow charge to flow (metals); insulators do not (plastic, glass, rubber). The gold-leaf electroscope detects charge and its sign: repulsion (more divergence) confirms like charge.
  • Coulomb's law: F = kq₁q₂/r², where k = 9 × 10⁹ N m² C⁻² (1/4πε₀). Force is proportional to the product of the charges and inversely proportional to the square of the separation.
  • Electric field is a region where a charge experiences a force. Field strength E = F/q (N/C or V/m); for a point charge E = kQ/r².
  • Field lines go from positive to negative charges, never cross, and are closer together where the field is stronger. A uniform field is shown by parallel, equally spaced lines.
  • Charge concentrates at sharp points on a conductor (point action), so it leaks away easily. This is the principle of the lightning conductor.
  • Potential at a distance r from point charge Q: V = kQ/r (volt). A capacitor stores charge, Q = CV, with C in farads; energy stored = ½CV².
Worked example: Two point charges of 2 μC and 3 μC are 0.3 m apart in air. F = kq₁q₂/r² = (9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶)/(0.3)² = 0.054/0.09 = 0.6 N, repulsive.
Exam trap: Forgetting to convert μC to C, and not squaring r. Candidates also think that positive charge is moved when a body is charged by friction, when electrons are what transfer.
Current Electricity SS2 · Physics

Current electricity deals with the steady flow of charge in conductors and circuits. It covers Ohm's law, resistance and electrical energy used and paid for in homes.

  • Electric current I = Q/t (ampere, A): rate of flow of charge. Potential difference V is energy per unit charge (volt, V = J/C). Resistance R = V/I (ohm, Ω).
  • Ohm's law: at constant temperature, the current through a conductor is proportional to the potential difference across it, V = IR. A V–I graph of an ohmic conductor is a straight line through the origin.
  • Resistivity ρ: R = ρL/A, so ρ = RA/L, unit Ω m. Resistance rises with length and falls with cross-sectional area. In metals, R increases with temperature.
  • Series: R = R₁ + R₂ + ...; the same current flows through each, and the voltages add up. Parallel: 1/R = 1/R₁ + 1/R₂ + ...; the voltage is the same, and the currents add. Total is less than the smallest.
  • A cell has e.m.f. E and internal resistance r: E = I(R + r), so terminal p.d. V = E − Ir. The e.m.f. equals the terminal p.d. when no current flows (open circuit).
  • Electrical power P = IV = I²R = V²/R (watts). Energy W = Pt = IVt (joules). Domestic energy unit: kilowatt-hour (kWh), 1 kWh = 3.6 × 10⁶ J.
  • Cost of electricity = energy in kWh × price per kWh, where energy in kWh = power (kW) × time (hours). Fuses (rated in A) protect appliances; a fuse rating is slightly above the normal current.
  • Ammeter is connected in series (low resistance); voltmeter is connected in parallel (high resistance). Shunts extend ammeter range; multipliers extend voltmeter range.
Worked example: A 6 Ω and a 3 Ω resistor in parallel are in series with a 4 Ω resistor across a 12 V supply. Parallel part = (6 × 3)/(6 + 3) = 2 Ω; total R = 2 + 4 = 6 Ω; I = 12/6 = 2 A. A 2 kW heater used 5 h at ₦60 per kWh costs 2 × 5 × 60 = ₦600.
Exam trap: Adding resistors in parallel as if they were in series, and using watts instead of kilowatts in kWh cost calculations. Also confusing e.m.f. with terminal p.d. when internal resistance is present.
Gravitational Field and Satellites SS3 · Physics

Gravitational field is the region around a mass where another mass feels an attractive force; it explains weight, planetary orbits and satellites. It is a core SS3 topic with frequent numerical questions.

  • Newton's law of gravitation: F = Gm₁m₂/r², where G = 6.67 × 10⁻¹¹ N m² kg⁻². The force is attractive, acts along the line joining the masses, and obeys the inverse-square law.
  • Gravitational field strength g = F/m (unit N/kg, same as m/s²). Around a point mass M, g = GM/r²; at Earth's surface g ≈ 9.8 m/s² (10 m/s² in many WAEC questions).
  • Gravitational field is a vector pointing toward the mass. g decreases with height as 1/r² (r measured from Earth's centre) and is slightly less at the equator than at the poles.
  • Relation between G and g: g = GM/R², so M = gR²/G. Mass of a body is constant everywhere, but its weight W = mg changes with the value of g.
  • Escape velocity: v = √(2GM/R) = √(2gR); for Earth it is about 11.2 km/s. It is independent of the mass of the escaping body.
  • Satellite in circular orbit: gravity supplies the centripetal force, GMm/r² = mv²/r, so v = √(GM/r). Orbital period T = 2πr/v; higher orbits mean lower speed and longer period.
  • Parking (geostationary) orbit: period 24 h, over the equator, moving west to east, height about 36,000 km. Used for communication and weather satellites. Weightlessness occurs because gravity supplies all the centripetal force.
  • Kepler's laws: planets move in ellipses with the Sun at one focus; the radius vector sweeps equal areas in equal times; T² ∝ r³.
Worked example: A satellite orbits Earth at 3.6 × 10⁶ m above the surface (R = 6.4 × 10⁶ m, M = 6.0 × 10²⁴ kg, G = 6.67 × 10⁻¹¹). Orbit radius r = 10.0 × 10⁶ m, so v = √(GM/r) = √(6.67×10⁻¹¹ × 6.0×10²⁴ / 1.0×10⁷) ≈ 6.3 × 10³ m/s, and T = 2πr/v ≈ 9.9 × 10³ s ≈ 2.8 h.
Exam trap: Using height instead of the distance from Earth's centre (r = R + h) in the inverse-square law. Also confusing G (universal constant) with g (field strength), and thinking mass changes with location.
Electric and Magnetic Forces SS3 · Physics

This topic covers the forces on charges in electric and magnetic fields and on current-carrying conductors in magnetic fields. It is the basis of motors, galvanometers and particle accelerators.

  • Magnetic flux density B (tesla, T) measures field strength; 1 T = 1 N/(A m) = 1 Wb/m². Magnetic flux Φ = BA cosθ, in weber (Wb), where θ is the angle between B and the normal to the area.
  • Force on a current-carrying conductor in a field: F = BIl sinθ. It is maximum (BIl) when the conductor is perpendicular to the field and zero when parallel. Direction is given by Fleming's left-hand rule.
  • Force on a moving charge in a magnetic field: F = Bqv sinθ. The force is perpendicular to velocity, so it does no work and the charge moves in a circle of radius r = mv/(Bq).
  • Force on a charge in an electric field: F = qE, where E = V/d for parallel plates (unit V/m = N/C). The force is along E for positive charges and opposite E for negative charges.
  • Parallel currents in the same direction attract; opposite directions repel. This defines the ampere: the force per metre between long wires 1 m apart is 2 × 10⁻⁷ N/m for 1 A each.
  • Torque on a rectangular coil of N turns in a field: τ = NBIA cosθ (θ from the plane of the coil to B), maximum when the plane is parallel to the field. This is the principle of the moving-coil meter and the d.c. motor.
  • Moving-coil galvanometer uses a radial field and a restoring hairspring, so deflection is proportional to current. It is converted to an ammeter with a low-resistance shunt and to a voltmeter with a high-resistance multiplier.
  • D.c. motor: a split-ring commutator reverses the coil current every half turn so the torque keeps the same sense. Speed and torque increase with larger current, stronger field, more turns and bigger area.
Worked example: An electron (q = 1.6 × 10⁻¹⁹ C) moves at 3.0 × 10⁶ m/s perpendicular to a field of 0.20 T. F = Bqv = 0.20 × 1.6×10⁻¹⁹ × 3.0×10⁶ = 9.6 × 10⁻¹⁴ N, directed perpendicular to both the velocity and the field.
Exam trap: Using the right-hand rule instead of Fleming's left-hand rule for motors, and forgetting that the force is zero when the conductor or velocity is parallel to B. Remember electron flow is opposite to conventional current.
Electromagnetic Induction and Transformers SS3 · Physics

Electromagnetic induction is the production of an e.m.f. by a changing magnetic flux. It underlies generators, transformers and power transmission.

  • Faraday's law: the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage, E = −N ΔΦ/Δt (volts). More turns, faster motion or a stronger field give a larger e.m.f.
  • Lenz's law: the induced current flows in a direction that opposes the change producing it. This follows from conservation of energy, and is the reason for the negative sign in Faraday's law.
  • E.m.f. in a straight conductor of length l moving at velocity v across a field B: E = Blv. The direction of the induced current is found with Fleming's right-hand (generator) rule.
  • A.c. generator (dynamo): a coil rotating in a magnetic field with slip rings and brushes gives an alternating e.m.f. E = E₀ sin ωt. A d.c. generator uses a split-ring commutator.
  • Self-inductance: a changing current in a coil induces a back e.m.f. E = −L ΔI/Δt, where L is in henry (H). Mutual induction links two coils and is the basis of the transformer and induction coil.
  • Transformer equation: Vₛ/Vₚ = Nₛ/Nₚ = Iₚ/Iₛ (ideal). Step-up has Nₛ > Nₚ; step-down has Nₛ < Nₚ. It works only with a.c. and uses a laminated soft-iron core.
  • Efficiency = (Vₛ Iₛ)/(Vₚ Iₚ) × 100%. Losses: copper (I²R heating), eddy currents (reduced by lamination), hysteresis (soft iron), and flux leakage.
  • Power is transmitted at very high voltage and low current to reduce I²R loss in the cables, then stepped down for consumers. The national grid uses step-up and step-down transformers.
Worked example: A transformer has 1000 primary turns on a 240 V supply and gives 12 V. Nₛ = Nₚ Vₛ/Vₚ = 1000 × 12/240 = 50 turns. If it is 80% efficient and delivers 5 A, output power = 60 W, input = 60/0.8 = 75 W, so Iₚ = 75/240 = 0.31 A.
Exam trap: Applying a transformer to d.c. (no induced e.m.f. in the secondary), and mixing up the ratio so that the current ratio is inverted: current is inversely proportional to turns. Also forgetting that Lenz's law gives opposition to change, not to the field.
Alternating Current Circuits SS3 · Physics

A.C. circuits involve currents and voltages that vary sinusoidally with time, and components that oppose current in different ways. They are essential for understanding power supply, radio tuning and filters.

  • Alternating current varies as I = I₀ sin ωt, with ω = 2πf. Peak value I₀; r.m.s. value = I₀/√2 ≈ 0.707 I₀ (the steady d.c. giving the same heating effect). Mains of 240 V r.m.s. has a peak of about 339 V.
  • Inductive reactance X_L = 2πfL (ohm); it increases with frequency, and the voltage leads the current by 90° in a pure inductor. It passes d.c. easily but chokes a.c.
  • Capacitive reactance X_C = 1/(2πfC) (ohm); it decreases with frequency, and the current leads the voltage by 90° in a pure capacitor. A capacitor blocks d.c. but passes a.c.
  • Resistance R keeps voltage and current in phase. In a series R-L-C circuit, impedance Z = √(R² + (X_L − X_C)²), and V = IZ with phase angle tan φ = (X_L − X_C)/R.
  • Series resonance occurs when X_L = X_C, so f₀ = 1/(2π√(LC)). Then Z = R is a minimum, the current is maximum and in phase with the voltage. This is used in radio tuning circuits.
  • Power in a.c.: P = I_rms V_rms cos φ, where cos φ is the power factor. At resonance cos φ = 1; in a pure L or C the average power is zero.
  • Rectification: a diode gives half-wave rectification; a bridge of four diodes gives full-wave. A smoothing capacitor reduces ripple. A choke (inductor) is used to limit a.c. without wasting power as heat.
  • Parallel resonance (rejector circuit) gives minimum line current and maximum impedance at f₀. Quality of tuning improves with low resistance (a sharper resonance curve).
Worked example: A series circuit has R = 30 Ω, X_L = 100 Ω and X_C = 60 Ω. Z = √(30² + (100 − 60)²) = √(900 + 1600) = 50 Ω. On a 100 V r.m.s. supply I = 100/50 = 2 A. For L = 0.5 H and C = 20 μF, f₀ = 1/(2π√(0.5 × 20×10⁻⁶)) ≈ 50 Hz.
Exam trap: Adding R, X_L and X_C arithmetically instead of using the phasor (Pythagoras) formula. Also mixing up which leads: remember CIVIL (in C, current leads V; in L, V leads current) and using peak values in power calculations.
Structure of the Atom and Photoelectric Effect SS3 · Physics

This topic traces atomic models, energy levels, spectra and the photoelectric effect, which showed that light behaves as particles (photons). It supports understanding of X-rays, lasers and semiconductors.

  • Models: Thomson (plum pudding, electrons in positive sphere); Rutherford (tiny dense positive nucleus, from the α-scattering experiment, with most of the atom empty); Bohr (electrons in fixed orbits with quantised energy).
  • Rutherford α-scattering: most α-particles pass straight through, a few are deflected through large angles and very few (about 1 in 8000) rebound. Rutherford's model could not explain atomic stability or line spectra.
  • Bohr: an electron moves in a stationary orbit without radiating; it emits or absorbs a photon only when jumping between levels, with hf = E₂ − E₁. Hydrogen levels: Eₙ = −13.6/n² eV; ground state −13.6 eV.
  • Line spectra: emission lines arise when electrons drop to lower levels (Lyman to n = 1, ultraviolet; Balmer to n = 2, visible). Absorption spectra show dark lines at the same wavelengths.
  • Photoelectric effect: electrons are emitted when light of frequency above the threshold frequency f₀ falls on a metal. Below f₀ there is no emission no matter how intense the light.
  • Einstein's equation: hf = W₀ + ½mv²max, where W₀ = hf₀ is the work function. Planck's constant h = 6.63 × 10⁻³⁴ J s; 1 eV = 1.6 × 10⁻¹⁹ J. Photon energy E = hf = hc/λ.
  • Intensity affects the number of photoelectrons (current), not their maximum kinetic energy; kinetic energy increases with frequency. Stopping potential V_s satisfies eV_s = ½mv²max. Wave theory could not explain the instant emission.
  • Uses: photocells, burglar alarms, automatic doors. X-rays form when fast electrons stop in a metal target; λ_min = hc/(eV). Ionisation energy removes an electron from the ground state.
Worked example: Light of frequency 1.0 × 10¹⁵ Hz falls on a metal of work function 3.0 × 10⁻¹⁹ J (h = 6.63 × 10⁻³⁴ J s). Photon energy = 6.63 × 10⁻¹⁹ J, so the maximum kinetic energy = 6.63 × 10⁻¹⁹ − 3.0 × 10⁻¹⁹ = 3.63 × 10⁻¹⁹ J (about 2.3 eV).
Exam trap: Thinking brighter light increases the maximum kinetic energy of photoelectrons (it only raises their number). Also forgetting to convert eV to joules, and applying the equation when the frequency is below the threshold.
Wave-Particle Duality SS3 · Physics

Wave-particle duality says that light and matter show both wave and particle behaviour depending on the experiment. It is a key idea of quantum physics and is used in electron microscopes.

  • Light shows wave behaviour (interference, diffraction, polarisation) and particle behaviour (photoelectric effect, Compton effect). Photon momentum p = h/λ and energy E = hf.
  • De Broglie hypothesis: every moving particle has an associated wave of wavelength λ = h/p = h/(mv), where h = 6.63 × 10⁻³⁴ J s. A faster or heavier particle has a shorter wavelength.
  • For a particle accelerated through potential difference V: ½mv² = eV, so λ = h/√(2meV). For an electron, λ ≈ 1.23/√V nm with V in volts.
  • Evidence for matter waves: Davisson and Germer showed that electrons are diffracted by a nickel crystal, and G. P. Thomson observed electron diffraction patterns through thin metal foils.
  • Everyday objects have wavelengths far too small to detect (a 1 kg ball at 1 m/s has λ ≈ 6.6 × 10⁻³⁴ m), so wave behaviour is observed only for atomic-sized particles.
  • Electron microscope: electrons accelerated through a high voltage have a very short wavelength, so resolution is much better than for an optical microscope, which is limited by the wavelength of light.
  • Heisenberg uncertainty principle: the position and momentum of a particle cannot both be known exactly, Δx Δp ≳ h/4π. It is a natural limit, not a fault of the apparatus.
  • Energy-time form: ΔE Δt ≳ h/4π. The principle explains why electrons cannot sit inside the nucleus and the nature of atomic orbitals as probability clouds.
Worked example: Find the de Broglie wavelength of an electron moving at 3.0 × 10⁶ m/s (m = 9.11 × 10⁻³¹ kg). λ = h/(mv) = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 3.0 × 10⁶) = 2.4 × 10⁻¹⁰ m, which is comparable to atomic spacing, so diffraction is observable.
Exam trap: Using the wrong momentum (forgetting mass) or kinetic energy in λ = h/mv, and concluding that large objects have a large wavelength (it is the opposite). Do not say light is only a wave or only a particle.
Radioactivity and Binding Energy SS3 · Physics

Radioactivity is the spontaneous disintegration of unstable nuclei with emission of radiation. It is important in medicine, dating of materials, power generation and radiation safety.

  • Nucleus: protons and neutrons (nucleons). Atomic number Z = protons, mass number A = protons + neutrons; isotopes have the same Z but different A. Notation: ᴬ_Z X.
  • α: helium nucleus ⁴₂He, strongly ionising, stopped by paper. β⁻: fast electron, stopped by a few mm of aluminium. γ: electromagnetic wave, weakly ionising, needs thick lead, undeflected by fields.
  • Decay equations: α-decay reduces A by 4 and Z by 2; β⁻ decay increases Z by 1 with A unchanged; γ emission changes neither. Example: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He.
  • Detectors: Geiger-Müller tube and ratemeter, cloud chamber (α tracks thick and straight, β thin and wavy), gold-leaf electroscope (discharged by ionisation), and scintillation counter.
  • Decay law: N = N₀e^(−λt); activity A = λN, in becquerel (Bq, one decay per second). Decay is random, spontaneous and not affected by temperature or pressure.
  • Half-life T½ is the time for half the radioactive nuclei (or the activity) to decay: λ = 0.693/T½. After n half-lives the remaining fraction is (½)ⁿ.
  • Mass defect Δm = (Zmₚ + (A − Z)mₙ) − m_nucleus; binding energy = Δm c², and 1 u = 931 MeV. Higher binding energy per nucleon means a more stable nucleus (peak near iron-56).
  • Uses: tracers, cancer treatment, thickness gauges, sterilisation, carbon-14 dating. Hazards: burns, cancer, genetic damage; use tongs, lead shielding, short exposure and distance.
Worked example: A sample has 64 g of a radioisotope of half-life 8 days. After 24 days, n = 24/8 = 3 half-lives, so the mass left = 64 × (½)³ = 8 g. The decay constant is λ = 0.693/8 = 0.087 per day.
Exam trap: Using the number of days instead of the number of half-lives in the (½)ⁿ expression, and confusing mass number with atomic number in decay equations (check that A and Z balance on both sides). Gamma emission does not change the nucleus's A or Z.
Nuclear Reactions SS3 · Physics

Nuclear reactions change the nuclei of atoms and release very large energy through fission and fusion. They are used to generate electricity, in medicine and in weapons, so safety and peaceful use matter.

  • Nuclear reactions obey conservation of mass number, atomic number (charge) and mass-energy. Energy released: E = Δm c², where Δm is the mass lost; 1 u = 931.5 MeV, and c = 3.0 × 10⁸ m/s.
  • Nuclear fission: a heavy nucleus such as ²³⁵U is split by a slow neutron into two medium nuclei, 2-3 neutrons and about 200 MeV. Example: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀n + energy.
  • Chain reaction: neutrons released cause further fissions. It is controlled in a reactor (control rods of cadmium or boron absorb neutrons; moderator of graphite or heavy water slows them) and uncontrolled in an atomic bomb.
  • Critical mass is the smallest mass of fissile material that sustains a chain reaction. A nuclear reactor has fuel rods, moderator, control rods, coolant, heat exchanger and thick concrete shielding.
  • Nuclear fusion: light nuclei join to form a heavier one, releasing energy, e.g. ²₁H + ³₁H → ⁴₂He + ¹₀n + 17.6 MeV. It needs very high temperatures (about 10⁷ to 10⁸ K) to overcome repulsion; it powers the Sun and stars and the hydrogen bomb.
  • Fusion gives more energy per kilogram than fission with less radioactive waste, but controlled fusion is not yet commercial.
  • Peaceful uses: electricity generation, medical diagnosis and radiotherapy, sterilisation of food and equipment, industrial tracers, agriculture (mutation breeding) and carbon-14 dating.
  • Hazards and safety: radiation sickness, cancer and genetic mutation, and radioactive waste disposal. Use lead and concrete shielding, film badges for monitoring, remote handling and safe waste storage.
Worked example: In a reaction the mass lost is 0.0304 u. Energy released = 0.0304 × 931.5 ≈ 28.3 MeV. In joules: 28.3 × 1.6 × 10⁻¹³ ≈ 4.5 × 10⁻¹² J, since 1 MeV = 1.6 × 10⁻¹³ J.
Exam trap: Mixing up fission (splitting heavy nucleus) and fusion (joining light nuclei), and forgetting that the moderator slows neutrons while control rods absorb them. In equations, check that the mass numbers and atomic numbers balance.
Chemistry and Society, Scientific Method, Safety SS1 · Chemistry

Chemistry is the study of the composition, structure, properties and changes of matter, and it underpins medicine, agriculture, energy and industry. Scientific method and laboratory safety are the habits that make chemical work reliable and safe.

  • Chemistry is the science that studies matter: its composition, properties, structure, the changes it undergoes and the energy involved in those changes.
  • Scientific method steps: observation, stating the problem, hypothesis, experiment, recording results, analysis, conclusion; a hypothesis that survives repeated tests may become a theory or law.
  • Careers: medicine and pharmacy, chemical engineering, petroleum and gas, food and brewing, agriculture, water treatment, forensic science, environmental science and teaching.
  • Industrial importance: Nigerian industries use chemistry for petroleum refining, cement, fertilisers (urea, NPK), soap, paints, plastics, textiles, cosmetics and pharmaceuticals.
  • Chemistry and society: it supplies drugs, clean water, fuels and materials, but misuse causes pollution, acid rain, oil spills, global warming and toxic waste.
  • Hazard symbols: flammable, corrosive, toxic, explosive, oxidising, irritant/harmful and radioactive; always read the label on a reagent bottle before use.
  • Safety rules: wear a lab coat and goggles, never taste chemicals, add acid to water (not water to acid), heat test tubes pointing away from people, and know where the fire extinguisher is.
  • First aid: wash acid or alkali splashes with plenty of running water; use a damp cloth or sand to smother small fires; report all accidents to the teacher at once.
Worked example: Why must concentrated sulphuric acid be diluted by adding the acid slowly to water? The dilution is highly exothermic; adding water to acid can make the water boil and spit acid out, whereas adding acid to the larger volume of water spreads the heat safely.
Exam trap: Students mix up the steps of the scientific method or say a hypothesis is 'proved' instead of 'tested'. Also note 'add acid to water', and do not confuse the corrosive and toxic hazard symbols.
Matter, Changes and Separation of Mixtures SS1 · Chemistry

Matter is anything that has mass and occupies space; it is classified as elements, compounds and mixtures and can undergo physical or chemical changes. Separation techniques use differences in physical properties to split mixtures.

  • Matter exists as solid, liquid or gas; an element has only one kind of atom (Fe, O₂), a compound has two or more elements chemically combined in fixed ratio (H₂O, NaCl).
  • A mixture has components physically combined in any proportion that keep their own properties; it may be homogeneous (salt solution, air) or heterogeneous (sand and water).
  • Physical change: no new substance, usually reversible (melting, boiling, dissolving, magnetising). Chemical change: new substance formed, usually irreversible (burning, rusting, souring, decomposition).
  • Compound versus mixture: a compound has fixed composition, fixed melting point and needs chemical means to separate; a mixture has variable composition and is separated physically.
  • Filtration separates an insoluble solid from a liquid; evaporation and crystallisation recover a dissolved solid; simple distillation separates a solvent from a solution (water from salt).
  • Fractional distillation separates miscible liquids with different boiling points (ethanol and water, crude oil fractions, liquid air); the lower boiling liquid distils first.
  • Separating funnel separates immiscible liquids (oil and water); sublimation separates NH₄Cl, iodine or solid CO₂ from non-subliming solids; magnet separates iron filings.
  • Chromatography separates substances by different solubilities and rates of movement on paper (Rf = distance moved by solute ÷ distance moved by solvent); decantation and sieving are simple methods too.
Worked example: How can a mixture of sand, salt and iron filings be separated? Use a magnet to remove the iron filings, add water to dissolve the salt, filter off the sand, then evaporate the filtrate to recover the salt.
Exam trap: Calling rusting, burning or cooking a physical change, and saying dissolving sugar is chemical. Also choosing filtration for a dissolved solid, or simple distillation where fractional distillation is needed.
Atomic Structure and Isotopes SS1 · Chemistry

The atom consists of a nucleus of protons and neutrons surrounded by electrons in energy levels. Knowing atomic number, mass number, isotopes and electronic configuration explains the properties and placement of every element.

  • Subatomic particles: proton (relative mass 1, charge +1), neutron (mass 1, charge 0), electron (mass about 1/1840, charge −1); protons and neutrons are in the nucleus.
  • Atomic number Z = number of protons (= electrons in a neutral atom); mass number A = protons + neutrons; neutrons = A − Z. Notation: ᴬ_Z X, e.g. ²³₁₁Na has 12 neutrons.
  • Isotopes are atoms of the same element with the same atomic number but different mass numbers (different neutrons), e.g. ¹²C and ¹⁴C, ³⁵Cl and ³⁷Cl; they have identical chemical properties.
  • Relative atomic mass = Σ(isotopic mass × % abundance) ÷ 100. It is a weighted average, so it is usually not a whole number (Cl = 35.5).
  • Electron shells hold at most 2n² electrons: K (2), L (8), M (18), N (32). Fill the shells in order, with the outer shell holding no more than 8 (2 for the first period).
  • Electronic configuration examples: Na 2,8,1; Cl 2,8,7; Ca 2,8,8,2. The outer-shell electrons are valence electrons and determine the element's group and valency.
  • Sub-levels fill in the order 1s 2s 2p 3s 3p 4s 3d 4p; s holds 2, p holds 6, d holds 10 electrons. Example: Fe 1s²2s²2p⁶3s²3p⁶4s²3d⁶.
  • Models: Dalton (indivisible sphere), Thomson (plum pudding), Rutherford (nuclear atom from the alpha scattering experiment), Bohr (electrons in fixed orbits); ions form by gain or loss of electrons.
Worked example: Chlorine has two isotopes, ³⁵Cl (75%) and ³⁷Cl (25%). Relative atomic mass = (35 × 75 + 37 × 25) ÷ 100 = (2625 + 925) ÷ 100 = 3550 ÷ 100 = 35.5.
Exam trap: Confusing mass number with atomic number, and forgetting that isotopes differ in neutrons, not protons. For ions, adjust electrons: Mg²⁺ has 10 electrons, not 12; also, the 4s sub-level fills before 3d.
The Periodic Table and Periodic Properties SS1 · Chemistry

The periodic table arranges elements by increasing atomic number so that elements with similar properties fall in the same group. Trends in atomic radius, ionisation energy and electron affinity let you predict how elements behave.

  • Development: Dobereiner (triads), Newlands (law of octaves), Mendeleev (arranged by atomic mass and left gaps for undiscovered elements), Moseley (arranged by atomic number: the modern periodic law).
  • Groups are vertical columns (same number of outer electrons); periods are horizontal rows (same number of shells). Group number equals outer electrons for groups 1 and 2, and group 13 to 18 have 3 to 8.
  • Blocks: s-block (groups 1 and 2), p-block (groups 13 to 18), d-block (transition metals), f-block (lanthanides and actinides); group 1 alkali metals, group 2 alkaline earth metals, group 17 halogens, group 18 noble gases.
  • Atomic radius decreases across a period (nuclear charge rises, same shell) and increases down a group (extra shells added).
  • First ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms: M(g) → M⁺(g) + e⁻. It increases across a period and decreases down a group.
  • Dips in ionisation energy: Be to B (the 2p electron is higher in energy) and N to O (pairing repulsion in the 2p orbital); noble gases have the highest values in their periods.
  • Electron affinity is the energy change when a gaseous atom gains one electron: X(g) + e⁻ → X⁻(g). Halogens have the most negative (highest) values; it generally increases across a period.
  • Electronegativity rises across a period and falls down a group (fluorine highest). Metallic character decreases across a period and increases down a group; oxides change from basic to acidic across a period.
Worked example: Why is the first ionisation energy of potassium lower than that of sodium? Potassium has an extra shell, so its outer electron is farther from the nucleus and better shielded, and is removed more easily.
Exam trap: Saying atomic radius increases across a period, or ignoring shielding when explaining group trends. Also mixing up ionisation energy (energy needed to remove an electron) with electron affinity (energy change on gaining one).
Chemical Bonding SS1 · Chemistry

Chemical bonding is the joining of atoms by transfer or sharing of electrons to attain a stable (usually octet) configuration. The type of bond determines melting point, conductivity and solubility of a substance.

  • Ionic (electrovalent) bond: transfer of electrons from a metal to a non-metal giving oppositely charged ions held by electrostatic attraction, e.g. Na⁺Cl⁻, Mg²⁺O²⁻.
  • Properties of ionic compounds: high melting and boiling points, hard but brittle, conduct electricity when molten or in aqueous solution but not as solids, often soluble in water.
  • Covalent bond: a shared pair of electrons between non-metal atoms, e.g. H₂, Cl₂, H₂O, CH₄, NH₃, CO₂ (double bonds), N₂ (triple bond). Properties: low melting points, poor conductors, often soluble in organic solvents.
  • Coordinate (dative) bond: both shared electrons come from one atom (donor) to an acceptor, as in NH₄⁺ (N to H⁺) and H₃O⁺; once formed it is identical to other covalent bonds.
  • Metallic bond: attraction between positive metal ions and a sea of delocalised electrons; it explains conductivity, malleability, ductility and lustre of metals.
  • Hydrogen bond: a strong dipole attraction between H bonded to F, O or N and a lone pair on another F, O or N molecule; it explains the high boiling point of H₂O, HF and NH₃ and ice being less dense than water.
  • Van der Waals forces are weak intermolecular attractions that act between all molecules and increase with molar mass; they explain the low boiling points of simple molecular substances such as I₂ and CH₄.
  • Shapes: CH₄ tetrahedral (109.5°), NH₃ pyramidal, H₂O bent (V-shaped), CO₂ linear, BeCl₂ linear. Giant covalent structures (diamond, silica) have very high melting points; graphite conducts.
Worked example: Describe the bonding in NH₄⁺. Three N–H bonds are normal covalent bonds. The fourth forms when the nitrogen lone pair is donated to an H⁺ ion with an empty orbital, forming a coordinate bond.
Exam trap: Treating hydrogen bonds as bonds inside a molecule rather than between molecules, and saying ionic solids conduct electricity. Do not call HCl ionic: it is covalent and only ionises in water.
Chemical Formulas and Equations SS1 · Chemistry

A chemical formula shows the elements and the ratio of atoms in a substance, while an equation shows reactants turning into products with the same atoms on both sides. These are the basis of every stoichiometric calculation.

  • Valency is the combining power of an element: Na, K, Ag = 1; Mg, Ca, Zn, O = 2; Al = 3; Fe = 2 or 3; Cu = 1 or 2. Radicals: OH⁻, NO₃⁻, SO₄²⁻, CO₃²⁻, PO₄³⁻, NH₄⁺.
  • Writing formulas: write the symbols with their charges and criss-cross the valencies, then simplify; Al³⁺ and SO₄²⁻ give Al₂(SO₄)₃; Ca²⁺ and OH⁻ give Ca(OH)₂.
  • Balancing equations obeys the law of conservation of mass: change only the coefficients in front of formulas, never the subscripts. Example: 2Mg + O₂ → 2MgO.
  • State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution. An ionic equation shows only the species that react, leaving out spectator ions, e.g. Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
  • Empirical formula is the simplest whole-number ratio of atoms in a compound; molecular formula is the actual number of atoms: molecular formula = (empirical formula)ₙ.
  • To find empirical formula: percentage or mass ÷ atomic mass, divide by the smallest value, and if needed multiply to get whole numbers (1.5 becomes 3:2 by doubling).
  • Find n = molar mass ÷ empirical formula mass. Related ideas: mole = mass ÷ molar mass, and 1 mole contains 6.02 × 10²³ particles (Avogadro's number).
  • Types of reaction: combination, decomposition, displacement, double decomposition (precipitation, neutralisation) and redox; Group formulae of common acids: HCl, HNO₃, H₂SO₄, H₂CO₃.
Worked example: A compound is 40.0% C, 6.7% H and 53.3% O by mass with molar mass 180 g/mol. Moles: C 40.0 ÷ 12 = 3.33, H 6.7 ÷ 1 = 6.7, O 53.3 ÷ 16 = 3.33. Ratio ÷ 3.33 = 1 : 2 : 1, so CH₂O (mass 30). n = 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆.
Exam trap: Balancing by changing subscripts, and forgetting to write the group formula in brackets (Ca(OH)₂ not CaOH₂). In empirical formula questions, do not round 1.5 to 2; multiply the whole ratio instead.
Kinetic Theory and Gas Laws SS1 · Chemistry

The kinetic theory explains the behaviour of solids, liquids and gases in terms of moving particles. The gas laws relate pressure, volume, temperature and amount of gas, and are used in many exam calculations.

  • Kinetic theory: matter is made of tiny particles in constant motion; the particles gain energy on heating. Solids vibrate in fixed positions, liquids slide over each other, and gas particles move freely and far apart.
  • Evidence: diffusion (spread of bromine vapour, ammonia smell) and Brownian motion. Lighter gases diffuse faster (Graham's law: rate ∝ 1/√molar mass), so NH₃ diffuses faster than HCl.
  • Boyle's law: at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure, so P₁V₁ = P₂V₂.
  • Charles's law: at constant pressure, volume is directly proportional to absolute temperature, V₁/T₁ = V₂/T₂. T(K) = θ(°C) + 273; absolute zero is −273 °C (0 K).
  • General gas law: P₁V₁/T₁ = P₂V₂/T₂. S.T.P. is 273 K (0 °C) and 760 mmHg (101.3 kPa or 1 atm). Temperatures must always be in kelvin.
  • Ideal gas equation PV = nRT, with R = 8.314 J/(mol·K), P in Pa and V in m³; molar volume of any gas at s.t.p. is 22.4 dm³/mol. An ideal gas has no intermolecular forces; real gases deviate at high pressure and low temperature.
  • Dalton's law of partial pressures: total pressure of a mixture of non-reacting gases is the sum of the partial pressures; partial pressure = mole fraction × total pressure.
  • Gay-Lussac's law of combining volumes and Avogadro's law (equal volumes of gases at the same temperature and pressure contain equal numbers of molecules) link volumes of gases in reactions.
Worked example: 500 cm³ of a gas at 27 °C and 750 mmHg is brought to s.t.p. T₁ = 300 K, T₂ = 273 K. V₂ = (750 × 500 × 273) ÷ (760 × 300) = 102 375 000 ÷ 228 000 ≈ 449 cm³.
Exam trap: Using °C instead of kelvin, and mixing pressure units (mmHg with kPa). Also assuming Boyle's law needs constant volume; it needs constant temperature, while Charles's law needs constant pressure.
The Mole Concept SS2 · Chemistry

The mole is the SI unit for amount of substance and links the masses of reacting particles to their numbers and gas volumes. It is the basis of every stoichiometric calculation in WAEC and JAMB chemistry.

  • One mole is the amount of substance containing 6.02 × 10²³ particles (Avogadro's number, L), the same as the number of atoms in 12 g of carbon-12.
  • Molar mass (g/mol) is the relative atomic or molecular mass in grams. Example: H₂SO₄ = 2(1) + 32 + 4(16) = 98 g/mol.
  • Moles = mass ÷ molar mass; number of particles = moles × 6.02 × 10²³; for gases, moles = volume ÷ molar volume.
  • Molar volume of any gas is 22.4 dm³ at s.t.p. (0 °C or 273 K and 760 mmHg or 101.325 kPa); at room temperature it is about 24 dm³.
  • Percentage composition by mass of an element = (number of atoms × atomic mass ÷ molar mass) × 100. In CaCO₃ (100), Ca = 40%, C = 12%, O = 48%.
  • Empirical formula: divide % or mass by atomic mass, divide by the smallest, and round to whole numbers. Molecular formula = (empirical formula)ₙ, n = molar mass ÷ empirical mass.
  • Stoichiometry: the balanced equation gives mole ratios. In 2H₂ + O₂ → 2H₂O, 2 mol H₂ reacts with 1 mol O₂ to give 2 mol H₂O; gas volumes follow the same ratio.
  • Limiting reagent is the reactant that is used up first and fixes the yield. Percentage yield = (actual yield ÷ theoretical yield) × 100.
Worked example: Find the volume at s.t.p. of 11 g of CO₂ (C = 12, O = 16). Molar mass = 12 + 32 = 44 g/mol; moles = 11 ÷ 44 = 0.25 mol; volume = 0.25 × 22.4 = 5.6 dm³.
Exam trap: Using the wrong molar mass (forgetting subscripts or brackets, e.g. Ca(OH)₂) and forgetting to balance the equation before using mole ratios. Also mixing cm³ and dm³ (1 dm³ = 1000 cm³).
Acid-Base Reactions, pH and Titration SS2 · Chemistry

Acids and bases are defined by proton or electron-pair behaviour and measured on the pH scale. Titration uses neutralisation to find unknown concentrations accurately.

  • Arrhenius: an acid produces H⁺ (H₃O⁺) in water; a base produces OH⁻ in water. An alkali is a base soluble in water, e.g. NaOH, KOH.
  • Brønsted-Lowry: an acid is a proton (H⁺) donor and a base is a proton acceptor. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, H₂O is the acid and NH₃ the base; conjugate pairs differ by one H⁺.
  • Lewis: an acid is an electron-pair acceptor (e.g. BF₃, H⁺) and a base is an electron-pair donor (e.g. NH₃, OH⁻).
  • pH = −log[H⁺]. pH < 7 is acidic, 7 neutral, > 7 alkaline; pH + pOH = 14 at 25 °C. A strong acid ionises completely (HCl); a weak acid only partly (CH₃COOH).
  • Indicators: litmus is red in acid, blue in alkali; methyl orange is red in acid, yellow in alkali; phenolphthalein is colourless in acid, pink in alkali. Use methyl orange for strong acid/weak base, phenolphthalein for weak acid/strong base.
  • Neutralisation: acid + base → salt + water. Salts may be normal (NaCl), acid (NaHSO₄) or basic (Zn(OH)Cl); amphoteric oxides such as ZnO and Al₂O₃ react with both acids and bases.
  • Titration: a pipette measures a fixed volume into a conical flask, the burette delivers the other solution to the end point. Rinse apparatus with the solution it will hold; repeat to get concordant titres.
  • Titration formula: (CₐVₐ)/(C_bV_b) = nₐ/n_b from the balanced equation. Concentration in g/dm³ = molarity × molar mass.
Worked example: 25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.100 mol/dm³ HCl (1:1 ratio). Moles HCl = 0.100 × 0.0200 = 0.00200 mol = moles NaOH; concentration = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.
Exam trap: Ignoring the mole ratio in the equation (e.g. H₂SO₄ with NaOH is 1:2) and choosing the wrong indicator. Students also forget that pH is a logarithmic scale, so pH 3 is ten times more acidic than pH 4.
Water and Solutions SS2 · Chemistry

Water is the commonest solvent, and the behaviour of solutes in it, including hardness, solubility and concentration, is central to chemistry and to water treatment.

  • A solution is a homogeneous mixture of solute and solvent. Saturated: holds the maximum solute at a given temperature; unsaturated: can dissolve more; supersaturated: holds more than normal and crystallises easily.
  • Solubility is the mass of solute (g) that saturates 100 g of water at a given temperature. A solubility curve plots solubility against temperature; most solids dissolve more as temperature rises, gases dissolve less.
  • Reading a curve: cooling a saturated solution from a higher to a lower temperature deposits crystals equal to the difference in solubility at the two temperatures.
  • Molarity (mol/dm³) = moles of solute ÷ volume in dm³. Concentration in g/dm³ = molarity × molar mass. Dilution: C₁V₁ = C₂V₂.
  • Soft water lathers readily with soap. Hard water does not, and forms scum, because of dissolved Ca²⁺ and Mg²⁺ salts.
  • Temporary hardness is caused by Ca(HCO₃)₂ and Mg(HCO₃)₂ and is removed by boiling: Ca(HCO₃)₂ → CaCO₃ + H₂O + CO₂. Permanent hardness is caused by sulphates and chlorides of Ca and Mg and survives boiling.
  • Permanent hardness is removed by adding washing soda (Na₂CO₃) to precipitate CaCO₃, by distillation, or by ion-exchange resins (permutit). Hard water wastes soap and causes boiler scale.
  • Colloids (e.g. milk, fog) have particles between a true solution and a suspension; they do not settle and scatter light (Tyndall effect). Water of crystallisation, efflorescence, deliquescence and hygroscopy are also tested.
Worked example: Calculate the molarity of a solution containing 4.0 g of NaOH (Na = 23, O = 16, H = 1) in 500 cm³. Moles = 4.0 ÷ 40 = 0.10 mol; volume = 0.500 dm³; molarity = 0.10 ÷ 0.500 = 0.20 mol/dm³.
Exam trap: Using cm³ instead of dm³ in molarity, and saying boiling removes all hardness (it removes only temporary hardness). Also misreading solubility curves by using the wrong temperature on the axis.
Energy Changes and Rates of Reaction SS2 · Chemistry

Chemical reactions absorb or release heat, and they proceed at different speeds. These ideas explain fuels, industrial conditions and why reactions can be sped up or slowed.

  • Exothermic reactions release heat to the surroundings (temperature rises, ΔH negative), e.g. combustion, neutralisation. Endothermic reactions absorb heat (temperature falls, ΔH positive), e.g. dissolving NH₄NO₃.
  • Enthalpy change ΔH = H(products) − H(reactants), in kJ/mol. Standard enthalpies include formation, combustion and neutralisation (about −57 kJ/mol for strong acid and strong base).
  • Heat from experiment: q = mcΔT, with c of water = 4.2 J/g/K (or 4.2 J g⁻¹ °C⁻¹). ΔH = −q ÷ moles of reacting substance.
  • Hess's law: the total enthalpy change of a reaction is the same whichever route is taken. Bond breaking absorbs energy; bond making releases it.
  • Energy profile diagrams: activation energy (Eₐ) is the minimum energy for effective collisions; in exothermic reactions products lie below reactants, in endothermic reactions above.
  • Rate of reaction = change in concentration (or mass/volume of product) per unit time, measured e.g. by gas volume collected or loss of mass.
  • Factors raising rate: higher concentration or pressure (gases), higher temperature, larger surface area (smaller particles), a catalyst, and light for photochemical reactions.
  • Collision theory: reaction needs collisions with energy ≥ Eₐ and correct orientation. A catalyst lowers Eₐ and is unchanged chemically at the end. Le Chatelier's principle covers equilibrium shifts.
Worked example: 50 g of water rises by 6.0 °C when 0.0300 mol of a solid dissolves (c = 4.2 J/g/°C). q = 50 × 4.2 × 6.0 = 1260 J; ΔH = −1260 ÷ 0.0300 = −42 000 J/mol = −42 kJ/mol (exothermic).
Exam trap: Giving ΔH the wrong sign (heat released means negative) and saying a catalyst changes ΔH or the yield. It only lowers activation energy and speeds both directions equally.
Non-Metals and their Compounds SS2 · Chemistry

This topic covers the preparation, properties and uses of hydrogen, oxygen, chlorine, nitrogen, carbon and sulphur and their key compounds. It supports industrial chemistry and many WAEC practical and theory questions.

  • Hydrogen: lab preparation by Zn + dilute H₂SO₄ (or HCl) → ZnSO₄ + H₂, collected over water or by downward displacement of air. It burns with a pop, is the lightest gas, and is used in the Haber process and margarine (hydrogenation).
  • Oxygen: lab preparation by heating KClO₃ with MnO₂ catalyst or by decomposing H₂O₂. Industrially from fractional distillation of liquid air. It relights a glowing splint; used in welding, respiration and steel making.
  • Chlorine: lab preparation by MnO₂ + conc. HCl with heating, dried with conc. H₂SO₄, collected by downward delivery. It is a greenish-yellow, poisonous gas that bleaches damp litmus; used for water treatment and bleach.
  • Nitrogen: obtained from liquid air; lab from heating NH₄Cl with NaNO₂. It is unreactive and used for ammonia (Haber process), fertilisers and inert atmospheres. Ammonia is alkaline and turns red litmus blue.
  • Carbon: allotropes are diamond (hard, non-conductor), graphite (soft, conducts, lubricant) and amorphous forms (charcoal, coke, soot). CO₂ turns limewater milky; CO is a toxic reducing agent.
  • Sulphur: allotropes are rhombic and monoclinic. It is mined by the Frasch process (superheated water and compressed air), and used in the Contact process for H₂SO₄ and in vulcanising rubber.
  • Contact process: S → SO₂ → SO₃ over V₂O₅ catalyst at about 450 °C, SO₃ absorbed in conc. H₂SO₄ to form oleum, then diluted. H₂SO₄ acts as an acid, a dehydrating agent and an oxidising agent.
  • Oxides and tests: SO₂ is acidic and bleaches, turns acidified K₂Cr₂O₇ green; NO₂ is brown. Ammonia is made in the Haber process (N₂ + 3H₂ ⇌ 2NH₃, Fe catalyst, 200 atm, 450 °C).
Worked example: State a use of the catalyst and the temperature in the Contact process. Vanadium(V) oxide, V₂O₅, converts SO₂ to SO₃ at about 450 °C; this is a compromise between rate and equilibrium yield.
Exam trap: Mixing up drying agents (conc. H₂SO₄ must not dry NH₃; use CaO) and collection methods. Candidates also confuse the Haber, Contact and Frasch processes.
Hydrocarbons and Nomenclature SS2 · Chemistry

Hydrocarbons are compounds of carbon and hydrogen only and form the foundation of organic chemistry. Systematic IUPAC naming lets any structure be named unambiguously.

  • Carbon is tetravalent (4 valence electrons, forms four covalent bonds) and catenates, forming chains and rings, which explains the huge number of organic compounds.
  • A homologous series has the same general formula and functional group, successive members differ by CH₂, and similar chemical properties with gradual change in physical properties.
  • Alkanes (CₙH₂ₙ₊₂, single bonds, saturated): methane, ethane, propane, butane (C₁ to C₄), pentane, hexane. They undergo substitution with Cl₂ in light and combust to CO₂ and H₂O.
  • Alkenes (CₙH₂ₙ, one C=C, unsaturated): ethene, propene, butene. They undergo addition reactions (Br₂, H₂, H₂O) and polymerise; bromine water is decolorised, which is the test for unsaturation.
  • Alkynes (CₙH₂ₙ₋₂, one C≡C): ethyne (acetylene), propyne. Ethyne is made from CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂ and used in oxy-acetylene welding.
  • IUPAC naming: choose the longest continuous chain (parent), number from the end giving substituents or double/triple bonds the lowest number, and name branches as prefixes, e.g. 2-methylpropane, but-1-ene, pent-2-yne.
  • Isomers have the same molecular formula but different structures: C₄H₁₀ has two (butane and 2-methylpropane); C₅H₁₂ has three. Alkyl groups: methyl CH₃–, ethyl C₂H₅–.
  • Sources: crude oil is separated by fractional distillation (refinery gas, petrol, kerosene, diesel, lubricating oil, bitumen); cracking breaks long chains into alkanes and alkenes. Natural gas is mostly methane.
Worked example: Name CH₃CH(CH₃)CH₂CH₃. The longest chain has 4 carbons (butane) with a methyl group on carbon 2, so it is 2-methylbutane (C₅H₁₂, an isomer of pentane).
Exam trap: Counting a chain that is not the longest and numbering from the wrong end. Candidates also give the wrong general formula (alkene CₙH₂ₙ, alkyne CₙH₂ₙ₋₂) or forget that alkanes undergo substitution, not addition.
Chemical Equilibrium SS3 · Chemistry

Chemical equilibrium describes reversible reactions in which forward and backward rates become equal so concentrations stay constant. It explains how to control yield in industrial processes such as the Haber and Contact processes.

  • A reversible reaction (⇌) can proceed in both directions, e.g. N₂ + 3H₂ ⇌ 2NH₃; it can reach equilibrium only in a closed system.
  • Dynamic equilibrium: forward and reverse rates are equal, so macroscopic properties (concentration, colour, pressure) stay constant although both reactions continue.
  • Le Chatelier's principle: if a system at equilibrium is disturbed, it shifts in the direction that opposes the change and restores equilibrium.
  • Concentration: adding reactant (or removing product) shifts equilibrium to the right; adding product shifts it to the left. Kc does not change.
  • Pressure affects only gases: raising pressure favours the side with fewer gas moles; if moles are equal (H₂ + I₂ ⇌ 2HI) there is no shift.
  • Temperature: heating favours the endothermic direction, cooling favours the exothermic one. Only temperature changes the value of K. A catalyst speeds both rates equally and does not shift equilibrium.
  • For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ) using equilibrium concentrations in mol/dm³; pure solids and liquids are omitted. Large Kc means products favoured.
  • Kp uses partial pressures of gases. Industrial uses: Haber process (450 °C, 200 atm, Fe catalyst) and Contact process (V₂O₅, about 450 °C) compromise rate and yield.
Worked example: For N₂ + 3H₂ ⇌ 2NH₃, equilibrium concentrations are N₂ 0.50, H₂ 0.20, NH₃ 0.40 mol/dm³. Kc = (0.40)² / (0.50 × 0.20³) = 0.16 / 0.0040 = 40 mol⁻² dm⁶.
Exam trap: Students say a catalyst increases yield or changes Kc; it does neither. Another error is forgetting to raise concentrations to the power of their coefficients or including solids in Kc.
Redox and Electrochemistry SS3 · Chemistry

This topic covers electron transfer reactions, electrochemical cells, electrolysis and corrosion. It underlies batteries, electroplating, metal refining and rust prevention.

  • Oxidation is loss of electrons or increase in oxidation number; reduction is gain of electrons or decrease in oxidation number (OIL RIG). An oxidising agent is reduced; a reducing agent is oxidised.
  • Oxidation number rules: free elements 0; simple ion = its charge; H usually +1, O usually −2 (peroxides −1); sum equals the overall charge. E.g. Mn in KMnO₄ is +7, Cr in Cr₂O₇²⁻ is +6.
  • Balancing redox equations: write half-equations, balance atoms and charge with electrons, equalise electrons lost and gained, then add. Oxidation number changes can also be balanced.
  • Electrochemical (galvanic) cell converts chemical energy to electrical energy; in Daniell cell Zn is the anode (oxidation, negative) and Cu the cathode (reduction, positive). A salt bridge completes the circuit.
  • Standard electrode potentials (E°) rank metals; E°cell = E°(cathode) − E°(anode). A metal higher in the electrochemical series displaces one below it from solution.
  • Electrolysis uses electricity to decompose electrolytes: cathode (negative) reduction, anode (positive) oxidation. Discharge depends on position in series, concentration and electrode type.
  • Faraday's 1st law: mass deposited ∝ quantity of charge, Q = It. 2nd law: same charge deposits masses in ratio of equivalent masses. 1 faraday = 96500 C mol⁻¹ of electrons.
  • Corrosion (rusting) needs iron, oxygen and water: Fe → Fe²⁺ + 2e⁻. Prevention: painting, greasing, galvanising (Zn), tin-plating, sacrificial anodes (Mg, Zn), alloying (stainless steel).
Worked example: Find mass of copper deposited by 2.0 A for 965 s from CuSO₄ (Cu = 64, 1 F = 96500 C). Q = It = 2.0 × 965 = 1930 C. Cu²⁺ + 2e⁻ → Cu, so 2 × 96500 = 193000 C gives 64 g. Mass = 1930/193000 × 64 = 0.64 g.
Exam trap: Mixing up anode and cathode between galvanic and electrolytic cells (oxidation is always at the anode). Also forgetting that oxidation numbers are per atom, not per formula, and omitting the charge on ions in half-equations.
Functional Groups SS3 · Chemistry

Functional groups are atoms or groups that give organic compounds their characteristic reactions. This topic studies alkanols, alkanoic acids, esters and amides, which appear widely in foods, drinks, soaps and medicines.

  • Alkanols (R–OH, general formula CₙH₂ₙ₊₁OH): primary, secondary and tertiary. Hydrogen bonding gives high boiling points and solubility of small members in water, e.g. ethanol C₂H₅OH.
  • Preparation of ethanol: fermentation of sugar by zymase (C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂) and hydration of ethene with steam (H₃PO₄ catalyst, 300 °C, high pressure).
  • Reactions of alkanols: burn to CO₂ and H₂O; react with sodium giving H₂; oxidation with acidified K₂Cr₂O₇ gives alkanal then alkanoic acid; dehydration with conc. H₂SO₄ gives ethene (170 °C).
  • Alkanoic acids (R–COOH, CₙH₂ₙ₊₁COOH) are weak acids: turn blue litmus red, react with metals, bases and carbonates (effervescence of CO₂). Examples: methanoic, ethanoic acid (vinegar).
  • Alkanoic acids are made by oxidising alkanols or alkanals and by hydrolysis of esters. Ethanoic acid is the sour component of vinegar; small members dissolve in water and form dimers.
  • Esters (R–COO–R′) form by esterification: acid + alcohol ⇌ ester + water, with conc. H₂SO₄ catalyst and heat. Ethanoic acid + ethanol gives ethyl ethanoate, with a fruity smell used in flavours and perfumes.
  • Hydrolysis of esters with acid gives acid and alcohol; with NaOH (saponification) gives sodium salt of the acid (soap from fats) and alcohol. Esters are insoluble in water.
  • Amides contain –CONH₂, e.g. ethanamide CH₃CONH₂, made from acid chloride or ester with ammonia. Hydrolysis gives acid and ammonia; the amide link joins amino acids in proteins and nylon.
Worked example: Name the product when ethanoic acid reacts with propan-1-ol using conc. H₂SO₄. Answer: propyl ethanoate, CH₃COOC₃H₇, plus water; the reaction is esterification (reversible).
Exam trap: Naming esters backwards: the alcohol-derived part comes first (ethyl ethanoate from ethanol and ethanoic acid). Candidates also forget that conc. H₂SO₄ in esterification is a catalyst and dehydrating agent, not a reactant.
Polymers and Biomolecules SS3 · Chemistry

Giant molecules are made of many repeating units joined together, and include plastics, fibres, carbohydrates, proteins and fats. They make up living tissue, food, clothing and packaging.

  • A polymer is a giant molecule formed by joining many small monomers; relative molecular mass = n × monomer mass. Natural polymers: starch, cellulose, proteins, rubber; synthetic: polythene, PVC, nylon, terylene.
  • Addition polymerisation joins unsaturated monomers with no by-product: ethene → polyethene, chloroethene → PVC, propene → polypropene, tetrafluoroethene → PTFE (Teflon).
  • Condensation polymerisation joins monomers with loss of small molecules (water): nylon from diamine + dicarboxylic acid, terylene (polyester) from diol + diacid, proteins and polysaccharides.
  • Plastics are thermoplastic (soften on heating, e.g. polythene, PVC) or thermosetting (do not soften, e.g. bakelite). Plastics are non-biodegradable, causing waste; they are recycled or burnt.
  • Carbohydrates CₓH₂ᵧOᵧ: monosaccharides (glucose, fructose, C₆H₁₂O₆), disaccharides (sucrose, maltose, C₁₂H₂₂O₁₁) and polysaccharides (starch, cellulose, glycogen). Tests: Benedict's/Fehling's for reducing sugars; iodine for starch.
  • Hydrolysis of starch with acid or enzymes gives glucose; sucrose gives glucose and fructose. Fermentation of glucose with zymase gives ethanol and CO₂.
  • Proteins are polymers of amino acids (NH₂–CHR–COOH) joined by peptide (amide) links; tests: biuret (violet) and Millon's. Hydrolysis gives amino acids. Sources: meat, fish, beans; used for growth and repair.
  • Fats and oils are esters of glycerol (propane-1,2,3-triol) and fatty acids; fats are solid (saturated), oils liquid (unsaturated). Saponification with NaOH gives soap and glycerol; hydrogenation hardens oil.
Worked example: A polyethene molecule has relative molecular mass 28000. With ethene C₂H₄ = 28, n = 28000 / 28 = 1000 monomer units.
Exam trap: Confusing addition and condensation polymers (only condensation gives a small molecule by-product). Another slip is saying fats and oils are carbohydrates or that unsaturated oils are solid.
Extraction of Metals SS3 · Chemistry

Extraction of metals is the process of obtaining metals from their ores, with the method chosen by the metal's position in the reactivity series. It is the basis of the metal industry and of alloy production.

  • Method depends on reactivity: very reactive metals (K, Na, Ca, Mg, Al) by electrolysis of molten compounds; moderately reactive (Zn, Fe, Sn, Pb) by carbon reduction; unreactive (Cu, Ag, Au) by heating or native.
  • Aluminium: ore bauxite (Al₂O₃·2H₂O), purified, then dissolved in molten cryolite (Na₃AlF₆, lowers melting point) and electrolysed at about 950 °C with carbon electrodes.
  • Aluminium electrodes: cathode Al³⁺ + 3e⁻ → Al; anode 2O²⁻ → O₂ + 4e⁻, and the oxygen burns away the carbon anodes (CO₂), which must be replaced. Alloys: duralumin; uses in cables and cans.
  • Iron: ores haematite (Fe₂O₃) and magnetite (Fe₃O₄) reduced in a blast furnace with coke, limestone and hot air. Reactions: C + O₂ → CO₂; CO₂ + C → 2CO; Fe₂O₃ + 3CO → 2Fe + 3CO₂.
  • In the blast furnace limestone (CaCO₃ → CaO + CO₂) removes silica as slag: CaO + SiO₂ → CaSiO₃. Pig iron (about 4% carbon) is brittle; steel is made by removing carbon and impurities.
  • Copper: ore chalcopyrite (CuFeS₂) concentrated by froth flotation, roasted and smelted to blister copper, then purified by electrolysis with impure copper anode and pure cathode in CuSO₄(aq). Uses: wires, brass, bronze.
  • Tin: ore cassiterite (SnO₂), crushed, washed, roasted to remove sulphur and arsenic, and reduced with carbon in a furnace: SnO₂ + 2C → Sn + 2CO. Uses: tin-plating, solder, bronze.
  • Alloys are mixtures of metals: brass (Cu + Zn), bronze (Cu + Sn), solder (Sn + Pb), stainless steel (Fe, Cr, Ni). Alloying improves hardness, strength and corrosion resistance.
Worked example: Why is cryolite added in aluminium extraction? Answer: Al₂O₃ melts at about 2050 °C, but dissolved in molten cryolite the mixture is electrolysed at about 950 °C, saving energy and improving conductivity.
Exam trap: Candidates write that aluminium is extracted by carbon reduction (wrong, it is too reactive) or forget slag formation from limestone in the iron furnace. Also mixing up ores and metal names.
Chemical Industries and Pollution SS3 · Chemistry

This topic covers industrial uses of biotechnology and chemistry, and the pollution these industries cause in air, water and land. It matters for health, environmental protection and sustainable waste management.

  • Biotechnology uses living organisms or enzymes: fermentation of sugar to ethanol (brewing, biofuel), yeast in bread, cheese and yoghurt, antibiotics, biogas from waste, and genetic engineering.
  • Air pollutants: CO (incomplete combustion, binds haemoglobin), SO₂ and NOₓ (acid rain), CO₂ (global warming), CFCs (ozone depletion), particulates and lead compounds from vehicles and industries.
  • Acid rain: SO₂ and NO₂ dissolve in rain to form H₂SO₃/H₂SO₄ and HNO₃, corroding buildings and metals, damaging crops and acidifying lakes. Reduced by scrubbing flue gases and catalytic converters.
  • Greenhouse effect: CO₂, CH₄ and water vapour trap heat, causing global warming. Control by reducing fossil-fuel burning, planting trees and using renewable energy.
  • Water pollution: sewage, oil spills, industrial effluents, pesticides, fertilisers (eutrophication, which lowers dissolved oxygen) and heavy metals such as Hg, Pb. Water is treated by filtration, chlorination and aeration.
  • Land pollution: refuse, plastics, mining wastes, agrochemicals and e-waste degrade soil and water. Gas flaring and oil exploration in the Niger Delta are examples of local pollution.
  • Waste management: reduce, reuse, recycle, composting, incineration and sanitary landfill; sewage treatment; biodegradable waste decomposes, non-biodegradable (plastics) persists.
  • Chemical industries: Haber (ammonia, fertiliser), Contact (H₂SO₄), Solvay (Na₂CO₃), soap, cement, petrochemicals and brewing. Sustainable practice includes pollution control, treating effluents and green chemistry.
Worked example: Explain how burning coal with sulphur impurities causes acid rain. Answer: S + O₂ → SO₂; SO₂ reacts with water and oxygen in the air to form H₂SO₄, which falls as acid rain with pH below 5.6.
Exam trap: Confusing the greenhouse effect (CO₂, CH₄) with ozone depletion (CFCs), and attributing acid rain to CO₂ alone. Candidates also forget to state a control measure when asked how to reduce pollution.
Characteristics of Living Things SS1 · Biology

Living things share a set of characteristics that separate them from non-living things, and plants and animals differ in how they show them. Knowing these helps in identifying and classifying organisms.

  • Seven characteristics of living things (MRS GREN): Movement, Respiration, Sensitivity (irritability), Growth, Reproduction, Excretion, Nutrition.
  • Movement: animals move the whole body (locomotion); plants move only parts, e.g. leaves folding in Mimosa pudica or shoots bending towards light.
  • Respiration releases energy from food in all living cells: glucose + oxygen → carbon dioxide + water + energy.
  • Sensitivity (irritability) is the ability to detect and respond to stimuli such as light, heat, touch and chemicals.
  • Growth is a permanent increase in size and dry weight; reproduction ensures continuity of species; excretion removes metabolic wastes such as CO₂, urea.
  • Nutrition: plants are autotrophic (make food by photosynthesis); animals are heterotrophic (feed on ready-made organic food).
  • Plants vs animals: plants have cellulose cell walls, chlorophyll, fixed position and growth at tips (meristems); animals lack walls and chlorophyll, move, and grow throughout the body.
  • Plants store food as starch, animals as glycogen and fat; animals have nervous systems and quick responses, plants respond slowly through hormones.
Worked example: Question: Name two ways a mango tree differs from a goat. Answer: the mango tree makes its own food by photosynthesis and is fixed in position, while the goat feeds on ready-made food and moves from place to place.
Exam trap: Students confuse excretion (removal of metabolic waste) with egestion (removal of undigested food), and forget that plants also respire and are sensitive to stimuli.
Classification of Living Things SS1 · Biology

Classification is the grouping of organisms based on similarities and differences. It makes the study of the millions of living species orderly and shows evolutionary relationships.

  • Taxonomy ranks from largest to smallest: Kingdom, Phylum (Division in plants), Class, Order, Family, Genus, Species (King Philip Came Over For Good Soup).
  • The species is the basic unit of classification: a group of organisms that interbreed to produce fertile offspring.
  • Binomial nomenclature (Linnaeus): genus name with capital letter, then species name in lower case, italicised or underlined, e.g. Homo sapiens, Zea mays.
  • Five kingdoms (Whittaker): Monera, Protista, Fungi, Plantae, Animalia.
  • Monera: prokaryotes without a true nucleus, e.g. bacteria and blue-green algae. Protista: mostly unicellular eukaryotes, e.g. Amoeba, Paramecium, Euglena.
  • Fungi: non-green, heterotrophic with chitin walls, e.g. mushroom, yeast, Rhizopus. Plantae: multicellular, photosynthetic with cellulose walls, e.g. mosses, ferns, flowering plants.
  • Animalia: multicellular heterotrophs without cell walls, e.g. insects (Arthropoda), fish, birds, mammals (Chordata).
  • Artificial classification uses few easily seen features; natural classification uses many features and shows evolutionary relationship.
Worked example: Question: Write the scientific name of the human correctly. Answer: Homo sapiens, with Homo capitalised and sapiens in lower case, and both underlined or italicised.
Exam trap: Common errors are writing the genus in lower case, putting the species name with a capital, and placing Euglena or Amoeba in Monera instead of Protista.
Levels of Organisation SS1 · Biology

Living organisms are built up in a hierarchy from cells to whole organisms. Understanding this order explains how specialised parts cooperate to keep an organism alive.

  • Order of organisation: cell → tissue → organ → organ system → organism.
  • A cell is the basic structural and functional unit of life; a tissue is a group of similar cells doing one function, e.g. muscle tissue, xylem.
  • An organ is a group of different tissues working together for a specific function, e.g. the heart, stomach, leaf, root.
  • An organ system is a group of organs performing a major body function, e.g. digestive system (mouth, stomach, intestine, liver).
  • Unicellular organisms (Amoeba, Chlamydomonas) carry out all life activities in one cell; multicellular ones show division of labour among specialised cells.
  • Colonial organisms such as Volvox are loose groups of similar cells with little specialisation. Specialised cells include red blood cells, root hair cells and nerve cells.
  • Plant tissues: epidermis, xylem, phloem, parenchyma, meristem. Animal tissues: epithelial, muscle, connective, nervous. Ecological levels continue upward: population, community, ecosystem.
Worked example: Question: Arrange in order: organ, cell, organism, tissue, system. Answer: cell → tissue → organ → system → organism.
Exam trap: Students swap tissue and organ, or call a heart a tissue; remember that an organ contains several different tissues.
The Cell SS1 · Biology

The cell is the basic unit of structure and function in all living things. Its parts and their functions are heavily tested, including differences between plant and animal cells.

  • Cell theory: all living things are made of cells, the cell is the unit of life, and new cells arise from existing cells.
  • Cell membrane: thin, partially permeable layer controlling movement of substances in and out. Cytoplasm: jelly-like fluid where metabolic reactions occur.
  • Nucleus contains chromosomes (DNA) and controls cell activities and division. Mitochondria are the sites of aerobic respiration, releasing energy (ATP).
  • Ribosomes make proteins; endoplasmic reticulum transports materials; Golgi apparatus packages and secretes substances; lysosomes digest waste.
  • Plant cell only: cellulose cell wall (support and shape), chloroplasts with chlorophyll (photosynthesis), large permanent central vacuole with cell sap.
  • Animal cell: no wall or chloroplast, small temporary vacuoles, irregular shape, centrioles present, stores glycogen; plant cells store starch.
  • Light microscope magnification = eyepiece lens × objective lens, e.g. 10 × 40 = 400. Electron microscopes show organelles in detail such as ribosomes.
  • Specialised cells: root hair cell (absorption, large surface area), red blood cell (carries oxygen), palisade cell (photosynthesis), sperm cell (motile, fertilisation).
Worked example: Question: A microscope has a 10× eyepiece and a 40× objective lens. Total magnification = 10 × 40 = 400 times.
Exam trap: Students attribute cell walls to animal cells, call mitochondria the site of photosynthesis, or add (instead of multiply) lens magnifications.
Diffusion, Osmosis and Plasmolysis SS1 · Biology

These are passive processes by which substances move into and out of cells. They explain water uptake by roots, gaseous exchange and why plant cells become firm or limp.

  • Diffusion: net movement of molecules or ions from a region of higher concentration to lower concentration down a concentration gradient; no energy needed.
  • Examples of diffusion: gaseous exchange in alveoli and stomata, spread of perfume, absorption of digested food. Rate rises with temperature, steeper gradient, small molecules, large surface area.
  • Osmosis: movement of water molecules from a region of higher water potential (dilute solution) to lower water potential (concentrated solution) through a partially permeable membrane.
  • Solutions: hypotonic is more dilute than the cell sap, hypertonic is more concentrated, isotonic has equal concentration; water enters cells in hypotonic surroundings.
  • Turgidity: a plant cell in dilute solution takes in water, swells and becomes turgid; turgor pressure gives support to herbaceous stems and opens stomata.
  • Plasmolysis: in a hypertonic solution a plant cell loses water, the cytoplasm shrinks away from the cell wall and the cell becomes flaccid. Reversal is deplasmolysis.
  • Animal cells lack walls: in pure water they swell and burst (lysis); in strong solution they shrink (crenation). Osmosis experiments use a potato osmometer or visking tubing.
  • Active transport differs: it moves substances against a concentration gradient using energy (ATP), e.g. mineral ion uptake by root hairs.
Worked example: Question: A potato cylinder is placed in concentrated salt solution. Answer: it loses water by osmosis, becomes shorter, softer and lighter (flaccid), because the solution has lower water potential than the cell sap.
Exam trap: Students say osmosis moves solutes, or that water moves to the region of high water potential; water moves from high to low water potential, and plasmolysis occurs only in plant cells.
Basic Ecological Concepts SS1 · Biology

Ecology is the study of the relationships between organisms and their environment. These concepts are the vocabulary for all later ecology topics.

  • Ecology: the study of the interrelationships between living organisms and their environment. Environment includes biotic (living) and abiotic (non-living) factors.
  • Biotic factors: plants, animals, microbes. Abiotic factors: temperature, light, water, soil, pH, wind, humidity.
  • Ecosystem: a community of living organisms interacting with each other and their physical environment, e.g. pond, forest, farm, aquarium.
  • Habitat is the place where an organism lives, e.g. a pond for a tilapia. Niche is the role or function of the organism in its habitat, such as what it eats and its position in the food chain.
  • Population: all organisms of one species in an area. Community: all populations of different species living together in a habitat.
  • Biosphere is the part of the Earth where life exists. Producers (green plants) make food, consumers feed on others, decomposers (bacteria, fungi) break down dead matter.
  • Food chain: a linear transfer of energy, e.g. grass → grasshopper → toad → snake. Interlinked chains form a food web; energy flow is one-way and only about 10% passes to each level.
  • Symbiotic relationships include mutualism, commensalism and parasitism; competition and predation also occur between organisms.
Worked example: Question: Distinguish between habitat and niche. Answer: habitat is the place where an organism lives, while niche is the function or role it plays in that place.
Exam trap: Candidates mix up habitat and niche, or population and community; a population is only one species.
Biomes and Habitats SS1 · Biology

A biome is a large natural region with characteristic climate, plants and animals. Habitats are the specific places within biomes where organisms live.

  • Biome: a large ecological region defined by climate and dominant vegetation, e.g. tropical rainforest, savanna, desert, tundra.
  • Tropical rainforest: high rainfall (over 1500 mm), high humidity, tall trees in layers (canopy), little undergrowth light, rich species diversity, e.g. mahogany, iroko, monkeys. In southern Nigeria.
  • Savanna (guinea, sudan, sahel) in northern and middle Nigeria: grasses with scattered trees, distinct wet and dry seasons, e.g. acacia, locust bean; animals include lion, antelope, giraffe.
  • Marine (salt-water) habitat: about 3.5% salinity, includes ocean, sea, estuary, mangrove swamp; organisms include seaweed, sharks, crabs, with adaptation to salt.
  • Freshwater habitat: ponds, lakes, rivers, streams with low salt content; organisms include water lily, tilapia, tadpoles, frogs and Spirogyra.
  • Terrestrial habitats are on land (forest, grassland, desert, soil); aquatic habitats are in water. Other habitats include mangrove swamps and the soil.
  • Adaptations: savanna grasses have deep roots and fire-resistant bark; aquatic plants have air spaces for buoyancy; desert animals conserve water.
Worked example: Question: Name two features of the savanna that differ from tropical rainforest. Answer: savanna has scattered trees with tall grass and distinct wet and dry seasons, while rainforest has dense multi-layered tall trees and rain throughout most of the year.
Exam trap: Students mix rainforest and savanna features, or forget that mangrove swamps are brackish rather than fully freshwater or marine.
Ecological Factors SS1 · Biology

Ecological factors are the environmental conditions that affect organisms. They are measured with standard instruments, which are commonly tested.

  • Abiotic factors include climatic (temperature, rainfall, humidity, light, wind), edaphic or soil (pH, texture, water content) and topographic (altitude, slope) factors.
  • Temperature affects enzyme activity, metabolism, and distribution of organisms; measured with a thermometer, and a maximum-minimum thermometer records daily extremes.
  • Rainfall: provides water for life and affects vegetation type; measured in mm with a rain gauge.
  • Humidity: amount of water vapour in the air; measured with a hygrometer or wet-and-dry bulb (psychrometer). High humidity reduces transpiration and evaporation.
  • Other instruments: anemometer for wind speed, wind vane for wind direction, barometer for atmospheric pressure, light meter for light intensity, Secchi disc for water turbidity.
  • Biotic factors include competition, predation, parasitism, and decomposers; edaphic factors include soil pH measured with a pH meter or indicator paper.
  • Light affects photosynthesis, flowering and animal behaviour; wind affects seed dispersal, transpiration and soil erosion.
Worked example: Question: Which instrument measures relative humidity and which one measures rainfall? Answer: hygrometer (or wet-and-dry bulb thermometer) for humidity, and rain gauge for rainfall.
Exam trap: Candidates confuse the anemometer (wind speed) with the barometer (pressure), or treat soil pH as a climatic rather than edaphic factor.
Carbon and Nitrogen Cycles SS1 · Biology

Nutrient cycles show how essential elements pass between living things and the environment. They keep the supply of carbon and nitrogen available for life.

  • Carbon cycle: plants take in CO₂ (about 0.04% of air) by photosynthesis to make carbohydrates; carbon passes to animals through feeding.
  • Carbon returns to the air by respiration of plants, animals and microbes, by decay of dead matter, and by combustion of fossil fuels and wood.
  • Carbon is also stored in fossil fuels, limestone and shells; deforestation and burning fuels raise CO₂ and increase the greenhouse effect.
  • Nitrogen makes up about 78% of air but plants cannot use it directly; they absorb nitrates (NO₃⁻) from the soil to make proteins.
  • Nitrogen fixation: by nitrogen-fixing bacteria (Rhizobium in legume root nodules, Azotobacter free in soil) and by lightning forming nitrogen oxides.
  • Nitrification: decay releases ammonia, Nitrosomonas converts ammonium to nitrites, and Nitrobacter converts nitrites to nitrates.
  • Denitrification by Pseudomonas denitrificans converts nitrates back to nitrogen gas, reducing soil fertility; putrefying bacteria decompose dead protein to ammonia.
  • Farmers keep soil nitrogen by crop rotation with legumes (beans, groundnut) and manure or fertilisers.
Worked example: Question: Name the bacteria in legume root nodules and their role. Answer: Rhizobium, which fixes atmospheric nitrogen into nitrogen compounds usable by the plant.
Exam trap: Students claim plants take in nitrogen gas directly, or mix up nitrification (ammonia to nitrates) with denitrification (nitrates to nitrogen gas).
Plant Nutrition and Photosynthesis SS2 · Biology

Plant nutrition is how green plants obtain and use raw materials to make food, mainly by photosynthesis. It underlies all food chains and oxygen supply.

  • Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, using light energy absorbed by chlorophyll; it occurs in chloroplasts.
  • Light stage (grana): light splits water (photolysis) releasing O₂, H⁺ and ATP. Dark stage (stroma): CO₂ is reduced to glucose using ATP and hydrogen; it needs no light directly.
  • Leaf structure: waxy cuticle, upper epidermis, palisade mesophyll (most chloroplasts), spongy mesophyll with air spaces, vascular bundle, lower epidermis with stomata and guard cells.
  • Leaf adaptations: broad flat blade traps light, thin for short diffusion distance, stomata admit CO₂, veins supply water and remove sugar, palisade cells lie near the light.
  • Factors affecting rate: light intensity, CO₂ concentration, temperature and water; at a limiting factor, raising other factors does not increase the rate.
  • Starch test: decolourise leaf in hot ethanol, rinse, add iodine; blue-black means starch present. Experiments show need for light, CO₂ and chlorophyll (variegated leaf).
  • Mineral elements: nitrogen (proteins, chlorophyll; lack gives yellowing), magnesium (chlorophyll), phosphorus (roots, ATP), potassium (flowering), iron, calcium (cell wall). Absorbed as ions.
  • Products: glucose is converted to starch for storage, sucrose for transport, cellulose for cell walls, and with nitrates to amino acids/proteins.
Worked example: A leaf kept in the dark for 48 hours is exposed to light with part covered by black paper, then tested with iodine. Only the uncovered part turns blue-black, showing light is needed for photosynthesis.
Exam trap: Students say oxygen is a raw material (it is a product) and forget that plants also respire day and night. Do not confuse the light stage (needs light) with the dark stage (enzyme-controlled, no direct light).
Animal Nutrition and Digestion SS2 · Biology

Animal nutrition is the intake and use of food to supply energy, growth and repair. Digestion breaks large insoluble molecules into small soluble ones that can be absorbed.

  • Food classes: carbohydrates (energy), proteins (growth, repair), fats and oils (energy store, insulation), vitamins and mineral salts (regulation), water and roughage (fibre for peristalsis).
  • Food tests: starch, iodine gives blue-black; reducing sugar, Benedict's gives brick-red on heating; protein, Biuret gives violet; fat, ethanol emulsion test gives cloudy white; vitamin C, DCPIP decolourised.
  • Balanced diet contains all classes in correct proportions. Deficiency: kwashiorkor (protein), marasmus (energy), rickets (vitamin D, calcium), scurvy (vitamin C), beriberi (B₁), goitre (iodine), anaemia (iron).
  • Mammalian gut: mouth, oesophagus, stomach, duodenum, ileum, colon, rectum, anus. Mouth: teeth chew, saliva amylase digests starch to maltose; peristalsis moves food.
  • Stomach: pepsin (with HCl, acid pH about 2) digests protein to peptides; rennin curdles milk. Duodenum: bile (from liver) emulsifies fats; pancreatic amylase, trypsin and lipase act.
  • Ileum: maltase, sucrase, lactase, peptidases and lipase complete digestion. Ileum is adapted for absorption by villi, thin walls, capillaries and lacteals; fatty acids and glycerol enter lacteals.
  • Enzymes are protein biological catalysts that are specific, work at optimum temperature and pH, and are destroyed by high heat (denaturation). Colon absorbs water; liver stores glycogen and deaminates amino acids.
  • Dentition: incisors cut, canines tear, premolars and molars grind. Dental formula of humans: I 2/2, C 1/1, PM 2/2, M 3/3 = 32 teeth.
Worked example: Starch is digested to maltose by amylase, maltose to glucose by maltase; glucose is absorbed in the ileum into blood capillaries. Final products: carbohydrate to glucose, protein to amino acids, fat to fatty acids and glycerol.
Exam trap: Do not say absorption of food occurs mainly in the stomach or colon; it is the ileum. Bile contains no enzymes; it only emulsifies fats and neutralises acid chyme.
Transport System SS2 · Biology

Transport systems carry water, food, gases and wastes through the body of an organism. They allow large organisms to supply every cell despite a small surface area to volume ratio.

  • Water uptake: root hairs absorb water by osmosis; water moves through cortex to xylem. Xylem (dead, lignified vessels) carries water and minerals upward from roots to leaves.
  • Mechanisms of water rise: root pressure, capillarity, cohesion and adhesion of water molecules, and transpiration pull (main force) from evaporation in leaves.
  • Phloem (sieve tubes with companion cells) carries manufactured food such as sucrose both up and down the plant; this is translocation. Ringing experiment shows food moves in phloem.
  • Blood components: plasma (water, nutrients, hormones, urea); red cells with haemoglobin carry O₂ (no nucleus, biconcave); white cells fight infection; platelets help clotting.
  • Blood vessels: arteries have thick elastic walls, carry blood away from the heart under high pressure; veins have valves, thin walls; capillaries are one cell thick for exchange.
  • Heart has four chambers: two atria and two ventricles; left ventricle has the thickest wall. Valves prevent backflow: bicuspid on left, tricuspid on right.
  • Double circulation: pulmonary (heart to lungs, deoxygenated blood in pulmonary artery) and systemic (heart to body; aorta carries oxygenated blood). Blood groups A, B, AB, O; O is universal donor, AB universal recipient.
  • Lymphatic system returns tissue fluid to blood and carries fats from lacteals. Other roles of blood: temperature regulation, defence by phagocytosis and antibodies, clotting.
Worked example: Trace blood from the body to the lungs: vena cava, right atrium, right ventricle, pulmonary artery, lungs, pulmonary vein, left atrium, left ventricle, aorta. The pulmonary vein is the only vein carrying oxygenated blood.
Exam trap: Candidates reverse the xylem and phloem roles or say arteries always carry oxygenated blood (pulmonary artery does not). Remember transpiration pull needs the water column to be unbroken.
Respiration and Respiratory Surfaces SS2 · Biology

Respiration is the release of energy from food in living cells, while gaseous exchange is the intake of O₂ and removal of CO₂. Different animals use different respiratory surfaces.

  • Aerobic respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (about 2900 kJ per mole, 38 ATP); occurs in mitochondria and releases much energy.
  • Anaerobic respiration without oxygen: in yeast, glucose → ethanol + CO₂ + little energy (fermentation, used in baking and brewing); in muscle, glucose → lactic acid + little energy.
  • Oxygen debt: lactic acid builds in muscle during vigorous exercise; extra O₂ is taken in afterwards (heavy breathing) to oxidise it. Anaerobic gives about 2 ATP against 38 in aerobic.
  • Features of a good respiratory surface: large surface area, thin, moist, well supplied with blood or tracheoles, permeable to gases. Diffusion drives exchange.
  • Insects breathe through spiracles into tracheae and tracheoles, which carry air directly to cells; body movements ventilate. Amphibians use moist skin, buccal cavity and lungs; the skin is used mostly in water.
  • Fish: gills have filaments and lamellae with many capillaries; water enters mouth, passes over gills, leaves via the operculum. Blood flows opposite to water (countercurrent) for efficient uptake.
  • Mammals: nostrils, trachea (C-shaped cartilage rings), bronchi, bronchioles, alveoli (thin, moist, capillary-rich). Inhalation: ribs rise, diaphragm flattens, volume rises, pressure falls.
  • Experiments: limewater turns milky with CO₂; hydrogencarbonate indicator turns yellow with CO₂; germinating seeds in a flask release heat and CO₂. Inhaled air has 21% O₂, 0.03% CO₂; exhaled 16% O₂, 4% CO₂.
Worked example: Why do mammalian alveoli make good gaseous exchange surfaces? They are very numerous (large area), one cell thick, moist, and surrounded by dense capillaries that maintain a steep concentration gradient.
Exam trap: Do not confuse breathing (a physical process) with respiration (chemical, in cells). Lactic acid is formed in animal muscle, ethanol in yeast and plants; do not swap them.
Excretion SS2 · Biology

Excretion is the removal of metabolic waste products such as CO₂, urea and excess water from the body. It prevents poisoning and keeps the internal environment constant.

  • Excretion differs from egestion: excretion removes metabolic wastes made in cells; egestion removes undigested food (faeces) that never entered cells.
  • Amoeba: CO₂ and ammonia diffuse out across the cell surface; excess water is removed by the contractile vacuole. Earthworm: nephridia remove urea and water; skin also gives off CO₂.
  • Insects: Malpighian tubules remove nitrogenous waste (uric acid) from the haemolymph into the gut, so water is conserved; CO₂ leaves through tracheae.
  • Mammalian kidneys: renal artery brings blood, renal vein carries it away; ureter carries urine to the bladder, urethra to the outside. Cortex, medulla and pelvis are the regions.
  • Nephron is the functional unit: Bowman's capsule and glomerulus (ultrafiltration), proximal tubule (selective reabsorption of glucose, water, salts), loop of Henle, distal tubule, collecting duct.
  • Urea is formed in the liver by deamination of excess amino acids. Kidneys also regulate water and salt balance (osmoregulation) under control of ADH from the pituitary gland.
  • Lungs excrete CO₂ and water vapour; skin excretes sweat (water, salts, a little urea) from sweat glands and also helps in cooling; liver excretes bile pigments.
  • Kidney failure leads to build-up of urea; treatment includes dialysis and kidney transplant. Diabetes is shown by glucose in urine.
Worked example: Name the part of the nephron where ultrafiltration occurs and what is filtered. Answer: glomerulus and Bowman's capsule; water, glucose, urea and salts are filtered, while blood cells and large proteins are not.
Exam trap: Candidates call defecation excretion; it is egestion. Glucose should not appear in normal urine because it is fully reabsorbed; protein in urine suggests kidney damage.
Support and Movement SS2 · Biology

Support and movement describe how organisms keep their shape and move. Skeletons and supporting tissues provide protection, attachment for muscles and body form.

  • Hydrostatic skeleton: fluid under pressure in body cavity supports soft-bodied animals such as earthworm and jellyfish; circular and longitudinal muscles change body shape to move.
  • Exoskeleton: hard external covering of chitin in insects (cuticle) giving protection and support and reducing water loss; it must be shed (ecdysis/moulting) for growth.
  • Endoskeleton: internal bone and cartilage in vertebrates; functions are support, protection (skull, ribcage), movement, blood cell production in bone marrow, mineral storage.
  • Axial skeleton: skull, vertebral column (cervical, thoracic, lumbar, sacral, caudal), ribs, sternum. Appendicular skeleton: pectoral and pelvic girdles and limbs.
  • Joints: ball-and-socket (shoulder, hip) allows rotation; hinge (elbow, knee) one plane; pivot (atlas-axis); fixed or immovable (skull). Ligaments join bone to bone; tendons join muscle to bone.
  • Movement by antagonistic muscles: biceps contracts and triceps relaxes to bend the arm; the reverse extends it. Muscles can only pull, so they work in pairs.
  • Plant support: turgor pressure in parenchyma cells keeps herbaceous plants firm (wilting when turgor is lost); collenchyma (thickened corners) and sclerenchyma (lignified, dead fibres) give strength; xylem also supports woody plants.
  • Vertebrate types: fish use fins and myotomes, birds have light hollow bones, and the skull and limbs vary with mode of life.
Worked example: When the biceps contracts, the arm bends at the elbow while the triceps relaxes. These muscles are antagonistic because their actions oppose each other.
Exam trap: Do not confuse ligaments (bone to bone) with tendons (muscle to bone). Insects have an exoskeleton, not an endoskeleton, and wilting is due to loss of turgor, not loss of lignin.
Population Studies and Sampling SS2 · Biology

Population studies estimate how many organisms of a species live in an area and how they are spread. Since counting everything is impractical, ecologists use sampling tools.

  • Population is all individuals of one species in a habitat. Population density = total number of individuals ÷ area (or volume) of the habitat.
  • Frequency is how often a species occurs in sampling units: percentage frequency = (number of quadrats containing the species ÷ total quadrats) × 100.
  • Quadrat: a square frame (often 1 m × 1 m) placed randomly to count plants or slow-moving animals. Estimated population = mean count per quadrat × total area ÷ quadrat area.
  • Transect: a line or belt across a habitat; organisms touching or near the line are recorded at intervals. It shows changes in distribution along a gradient such as slope or moisture.
  • Capture-mark-release-recapture for mobile animals: population = (number first marked × total second catch) ÷ number of marked recaptured. Marks must be harmless and not affect survival.
  • Assumptions: random sampling, enough samples to be representative, no migration, birth or death in the interval, marked animals mix evenly.
  • Population growth is affected by natality, mortality, immigration and emigration; limiting factors include food, space, disease, predation and competition.
  • Abundance may be expressed as count, density, frequency or percentage cover; use of a larger number of quadrats improves reliability.
Worked example: A farmer places 10 quadrats of 1 m² and counts 4, 6, 5, 3, 7, 5, 4, 6, 5, 5 plants (total 50, mean 5). For a 400 m² field, estimated population = 5 × 400 = 2000 plants.
Exam trap: Wrongly dividing by the quadrat count instead of finding the mean, or ignoring units of area. In recapture, multiply first and second catches then divide by recaptured, not add them.
Ecological Succession SS2 · Biology

Ecological succession is the gradual, orderly replacement of one community by another in an area until a stable climax is reached. It explains how bare land becomes forest.

  • Succession is a directional change in species composition over time; each community modifies the environment, making it suitable for the next one.
  • Primary succession begins on a bare area never occupied before (bare rock, new sand dune, lava, cooled volcanic land); it is very slow because soil must first form.
  • Secondary succession occurs where an existing community was destroyed but soil remains (abandoned farmland, after bush fire or flooding); it is faster than primary succession.
  • Primary sequence on rock: pioneer species (lichens, algae, mosses) weather rock and add humus, then herbs and grasses, then shrubs, then trees. Pioneers are the first colonisers.
  • Seral stages are the intermediate communities; the final stable community in balance with climate is the climax community (for example tropical rainforest in southern Nigeria, savanna in the north).
  • Trends during succession: increase in species diversity, biomass, soil depth and humus, and complexity of food webs; physical conditions become less extreme.
  • Hydrosere (aquatic): plankton and submerged plants, floating plants, reeds and sedges, then marsh and land plants as the pond fills with silt.
  • Human activities such as bush burning, farming and deforestation can stop or reverse succession (a plagioclimax).
Worked example: Sequence on a bare rock: lichens → mosses → herbs/grasses → shrubs → forest trees (climax). Lichens are the pioneer species because they can survive on bare rock with little water.
Exam trap: Candidates mix up primary and secondary succession; soil already present means secondary. The pioneer community is not the climax community, and succession proceeds toward more stability.
Homeostasis: Kidney, Liver and Skin SS3 · Biology

Homeostasis is the maintenance of a constant internal environment (water, temperature, glucose, pH, osmotic pressure). The kidney, liver and skin are the main organs that regulate and excrete.

  • Homeostasis keeps internal conditions steady despite external changes; it works by negative feedback, in which a change triggers a response that reverses it.
  • Kidney structure: cortex, medulla and pelvis; the nephron (Bowman's capsule with glomerulus, proximal tubule, loop of Henle, distal tubule, collecting duct) is the functional unit.
  • Ultrafiltration in the glomerulus forces water, glucose, salts and urea into Bowman's capsule; selective reabsorption in the proximal tubule returns all glucose and most water and salts to blood.
  • Osmoregulation: when blood is too concentrated, the pituitary releases more ADH, making the collecting duct more permeable, so more water is reabsorbed and little concentrated urine is produced.
  • Liver functions: converts excess amino acids to urea (deamination), stores glycogen (insulin/glucagon control), detoxifies drugs and alcohol, makes bile, bile pigments and plasma proteins, stores iron and vitamins.
  • Skin: sweat glands lose water and heat by evaporation, hair erector muscles and vasodilation/vasoconstriction of skin capillaries regulate heat; fat layer insulates; the skin also excretes urea and salts in sweat.
  • Cold response: vasoconstriction, shivering, hairs erect, less sweating. Hot response: vasodilation, sweating, hairs lie flat. The hypothalamus is the body's thermostat.
  • Excretion removes metabolic waste: urea (kidney, liver), CO₂ (lungs), bile pigments (liver) and water, salts and urea (skin). Kidney failure may be treated by dialysis or transplant.
Worked example: A patient passes glucose in urine. Reason: the glucose in the glomerular filtrate exceeded the amount the proximal tubule can reabsorb, as in diabetes mellitus where insulin is deficient and blood glucose is high.
Exam trap: Do not confuse excretion (removal of metabolic waste) with egestion (removal of undigested food) or secretion. Students also say the skin controls temperature by 'capillaries moving closer to the surface' when capillaries do not move; they dilate.
Hormonal Coordination SS3 · Biology

Hormones are chemical messengers made by endocrine glands or plant tissues that regulate growth, metabolism and reproduction. They coordinate slowly but with long-lasting effects.

  • Plant hormones: auxins (IAA) made in shoot and root tips promote cell elongation, apical dominance, tropisms and root formation in cuttings; high concentrations act as selective weedkillers (2,4-D).
  • Gibberellins promote stem elongation, break seed dormancy, stimulate germination and flowering and form seedless fruits; other plant hormones include cytokinins, abscisic acid and ethylene (fruit ripening).
  • Phototropism: auxin moves to the shaded side of a shoot, causing greater elongation there so the shoot bends towards light; roots are positively geotropic and shoots negatively geotropic.
  • Mammalian endocrine glands are ductless and release hormones directly into blood; hormones act only on target organs with specific receptors.
  • Pituitary (master gland): growth hormone, ADH, FSH, LH, TSH. Thyroid: thyroxine controls metabolic rate (lack of iodine gives goitre). Adrenal: adrenaline for fight or flight, raising heart rate and blood sugar.
  • Pancreas (islets of Langerhans): insulin lowers blood glucose by converting it to glycogen; glucagon raises it. Lack of insulin causes diabetes mellitus.
  • Gonads: testes secrete testosterone (male secondary sexual characters); ovaries secrete oestrogen (female characters, uterine lining repair) and progesterone (maintains pregnancy and uterine lining).
  • Nervous versus hormonal control: nerves give fast, short-lived, localised response by impulses; hormones give slow, long-lasting, widespread response by blood transport.
Worked example: Name the hormone that lowers blood sugar and its source. Answer: insulin, from the beta cells of the islets of Langerhans in the pancreas.
Exam trap: Do not say hormones are carried by ducts or nerves; endocrine glands are ductless and use blood. Do not mix insulin (lowers glucose) with glucagon (raises it), or auxin with gibberellin functions.
Nervous Coordination SS3 · Biology

The nervous system detects stimuli and sends electrical impulses to produce rapid responses. It consists of the central and peripheral nervous systems.

  • The central nervous system (CNS) is the brain and spinal cord; the peripheral nervous system (PNS) is the cranial and spinal nerves, which carry impulses to and from the CNS.
  • The neurone has a cell body, dendrites, an axon with myelin sheath and nodes of Ranvier, and axon terminals. Types: sensory (to CNS), relay/intermediate, motor (to effector).
  • Brain parts: cerebrum (thinking, memory, voluntary action, senses), cerebellum (balance and coordination), medulla oblongata (heart rate, breathing), hypothalamus (temperature, water balance), pituitary below.
  • Spinal cord: grey matter inside (H-shaped) and white matter outside; it conducts impulses to the brain and coordinates reflex actions.
  • A synapse is the junction between two neurones; impulses cross by release of a chemical transmitter (acetylcholine) from the terminal, which diffuses across the gap. Impulses pass one way only.
  • Reflex arc: stimulus, receptor, sensory neurone, relay neurone in spinal cord, motor neurone, effector (muscle/gland), response. Reflexes are rapid, automatic and protective, e.g. knee jerk, withdrawing a hand from heat.
  • Voluntary actions are controlled by the cerebrum; involuntary actions such as peristalsis and heartbeat are controlled by the autonomic nervous system (sympathetic and parasympathetic).
  • Disorders and damage: poliomyelitis, meningitis, epilepsy and stroke affect the nervous system; alcohol and drugs slow reaction time.
Worked example: Describe the path of a reflex action when a hand touches a hot pot. Answer: heat is detected by skin receptors, sensory neurone carries impulse to the spinal cord, relay neurone passes it to a motor neurone, and the arm muscle (effector) contracts to withdraw the hand.
Exam trap: Do not say reflex action involves the brain first; the brain is informed afterwards. Do not confuse sensory (towards CNS) with motor (away from CNS) neurones, or reflex arc with reflex action.
Sense Organs SS3 · Biology

Sense organs contain receptors that detect stimuli such as light, sound, chemicals, touch and temperature. The eye and ear are the two most examined.

  • Eye parts: sclera, cornea (refracts light), choroid, iris (controls pupil size), lens (focuses), ciliary muscles and suspensory ligaments, aqueous and vitreous humours, retina (rods and cones), blind spot, optic nerve.
  • Rods work in dim light and see black and white; cones work in bright light and see colour. The fovea has the most cones and gives sharpest vision.
  • Accommodation: for near objects ciliary muscles contract, suspensory ligaments slacken and the lens becomes thicker and rounder; for distant objects the reverse happens.
  • Pupil reflex: in bright light circular iris muscles contract and the pupil narrows; in dim light radial muscles contract and the pupil widens.
  • Eye defects: short-sight (myopia, image forms in front of retina) is corrected by a concave lens; long-sight (hypermetropia, image behind retina) by a convex lens; astigmatism by cylindrical lens; cataract by surgery.
  • Ear parts: pinna, auditory canal, eardrum, ear ossicles (malleus, incus, stapes), oval window, cochlea (hearing receptors), semicircular canals (balance), Eustachian tube (equalises pressure), auditory nerve.
  • Hearing: sound waves vibrate the eardrum, ossicles amplify and pass the vibrations to the oval window, fluid in the cochlea moves, and sensory hair cells send impulses along the auditory nerve.
  • Ear defects include deafness, ear infections and wax blockage; hearing aids and avoiding loud noise help. Other receptors: tongue (taste buds), nose (smell), skin (touch, pain, heat, cold).
Worked example: A student cannot see distant objects clearly. Answer: short-sightedness (myopia); the eyeball is too long or the lens too strong, so the image falls in front of the retina; wear a concave (diverging) lens.
Exam trap: Do not mix up the concave lens for myopia with the convex lens for long-sight. Do not say the lens is the main refractor; most refraction occurs at the cornea, and the lens does fine focusing.
Reproduction in Plants SS3 · Biology

Plants reproduce asexually (one parent, identical offspring) and sexually (flowers, gametes, seeds). Understanding both is key to farming and crop improvement.

  • Asexual methods: vegetative propagation by runners (grass), rhizomes (ginger), tubers (yam), bulbs (onion), suckers (banana), stem cuttings (cassava) and budding (yeast); offspring are genetically identical to the parent.
  • Asexual reproduction is fast and keeps good traits but gives no variation, so crops are all vulnerable to the same disease.
  • Flower parts: sepals (calyx), petals (corolla), stamen (anther and filament, male) and carpel/pistil (stigma, style, ovary with ovules, female). Receptacle is the base.
  • Pollination is the transfer of pollen from anther to stigma; self-pollination is within the same flower or plant, cross-pollination is between different plants of the same species and gives more variation.
  • Insect-pollinated flowers have large coloured petals, scent, nectar and sticky pollen; wind-pollinated flowers have small dull petals, long feathery stigmas and light abundant pollen, e.g. maize and grasses.
  • Fertilisation: the pollen grain forms a pollen tube down the style to the ovule through the micropyle; the male nucleus fuses with the egg cell to form a zygote, and a second male nucleus fuses with polar nuclei (double fertilisation).
  • After fertilisation the ovule becomes the seed, the ovary becomes the fruit, and other floral parts wither; the zygote forms the embryo.
  • Fruits and seeds are dispersed by wind, water, animals and explosive mechanism; this reduces competition and spreads the species.
Worked example: Name the part of the flower that develops into a fruit after fertilisation. Answer: the ovary; the ovules in it become seeds.
Exam trap: Do not confuse pollination (transfer of pollen) with fertilisation (fusion of gametes). Do not say the whole flower becomes the fruit, or that stigma and style are the male parts.
Reproduction in Animals SS3 · Biology

Animal reproduction involves male and female reproductive systems, gamete formation, fertilisation and development of the embryo. It ensures continuity of species.

  • Male system: testes (in scrotum, make sperm and testosterone), epididymis, sperm duct (vas deferens), seminal vesicle and prostate gland (seminal fluid), urethra and penis.
  • Female system: ovaries (make ova and hormones), Fallopian tubes (oviducts, site of fertilisation), uterus (womb), cervix, vagina.
  • Gametogenesis: spermatogenesis in the testes produces many small motile sperm with head, middle piece and tail; oogenesis in the ovaries produces a large, nutrient-rich, non-motile ovum. Both involve meiosis.
  • Menstrual cycle lasts about 28 days: ovulation occurs about day 14, oestrogen rebuilds the uterine lining, progesterone maintains it; if no fertilisation, menstruation occurs.
  • Fertilisation: a sperm fuses with an ovum in the Fallopian tube to form a diploid zygote, which divides by mitosis to form a ball of cells (blastocyst) and implants in the uterine wall.
  • Embryonic development: the embryo is protected by the amniotic sac and amniotic fluid; the placenta supplies oxygen and food and removes waste via the umbilical cord. Gestation in humans is about 9 months.
  • Parental care and birth: labour is stimulated by hormones, contractions push the baby out; breast milk gives nutrition and antibodies. Twins: identical (one zygote) or fraternal (two ova).
  • Contraception and STIs: condoms, pills, IUD and abstinence prevent pregnancy; condoms also reduce spread of HIV/AIDS, gonorrhoea and syphilis.
Worked example: State where fertilisation occurs in humans and what the zygote does next. Answer: in the Fallopian tube (oviduct); it divides repeatedly while moving to the uterus, where it implants.
Exam trap: Do not say fertilisation happens in the uterus, or that the placenta mixes mother's and baby's blood; materials pass by diffusion across the placental membrane. Do not confuse sperm duct with urethra.
Heredity and Variation SS3 · Biology

Heredity is the passing of characters from parents to offspring through genes; variation is the differences between individuals. Mendel's work explains the rules of inheritance.

  • Key terms: gene, allele, dominant, recessive, genotype, phenotype, homozygous (pure), heterozygous (hybrid), F₁ and F₂ generations, gamete, chromosome, DNA.
  • Mendel's first law (segregation): each character is controlled by a pair of factors that separate during gamete formation so each gamete carries only one.
  • Monohybrid cross: Tt × Tt gives genotypes 1 TT : 2 Tt : 1 tt and phenotypic ratio 3 dominant : 1 recessive in F₂. A test cross with the homozygous recessive shows if an organism is TT or Tt.
  • Mendel's second law (independent assortment): factors for different characters assort independently. Dihybrid cross of RrYy × RrYy gives 9 : 3 : 3 : 1 phenotypes.
  • Incomplete dominance (red × white giving pink) and sex determination: XX female, XY male; the father's sperm decides the sex of the child, with a 50 percent chance each.
  • Sex-linked traits such as colour blindness and haemophilia are carried on the X chromosome, so they appear more often in males.
  • Discontinuous variation has distinct categories with no intermediates (blood group, tongue rolling, sex) and is genetic; continuous variation shows a range (height, weight, skin colour) and is affected by genes and environment.
  • Causes of variation: meiosis (crossing over, independent assortment), random fertilisation and mutation; mutagens include X-rays and chemicals.
Worked example: Cross two heterozygous tall pea plants (Tt × Tt). Gametes T or t from each; offspring TT, Tt, Tt, tt, so 3 tall : 1 short, a 3 : 1 ratio, and 25 percent are short.
Exam trap: Do not confuse genotype with phenotype, or allele with gene. Remember that 3:1 is a phenotypic ratio whereas 1:2:1 is the genotypic ratio, and F₂ ratios are only expected from large samples.
Applications of Genetics SS3 · Biology

Genetics is applied in medicine, forensics and agriculture to predict inheritance of blood groups, diseases and to identify individuals. It helps in counselling and crime detection.

  • ABO blood groups are controlled by three alleles: Iᴬ and Iᴮ are codominant and i is recessive. Group A is IᴬIᴬ or Iᴬi, B is IᴮIᴮ or Iᴮi, AB is IᴬIᴮ, O is ii.
  • Blood transfusion: group O is the universal donor (no A or B antigens), AB the universal recipient; mismatches cause agglutination (clumping) of red cells.
  • Rhesus factor: Rh-positive people have the Rh antigen (dominant); Rh-negative lack it. A Rh-negative mother carrying a Rh-positive baby may form antibodies, endangering later Rh-positive babies.
  • Sickle cell: caused by a mutated haemoglobin gene (HbS). HbAA is normal, HbAS is the carrier (trait, mostly healthy, resists malaria), HbSS has sickle cell anaemia.
  • Two carriers (AS × AS) have 25 percent AA, 50 percent AS and 25 percent SS children; marriage between AS partners is discouraged, and genotype testing is advised before marriage.
  • DNA fingerprinting (profiling) compares unique DNA banding patterns, used in paternity tests, crime scene identification and tracing relatives.
  • Other applications: selective breeding, hybrid vigour in crops and livestock, genetic engineering (insulin production), and genetic counselling. Inherited diseases include albinism, haemophilia and colour blindness.
  • Human genotypes in blood: a child's blood group cannot contain an allele absent from both parents, so groups can exclude, but never prove, paternity.
Worked example: A man of genotype HbAS marries a woman HbAS. Probability of an HbSS child: gametes A or S from each; offspring AA, AS, AS, SS; so 1 in 4, or 25 percent.
Exam trap: Do not say group O has no alleles, or that sickle cell trait (AS) is the disease; only SS has the disease. Do not forget that groups A and B can be heterozygous, so two A parents can have an O child.
Evolution SS3 · Biology

Evolution is the gradual change in living organisms over generations, giving rise to new species. Lamarck and Darwin gave the main theories, and fossils, anatomy and DNA give the evidence.

  • Lamarck's theory: use and disuse of organs and inheritance of acquired characters, e.g. the giraffe stretching its neck and passing the longer neck to offspring. It is rejected because acquired traits are not inherited.
  • Darwin's theory of natural selection: overproduction of offspring, variation, struggle for existence, survival of the fittest, and passing of favourable traits to the next generation.
  • Darwin's example: giraffes with naturally longer necks reached more food, survived and bred more, so the long-neck trait became common over many generations.
  • Modern examples: antibiotic-resistant bacteria, pesticide-resistant insects and industrial melanism in peppered moths show natural selection in action.
  • Evidence for evolution: fossils (e.g. Archaeopteryx), homologous structures (pentadactyl limb) showing divergent evolution, vestigial organs (appendix), comparative embryology, biochemistry and DNA similarity.
  • Mutations, isolation and genetic drift supply new variation; species form when populations are isolated so they cannot interbreed (speciation).
  • Adaptive radiation and convergent evolution: related organisms adapt to different niches (Darwin's finches); unrelated organisms develop similar analogous structures, e.g. wings of bird and insect.
  • Artificial selection by humans (cattle, maize, dogs) is similar to natural selection but with humans choosing the traits.
Worked example: Explain how bacteria become resistant to an antibiotic. Answer: random mutation produces a few resistant bacteria; the antibiotic kills the others, resistant ones survive and reproduce, and the resistant gene becomes common (natural selection).
Exam trap: Do not attribute 'inheritance of acquired characters' to Darwin; it is Lamarck's idea. Do not say individuals evolve; populations evolve over generations, and 'survival of the fittest' means best adapted, not strongest.
Alkenes and Addition Reactions SS2 · Chemistry

Alkenes are unsaturated hydrocarbons; the C=C double bond makes them reactive.

  • Ethene C2H4 is made by dehydrating ethanol with concentrated H2SO4 at 170 C.
  • Test for unsaturation: it decolourises bromine water (reddish-brown to colourless) and acidified KMnO4 (purple to colourless).
  • Addition reactions: hydrogenation (Ni catalyst), halogenation, hydration.
  • Many ethene molecules join by addition polymerisation to make polyethene.
Exam trap: Bromine water is decolourised by alkenes and alkynes but not by alkanes. Do not say it turns the solution 'clear' without saying colourless.
Benzene and Aromatic Compounds SS3 · Chemistry

Benzene (C6H6) is a planar ring with delocalised electrons, which makes it unusually stable.

  • All C-C bonds in benzene are equal in length; the structure is a resonance hybrid.
  • It mainly undergoes substitution (nitration, halogenation, alkylation), not addition.
  • Nitration uses concentrated HNO3 with concentrated H2SO4 to give nitrobenzene.
  • Addition needs harsh conditions, e.g. hydrogenation to cyclohexane with a Ni catalyst.
Exam trap: Do not show benzene decolourising bromine water. It does not, because it is not a simple alkene. Remember it burns with a smoky, sooty flame.
Toluene and Benzene Derivatives SS3 · Chemistry

Toluene is methylbenzene, benzene with one hydrogen replaced by a CH3 group.

  • Formula C6H5CH3 (C7H8).
  • The methyl group makes the ring more reactive than benzene in substitution.
  • It is a common solvent and the starting point for making TNT and other chemicals.
  • Compare: phenol is C6H5OH and benzoic acid is C6H5COOH.
Exam trap: Toluene is not an alkane or alkene, it is an aromatic compound. Do not mix up toluene (C6H5CH3) with phenol (C6H5OH).
Viruses and the Immune System SS3 · Biology

Viruses are tiny non-cellular particles that reproduce only inside living host cells.

  • A virus has genetic material (DNA or RNA) in a protein coat.
  • HIV attacks CD4 helper T cells, weakening immunity and leading to AIDS.
  • HIV spreads through blood, sexual contact, and from mother to child.
  • Vaccines contain weakened or dead antigens that trigger memory cells.
Exam trap: HIV is the virus; AIDS is the syndrome it causes. Antibiotics kill bacteria, not viruses.
Binomial Expansion and Pascal's Triangle SS2 · Mathematics

Pascal's triangle gives the coefficients when you expand (a + b)^n.

  • Each number is the sum of the two numbers above it; rows start 1, 1 1, 1 2 1, 1 3 3 1 ...
  • (a + b)^n has n + 1 terms; powers of a fall from n to 0 while powers of b rise from 0 to n.
  • The powers in every term add up to n.
  • Pascal's pyramid extends the idea to three terms, (a + b + c)^n.
Exam trap: Row n of the triangle has n + 1 numbers (the top row is row 0). When b is negative, the signs alternate.