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1Getting to know the examBeginner

What is the CSCA? Structure, registration and scores

Open lesson

2Mathematics: sets, inequalities and functionsIntermediate

Sets and set operations

Open lesson
subsets: 2ⁿ proper subsets: 2ⁿ − 1 non-empty proper subsets: 2ⁿ − 2
where:
  • nthe number of elements in the set

Each element is either in the subset or out of it: 2 choices per element, 2ⁿ in total. For proper subsets remove the set itself; for non-empty proper subsets remove ∅ as well.

A ∪ B = {x | x ∈ A or x ∈ B} A ∩ B = {x | x ∈ A and x ∈ B} ∁ᵤA = {x | x ∈ U, x ∉ A}
where:
  • A ∪ Bunion (并集): the elements of at least one of the sets
  • A ∩ Bintersection (交集): the elements of both sets
  • ∁ᵤAthe complement (补集) of A in the universal set U (全集); other books write Ā or Aᶜ

Union means “or”, intersection “and”, complement “not”. A complement is always taken in U; for sets of numbers with no U given, U = R. Always A ∩ B ⊆ A ⊆ A ∪ B, and both A ∩ B = A and A ∪ B = B mean A ⊆ B.

∁ᵤ(A ∪ B) = (∁ᵤA) ∩ (∁ᵤB) ∁ᵤ(A ∩ B) = (∁ᵤA) ∪ (∁ᵤB)
where:
  • ∁ᵤthe complement in the universal set U

De Morgan’s laws: the complement turns a union into an intersection and an intersection into a union. “Not (A or B)” = “not A and not B”.

|A ∪ B| = |A| + |B| − |A ∩ B|
where:
  • |A|the number of elements of a finite set A (also written n(A) or card(A))

Adding |A| and |B| counts the common part twice, so it is subtracted once.

|A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|
where:
  • |A ∩ B ∩ C|the number of elements in all three sets

Add the singles, subtract the pairs, add back the triple: the centre is added 3 times and subtracted 3 times, so it must be added once more.

P ⊆ Q ⇔ (p ⇒ q)
where:
  • Pthe set of all x that satisfy p: P = {x | p(x)}
  • Qthe set of all x that satisfy q: Q = {x | q(x)}

The set picture: a condition with a smaller solution set implies a condition with a larger one. “Small ⇒ large”: the smaller set is the sufficient condition, the larger set the necessary one (in Chinese textbooks 小范围推出大范围).

Inequalities: properties, quadratic and rational inequalities

Open lesson
a > b, c > 0 ⇒ ac > bc; a > b, c < 0 ⇒ ac < bc
where:
  • a, bthe two sides of the inequality
  • cthe number both sides are multiplied (or divided) by

The only rule that reverses the sign: multiplying or dividing by a negative number. You may not multiply by an expression of unknown sign, such as x − 3, at all.

a > 0, D > 0, x₁ < x₂: ax² + bx + c > 0 ⇔ x < x₁ or x > x₂; ax² + bx + c < 0 ⇔ x₁ < x < x₂
where:
  • athe coefficient of x² (opens upwards when a > 0)
  • x₁, x₂the roots of ax² + bx + c = 0
  • Dthe discriminant, D = b² − 4ac (判别式)

“Greater — outside the roots, less — between the roots” (when a > 0). If a < 0, first multiply the inequality by −1 and reverse the sign.

ax² + bx + c > 0 for every x ∈ R ⇔ a > 0 and D < 0 (or a = b = 0, c > 0)
where:
  • Dthe discriminant: D < 0 means the parabola does not meet the axis
  • a = 0the inequality becomes linear and is checked separately

For “≥ 0 for every x”, D ≤ 0 (touching is allowed). For “< 0 for every x”, a < 0 and D < 0.

f(x)/g(x) > 0 ⇔ f(x) · g(x) > 0; f(x)/g(x) ≥ 0 ⇔ f(x) · g(x) ≥ 0 and g(x) ≠ 0f(x)/g(x) > 0 ⇔ f(x) · g(x) > 0; f(x)/g(x) ≥ 0 ⇔ f(x) · g(x) ≥ 0 and g(x) ≠ 0
where:
  • f(x)the numerator
  • g(x)the denominator; its zeros are never in the answer

A quotient and a product have the same sign, so a fractional inequality (分式不等式) is replaced by a product; then the interval method (穿针引线法, “threading the needle”) finishes the job.

|x| < a ⇔ −a < x < a; |x| > a ⇔ x < −a or x > a (a > 0)
where:
  • |x − c|the distance between x and c on the number line (绝对值, absolute value)
  • aa positive number: the allowed distance

|x − c| < a: x is less than a away from c; |x − c| > a: x is more than a away from c. For |f(x)| < a solve the double inequality −a < f(x) < a.

(a + b)/2 ≥ √(ab) ⇔ a + b ≥ 2√(ab) (a > 0, b > 0)(a + b)/2 ≥ √(ab) ⇔ a + b ≥ 2√(ab) (a > 0, b > 0)
where:
  • (a + b)/2(a + b)/2the arithmetic mean (算术平均数)
  • √(ab)the geometric mean (几何平均数)

The “basic inequality” (基本不等式) of Chinese textbooks, also called the Cauchy or AM–GM inequality. It follows from (√a − √b)² ≥ 0, and equality holds only when a = b. A fixed product gives the sum a minimum of 2√(ab); a fixed sum s gives the product a maximum of (s/2)².

Functions and their properties

Open lesson
u/v: v ≠ 0 √u, ⁴√u, …: u ≥ 0 logₐ u: u > 0 u⁰: u ≠ 0u/v: v ≠ 0 √u, ⁴√u, …: u ≥ 0 logₐ u: u > 0 u⁰: u ≠ 0
where:
  • u, vexpressions containing x
  • √u, ⁴√uroots of even degree; odd roots such as ∛u accept any u
  • logₐ uthe argument of a logarithm (see “Logarithms and logarithmic functions”)

Write every condition, then take their intersection and give the answer as a set or intervals. In a word problem the situation adds its own conditions: lengths are positive, numbers of objects are whole.

y = (ax + b)/(cx + d) (c ≠ 0, ad − bc ≠ 0): x ≠ −d/c, y ≠ a/cy = (ax + b)/(cx + d) (c ≠ 0, ad − bc ≠ 0): x ≠ −d/c, y ≠ a/c
where:
  • a/ca/cthe ratio of the leading coefficients: the horizontal asymptote y = a/c
  • −d/c−d/cthe zero of the denominator: the vertical asymptote

Why y ≠ a/c: solving y = (ax + b)/(cx + d) for x gives x = (b − dy)/(cy − a), which exists for every y except a/c. The graph is the hyperbola y = k/x moved so that its centre is (−d/c, a/c).

(f ∘ g)(x) = f(g(x)), x ∈ D(g), g(x) ∈ D(f)
where:
  • gthe inner function: it acts first
  • fthe outer function: it acts on the value g(x)
  • D(f), D(g)the domains

A composite function (复合函数): the order matters, f(g(x)) ≠ g(f(x)) in general. If D(f) = [m, n], the domain of f(g(x)) is the solution set of m ≤ g(x) ≤ n.

x₁ < x₂ ⇒ f(x₁) < f(x₂) — increasing; x₁ < x₂ ⇒ f(x₁) > f(x₂) — decreasing
where:
  • x₁, x₂any two numbers of the interval I
  • Ia monotonic interval (单调区间): the function increases (增函数) or decreases (减函数) on the whole of it

Increasing + increasing is increasing; −f reverses the direction, and so does 1/f when f keeps one sign. For y = f(g(x)): if f and g change in the same direction the composite increases, if in opposite directions it decreases — 同增异减. The derivative gives the monotonicity of any function: “Applications of the derivative: monotonicity and extrema”.

f(−x) = f(x) — even; f(−x) = −f(x) — odd; odd and 0 ∈ D ⇒ f(0) = 0
where:
  • Dthe domain; it must be symmetric about 0, otherwise f is neither even nor odd
  • f(0) = 0from f(−0) = −f(0): an odd function defined at 0 passes through the origin

odd ± odd = odd, even ± even = even; odd · odd = even, even · even = even, odd · even = odd — like the signs + and −. A polynomial with only even powers (and a constant) is even; with only odd powers, odd.

f(x + T) = f(x); f(x + a) = −f(x) ⇒ T = 2a; f(x + a) = 1/f(x) ⇒ T = 2af(x + T) = f(x); f(x + a) = −f(x) ⇒ T = 2a; f(x + a) = 1/f(x) ⇒ T = 2a
where:
  • Ta period
  • aa constant shift, a ≠ 0

Proof of the second rule: f(x + 2a) = −f(x + a) = −(−f(x)) = f(x). To find f at a large argument, subtract whole periods (look at the remainder), then use parity if needed. The periods of sin and cos are in “Graphs of trigonometric functions”.

y = f(x − a) + b; y = A·f(x); y = f(ωx); y = −f(x); y = f(−x); y = −f(−x); y = |f(x)|; y = f(|x|)
where:
  • a, bshift: a > 0 to the right, b > 0 upwards (a < 0 left, b < 0 down)
  • A, ωvertical stretch by A: (x, y) → (x, Ay); for ω > 1 horizontal compression by ω: (x, y) → (x/ω, y)
  • −f(x), f(−x), −f(−x)reflections in the x-axis, in the y-axis and in the origin
  • |f(x)|, f(|x|)|f(x)|: the parts below the x-axis are flipped up; f(|x|): the right half is kept and mirrored to the left

Changes inside the brackets act horizontally and “the other way round” (x − 2 moves right, 2x compresses); changes outside act vertically and “as written”. Chinese students remember the shifts as 左加右减,上加下减: “left add, right subtract; up add, down subtract”.

y = f(x) ⇔ x = f⁻¹(y); D(f⁻¹) = E(f), E(f⁻¹) = D(f); f(a) = b ⇔ f⁻¹(b) = a
where:
  • D(f), E(f)the domain and the range of f; for the inverse they swap
  • (a, b)a point of the graph of f; then (b, a) is on the graph of f⁻¹

The graphs of f and f⁻¹ are symmetric about the line y = x. The most important pair, y = aˣ and y = logₐ x, is in “Logarithms and logarithmic functions”. Note that f⁻¹(x) is not 1/f(x).

f even, increasing on [0, +∞): f(x₁) < f(x₂) ⇔ |x₁| < |x₂|
where:
  • |x₁|, |x₂|distances from 0: an even function takes equal values at equal distances
  • f oddif it increases on [0, +∞), it increases on the whole of R, so f(x₁) < f(x₂) ⇔ x₁ < x₂

If the even function decreases on [0, +∞), reverse the inequality: f(x₁) < f(x₂) ⇔ |x₁| > |x₂|.

Power and exponential functions

Open lesson
a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ, a^(−m/n) = 1 / a^(m/n), a⁰ = 1a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ, a^(−m/n) = 1 / a^(m/n), a⁰ = 1
where:
  • athe base: a > 0 (0 to a positive power is 0; 0⁰ and 0 to a negative power are undefined)
  • m, npositive integers, n > 1 (Chinese textbooks: m, n ∈ N*)

The denominator of the exponent is the root, the numerator is the power. Take the root first to keep numbers small: 27^(2/3) = (∛27)² = 9, not ∛729. A negative exponent means the reciprocal — it never makes the number negative.

aʳ · aˢ = aʳ⁺ˢ aʳ / aˢ = aʳ⁻ˢ (aʳ)ˢ = aʳˢ (ab)ʳ = aʳbʳaʳ · aˢ = aʳ⁺ˢ aʳ / aˢ = aʳ⁻ˢ (aʳ)ˢ = aʳˢ (ab)ʳ = aʳbʳ
where:
  • a, bpositive bases
  • r, sany rational (and, later, real) exponents

The same laws as for whole exponents. There is no rule for a sum: aʳ + aˢ is not a^(r + s), and (a + b)ʳ is not aʳ + bʳ.

y = x^α: (1, 1) on the graph; α > 0 ⇒ increasing on [0, +∞), through (0, 0); α < 0 ⇒ decreasing on (0, +∞)
where:
  • αthe constant exponent (any real number)
  • x > 1, 0 < x < 1for x > 1 a larger α gives a higher graph; for 0 < x < 1 a larger α gives a lower one

In the first quadrant all power curves cross at (1, 1): “above 1, big exponents win; below 1, they lose” (2³ > 2², but 0.5³ < 0.5²). Parity depends on α = p/q in lowest terms: q odd, p even — even (x^(2/3)); q odd, p odd — odd (x^(3/5)); q even — defined only for x ≥ 0, neither (x^(3/4)).

y = aˣ (a > 0, a ≠ 1): a⁰ = 1; a > 1 ⇒ increasing; 0 < a < 1 ⇒ decreasing; aˣ > 0 for all x
where:
  • athe base
  • a⁰ = 1every graph passes through (0, 1): the key to “fixed point” (过定点) questions

For y = a^(kx + m) + n, the point where the exponent is 0 does not depend on a: kx + m = 0, y = 1 + n.

a^f(x) = a^g(x) ⇔ f(x) = g(x); a > 1: a^f(x) > a^g(x) ⇔ f(x) > g(x); 0 < a < 1: a^f(x) > a^g(x) ⇔ f(x) < g(x)
where:
  • aa common base, a > 0, a ≠ 1: write 8 = 2³, 0.25 = 2⁻², √2 = 2^(1/2)
  • f(x), g(x)the exponents: after the base is removed they are compared directly

Equal bases ⇒ equal exponents, because y = aˣ is monotonic. In an inequality, a base above 1 keeps the sign and a base between 0 and 1 reverses it. Unlike logarithms, there is no domain step: aˣ is defined for every x. If the bases cannot be made equal (2ˣ = 5), the answer is a logarithm, x = log₂ 5 — see “Logarithms and logarithmic functions”.

aˣ > 1 ⇔ (a − 1)·x > 0; 0 < aˣ < 1 ⇔ (a − 1)·x < 0 (a > 0)
where:
  • a − 1its sign says whether the base is above or below 1
  • xthe exponent; its sign says whether we are to the right or to the left of 0

“Same side ⇒ above 1”: 1.2^0.3 > 1 (both factors positive), 0.7^(−2) > 1 (both negative), 0.7^0.3 < 1 and 1.2^(−0.3) < 1 (opposite signs).

Logarithms and logarithmic functions

Open lesson
logₐ N = b ⇔ aᵇ = N logₐ 1 = 0 logₐ a = 1 a^(logₐ N) = N logₐ aᵇ = b
where:
  • athe base: a > 0, a ≠ 1
  • Nthe argument, always N > 0
  • lg N, ln Nlg N = log₁₀ N (common logarithm), ln N = logₑ N (natural logarithm, e ≈ 2.718)

The definition and the basic identities. a^(logₐ N) = N is the logarithmic identity (对数恒等式): raising to a power and taking a logarithm undo each other. Chinese textbooks and CSCA write lg and ln.

logₐ(MN) = logₐ M + logₐ N logₐ(M/N) = logₐ M − logₐ N logₐ Mⁿ = n · logₐ Mlogₐ(MN) = logₐ M + logₐ N logₐ(M/N) = logₐ M − logₐ N logₐ Mⁿ = n · logₐ M
where:
  • M, Npositive numbers
  • nany real number

The product, quotient and power laws. They hold only for positive M and N and one common base a.

logₐ b = log_c b / log_c a = lg b / lg a = ln b / ln a logₐ b · log_b a = 1 logₐ b · log_b c = logₐ c log_(aᵐ) bⁿ = (n/m) · logₐ blogₐ b = log_c b / log_c a = lg b / lg a = ln b / ln a logₐ b · log_b a = 1 logₐ b · log_b c = logₐ c log_(aᵐ) bⁿ = (n/m) · logₐ b
where:
  • cany new base (c > 0, c ≠ 1)
  • a, ba > 0, a ≠ 1, b > 0 (in the later formulas also b ≠ 1)
  • m, nthe exponents of the base and of the argument (m ≠ 0)

The change-of-base formula (换底公式) and its three consequences. Proof: if logₐ b = x, then aˣ = b; take log_c of both sides: x · log_c a = log_c b.

a > 1: logₐ f(x) > logₐ g(x) ⇔ f(x) > g(x) > 0 0 < a < 1: logₐ f(x) > logₐ g(x) ⇔ 0 < f(x) < g(x)
where:
  • f(x), g(x)the arguments
  • athe common base

In both cases the smaller argument must be positive; the larger one is then positive automatically. First write a number as a logarithm: 3 = log₂ 8, −1 = log_(1/3) 3.

Trigonometric functions and identities

Open lesson
α (rad) = α° · π/180 l = |α| · r S = ½ · l · r = ½ · |α| · r²α (rad) = α° · π/180 l = |α| · r S = ½ · l · r = ½ · |α| · r²
where:
  • αthe angle in radians
  • lthe arc length
  • rthe radius
  • Sthe area of the sector

In radians the arc and sector formulas need no 360°: this is why calculus and Chinese textbooks (弧度制) prefer radians.

sin α = y/r cos α = x/r tan α = y/x r = √(x² + y²)sin α = y/r cos α = x/r tan α = y/x r = √(x² + y²)
where:
  • (x, y)any point on the terminal side of α (other than the origin)
  • rthe distance from the point to the origin

On the unit circle r = 1, so the point is simply (cos α, sin α). The signs of x and y give the signs by quadrant: I all positive, II only sin, III only tan, IV only cos.

sin²α + cos²α = 1 tan α = sin α / cos αsin²α + cos²α = 1 tan α = sin α / cos α
where:
  • αany angle (for tan, cos α ≠ 0)

The basic identities of one angle (同角三角函数关系). The first is Pythagoras on the unit circle, x² + y² = 1. From one value you get the other two; the quadrant decides the sign.

sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β tan(α ± β) = (tan α ± tan β) / (1 ∓ tan α tan β)sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β tan(α ± β) = (tan α ± tan β) / (1 ∓ tan α tan β)
where:
  • α, βany angles (for tan, all tangents and the denominator must be defined)

The sum and difference formulas (两角和与差公式). In the cosine formula the sign between the products is the opposite of the sign in the bracket. They turn 15°, 75° and 105° into 45° ± 30° or 60° ± 45°.

sin 2α = 2 sin α cos α cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²α tan 2α = 2 tan α / (1 − tan²α) cos²α = (1 + cos 2α)/2 sin²α = (1 − cos 2α)/2sin 2α = 2 sin α cos α cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²α tan 2α = 2 tan α / (1 − tan²α) cos²α = (1 + cos 2α)/2 sin²α = (1 − cos 2α)/2
where:
  • αany angle (for tan 2α, tan α ≠ ±1)

The double-angle formulas (二倍角公式) are the sum formulas with β = α. Read from right to left, the formula for cos 2α gives the power-reduction formulas (降幂公式), which replace squares by the first power of cos 2α.

a sin x + b cos x = R sin(x + φ), R = √(a² + b²), cos φ = a/R, sin φ = b/Ra sin x + b cos x = R sin(x + φ), R = √(a² + b²), cos φ = a/R, sin φ = b/R
where:
  • Rthe amplitude: maximum R, minimum −R
  • φthe auxiliary angle; its quadrant is chosen from the signs of a and b

The period and the graph of R sin(ωx + φ) are in «Graphs of trigonometric functions»; here we only need the rewriting and the range [−R, R].

Graphs of trigonometric functions

Open lesson
axis of symmetry: x = x₀ ⇔ f(x₀) = ±1 centre of symmetry: (x₀, 0) ⇔ f(x₀) = 0
where:
  • x₀the x-coordinate of a point of the curve y = sin x or y = cos x (or of any sine wave)
  • ±1the largest or the smallest value: axes pass through the peaks and the troughs
  • 0the value on the midline: centres of symmetry are the zeros of the curve

For sin x the axes are x = π/2 + kπ and the centres (kπ, 0); for cos x the axes are x = kπ and the centres (π/2 + kπ, 0). Neighbouring axes are T/2 apart; an axis and the nearest centre are T/4 apart. The tangent curve has no axes, and its centres are (kπ/2, 0).

T = 2π/ω ymax = b + A ymin = b − AT = 2π/ω ymax = b + A ymin = b − A
where:
  • Aamplitude (A > 0); for A < 0 use |A| — the maximum and the minimum swap places
  • ωangular frequency (ω > 0); it alone sets the period
  • φinitial phase; it moves the graph sideways but does not change T
  • bvertical shift: the midline is y = b

In general T = 2π/|ω|. The same formulas hold for y = A cos(ωx + φ) + b; for y = A tan(ωx + φ) the period is π/|ω|.

increasing: −π/2 + 2kπ ≤ ωx + φ ≤ π/2 + 2kπ axis: ωx + φ = π/2 + kπ centre: ωx + φ = kπincreasing: −π/2 + 2kπ ≤ ωx + φ ≤ π/2 + 2kπ axis: ωx + φ = π/2 + kπ centre: ωx + φ = kπ
where:
  • ωx + φthe phase, treated as one variable t
  • kany integer (k ∈ Z)

For A > 0, ω > 0. Decreasing: π/2 + 2kπ ≤ ωx + φ ≤ 3π/2 + 2kπ. For a cosine use the cosine conditions: increasing −π + 2kπ ≤ ωx + φ ≤ 2kπ, axis ωx + φ = kπ, centre ωx + φ = π/2 + kπ.

y = sin x → y = sin(x + φ) → y = sin(ωx + φ) → y = A sin(ωx + φ) + b
where:
  • x + φshift left by φ if φ > 0, right by |φ| if φ < 0: “left add, right subtract” (左加右减)
  • ωxthe x-coordinates are multiplied by 1/ω (y unchanged): a compression if ω > 1, a stretch if 0 < ω < 1
  • A, bthe y-coordinates are multiplied by A, then the graph moves up by b: “up add, down subtract” (上加下减)

The other order — stretch first, shift second — also works, but then the shift is φ/ω, because sin(ωx + φ) = sin ω(x + φ/ω): a shift always acts on x itself, never on ωx.

sin x = a: x = arcsin a + 2kπ or x = π − arcsin a + 2kπ cos x = a: x = ±arccos a + 2kπ tan x = a: x = arctan a + kπ
where:
  • athe given number; for sine and cosine a root exists only if |a| ≤ 1
  • arcsin athe angle in [−π/2, π/2] whose sine is a (arccos a ∈ [0, π], arctan a ∈ (−π/2, π/2))
  • kany integer (k ∈ Z)

For |a| < 1, sin x = a has two roots per period, symmetric about an axis x = π/2 + 2kπ, and cos x = a has two roots symmetric about x = 2kπ; tan x = a has one root in every period π.

Solving triangles: the law of sines and the law of cosines

Open lesson
a / sin A = b / sin B = c / sin C = 2Ra / sin A = b / sin B = c / sin C = 2R
where:
  • a, b, cthe sides opposite the angles A, B, C
  • A, B, Cthe angles of the triangle
  • Rthe radius of the circumscribed circle (外接圆半径)

Consequences: a = 2R sin A, sin A = a/(2R) and a : b : c = sin A : sin B : sin C. Use it when two angles and a side are known, or two sides and the angle opposite one of them.

a² = b² + c² − 2bc cos A cos A = (b² + c² − a²) / (2bc)a² = b² + c² − 2bc cos A cos A = (b² + c² − a²) / (2bc)
where:
  • athe side opposite A
  • b, cthe two sides that form the angle A
  • Athe angle between b and c, the included angle (夹角)

The same holds for the other sides: b² = a² + c² − 2ac cos B, c² = a² + b² − 2ab cos C. For A = 90°, cos A = 0 and the law becomes Pythagoras' theorem. The first form finds a side from two sides and the included angle, the second an angle from three sides.

S = ½ab sin C = ½bc sin A = ½ca sin B
where:
  • a, btwo sides
  • Cthe angle between them
  • Sthe area of the triangle

Half the product of two sides and the sine of the angle between them. Together with the law of sines it also gives S = abc/(4R).

h = b sin A
where:
  • hthe distance from C to the other side of the angle A (the height)
  • athe side opposite A
  • bthe side next to A

A acute: a < h — no triangle; a = h — one right triangle; h < a < b — two triangles; a ≥ b — one triangle. A right or obtuse: one triangle if a > b, none if a ≤ b.

c² < a² + b² ⇔ C < 90° c² = a² + b² ⇔ C = 90° c² > a² + b² ⇔ C > 90°
where:
  • cthe longest side
  • Cthe largest angle (opposite c)

It follows from cos C = (a² + b² − c²)/(2ab): the sign of the numerator is the sign of cos C. The triangle is acute if its largest angle is acute, so check only the longest side.

Arithmetic sequences

Open lesson
aₙ = a₁ + (n − 1)d aₙ = aₘ + (n − m)d d = (aₙ − aₘ)/(n − m)aₙ = a₁ + (n − 1)d aₙ = aₘ + (n − m)d d = (aₙ − aₘ)/(n − m)
where:
  • a₁the first term (首项)
  • dthe common difference
  • n, mpositions of terms (positive integers)

From a₁ to aₙ there are n − 1 steps of size d; from aₘ to aₙ there are n − m steps. Written as aₙ = dn + (a₁ − d), the general term is a linear function of n: the points (n, aₙ) lie on a line with slope d.

Sₙ = n(a₁ + aₙ)/2 Sₙ = na₁ + n(n − 1)d/2Sₙ = n(a₁ + aₙ)/2 Sₙ = na₁ + n(n − 1)d/2
where:
  • Sₙthe sum of the first n terms (前n项和)
  • aₙthe last term added
  • nthe number of terms (项数)

The two forms of the sum. In words: sum = number of terms × the mean of the first and last terms. So Sₙ/n = (a₁ + aₙ)/2 is the average term.

m + n = p + t ⇒ aₘ + aₙ = aₚ + aₜ 2aₙ = aₙ₋ₖ + aₙ₊ₖ (S₂ₖ − Sₖ) − Sₖ = k²d
where:
  • m, n, p, tpositions with equal sums
  • kthe number of steps to the left and right of aₙ; the length of a block
  • Sₖ, S₂ₖ − Sₖ, S₃ₖ − S₂ₖsums of consecutive blocks of k terms

Terms whose positions have the same sum have the same sum (等差数列的性质). Both sides need the same number of terms: a₁ + a₉ = a₄ + a₆ = 2a₅, but a₁ + a₉ ≠ a₁₀. Consecutive blocks of k terms, Sₖ, S₂ₖ − Sₖ, S₃ₖ − S₂ₖ, again form an arithmetic sequence, with difference k²d.

Sₙ = (d/2)n² + (a₁ − d/2)n n₀ = 1/2 − a₁/dSₙ = (d/2)n² + (a₁ − d/2)n n₀ = 1/2 − a₁/d
where:
  • d/2d/2the leading coefficient: for d < 0 the parabola opens downward and Sₙ has a largest value; for d > 0 it has a smallest value
  • n₀the vertex of the parabola (its axis of symmetry)

Sₙ as a quadratic in n. The answer is the positive integer nearest to n₀; if n₀ ends in .5, both neighbours are answers. Conversely, Sₙ = An² + Bn (no constant term) is always the sum of an arithmetic sequence with d = 2A (the method aₙ = Sₙ − Sₙ₋₁ that proves it is in the lesson «Geometric sequences»).

1/(aₖaₖ₊₁) = (1/d)(1/aₖ − 1/aₖ₊₁) 1/(k(k + 1)) = 1/k − 1/(k + 1) 1/(√(k + 1) + √k) = √(k + 1) − √k1/(aₖaₖ₊₁) = (1/d)(1/aₖ − 1/aₖ₊₁) 1/(k(k + 1)) = 1/k − 1/(k + 1) 1/(√(k + 1) + √k) = √(k + 1) − √k
where:
  • aₖ, aₖ₊₁neighbouring terms of an arithmetic sequence with d ≠ 0 and no zero terms
  • kthe position of the term in the sum

Split, then cancel. Check the factor 1/d by putting the right-hand side over a common denominator: 1/aₖ − 1/aₖ₊₁ = (aₖ₊₁ − aₖ)/(aₖaₖ₊₁) = d/(aₖaₖ₊₁).

Geometric sequences

Open lesson
aₙ = a₁qⁿ⁻¹ aₙ = aₘqⁿ⁻ᵐ qⁿ⁻ᵐ = aₙ/aₘaₙ = a₁qⁿ⁻¹ aₙ = aₘqⁿ⁻ᵐ qⁿ⁻ᵐ = aₙ/aₘ
where:
  • a₁the first term (首项)
  • qthe common ratio, q ≠ 0
  • n, mpositions of terms

From a₁ to aₙ you multiply by q exactly n − 1 times. Two known terms give qⁿ⁻ᵐ; when n − m is even, q is found only up to its sign.

Sₙ = a₁(1 − qⁿ)/(1 − q) = (a₁ − aₙq)/(1 − q) (q ≠ 1) Sₙ = na₁ (q = 1)Sₙ = a₁(1 − qⁿ)/(1 − q) = (a₁ − aₙq)/(1 − q) (q ≠ 1) Sₙ = na₁ (q = 1)
where:
  • Sₙthe sum of the first n terms (前n项和)
  • aₙthe last term added
  • qthe common ratio

The sum of a geometric sequence. For q > 1 the same formula written as a₁(qⁿ − 1)/(q − 1) avoids minus signs. When q is unknown, always ask whether q = 1 is possible.

m + n = p + t ⇒ aₘaₙ = aₚaₜ aₙ² = aₙ₋ₖaₙ₊ₖ (S₂ₖ − Sₖ) : Sₖ = qᵏ
where:
  • m, n, p, tpositions with equal sums
  • kthe number of steps to the left and right of aₙ; the length of a block
  • qᵏthe ratio of neighbouring blocks of k terms

The product version of the index property of arithmetic sequences. Consecutive blocks of k terms, Sₖ, S₂ₖ − Sₖ, S₃ₖ − S₂ₖ, form a geometric sequence with ratio qᵏ (when Sₖ ≠ 0). Also a₁a₂…a₂ₖ₋₁ = aₖ²ᵏ⁻¹: the product of an odd number of terms is the middle term to that power.

S = a₁/(1 − q) (0 < |q| < 1)S = a₁/(1 − q) (0 < |q| < 1)
where:
  • Sthe sum of all terms, the limit of Sₙ
  • a₁the first term
  • qthe common ratio, |q| < 1

The sum of an infinite decreasing geometric series. With q < 0 the terms alternate in sign, and the formula still works. Sanity check: an infinite sum of positive terms is always larger than a₁.

aₙ = S₁ (n = 1); aₙ = Sₙ − Sₙ₋₁ (n ≥ 2)
where:
  • Sₙthe sum of the first n terms, given as a formula
  • Sₙ₋₁the same formula with n − 1 in place of n

This works for every sequence, not only geometric ones. Sₙ − Sₙ₋₁ is the last term aₙ, but only for n ≥ 2, because S₀ is not part of the rule. Always check whether the formula for n ≥ 2 also gives a₁ = S₁; if not, the answer is piecewise.

The derivative: definition, geometric meaning and rules

Open lesson
Δy/Δx = [f(x₀ + Δx) − f(x₀)] / ΔxΔy/Δx = [f(x₀ + Δx) − f(x₀)] / Δx
where:
  • Δxthe increment of the argument (Δx ≠ 0; it may be negative)
  • Δythe increment of the function

Average rate of change on [x₀, x₀ + Δx] = slope of the secant. On an interval [a, b]: [f(b) − f(a)]/(b − a).

f′(x₀) = lim (Δx→0) [f(x₀ + Δx) − f(x₀)] / Δxf′(x₀) = lim (Δx→0) [f(x₀ + Δx) − f(x₀)] / Δx
where:
  • f′(x₀)the derivative at x₀ = the instantaneous rate of change
  • lim (Δx→0)the limit as Δx tends to 0 (from both sides)

The definition of the derivative. The name of the step does not matter — h, Δx or 2Δx; what counts is that the same step stands in the argument and in the denominator.

(C)′ = 0 (xⁿ)′ = n·xⁿ⁻¹ (sin x)′ = cos x (cos x)′ = −sin x
where:
  • Cany constant (a number such as 5, π, e², ln 3)
  • nany real exponent: √x = x^(1/2), 1/x = x⁻¹, 1/x² = x⁻²
  • xin sin and cos the angle is in radians

Power and trigonometric functions. Special cases worth knowing by heart: (√x)′ = 1/(2√x), (1/x)′ = −1/x².

(aˣ)′ = aˣ ln a (eˣ)′ = eˣ (logₐ x)′ = 1/(x ln a) (ln x)′ = 1/x(aˣ)′ = aˣ ln a (eˣ)′ = eˣ (logₐ x)′ = 1/(x ln a) (ln x)′ = 1/x
where:
  • athe base: a > 0, a ≠ 1
  • ee ≈ 2.718, the base with ln e = 1, for which the extra factor disappears
  • xx > 0 in the logarithms

Exponential and logarithmic functions (see «Power and exponential functions» and «Logarithms and logarithmic functions»). eˣ is its own derivative — that is why e is the favourite base of calculus.

(u ± v)′ = u′ ± v′ (Cu)′ = Cu′ (uv)′ = u′v + uv′ (u/v)′ = (u′v − uv′)/v²(u ± v)′ = u′ ± v′ (Cu)′ = Cu′ (uv)′ = u′v + uv′ (u/v)′ = (u′v − uv′)/v²
where:
  • u, vdifferentiable functions of x; v ≠ 0 in the quotient
  • Ca constant factor

The rules of differentiation (求导法则). The derivative of a product is NOT the product of the derivatives; in the quotient rule the numerator starts with u′v.

[f(g(x))]′ = f′(g(x)) · g′(x); [f(ax + b)]′ = a · f′(ax + b)
where:
  • gthe inner function
  • fthe outer function; its derivative is taken at the inner value g(x)
  • a, bnumbers, a ≠ 0

The chain rule for a composite function (复合函数): outer derivative × inner derivative. The Chinese school programme limits composite functions to the form f(ax + b), so expect this case first of all.

y − f(x₀) = f′(x₀) · (x − x₀)
where:
  • (x₀, f(x₀))the point of tangency — it lies on the curve and on the tangent
  • f′(x₀)the slope of the tangent, k = tan α

The equation of the tangent line at x₀ (point–slope form).

v(t) = s′(t) a(t) = v′(t)
where:
  • s(t)the position (coordinate) at time t, m
  • v(t)the instantaneous velocity (瞬时速度), m/s; its sign shows the direction
  • a(t)the acceleration, m/s²

The physical meaning of the derivative: velocity is the rate of change of position, acceleration the rate of change of velocity (see «Kinematics: displacement, velocity, acceleration, free fall»).

Applications of the derivative: monotonicity and extrema

Open lesson
f′(x) > 0 on (a, b) ⇒ f increases on (a, b); f′(x) < 0 on (a, b) ⇒ f decreases on (a, b)
where:
  • f′(x)the slope of the tangent at x
  • (a, b)an interval inside the domain

A sufficient condition, not a necessary one: y = x³ increases on the whole line although y′(0) = 0. In a parameter question («f is increasing on I») require f′(x) ≥ 0 on I (zero only at isolated points).

f′: + → − at x₀ ⇒ local maximum; f′: − → + at x₀ ⇒ local minimum; no sign change ⇒ no extremum
where:
  • x₀a critical point
  • + → −f′ is positive to the left of x₀ and negative to the right: f rises, then falls

The first-derivative test. f′(x₀) = 0 alone is not enough: y = x³ has y′(0) = 0, but y′ > 0 on both sides, so there is no extremum.

greatest value of f on [a, b] = the largest of f(a), f(b), f(x₁), …, f(xₖ); least value = the smallest of them
where:
  • [a, b]a closed interval on which f is continuous
  • x₁, …, xₖthe critical points inside (a, b)

A function continuous on a closed interval always reaches its greatest value (最大值) and least value (最小值) — at a critical point or at an end. No sign chart is needed: just compare the numbers.

∫ₐᵇ f(x) dx = F(b) − F(a), where F′(x) = f(x)
where:
  • Fan antiderivative of f: (xⁿ⁺¹/(n + 1))′ = xⁿ, (−cos x)′ = sin x
  • a, bthe limits of integration

The Newton–Leibniz formula, the fundamental theorem of calculus (微积分基本定理). For f ≥ 0 on [a, b] the integral is the area under the curve; for a velocity v(t) ≥ 0 it is the distance travelled.

3Mathematics: geometry and algebraIntermediate

Lines and circles in the coordinate plane

Open lesson
k = tan α = (y₂ − y₁)/(x₂ − x₁)k = tan α = (y₂ − y₁)/(x₂ − x₁)
where:
  • kslope
  • αangle of inclination, 0° ≤ α < 180°, α ≠ 90°
  • (x₁, y₁), (x₂, y₂)two points of the line, x₁ ≠ x₂

k > 0: the line rises (0° < α < 90°); k < 0: it falls (90° < α < 180°); k = 0: it is horizontal. If x₁ = x₂, the line is vertical and has no slope.

Ax + By + C = 0: k = −A/B, b = −C/B (B ≠ 0)Ax + By + C = 0: k = −A/B, b = −C/B (B ≠ 0)
where:
  • A, B, Ccoefficients; A and B are not both zero
  • kslope
  • bthe y-intercept: the line meets the y-axis at (0, b)

An intercept is a coordinate, not a length, so it can be negative. The x-intercept of Ax + By + C = 0 is −C/A (A ≠ 0). If B = 0, the line is vertical: x = −C/A.

l₁ ∥ l₂ ⇔ k₁ = k₂ and b₁ ≠ b₂ l₁ ⊥ l₂ ⇔ k₁ · k₂ = −1
where:
  • k₁, k₂slopes of the lines y = k₁x + b₁ and y = k₂x + b₂
  • b₁, b₂their y-intercepts

Perpendicular slopes are negative reciprocals of each other: 2 and −1/2, −3/4 and 4/3. The test needs both slopes to exist: a vertical line is perpendicular to every horizontal one, although it has no slope.

A₁A₂ + B₁B₂ = 0 ⇔ l₁ ⊥ l₂ A₁B₂ − A₂B₁ = 0 ⇔ l₁ ∥ l₂ or the lines coincide
where:
  • A₁, B₁, A₂, B₂coefficients of l₁: A₁x + B₁y + C₁ = 0 and l₂: A₂x + B₂y + C₂ = 0

These conditions work for all lines, vertical ones included, so they are the safest choice in parameter questions. A parallel line keeps A and B and changes only C; a perpendicular line swaps A and B and changes one sign: Ax + By + C = 0 → Bx − Ay + C′ = 0.

d = |Ax₀ + By₀ + C| / √(A² + B²) d = |C₁ − C₂| / √(A² + B²)d = |Ax₀ + By₀ + C| / √(A² + B²) d = |C₁ − C₂| / √(A² + B²)
where:
  • (x₀, y₀)the point
  • A, B, Ccoefficients of the general equation of the line
  • C₁, C₂free terms of the parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0, which must have the same A and B

The distance is the length of the perpendicular dropped from the point to the line. The second formula is the first one applied to a point of one of the lines, so first make A and B of both equations equal.

(x − a)² + (y − b)² = r²
where:
  • (a, b)centre
  • rradius, r > 0
  • (x, y)any point of the circle

This is the distance formula |PC| = √((x − a)² + (y − b)²) = r, squared. A point P lies inside, on or outside the circle when |PC| is less than, equal to or greater than r. Some English books write the centre as (h, k); CSCA and Chinese textbooks use (a, b).

x² + y² + Dx + Ey + F = 0 (D² + E² − 4F > 0): centre (−D/2, −E/2), r = √(D² + E² − 4F)/2x² + y² + Dx + Ey + F = 0 (D² + E² − 4F > 0): centre (−D/2, −E/2), r = √(D² + E² − 4F)/2
where:
  • D, E, Fcoefficients of the general equation

It is the standard equation with the brackets opened. The coefficients of x² and y² must be equal, and there must be no xy term. If D² + E² − 4F = 0, the equation describes a single point; if it is negative, no point at all.

d < r: two common points d = r: tangent d > r: no common points
where:
  • ddistance from the centre (a, b) to the line: d = |Aa + Bb + C|/√(A² + B²)
  • rradius of the circle

In Chinese the three cases are 相交 (intersect), 相切 (tangent) and 相离 (separate). The algebraic check gives the same result: substituting the line into the circle gives a quadratic equation, and Δ > 0, Δ = 0 and Δ < 0 correspond to the three cases.

|AB| = 2√(r² − d²)
where:
  • |AB|length of the chord
  • ddistance from the centre to the line

The perpendicular from the centre to a chord bisects it, so r, d and half the chord form a right triangle (see the drawing).

x₀x + y₀y = r² (x₀ − a)(x − a) + (y₀ − b)(y − b) = r² |PT| = √(|PC|² − r²)
where:
  • (x₀, y₀)the point of contact on the circle (first formula: centre at O)
  • |PT|length of the tangent segment from an outside point P to the point of contact T
  • |PC|distance from P to the centre

The first two formulas give the tangent at a point of the circle (“replace one x by x₀ and one y by y₀”). They work because the tangent is perpendicular to the radius at the point of contact; the same right angle gives |PT| by Pythagoras. From a point outside the circle there are exactly two tangents, from a point on it one, from a point inside none.

(D₁ − D₂)x + (E₁ − E₂)y + F₁ − F₂ = 0
where:
  • D₁, E₁, F₁; D₂, E₂, F₂coefficients of the general equations of the two circles

If two circles intersect, subtracting their general equations removes x² and y² and leaves a line through both common points: the line of the common chord (公共弦).

The ellipse

Open lesson
x²/a² + y²/b² = 1 y²/a² + x²/b² = 1 (a > b > 0, b² = a² − c²)x²/a² + y²/b² = 1 y²/a² + x²/b² = 1 (a > b > 0, b² = a² − c²)
where:
  • asemi-major axis (half of the major axis)
  • bsemi-minor axis
  • chalf of the focal distance: foci (±c, 0) in the first equation and (0, ±c) in the second

The foci lie on the axis of the larger denominator: x²/16 + y²/25 = 1 has 25 under y², so its foci are on the y-axis. In Chinese textbooks a is always the larger number (a > b > 0), whichever variable it stands under.

a² = b² + c²
where:
  • asemi-major axis: the distance from an end of the minor axis to a focus
  • bsemi-minor axis
  • cdistance from the centre to a focus

The end B₂(0, b) of the minor axis is equally far from both foci, and the two distances add up to 2a, so |B₂F₁| = |B₂F₂| = a. The right triangle OB₂F₂ has legs b and c and hypotenuse a: in an ellipse a is the largest of the three numbers.

e = c/a = √(1 − b²/a²), 0 < e < 1e = c/a = √(1 − b²/a²), 0 < e < 1
where:
  • eeccentricity (离心率)
  • a, b, cas above

e measures how flat the ellipse is. As e → 0, c → 0 and b → a: the ellipse becomes a circle. As e → 1, b → 0 and it flattens towards a segment. The ratio of the axes is b/a = √(1 − e²).

perimeter of △PF₁F₂ = 2a + 2c perimeter of △ABF₂ = 4a (chord AB through F₁)
where:
  • Pany point of the ellipse (not an end of the major axis)
  • ABa chord through the focus F₁

For the chord: |AF₁| + |AF₂| = 2a and |BF₁| + |BF₂| = 2a, while |AF₁| + |BF₁| = |AB|. Adding gives |AB| + |AF₂| + |BF₂| = 4a, whatever the direction of the chord.

a − c ≤ |PF₁| ≤ a + c |PF₁| = a + ex₀, |PF₂| = a − ex₀
where:
  • P(x₀, y₀)a point of the ellipse x²/a² + y²/b² = 1
  • F₁(−c, 0), F₂(c, 0)left and right foci
  • eeccentricity

The distance from a point of the ellipse to a focus is called a focal radius (焦半径). Substituting y₀² = b²(1 − x₀²/a²) into |PF₂|² = (x₀ − c)² + y₀² gives (a − ex₀)², so the focal radius is linear in x₀: it is smallest, a − c, at the vertex nearest to the focus and largest, a + c, at the far vertex.

S = b² · tan(θ/2), θ = ∠F₁PF₂S = b² · tan(θ/2), θ = ∠F₁PF₂
where:
  • Sarea of triangle PF₁F₂
  • θthe angle at P
  • bsemi-minor axis

Derivation: let m = |PF₁| and n = |PF₂|. The law of cosines gives 4c² = m² + n² − 2mn cos θ = (m + n)² − 2mn(1 + cos θ) = 4a² − 2mn(1 + cos θ), so mn = 2b²/(1 + cos θ). Then S = ½mn sin θ = b² sin θ/(1 + cos θ) = b² tan(θ/2). For θ = 90°: S = b².

|AB| = √(1 + k²) · |x₁ − x₂|
where:
  • kslope of the line
  • x₁, x₂x-coordinates of A and B

The same chord formula works for any curve, the parabola and the hyperbola included; |x₁ − x₂| comes from Vieta’s formulas without solving the equation.

The hyperbola

Open lesson
||PF₁| − |PF₂|| = 2a, 0 < 2a < 2c
where:
  • Pany point of the hyperbola
  • F₁, F₂the foci; |F₁F₂| = 2c
  • 2athe length of the real axis (the distance between the vertices)

On the branch nearer to F₂: |PF₁| − |PF₂| = 2a; on the branch nearer to F₁ it is the other way round. The closest a point of the hyperbola gets to a focus is c − a (at a vertex).

x²/a² − y²/b² = 1 y²/a² − x²/b² = 1 c² = a² + b²x²/a² − y²/b² = 1 y²/a² − x²/b² = 1 c² = a² + b²
where:
  • a > 0the real semi-axis: vertices (±a, 0) or (0, ±a)
  • b > 0the imaginary semi-axis
  • cthe distance from the centre to a focus: foci (±c, 0) or (0, ±c)

Left: foci on the x-axis; middle: foci on the y-axis. In a hyperbola c is the largest of the three numbers, and a² always stands under the positive square, whichever denominator is bigger.

x²/a² − y²/b² = 1 ⇒ y = ±(b/a)x y²/a² − x²/b² = 1 ⇒ y = ±(a/b)xx²/a² − y²/b² = 1 ⇒ y = ±(b/a)x y²/a² − x²/b² = 1 ⇒ y = ±(a/b)x
where:
  • b/a, a/bb/a, a/bthe slopes of the asymptotes: always √(denominator under y²) ÷ √(denominator under x²)

Quick way: replace the 1 on the right by 0 and solve for y: x²/a² − y²/b² = 0 ⇒ y = ±(b/a)x.

e = c/a = √(1 + b²/a²) > 1 b/a = √(e² − 1)e = c/a = √(1 + b²/a²) > 1 b/a = √(e² − 1)
where:
  • ethe eccentricity (离心率)
  • b/ab/athe slope of an asymptote when the foci are on the x-axis

Because c > a, always e > 1. The larger e, the steeper the asymptotes (foci on the x-axis) and the wider the branches open.

The parabola

Open lesson
y² = 2px (p > 0), F(p/2, 0), l: x = −p/2y² = 2px (p > 0), F(p/2, 0), l: x = −p/2
where:
  • pthe distance from the focus to the directrix (the parameter of the parabola)
  • Fthe focus
  • lthe directrix

The vertex O(0, 0) is halfway between the focus and the directrix. Chinese textbooks write y² = 2px; some English books write the same curve as y² = 4ax with focus (a, 0), so a = p/2 — check which form you are given.

|PF| = x₀ + p/2 (y² = 2px) |PF| = y₀ + p/2 (x² = 2py)|PF| = x₀ + p/2 (y² = 2px) |PF| = y₀ + p/2 (x² = 2py)
where:
  • P(x₀, y₀)a point of the parabola
  • p/2p/2the distance from the vertex to the focus (and to the directrix)
  • |PF|the focal radius

For parabolas opening to the left or downwards use |x₀| + p/2 or |y₀| + p/2. The vertex is the point closest to the focus: always |PF| ≥ p/2.

|AB| = x₁ + x₂ + p |AB| = 2p/sin²α ≥ 2p y₁y₂ = −p², x₁x₂ = p²/4|AB| = x₁ + x₂ + p |AB| = 2p/sin²α ≥ 2p y₁y₂ = −p², x₁x₂ = p²/4
where:
  • A(x₁, y₁), B(x₂, y₂)the ends of a chord of y² = 2px through the focus
  • αthe inclination angle of the chord
  • 2pthe length of the latus rectum

Only for chords through the focus. With α = 90° the second formula gives the smallest value, 2p: a focal chord shorter than 2p does not exist. For x² = 2py swap the roles of x and y: |AB| = y₁ + y₂ + p, x₁x₂ = −p², y₁y₂ = p²/4.

|PF| / d(P, l) = e: e < 1 ellipse, e = 1 parabola, e > 1 hyperbola|PF| / d(P, l) = e: e < 1 ellipse, e = 1 parabola, e > 1 hyperbola
where:
  • ethe eccentricity: e = c/a for the ellipse and the hyperbola, e = 1 for every parabola
  • lthe directrix belonging to the focus: x = a²/c for the ellipse and the hyperbola (focus (c, 0)), x = −p/2 for y² = 2px

A circle is the limiting case e = 0. All parabolas have the same eccentricity and the same shape — like all circles, they differ only in size.

Plane vectors

Open lesson
AB⃗ + BC⃗ = AC⃗ AB⃗ + AD⃗ = AC⃗ (ABCD is a parallelogram) AB⃗ − AC⃗ = CB⃗ |λa⃗| = |λ| · |a⃗|
where:
  • AB⃗ + BC⃗triangle rule: the end of the first vector is the start of the second
  • AB⃗ + AD⃗parallelogram rule: a common start, the sum is the diagonal
  • AB⃗ − AC⃗a common start: the difference points from the end of the subtracted vector to the end of the other
  • λa⃗λ > 0: the direction of a⃗; λ < 0: the opposite direction; λ = 0: 0⃗

The chain of arrows: AB⃗ + BC⃗ + CD⃗ + … + YZ⃗ = AZ⃗ — the inner letters cancel. The usual laws hold: a⃗ + b⃗ = b⃗ + a⃗, λ(a⃗ + b⃗) = λa⃗ + λb⃗.

AB⃗ = (x₂ − x₁, y₂ − y₁) a⃗ ± b⃗ = (x₁ ± x₂, y₁ ± y₂) λa⃗ = (λx₁, λy₁) |a⃗| = √(x₁² + y₁²)
where:
  • A(x₁, y₁), B(x₂, y₂)the start and the end of the vector: the coordinates are “end minus start”
  • a⃗ = (x₁, y₁), b⃗ = (x₂, y₂)the coordinates of the vectors; every operation works coordinate by coordinate
  • |a⃗|the length of the vector (Pythagoras); |AB⃗| is the distance between A and B

The midpoint of AB is ((x₁ + x₂)/2, (y₁ + y₂)/2). Equal vectors have equal coordinates — that is how the fourth vertex of a parallelogram is found.

a⃗ ∥ b⃗ (b⃗ ≠ 0⃗) ⇔ a⃗ = λb⃗ ⇔ x₁y₂ − x₂y₁ = 0
where:
  • λthe number with a⃗ = λb⃗: λ > 0 — same direction, λ < 0 — opposite directions
  • x₁y₂ − x₂y₁the “cross” difference: it is zero exactly when the coordinates are proportional

Three points A, B, C lie on one line ⇔ AB⃗ ∥ AC⃗. Prefer x₁y₂ − x₂y₁ = 0 to the proportion x₁/x₂ = y₁/y₂: it also works when a coordinate is 0.

a⃗ · b⃗ = |a⃗| · |b⃗| · cos θ = x₁x₂ + y₁y₂ cos θ = (x₁x₂ + y₁y₂) / (|a⃗| · |b⃗|) a⃗ ⊥ b⃗ ⇔ x₁x₂ + y₁y₂ = 0a⃗ · b⃗ = |a⃗| · |b⃗| · cos θ = x₁x₂ + y₁y₂ cos θ = (x₁x₂ + y₁y₂) / (|a⃗| · |b⃗|) a⃗ ⊥ b⃗ ⇔ x₁x₂ + y₁y₂ = 0
where:
  • a⃗ · b⃗the dot (scalar) product (数量积) — a number, not a vector
  • θthe angle between the vectors, 0 ≤ θ ≤ π
  • x₁x₂ + y₁y₂the dot product in coordinates: multiply matching coordinates and add

The sign tells you the angle: a⃗ · b⃗ > 0 — θ is acute (or 0), = 0 — a right angle, < 0 — θ is obtuse (or π). Also a⃗ · a⃗ = |a⃗|², so every question about a length becomes a dot-product question.

|a⃗ ± b⃗|² = |a⃗|² ± 2a⃗ · b⃗ + |b⃗|² (a⃗ + b⃗) · (a⃗ − b⃗) = |a⃗|² − |b⃗|²
where:
  • |a⃗ ± b⃗|the length of the sum (difference) — the diagonals of the parallelogram
  • 2a⃗ · b⃗the middle term; it vanishes for θ = 90°, and Pythagoras remains

Square the length and expand it like (x ± y)² — this is the law of cosines in vector form (see “Solving triangles: the law of sines and the law of cosines”).

projection of a⃗ on b⃗ (a number): |a⃗| cos θ = a⃗ · b⃗ / |b⃗| projection vector: (a⃗ · b⃗ / |b⃗|²) · b⃗projection of a⃗ on b⃗ (a number): |a⃗| cos θ = a⃗ · b⃗ / |b⃗| projection vector: (a⃗ · b⃗ / |b⃗|²) · b⃗
where:
  • |a⃗| cos θthe signed length of the “shadow” of a⃗ on the line of b⃗; negative when θ is obtuse
  • (a⃗ · b⃗ / |b⃗|²) · b⃗(a⃗ · b⃗ / |b⃗|²) · b⃗the same shadow as a vector: collinear with b⃗, of length ||a⃗| cos θ|

The geometric meaning of the dot product: a⃗ · b⃗ = |b⃗| × (the projection of a⃗ on b⃗). Read the question carefully: is a number or a vector asked? Older Chinese textbooks call the number |a⃗| cos θ 投影; the current ones use the projection vector, 投影向量.

D is the midpoint of BC: AD⃗ = ½(AB⃗ + AC⃗) D on BC, BD⃗ = t · BC⃗: AD⃗ = (1 − t)AB⃗ + t · AC⃗ G is the centroid: GA⃗ + GB⃗ + GC⃗ = 0⃗, AG⃗ = ⅓(AB⃗ + AC⃗) OP⃗ = λOA⃗ + μOB⃗: P lies on line AB ⇔ λ + μ = 1
where:
  • tthe ratio BD : BC (for 0 ≤ t ≤ 1, D lies on the segment)
  • Gthe intersection of the medians; it divides each median 2 : 1 counting from the vertex
  • Oany point not on line AB

The coefficients are “crossed”: the closer D is to C, the larger the coefficient of AC⃗. For the midpoint t = ½; the coordinates of the centroid are the means of the vertices' coordinates.

Complex numbers

Open lesson
z = a + bi (a, b ∈ R): real ⇔ b = 0; imaginary ⇔ b ≠ 0; pure imaginary ⇔ a = 0 and b ≠ 0 a + bi = c + di ⇔ a = c and b = d
where:
  • a = Re zthe real part
  • b = Im zthe imaginary part — a real number, written without i
  • a = c, b = done complex equation gives two real equations

Chinese textbooks: 实数 (b = 0), 虚数 (b ≠ 0), 纯虚数 (a = 0, b ≠ 0). Careful: many English books use “imaginary number” only for numbers of the form bi, while in the Chinese sense 3 + 2i is already 虚数. Complex numbers can be equal or unequal, but unless both are real they cannot be compared with < or >.

z = a + bi ↔ Z(a, b) ↔ OZ⃗ z̄ = a − bi |z| = |OZ⃗| = √(a² + b²) |z₁ − z₂| = |Z₁Z₂|
where:
  • z̄the conjugate (共轭复数): the mirror image in the real axis
  • |z|the modulus (模): the distance from O to Z; |z̄| = |z|
  • |z₁ − z₂|the distance between the points of z₁ and z₂

Quadrants: I — a > 0, b > 0; II — a < 0, b > 0; III — a < 0, b < 0; IV — a > 0, b < 0. A number on an axis lies in no quadrant.

(a + bi) ± (c + di) = (a ± c) + (b ± d)i (a + bi)(c + di) = (ac − bd) + (ad + bc)i (a + bi)(a − bi) = a² + b²
where:
  • ±addition and subtraction: real parts separately, imaginary parts separately
  • ac − bdmultiplication: expand like binomials, bi · di = bd · i² = −bd
  • z · z̄ = a² + b²= |z|², always a non-negative real number

Squares worth remembering: (a + bi)² = (a² − b²) + 2abi, (1 + i)² = 2i, (1 − i)² = −2i.

(a + bi)/(c + di) = (a + bi)(c − di) / ((c + di)(c − di)) = ((ac + bd) + (bc − ad)i) / (c² + d²)(a + bi)/(c + di) = (a + bi)(c − di) / ((c + di)(c − di)) = ((ac + bd) + (bc − ad)i) / (c² + d²)
where:
  • c − dithe conjugate of the denominator
  • c² + d²a real denominator: |c + di|²

Multiplying by the conjugate makes the denominator real — like removing a root from a denominator. Moduli behave simply: |z₁z₂| = |z₁| · |z₂| and |z₁/z₂| = |z₁|/|z₂|.

i¹ = i, i² = −1, i³ = −i, i⁴ = 1 i⁴ᵏ⁺ʳ = iʳ iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0
where:
  • rthe remainder of n divided by 4 (0, 1, 2, 3); i⁰ = 1
  • kan integer

The powers go round the cycle i → −1 → −i → 1 with period 4, so four consecutive powers always add up to 0. Negative exponents too: i⁻¹ = −i, i⁻² = −1.

ax² + bx + c = 0 (a, b, c ∈ R), D = b² − 4ac < 0: x₁,₂ = (−b ± i√(−D)) / (2a)ax² + bx + c = 0 (a, b, c ∈ R), D = b² − 4ac < 0: x₁,₂ = (−b ± i√(−D)) / (2a)
where:
  • Dthe discriminant; for D < 0 there are no real roots but two conjugate complex roots
  • √(−D)an ordinary square root, since −D > 0

Vieta's formulas still hold: x₁ + x₂ = −b/a, x₁x₂ = c/a. If a + bi is a root of an equation with real coefficients, so is a − bi.

Solid geometry and the coordinate system in space

Open lesson
V(prism) = S · h V(cylinder) = πr²h V(pyramid) = ⅓ · S · h V(cone) = ⅓ · πr²h V(ball) = 4πR³/3V(prism) = S · h V(cylinder) = πr²h V(pyramid) = ⅓ · S · h V(cone) = ⅓ · πr²h V(ball) = 4πR³/3
where:
  • Sarea of the base
  • hheight: the perpendicular distance between the bases (prism, cylinder) or from the apex to the base (pyramid, cone)
  • rradius of the base
  • Rradius of the ball

Prisms and cylinders: base × height; pyramids and cones: one third of that. The height is always the perpendicular distance, never a slanting edge.

S(cylinder) = 2πr² + 2πrh S(lateral, cone) = πrl S(cone) = πr² + πrl S(sphere) = 4πR²
where:
  • r, hradius of the base and height
  • lslant height of the cone, l² = r² + h²
  • Rradius of the sphere

Unroll the lateral surface: a cylinder gives a rectangle 2πr × h; a cone gives a sector of radius l with arc length 2πr, whose central angle is 360° · r/l. For prisms and pyramids, simply add the areas of all faces.

|AB| = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²) M((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2) |OP| = √(x² + y² + z²)|AB| = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²) M((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2) |OP| = √(x² + y² + z²)
where:
  • A(x₁, y₁, z₁), B(x₂, y₂, z₂)two points
  • Mmidpoint of AB
  • |OP|distance from P(x, y, z) to the origin

The plane formulas from “Lines and circles in the coordinate plane” with a third term. The distance from P(x, y, z) to the z-axis is √(x² + y²), and to the plane xOy it is |z|.

d = √(a² + b² + c²) cube: d = a√3, face diagonal a√2 R = d/2 (sphere through all vertices) r = a/2 (ball inscribed in a cube)d = √(a² + b² + c²) cube: d = a√3, face diagonal a√2 R = d/2 (sphere through all vertices) r = a/2 (ball inscribed in a cube)
where:
  • a, b, cedges of a rectangular box
  • dspace diagonal
  • Rradius of the circumscribed sphere (外接球)
  • rradius of the ball inscribed in a cube (内切球)

Put the box in coordinates with one vertex at O: the opposite vertex is (a, b, c), so d = |OP|, the distance formula again. The centre of the circumscribed sphere is the midpoint of the diagonal.

a · b = x₁x₂ + y₁y₂ + z₁z₂ |a| = √(x₁² + y₁² + z₁²) cos⟨a, b⟩ = a · b / (|a| · |b|) a ⊥ b ⇔ a · b = 0 a ∥ b ⇔ b = λaa · b = x₁x₂ + y₁y₂ + z₁z₂ |a| = √(x₁² + y₁² + z₁²) cos⟨a, b⟩ = a · b / (|a| · |b|) a ⊥ b ⇔ a · b = 0 a ∥ b ⇔ b = λa
where:
  • a = (x₁, y₁, z₁), b = (x₂, y₂, z₂)vectors in space (空间向量)
  • ⟨a, b⟩the angle between the vectors, from 0° to 180°
  • λa number

For two lines with direction vectors a and b, the angle θ is between 0° and 90°: cos θ = |a · b| / (|a| · |b|). Parallel vectors have proportional coordinates.

sin θ = |a · n| / (|a| · |n|)sin θ = |a · n| / (|a| · |n|)
where:
  • adirection vector of the line
  • nnormal vector (法向量) of the plane: a nonzero vector perpendicular to the plane
  • θangle between the line and the plane (0° ≤ θ ≤ 90°)

The angle between a and n is 90° − θ, which is why the formula has sin, not cos. The coordinate planes have simple normals: n = (0, 0, 1) for xOy, (1, 0, 0) for yOz and (0, 1, 0) for xOz.

4Mathematics: probability and statisticsIntermediate

Classical probability

Open lesson
P(A) = m / n, 0 ≤ P(A) ≤ 1, P(Ω) = 1, P(∅) = 0P(A) = m / n, 0 ≤ P(A) ≤ 1, P(Ω) = 1, P(∅) = 0
where:
  • nnumber of all equally likely outcomes (the size of Ω)
  • mnumber of outcomes favourable to A
  • P(A)probability of A (概率)

First make the outcomes equally likely, then count; m and n must be counted in the same way (both ordered or both unordered).

N = n₁ + n₂ + … + nₖ (one of several classes) N = n₁ · n₂ · … · nₖ (several steps)
where:
  • n₁, n₂, …, nₖnumber of ways in each class or at each step
  • Ntotal number of ways

Ask yourself: is the task finished after one choice (add) or only after all the steps (multiply)?

Aₙᵏ = n(n − 1)…(n − k + 1) = n! / (n − k)! Aₙⁿ = n! Cₙᵏ = Aₙᵏ / k! = n! / (k!(n − k)!) Cₙᵏ = Cₙⁿ⁻ᵏAₙᵏ = n(n − 1)…(n − k + 1) = n! / (n − k)! Aₙⁿ = n! Cₙᵏ = Aₙᵏ / k! = n! / (k!(n − k)!) Cₙᵏ = Cₙⁿ⁻ᵏ
where:
  • n!1 · 2 · … · n (0! = 1)
  • Aₙᵏarrangements (permutations, 排列) of k objects chosen from n: order matters
  • Cₙᵏcombinations (组合) of k objects chosen from n: order does not matter

Each combination of k objects can be ordered in k! ways, so Aₙᵏ = k! · Cₙᵏ. Chinese textbooks use the same symbols Aₙᵏ and Cₙᵏ; some books write P(n, k) and C(n, k).

P(Ā) = 1 − P(A) P(A ∪ B) = P(A) + P(B) (A ∩ B = ∅) P(A ∪ B) = P(A) + P(B) − P(AB)
where:
  • Āthe complement of A (“A does not happen”)
  • A ∪ BA or B (at least one of them)
  • AB = A ∩ BA and B together

The last formula is inclusion–exclusion from “Sets and set operations”: the common part AB would otherwise be counted twice. Complementary events are always mutually exclusive, but not the other way round.

P(AB) = P(A) · P(B) P(at least one) = 1 − (1 − p₁)(1 − p₂)…(1 − pₙ)
where:
  • p₁, …, pₙprobabilities of independent events
  • 1 − pᵢprobability that the i-th event does not happen

If A and B are independent, so are Ā and B, A and B̄, and Ā and B̄. Exactly one of two: P = p₁(1 − p₂) + (1 − p₁)p₂.

P(A) = (length, area or volume of the region A) / (length, area or volume of Ω)P(A) = (length, area or volume of the region A) / (length, area or volume of Ω)
where:
  • Ωthe whole segment, region or solid
  • Athe part where the event happens

A single point has length 0, so its probability is 0; that is why “<” and “≤” give the same answer here.

Numerical characteristics of data

Open lesson
nᵢ = n · Nᵢ / Nnᵢ = n · Nᵢ / N
where:
  • nᵢnumber of individuals taken from stratum i
  • Nᵢsize of stratum i
  • n, Nsample size and population size

Proportional allocation: the sampling fraction n/N is the same in every stratum, and it is also the probability that any given individual is chosen.

wᵢ = hᵢ · d, (h₁ + h₂ + … + hₖ) · d = 1, x̄ ≈ m₁w₁ + m₂w₂ + … + mₖwₖ
where:
  • hᵢheight of bar i (relative frequency / class width)
  • dclass width
  • wᵢrelative frequency of class i (the area of its bar)
  • mᵢmidpoint of class i

Estimates from a histogram: mean ≈ Σ midpoint × relative frequency; mode ≈ midpoint of the tallest bar; median = the point x where the area to the left of x is 0.5.

x̄ = (x₁ + x₂ + … + xₙ)/n, x̄ = (x₁f₁ + x₂f₂ + … + xₖfₖ)/nx̄ = (x₁ + x₂ + … + xₙ)/n, x̄ = (x₁f₁ + x₂f₂ + … + xₖfₖ)/n
where:
  • xᵢthe values (in the second form, the different values)
  • fᵢtheir frequencies, f₁ + … + fₖ = n
  • nthe number of all values

The second form (the weighted mean) is used for a frequency table or a bar chart. Remember also: the sum of all values = n · x̄.

R = xₘₐₓ − xₘᵢₙ, s² = [(x₁ − x̄)² + (x₂ − x̄)² + … + (xₙ − x̄)²]/n = (x₁² + x₂² + … + xₙ²)/n − x̄², s = √s²R = xₘₐₓ − xₘᵢₙ, s² = [(x₁ − x̄)² + (x₂ − x̄)² + … + (xₙ − x̄)²]/n = (x₁² + x₂² + … + xₙ²)/n − x̄², s = √s²
where:
  • Rrange (极差)
  • s²variance (方差)
  • sstandard deviation (标准差), in the units of the data
  • xᵢ − x̄deviation of a value from the mean

Chinese school textbooks, and therefore CSCA, divide by n. (University statistics also uses a «sample variance» with n − 1; do not use it here.) The second form — the mean of the squares minus the square of the mean — is faster when x̄ is not a whole number.

yᵢ = axᵢ + b ⇒ ȳ = a·x̄ + b, sy² = a²·sx², sy = |a|·sx
where:
  • ascale factor
  • bshift
  • x̄, sx²mean and variance of the original data
  • ȳ, sy²mean and variance of the new data

The median, the mode and the percentiles change like the mean (for a > 0); the range and the standard deviation are multiplied by |a|. The shift b never changes the spread.

w̄ = (m·x̄ + n·ȳ)/(m + n), s² = {m·[s₁² + (x̄ − w̄)²] + n·[s₂² + (ȳ − w̄)²]}/(m + n)w̄ = (m·x̄ + n·ȳ)/(m + n), s² = {m·[s₁² + (x̄ − w̄)²] + n·[s₂² + (ȳ − w̄)²]}/(m + n)
where:
  • m, nsizes of the two groups (strata)
  • x̄, s₁²mean and variance of the first group
  • ȳ, s₂²mean and variance of the second group
  • w̄, s²mean and variance of all m + n values

Used for stratified samples: each group brings its own variance plus the squared distance of its mean from the overall mean. The overall variance is never just the weighted mean of s₁² and s₂².

The normal distribution

Open lesson
E(X) = x₁p₁ + x₂p₂ + … + xₙpₙ, D(X) = (x₁ − E(X))²p₁ + … + (xₙ − E(X))²pₙ = E(X²) − [E(X)]², σ(X) = √D(X)
where:
  • E(X)expectation (mean value, 数学期望)
  • D(X)variance (方差); English books often write Var(X)
  • σ(X)standard deviation
  • E(X²)x₁²p₁ + x₂²p₂ + … + xₙ²pₙ

E(X) is the long-run average value of X and D(X) measures how far X scatters around it. These are the formulas of the data lesson with the relative frequencies replaced by probabilities. Chinese textbooks write E(X) and D(X).

E(aX + b) = a·E(X) + b, D(aX + b) = a²·D(X)
where:
  • a, bconstants

The same rules as for data (see «Numerical characteristics of data»): the shift b moves the expectation but never changes the variance.

X ~ B(1, p): E(X) = p, D(X) = p(1 − p); X ~ B(n, p): E(X) = np, D(X) = np(1 − p)
where:
  • nnumber of independent trials
  • pprobability of success in one trial
  • 1 − pprobability of failure in one trial

B(1, p) is the two-point distribution. For a binomial variable D/E = 1 − p — a quick way to recover p and n.

f(x) = 1/(σ√(2π)) · e^(−(x − μ)²/(2σ²)), x ∈ ℝf(x) = 1/(σ√(2π)) · e^(−(x − μ)²/(2σ²)), x ∈ ℝ
where:
  • μthe mean: the axis of symmetry of the curve and the position of its peak
  • σthe standard deviation: the width of the bell; in N(μ, σ²) the second number is σ², not σ
  • 1/(σ√(2π))1/(σ√(2π))the height of the peak

In the CSCA you never integrate this function; you need its shape and a few areas.

P(μ − σ ≤ X ≤ μ + σ) ≈ 0.6827, P(μ − 2σ ≤ X ≤ μ + 2σ) ≈ 0.9545, P(μ − 3σ ≤ X ≤ μ + 3σ) ≈ 0.9973
where:
  • μ ± kσthe interval of k standard deviations around the mean
  • μ, σthe parameters of the normal distribution

Chinese textbooks give exactly these values (older books: 0.6826, 0.9544, 0.9974); CSCA items usually print them in the question. They hold for every normal distribution, whatever μ and σ are.

Z = (X − μ)/σ ~ N(0, 1), P(X < x) = Φ((x − μ)/σ), Φ(−z) = 1 − Φ(z), Φ(0) = 0.5Z = (X − μ)/σ ~ N(0, 1), P(X < x) = Φ((x − μ)/σ), Φ(−z) = 1 − Φ(z), Φ(0) = 0.5
where:
  • Zthe standardised variable (z-score): how many σ a value lies from μ
  • Φ(z)P(Z < z), the distribution function of N(0, 1)

A z-score of 2 means «two standard deviations above the mean», so the 68–95–99.7 rule is really a statement about Z: P(|Z| < 1) ≈ 0.6827, P(|Z| < 2) ≈ 0.9545, P(|Z| < 3) ≈ 0.9973.

5Physics: mechanicsIntermediate

Kinematics: displacement, velocity, acceleration, free fall

Open lesson
v̄ = Δx / Δt average speed = s / tv̄ = Δx / Δt average speed = s / t
where:
  • v̄average velocity (its sign gives the direction), m/s
  • Δxdisplacement, m
  • sdistance travelled, m
  • Δt, ttime interval, s

The two are equal only if the body moves in one direction without turning back. Units: 1 m/s = 3.6 km/h, so 72 km/h = 20 m/s and 54 km/h = 15 m/s.

vAB = vA − vB
where:
  • vA, vBvelocities of A and B relative to the ground (with signs), m/s
  • vABvelocity of A relative to B (as seen from B), m/s

Same direction: the speeds subtract; opposite directions: they add. A gap L between two bodies closes in t = L / |vAB|.

v = v₀ + at x = v₀t + at²/2 v² − v₀² = 2ax x = (v₀ + v)t / 2v = v₀ + at x = v₀t + at²/2 v² − v₀² = 2ax x = (v₀ + v)t / 2
where:
  • v₀initial velocity, m/s
  • vvelocity at time t, m/s
  • aconstant acceleration, m/s²
  • ttime, s
  • xdisplacement during the time t, m

Five quantities, four equations: each equation leaves one quantity out (see the table). Take the one without the quantity you are neither given nor asked for.

Δx = aT² v(t/2) = v̄ = (v₀ + v) / 2Δx = aT² v(t/2) = v̄ = (v₀ + v) / 2
where:
  • Δxdifference between the displacements in two successive equal intervals, m
  • Tlength of each interval, s
  • v(t/2)v(t/2)velocity at the middle moment of an interval, m/s

Two shortcuts that Chinese textbooks use all the time (逐差法): displacements in successive equal intervals differ by aT², and the velocity at the middle moment equals the average velocity. Starting from rest, the displacements in successive equal intervals are in the ratio 1 : 3 : 5 : 7 …

v = slope of the x–t graph a = slope of the v–t graph Δx = area under the v–t graph
where:
  • slopeΔ(vertical) / Δ(horizontal) between two points of a straight part; for a curve, the slope of the tangent
  • areaarea between the graph and the t-axis, m

Area above the t-axis is positive displacement, area below it is negative. Displacement = the sum with signs; distance = the sum of the absolute values.

v = gt h = gt²/2 v² = 2ghv = gt h = gt²/2 v² = 2gh
where:
  • hheight fallen, m
  • gfree-fall acceleration, m/s²
  • ttime since release, s
  • vspeed, m/s

These are the four equations with v₀ = 0 and a = g. With g = 10 m/s², after 1, 2, 3, 4, 5 s: speed 10, 20, 30, 40, 50 m/s; height fallen 5, 20, 45, 80, 125 m.

v = v₀ − gt h = v₀t − gt²/2 H = v₀² / (2g) t↑ = v₀ / gv = v₀ − gt h = v₀t − gt²/2 H = v₀² / (2g) t↑ = v₀ / g
where:
  • v₀launch speed, upward, m/s
  • hheight above the launch point (up is positive), m
  • Hmaximum height, m
  • t↑time to the top, s

In a vertical throw (竖直上抛运动) a = −g throughout, even at the top, where v = 0 but the acceleration is still g. The motion is symmetric: the fall takes as long as the rise, and at every height the speed is the same going up and coming down.

Projectile motion

Open lesson
x = vₓt vy = v₀ᵧ − gt y = v₀ᵧt − gt²/2 v = √(vₓ² + vy²)x = vₓt vy = v₀ᵧ − gt y = v₀ᵧt − gt²/2 v = √(vₓ² + vy²)
where:
  • vₓhorizontal component of the velocity (constant), m/s
  • v₀ᵧ, vyvertical component of the velocity at the start and at time t (up is positive), m/s
  • x, yhorizontal and vertical displacement from the launch point, m
  • vspeed (magnitude of the velocity), m/s

Split the initial velocity into components, solve each direction on its own, and add the results as vectors (Pythagoras) only at the end.

t = √(2h/g) x = v₀t = v₀√(2h/g) vy = gt = √(2gh) v = √(v₀² + 2gh) tan θ = gt / v₀t = √(2h/g) x = v₀t = v₀√(2h/g) vy = gt = √(2gh) v = √(v₀² + 2gh) tan θ = gt / v₀
where:
  • hlaunch height, m
  • v₀horizontal launch speed, m/s
  • xhorizontal range, m
  • θangle between the velocity and the horizontal

The time does not depend on v₀; the range grows in proportion to v₀ and to √h.

tan θ = gt / v₀ tan α = y / x = gt / (2v₀) tan θ = 2 tan αtan θ = gt / v₀ tan α = y / x = gt / (2v₀) tan θ = 2 tan α
where:
  • θangle of the velocity with the horizontal
  • αangle of the displacement with the horizontal
  • x, yhorizontal and vertical (downward) distance from the launch point, m

A consequence: the velocity vector, extended backwards, passes through the midpoint of the horizontal displacement.

v₀ₓ = v₀ cos θ v₀ᵧ = v₀ sin θ T = 2v₀ sin θ / g H = v₀² sin²θ / (2g) R = v₀² sin 2θ / gv₀ₓ = v₀ cos θ v₀ᵧ = v₀ sin θ T = 2v₀ sin θ / g H = v₀² sin²θ / (2g) R = v₀² sin 2θ / g
where:
  • θlaunch angle above the horizontal
  • Ttime of flight (back to the launch level), s
  • Hmaximum height, m
  • Rrange on level ground, m

R = v₀ cos θ · T. Useful values: sin 30° = 1/2, sin 37° = 0.6, sin 45° = √2/2, sin 53° = 0.8, sin 60° = √3/2 (cos 37° = 0.8, cos 53° = 0.6).

tan θ = y / x = gt / (2v₀) ⇒ t = 2v₀ tan θ / gtan θ = y / x = gt / (2v₀) ⇒ t = 2v₀ tan θ / g
where:
  • θangle of the slope with the horizontal
  • ttime of flight of a body thrown horizontally from the top of the slope until it lands on the slope, s

Landing on a slope (斜面上的平抛): the displacement lies along the slope, so the displacement angle α equals θ. t is proportional to v₀, and the distance along the slope to v₀².

Δy = gT² v₀ = Δx / T vy = (y₁ + y₂) / (2T)Δy = gT² v₀ = Δx / T vy = (y₁ + y₂) / (2T)
where:
  • Δydifference between two successive vertical steps, m
  • Ttime between two images, s
  • Δxone horizontal step, m
  • y₁, y₂the vertical steps before and after the middle image, m
  • vyvertical component of the velocity at the middle image, m/s

First find T from the difference of the vertical steps, then v₀ from a horizontal step; vy at the middle image is the average velocity over the two neighbouring steps.

Forces and equilibrium

Open lesson
F = kx
where:
  • Fthe spring force, N
  • kthe spring constant (stiffness), N/m (劲度系数)
  • xthe extension or compression — the change of length, not the length itself, m

Hooke's law (胡克定律) holds within the elastic limit. The force always points back toward the natural length.

0 ≤ fₛ ≤ fₘₐₓ fₖ = μN
where:
  • fₛstatic friction: exactly what keeps the body at rest
  • fₘₐₓmaximum static friction; CSCA items take it as μN unless told otherwise
  • fₖkinetic (sliding) friction, N
  • μcoefficient of friction, no unit (动摩擦因数)
  • Nthe normal force (Chinese textbooks: FN); not always mg!

Friction depends on N and μ; it does not depend on the contact area, and kinetic friction does not depend on the speed.

R = √(F₁² + F₂² + 2F₁F₂ cos θ) |F₁ − F₂| ≤ R ≤ F₁ + F₂
where:
  • F₁, F₂two forces acting at one point, N
  • θthe angle between the forces
  • Rthe resultant (合力): the diagonal of the parallelogram built on the two forces

The parallelogram rule (平行四边形定则). As θ grows from 0° to 180°, R decreases from F₁ + F₂ to |F₁ − F₂|.

Fx = F cos α Fy = F sin α F = √(Fx² + Fy²)
where:
  • αthe angle between the force and the x-axis
  • Fx, Fythe components of the force along the axes (分力), N

Orthogonal decomposition (正交分解): replace a slanted force by two perpendicular components, then balance each axis separately.

N = mg cos θ fₛ = mg sin θ sliding at constant speed: μ = tan θ
where:
  • θthe angle of the incline with the horizontal
  • mg sin θthe component of gravity down the slope
  • mg cos θthe component of gravity into the slope; N balances it

A block on an incline with no other forces. If it slides down at constant speed, μmg cos θ = mg sin θ, so μ = tan θ, whatever the mass.

Newton's laws of motion and their applications

Open lesson
ΣF = ma
where:
  • ΣFthe net (resultant) force of all forces on the body (合外力), N
  • mmass, kg
  • aacceleration, m/s², always in the direction of ΣF

Newton's second law (牛顿第二定律). It is a vector law and holds at every instant: when the net force changes, the acceleration changes at the same moment. 1 N = 1 kg · m/s².

down: a = g(sin θ − μ cos θ) up: a = g(sin θ + μ cos θ)
where:
  • θthe angle of the incline
  • μthe coefficient of kinetic friction
  • athe size of the acceleration; in both cases it points down the slope

A block on a rough incline with no other forces. Sliding down, friction points up the slope and is subtracted; sliding up, both the gravity component and friction point down the slope and add up. If μ ≥ tan θ, a block at rest does not start to slide.

a = F / (m₁ + m₂) T = m₂F / (m₁ + m₂)a = F / (m₁ + m₂) T = m₂F / (m₁ + m₂)
where:
  • Fthe external pull on the front body m₁, N
  • m₂the mass of the body pulled by the string (behind), kg
  • Tthe tension in the string between the bodies, N

Two blocks on a smooth horizontal floor, F pulling the front block m₁, a string to m₂. The internal force is shared in proportion to the mass that is pulled. T stays the same if both blocks have the same μ, and also along an incline.

a = (m₁ − m₂)g / (m₁ + m₂) T = 2m₁m₂g / (m₁ + m₂)a = (m₁ − m₂)g / (m₁ + m₂) T = 2m₁m₂g / (m₁ + m₂)
where:
  • m₁ > m₂the masses at the two ends of a light string over a smooth fixed pulley (定滑轮)
  • Tthe tension in the string, N

The Atwood machine. Check: equal masses give a = 0 and T = mg; as m₂ → 0, a → g and T → 0.

a up: N = m(g + a) a down: N = m(g − a) free fall: N = 0
where:
  • Napparent weight: the reading of the scale (or the tension of a spring balance or rope), N
  • athe size of the lift's acceleration, m/s²

Gravity mg does not change in a lift; only the support force does.

Work, power and energy

Open lesson
W = F · s · cos α
where:
  • Wwork of the force, J
  • Fmagnitude of the constant force, N
  • smagnitude of the displacement, m
  • αangle between the force and the displacement

Chinese textbooks write W = Fl cos α (l = displacement); the CSCA uses W for work and P for power, while some school textbooks use A and N. Only the component F cos α along the motion does work.

Wtotal = W₁ + W₂ + … = Fnet · s · cos α
where:
  • Wtotaltotal work of all forces on the body (with signs)
  • Fnetnet force (for constant forces)

Works are scalars: add them as signed numbers, not as vectors.

P = W / t P = F · v · cos αP = W / t P = F · v · cos α
where:
  • Ppower, W
  • ttime taken, s
  • vspeed of the body, m/s
  • αangle between the force F and the velocity v

W/t gives the average power. F · v gives the instantaneous power with the instantaneous speed, or the average power with the average speed.

F = P / v a = (P/v − f) / m vmax = P / fF = P / v a = (P/v − f) / m vmax = P / f
where:
  • Ftraction force of the engine, N
  • fconstant resistive force, N
  • vmaxmaximum speed on a level road, m/s

At vmax the acceleration is zero: the traction force exactly balances the resistance (Chinese textbooks call this topic 机车启动, «a vehicle starting off»).

Wtotal = Ek₂ − Ek₁ = mv₂²/2 − mv₁²/2Wtotal = Ek₂ − Ek₁ = mv₂²/2 − mv₁²/2
where:
  • Wtotaltotal work of all forces (gravity and friction included), J
  • Ek₁, Ek₂initial and final kinetic energy
  • v₁, v₂initial and final speed, m/s

The work–energy theorem (Chinese 动能定理, «kinetic energy theorem») also holds on curved paths and for changing forces, as long as you can find their work.

Ep = mgh WG = mg(h₁ − h₂) = −ΔEp
where:
  • hheight above the reference plane, m
  • WGwork of gravity when the body moves from height h₁ to h₂
  • ΔEpchange of potential energy: Ep₂ − Ep₁

The work of gravity depends only on the start and end heights, never on the path: going down it is positive and Ep decreases; going up it is negative and Ep increases.

Ep = kx²/2Ep = kx²/2
where:
  • kspring constant (stiffness), N/m
  • xdeformation (extension or compression), m

Doubling the deformation stores four times the energy. The spring force kx itself belongs to the lesson «Forces and equilibrium».

Ek₁ + Ep₁ = Ek₂ + Ep₂ (mv₁²/2 + mgh₁ = mv₂²/2 + mgh₂)Ek₁ + Ep₁ = Ek₂ + Ep₂ (mv₁²/2 + mgh₁ = mv₂²/2 + mgh₂)
where:
  • Ek₁ + Ep₁mechanical energy in the first state
  • Ek₂ + Ep₂mechanical energy in the second state

Condition: only gravity and elastic (spring) forces do work. Other forces may act if they do no work: the normal force of a smooth surface, the tension of a pendulum string.

Wother = E₂ − E₁ Q = f · srel
where:
  • Wotherwork of the forces other than gravity and spring forces, J
  • Emechanical energy: E = Ek + Ep
  • Qheat produced by kinetic friction, J
  • fkinetic friction force, N
  • srelsliding distance of the two surfaces relative to each other (相对位移), m

The heat depends on the relative sliding distance, not on the distance measured from the ground. For a block sliding on a fixed floor, srel is simply the block's path.

Momentum, impulse and conservation of momentum

Open lesson
p = mv Δp = mv₂ − mv₁ Ek = p²/(2m)p = mv Δp = mv₂ − mv₁ Ek = p²/(2m)
where:
  • pmomentum, kg · m/s
  • v₁, v₂initial and final velocity (along a line, with signs), m/s
  • Δpchange of momentum (a vector difference)

In one dimension choose a positive direction: velocities and momenta against it are negative. Ek = p²/(2m): for the same momentum the lighter body has more kinetic energy.

I = F · Δt I = area under the F–t graph
where:
  • Iimpulse of the force, N · s
  • Fconstant force, N
  • Δttime during which the force acts, s

Unlike work, impulse is not zero when the body stays in place: what counts is time, not distance.

Fnet · Δt = mv₂ − mv₁ = Δp
where:
  • Fnetnet force of all forces (gravity included); for a short impact its average value, N
  • Δtduration of the interaction, s
  • Δpchange of the body's momentum, kg · m/s

Chinese: 动量定理 («momentum theorem»). It is Newton's second law in the form F = Δp/Δt: the impulse of the net force equals the change of the body's momentum.

m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′
where:
  • m₁, m₂masses of the bodies, kg
  • v₁, v₂velocities before the interaction (with signs), m/s
  • v₁′, v₂′velocities after the interaction, m/s

Conditions: 1) the net external force is zero (a smooth horizontal surface: gravity and the normal force balance); 2) in a short collision or explosion the external forces are much smaller than the internal ones, so momentum is conserved approximately; 3) if the net external force is zero along one direction, only the momentum component along it is conserved.

v = (m₁v₁ + m₂v₂)/(m₁ + m₂) ΔE_lost = m₂/(m₁ + m₂) · Ek₀ (v₂ = 0)v = (m₁v₁ + m₂v₂)/(m₁ + m₂) ΔE_lost = m₂/(m₁ + m₂) · Ek₀ (v₂ = 0)
where:
  • vcommon velocity of the stuck bodies, m/s
  • ΔE_lostkinetic energy turned into heat, J
  • Ek₀initial kinetic energy of the moving body (the other at rest)

Perfectly inelastic collision. The heavier the target, the larger the share of energy lost: half for m₂ = m₁, 80% for m₂ = 4m₁.

v₁′ = (m₁ − m₂)v₁/(m₁ + m₂) v₂′ = 2m₁v₁/(m₁ + m₂)v₁′ = (m₁ − m₂)v₁/(m₁ + m₂) v₂′ = 2m₁v₁/(m₁ + m₂)
where:
  • v₁velocity of body 1 before the collision (body 2 at rest), m/s
  • v₁′, v₂′velocities after the collision, m/s

Perfectly elastic head-on collision (对心碰撞, «one moving, one at rest»): it follows from conservation of both momentum and kinetic energy.

mv₀ = (m + M)V Q = f · d = mv₀²/2 − (m + M)V²/2 = M/(m + M) · mv₀²/2mv₀ = (m + M)V Q = f · d = mv₀²/2 − (m + M)V²/2 = M/(m + M) · mv₀²/2
where:
  • m, v₀mass and speed of the bullet
  • M, Vmass of the block and the common speed after the bullet is embedded
  • Qheat produced, J
  • f, daverage resistive force in the block and the penetration depth (the relative displacement)

The block is free on a smooth surface. The heat follows the rule Q = f · srel from «Work, power and energy»: d is the bullet's path relative to the block.

Circular motion

Open lesson
v = 2πr/T ω = 2π/T = 2πf v = ωr T = 1/fv = 2πr/T ω = 2π/T = 2πf v = ωr T = 1/f
where:
  • vlinear speed, m/s (along the tangent)
  • ωangular velocity, rad/s
  • Tperiod, s
  • ffrequency, Hz (revolutions per second)
  • rradius (distance of the point from the axis), m

In one revolution the point travels 2πr and the radius turns through 2π rad, both in time T. That gives v = ωr.

belt, chain, gears: v₁ = v₂ ⇒ ω₁r₁ = ω₂r₂ ⇒ ω₁ : ω₂ = r₂ : r₁ same axle: ω₁ = ω₂ ⇒ v₁ : v₂ = r₁ : r₂
where:
  • r₁, r₂radii of the wheels or distances of the points from the axis; for gears the radius is proportional to the number of teeth

On a belt the smaller wheel spins faster; on one axle the point farther from the axis moves faster.

a = v²/r = ω²r = ωv = 4π²r/T² F = ma = mv²/r = mω²r = 4π²mr/T²a = v²/r = ω²r = ωv = 4π²r/T² F = ma = mv²/r = mω²r = 4π²mr/T²
where:
  • acentripetal acceleration, m/s², always toward the centre
  • Fcentripetal force — the resultant force toward the centre, N
  • mmass, kg

The directions of a and F change every instant, so uniform circular motion is not uniformly accelerated motion. The force is perpendicular to the velocity, so it does no work and the kinetic energy stays constant.

flat turn: μmg ≥ mv²/r ⇒ vmax = √(μgr) banked turn (no friction): tan θ = v²/(gr) conical pendulum: a = g tan θ, r = L sin θ, ω = √(g/(L cos θ)) = √(g/h)flat turn: μmg ≥ mv²/r ⇒ vmax = √(μgr) banked turn (no friction): tan θ = v²/(gr) conical pendulum: a = g tan θ, r = L sin θ, ω = √(g/(L cos θ)) = √(g/h)
where:
  • μcoefficient of static friction
  • θbank angle of the road, or the angle of the string to the vertical
  • Llength of the string, m
  • hheight of the cone h = L cos θ — vertical distance from the pivot to the plane of the circle

On a banked turn and in the conical pendulum the resultant of mg and the normal force (tension) is horizontal: F = mg tan θ, so a = g tan θ.

top: F + mg = mv²/r bottom: F − mg = mv²/r string (inside of a loop): vtop ≥ √(gr) rod (tube): vtop ≥ 0top: F + mg = mv²/r bottom: F − mg = mv²/r string (inside of a loop): vtop ≥ √(gr) rod (tube): vtop ≥ 0
where:
  • Fforce of the string, rod or track on the body, taken positive toward the centre
  • rradius of the circle, m

A string can only pull: at the top F ≥ 0, so mg ≤ mv²/r. A rod can pull or push: the ball can pass the top at any v ≥ 0; at v = √(gr) the rod exerts no force.

Universal gravitation and satellites

Open lesson
a³/T² = ka³/T² = k
where:
  • asemi-major axis of the ellipse (the radius r for a circular orbit), m
  • Torbital period, s
  • ka constant that depends only on the mass of the central body: k = GM/(4π²)

Compare only bodies that orbit the same centre: planets of the Sun with each other, moons of Jupiter with each other — never the Moon with the Earth.

F = G · m₁m₂/r²F = G · m₁m₂/r²
where:
  • Ggravitational constant, 6.67 × 10⁻¹¹ N · m²/kg²
  • m₁, m₂masses of the bodies, kg
  • rdistance between the centres, m

The force is proportional to the product of the masses and inversely proportional to the square of the distance: double r and F becomes four times smaller.

g = GM/R² GM = gR² gh = GM/(R + h)² = g · R²/(R + h)²g = GM/R² GM = gR² gh = GM/(R + h)² = g · R²/(R + h)²
where:
  • M, Rmass and radius of the planet
  • hheight above the surface; the distance from the centre is R + h
  • ghfree-fall acceleration at height h

g falls with the square of the distance from the centre: g/4 at 2R, g/9 at 3R.

M = gR²/G M = 4π²r³/(GT²) ρ = M/(4πR³/3) = 3πr³/(GT²R³); r = R: ρ = 3π/(GT²)M = gR²/G M = 4π²r³/(GT²) ρ = M/(4πR³/3) = 3πr³/(GT²R³); r = R: ρ = 3π/(GT²)
where:
  • r, Tradius and period of the orbit of a moon or satellite
  • ρmean density of the planet, kg/m³

The satellite’s own mass cancels — only the central body can be “weighed” this way. For an orbit just above the surface, the density needs only T.

v = √(GM/r) ω = √(GM/r³) T = 2π√(r³/(GM)) a = GM/r²v = √(GM/r) ω = √(GM/r³) T = 2π√(r³/(GM)) a = GM/r²
where:
  • rorbit radius r = R + h (from the centre of the planet)
  • Mmass of the central body (GM = gR²)

The higher the orbit, the smaller v, ω and a, and the longer T.

v₁ = √(GM/R) = √(gR)v₁ = √(GM/R) = √(gR)
where:
  • Rradius of the planet, m
  • gfree-fall acceleration at the planet’s surface

mg = mv²/R ⇒ v₁ = √(gR): at the surface gravity supplies the centripetal force.

Simple harmonic motion

Open lesson
F = −kx a = −(k/m)·xF = −kx a = −(k/m)·x
where:
  • Frestoring force, N (the net force along the line of motion)
  • xdisplacement from the equilibrium position, m (it can be negative)
  • ka positive constant, N/m; for a spring it is the spring constant
  • aacceleration, m/s²; proportional to x and opposite to it

The minus sign is the whole idea: the force and the acceleration always point toward equilibrium, opposite to the displacement. Any motion that obeys a = −(constant)·x is SHM.

x = A sin(ωt + φ) ω = 2π/T = 2πf f = 1/Tx = A sin(ωt + φ) ω = 2π/T = 2πf f = 1/T
where:
  • xdisplacement at time t
  • Aamplitude (always positive)
  • ωangular frequency, rad/s
  • φinitial phase, rad
  • T, fperiod (s) and frequency (Hz)

The graph of x against t is a sine curve. Some books write x = A cos(ωt + φ); that is the same motion with the phase shifted by π/2. Always work in radians inside the sine.

T = 2π√(m/k)T = 2π√(m/k)
where:
  • mmass of the oscillating body, kg
  • kspring constant, N/m (劲度系数)

It follows from a = −(k/m)x, so ω = √(k/m). The period does not depend on the amplitude, and it is the same for a vertical spring: gravity only moves the equilibrium point down by mg/k.

T = 2π√(L/g)T = 2π√(L/g)
where:
  • Llength from the pivot to the centre of the bob, m
  • gfree-fall acceleration, m/s² (10 m/s² in CSCA items unless stated otherwise)

Valid for small swings (up to about 5°): then the restoring force mg sin θ ≈ mg·x/L is proportional to x. The period depends on neither the mass nor the amplitude. In a lift accelerating upward use g + a instead of g; in one accelerating downward, g − a.

E = ½kA² = ½mv² + ½kx² vmax = ωA amax = ω²A
where:
  • Etotal mechanical energy, J
  • ½mv², ½kx²kinetic and elastic potential energy at displacement x
  • vmaxgreatest speed, reached at equilibrium
  • amaxgreatest acceleration, reached at the extremes

For a pendulum the potential energy is mgh, but the picture is the same. The kinetic and the potential energy each repeat twice per oscillation, so they change with period T/2 (frequency 2f).

Mechanical waves

Open lesson
v = λ/T = λfv = λ/T = λf
where:
  • vwave speed, m/s (波速)
  • λwavelength: the distance between the two nearest points that vibrate in phase, m (波长)
  • T, fperiod and frequency of the vibration — the same as the source's

In one period each particle makes one full vibration and the wave moves on by exactly one wavelength. The medium sets v and the source sets f; when a wave passes into another medium, f stays the same and λ changes in proportion to v.

Δr = nλ → maximum Δr = (2n + 1)·λ/2 → minimum (n = 0, 1, 2, …)Δr = nλ → maximum Δr = (2n + 1)·λ/2 → minimum (n = 0, 1, 2, …)
where:
  • Δrpath difference |r₁ − r₂|, m
  • λwavelength, m
  • na whole number: 0, 1, 2, …

For sources vibrating in phase. If the sources are in antiphase, the two conditions swap. A whole number of wavelengths means crest meets crest; an odd number of half-wavelengths means crest meets trough.

L = n·λ/2 fₙ = n·v/(2L) (n = 1, 2, 3, …)L = n·λ/2 fₙ = n·v/(2L) (n = 1, 2, 3, …)
where:
  • Llength of the string fixed at both ends, m
  • nnumber of loops (antinodes); n = 1 is the fundamental
  • vspeed of the travelling waves on the string, m/s

Both ends are nodes, so a whole number of half-wavelengths must fit on the string. The lowest frequency (n = 1, λ = 2L) is the fundamental; the others are whole multiples of it.

6Physics: electromagnetismIntermediate

Coulomb's law and the electric field

Open lesson
q = n · e
where:
  • qcharge of the body, C
  • nnumber of extra (or missing) electrons, a whole number
  • eelementary charge, 1.6 × 10⁻¹⁹ C

Charge is quantized: 1 C corresponds to 1/e = 6.25 × 10¹⁸ electrons.

q₁′ = q₂′ = (q₁ + q₂) / 2q₁′ = q₂′ = (q₁ + q₂) / 2
where:
  • q₁, q₂charges of the spheres before contact, with their signs
  • q₁′, q₂′charges after they are separated

Only for two identical metal spheres: opposite charges first cancel, then the rest is shared equally (Chinese textbooks: 先中和后平分, “first neutralize, then share equally”).

F = k · q₁q₂ / r²F = k · q₁q₂ / r²
where:
  • FCoulomb force (electrostatic force), N
  • q₁, q₂magnitudes of the charges, C
  • rdistance between the charges, m
  • k9.0 × 10⁹ N·m²/C² (k = 1/(4πε₀))

In vacuum (air gives practically the same). Put in magnitudes; the force acts along the line joining the charges: like charges repel, unlike attract. The two charges act on each other with equal and opposite forces (Newton's third law), even if the charges are different.

x / (L − x) = √(q₁ / q₂)x / (L − x) = √(q₁ / q₂)
where:
  • xdistance from the smaller charge q₁ to the equilibrium point, m
  • Ldistance between the two fixed charges, m
  • q₁, q₂magnitudes of the fixed charges (q₁ < q₂)

From kq₁q/x² = kq₂q/(L − x)² for like charges. For unlike charges the point is outside, beyond q₁: x/(L + x) = √(q₁/q₂). At the same point the field of the two charges is zero.

E = F / qE = F / q
where:
  • Efield strength at the point, N/C
  • Fforce on the test charge, N
  • qtest charge, C

E belongs to the point of the field, not to the test charge: doubling q doubles F, and E stays the same. Read backwards, F = qE gives the force on any charge.

E = k · Q / r²E = k · Q / r²
where:
  • Qsource point charge (magnitude), C
  • rdistance from Q to the point, m

Field of a point charge: directed away from a positive Q and towards a negative Q; it falls as 1/r² — twice as far, a quarter of the field.

E = E₁ + E₂ + E₃ + …
where:
  • E₁, E₂, …fields of the individual charges at the point (vectors)

A vector sum: same direction — add the magnitudes; opposite — subtract; perpendicular — E = √(E₁² + E₂²).

a = qE / m v² = 2ad = 2qEd / ma = qE / m v² = 2ad = 2qEd / m
where:
  • aacceleration of the particle, m/s²
  • q/mq/mcharge-to-mass ratio (specific charge), C/kg
  • ddistance travelled along the field from rest, m
  • vspeed after the distance d, m/s

For a particle that starts from rest and moves along the field lines. The lesson “Electric potential, potential difference and capacitors” gets the same speed from energy in one line: qU = mv²/2.

t = L / v₀ y = qEL² / (2mv₀²) tan θ = qEL / (mv₀²)t = L / v₀ y = qEL² / (2mv₀²) tan θ = qEL / (mv₀²)
where:
  • v₀entry speed, perpendicular to the field, m/s
  • Llength of the plates, m
  • ysideways deflection at the exit, m
  • θangle between the exit velocity and the initial direction

A particle entering perpendicular to the field moves like a horizontally thrown stone (Chinese: 类平抛运动, “projectile-like motion”): uniformly along the plates, uniformly accelerated across them. It is the algebra of the lesson “Projectile motion” with g replaced by qE/m. Useful: tan θ = 2y/L.

Electric potential, potential difference and capacitors

Open lesson
W = qEd
where:
  • Wwork done by the field, J
  • qcharge, C
  • Estrength of the uniform field, N/C = V/m
  • ddisplacement along the field lines (the projection of the path), m

Uniform field only. The work is positive when the charge moves the way the field pushes it.

WAB = EpA − EpB
where:
  • WABwork done by the field as the charge moves from A to B, J
  • EpA, EpBpotential energy of the charge at A and at B, J

The field's work equals the decrease of potential energy: positive work → Ep falls; negative work (the charge is forced against the field) → Ep rises, like a stone being lifted.

φ = Ep / q Ep = qφφ = Ep / q Ep = qφ
where:
  • φpotential at the point, V
  • Eppotential energy of the charge there, J
  • qcharge, with its sign, C

Substitute q with its sign. A positive charge has a large Ep where φ is high; a negative charge where φ is low. With the zero at infinity, a point charge gives φ = kQ/r: positive around +Q, negative around −Q.

UAB = φA − φB WAB = q · UAB
where:
  • UABvoltage between A and B, V
  • WABwork of the field when q moves from A to B, J
  • qcharge, with its sign, C

Chinese textbooks put the signs of both q and UAB into the formula; the sign of W then tells whether Ep falls (W > 0) or rises (W < 0). Many other textbooks write the same law as A = qU.

qU = ½mv² − ½mv₀² 1 eV = 1.6 × 10⁻¹⁹ J
where:
  • qUwork of the field between the two points (q and U with signs), J
  • mmass of the particle, kg
  • v₀, vspeeds at the start and at the end, m/s
  • eVelectronvolt: the energy an electron gains through 1 V

A particle of charge e accelerated from rest through U volts gains U electronvolts; a charge of 2e gains 2U eV.

U = E · d E = U / dU = E · d E = U / d
where:
  • Uvoltage between two points, V
  • Estrength of the uniform field, V/m
  • ddistance between the points measured along the field lines (the projection), m

Uniform fields only. It shows why N/C = V/m, and why E points from high to low potential: “the field is the drop of potential per metre”.

C = Q / U = ΔQ / ΔUC = Q / U = ΔQ / ΔU
where:
  • Ccapacitance, farad (F); 1 μF = 10⁻⁶ F, 1 pF = 10⁻¹² F
  • Qcharge of the capacitor, C
  • Uvoltage between the plates, V

C is set by the construction of the capacitor, not by Q or U: double the voltage and the charge doubles, while C stays the same.

C = εᵣS / (4πkd) = ε₀εᵣS / dC = εᵣS / (4πkd) = ε₀εᵣS / d
where:
  • εᵣrelative permittivity (dielectric constant) of the insulator: 1 for vacuum and air
  • Sarea of overlap of the plates, m²
  • ddistance between the plates, m
  • k, ε₀k = 9 × 10⁹ N·m²/C²; ε₀ = 1/(4πk) ≈ 8.85 × 10⁻¹² F/m

The parallel-plate capacitor. The first form is the one in Chinese textbooks (C = εᵣS/4πkd); both forms are the same formula, because 1/(4πk) = ε₀. C grows with S and εᵣ and falls with d.

Direct current circuits

Open lesson
I = q/tI = q/t
where:
  • Icurrent, A
  • qcharge through a cross-section, C
  • ttime, s

Chinese textbooks also give the microscopic form I = nqSv: n carriers per cubic metre, each with charge q, drifting at speed v through a cross-section of area S.

I = U/RI = U/R
where:
  • Uvoltage across that part of the circuit, V
  • Rresistance of that part, Ω
  • Icurrent through it, A

Ohm's law for a part of a circuit (部分电路欧姆定律). It holds for metals and electrolytes at constant temperature, not for gas discharges or a turning motor.

R = ρL/SR = ρL/S
where:
  • ρresistivity, Ω·m (depends on the material and the temperature)
  • Llength of the wire, m
  • Scross-sectional area, m²

A longer wire has more resistance, a thicker one less. In metals ρ grows with temperature (copper has ρ ≈ 1.7 × 10⁻⁸ Ω·m at room temperature, which is why wires are made of copper).

Series: R = R₁ + R₂ + … Parallel: 1/R = 1/R₁ + 1/R₂ + …Series: R = R₁ + R₂ + … Parallel: 1/R = 1/R₁ + 1/R₂ + …
where:
  • Requivalent (total) resistance
  • R₁, R₂resistances of the separate elements

The series total is larger than the largest element; the parallel total is smaller than the smallest one.

Ammeter: Rsh = Ig·Rg/(I − Ig) Voltmeter: Rs = U/Ig − RgAmmeter: Rsh = Ig·Rg/(I − Ig) Voltmeter: Rs = U/Ig − Rg
where:
  • Ig, Rgfull-scale current and resistance of the galvanometer
  • Inew full-scale current of the ammeter
  • Unew full-scale voltage of the voltmeter
  • Rsh, Rsparallel shunt and series resistor

For an n times larger range: shunt Rg/(n − 1), series resistor (n − 1)·Rg.

I = E/(R + r) U = E − IrI = E/(R + r) U = E − Ir
where:
  • EEMF of the source, V
  • Rexternal resistance, Ω
  • rinternal resistance of the source, Ω
  • Uterminal voltage (路端电压): the voltage across the external circuit, U = IR

Ohm's law for the closed circuit (闭合电路欧姆定律). Open circuit (I = 0): U = E. Short circuit (R = 0): I = E/r, dangerously large. The U–I graph of a source is a straight line: it cuts the U axis at E, and the magnitude of its slope is r.

P = UI Q = I²Rt
where:
  • Ppower, W
  • QJoule heat, J
  • ttime, s

Pure resistor: P = I²R = U²/R. Motor with coil resistance rM: input power UI, heating power I²rM, mechanical power UI − I²rM.

Pout = E²R/(R + r)² Pmax = E²/(4r) when R = r η = R/(R + r)Pout = E²R/(R + r)² Pmax = E²/(4r) when R = r η = R/(R + r)
where:
  • Poutoutput power (in the external circuit)
  • Pmaxgreatest possible output power
  • ηefficiency

The output power is greatest when the external resistance equals the internal one; the efficiency is then only 50 %. Two different loads give the same output power when R₁R₂ = r².

Magnetic field: Ampère force and Lorentz force

Open lesson
Φ = BS sin θ (S ⟂ B: Φ = BS)
where:
  • Φmagnetic flux, Wb (weber), 1 Wb = 1 T·m²
  • Bmagnetic induction, T
  • Sarea of the surface, m²
  • θangle between the plane of the surface and B

If the angle α is measured between B and the normal to the surface, the same flux is Φ = BS cos α. Φ does not depend on the number of turns of a coil. When a loop is turned over (through 180°), the flux changes sign: ΔΦ = 2BS. Changes of flux drive the induced currents of the lesson “Electromagnetic induction: Faraday's law and Lenz's law”.

F = BIL sin θ
where:
  • FAmpère force, N
  • Icurrent, A
  • Llength of the wire inside the field, m
  • θangle between the wire (the current) and B

The force is greatest, F = BIL, when the wire is perpendicular to B, and zero when the wire is parallel to B. It is always perpendicular to both the wire and B.

F = qvB sin θ
where:
  • qcharge of the particle, C
  • vspeed of the particle, m/s
  • θangle between v and B

v ⟂ B: F = qvB (maximum); v ∥ B: F = 0, and the particle moves on in a straight line. Direction: the left-hand rule with the four fingers along the velocity of a positive charge; for a negative charge (an electron) point them against the velocity.

r = mv/(qB) T = 2πr/v = 2πm/(qB) t = (θ/360°) · Tr = mv/(qB) T = 2πr/v = 2πm/(qB) t = (θ/360°) · T
where:
  • rradius of the circle, m
  • mmass of the particle, kg
  • Tperiod of revolution, s
  • t, θtime in the field and the central angle of the arc (= the angle through which the velocity turns)

The radius grows with the speed (with the momentum mv), but the period depends neither on v nor on r: faster particles run along bigger circles in the same time. With the kinetic energy: r = √(2mE_k)/(qB).

v = E/Bv = E/B
where:
  • Eelectric field strength, V/m (= N/C)
  • Bmagnetic induction, T
  • vspeed that passes undeflected

The selected speed depends neither on the charge (or its sign) nor on the mass. A faster particle is pushed toward the side of the magnetic force (qvB > qE), a slower one toward the side of the electric force.

r = (1/B) · √(2mU/q) m = qB²r²/(2U)r = (1/B) · √(2mU/q) m = qB²r²/(2U)
where:
  • Uaccelerating voltage, V
  • rradius of the half-circle (the spot is 2r from the slit)
  • m, qmass and charge of the ion

At the same U and B, r ∝ √(m/q): isotopes (same q, different m) land at different places, the heavier ones farther out.

f = qB/(2πm) Ek max = q²B²R²/(2m)f = qB/(2πm) Ek max = q²B²R²/(2m)
where:
  • ffrequency of the alternating voltage (equal to the particle's revolution frequency)
  • Rradius of the D's
  • Ek maxmaximum kinetic energy of the particles leaving the cyclotron

From R = mvmax/(qB). The final energy depends on B and R, not on the accelerating voltage: a larger voltage only means fewer turns.

Electromagnetic induction: Faraday's law and Lenz's law

Open lesson
E = n · ΔΦ/ΔtE = n · ΔΦ/Δt
where:
  • Einduced EMF (average over Δt), V
  • nnumber of turns
  • ΔΦ = Φ₂ − Φ₁change of flux through one turn, Wb
  • Δttime of the change, s

Faraday's law (法拉第电磁感应定律): the EMF is proportional to the rate of change of the flux, not to the flux itself (1 Wb/s = 1 V). On a Φ–t graph the EMF is the slope.

E = BLv E = BLv sin α
where:
  • Bmagnetic induction, T
  • Llength of the rod in the field (perpendicular to v), m
  • vspeed of the rod, m/s
  • αangle between v and B: only the part of v perpendicular to B counts

Direction by the right-hand rule (右手定则): field lines enter the palm, the thumb points along v, the four fingers show the current in the rod. The rod is the source, so the fingers point to its higher-potential end (inside a source the current flows from − to +).

I = BLv/(R + r) FA = BIL = B²L²v/(R + r) vmax = F(R + r)/(B²L²)I = BLv/(R + r) FA = BIL = B²L²v/(R + r) vmax = F(R + r)/(B²L²)
where:
  • R, rexternal resistance and resistance of the rod, Ω
  • FAAmpère force (安培力) on the rod, directed against v, N
  • Fconstant pulling force (on vertical rails F = mg, on inclined rails F = mg sin θ), N
  • vmaxterminal (maximum) speed, reached when F = FA, m/s

The braking force grows with speed, so the rod’s acceleration falls until the forces balance; after that the rod moves uniformly at vmax.

e = Eₘ sin ωt Eₘ = nBSω E = Eₘ/√2e = Eₘ sin ωt Eₘ = nBSω E = Eₘ/√2
where:
  • einstantaneous EMF, V
  • Eₘpeak (maximum) EMF, V
  • ω = 2πfangular velocity of the coil, rad/s (f — frequency, Hz)
  • Eeffective (rms) value (有效值), V

Time is counted from the neutral plane. Chinese household mains are 220 V, 50 Hz: 220 V is the effective value, and the peak is 220√2 ≈ 311 V.

U₁/U₂ = n₁/n₂ I₁/I₂ = n₂/n₁ P₁ = P₂U₁/U₂ = n₁/n₂ I₁/I₂ = n₂/n₁ P₁ = P₂
where:
  • U₁, n₁voltage and number of turns of the primary coil (原线圈)
  • U₂, n₂the same for the secondary coil (副线圈)
  • I₁, I₂currents in the primary and secondary coils

Ideal transformer (no losses): power in = power out, so stepping the voltage up steps the current down. The load decides: the primary current is set by what is connected to the secondary.

7Physics: heat, optics and modern physicsIntermediate

Kinetic theory of gases and the ideal gas equation

Open lesson
n = N/NA = m/M m₀ = M/NAn = N/NA = m/M m₀ = M/NA
where:
  • namount of substance, mol
  • Nnumber of molecules
  • m, Mmass of the sample and molar mass (kg/mol)
  • m₀mass of one molecule, kg

Without a calculator use NA ≈ 6 × 10²³ mol⁻¹. Remember to convert g/mol to kg/mol: water has M = 18 g/mol = 1.8 × 10⁻² kg/mol.

Ēₖ = (3/2)kT T = t + 273 KĒₖ = (3/2)kT T = t + 273 K
where:
  • Ēₖaverage kinetic energy of the translational motion of one molecule, J
  • kBoltzmann constant, 1.38 × 10⁻²³ J/K (k = R/NA)
  • T, tabsolute temperature (K) and Celsius temperature (°C)

Ēₖ ∝ T. From ½m₀v² = (3/2)kT the typical (root-mean-square) speed is v = √(3kT/m₀) = √(3RT/M), so v ∝ √(T/M).

p = (2/3)n₀Ēₖ = n₀kTp = (2/3)n₀Ēₖ = n₀kT
where:
  • ppressure of the gas, Pa
  • n₀ = N/Vn₀ = N/Vnumber density (molecules per unit volume), m⁻³

Multiply by V: pV = NkT = nRT, because Nk = nNA k = nR. The molecular picture gives the ideal gas equation of the next section.

pV = nRT p₁V₁/T₁ = p₂V₂/T₂pV = nRT p₁V₁/T₁ = p₂V₂/T₂
where:
  • ppressure, Pa
  • Vvolume, m³ (1 L = 10⁻³ m³)
  • namount of gas, mol
  • Rgas constant, 8.31 J/(mol · K)
  • Tabsolute temperature, K

pV = nRT is the ideal gas equation of state (理想气体状态方程). For a fixed amount of gas pV/T = const (the second form). At 0 °C and 1.013 × 10⁵ Pa one mole of any gas occupies 22.4 L; the lesson “The ideal gas law in chemistry” uses the same law for molar masses and mixtures.

p = p₀ ± h (cmHg) p = p₀ ± ρgh p = p₀ ± mg/Sp = p₀ ± h (cmHg) p = p₀ ± ρgh p = p₀ ± mg/S
where:
  • p₀atmospheric pressure
  • hvertical height of the mercury column (in a tilted tube not its length: h = l sin α)
  • ρghpressure of a liquid column in Pa (mercury: ρ = 13.6 × 10³ kg/m³)
  • m, Smass and area of the piston

“+” when the column or piston presses on the gas from above; “−” when the gas is above and the column hangs below it (open end down).

The first law of thermodynamics

Open lesson
U = (3/2)nRT = (3/2)pVU = (3/2)nRT = (3/2)pV
where:
  • Uinternal energy of a monatomic ideal gas (He, Ne, Ar), in J
  • namount of gas, in mol
  • Rthe gas constant, 8.31 J/(mol · K)
  • Tabsolute temperature, in K
  • pVpressure × volume (pV = nRT), in Pa · m³ = J

For a monatomic ideal gas. CSCA items state this formula when they need it; the idea to remember is ΔU ∝ ΔT. Handy unit: 10⁵ Pa · 1 L = 10⁵ · 10⁻³ J = 100 J.

ΔU = Q + W
where:
  • ΔUchange in internal energy: + if U grows (for an ideal gas, T rises)
  • Qheat: + if the gas absorbs it (吸热), − if it gives it off (放热)
  • Wwork done ON the gas by the surroundings: + when the gas is compressed, − when it expands, 0 at constant volume

The form used in Chinese textbooks and in CSCA items (热力学第一定律). Every quantity is signed from the gas's point of view: whatever comes in is positive.

Q = ΔU + A
where:
  • Awork done BY the gas, A = −W: + when it expands, − when it is compressed
  • Qheat received by the gas (the same Q as above)

The same law in the school form used in Azerbaijan and Russia: the heat received goes partly into internal energy and partly into the work the gas does. Many English textbooks write ΔU = Q − W, where their W is the work done by the gas. Different letters, the same physics.

A = pΔV (p = const); |A| = area under the p–V graph
where:
  • Awork done by the gas, in J (W = −A)
  • ppressure, in Pa
  • ΔVchange in volume V₂ − V₁, in m³ (1 L = 10⁻³ m³)

Moving right on the graph (expansion): the gas does work, W < 0. Moving left (compression): W > 0. A vertical segment (V constant) has zero area: no work.

η = A/Q₁ = (Q₁ − Q₂)/Q₁ = 1 − Q₂/Q₁η = A/Q₁ = (Q₁ − Q₂)/Q₁ = 1 − Q₂/Q₁
where:
  • ηefficiency (效率), as a fraction or a percentage
  • Auseful work per cycle, A = Q₁ − Q₂
  • Q₁heat received from the hot source per cycle
  • Q₂heat given to the cold sink per cycle (a positive number)

Q₂ can never be zero, so η < 1 (100%) for every heat engine.

Geometrical optics: reflection and refraction

Open lesson
i′ = i
where:
  • iangle of incidence (from the normal)
  • i′angle of reflection

Consequences: the angle between the incident and reflected rays is 2i; if the mirror turns by θ while the incident ray stays fixed, the reflected ray turns by 2θ. Rough surfaces scatter light in all directions (diffuse reflection, 漫反射), but every tiny piece still obeys i′ = i.

n = sin i / sin r = c / vn = sin i / sin r = c / v
where:
  • nrefractive index of the medium (折射率), always > 1; air ≈ 1, water ≈ 4/3, glass ≈ 1.5
  • iangle of incidence in air (vacuum), from the normal
  • rangle of refraction in the medium
  • cspeed of light in vacuum, 3 × 10⁸ m/s
  • vspeed of light in the medium

Chinese textbooks define n for a ray coming from air (vacuum): n = sin i / sin r. Inside the medium light travels at c/n.

n₁ sin θ₁ = n₂ sin θ₂ λ = λ₀ / n h′ = h / nn₁ sin θ₁ = n₂ sin θ₂ λ = λ₀ / n h′ = h / n
where:
  • n₁, n₂refractive indices of the first and second media
  • θ₁, θ₂angles from the normal in each medium
  • λ₀, λwavelength in vacuum and in the medium; the frequency does not change
  • h, h′real and apparent depth when you look almost straight down from air

The speed ratio follows from n = c/v: v₁/v₂ = n₂/n₁. The colour of light is set by its frequency, which stays the same in every medium.

sin C = 1/n (between two media: sin C = n₂/n₁)sin C = 1/n (between two media: sin C = n₂/n₁)
where:
  • Ccritical angle (临界角), measured from the normal inside the denser medium
  • nrefractive index of the medium, with air on the other side
  • n₁ > n₂the denser and the less dense medium

It follows from n sin C = 1 · sin 90°. The larger n, the smaller C, and the easier total internal reflection becomes.

1/u + 1/v = 1/f m = v/u1/u + 1/v = 1/f m = v/u
where:
  • uobject distance (物距)
  • vimage distance (像距): + for a real image behind the lens, − for a virtual image on the object's side
  • ffocal length (焦距): + for a converging lens, − for a diverging lens
  • mmagnification: image height ÷ object height (taken without sign)

The Chinese form, with real distances positive. Azerbaijani and Russian textbooks write the same law as 1/d + 1/f = 1/F (d to the object, f to the image, F the focal length), so watch the letters.

Physical optics: interference and diffraction

Open lesson
bright: Δr = kλ dark: Δr = (2k + 1) · λ/2 (k = 0, 1, 2, …)bright: Δr = kλ dark: Δr = (2k + 1) · λ/2 (k = 0, 1, 2, …)
where:
  • Δrpath difference: the difference between the distances from the point to the two sources
  • λwavelength of the light
  • korder of the fringe (0 = the central one)

For two sources vibrating in phase: a whole number of wavelengths gives constructive interference, an odd number of half-wavelengths gives destructive interference.

Δx = Lλ/dΔx = Lλ/d
where:
  • Δxfringe spacing: the distance between neighbouring bright (or dark) fringes
  • Ldistance from the slits to the screen
  • ddistance between the two slits
  • λwavelength

The fringes are equally spaced and almost equally bright. Chinese textbooks write Δx = (l/d)λ. Red light (λ ≈ 700 nm) gives wider fringes than violet (λ ≈ 400 nm); in white light the central fringe is white and the others are coloured, violet on the inside and red on the outside.

dₘᵢₙ = λ/(4n)dₘᵢₙ = λ/(4n)
where:
  • dₘᵢₙsmallest thickness of the coating
  • λwavelength in vacuum (air)
  • nrefractive index of the coating (smaller than that of the glass)

The two reflected waves must differ by half a wavelength: 2d = λ/(2n). In Chinese-textbook form: the coating is a quarter of the wavelength inside the film, d = λfilm/4 with λfilm = λ/n. No energy is destroyed: what is not reflected is transmitted.

d · sin θ = kλ
where:
  • dgrating period: the distance between neighbouring slits, d = 1/N for N slits per unit length
  • θangle between the k-th order line and the straight-through direction
  • korder (0, 1, 2, …)

Since sin θ ≤ 1, only orders with k < d/λ can be seen. A longer wavelength is deviated more: in a grating spectrum red lies farthest out — the opposite of a prism, which bends violet most.

I = I₀ · cos²θ
where:
  • I₀intensity of the polarised light falling on the analyser
  • θangle between the axes of the polariser and the analyser
  • Itransmitted intensity

Malus's law. Unpolarised light loses half of its intensity in the first polariser, whatever the direction of its axis.

c = λν
where:
  • cspeed of light in vacuum, 3.0 × 10⁸ m/s
  • λwavelength
  • νfrequency

A shorter wavelength means a higher frequency — and, as the lesson «The photoelectric effect» shows, more energetic photons.

The photoelectric effect

Open lesson
E = hν = hc/λE = hν = hc/λ
where:
  • hPlanck's constant, 6.63 × 10⁻³⁴ J·s (often rounded to 6.6 × 10⁻³⁴)
  • νfrequency of the light
  • λ, cwavelength; c = 3.0 × 10⁸ m/s

Atomic energies are tiny, so they are given in electronvolts: 1 eV = 1.6 × 10⁻¹⁹ J. A photon of visible light carries about 1.6–3.1 eV.

Eₖ = hν − W₀ W₀ = hνc = hc/λcEₖ = hν − W₀ W₀ = hνc = hc/λc
where:
  • Eₖmaximum kinetic energy of the photoelectrons
  • hνenergy of one photon
  • W₀work function of the metal
  • νc, λcthreshold frequency and threshold wavelength

Emission happens only if ν ≥ νc (λ ≤ λc). Eₖ grows linearly with ν but is not proportional to it.

eUc = Eₖ = hν − W₀ ⇒ Uc = (h/e) · ν − W₀/eeUc = Eₖ = hν − W₀ ⇒ Uc = (h/e) · ν − W₀/e
where:
  • Ucstopping (cut-off) voltage
  • eelementary charge, 1.6 × 10⁻¹⁹ C
  • h/eh/eslope of the Uc–ν graph

The work of the field eUc takes away all the kinetic energy of the fastest electron. In numbers: Uc in volts = Eₖ in electronvolts.

λ = h/p = h/(mv) = h/√(2mEₖ)λ = h/p = h/(mv) = h/√(2mEₖ)
where:
  • λde Broglie wavelength
  • pmomentum of the particle
  • m, vmass and speed
  • Eₖkinetic energy (for an electron accelerated from rest through a voltage U: Eₖ = eU)

A photon has the momentum p = h/λ as well. A larger momentum means a shorter wavelength; at the same kinetic energy a heavier particle has the shorter wavelength.

Atomic structure and spectra

Open lesson
r(atom) : r(nucleus) ≈ 10⁻¹⁰ m : 10⁻¹⁵ m = 10⁵ : 1
where:
  • r(atom)radius of the atom (the size of the electron cloud), m
  • r(nucleus)radius of the nucleus, m

Lengths differ by 10⁵, so areas differ by 10¹⁰ and volumes by 10¹⁵. That is why most α particles pass without ever coming close to a nucleus.

1/λ = R(1/2² − 1/n²), n = 3, 4, 5, …1/λ = R(1/2² − 1/n²), n = 3, 4, 5, …
where:
  • λwavelength of the line, m
  • Rthe Rydberg constant, R ≈ 1.10 × 10⁷ m⁻¹
  • na whole number from 3 up; each n gives one line

Balmer's formula (巴耳末公式 in Chinese textbooks). As n grows, the lines crowd together and approach the series limit at n → ∞.

En = E₁/n², rn = n²r₁ (E₁ = −13.6 eV, r₁ = 0.53 × 10⁻¹⁰ m)En = E₁/n², rn = n²r₁ (E₁ = −13.6 eV, r₁ = 0.53 × 10⁻¹⁰ m)
where:
  • nthe principal quantum number: 1, 2, 3, …
  • Enenergy of the atom in level n, eV; zero means the electron at rest infinitely far from the nucleus, so all bound levels are negative
  • E₁ground-state energy, −13.6 eV (1 eV = 1.6 × 10⁻¹⁹ J)
  • rn, r₁radius of orbit n and the first (Bohr) radius

Energy goes as 1/n², radius as n². As n grows, the levels approach zero and crowd together.

Ek = ke²/(2r_n) = −En, Ep = −ke²/rn = 2E_nEk = ke²/(2r_n) = −En, Ep = −ke²/rn = 2E_n
where:
  • Ekkinetic energy of the electron
  • Eppotential energy of the electron–nucleus system (zero at infinity)
  • kCoulomb's constant, 9 × 10⁹ N · m²/C²

As n increases: r grows, v falls, Ek falls, Ep rises and the total energy En rises (towards zero).

hν = Em − En, λ = hc/(Em − En) (m > n)hν = Em − En, λ = hc/(Em − En) (m > n)
where:
  • hPlanck's constant, 6.63 × 10⁻³⁴ J · s
  • ν, λfrequency and wavelength of the emitted or absorbed photon
  • Em, Enenergies of the upper and the lower level
  • cspeed of light, 3 × 10⁸ m/s

A jump down emits a photon; a jump up absorbs one. Because the photon energy is fixed exactly, the spectrum consists of lines.

N = C(n, 2) = n(n − 1)/2N = C(n, 2) = n(n − 1)/2
where:
  • nthe highest level the atoms are excited to
  • Nnumber of different lines (frequencies) that many atoms emit while returning to the ground state

Every pair of levels gives one line. A single atom, however, can emit at most n − 1 photons (going down one step at a time).

Eion = 0 − En = 13.6/n² eV, Ek = hν − EionEion = 0 − En = 13.6/n² eV, Ek = hν − Eion
where:
  • Eionionisation energy from level n
  • Ekkinetic energy of the electron after ionisation

An excited atom is easier to ionise: its electron already sits on a level closer to zero.

Fundamentals of nuclear physics

Open lesson
A = Z + N, q(nucleus) = Ze
where:
  • Amass number (number of nucleons; roughly the mass in u)
  • Zcharge number = number of protons
  • Nnumber of neutrons
  • eelementary charge, 1.6 × 10⁻¹⁹ C

A neutral atom has Z electrons. Only Z decides which chemical element a nucleus belongs to.

α: (A, Z) → (A − 4, Z − 2) + ⁴₂He β⁻: (A, Z) → (A, Z + 1) + ⁰₋₁e
where:
  • (A, Z)parent nucleus with mass number A and charge number Z
  • ⁴₂Hethe α particle
  • ⁰₋₁ethe β particle (an electron)

In every decay and every nuclear reaction the sum of the mass numbers and the sum of the charges are conserved. γ emission changes neither A nor Z: an excited daughter nucleus simply gives off its extra energy as a photon.

x = (A − A′)/4, y = 2x − (Z − Z′)x = (A − A′)/4, y = 2x − (Z − Z′)
where:
  • x, ythe numbers of α and β⁻ decays
  • A, Z → A′, Z′mass and charge numbers of the first and the last nucleus

Only α changes A, so count the α decays first. x α decays lower Z by 2x; the β⁻ decays make up the rest.

N = N₀ · (1/2)^(t/T), m = m₀ · (1/2)^(t/T)N = N₀ · (1/2)^(t/T), m = m₀ · (1/2)^(t/T)
where:
  • N₀, m₀initial number and mass of the radioactive nuclei
  • N, mthe nuclei not yet decayed after time t
  • t/Tt/Tthe number of half-lives that have passed

The decayed part is N₀ − N: after 3 half-lives 1/8 remains and 7/8 has decayed. The activity (decays per second) falls by the same law.

ΔE = Δm · c², 1 u ↔ 931.5 MeV
where:
  • Δmmass defect (kg or u)
  • c²9 × 10¹⁶ m²/s²
  • uatomic mass unit, 1 u = 1.66 × 10⁻²⁷ kg; 1 MeV = 1.6 × 10⁻¹³ J

If Δm is in kilograms, multiply by c² to get joules; if it is in u, multiply by 931.5 to get MeV.

8Chemistry: basic concepts and calculationsIntermediate

Classification of matter and changes of state

Open lesson
3O₂ → 2O₃ (electric discharge)
where:
  • O₂dioxygen, the colourless gas of the air
  • O₃ozone: a pale blue gas with a sharp smell and a stronger oxidant than O₂; the ozone layer absorbs ultraviolet light
  • 3 : 2at the same temperature and pressure 3 volumes of gas become 2, but the mass does not change

Converting one allotrope into another is a chemical change (a new substance forms), but not a redox reaction: the oxidation number of oxygen stays 0. It happens in lightning and in ozone generators.

solution: d < 1 nm colloid: 1 nm ≤ d ≤ 100 nm suspension, emulsion: d > 100 nm
where:
  • ddiameter of the dispersed particles
  • nmnanometre: 1 nm = 10⁻⁹ m (a millionth of a millimetre)

Particle size is the essential difference between the three kinds of dispersion (分散质粒子直径). The Tyndall effect is only the method used to tell a colloid from a solution.

FeCl₃ + 3H₂O →(△) Fe(OH)₃ (colloid) + 3HCl
where:
  • FeCl₃saturated iron(III) chloride solution, added drop by drop
  • H₂Oboiling distilled water
  • △heating (the sign Chinese textbooks write over the arrow)
  • (colloid)written with the word «colloid» and no ↓: the particles stay dispersed and do not settle

Preparing iron(III) hydroxide colloid (Fe(OH)₃胶体), the standard colloid of Chinese textbooks and CSCA items.

Q = n · ΔH
where:
  • Qheat absorbed or released, kJ
  • namount of substance, mol (n = m/M)
  • ΔHmolar heat of the change of state; for water: melting +6.0 kJ/mol at 0 °C, vaporisation +40.7 kJ/mol at 100 °C; the reverse changes have the same size with a minus sign

While a pure substance melts or boils, all the heat goes into pulling the particles apart, so the temperature stays constant — a flat step on the heating curve.

Chemical notation and writing equations

Open lesson
AₓBᵧ: x · a = y · b
where:
  • a, bthe sizes of the valences of A and B (without signs)
  • x, ythe subscripts: the smallest whole numbers that make the charges cancel

The criss-cross rule: x = b and y = a, then cancel any common factor. A polyatomic ion taken more than once keeps its brackets: Ca₃(PO₄)₂, not Ca₃P₂O₈.

for every element: number of atoms on the left = number of atoms on the right
where:
  • coefficientthe number in front of a formula; the smallest whole numbers are used (CSCA: «balanced with the smallest whole numbers»)
  • subscriptfixed by the formula of the substance: never changed while balancing

The law of conservation of mass (质量守恒定律) written for atoms. Thermochemical and ionic equations obey it too, and an ionic equation must also balance the charge.

total increase in valence = total decrease in valence
where:
  • ↑for the element whose valence rises: (change per atom) × (number of such atoms in its formula)
  • ↓for the element whose valence falls, counted the same way

First balance the changing elements with these two numbers (cross-multiply to their LCM), then finish the rest — usually H and O — by inspection.

Σ charges on the left = Σ charges on the right
where:
  • Σsum over all particles: coefficient × charge of each ion (neutral formulas count 0)

An ionic equation must balance both atoms and charge. 2Fe³⁺ + Fe → 3Fe²⁺ is right (+6 = +6); Fe³⁺ + Fe → 2Fe²⁺ has balanced atoms, but +3 ≠ +4.

Q = n · |ΔH|
where:
  • Qheat released or absorbed, kJ
  • nmoles of reaction: the moles of a substance ÷ its coefficient in the equation
  • ΔHenthalpy change of the equation as written, kJ/mol

Heat is proportional to the amount that reacts. Two definitions CSCA likes: the heat of combustion (燃烧热) refers to 1 mol of fuel burning completely to stable products (CO₂(g), H₂O(l)); the heat of neutralisation (中和热) of a strong acid and a strong base is 57.3 kJ per mole of H₂O formed.

ΔH = ΔH₁ + ΔH₂ + … ΔH = ΣE(bonds broken) − ΣE(bonds formed)
where:
  • ΔH₁, ΔH₂ΔH of the given equations after multiplying or reversing them as needed
  • Ebond energy, kJ/mol: the energy needed to break 1 mol of that bond in the gas phase

Reactant bonds are broken and product bonds are formed — «broken minus formed». If the products’ bonds are stronger, ΔH < 0.

Amount of substance: the mole

Open lesson
N = n · Nₐ n = N/NₐN = n · Nₐ n = N/Nₐ
where:
  • namount of substance, mol
  • Nnumber of particles (atoms, molecules, ions, electrons)
  • NₐAvogadro’s constant, 6.02 × 10²³ mol⁻¹ (阿伏加德罗常数)

Particles inside particles: 1 mol of H₂SO₄ contains 2 mol of H atoms, 4 mol of O atoms and 7 mol of atoms in total — multiply by the number of atoms (ions, electrons) in one formula unit. Answers are often written as multiples of Nₐ, such as 0.5Nₐ.

n = m/M m = n · Mn = m/M m = n · M
where:
  • mmass of the substance, g
  • Mmolar mass, g/mol

Relative atomic masses used in this course (standard rounded values): H 1, C 12, N 14, O 16, Na 23, Mg 24, Al 27, S 32, Cl 35.5, K 39, Ca 40, Fe 56, Cu 64, Zn 65. CSCA items print the values they need next to the question.

n = V/Vₘ Vₘ = 22.4 L/mol (STP)n = V/Vₘ Vₘ = 22.4 L/mol (STP)
where:
  • Vvolume of the gas, L
  • Vₘmolar volume, L/mol

Only for gases, and only at STP. At 25 °C and 101 kPa Vₘ ≈ 24.5 L/mol; for any other conditions use pV = nRT from the lesson “The ideal gas law in chemistry”.

aA + bB → cC: n(A)/a = n(B)/b = n(C)/caA + bB → cC: n(A)/a = n(B)/b = n(C)/c
where:
  • a, b, ccoefficients in the balanced equation
  • n(X)amount of X that reacts or forms, mol

The amounts that react or form are in the ratio of the coefficients. Use it with moles only — never put grams straight into this ratio.

purity = m(pure substance)/m(sample) · 100% yield = m(actual)/m(theoretical) · 100%purity = m(pure substance)/m(sample) · 100% yield = m(actual)/m(theoretical) · 100%
where:
  • m(theoretical)the mass calculated from the equation (from the limiting reagent)
  • m(actual)the mass really obtained

Only the pure part of a sample reacts, so multiply by the purity before using the equation. The yield is applied after it: actual mass = theoretical mass · yield.

w(E) = x · Ar(E)/Mr m(E) = w(E) · m(sample)w(E) = x · Ar(E)/Mr m(E) = w(E) · m(sample)
where:
  • w(E)mass fraction of element E (× 100 for per cent)
  • xnumber of E atoms in the formula
  • Mrrelative formula mass

The fraction is the same for any sample of a pure compound, from one gram to one tonne. Chinese textbooks write the mass fraction as w; many Russian books use ω.

x : y = w(A)/Ar(A) : w(B)/Ar(B) k = M/M(empirical)x : y = w(A)/Ar(A) : w(B)/Ar(B) k = M/M(empirical)
where:
  • x : yratio of the numbers of A and B atoms
  • ka whole number: molecular formula = (empirical formula)ₖ

Masses work as well as fractions: x : y = m(A)/Ar(A) : m(B)/Ar(B).

Solution concentration and pH calculations

Open lesson
w = m(solute)/m(solution) · 100% m(solution) = m(solute) + m(solvent) = ρVw = m(solute)/m(solution) · 100% m(solution) = m(solute) + m(solvent) = ρV
where:
  • wmass fraction of the solute (质量分数)
  • ρ, Vdensity (g/mL = g/cm³) and volume (mL) of the solution

For a saturated solution with solubility S (grams of solute per 100 g of water): w = S/(100 + S). NaCl at 20 °C has S ≈ 36 g, so w ≈ 36/136 ≈ 26.5% — the most concentrated NaCl solution possible at that temperature.

c = n/V n = c · Vc = n/V n = c · V
where:
  • cmolar concentration, mol/L
  • namount of solute, mol
  • Vvolume of the solution, L (mL ÷ 1000)

Ions follow the formula: in 0.1 mol/L Fe₂(SO₄)₃, c(Fe³⁺) = 0.2 mol/L and c(SO₄²⁻) = 0.3 mol/L. Pouring out part of a solution changes n and V, but not c.

c = 1000ρw/M w = cM/(1000ρ)c = 1000ρw/M w = cM/(1000ρ)
where:
  • ρdensity of the solution, g/cm³
  • wmass fraction as a decimal (36.5% → 0.365)
  • Mmolar mass of the solute, g/mol
  • 10001 L = 1000 mL (cm³)

Where it comes from: take 1 L of solution. Its mass is 1000ρ g, the solute in it is 1000ρw g, which is 1000ρw/M mol — in one litre.

c₁V₁ = c₂V₂ m₁w₁ = m₂w₂
where:
  • c₁, V₁concentration and volume before dilution
  • c₂, V₂concentration and volume after dilution
  • m, wmass of the solution and its mass fraction

On dilution the amount (and the mass) of solute stays the same. V₁ and V₂ may both be in mL — the unit cancels.

c = (c₁V₁ + c₂V₂)/(V₁ + V₂)c = (c₁V₁ + c₂V₂)/(V₁ + V₂)
where:
  • cconcentration of the mixture

Mixing two solutions of the same solute: add the moles, then divide by the total volume. Volumes are added only when the problem says so (“assume the volumes are additive”); strictly, only masses add.

pH = −lg c(H⁺) pH = a ⇔ c(H⁺) = 10⁻ᵃ mol/LpH = −lg c(H⁺) pH = a ⇔ c(H⁺) = 10⁻ᵃ mol/L
where:
  • c(H⁺)molar concentration of hydrogen ions, mol/L

Strong acids (HCl, HNO₃, H₂SO₄) and strong bases (NaOH, KOH, Ba(OH)₂) ionise completely, so c(H⁺) or c(OH⁻) comes straight from the formula: 0.05 mol/L H₂SO₄ gives 0.1 mol/L of H⁺.

Kw = c(H⁺) · c(OH⁻) = 1.0 × 10⁻¹⁴ (25 °C) pH + pOH = 14
where:
  • Kwionic product of water (水的离子积)
  • pOH−lg c(OH⁻)

c(H⁺) · c(OH⁻) = Kw holds in every aqueous solution, acidic or basic. Kw grows with temperature: near 100 °C it is of the order of 10⁻¹² (problems give Kw = 1 × 10⁻¹²), so neutral water has pH ≈ 6 there — still neutral, because c(H⁺) = c(OH⁻). Where Kw comes from, the ionisation of water, is part of “Theories of electrolyte solutions”.

strong acid diluted 10ⁿ times: pH → pH + n strong base: pH → pH − n (never across 7)
where:
  • nthe power of ten of the dilution (100 times → n = 2)

Dilution divides c(H⁺) of an acid (or c(OH⁻) of a base) by 10ⁿ. Very dilute solutions approach pH 7 from their own side: water itself supplies 10⁻⁷ mol/L of H⁺ and of OH⁻.

acid + base: c(H⁺ or OH⁻ left over) = |n(H⁺) − n(OH⁻)|/(V₁ + V₂)acid + base: c(H⁺ or OH⁻ left over) = |n(H⁺) − n(OH⁻)|/(V₁ + V₂)
where:
  • V₁ + V₂total volume of the mixture, L

First neutralise H⁺ with OH⁻ in moles, then divide what is left by the total volume and take the pH. Never average pH values.

The ideal gas law in chemistry

Open lesson
pV = nRT Vₘ = V/n = RT/ppV = nRT Vₘ = V/n = RT/p
where:
  • ppressure of the gas, kPa (or Pa)
  • Vvolume, L (with kPa) or m³ (with Pa)
  • namount of gas, mol
  • Rgas constant, 8.31 J/(mol · K) = 8.31 kPa · L/(mol · K)
  • Tabsolute temperature, K: T = t + 273
  • Vₘmolar volume (气体摩尔体积), L/mol — the volume of one mole of gas; it depends only on T and p, not on the kind of gas

pV = nRT is the ideal gas equation of state (理想气体状态方程). With p in kPa and V in litres the same number R = 8.31 works, because 1 kPa · L = 1 J. At STP (0 °C, 101 kPa) Vₘ = 22.4 L/mol; at 25 °C and 101 kPa Vₘ ≈ 24.5 L/mol.

Same T and p: V₁/V₂ = n₁/n₂ = N₁/N₂ ρ₁/ρ₂ = M₁/M₂Same T and V: p₁/p₂ = n₁/n₂Same T and p: V₁/V₂ = n₁/n₂ = N₁/N₂ ρ₁/ρ₂ = M₁/M₂Same T and V: p₁/p₂ = n₁/n₂
where:
  • Nnumber of molecules
  • ρdensity of the gas, g/L
  • Mmolar mass, g/mol

All three follow from pV = nRT by keeping two quantities fixed. A fourth form is often useful: at the same T and p, equal masses occupy volumes in the inverse ratio of their molar masses, V₁/V₂ = M₂/M₁.

p₀V₀/T₀ = pV/T ⇒ V₀ = V · (p/p₀) · (T₀/T)p₀V₀/T₀ = pV/T ⇒ V₀ = V · (p/p₀) · (T₀/T)
where:
  • V₀volume of the gas reduced to STP
  • p₀, T₀STP: 101 kPa and 273 K
  • p, V, Tpressure, volume and absolute temperature at which the gas was measured

This is the law pV/T = const for a fixed amount of gas; its three special cases are explained in the physics lesson. In chemistry it is used mainly to reduce a volume to STP: n = V₀ / 22.4.

M = mRT/(pV) = ρRT/p M = ρ · Vₘ (STP: M = 22.4ρ)D = ρ₁/ρ₂ = M₁/M₂: M = 2D(H₂) = 29D(air)M = mRT/(pV) = ρRT/p M = ρ · Vₘ (STP: M = 22.4ρ)D = ρ₁/ρ₂ = M₁/M₂: M = 2D(H₂) = 29D(air)
where:
  • ρdensity of the gas, g/L
  • Drelative density (相对密度) of gas 1 with respect to gas 2 — a pure number, taken at the same T and p
  • 29average molar mass of air, g/mol (≈ 78 % N₂, 21 % O₂, 1 % Ar)

Gases with M > 29 g/mol are heavier than air (CO₂, Cl₂, SO₂); those with M < 29 g/mol are lighter (H₂, CH₄, NH₃). A molar mass alone does not always name a gas: N₂, CO and C₂H₄ all have M = 28 g/mol, and CO₂, N₂O and C₃H₈ all have 44 g/mol.

M̄ = m(total) / n(total) = M₁x₁ + M₂x₂ + … xᵢ = nᵢ/n = Vᵢ/V = φᵢM̄ = m(total) / n(total) = M₁x₁ + M₂x₂ + … xᵢ = nᵢ/n = Vᵢ/V = φᵢ
where:
  • M̄average molar mass of the mixture (平均摩尔质量), g/mol
  • xᵢmole fraction of component i
  • φᵢvolume fraction (体积分数) — for gases equal to the mole fraction, not to the mass fraction

Each component also behaves as if it were alone: its partial pressure (分压) is pᵢ = xᵢ · p. The average molar mass always lies between the smallest and the largest Mᵢ.

aA(g) + bB(g) → cC(g): V(A) : V(B) : V(C) = a : b : cΔV = V(before) − V(after)Rigid vessel, constant T: p₂/p₁ = n₂/n₁aA(g) + bB(g) → cC(g): V(A) : V(B) : V(C) = a : b : cΔV = V(before) − V(after)Rigid vessel, constant T: p₂/p₁ = n₂/n₁
where:
  • a, b, ccoefficients of the gases in the equation
  • ΔVdecrease of the gas volume at the same T and p
  • p₁, p₂; n₁, n₂pressure and total amount of gas before and after the reaction — count gases only

The volume ratio is the mole ratio, so the whole “mole bridge” of stoichiometry works directly with litres or millilitres, as long as all volumes are measured at the same T and p.

9Chemistry: properties and reactions of substancesIntermediate

Oxides, acids, bases and salts

Open lesson
CaO + H₂O → Ca(OH)₂ CuO + H₂SO₄ → CuSO₄ + H₂O CaO + CO₂ → CaCO₃
where:
  • CaO + H₂Oonly the oxides of K, Ca, Na and Ba give an alkali with water (this is how the quicklime in a drying packet works)
  • CuO + H₂SO₄every basic oxide + acid → salt + water; the metal keeps its oxidation state
  • CaO + CO₂basic oxide + acidic oxide → salt (no water forms)

The three typical reactions of basic oxides. Chinese textbooks write balanced equations with «=» instead of «→»; the meaning is the same.

SO₃ + H₂O → H₂SO₄ CO₂ + 2NaOH → Na₂CO₃ + H₂O SiO₂ + CaO → CaSiO₃ (t°)
where:
  • SO₃ + H₂Oacidic oxide + water → acid (P₂O₅ → H₃PO₄, SO₂ → H₂SO₃); SiO₂ is the exception
  • CO₂ + 2NaOHacidic oxide + base → salt + water; this is why NaOH solution left open absorbs CO₂ from the air and goes bad
  • SiO₂ + CaOacidic oxide + basic oxide → salt (at high temperature)

The three typical reactions of acidic oxides; t° means heating.

Zn + H₂SO₄ → ZnSO₄ + H₂↑ MgO + 2HCl → MgCl₂ + H₂O CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂↑
where:
  • Zn + H₂SO₄acid + a metal before hydrogen in the activity series → salt + H₂ (dilute HCl, dilute H₂SO₄); iron gives an Fe²⁺ salt; Cu and Ag do not react
  • MgO + 2HClacid + basic oxide → salt + water
  • CaCO₃ + 2HClacid + salt → new salt + new acid: a stronger acid displaces a weaker one (H₂CO₃ → H₂O + CO₂), or a precipitate forms (H₂SO₄ + BaCl₂ → BaSO₄↓ + 2HCl)

Acid + base is neutralisation, covered in the next section. HNO₃ and concentrated H₂SO₄ react with metals without releasing H₂ — see «Common non-metals and their compounds».

2NaOH + SO₂ → Na₂SO₃ + H₂O CuSO₄ + 2NaOH → Cu(OH)₂↓ + Na₂SO₄ Cu(OH)₂ → CuO + H₂O (t°)
where:
  • 2NaOH + SO₂alkali + acidic oxide → salt + water (this is how SO₂ and CO₂ are absorbed)
  • CuSO₄ + 2NaOHalkali + soluble salt → new base + new salt, only if a precipitate or a gas forms (NH₄Cl + NaOH → NaCl + NH₃↑ + H₂O on warming)
  • Cu(OH)₂ → CuO + H₂Oinsoluble bases decompose on heating into the oxide and water (blue → black); alkalis such as NaOH do not

Alkalis turn litmus blue and phenolphthalein pink. Insoluble bases do not change indicators, but they still react with acids.

n(H⁺) = n(OH⁻) ⇔ a · c(acid) · V(acid) = b · c(base) · V(base)
where:
  • anumber of H⁺ given by one formula of the acid (HCl 1, H₂SO₄ 2)
  • bnumber of OH⁻ in one formula of the base (NaOH 1, Ba(OH)₂ 2)
  • cmolar concentration, mol/L
  • Vvolume; both volumes in the same unit (mL is fine)

It works for any strong acid and base and saves writing the equation. With masses, first convert to moles: n = m/M (see «Amount of substance: the mole»).

CO₂ + 2NaOH → Na₂CO₃ + H₂O CO₂ + NaOH → NaHCO₃
where:
  • n(NaOH) : n(CO₂) ≥ 2only Na₂CO₃
  • 1 < n(NaOH) : n(CO₂) < 2a mixture of Na₂CO₃ and NaHCO₃
  • n(NaOH) : n(CO₂) ≤ 1only NaHCO₃

The same logic works for H₂SO₄ + NaOH (NaHSO₄ or Na₂SO₄) and for H₃PO₄ (NaH₂PO₄, Na₂HPO₄, Na₃PO₄). The properties of Na₂CO₃ and NaHCO₃ themselves are in «Common metals and their compounds».

Fe + CuSO₄ → FeSO₄ + Cu Na₂CO₃ + CaCl₂ → CaCO₃↓ + 2NaCl CaCO₃ → CaO + CO₂↑ (t°)
where:
  • Fe + CuSO₄salt + metal: a more active metal displaces a less active one from a solution of its salt (Cu + ZnSO₄: no reaction)
  • Na₂CO₃ + CaCl₂salt + salt (and salt + alkali): both reactants must be soluble and a precipitate must form
  • CaCO₃ → CaO + CO₂some salts decompose on heating: carbonates such as CaCO₃, ammonium salts, hydrogencarbonates

Exchange reactions between acids, bases and salts (复分解反应, double decomposition) go only if a precipitate, a gas or water forms; the ionic view of this rule is in «Ionic reactions and tests for ions».

Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O Al(OH)₃ + NaOH → Na[Al(OH)₄]
where:
  • Al(OH)₃ + 3H⁺ → Al³⁺ + 3H₂Oacts as a base
  • Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻acts as an acid; older Chinese textbooks write Al(OH)₃ + NaOH → NaAlO₂ + 2H₂O (sodium metaaluminate, 偏铝酸钠)
  • Al₂O₃the oxide does the same: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O; Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]

Al(OH)₃ dissolves only in strong bases: not in ammonia solution and not in carbonic acid. ZnO and Zn(OH)₂ react the same way (giving ZnCl₂ or Na₂[Zn(OH)₄]). Aluminium metal itself is in «Common metals and their compounds».

Ca → CaO → Ca(OH)₂ → CaCO₃ S → SO₂ → SO₃ → H₂SO₄ → BaSO₄
where:
  • Ca → CaO → Ca(OH)₂ → CaCO₃2Ca + O₂ → 2CaO; CaO + H₂O → Ca(OH)₂; Ca(OH)₂ + CO₂ → CaCO₃↓ + H₂O
  • S → SO₂ → SO₃ → H₂SO₄ → BaSO₄S + O₂ → SO₂; 2SO₂ + O₂ ⇌ 2SO₃ (catalyst, heating); SO₃ + H₂O → H₂SO₄; H₂SO₄ + BaCl₂ → BaSO₄↓ + 2HCl. Burning sulfur never gives SO₃ directly.

A one-step «oxide → base» or «oxide → acid» works only when the oxide reacts with water: CuO → Cu(OH)₂, Fe₂O₃ → Fe(OH)₃ and SiO₂ → H₂SiO₃ need two steps (make a salt first).

Common non-metals and their compounds

Open lesson
2Na + Cl₂ → 2NaCl 2Fe + 3Cl₂ → 2FeCl₃ Cu + Cl₂ → CuCl₂ H₂ + Cl₂ → 2HCl
where:
  • 2Na + Cl₂sodium burns with a yellow flame, giving white smoke (NaCl)
  • 2Fe + 3Cl₂iron burns with brown smoke and gives only FeCl₃, even with excess iron: chlorine takes metals to their higher oxidation state (HCl gives FeCl₂)
  • Cu + Cl₂copper burns with brown-yellow smoke; CuCl₂ solution is blue-green
  • H₂ + Cl₂H₂ burns quietly in Cl₂ with a pale flame, but a mixture explodes in strong light; HCl forms a white mist in moist air

Chinese textbooks ask a lot about observations (colour, smoke, mist, flame): «smoke» (烟) is solid particles, «mist» (雾) is liquid droplets.

Cl₂ + H₂O ⇌ HCl + HClO Cl₂ + 2NaOH → NaCl + NaClO + H₂O 2Cl₂ + 2Ca(OH)₂ → CaCl₂ + Ca(ClO)₂ + 2H₂O
where:
  • Cl₂ + H₂Ochlorine water; HClO bleaches and disinfects and decomposes in light
  • Cl₂ + 2NaOHabsorbing waste chlorine; NaClO is the active part of «84» disinfectant (household bleach)
  • 2Cl₂ + 2Ca(OH)₂bleaching powder (漂白粉), active part Ca(ClO)₂; in air Ca(ClO)₂ + CO₂ + H₂O → CaCO₃↓ + 2HClO — this is how it works and why it spoils when left open

Mixing bleach with an acidic cleaner is dangerous: NaClO + 2HCl → NaCl + Cl₂↑ + H₂O. The laboratory preparation of Cl₂ is in «Preparation and identification of common gases», the chlor-alkali industry in «Industrial chemical processes».

Cl₂ + 2KBr → 2KCl + Br₂ Cl₂ + 2KI → 2KCl + I₂ Br₂ + 2KI → 2KBr + I₂
where:
  • F₂ > Cl₂ > Br₂ > I₂oxidising power (non-metallic activity)
  • I⁻ > Br⁻ > Cl⁻reducing power of the halide ions

The reverse reactions (Br₂ + 2KCl, I₂ + 2KBr) do not occur. The tests for Cl⁻, Br⁻ and I⁻ with AgNO₃ are in «Ionic reactions and tests for ions».

SO₂ + 2NaOH → Na₂SO₃ + H₂O SO₂ + Br₂ + 2H₂O → H₂SO₄ + 2HBr SO₂ + 2H₂S → 3S↓ + 2H₂O
where:
  • acidic oxideSO₂ + H₂O ⇌ H₂SO₃; gives salts with alkalis; turns limewater milky just like CO₂ (CaSO₃↓), so limewater cannot tell CO₂ from SO₂; turns moist blue litmus red but does not bleach it
  • bleaching agentturns magenta (品红) into a colourless compound; heating brings the colour back — the bleaching is reversible
  • reducing agentS +4 → +6: decolourises bromine water and acidified KMnO₄; 2SO₂ + O₂ ⇌ 2SO₃ (catalyst)
  • oxidising agentS +4 → 0: gives a yellow precipitate of sulfur with H₂S

The four faces of SO₂. Decolourising KMnO₄ or bromine water is reduction, not bleaching; «bleaching» means only the magenta test.

Cu + 2H₂SO₄(conc.) → CuSO₄ + SO₂↑ + 2H₂O (t°) C + 2H₂SO₄(conc.) → CO₂↑ + 2SO₂↑ + 2H₂O (t°)
where:
  • Cu + 2H₂SO₄of the two H₂SO₄, one is the oxidising agent (becomes SO₂) and the other acts as an acid (forms the salt); as the acid is used up and diluted, the reaction stops
  • C + 2H₂SO₄with a non-metal H₂SO₄ acts only as an oxidising agent; no salt forms

Copper with concentrated H₂SO₄ needs heating; with dilute H₂SO₄ copper does not react at all.

N₂ + O₂ → 2NO 2NO + O₂ → 2NO₂ 3NO₂ + H₂O → 2HNO₃ + NO
where:
  • N₂ + O₂ → 2NOin lightning or an engine; the Chinese saying «thunderstorms nourish the crops» (雷雨发庄稼)
  • 2NO + O₂ → 2NO₂the colourless gas turns brown in air — the way to recognise NO
  • 3NO₂ + H₂O → 2HNO₃ + NOin a tube of NO₂ inverted in water, the water rises to 2/3 of the tube and NO fills the last 1/3
  • 4NO₂ + O₂ + 2H₂O → 4HNO₃, 4NO + 3O₂ + 2H₂O → 4HNO₃mixed with oxygen in these volume ratios, the gases dissolve completely in water and nothing is left

NO and NO₂ are among the causes of smog and acid rain. The equilibrium 2NO₂ ⇌ N₂O₄ is treated in «Chemical equilibrium».

NH₃ + H₂O ⇌ NH₃·H₂O ⇌ NH₄⁺ + OH⁻ NH₃ + HCl → NH₄Cl 4NH₃ + 5O₂ → 4NO + 6H₂O (catalyst, t°)
where:
  • NH₃ + H₂Oammonia solution is a weak base: it turns moist red litmus blue
  • NH₃ + HClwhite smoke (NH₄Cl) appears when the open bottles of concentrated ammonia and concentrated hydrochloric acid are brought close; with non-volatile H₂SO₄ there is no smoke
  • 4NH₃ + 5O₂catalytic oxidation, the first step in making nitric acid («Industrial chemical processes»)

Ammonium salts are soluble white crystals. They decompose on heating (NH₄Cl → NH₃↑ + HCl↑, which recombine to NH₄Cl in the cool part of the tube — it looks like sublimation but is a chemical change; NH₄HCO₃ → NH₃↑ + CO₂↑ + H₂O), and with alkalis on warming they release NH₃: NH₄⁺ + OH⁻ → NH₃↑ + H₂O. That is why ammonium fertilisers must not be mixed with alkaline materials such as slaked lime or plant ash. The NH₄⁺ test is in «Ionic reactions and tests for ions», the preparation of NH₃ in «Preparation and identification of common gases».

Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂↑ + 2H₂O 3Cu + 8HNO₃(dilute) → 3Cu(NO₃)₂ + 2NO↑ + 4H₂O
where:
  • concentratedred-brown NO₂ is released; 2 of the 4 HNO₃ are reduced
  • dilutecolourless NO is released and turns brown at the mouth of the tube; 2 of the 8 HNO₃ are reduced

In both cases the solution turns blue from Cu²⁺. When copper is in excess, the concentrated acid becomes dilute as it is used up and starts giving NO instead of NO₂.

CO₂ + Ca(OH)₂ → CaCO₃↓ + H₂O CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂
where:
  • CO₂ + Ca(OH)₂limewater turns milky — the test for CO₂
  • CaCO₃ + CO₂ + H₂Owith excess CO₂ the precipitate dissolves and the milkiness disappears; heating Ca(HCO₃)₂ gives CaCO₃ back (kettle scale, stalactites in caves)

Per mol of Ca(OH)₂: CO₂ ≤ 1 mol — the precipitate equals the CO₂; 1 < CO₂ < 2 mol — the precipitate is (2 − n(CO₂)) mol; CO₂ ≥ 2 mol — no precipitate is left.

SiO₂ + 2NaOH → Na₂SiO₃ + H₂O SiO₂ + 4HF → SiF₄↑ + 2H₂O Na₂SiO₃ + CO₂ + H₂O → H₂SiO₃↓ + Na₂CO₃
where:
  • SiO₂ + 2NaOHNaOH solution is kept in bottles with rubber stoppers: the Na₂SiO₃ formed (its solution is «water glass», a fire-proofing agent and glue) would stick a glass stopper fast
  • SiO₂ + 4HFHF etches glass (used to mark scales on glassware), so it is kept in plastic bottles
  • Na₂SiO₃ + CO₂ + H₂Ocarbonic acid is stronger than silicic acid: carbon is more non-metallic than silicon

In glass making at high temperature: Na₂CO₃ + SiO₂ → Na₂SiO₃ + CO₂↑ and CaCO₃ + SiO₂ → CaSiO₃ + CO₂↑.

Common metals and their compounds

Open lesson
K Ca Na Mg Al Zn Fe Sn Pb (H) Cu Hg Ag Pt Au
where:
  • to the leftmore active metals: stronger reducing agents, harder to extract from their compounds
  • (H)metals before (H) release H₂ from dilute HCl and dilute H₂SO₄; metals after it do not
  • K, Ca, Nareact with cold water; in a salt solution they react with the water first

The activity series of metals: activity decreases from left to right.

Fe + CuSO₄ → FeSO₄ + Cu: Δm(solid) = 64 − 56 = +8 g per 1 mol Fe
where:
  • Δmchange in the mass of the plate (nail) = mass of metal deposited − mass of metal dissolved; positive means the plate gets heavier
  • Mmolar masses: Fe 56, Cu 64, Zn 65, Ag 108 g/mol

The «difference method» (差量法): the change in mass is proportional to the amount that has reacted.

2Na₂O₂ + 2H₂O → 4NaOH + O₂↑ 2Na₂O₂ + 2CO₂ → 2Na₂CO₃ + O₂
where:
  • O: −1 → −2, 0Na₂O₂ is both the oxidising and the reducing agent; 2 mol Na₂O₂ → 1 mol O₂, with 2 mol e⁻ transferred
  • Δmthe solid gains the mass of «CO» (28 g) per mol of CO₂ absorbed, or of «H₂» (2 g) per mol of H₂O

Sodium peroxide releases oxygen — it is the oxygen source in breathing apparatus and submarines.

2NaHCO₃ → Na₂CO₃ + H₂O + CO₂↑ (t°)
where:
  • t°heating
  • 168 → 1062 · 84 g of NaHCO₃ leave 106 g of Na₂CO₃: the loss of mass is 62 g (H₂O + CO₂)

Thermal decomposition of sodium hydrogen carbonate — used to tell the two solids apart and to remove NaHCO₃ from Na₂CO₃.

2Al + 6H⁺ → 2Al³⁺ + 3H₂↑ 2Al + 2OH⁻ + 6H₂O → 2[Al(OH)₄]⁻ + 3H₂↑
where:
  • 2Al → 3H₂the same amount of Al gives the same amount of H₂ with excess acid or with excess alkali
  • [Al(OH)₄]⁻tetrahydroxidoaluminate (四羟基合铝酸钠 = Na[Al(OH)₄]); older books and many test items write AlO₂⁻ (NaAlO₂, 偏铝酸钠): 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂↑

Aluminium with acids and with alkalis.

Al³⁺ + 3OH⁻ → Al(OH)₃↓ Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻
where:
  • 3 : 1NaOH used up to the maximum precipitate : NaOH that dissolves it again
  • n(OH⁻) = 4n(Al³⁺)the precipitate has just disappeared

NaOH added drop by drop to an Al³⁺ solution: the white precipitate first grows, then dissolves.

2Fe³⁺ + Fe → 3Fe²⁺ 2Fe³⁺ + Cu → 2Fe²⁺ + Cu²⁺ 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻
where:
  • Fe³⁺ → Fe²⁺reducing agents: Fe, Cu, I⁻ (2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂)
  • Fe²⁺ → Fe³⁺oxidising agents: Cl₂, H₂O₂, oxygen of the air, acidified KMnO₄

Conversions between Fe²⁺ and Fe³⁺: the total charge must be equal on both sides.

(−) Fe − 2e⁻ → Fe²⁺ (+) O₂ + 2H₂O + 4e⁻ → 4OH⁻
where:
  • (+) in acidic water2H⁺ + 2e⁻ → H₂↑ — hydrogen-evolution corrosion (析氢腐蚀); in neutral or weakly acidic water — oxygen-absorption corrosion (吸氧腐蚀), the usual case
  • rustFe²⁺ + 2OH⁻ → Fe(OH)₂ → Fe(OH)₃ → Fe₂O₃·xH₂O

Electrochemical corrosion of steel (Chinese textbooks write Fe − 2e⁻ = Fe²⁺).

Redox reactions

Open lesson
Σ nᵢ · xᵢ = z
where:
  • nᵢnumber of atoms of element i in the formula
  • xᵢits oxidation number
  • zcharge of the particle (0 for a neutral substance)

Put in the known values and solve for the one unknown. If the atoms of one element are in different states (Fe₃O₄, Na₂S₂O₃), you get their average, which may be a fraction.

n(e⁻) = n₁ · a · Δx₁ = n₂ · b · Δx₂
where:
  • n₁, n₂amounts of the reducing agent and of the oxidising agent, mol
  • a, batoms of the changing element in one formula unit
  • Δx₁, Δx₂rise and fall of the oxidation number per atom

Count per formula, not per atom: K₂Cr₂O₇ → 2Cr³⁺ takes 2 · 3 = 6 electrons. In an equation, multiply the per-formula changes up to their least common multiple.

A + B → a + b: A > b (oxidising strength), B > a (reducing strength)
where:
  • A, athe oxidising agent and its reduction product
  • B, bthe reducing agent and its oxidation product

The oxidising agent is stronger than the oxidation product, the reducing agent stronger than the reduction product. So two or three given reactions chain into a full order.

n(e⁻) = 2n(Cu) = 2n(H₂) = 2n(Cl₂) = 4n(O₂) = n(Ag)
where:
  • n(e⁻)amount of electrons passed, mol — the same at every electrode connected in series
  • n(X)amount of the substance formed at an electrode, mol

The factors come from the electrode reactions: Cu²⁺ + 2e⁻ → Cu, 2Cl⁻ → Cl₂↑ + 2e⁻, 2H₂O → O₂↑ + 4H⁺ + 4e⁻ (or 4OH⁻ → O₂↑ + 2H₂O + 4e⁻), Ag⁺ + e⁻ → Ag.

Ionic reactions and tests for ions

Open lesson
Na₂SO₄ → 2Na⁺ + SO₄²⁻ CH₃COOH ⇌ CH₃COO⁻ + H⁺
where:
  • →complete ionisation (strong electrolyte)
  • ⇌partial, reversible ionisation (weak electrolyte)

The charges on the right add up to zero. Acid salts: NaHCO₃ → Na⁺ + HCO₃⁻ (HCO₃⁻ is not split), but NaHSO₄ → Na⁺ + H⁺ + SO₄²⁻ in solution. Polyprotic weak acids ionise step by step: first H₂CO₃ ⇌ H⁺ + HCO₃⁻.

Σ z (left side) = Σ z (right side)
where:
  • zthe charge of each ion multiplied by its coefficient

Together with the balance of atoms. In a redox ionic equation the electrons lost must also equal the electrons gained; if you check only the atoms, many wrong options look right.

Hydrocarbons

Open lesson
Alkanes CₙH₂ₙ₊₂ Alkenes CₙH₂ₙ (n ≥ 2) Alkynes CₙH₂ₙ₋₂ (n ≥ 2) Benzene homologues CₙH₂ₙ₋₆ (n ≥ 6)
where:
  • nnumber of carbon atoms in the molecule
  • −2Heach C=C bond or ring removes 2 H atoms from CₙH₂ₙ₊₂, each C≡C removes 4

General formulas of the hydrocarbon series (CₙH₂ₙ also fits cycloalkanes, CₙH₂ₙ₋₂ also fits dienes).

CH₄ + Cl₂ → CH₃Cl + HCl (light) CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄
where:
  • 1 Cl₂ per Hevery H atom replaced uses one Cl₂ and gives one HCl
  • productsall four chloromethanes form together, whatever the ratio; HCl is the product formed in the greatest amount

Substitution (取代反应), the characteristic reaction of alkanes.

CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br n CH₂=CH₂ → [–CH₂–CH₂–]ₙ
where:
  • Br₂bromine water or a solution of Br₂ in CCl₄ loses its orange colour; 1 mol of C=C adds 1 mol of Br₂ (a C≡C adds 2 mol)
  • ndegree of polymerisation; –CH₂–CH₂– is the repeating unit (链节), CH₂=CH₂ the monomer (单体)

Addition (加成反应) and addition polymerisation (加聚反应).

C₆H₆ + Br₂ → C₆H₅Br + HBr (FeBr₃) C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (conc. H₂SO₄, 50–60 °C) C₆H₆ + 3H₂ → C₆H₁₂ (Ni, t°)
where:
  • Br₂pure liquid bromine, not bromine water; iron filings form the catalyst FeBr₃
  • C₆H₅NO₂nitrobenzene — an oily liquid, denser than water, smelling of bitter almonds
  • C₆H₁₂cyclohexane — the one common addition of benzene; it needs 3 mol of H₂

Reactions of benzene: two substitutions and one addition.

CₓHᵧ + (x + y/4) O₂ → x CO₂ + (y/2) H₂OCₓHᵧ + (x + y/4) O₂ → x CO₂ + (y/2) H₂O
where:
  • x, ynumbers of C and H atoms in one molecule
  • n(C) = n(CO₂), n(H) = 2n(H₂O)the way back from combustion data to the formula
  • ΔV = y/4 − 1ΔV = y/4 − 1change in gas volume per volume of hydrocarbon when water is a gas (above 100 °C); zero when y = 4

Complete combustion of a hydrocarbon.

Derivatives of hydrocarbons

Open lesson
2CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂↑n(H₂) = ½ · n(–OH)
where:
  • CH₃CH₂ONasodium ethoxide: the H of the –OH group has been replaced by Na
  • n(–OH)amount of hydroxyl groups that react, mol
  • n(H₂)hydrogen released, mol: every two –OH groups give one H₂ molecule

Sodium replaces only the hydrogen of the –OH group, never an H on carbon. The reaction is gentler than with water: the O–H hydrogen of ethanol is less active. The same ratio holds for –COOH: 2 –COOH ~ H₂.

2CH₃CH₂OH + O₂ →(Cu, Δ) 2CH₃CHO + 2H₂O
where:
  • Cucatalyst (Ag also works): copper first becomes black CuO, which ethanol reduces back to red Cu
  • Δheating
  • CH₃CHOethanal (acetaldehyde), the first oxidation step

Other oxidations of ethanol: it burns (C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O), and strong oxidants — acidified K₂Cr₂O₇ (orange → green) or KMnO₄ (purple → colourless) — take it on to acetic acid.

CH₃CHO + 2[Ag(NH₃)₂]OH →(Δ) CH₃COONH₄ + 2Ag↓ + 3NH₃ + H₂OCH₃CHO + 2Cu(OH)₂ + NaOH →(Δ) CH₃COONa + Cu₂O↓ + 3H₂O
where:
  • [Ag(NH₃)₂]OHTollens’ reagent (ammoniacal silver hydroxide); warmed in a water bath, it leaves a silver mirror on the wall of the tube
  • Cu(OH)₂a freshly precipitated suspension with excess NaOH; on boiling it gives a brick-red Cu₂O precipitate
  • ratios1 –CHO ~ 2Ag ~ 1Cu₂O; HCHO counts as two –CHO groups: 1 HCHO ~ 4Ag

In both reactions the aldehyde is oxidised (it is the reducing agent). They are the tests for the –CHO group; glucose, formic acid and formates give them too.

2CH₃COOH + CaCO₃ → (CH₃COO)₂Ca + H₂O + CO₂↑Activity of the O–H hydrogen: CH₃COOH > H₂CO₃ > H₂O > C₂H₅OH
where:
  • CaCO₃kettle scale or marble; vinegar dissolves it
  • (CH₃COO)₂Cacalcium acetate, a soluble salt
  • ordera stronger acid displaces a weaker one from its salt

Na reacts with all four substances, NaOH only with the acids, and NaHCO₃ only with acids stronger than carbonic acid, that is with –COOH.

CH₃COOH + HO–CH₂CH₃ ⇌(conc. H₂SO₄, Δ) CH₃COO–CH₂CH₃ + H₂O
where:
  • conc. H₂SO₄catalyst and water absorber (it shifts the equilibrium to the right)
  • ⇌the reaction is reversible, so the acid is never used up completely
  • mechanism«the acid loses –OH, the alcohol loses H» (酸脱羟基醇脱氢): the alcohol’s oxygen stays in the ester

Naming: alcohol part + acid part «-ate» — CH₃COOC₂H₅ is ethyl acetate, HCOOCH₃ methyl formate. Esterification is a kind of substitution reaction.

CH₃COOC₂H₅ + H₂O ⇌(dilute H₂SO₄, Δ) CH₃COOH + C₂H₅OHCH₃COOC₂H₅ + NaOH →(Δ) CH₃COONa + C₂H₅OH
where:
  • acidic hydrolysisthe reverse of esterification, reversible
  • basic hydrolysisgoes to completion: NaOH turns the acid formed into a salt and removes it from the equilibrium
  • ratio1 ester group –COO– ~ 1 NaOH

An ester always splits at the same C–O bond: the acid part takes back –OH and the alcohol part takes back H.

(C₁₇H₃₅COO)₃C₃H₅ + 3NaOH →(Δ) 3C₁₇H₃₅COONa + C₃H₅(OH)₃
where:
  • (C₁₇H₃₅COO)₃C₃H₅tristearin (glyceryl tristearate), M = 890 g/mol
  • C₁₇H₃₅COONasodium stearate, the main part of soap
  • C₃H₅(OH)₃glycerol, M = 92 g/mol

Saponification (皂化反应) is the basic hydrolysis of a fat: 1 mol of fat uses 3 mol of NaOH and gives 1 mol of glycerol.

C₆H₁₂O₆ →(enzymes) 2C₂H₅OH + 2CO₂↑(C₆H₁₀O₅)ₙ + nH₂O →(acid or enzyme) nC₆H₁₂O₆
where:
  • C₆H₁₂O₆glucose, M = 180 g/mol
  • C₆H₁₀O₅the repeating unit of starch and cellulose, M = 162 g/mol
  • 162 : 180on hydrolysis the mass grows by a factor 10/9 (water is added)

Alcoholic fermentation turns 1 mol of glucose into 2 mol of ethanol and 2 mol of CO₂.

n = M(polymer) / M(repeating unit)n = M(polymer) / M(repeating unit)
where:
  • ndegree of polymerisation (an average value)
  • M(polymer)average molar mass of the polymer, g/mol
  • M(repeating unit)mass of one repeating unit (162 for C₆H₁₀O₅)

The molecules in a polymer sample have different n, so a polymer is a mixture with no sharp melting point. For the same reason starch and cellulose are not isomers.

10Chemistry: theories and lawsIntermediate

Atomic structure and the periodic law

Open lesson
A = Z + N number of electrons = Z − q
where:
  • Amass number (protons + neutrons)
  • Zatomic number = number of protons
  • Nnumber of neutrons
  • qcharge of the particle with its sign (0 for an atom, +2 for Mg²⁺, −1 for Cl⁻)

Forming an ion changes only the electrons: a cation has lost q electrons, an anion has gained |q|. The nucleus (Z and A) stays the same.

Aᵣ = A₁x₁ + A₂x₂ + …, x₁ + x₂ + … = 1
where:
  • Aᵣrelative atomic mass of the element (the number in the periodic table)
  • A₁, A₂mass numbers of the isotopes (strictly their relative masses; the mass numbers are close enough for CSCA)
  • x₁, x₂abundances: the fraction of the atoms that are each isotope (75 % → 0.75)

The table value is a weighted average, which is why chlorine has 35.5 although no chlorine atom has a mass of 35.5. Turning it into grams per mole is the job of the lesson «Amount of substance: the mole».

shell n holds at most 2n² electrons (2, 8, 18, 32); s² p⁶ d¹⁰; filling order: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p
where:
  • nshell number (principal quantum number)
  • s, p, dsubshells of 1, 3 and 5 orbitals, each orbital holding 2 electrons

Extra rules for main-group atoms: the outer shell never holds more than 8 electrons (the K shell 2), and 4s fills before 3d. That is why potassium is 2, 8, 8, 1 and not 2, 8, 9.

more shells ⇒ larger; same number of shells ⇒ larger Z, smaller; r(cation) < r(atom) < r(anion)
where:
  • rradius of the particle (atom or ion)
  • Znuclear charge (number of protons)

Ions with the same electron configuration (isoelectronic ions) are ordered by Z alone: O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺ — all have 10 electrons, and more protons pull them tighter.

highest oxidation number = +G; lowest = G − 8 (non-metals)
where:
  • Gnumber of the main group (IA → 1 … VIIA → 7) = number of valence electrons

For groups IVA–VIIA: |highest| + |lowest| = 8. Exceptions: fluorine has no positive oxidation number, and oxygen never reaches +6. Metals have no negative oxidation numbers.

Chemical bonds and intermolecular forces

Open lesson
Δχ = |χ(A) − χ(B)|
where:
  • χ(A), χ(B)electronegativities of the two atoms (Pauling scale)
  • Δχtheir difference: a measure of how polar the bond is

Δχ = 0: non-polar covalent; 0 < Δχ < about 1.7: polar covalent (the larger Δχ, the more polar); Δχ > about 1.7 between a metal and a non-metal: mainly ionic. The border is only a guide: H–F (Δχ = 1.9) is a very polar covalent bond, and AlCl₃ (Δχ = 1.5) is a covalent compound.

number of covalent bonds of a non-metal atom = 8 − G (H: 1)
where:
  • Gmain-group number = number of valence electrons

C forms 4 bonds, N 3, O 2 and the halogens 1: as many as the electrons still missing from the octet. The remaining valence electrons stay as lone pairs (孤电子对): N has 1, O 2, Cl 3.

lone pairs on A = ½(a − x·b); domains = x + lone pairs
where:
  • Athe central atom of a molecule ABₓ
  • avalence electrons of A (for a cation subtract its charge, for an anion add it)
  • xnumber of atoms bonded to A
  • belectrons each outer atom still needs: H and the halogens 1, O and S 2

This is the formula of Chinese textbooks (中心原子上的孤电子对数). 2 domains → linear, 180°; 3 → planar, 120°; 4 → tetrahedral, 109.5°; lone pairs make the angles smaller.

Reaction rate

Open lesson
v(B) = Δc(B)/Δt = Δn(B)/(V · Δt)v(B) = Δc(B)/Δt = Δn(B)/(V · Δt)
where:
  • v(B)rate expressed with substance B, in mol/(L·s) or mol/(L·min)
  • Δc(B)change in the molar concentration of B (taken as positive), in mol/L
  • Δn(B)change in the amount of B, in mol
  • Vvolume of the vessel or the solution, in L
  • Δttime interval, in s or min

This is the average rate over Δt, not the rate at one instant. Solids and pure liquids (CaCO₃(s), H₂O(l)) have a constant «concentration», so rates are never expressed with them. 1 mol/(L·min) = 1/60 mol/(L·s).

aA + bB → cC + dD: v(A) : v(B) : v(C) : v(D) = a : b : c : d
where:
  • a, b, c, dcoefficients of the balanced equation
  • v(X)rate expressed with substance X; all in the same unit

The same ratio holds for Δc and, in one vessel, for Δn. Dividing any v(X) by its coefficient gives one number for the whole reaction, which is how rates written with different substances are compared.

v₂ = v₁ · γ^((t₂ − t₁)/10)
where:
  • v₁, v₂rates at the temperatures t₁ and t₂
  • t₁, t₂temperatures, in °C
  • γtemperature coefficient, usually 2–4

An approximate rule (van ’t Hoff). Chinese textbooks state it in words: every 10 °C rise usually makes the rate 2–4 times larger. The time needed for the same change shrinks by the same factor: τ₂ = τ₁ / γ^((t₂ − t₁)/10).

ΔH = E₁ − E₂
where:
  • E₁activation energy of the forward reaction, in kJ/mol
  • E₂activation energy of the reverse reaction, in kJ/mol
  • ΔHenthalpy change of the reaction (焓变): < 0 exothermic, > 0 endothermic

On an energy diagram the reactants and the products are two levels, and the curve climbs a «hill» between them. E₁ is the height of the hill above the reactants, E₂ its height above the products. In an exothermic reaction the products lie lower and E₁ < E₂. A catalyst lowers the top of the hill, so E₁ and E₂ fall by the same amount and ΔH stays the same.

Chemical equilibrium

Open lesson
aA(g) + bB(g) ⇌ cC(g) + dD(g): K = cᶜ(C) · cᵈ(D) / (cᵃ(A) · cᵇ(B))aA(g) + bB(g) ⇌ cC(g) + dD(g): K = cᶜ(C) · cᵈ(D) / (cᵃ(A) · cᵇ(B))
where:
  • c(X)equilibrium molar concentration of X, in mol/L (also written [X])
  • a, b, c, dcoefficients, used as powers
  • Kthe equilibrium constant (化学平衡常数); it depends only on temperature

Products over reactants, each raised to the power of its coefficient. Solids and pure liquids (and water as the solvent) are left out. K does not depend on the concentrations, the pressure or a catalyst — only on temperature. CSCA items usually give K without a unit.

Q < K: forward; Q = K: equilibrium; Q > K: reverse
where:
  • Qthe reaction quotient (浓度商): the same expression as K, filled with the concentrations at the given moment
  • Kthe equilibrium constant at this temperature

Q < K: there is «too little product», so v(forward) > v(reverse) until Q grows to K. Q > K: too much product, and the reverse reaction wins.

α(A) = Δn(A)/n₀(A) · 100% = Δc(A)/c₀(A) · 100%α(A) = Δn(A)/n₀(A) · 100% = Δc(A)/c₀(A) · 100%
where:
  • α(A)conversion of A (转化率)
  • Δn(A), Δc(A)amount (mol) or concentration (mol/L) of A that has reacted
  • n₀(A), c₀(A)initial amount or concentration of A

Yield (产率) = actual amount of product / amount expected if the limiting reactant reacted completely · 100%. At equilibrium both are below 100%. At constant T and V the pressure is proportional to the total amount of gas: p(eq)/p(start) = n(eq)/n(start), and the volume fraction of a gas equals its mole fraction.

Theories of electrolyte solutions

Open lesson
HA ⇌ H⁺ + A⁻ Ka = c(H⁺) · c(A⁻) / c(HA)α = n(ionised) / n(total) · 100% c(H⁺) = c · α ≈ √(Ka · c)HA ⇌ H⁺ + A⁻ Ka = c(H⁺) · c(A⁻) / c(HA)α = n(ionised) / n(total) · 100% c(H⁺) = c · α ≈ √(Ka · c)
where:
  • c(H⁺), c(A⁻), c(HA)equilibrium molar concentrations, mol/L
  • Kaionisation constant of the acid (电离常数): the larger Ka, the stronger the acid; it depends only on temperature (for a base: Kb)
  • αdegree of ionisation (电离度): the share of molecules that have ionised; it changes with concentration
  • cinitial concentration of the acid, mol/L

When α is below about 5 %, c(HA) ≈ c, so Ka ≈ cα² and c(H⁺) ≈ √(Ka · c). At 25 °C: Ka(HF) ≈ 6.3 × 10⁻⁴, Ka(CH₃COOH) ≈ 1.8 × 10⁻⁵, Ka₁(H₂CO₃) ≈ 4.5 × 10⁻⁷, Ka(HClO) ≈ 4 × 10⁻⁸, Ka₂(H₂CO₃) ≈ 4.7 × 10⁻¹¹ — the acid strength falls in this order.

Kw = c(H⁺) · c(OH⁻) = 1.0 × 10⁻¹⁴ (25 °C)
where:
  • Kwionic product of water (水的离子积); it depends only on temperature: about 1.0 × 10⁻¹⁴ at 25 °C and about 5.5 × 10⁻¹³ at 100 °C (problems often give a rounded “Kw = 1 × 10⁻¹² at a higher temperature”)
  • c(H⁺), c(OH⁻)total concentrations of the ions in the solution, whatever their source, mol/L

Kw holds in pure water and in every dilute aqueous solution — acidic, basic or neutral. Added H⁺ or OH⁻ (an acid or a base) suppresses the ionisation of water; ions that bind H⁺ or OH⁻ (hydrolysing salts) promote it.

A⁻ + H₂O ⇌ HA + OH⁻ Kh = Kw / Kac(OH⁻) ≈ √(Kh · c)A⁻ + H₂O ⇌ HA + OH⁻ Kh = Kw / Kac(OH⁻) ≈ √(Kh · c)
where:
  • Khhydrolysis constant of the anion A⁻ (for the cation of a weak base: Kh = Kw / Kb)
  • Kaionisation constant of the matching acid HA
  • cconcentration of the salt, mol/L

The smaller Ka, the larger Kh: “the weaker the acid, the more basic its salt” in numbers. Since Kh(CH₃COO⁻) = 10⁻¹⁴ / (1.8 × 10⁻⁵) ≈ 5.6 × 10⁻¹⁰ is far smaller than Ka, a mixture of equal amounts of CH₃COOH and CH₃COONa is acidic: ionisation beats hydrolysis.

c(Na⁺) + c(H⁺) = 2c(CO₃²⁻) + c(HCO₃⁻) + c(OH⁻)c(Na⁺) = 2[c(CO₃²⁻) + c(HCO₃⁻) + c(H₂CO₃)]c(OH⁻) = c(H⁺) + c(HCO₃⁻) + 2c(H₂CO₃)
where:
  • 1charge balance (电荷守恒): the solution is electrically neutral; every ion is multiplied by its charge, so CO₃²⁻ counts twice
  • 2material balance (物料守恒): the formula Na₂CO₃ fixes n(Na) = 2n(C), and the carbon is shared by CO₃²⁻, HCO₃⁻ and H₂CO₃
  • 3proton balance (质子守恒): the OH⁻ released by water equals the free H⁺ plus the H⁺ taken up by CO₃²⁻ (one in HCO₃⁻, two in H₂CO₃); it is line 1 minus line 2

The three balances of a Na₂CO₃ solution. Order of concentrations: c(Na⁺) > c(CO₃²⁻) > c(OH⁻) > c(HCO₃⁻) > c(H⁺) — OH⁻ comes from both hydrolysis steps and from water, so it exceeds HCO₃⁻.

AₘBₙ(s) ⇌ mAⁿ⁺ + nBᵐ⁻ Ksp = cᵐ(Aⁿ⁺) · cⁿ(Bᵐ⁻)Q > Ksp: a precipitate forms Q = Ksp: saturated Q < Ksp: more dissolves
where:
  • Kspsolubility product (溶度积); for a given substance it depends only on temperature; the solid is not written
  • Qion product: the same expression with the actual concentrations at any moment, not necessarily at equilibrium
  • ssolubility in mol/L; for a 1 : 1 salt AB in pure water s = √Ksp

Ksp compares solubilities directly only for salts of the same type (AgCl, AgBr and AgI are all 1 : 1). Ag₂CrO₄ has a smaller Ksp than AgCl but is more soluble. At 25 °C: Ksp(AgCl) = 1.8 × 10⁻¹⁰, Ksp(AgI) = 8.5 × 10⁻¹⁷, Ksp(BaSO₄) = 1.1 × 10⁻¹⁰.

11Chemistry: experiments and applicationsIntermediate

Laboratory safety and apparatus

Open lesson
V(delivered) = V₂ − V₁
where:
  • V₁initial burette reading, mL (two decimals, e.g. 0.50)
  • V₂final reading, mL; the numbers increase downwards, so V₂ > V₁

Every reading is taken at the bottom of the meniscus with the eye level with it. The result is always written to 0.01 mL: 22.30 mL, not 22.3 mL.

m(left pan) = m(right pan) + m(rider)m(object) = m(weights) + m(rider); pans swapped: m(object) = m(weights) − m(rider)
where:
  • m(rider)the reading of the rider on the beam scale; it always adds to the right pan
  • m(weights)total of the weights on the pan, g; weights are moved with tweezers, not fingers

The rule is «object left, weights right» (左物右码). If the rider is not used, swapping the pans does not change the result.

c(B) = c(A) · V(A) / V(B)c(B) = c(A) · V(A) / V(B)
where:
  • c(A), V(A)concentration and used volume of the standard solution in the burette (e.g. HCl)
  • c(B), V(B)concentration and pipetted volume of the unknown solution in the flask (e.g. NaOH)

For a 1 : 1 reaction (HCl + NaOH). The key to error analysis: c(A) and V(B) are taken as fixed, so every error acts through the V(A) read from the burette — if V(A) is read too large, c(B) comes out too high.

Preparation and identification of common gases

Open lesson
D(air) = M(gas) / 29D(air) = M(gas) / 29
where:
  • M(gas)molar mass of the gas, g/mol
  • 29average molar mass of air, g/mol (≈ 78 % N₂ and 21 % O₂)
  • D(air)relative density of the gas with respect to air

D > 1: the gas sinks in air — use an upright jar with the delivery tube reaching the bottom. D < 1: the gas rises — use an inverted jar. If M is close to 29 (N₂, CO, C₂H₄) or the gas reacts with air (NO), collect it over water only.

2KMnO₄ → K₂MnO₄ + MnO₂ + O₂↑ (t°)2KClO₃ → 2KCl + 3O₂↑ (MnO₂, t°)2H₂O₂ → 2H₂O + O₂↑ (MnO₂)
where:
  • KMnO₄potassium permanganate, a purple solid: a solid is heated; a loose plug of cotton wool in the mouth of the tube stops the powder entering the delivery tube
  • KClO₃ + MnO₂potassium chlorate with the catalyst MnO₂: solid + solid, heated; without MnO₂ a much higher temperature is needed
  • H₂O₂ + MnO₂hydrogen peroxide solution poured onto MnO₂: solid + liquid, no heating — the most convenient and the “greenest” method

MnO₂ is a catalyst in the last two reactions: its mass and chemical properties are the same after the reaction, and it can be recovered by filtering. In the first reaction MnO₂ is a product. Collect O₂ over water (it is poorly soluble) or in an upright jar (M = 32 > 29). Test: a glowing splint relights.

Zn + H₂SO₄(dilute) → ZnSO₄ + H₂↑
where:
  • Znzinc granules: a convenient, steady rate; Na or K would react violently, Cu does not react at all
  • H₂SO₄(dilute)not concentrated H₂SO₄ and not HNO₃ of any concentration: these oxidising acids give SO₂ or NO/NO₂ instead of H₂; dilute HCl also works, but it is volatile and adds HCl gas to the hydrogen

Collect H₂ over water or in an inverted jar (M = 2). Crude zinc reacts faster than pure zinc, and a few drops of CuSO₄ solution speed the reaction up: copper deposited on the zinc forms tiny galvanic cells (see «Redox reactions»).

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂↑CO₂ + Ca(OH)₂ → CaCO₃↓ + H₂O
where:
  • CaCO₃marble or limestone in lumps: a steady rate; powdered CaCO₃ or Na₂CO₃ reacts too fast to control
  • HCldilute hydrochloric acid; not H₂SO₄ — slightly soluble CaSO₄ coats the marble and the reaction soon stops; not concentrated HCl — it is volatile and the CO₂ would contain much HCl
  • Ca(OH)₂limewater (second line — the identification test): it turns milky; with excess CO₂ it clears again, CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂

CO₂ is denser than air (M = 44) and fairly soluble in water, so it is usually collected in an upright jar. To check that the jar is full, hold a burning splint at its mouth (not inside): it goes out.

MnO₂ + 4HCl(conc.) → MnCl₂ + Cl₂↑ + 2H₂O (t°)
where:
  • MnO₂the oxidising agent (Mn: +4 → +2)
  • 4HClonly 2 of the 4 HCl are oxidised (Cl: −1 → 0); the other 2 act as an acid and end up in MnCl₂
  • conc.dilute hydrochloric acid does not react with MnO₂: as the acid is used up and diluted, the reaction stops, so part of the HCl always remains

Without heating, KMnO₄ (or KClO₃) also gives Cl₂ with concentrated HCl, for example 2KMnO₄ + 16HCl(conc.) → 2KCl + 2MnCl₂ + 5Cl₂↑ + 8H₂O. Chlorine is toxic: the set-up must end with a tail-gas absorber.

2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃↑ + 2H₂O (t°)
where:
  • NH₄Cl + Ca(OH)₂both solids, mixed and heated in a test tube with its mouth tilted slightly down (the same generator as for KMnO₄)
  • not NH₄Cl aloneit decomposes into NH₃ and HCl, which recombine into NH₄Cl near the cool mouth of the tube
  • not NaOHNaOH and KOH attack hot glass and make the mixture cake, so Ca(OH)₂ is used

Quick method without heating: drip concentrated ammonia solution onto solid CaO or NaOH. NH₃ is very soluble (≈ 700 volumes in 1 volume of water) and lighter than air: collect it only in a dry inverted vessel, with a loose plug of cotton wool at the mouth to stop air mixing in.

Separation and purification of substances

Open lesson
m(crystals) = m(water) · (S₁ − S₂) / 100m(crystals) = m(water) · (S₁ − S₂) / 100
where:
  • S₁solubility at the higher temperature, g per 100 g of water
  • S₂solubility at the lower temperature
  • m(water)mass of water in the solution, g

Valid for a solution saturated at the higher temperature when no water evaporates. The liquid left over (the mother liquor) is still saturated at the lower temperature, so the crystals are never 100 % of the solute.

KD = c(organic layer) / c(water)q = Vw / (Vw + KD · Vo); after n portions qⁿKD = c(organic layer) / c(water)q = Vw / (Vw + KD · Vo); after n portions qⁿ
where:
  • KDdistribution coefficient: a constant for the solute and the two solvents at a given temperature
  • Vw, Vovolume of the aqueous solution and of one portion of the organic solvent
  • qfraction of the solute left in the water after one portion

Several small portions of the same total volume leave less solute behind than one large portion — the rule “little and often” (Chinese 少量多次).

Rf = distance moved by the substance / distance moved by the solvent frontRf = distance moved by the substance / distance moved by the solvent front
where:
  • Rfretardation factor (no unit)
  • distancesboth measured from the start line (to the centre of the spot)

0 < Rf < 1. For a given solvent, paper and temperature, Rf is a constant of the substance: equal Rf values point to the same substance, and a pure substance gives a single spot.

Industrial chemical processes

Open lesson
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92.4 kJ/molN₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92.4 kJ/mol
where:
  • ΔH < 0exothermic: a low temperature favours NH₃ at equilibrium
  • 4 → 2moles of gas fall from 4 to 2: a high pressure favours NH₃
  • ⇌reversible: the reaction never goes to completion, so unreacted N₂ and H₂ always remain

Ammonia synthesis (合成氨), the Haber–Bosch process. The value of ΔH is the one given in Chinese textbooks; CSCA items usually need only its sign.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH < 0 (V₂O₅, 400–500 °C, ordinary pressure)
where:
  • V₂O₅vanadium(V) oxide catalyst: speeds the reaction up, does not move the equilibrium
  • O₂taken in excess (air) to raise the conversion of the more valuable SO₂
  • 3 → 2the gas moles fall, so pressure would help — but the conversion is already high

Why ordinary pressure, unlike ammonia synthesis? At atmospheric pressure the conversion of SO₂ is already over 90%, so costly high-pressure equipment would add little.

FeS₂ ~ 2H₂SO₄ (120 : 196) S ~ H₂SO₄ (32 : 98) NH₃ ~ HNO₃ (17 : 63)
where:
  • ~«corresponds to»: the same number of atoms of the key element (S or N) on both sides
  • 120 : 196corresponding masses (in g, kg or t): M(FeS₂) = 120 g/mol, 2M(H₂SO₄) = 196 g/mol

The relationship method (关系式法): link the first raw material with the final product through one conserved element and skip the intermediate equations. It assumes no losses — losses are included through the yield below.

η = actual amount / theoretical amount × 100% η(overall) = η₁ · η₂ · η₃ …η = actual amount / theoretical amount × 100% η(overall) = η₁ · η₂ · η₃ …
where:
  • ηpercentage yield (产率); the «theoretical» amount is calculated from the equation
  • η₁, η₂ …yields of the successive steps (as fractions); a loss of x% of the key element in a step means η = (100 − x)%

Conversion (转化率) refers to a reactant (what share of it has reacted), yield (产率) to a product. For a multi-step process multiply the step yields — never add or average them.

4NH₃ + 5O₂ → 4NO + 6H₂O (Pt–Rh, heat) 2NO + O₂ → 2NO₂ 3NO₂ + H₂O → 2HNO₃ + NO
where:
  • 4NH₃ + 5O₂catalytic oxidation of ammonia (氨的催化氧化); NH₃ is the reducing agent
  • 3NO₂ + H₂Oabsorption: N goes from +4 to +5 (HNO₃) and to +2 (NO) — a disproportionation
  • NH₃ ~ HNO₃overall, when the NO is recycled: NH₃ + 2O₂ → HNO₃ + H₂O, one N atom per HNO₃ (17 : 63)

Nitric acid from ammonia. Blowing extra air into the absorption tower lets all the nitrogen end up in the acid: 4NO₂ + O₂ + 2H₂O → 4HNO₃.

2NaCl + 2H₂O → 2NaOH + H₂↑ + Cl₂↑ (electrolysis)
where:
  • anode (+)2Cl⁻ − 2e⁻ → Cl₂↑: chloride ions are oxidised
  • cathode (−)2H₂O + 2e⁻ → H₂↑ + 2OH⁻: water is reduced and OH⁻ accumulates here, so NaOH forms at the cathode
  • membranea cation-exchange membrane lets only cations (Na⁺) pass from the anode side to the cathode side

The chlor-alkali industry (氯碱工业) with an ion-exchange membrane cell. Chinese textbooks write electrode equations with «− 2e⁻» on the left and «=»: 2Cl⁻ − 2e⁻ = Cl₂↑.

atom economy = a · M(product) / Σ b · M(reactant) × 100%atom economy = a · M(product) / Σ b · M(reactant) × 100%
where:
  • a · M(product)coefficient × molar mass of the desired product
  • Σ b · M(reactant)sum over all reactants of coefficient × molar mass (by conservation of mass, equal to the total mass of all products)

Chinese: 原子经济性 or 原子利用率. It is read from the balanced equation alone and says nothing about the yield: a reaction can have an atom economy of 100% and a yield of 15%, or the other way round.

12Professional Chinese and score strategiesAdvanced

Professional Chinese: the test, vocabulary and reading strategies

Open lesson

Test strategy, glossary of terms and study plan

Open lesson