IB AQA Science: Calculation Drill Training | IB AQA 科学：计算题专项训练

📚 IB AQA Science: Calculation Drill Training | IB AQA 科学：计算题专项训练

Mastering calculations is essential for success in IB and AQA science courses. Whether you’re balancing chemical equations, analysing motion graphs, or determining half-lives, a solid grasp of quantitative methods separates top candidates from the rest. This revision guide walks you through the most common calculation styles found in Physics, Chemistry, and Biology, with clear step-by-step approaches and plenty of practice points.

在 IB 和 AQA 科学课程中，掌握计算是取得高分的关键。无论是配平化学方程式、分析运动图像，还是计算半衰期，扎实的定量分析能力往往能让考生脱颖而出。这份复习指南将带你梳理物理、化学和生物中最常见的计算题型，通过清晰的步骤讲解和配套训练要点，帮助你在计算题上稳稳拿分。

1. Understanding Units and Conversions | 理解单位与换算

Every scientific calculation starts with getting the units right. In IB and AQA exams, you are frequently expected to convert between prefixes such as kilo (k, 10³), centi (c, 10⁻²), milli (m, 10⁻³), micro (µ, 10⁻⁶), and nano (n, 10⁻⁹). Always write down the quantity with its unit before plugging numbers into a formula. A simple table can help you internalise common conversions.

科学计算的第一步是确保单位正确。在 IB 和 AQA 考试中，你需要熟练转换单位前缀，如千 (k, 10³)、厘 (c, 10⁻²)、毫 (m, 10⁻³)、微 (µ, 10⁻⁶) 和纳 (n, 10⁻⁹)。在代入公式之前，务必写下带有单位的物理量。牢记以下常见换算表，会让你的计算顺畅不少。

Prefix	Symbol	Factor	Example
kilo	k	10³	1 km = 1000 m
centi	c	10⁻²	1 cm = 0.01 m
milli	m	10⁻³	1 mA = 0.001 A
micro	µ	10⁻⁶	1 µm = 10⁻⁶ m
nano	n	10⁻⁹	1 nm = 10⁻⁹ m

When converting derived units such as density (g cm⁻³ to kg m⁻³), break the conversion into base units first. For example, 1 g cm⁻³ = 1000 kg m⁻³. Practise with speed (m s⁻¹ to km h⁻¹) and pressure (Pa to kPa) until these become automatic.

在转换导出单位（如密度 g cm⁻³ 转为 kg m⁻³）时，先拆解为基本单位再换算。例如，1 g cm⁻³ = 1000 kg m⁻³。多练习速度 (m s⁻¹ 转为 km h⁻¹) 和压强 (Pa 转为 kPa) 的转换，直到形成条件反射。

2. Moles and Molar Calculations | 摩尔与摩尔计算

The mole is the chemist’s counting unit. The core relationship is n = m / M, where n is amount in moles, m is mass in grams, and M is molar mass in g mol⁻¹. Always calculate M from the periodic table using relative atomic masses. For gases at room temperature and pressure (RTP), one mole occupies 24 dm³; use this only when the question specifies RTP or you are asked for gas volumes.

摩尔是化学家的计数单位。核心关系式为 n = m / M，其中 n 为物质的量（单位 mol），m 为质量（g），M 为摩尔质量（g mol⁻¹）。务必根据相对原子质量表计算 M。在常温常压 (RTP) 下，1 摩尔气体占据约 24 dm³，但仅当题目明确说明 RTP 或要求计算气体体积时才可使用。

n = m ÷ M | n = V(gas) ÷ 24 dm³ mol⁻¹ (at RTP)

A typical exam problem: ‘Calculate the mass of CO₂ produced when 10.0 g of CaCO₃ decomposes.’ Step 1: write the balanced equation CaCO₃ → CaO + CO₂. Step 2: find moles of CaCO₃ (M = 100.1 g mol⁻¹, n = 10.0/100.1 = 0.0999 mol). Step 3: mole ratio 1:1 gives 0.0999 mol CO₂. Step 4: mass of CO₂ = n × M = 0.0999 × 44.0 = 4.40 g. Always show working clearly.

经典考题：“计算 10.0 g CaCO₃ 分解时产生 CO₂ 的质量”。第一步：写出配平方程式 CaCO₃ → CaO + CO₂。第二步：计算 CaCO₃ 的物质的量 (M = 100.1 g mol⁻¹, n = 10.0/100.1 = 0.0999 mol)。第三步：摩尔比 1:1，得 CO₂ 为 0.0999 mol。第四步：CO₂ 质量 = n × M = 0.0999 × 44.0 = 4.40 g。务必清晰展示每一步推理。

3. Concentration and Titration | 浓度与滴定计算

Concentration calculations feature heavily in both IB and AQA chemistry. The key formula is n = c × V, where c is concentration in mol dm⁻³ and V is volume in dm³. Remember that 1 dm³ = 1000 cm³, so converting cm³ to dm³ is a common source of error. In titrations, use the balanced equation to deduce the mole ratio between the acid and the base.

浓度计算在 IB 和 AQA 化学中占比很高。核心公式为 n = c × V，其中 c 为浓度 (mol dm⁻³)，V 为体积 (dm³)。注意 1 dm³ = 1000 cm³，因此把 cm³ 换算为 dm³ 是常出错的地方。在滴定中，利用配平方程推导酸碱之间的摩尔比。

c₁V₁ / n₁ = c₂V₂ / n₂ (for reactions aA + bB → products)

For example, 25.0 cm³ of NaOH is neutralised by 20.0 cm³ of HCl (c = 0.100 mol dm⁻³). Moles HCl = 0.100 × 0.0200 = 0.00200 mol. Since NaOH + HCl → NaCl + H₂O, ratio is 1:1, so moles NaOH = 0.00200 mol. Concentration of NaOH = 0.00200 / 0.0250 = 0.0800 mol dm⁻³. Always include units in each step to catch mistakes.

例如，25.0 cm³ NaOH 溶液被 20.0 cm³ HCl (浓度 0.100 mol dm⁻³) 中和。HCl 的物质的量 = 0.100 × 0.0200 = 0.00200 mol。由 NaOH + HCl → NaCl + H₂O 知摩尔比为 1:1，故 NaOH 为 0.00200 mol。NaOH 浓度 = 0.00200 / 0.0250 = 0.0800 mol dm⁻³。每一步都带上单位，能有效避免计算错误。

4. Empirical and Molecular Formulae | 经验式与分子式计算

Determining empirical formulae from percentage composition is a routine skill. Divide the percentage of each element by its relative atomic mass, then divide all results by the smallest value to get the simplest whole-number ratio. If you obtain a ratio like 1:1.5, multiply by 2 to clear the fraction.

由质量百分比推算经验式是一项必备技能。将各元素的质量百分数除以各自的相对原子质量，再将所有结果除以其中的最小值，得到最简整数比。若得到类似 1:1.5 的比例，乘以 2 化为整数。

A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Moles: C = 40.0/12.0 = 3.33, H = 6.7/1.0 = 6.7, O = 53.3/16.0 = 3.33. Divide by 3.33 gives C:H:O = 1:2:1, so empirical formula is CH₂O. If the molar mass is found to be 180 g mol⁻¹, the molecular formula is C₆H₁₂O₆. Always check that your final ratio makes chemical sense.

某化合物含碳 40.0%、氢 6.7%、氧 53.3%。摩尔数：C = 40.0/12.0 = 3.33，H = 6.7/1.0 = 6.7，O = 53.3/16.0 = 3.33。除以 3.33 得 C:H:O = 1:2:1，经验式为 CH₂O。若测得其摩尔质量为 180 g mol⁻¹，则分子式为 C₆H₁₂O₆。算完后检查所得比例是否化学上合理。

5. Gas Laws and Ideal Gas Equation | 气体定律与理想气体方程

The ideal gas equation pV = nRT links pressure, volume, amount, and temperature. Use p in Pa (or kPa if R is adjusted), V in m³, n in mol, T in K, and R = 8.31 J mol⁻¹ K⁻¹. The combined gas law (p₁V₁/T₁ = p₂V₂/T₂) is useful when the amount of gas is constant. IB and AQA both require conversions to absolute temperature (K = °C + 273) and standard pressure units.

理想气体方程 pV = nRT 将压强、体积、物质的量和温度联系在一起。使用 p (Pa)、V (m³)、n (mol)、T (K)，R = 8.31 J mol⁻¹ K⁻¹。当气体物质的量不变时，可用联合气体定律 (p₁V₁/T₁ = p₂V₂/T₂)。IB 和 AQA 都要求掌握开尔文温度换算 (K = °C + 273) 及标准压强单位。

pV = nRT and p₁V₁ / T₁ = p₂V₂ / T₂

A typical problem: ‘Calculate the volume of 0.500 mol of an ideal gas at 100 kPa and 25 °C.’ T = 25 + 273 = 298 K, p = 100 000 Pa. V = nRT/p = (0.500 × 8.31 × 298) / 100 000 = 0.0124 m³ = 12.4 dm³. Pay close attention to units: using kPa with R = 8.31 requires V in dm³ if you adjust R, but it is safer to use SI units throughout.

典型题目：“计算 0.500 mol 理想气体在 100 kPa、25 °C 下的体积”。T = 298 K，p = 100 000 Pa，V = nRT/p = (0.500 × 8.31 × 298) / 100 000 = 0.0124 m³ = 12.4 dm³。注意单位一致：使用 kPa 时若修正 R 值较为麻烦，推荐全程使用国际单位制。

6. Enthalpy Changes and Calorimetry | 焓变与量热法

Calorimetry calculations rely on q = mcΔT, where q is heat energy (J), m is mass of water or solution (g), c is specific heat capacity (usually 4.18 J g⁻¹ °C⁻¹ for water), and ΔT is temperature change (°C). For enthalpy change per mole, use ΔH = −q / n, remembering the negative sign for exothermic reactions.

量热计算基于 q = mcΔT，其中 q 为热量 (J)，m 为水或溶液的质量 (g)，c 为比热容（水通常取 4.18 J g⁻¹ °C⁻¹），ΔT 为温度变化 (°C)。要计算摩尔焓变，用 ΔH = −q / n，注意放热反应为负值。

q = m c ΔT | ΔH = −q / n

In a typical experiment, 0.0500 mol of a substance is burned, raising the temperature of 200 g of water by 12.5 °C. q = 200 × 4.18 × 12.5 = 10 450 J. ΔH = −10 450 J / 0.0500 mol = −209 000 J mol⁻¹ = −209 kJ mol⁻¹. Common pitfalls include using mass of fuel instead of water, and forgetting unit conversions between J and kJ.

实验中，0.0500 mol 物质燃烧使 200 g 水升温 12.5 °C。q = 200 × 4.18 × 12.5 = 10 450 J。ΔH = −10 450 J / 0.0500 mol = −209 000 J mol⁻¹ = −209 kJ mol⁻¹。常见错误包括误用燃料质量代替水的质量，以及在 J 和 kJ 之间转换出错。

7. Kinematics: Motion in a Straight Line | 运动学：直线运动

Kinematics problems in physics use the SUVAT equations. You need three known quantities to find the fourth. The four equations are: v = u + at, s = ut + ½at², v² = u² + 2as, and s = ½(u + v)t. Always define the positive direction first, and remember that acceleration due to gravity near Earth’s surface is g = 9.81 m s⁻² (downwards).

物理中的运动学问题依赖 SUVAT 方程。需要已知三个量才能求第四个。四个方程为：v = u + at、s = ut + ½at²、v² = u² + 2as 和 s = ½(u + v)t。首先确定正方向，牢记地表重力加速度 g = 9.81 m s⁻²（方向向下）。

v = u + at | s = ut + ½at² | v² = u² + 2as | s = ½(u + v)t

Example: a ball is thrown vertically upwards at 20 m s⁻¹. Find maximum height. Take up as positive, a = −9.81 m s⁻². At the highest point v = 0. Using v² = u² + 2as: 0 = (20)² + 2(−9.81)s → s = 400 / 19.62 = 20.4 m. Always substitute signs carefully; a common mistake is ignoring the negative acceleration when an object moves against gravity.

例题：一球以 20 m s⁻¹ 竖直上抛，求最大高度。取上为正，a = −9.81 m s⁻²，最高点 v = 0。由 v² = u² + 2as：0 = (20)² + 2(−9.81)s → s = 400 / 19.62 = 20.4 m。务必小心代入正负号；常见错误是物体逆重力运动时忽略加速度的负号。

8. Forces, Work, and Energy | 力、功与能量

Newton’s second law, F = ma, is fundamental. Work done is W = Fs cosθ, and kinetic energy is Ek = ½mv². Gravitational potential energy change is ΔEp = mgΔh. In an ideal system, energy is conserved, so loss in Ep = gain in Ek. Use these relationships to solve problems involving inclined planes, falling objects, and braking forces.

牛顿第二定律 F = ma 是基础。做功 W = Fs cosθ，动能 Ek = ½mv²，重力势能变化 ΔEp = mgΔh。理想系统中能量守恒，故 Ep 的减少等于 Ek 的增加。利用这些关系可以解决斜面、落体和刹车力等问题。

F = m a | W = F s cosθ | Ek = ½ m v² | Ep = m g h

A 1200 kg car accelerates from rest to 15 m s⁻¹. The kinetic energy gained is Ek = ½ × 1200 × (15)² = 135 000 J. If the driving force does 180 000 J of work, the efficiency is (useful output/input) × 100 = (135 000/180 000) × 100 = 75%. Always express efficiency as a percentage and link it to energy dissipation by friction or air resistance.

一辆 1200 kg 的汽车从静止加速到 15 m s⁻¹，动能增加 Ek = ½ × 1200 × (15)² = 135 000 J。若驱动力做功 180 000 J，效率 = (135 000/180 000) × 100 = 75%。效率要用百分比表示，并联系到摩擦或空气阻力引起的能量耗散。

9. Electricity and Circuits | 电学与电路

Key relationships include Ohm’s law V = IR, power P = IV = I²R = V²/R, and combinations of resistors. In series, R_total = R₁ + R₂ + …; in parallel, 1/R_total = 1/R₁ + 1/R₂ + … . Pay attention to potential dividers and internal resistance of batteries: terminal p.d. V = ε − Ir, where ε is emf and r is internal resistance.

核心关系包括欧姆定律 V = IR、功率 P = IV = I²R = V²/R 及电阻组合。串联时 R_total = R₁ + R₂ + …；并联时 1/R_total = 1/R₁ + 1/R₂ + …。注意分压器和电池内阻：端电压 V = ε − Ir，其中 ε 为电动势，r 为内阻。

V = I R | P = I V = I² R = V² / R | V = ε − I r

A 12 V battery with internal resistance 0.50 Ω is connected to a 5.0 Ω resistor. Total circuit resistance = 5.0 + 0.50 = 5.5 Ω. Current I = ε / R_total = 12 / 5.5 = 2.18 A. Terminal p.d. = ε − Ir = 12 − (2.18 × 0.50) = 10.9 V. Using V = IR across the external resistor gives 2.18 × 5.0 = 10.9 V as a check. Always double-check that the current is the same throughout a series circuit.

一节 12 V 电池内阻 0.50 Ω，外接 5.0 Ω 电阻。电路总电阻 = 5.0 + 0.50 = 5.5 Ω。电流 I = ε / R_total = 12 / 5.5 = 2.18 A。端电压 V = ε − Ir = 12 − (2.18 × 0.50) = 10.9 V。用外电阻的 V = IR 验证：2.18 × 5.0 = 10.9 V。注意串联电路中电流处处相等，可作交叉检验。

10. Waves: Frequency, Wavelength, Speed | 波：频率、波长、波速

The wave equation v = fλ connects speed (m s⁻¹), frequency (Hz), and wavelength (m). For electromagnetic waves, v = c = 3.00 × 10⁸ m s⁻¹ in a vacuum. Learn to read values directly from graphs: distance between two consecutive crests gives λ; time for one complete oscillation gives period T, then f = 1/T. In ripple tank or sound experiments, measure distances and use timing to determine speed.

波动方程 v = fλ 连接波速 (m s⁻¹)、频率 (Hz) 和波长 (m)。电磁波在真空中 v = c = 3.00 × 10⁸ m s⁻¹。学会从图像直接读取数值：相邻两波峰的距离即为 λ；一次完整振动的时间为周期 T，则 f = 1/T。在波纹槽或声波实验中，测量距离并计时以确定波速。

v = f λ | f = 1 / T

A water wave has a wavelength of 0.40 m and frequency 5.0 Hz. Speed v = 5.0 × 0.40 = 2.0 m s⁻¹. If the same wave enters deeper water and its speed increases to 2.8 m s⁻¹ while frequency remains constant, the new wavelength is λ = v / f = 2.8 / 5.0 = 0.56 m. Remember that frequency depends on the source and does not change when the wave changes medium.

水波波长 0.40 m，频率 5.0 Hz，波速 v = 5.0 × 0.40 = 2.0 m s⁻¹。若同一列波进入较深水域，波速增至 2.8 m s⁻¹，频率不变，则新波长 λ = v / f = 2.8 / 5.0 = 0.56 m。记住频率由波源决定，波进入不同介质时频率保持不变。

11. Radioactive Decay and Half-life | 放射性衰变与半衰期

Half-life (t½) is the time taken for half the radioactive nuclei in a sample to decay. You can solve problems using the decay equation N = N₀ (½)^(t / t½) or by repeatedly halving. Activity A follows the same pattern. If a sample has an initial count rate of 800 counts per minute and a half-life of 3 days, after 9 days the count rate is 800 → 400 → 200 → 100 counts per minute.

半衰期 (t½) 是样品中一半放射性原子核发生衰变所需的时间。可用衰变方程 N = N₀ (½)^(t / t½) 求解，或逐步折半计算。放射性活度 A 也遵循相同规律。若样品初始计数率为 800 次/分，半衰期 3 天，则 9 天后计数率依次衰减为 800 → 400 → 200 → 100 次/分。

N = N₀ (½)^(t / t½) | A = A₀ (½)^(t / t½)

In IB and AQA, you may also need to find half-life from a graph. Locate the time when activity drops to half its initial value; this time interval should be constant for a random decay process. Be comfortable with logarithmic forms if required, but halving methods are often sufficient. Always subtract background count before calculating.

在 IB 和 AQA 考试中，也可能要求从图像求半衰期。找到活度减半所用的时间；对于一个随机衰变过程，该时间间隔应恒定。如需使用对数形式也应掌握，但逐次折半的方法通常够用。计算前务必先扣除本底计数。

12. Data Handling and Graphical Analysis | 数据处理与图像分析

Science calculations are not only about equations; you must extract information from tables and graphs. For a linear graph, gradient = Δy / Δx and equation y = mx + c. Use large triangles to calculate gradient, covering at least half the line. Intercepts can give important physical quantities, such as internal resistance from a V–I graph (y-intercept = ε, gradient = −r).

科学计算不只在于套用公式，还要从表格和图像中提取信息。对于线性图像，斜率 = Δy / Δx，方程形式为 y = mx + c。选取覆盖图像一半以上的大三角形来计算斜率。截距可提供重要的物理量，例如 V–I 图像中 y 截距 = ε，斜率 = −r。

gradient = (y₂ − y₁) / (x₂ − x₁) | y = m x + c

Uncertainty handling is also crucial. For IB, you may report absolute and percentage uncertainty. The uncertainty in a reading is half the smallest scale division; for a digital instrument, it is the last stable digit. When multiplying or dividing quantities, add percentage uncertainties. Practice with data sets from required practicals, such as pendulum periods or extension-load for springs.

不确定度的处理同样重要。在 IB 中，可能需要报告绝对不确定度和百分比不确定度。读数的绝对不确定度取最小分度值的一半；对于数字仪表，则取最后一位稳定数字。当乘除物理量时，将百分比不确定度相加。要多用必做实验的数据来练习，比如单摆周期或弹簧的伸长–负载数据。

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IB AQA Science: Calculation Drill Training | IB AQA 科学：计算题专项训练