GCSE AQA Science: Typical Worked Examples | GCSE AQA 科学：典型例题详解

📚 GCSE AQA Science: Typical Worked Examples | GCSE AQA 科学：典型例题详解

This article provides detailed worked examples covering key topics in GCSE AQA Combined Science: Trilogy, including Physics, Chemistry, and Biology. Each example is broken down step by step to help you understand how to apply knowledge and achieve full marks in exams.

本文提供了涵盖 GCSE AQA 综合科学（三部曲）中物理、化学和生物关键主题的详细例题解析。每个例题都逐步分解，帮助你理解如何应用知识并在考试中获得满分。

1. Using Ohm’s Law to Find Resistance | 使用欧姆定律求电阻

A typical question: A resistor has a current of 0.25 A flowing through it when a potential difference of 6.0 V is applied. Calculate the resistance.

典型题目：一个电阻在施加6.0 V电势差时，流过0.25 A的电流。计算该电阻的阻值。

Step 1: Write the equation from the required practical. Ohm’s Law states that potential difference equals current multiplied by resistance, provided temperature is constant:
V = I × R
where V is in volts (V), I in amperes (A), and R in ohms (Ω).

步骤1：写出实验所得的方程。欧姆定律指出，在温度不变的条件下，电势差等于电流乘以电阻：
V = I × R
其中V单位为伏特（V），I为安培（A），R为欧姆（Ω）。

Step 2: Rearrange the equation to make R the subject. Divide both sides by I:
R = V ÷ I

步骤2：重新排列方程，将R作为主体。两边同时除以I：
R = V ÷ I

Step 3: Substitute the values and calculate. R = 6.0 V ÷ 0.25 A = 24 Ω. Remember to include the unit.

步骤3：代入数值计算。R = 6.0 V ÷ 0.25 A = 24 Ω。务必带上单位。

---

2. Calculating Energy Transferred by an Appliance | 计算用电器转移的能量

A common problem: A 60 W lamp is left on for 5 minutes. Calculate the energy transferred. Use the equation E = P × t.

常见问题：一盏60 W的灯开了5分钟。计算转移的能量。使用方程 E = P × t。

Step 1: Convert time to seconds because the watt is joules per second. 5 minutes = 5 × 60 = 300 s.

步骤1：将时间转换为秒，因为瓦特表示焦耳每秒。5分钟 = 5 × 60 = 300秒。

Step 2: Write the formula and substitute:
E = P × t
E = 60 W × 300 s.

步骤2：写出公式并代入：
E = P × t
E = 60 W × 300 s。

Step 3: Calculate: E = 18 000 J (or 18 kJ). Give the answer in joules unless otherwise requested.

步骤3：计算：E = 18 000 J（或18 kJ）。除非另有要求，答案以焦耳给出。

---

3. Mole Calculations Using Mass and Mᵣ | 利用质量和相对分子质量的摩尔计算

Exam question: How many moles are present in 5.85 g of sodium chloride, NaCl? The relative formula mass Mᵣ of NaCl is 58.5.

考题：5.85 g氯化钠（NaCl）中含有多少摩尔？NaCl的相对分子质量Mᵣ为58.5。

Step 1: Recall the mole equation: number of moles (n) = mass (m) ÷ Mᵣ.
n = m ÷ Mᵣ

步骤1：回忆摩尔方程式：摩尔数 (n) = 质量 (m) ÷ 相对分子质量 (Mᵣ)。
n = m ÷ Mᵣ

Step 2: Substitute the values: n = 5.85 g ÷ 58.5 g mol⁻¹ = 0.10 mol. The units of Mᵣ are g mol⁻¹, so moles are correctly obtained.

步骤2：代入数值：n = 5.85 g ÷ 58.5 g mol⁻¹ = 0.10 mol。Mᵣ的单位是g mol⁻¹，因此正确得到摩尔数。

---

4. Predicting Products of Electrolysis | 预测电解产物

Consider the electrolysis of aqueous copper(II) chloride using inert electrodes. Predict the products at the anode and cathode.

考虑使用惰性电极电解氯化铜水溶液。预测阳极和阴极的产物。

Step 1: Identify the ions present in the solution. From CuCl₂: Cu²⁺ and 2Cl⁻. Water also ionises slightly giving H⁺ and OH⁻.

步骤1：识别溶液中存在的离子。来自CuCl₂：Cu²⁺和2Cl⁻。水也会轻微电离产生H⁺和OH⁻。

Step 2: At the cathode (negative electrode), cations are reduced. Copper is less reactive than hydrogen, so Cu²⁺ ions gain electrons: Cu²⁺ + 2e⁻ → Cu (s). Copper metal deposits.

步骤2：在阴极（负极），阳离子被还原。铜不如氢活泼，因此Cu²⁺离子得到电子：Cu²⁺ + 2e⁻ → Cu（固体）。析出铜金属。

Step 3: At the anode (positive electrode), anions are oxidised. Chloride ions are halide ions and are preferentially oxidised over hydroxide ions when concentrated enough: 2Cl⁻ → Cl₂ (g) + 2e⁻. Chlorine gas is produced.

步骤3：在阳极（正极），阴离子被氧化。氯离子是卤素离子，在足够浓时优先于氢氧根被氧化：2Cl⁻ → Cl₂（气体） + 2e⁻。产生氯气。

---

5. Explaining Enzyme Activity vs. Temperature | 解释酶活性与温度的关系

A graph of reaction rate against temperature shows an increase up to an optimum, then a sharp decrease. Explain why the rate falls above 40 °C for many human enzymes.

反应速率对温度的曲线显示在达到最适温度之前上升，然后急剧下降。解释为什么对于许多人体酶，速率在40°C以上会下降。

Step 1: State that enzymes are proteins with a

Published by TutorHao | GCSE Science Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）