📚 SAT Subject Test Math & Chemistry: Common Mistakes and Pitfalls | SAT2 数学与化学：常见易错题汇总
The SAT Subject Tests in Mathematics and Chemistry demand both conceptual understanding and attention to detail. Many high-scoring students lose points not because they lack knowledge, but because they fall into predictable traps. This article compiles the most frequent errors seen on SAT2 Math Level 2 and SAT Chemistry, with paired examples to sharpen your test-taking instincts.
SAT2 数学与化学科目考试既要求对概念的深入理解，也考验对细节的关注。许多高分学生丢分并非因为知识不足，而是落入了出题人预设的陷阱。本文汇总了 SAT2 数学 Level 2 和 SAT 化学中最常见的错误，并提供中英对照的示例，帮助你磨砺应试直觉。
1. Algebra: Sign Reversal and Absolute Value Traps | 代数：符号反转与绝对值陷阱
When solving inequalities, the most common mistake is forgetting to flip the inequality sign after multiplying or dividing by a negative number. For example, given -2x > 6, you must write x < -3, not x > -3.
在解不等式时，最常见的错误是在乘以或除以负数后忘掉反转不等号。例如，对于 -2x > 6，必须写出 x < -3，而不是 x > -3。
Absolute value equations such as |x – 3| = 5 lead to two linear equations: x – 3 = 5 and x – 3 = -5, giving x = 8 and x = -2. Failing to consider the negative branch loses half the solution set.
对于绝对值方程如 |x – 3| = 5，会产生两个一元一次方程：x – 3 = 5 和 x – 3 = -5，解得 x = 8 与 x = -2。不考虑负分支会丢掉一半的解集。
2. Functions: Composition and Domain Errors | 函数：复合与定义域错误
A classic pitfall in function composition is confusing f(g(x)) with g(f(x)). For f(x) = 2x and g(x) = x + 1, f(g(x)) = 2(x + 1) = 2x + 2, whereas g(f(x)) = 2x + 1. The order matters greatly.
复合函数的经典陷阱是把 f(g(x)) 与 g(f(x)) 搞混。若 f(x) = 2x，g(x) = x + 1，则 f(g(x)) = 2(x + 1) = 2x + 2，而 g(f(x)) = 2x + 1，顺序至关重要。
Many students neglect to state the domain of a composite function. Even if f(x) = √x and g(x) = x – 4, the inner function g(x) must keep the radicand non-negative: x – 4 ≥ 0, so x ≥ 4. Simply writing ‘all real numbers’ is incorrect.
许多学生常忽略写出复合函数的定义域。即使 f(x) = √x，g(x) = x – 4，内层函数 g(x) 仍需保证被开方数非负：x – 4 ≥ 0，即 x ≥ 4。简单写成“全体实数”是错误的。
3. Geometry: Similar Triangles and the Pythagorean Theorem | 几何：相似三角形与勾股定理
SAT2 Math frequently tests the recognition of similar triangles in overlapping figures. A common mistake is to assume that segments are proportional without first establishing the correct corresponding vertices. Always check that the angle correspondences match before setting up a proportion.
SAT2 数学经常考查在重叠图形中识别相似三角形。常见错误是没有先确定对应顶点就直接假设线段成比例。在列出比例式之前，务必核对角对应是否一致。
The converse of the Pythagorean Theorem is equally important: if a triangle has sides of lengths 5, 12, and 13, it is a right triangle because 5² + 12² = 13². Many students forget to square the longest side and check equality, leading to wrong classification.
勾股定理的逆定理同样重要：若三角形三边长分别为 5、12、13，因为 5² + 12² = 13²，所以它是直角三角形。很多学生忘记对最长边平方并检验等式，从而错误分类。
4. Trigonometry: Radian Confusion and Identities | 三角学：弧度混淆与恒等式错误
A major source of errors is using degree mode when the problem is in radians. If the question asks for sin(π/6), the answer is 1/2, but if the calculator is in degrees, sin(π/6) ≈ sin(0.5236°) gives a tiny number, which is completely wrong.
一个重大错误源是当题目使用弧度时却在角度模式下计算。若求 sin(π/6)，答案应为 1/2，但若计算器处于角度模式，sin(π/6) ≈ sin(0.5236°) 会得出极小的数，完全错误。
In identities, students mistakenly apply sin(a + b) = sin a + sin b, but the correct identity is sin(a + b) = sin a cos b + cos a sin b. Memorize the exact formulas; the shortcuts do not work.
在恒等式中，学生会错误地使用 sin(a + b) = sin a + sin b，而正确恒等式为 sin(a + b) = sin a cos b + cos a sin b。请熟记精确公式，投机取巧行不通。
5. Probability: ‘At Least One’ and Conditional Pitfalls | 概率：“至少一个”与条件陷阱
For ‘at least one’ problems, using the complement is much safer. For instance, the probability of getting at least one head when flipping a coin three times is 1 – P(all tails) = 1 – (½)³ = 7/8. Direct enumeration often leads to missing cases.
处理“至少一个”的问题时，使用补集更为稳妥。例如，掷一枚硬币三次，至少出现一次正面的概率为 1 – P(全是反面) = 1 – (½)³ = 7/8。直接枚举常常漏掉情况。
Conditional probability problems trip up students who swap the given event. If asked P(A|B), the formula is P(A ∩ B) / P(B). A typical error is to divide by P(A) instead. Read the condition after the vertical bar carefully as the denominator’s event.
条件概率题让那些混淆给发事件的学生栽跟头。若问 P(A|B)，公式为 P(A ∩ B) / P(B)。典型错误是除以 P(A) 而不是 P(B)。请仔细识别竖线后的事件作为分母事件。
6. Data Analysis: Mean, Median, and Standard Deviation Misconceptions | 数据分析：平均数、中位数和标准差的误解
When a data set contains an extreme outlier, the mean shifts dramatically while the median remains robust. For the set {3, 4, 5, 6, 100}, the mean is 23.6, but the median is 5. Students frequently cite the mean as the measure of center without checking for skew.
当数据集包含极端异常值时，平均数会大幅偏移，而中位数保持稳健。对于集合 {3, 4, 5, 6, 100}，平均数为 23.6，而中位数为 5。学生常不加检查偏态就直接引用平均数作为中心度量。
Standard deviation measures spread, not central tendency. Adding a constant to every data point does not change the standard deviation, but multiplying by a constant multiplies the standard deviation by the absolute value of that constant. These transformations are often confused.
标准差衡量离散程度，而非集中趋势。给每个数据点加一个常数不改变标准差，但乘以一个常数会使标准差乘以该常数的绝对值。这些变换经常被混淆。
7. Stoichiometry: Balancing Equations and Mole Ratios | 化学计量：配平方程和摩尔比
One of the most pervasive mistakes in SAT Chemistry is using mole ratios directly from the unbalanced equation. For example, in the reaction H₂ + O₂ → H₂O, an unbalanced ratio would suggest 1 mol H₂ : 1 mol O₂ : 1 mol H₂O, but the balanced equation 2H₂ + O₂ → 2H₂O shows the correct ratio is 2:1:2.
SAT 化学中最普遍的失误之一是直接从未配平的方程中使用摩尔比。例如，在反应 H₂ + O₂ → H₂O 中，未配平的摩尔比暗示为 1:1:1，而配平后的方程 2H₂ + O₂ → 2H₂O 表明正确比例为 2:1:2。
When performing limiting reactant calculations, students often pick the reactant with the smaller mass as the limiting one. However, you must convert masses to moles and compare the molar ratios from the balanced equation. A 10 g sample of a substance with a low molar mass may supply more moles than a 12 g sample of a high-molar-mass reactant.
进行限制试剂计算时，学生常以为质量较小的反应物就是限制试剂。但必须先将质量转换为物质的量，再根据配平方程比较摩尔比。某物质 10 g 若摩尔质量低，可能比摩尔质量高的 12 g 反应物提供更多的摩尔数。
8. Thermochemistry: Sign Conventions and Hess’s Law | 热化学：符号规则与盖斯定律
Exothermic reactions release energy, so the enthalpy change ΔH is negative. Many test-takers confuse this with bond-breaking energies and mistakenly assign a positive ΔH to a combustion reaction that releases heat. Always check the system’s point of view: ΔH < 0 for exothermic.
放热反应释放能量，因此焓变 ΔH 为负。许多考生将此与断键能混淆，错误地对释放热量的燃烧反应赋予正的 ΔH。永远要从体系的角度检查：放热时 ΔH < 0。
When applying Hess’s Law, flipping an equation reverses the sign of its ΔH, and multiplying coefficients scales the ΔH value by the same factor. A common slip is to reverse an equation but keep the original sign, or to scale coefficients without scaling the corresponding enthalpy.
应用盖斯定律时，反转方程会翻转 ΔH 的符号，系数加倍则 ΔH 也要乘以相同倍数。常见的疏漏是反转方程却保留原符号，或只缩放系数但不缩放相应的焓值。
9. Equilibrium: Misapplying Le Chatelier’s Principle | 化学平衡：误用勒夏特列原理
Adding an inert gas at constant volume does not shift the equilibrium position because it does not change the partial pressures of the reacting species. Many students mistakenly believe any addition of gas shifts the equilibrium, forgetting that the principle relies on changes in concentration or partial pressure of reactants/products.
在恒容条件下加入惰性气体不会移动平衡位置，因为它不会改变反应物种的分压。许多学生误以为加入任何气体都会导致平衡移动，忘记了该原理依赖于反应物或产物浓度（分压）的变化。
Catalysts lower activation energy and speed up both forward and reverse reactions equally; they do not shift the equilibrium constant or the equilibrium position. A frequent error is to claim that a catalyst increases the yield of the products by favouring the forward reaction.
催化剂降低活化能，同等加快正逆反应速率；它不改变平衡常数，也不移动平衡位置。常见错误是宣称催化剂通过促进正反应提高了产物的产率。
10. Acids and Bases: pH Calculations and Strong/Weak Identification | 酸碱：pH 计算与强弱识别
A strong acid completely dissociates, so [H⁺] equals the initial acid concentration for a monoprotic acid. With a 0.01 M HCl solution, pH = -log(0.01) = 2. For a weak acid like acetic acid, you must use the Ka expression and an ICE table; assuming full dissociation will overestimate [H⁺] and give a much lower pH.
强酸完全电离，因此对于一元强酸，[H⁺] 等于酸的初始浓度。对于 0.01 M HCl 溶液，pH = -log(0.01) = 2。对于醋酸等弱酸，必须使用 Ka 表达式和 ICE 表格；若假定完全电离，会高估 [H⁺]，得出低得离谱的 pH。
When converting pH to [H⁺], the formula is [H⁺] = 10⁻ᵖᴴ. If pH = 4.7, [H⁺] = 10⁻⁴·⁷ ≈ 2.0 × 10⁻⁵ M. A slip often occurs when students treat pH as a linear scale and estimate [H⁺] as 4.7 × 10⁻⁵, which is wrong by a factor of two.
将 pH 转换为 [H⁺] 时，公式为 [H⁺] = 10⁻ᵖᴴ。若 pH = 4.7，[H⁺] = 10⁻⁴·⁷ ≈ 2.0 × 10⁻⁵ M。常见失误是学生把 pH 当作线性标度，估算 [H⁺] 为 4.7 × 10⁻⁵，这导致了两倍左右的误差。
11. Redox: Assigning Oxidation Numbers and Half-Reactions | 氧化还原：确定氧化数与半反应
The oxidation number of an element in its standard state (like O₂, Na, S₈) is always zero. A frequent mistake is assigning oxygen in O₂ an oxidation number of -2. The rule (-2 for oxygen) only applies in compounds, except in peroxides where it is -1.
单质状态下（如 O₂、Na、S₈）元素的氧化数总为零。常见错误是给 O₂ 中的氧分配 -2 氧化数。规则 (-2 通常用于氧) 只在化合物中适用，而过氧化物中氧为 -1。
In balancing half-reactions, the number of electrons lost in oxidation must equal the number gained in reduction. When combining the two half-equations, some students forget to scale the half-reactions by appropriate factors before adding, resulting in an unbalanced net ionic equation.
在配平半反应时，氧化过程失去的电子数必须等于还原过程得到的电子数。合并两个半方程时，有些学生忘记先乘上合适的系数再加合，导致净离子方程式不平衡。
12. Organic Chemistry: Naming and Functional Group Priority | 有机化学：命名与官能团优先级
When numbering the parent chain, the highest priority functional group should receive the lowest possible number. For example, in 3-hydroxybutanoic acid, the carboxylic acid group takes precedence over the alcohol group. A typical mistake is to name it 2-hydroxybutanoic acid, which violates IUPAC rules.
给主链编号时，优先级最高的官能团应获得尽可能小的编号。例如，在 3-羟基丁酸中，羧基的优先序高于醇羟基。典型错误是将其命名为 2-羟基丁酸，这不符合 IUPAC 规则。
In alkenes and alkynes, the double or triple bond must be included in the longest chain and given the lowest number. For a molecule with a six-carbon chain containing a double bond between C2 and C3, the correct name is hex-2-ene, not hex-4-ene. Students often number from the wrong end.
对于烯烃和炔烃，双键或三键必须包含在最长碳链中并给予最低编号。对于含有一个六碳链且双键位于 C2 与 C3 之间的分子，正确命名为 hex-2-ene，而非 hex-4-ene。学生常从错误的一端开始编号。
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