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Solutions详解

Chemical Reactions and Equations · Solutions化学反应与方程式 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 25 marksAP 选择题 + 安/卑省考短答 · 共 25 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Evidence of reaction反应证据 · HS-PS1-2 [3 marks][3 分]

A student mixes two clear, colourless solutions. Which observation provides the STRONGEST evidence that a chemical reaction has occurred?学生将两种透明无色溶液混合。以下哪种观察现象为化学反应发生提供了最有力的证据？

Answer:答案：  (B)  A yellow solid forms that cannot be filtered back into either original solution形成黄色固体，无法过滤还原为任一原溶液

Evaluate each option逐项分析 M1·A1·A1

A new substance with different properties (the yellow solid) is the hallmark of a chemical change. Option (B) describes formation of a precipitate from two dissolved reactants; because the solid cannot be redissolved into the original solutions, a new chemical substance has been produced. Options (A), (C), and (D) can all occur without a chemical reaction.生成具有不同性质的新物质（黄色固体）是化学变化的标志。选项 (B) 描述了两种溶质反应生成沉淀；该固体无法溶回原溶液，说明生成了新的化学物质。选项 (A)、(C)、(D) 均可在无化学反应的情况下出现。
Why the distractors fail.干扰项分析。
(A) Slight warming can be caused by mixing without reaction (e.g. dissolving a salt); temperature change alone is not conclusive.轻微变热可由混合本身引起（如溶解某些盐），温度变化本身不足为证。
(C) Volume increase is a physical mixing effect, not evidence of a new substance.体积增大是物理混合效应，不能说明有新物质生成。
(D) Cloudiness that clears on stirring suggests a temporary physical dispersion, not a new substance.搅拌后变清的混浊是暂时的物理分散，不是化学反应。
Formation of a new substance is the defining criterion for a chemical reaction.生成新物质是化学反应的判断标准。 The six classic indicators are: precipitate formation, gas production, colour change, temperature change (energy released or absorbed), light emission, and odour change. Of these, only the first three are unambiguous on their own because they confirm a new substance with new physical properties. Temperature change and odour change can accompany physical processes (dissolving, evaporation), so they are supporting evidence rather than proof. When the question asks for the STRONGEST single piece of evidence, look for formation of a substance that cannot be reversed by simple physical means.化学反应的六大经典证据：生成沉淀、产生气体、颜色改变、温度变化（放热或吸热）、发光、气味改变。其中，前三条本身已能确认存在具有新物理性质的新物质，是最直接的证据。温度变化和气味改变也可伴随物理过程（溶解、蒸发），因此只能作为辅助证据而非确证。题目问最有力的单一证据时，应寻找无法通过简单物理手段逆转的新物质生成。
Q2EASY易 🇨🇦 ON安 AP-style MCQAP 风格选择题 §2 Balancing equations配平方程式 · SCH3U C2.2 [3 marks][3 分]

Which is the correctly balanced equation for aluminium reacting with oxygen to form $\text{Al}_2\text{O}_3$?下列哪个是铝与氧气反应生成 $\text{Al}_2\text{O}_3$ 的正确配平方程式？

Answer:答案：  (C)  $4\,\text{Al(s)} + 3\,\text{O}_2\text{(g)} \to 2\,\text{Al}_2\text{O}_3\text{(s)}$

Atom count verification for option (C)验证选项 (C) 的原子守恒 M1·A1·A1

Left side: 4 Al atoms, $3 \times 2 = 6$ O atoms. Right side: $2 \times 2 = 4$ Al atoms, $2 \times 3 = 6$ O atoms. Both sides balance. $\checkmark$左侧：4 个 Al 原子，$3 \times 2 = 6$ 个 O 原子。右侧：$2 \times 2 = 4$ 个 Al 原子，$2 \times 3 = 6$ 个 O 原子。两边均衡。$\checkmark$
Why the distractors fail.干扰项分析。
(A) 1 Al and 2 O left vs 2 Al and 3 O right: unbalanced on both elements.左边 1 Al、2 O，右边 2 Al、3 O：两种元素均不平衡。
(B) 2 Al and 2 O left vs 2 Al and 3 O right: Al balances but O does not.左边 2 Al、2 O，右边 2 Al、3 O：Al 平衡但 O 不平衡。
(D) 2 Al and 6 O left vs 2 Al and 3 O right: Al balances but O does not (extra O left).左边 2 Al、6 O，右边 2 Al、3 O：Al 平衡但 O 不平衡（左边 O 多余）。
Balancing Al with O requires the LCM of their valences: Al is +3, O is -2, LCM(3,2) = 6.配平铝和氧需要用到化合价的最小公倍数：Al 为 +3，O 为 -2，LCM(3,2) = 6。 Each Al atom contributes 3 bonds; each O atom accepts 2 bonds. To satisfy both, you need 4 Al atoms (giving 12 bond-equivalents) and 6 O atoms (also 12 bond-equivalents), which package into 2 formula units of Al$_2$O$_3$. The LCM approach avoids trial and error: find LCM(3, 2) = 6; that means 6 O in total, so $3\,\text{O}_2$ molecules; and 4 Al to balance. This is also a synthesis reaction ($\text{element} + \text{element} \to \text{compound}$).每个 Al 原子贡献 3 个化合键等价量，每个 O 原子接受 2 个。为同时满足两者，需要 4 个 Al 原子（共 12 个化合键等价量）和 6 个 O 原子（同样 12 个），封装成 2 个 Al$_2$O$_3$ 化学式单位。最小公倍数法可避免反复试凑：LCM(3,2) = 6，即共 6 个 O 原子，写成 $3\,\text{O}_2$；再配 4 个 Al 即可平衡。这同时也是一个化合反应（元素 + 元素 → 化合物）。
Q3EASY易 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Synthesis and decomposition化合与分解 · Science 10 [5 marks][5 分]

Calcium carbonate, $\text{CaCO}_3$, decomposes when strongly heated.碳酸钙 $\text{CaCO}_3$ 在强热下分解。

Answer:答案：  (a) $\text{CaCO}_3\text{(s)} \to \text{CaO(s)} + \text{CO}_2\text{(g)}$  · (b) calcium oxide; carbon dioxide氧化钙；二氧化碳  · (c) gas bubbles / CO$_2$ produced产生气泡 / 生成 CO$_2$

(a) Balanced decomposition equation with state symbols含状态符号的配平分解方程式 M1·A1

$$\text{CaCO}_3\text{(s)} \xrightarrow{\Delta} \text{CaO(s)} + \text{CO}_2\text{(g)}$$ Atom check: Ca 1=1, C 1=1, O 3=1+2. $\checkmark$ State symbols: solid reactant, solid CaO product, gaseous CO$_2$ product.原子验证：Ca 1=1，C 1=1，O 3=1+2。$\checkmark$ 状态符号：固体反应物，固体产物 CaO，气体产物 CO$_2$。

(b) Name the two products命名两种产物 A1·A1

Product 1: calcium oxide (CaO, also called quicklime). Product 2: carbon dioxide (CO$_2$).产物 1：氧化钙（CaO，又称生石灰）。产物 2：二氧化碳（CO$_2$）。

(c) Observable evidence可观察证据 A1

Gas bubbles (effervescence) are produced as CO$_2$ escapes. Alternatively: the white solid gradually decreases in mass (loss of CO$_2$); or the residual solid (CaO) turns milky white and may glow (incandescence at high temperature).产生气泡（逸出 CO$_2$）。或：白色固体质量逐渐减小（失去 CO$_2$）；或残余固体（CaO）变为石灰白色，高温时可能发光（白热现象）。
Thermal decomposition of carbonates is a cornerstone reaction in both industry and geology.碳酸盐的热分解是工业和地质学的基础反应。 This reaction (calcination) is how lime (CaO) is manufactured commercially: limestone (CaCO$_3$) is heated above 840 degrees C in a kiln. It is also the reverse of the mineralisation process by which organisms build shells. In an exam, always include the $\Delta$ (heat) symbol above the reaction arrow when the reaction requires heating. The pattern for all metal carbonate decompositions is: $\text{MCO}_3 \to \text{MO} + \text{CO}_2$. The state symbols matter: CaO is solid and CO$_2$ is gaseous at reaction temperature.该反应（煅烧）是工业制石灰（CaO）的方法：将石灰石（CaCO$_3$）在窑炉中加热到 840 摄氏度以上。它也是生物体生成贝壳的矿化过程的逆过程。考试中，当反应需要加热时，务必在反应箭头上方标注 $\Delta$ 符号。所有金属碳酸盐分解的通式为：$\text{MCO}_3 \to \text{MO} + \text{CO}_2$。状态符号不可忽略：在反应温度下 CaO 为固体，CO$_2$ 为气体。
Q4MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §4 Single replacement单置换 · Chem 20 D GO1 [6 marks][6 分]

Zinc metal is placed into copper(II) sulfate solution. A red-brown solid forms and the blue colour fades.锌金属放入硫酸铜溶液中，形成红棕色固体，蓝色褪去。

Answer:答案：  (a) $\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \to \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$  · (b) single replacement; A + BC单置换；A + BC $\to$ AC + BAC + B  · (c) Zn is above Cu in activity seriesZn 在活动性顺序中排于 Cu 上方

(a) Balanced molecular equation with state symbols含状态符号的配平分子方程式 M1·A1

$$\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \to \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$$ Atom check: Zn 1=1, Cu 1=1, S 1=1, O 4=4. $\checkmark$ The red-brown solid is copper metal; the blue $\text{Cu}^{2+}$ ions are consumed, so the solution fades.原子验证：Zn 1=1，Cu 1=1，S 1=1，O 4=4。$\checkmark$ 红棕色固体为铜金属；蓝色 $\text{Cu}^{2+}$ 离子被消耗，故溶液褪色。

(b) Reaction type and general pattern反应类型和通式 A1·A1

This is a single replacement (single displacement) reaction. General pattern: $\text{A} + \text{BC} \to \text{AC} + \text{B}$, where a more active element A displaces a less active element B from a compound BC.这是单置换（单取代）反应。通式：$\text{A} + \text{BC} \to \text{AC} + \text{B}$，其中活动性较强的元素 A 将活动性较弱的元素 B 从化合物 BC 中置换出来。

(c) Activity series explanation活动性顺序解释 M1·A1

In the standard activity series, Zn appears above Cu, meaning Zn is a stronger reducing agent (loses electrons more readily). Zn can therefore displace Cu$^{2+}$ from solution: $\text{Zn} \to \text{Zn}^{2+} + 2e^-$ (oxidation), $\text{Cu}^{2+} + 2e^- \to \text{Cu}$ (reduction). If Cu were placed in ZnSO$_4$ solution, Cu is below Zn in the activity series and cannot reduce Zn$^{2+}$; no reaction would occur (NR).在标准活动性顺序中，Zn 排在 Cu 上方，说明 Zn 是更强的还原剂（更容易失去电子）。因此 Zn 能将 Cu$^{2+}$ 从溶液中置换出来：$\text{Zn} \to \text{Zn}^{2+} + 2e^-$（氧化），$\text{Cu}^{2+} + 2e^- \to \text{Cu}$（还原）。若将 Cu 放入 ZnSO$_4$ 溶液中，由于 Cu 排在 Zn 下方，无法还原 Zn$^{2+}$，故不发生反应（NR）。
Single replacement reactions are spontaneous only when the free element is higher in the activity series than the ion it displaces.单置换反应仅在游离元素的活动性高于被置换离子时才能自发发生。 The activity series (from most to least active for common metals): K, Ca, Na, Mg, Al, Zn, Fe, Ni, Sn, Pb, H$_2$, Cu, Hg, Ag, Au. A metal can displace any ion below it in the list. The Zn/Cu pair is the classic galvanic cell demonstration: this reaction releases electrical energy in a voltaic cell (standard cell potential $+1.10$ V). The colour change from blue to colourless as Cu$^{2+}$ is consumed, combined with the red-brown copper deposit, provides two visible pieces of evidence that the reaction is occurring.常见金属活动性顺序（从强到弱）：K、Ca、Na、Mg、Al、Zn、Fe、Ni、Sn、Pb、H$_2$、Cu、Hg、Ag、Au。金属可置换顺序表中排在其下方的任何离子。Zn/Cu 组合是经典原电池演示实验：该反应在伏打电池中释放电能（标准电池电势 $+1.10$ V）。溶液由蓝色变为无色（Cu$^{2+}$ 被消耗）并伴随红棕色铜的沉积，为反应正在进行提供了两个可见证据。
Q5MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §5 Combustion燃烧 · HS-PS1-7 [8 marks][8 分]

Propane, $\text{C}_3\text{H}_8$, undergoes complete combustion in excess oxygen.丙烷 $\text{C}_3\text{H}_8$ 在过量氧气中完全燃烧。

Answer:答案：  (a) $\text{C}_3\text{H}_8\text{(g)} + 5\,\text{O}_2\text{(g)} \to 3\,\text{CO}_2\text{(g)} + 4\,\text{H}_2\text{O(g)}$  · (b) CO$_2$ and H$_2$OCO$_2$ 和 H$_2$O  · (c) CO (carbon monoxide); toxic air pollutantCO（一氧化碳）；有毒空气污染物

(a) Balanced combustion equation with state symbols含状态符号的配平燃烧方程式 M1·A1·A1

$$\text{C}_3\text{H}_8\text{(g)} + 5\,\text{O}_2\text{(g)} \to 3\,\text{CO}_2\text{(g)} + 4\,\text{H}_2\text{O(g)}$$ Strategy: balance C first (3 on left, so 3 CO$_2$ on right), then H (8 on left, so 4 H$_2$O on right), then O (3$\times$2 + 4$\times$1 = 10 O on right, so 5 O$_2$ on left). Check: left 3C 8H 10O = right 3C 8H 10O. $\checkmark$策略：先配平 C（左边 3 个，故右边 3 个 CO$_2$），再配平 H（左边 8 个，故右边 4 个 H$_2$O），最后配平 O（右边 $3\times2 + 4\times1 = 10$ 个 O，故左边 5 个 O$_2$）。验证：左边 3C 8H 10O = 右边 3C 8H 10O。$\checkmark$

(b) Products of complete combustion of any hydrocarbon任何烃类完全燃烧的产物 A1·A1

Carbon dioxide, CO$_2$ (from all carbon atoms) and water, H$_2$O (from all hydrogen atoms). These are the only products when combustion is complete and oxygen is in excess.二氧化碳 CO$_2$（来自所有碳原子）和水 H$_2$O（来自所有氢原子）。氧气充足、燃烧完全时，这是唯一的两种产物。

(c) Product of incomplete combustion and environmental concern不完全燃烧产物及环境危害 M1·A1·A1

Carbon monoxide (CO) forms when insufficient oxygen is available and carbon is only partially oxidised. Environmental concern: CO is a colourless, odourless, highly toxic gas; it binds to haemoglobin 200 times more strongly than O$_2$, preventing oxygen transport and causing poisoning or death. It is also a contributor to smog formation. (Unburned carbon / soot is also an acceptable answer: fine particulate matter harms respiratory health.)一氧化碳（CO）在氧气不足、碳仅被部分氧化时生成。环境危害：CO 是无色无味的高毒气体，与血红蛋白的结合能力是 O$_2$ 的 200 倍，阻碍氧气输运，可导致中毒甚至死亡。CO 也是烟雾形成的成因之一。（未燃烧的碳/煤烟也可接受：细颗粒物危害呼吸系统健康。）
Balance hydrocarbon combustion systematically: C first, H second, O last.系统地配平烃的燃烧方程式：先配 C，再配 H，最后配 O。 For a general hydrocarbon $\text{C}_x\text{H}_y$: $\text{C}_x\text{H}_y + (x + y/4)\,\text{O}_2 \to x\,\text{CO}_2 + (y/2)\,\text{H}_2\text{O}$. For propane ($x=3, y=8$): coefficient of O$_2$ = $3 + 8/4 = 3 + 2 = 5$. If this gives a fractional coefficient, multiply all coefficients by 2. The reason O is balanced last is that it appears in both products (CO$_2$ and H$_2$O), so fixing C and H first determines the oxygen demand exactly. This systematic approach works for all complete combustion problems at this level.对于通式 $\text{C}_x\text{H}_y$ 的烃：$\text{C}_x\text{H}_y + (x + y/4)\,\text{O}_2 \to x\,\text{CO}_2 + (y/2)\,\text{H}_2\text{O}$。对丙烷（$x=3, y=8$）：O$_2$ 系数 $= 3 + 8/4 = 3 + 2 = 5$。若系数出现分数，将所有系数乘以 2。O 最后配平是因为它出现在两种产物中（CO$_2$ 和 H$_2$O），先固定 C 和 H 才能精确确定耗氧量。该系统方法适用于本阶段所有完全燃烧题目。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6EASY易 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §2 Balancing (multi-step)配平（多步） · HS-PS1-7 [7 marks][7 分]

Balance each equation by inspection and show your atom count check.用观察法配平各方程式，并写出原子计数核验。

Answer:答案：  (a) $4\,\text{Fe(s)} + 3\,\text{O}_2\text{(g)} \to 2\,\text{Fe}_2\text{O}_3\text{(s)}$  · (b) $2\,\text{Na(s)} + 2\,\text{H}_2\text{O(l)} \to 2\,\text{NaOH(aq)} + \text{H}_2\text{(g)}$  · (c) $2\,\text{C}_4\text{H}_{10}\text{(g)} + 13\,\text{O}_2\text{(g)} \to 8\,\text{CO}_2\text{(g)} + 10\,\text{H}_2\text{O(g)}$

(a) Balancing iron combustion配平铁的燃烧 M1·A1

Target: $\text{Fe}_2\text{O}_3$ has 2 Fe and 3 O. LCM(2,3) = 6, so need 6 O (i.e., $3\,\text{O}_2$) and 4 Fe (to give $2\,\text{Fe}_2\text{O}_3$).目标：$\text{Fe}_2\text{O}_3$ 含 2 Fe 和 3 O。LCM(2,3) = 6，故需 6 个 O（即 $3\,\text{O}_2$）和 4 个 Fe（生成 $2\,\text{Fe}_2\text{O}_3$）。 $$4\,\text{Fe(s)} + 3\,\text{O}_2\text{(g)} \to 2\,\text{Fe}_2\text{O}_3\text{(s)}$$ Check: Fe $4 = 4\,\checkmark$; O $6 = 6\,\checkmark$.验证：Fe $4 = 4\,\checkmark$；O $6 = 6\,\checkmark$。

(b) Balancing sodium and water配平钠与水的反应 M1·A1

Start with 2 Na. Each Na produces 1 NaOH, so 2 NaOH. The 2 H$_2$O supply 4 H; 2 go into NaOH, 2 remain, forming 1 H$_2$.从 2 个 Na 开始。每个 Na 生成 1 个 NaOH，即 2 NaOH。2 个 H$_2$O 共提供 4 个 H；2 个进入 NaOH，剩余 2 个生成 1 个 H$_2$。 $$2\,\text{Na(s)} + 2\,\text{H}_2\text{O(l)} \to 2\,\text{NaOH(aq)} + \text{H}_2\text{(g)}$$ Check: Na $2=2\,\checkmark$; H $4=2+2=4\,\checkmark$; O $2=2\,\checkmark$.验证：Na $2=2\,\checkmark$；H $4=2+2=4\,\checkmark$；O $2=2\,\checkmark$。

(c) Balancing butane combustion配平丁烷的燃烧 M1·A1·A1

Using $\text{C}_x\text{H}_y$ formula with $x=4, y=10$: O$_2$ coefficient $= 4 + 10/4 = 6.5$. Fractional, so multiply all by 2:用 $\text{C}_x\text{H}_y$ 通式，$x=4, y=10$：O$_2$ 系数 $= 4 + 10/4 = 6.5$。系数为分数，故全部乘以 2： $$2\,\text{C}_4\text{H}_{10}\text{(g)} + 13\,\text{O}_2\text{(g)} \to 8\,\text{CO}_2\text{(g)} + 10\,\text{H}_2\text{O(g)}$$ Check: C $8=8\,\checkmark$; H $20=20\,\checkmark$; O $26 = 16+10 = 26\,\checkmark$.验证：C $8=8\,\checkmark$；H $20=20\,\checkmark$；O $26 = 16+10 = 26\,\checkmark$。
Always write the atom-count check after balancing; it is a required step in provincial and AP marking.配平后务必写出原子计数核验；这在省考和 AP 评分中是必要步骤。 The LCM strategy generalises: whenever a product formula contains two elements in a ratio $p:q$ and neither element appears in another product, you need a coefficient of $\text{LCM}(p,q)/p$ in front of that product. For (a), Fe$_2$O$_3$ has ratio 2:3, so LCM(2,3)=6 total oxygen, requiring 2 formula units of Fe$_2$O$_3$ and 4 Fe. For (c), fractional O$_2$ coefficients are common with odd-carbon or odd-hydrogen hydrocarbons; always scale up to integers. Reactions (a) and (b) are also single-replacement or redox if framed at a higher level.LCM 策略具有普遍性：当某产物的化学式中两种元素之比为 $p:q$，且这两种元素均不出现在其他产物中时，该产物前的系数应为 $\text{LCM}(p,q)/p$。对 (a)，Fe$_2$O$_3$ 的比值为 2:3，LCM(2,3)=6 个 O，需 2 个 Fe$_2$O$_3$ 化学式单位和 4 个 Fe。对 (c)，奇数碳或奇数氢的烃类常出现分数 O$_2$ 系数，必须放大为整数。从更高层次看，(a) 和 (b) 也可归类为单置换或氧化还原反应。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §6 Solubility and precipitation溶解度与沉淀 · SCH3U C3.2 [8 marks][8 分]

Lead(II) nitrate solution mixed with potassium iodide solution; bright yellow precipitate forms.硝酸铅溶液与碘化钾溶液混合，立即生成亮黄色沉淀。

Answer:答案：  (a) lead(II) iodide, PbI$_2$(s)碘化铅，PbI$_2$(s)  · (b) $\text{Pb(NO}_3)_2\text{(aq)} + 2\,\text{KI(aq)} \to \text{PbI}_2\text{(s)} + 2\,\text{KNO}_3\text{(aq)}$  · (c) $\text{Pb}^{2+}\text{(aq)} + 2\,\text{NO}_3^-\text{(aq)} + 2\,\text{K}^+\text{(aq)} + 2\,\text{I}^-\text{(aq)} \to \text{PbI}_2\text{(s)} + 2\,\text{K}^+\text{(aq)} + 2\,\text{NO}_3^-\text{(aq)}$  · (d) spectators: K$^+$, NO$_3^-$; net: Pb$^{2+}$(aq) + 2I$^-$(aq)旁观离子：K$^+$、NO$_3^-$；净离子：Pb$^{2+}$(aq) + 2I$^-$(aq) $\to$ $\text{PbI}_2\text{(s)}$

(a) Identify the precipitate using solubility rules利用溶解度规则确定沉淀 M1·A1

The two possible exchange products are PbI$_2$ and KNO$_3$. By solubility rules: KNO$_3$ is soluble (all nitrates are soluble; all potassium salts are soluble). PbI$_2$ is insoluble (lead(II) halides except fluoride are insoluble). Therefore the precipitate is lead(II) iodide, PbI$_2$(s), which is bright yellow.两种可能的交换产物为 PbI$_2$ 和 KNO$_3$。根据溶解度规则：KNO$_3$ 可溶（所有硝酸盐可溶；所有钾盐可溶）。PbI$_2$ 不溶（铅(II)卤化物除氟化物外均不溶）。因此沉淀为碘化铅 PbI$_2$(s)，呈亮黄色。

(b) Balanced molecular equation with state symbols含状态符号的配平分子方程式 M1·A1

$$\text{Pb(NO}_3)_2\text{(aq)} + 2\,\text{KI(aq)} \to \text{PbI}_2\text{(s)} + 2\,\text{KNO}_3\text{(aq)}$$ Check: Pb $1=1\,\checkmark$; N $2=2\,\checkmark$; O $6=6\,\checkmark$; K $2=2\,\checkmark$; I $2=2\,\checkmark$.验证：Pb $1=1\,\checkmark$；N $2=2\,\checkmark$；O $6=6\,\checkmark$；K $2=2\,\checkmark$；I $2=2\,\checkmark$。

(c) Complete ionic equation完全离子方程式 M1·A1

Expand all aqueous soluble compounds into their component ions; keep the solid precipitate intact:将所有可溶水溶液化合物展开为离子；保留固体沉淀不拆分： $$\text{Pb}^{2+}\text{(aq)} + 2\,\text{NO}_3^-\text{(aq)} + 2\,\text{K}^+\text{(aq)} + 2\,\text{I}^-\text{(aq)} \to \text{PbI}_2\text{(s)} + 2\,\text{K}^+\text{(aq)} + 2\,\text{NO}_3^-\text{(aq)}$$

(d) Net ionic equation净离子方程式 M1·A1

Cancel spectator ions (K$^+$ and NO$_3^-$ appear on both sides unchanged):消去旁观离子（K$^+$ 和 NO$_3^-$ 两侧相同，消去）： $$\text{Pb}^{2+}\text{(aq)} + 2\,\text{I}^-\text{(aq)} \to \text{PbI}_2\text{(s)}$$
The net ionic equation reveals the essential chemistry; the spectator ions are just along for the ride.净离子方程式揭示了反应的本质；旁观离子只是"搭便车"。 The same net ionic equation $\text{Pb}^{2+}\text{(aq)} + 2\,\text{I}^-\text{(aq)} \to \text{PbI}_2\text{(s)}$ would result from mixing any lead(II) salt solution with any iodide salt solution, regardless of what the spectator counter-ions are. This is why net ionic equations are the most compact and general form. On AP Chemistry and Ontario Grade 12 exams, writing only the molecular equation when the question asks for the net ionic earns zero marks for that part. The stepwise procedure is: (1) write the balanced molecular equation, (2) dissociate all strong electrolytes, (3) cancel spectators. Solids, weak acids, and gases are never dissociated.相同的净离子方程式 $\text{Pb}^{2+}\text{(aq)} + 2\,\text{I}^-\text{(aq)} \to \text{PbI}_2\text{(s)}$ 适用于任何铅(II)盐溶液与任何碘化物溶液的混合，无论旁观离子为何。这正是净离子方程式最简洁、最通用的原因。在 AP 化学和安大略省 12 年级考试中，当题目要求净离子方程式时，仅写分子方程式该部分得零分。步骤：(1) 写出配平的分子方程式，(2) 将所有强电解质拆分为离子，(3) 消去旁观离子。固体、弱酸和气体绝不拆分。
Q8MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Double replacement双置换 · Science 10 [7 marks][7 分]

Sodium sulfate solution combined with barium chloride solution.硫酸钠溶液与氯化钡溶液混合。

Answer:答案：  (a) yes, BaSO$_4$(s) precipitates是，生成 BaSO$_4$(s) 沉淀  · (b) $\text{Na}_2\text{SO}_4\text{(aq)} + \text{BaCl}_2\text{(aq)} \to \text{BaSO}_4\text{(s)} + 2\,\text{NaCl(aq)}$  · (c) $\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \to \text{BaSO}_4\text{(s)}$  · (d) double replacement (double displacement / metathesis)双置换（复分解）反应

(a) Predict precipitate using solubility rules利用溶解度规则预测沉淀 M1·A1

Possible exchange products: BaSO$_4$ and NaCl. Solubility check: NaCl is soluble (all sodium salts are soluble; all chlorides are soluble except Ag$^+$, Pb$^{2+}$, Hg$_2^{2+}$). BaSO$_4$ is insoluble (barium sulfate is one of the most insoluble salts; sulfates of Ba, Pb, Ca, and Hg are insoluble). Therefore a precipitate forms: barium sulfate, BaSO$_4$(s), a dense white solid.可能的交换产物：BaSO$_4$ 和 NaCl。溶解性检验：NaCl 可溶（所有钠盐可溶；所有氯化物可溶，除 Ag$^+$、Pb$^{2+}$、Hg$_2^{2+}$ 外）。BaSO$_4$ 不溶（硫酸钡是最难溶的盐之一；Ba、Pb、Ca 和 Hg 的硫酸盐不溶）。因此生成沉淀：硫酸钡 BaSO$_4$(s)，为致密白色固体。

(b) Balanced molecular equation with state symbols含状态符号的配平分子方程式 M1·A1

$$\text{Na}_2\text{SO}_4\text{(aq)} + \text{BaCl}_2\text{(aq)} \to \text{BaSO}_4\text{(s)} + 2\,\text{NaCl(aq)}$$ Check: Na $2=2\,\checkmark$; S $1=1\,\checkmark$; O $4=4\,\checkmark$; Ba $1=1\,\checkmark$; Cl $2=2\,\checkmark$.验证：Na $2=2\,\checkmark$；S $1=1\,\checkmark$；O $4=4\,\checkmark$；Ba $1=1\,\checkmark$；Cl $2=2\,\checkmark$。

(c) Net ionic equation净离子方程式 M1·A1

Spectator ions are Na$^+$ and Cl$^-$ (both present unchanged on both sides). Cancel them:旁观离子为 Na$^+$ 和 Cl$^-$（两侧均相同，消去）： $$\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \to \text{BaSO}_4\text{(s)}$$

(d) Reaction type反应类型 A1

This is a double replacement (double displacement, or metathesis) reaction. General pattern: AB + CD $\to$ AD + CB, where the cations exchange partners. Both products must be identified (one precipitate, one soluble salt) to confirm the classification.这是双置换（复分解）反应。通式：AB + CD $\to$ AD + CB，两种阳离子互换配对。需确认两种产物（一种沉淀，一种可溶盐）才能确认分类。
BaSO$_4$ precipitation is used clinically as a "barium meal" for gastrointestinal X-ray imaging.BaSO$_4$ 沉淀反应在临床上用作胃肠道 X 射线造影的"钡餐"。 Barium sulfate is so insoluble ($K_{sp} = 1.1 \times 10^{-10}$) that swallowing a suspension of it is safe even though soluble barium salts are toxic. The white BaSO$_4$ is opaque to X-rays, outlining the digestive tract. This is a real-world context that connects the solubility rules to medicine. Double replacement reactions driven by precipitate formation, gas evolution, or water formation are collectively called metathesis reactions; the driving force is the removal of ions from solution to a more stable phase.硫酸钡溶解度极低（$K_{sp} = 1.1 \times 10^{-10}$），即使可溶钡盐有毒，吞服其悬浮液也是安全的。白色 BaSO$_4$ 对 X 射线不透明，可勾勒出消化道轮廓。这是将溶解度规则与医学联系起来的真实案例。由沉淀生成、气体逸出或水生成驱动的双置换反应统称为复分解反应；其驱动力是离子从溶液转移到更稳定相态。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Net ionic equations (synthesis)净离子方程式（综合） · HS-PS1-2 [8 marks][8 分]

For each pair of aqueous solutions, predict whether a reaction occurs and write all three equations, or write NR and explain.对于每对水溶液，预测是否反应，书写三种方程式，或写 NR 并解释。

Answer:答案：  (a) Reaction occurs; net: Ag$^+$(aq) + Cl$^-$(aq)发生反应；净离子：Ag$^+$(aq) + Cl$^-$(aq) $\to$ AgCl(s)  · (b) NR; all possible products (KCl, NaNO$_3$) are solubleNR；所有可能产物（KCl、NaNO$_3$）均可溶

(a) AgNO$_3$(aq) + NaCl(aq)AgNO$_3$(aq) + NaCl(aq) M1·A1·M1·A1

Possible products: AgCl and NaNO$_3$. By solubility rules: NaNO$_3$ is soluble (all nitrates and all sodium salts are soluble). AgCl is insoluble (silver chloride is the classic insoluble halide; all Ag halides except AgF are insoluble). Reaction occurs.可能的产物：AgCl 和 NaNO$_3$。溶解度规则：NaNO$_3$ 可溶（所有硝酸盐和钠盐可溶）。AgCl 不溶（氯化银是经典难溶卤化物；除 AgF 外所有 Ag 卤化物均难溶）。反应发生。

Molecular equation:分子方程式：

$$\text{AgNO}_3\text{(aq)} + \text{NaCl(aq)} \to \text{AgCl(s)} + \text{NaNO}_3\text{(aq)}$$

Complete ionic equation:完全离子方程式：

$$\text{Ag}^+\text{(aq)} + \text{NO}_3^-\text{(aq)} + \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \to \text{AgCl(s)} + \text{Na}^+\text{(aq)} + \text{NO}_3^-\text{(aq)}$$

Net ionic equation (cancel Na$^+$ and NO$_3^-$ spectators):净离子方程式（消去旁观离子 Na$^+$ 和 NO$_3^-$）：

$$\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \to \text{AgCl(s)}$$

(b) KNO$_3$(aq) + NaCl(aq)KNO$_3$(aq) + NaCl(aq) M1·A1·M1·A1

Possible exchange products: KCl and NaNO$_3$. By solubility rules: KCl is soluble (all potassium salts are soluble; all chlorides except Ag$^+$, Pb$^{2+}$, Hg$_2^{2+}$ are soluble). NaNO$_3$ is soluble (all nitrates and sodium salts are soluble). No precipitate, gas, or water forms. All four ions remain in solution unchanged. NR.可能的交换产物：KCl 和 NaNO$_3$。溶解度规则：KCl 可溶（所有钾盐可溶；除 Ag$^+$、Pb$^{2+}$、Hg$_2^{2+}$ 外所有氯化物可溶）。NaNO$_3$ 可溶（所有硝酸盐和钠盐可溶）。无沉淀、气体或水生成。四种离子均保持不变留在溶液中。NR。

Explanation: because all possible products are soluble, there is no thermodynamic driving force to remove any ion pair from solution. The four ions (K$^+$, NO$_3^-$, Na$^+$, Cl$^-$) are all spectators; the "net ionic equation" would be blank (no reaction).解释：由于所有可能产物均可溶，没有热力学驱动力将任何离子对从溶液中移除。四种离子（K$^+$、NO$_3^-$、Na$^+$、Cl$^-$）均为旁观离子；"净离子方程式"为空（无反应）。

A reaction in solution needs a driving force: precipitate, gas, or weak electrolyte (water). Without one, NR.溶液中的反应需要驱动力：沉淀、气体或弱电解质（水）。没有驱动力则 NR。 This is the most important principle for predicting ionic reactions in solution. When evaluating any ion-combination question, systematically identify the two possible exchange products and check each against solubility rules (and whether a gas or water would form). Only if at least one product is insoluble, volatile, or a weak electrolyte does a reaction occur. The AgCl test in (a) is the classic qualitative test for Cl$^-$ ions: a white precipitate that turns purple in sunlight (photodecomposition) is diagnostic. In (b), the ions simply co-exist in solution; the mixture is indistinguishable from a solution of the same ions prepared any other way.这是预测溶液中离子反应最重要的原理。评估任何离子组合题时，系统地确定两种可能的交换产物，对照溶解度规则逐一检验（以及是否会生成气体或水）。只有当至少一种产物不溶、挥发或为弱电解质时，反应才会发生。(a) 中的 AgCl 检验是检测 Cl$^-$ 离子的经典定性实验：白色沉淀在阳光下变紫（光分解）为其特征。在 (b) 中，离子只是在溶液中共存；该混合物与用任何其他方式制备的相同离子溶液无从区分。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 26 marks阿省毕业考 + 通用题型 · 共 26 分

Section C · Worked SolutionsC 部分 · 详细解答

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §3 + §5 Reaction types (applied)反应类型（应用） · Chem 20 D GO1 [8 marks][8 分]

Four lab observations: classify the reaction type and write the balanced equation for each.四种实验室观察：各确定反应类型并书写配平方程式。

Answer:答案：  (a) synthesis; $2\,\text{Mg(s)} + \text{O}_2\text{(g)} \to 2\,\text{MgO(s)}$化合；$2\,\text{Mg(s)} + \text{O}_2\text{(g)} \to 2\,\text{MgO(s)}$  · (b) decomposition; $2\,\text{H}_2\text{O}_2\text{(aq)} \to 2\,\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}$分解；$2\,\text{H}_2\text{O}_2\text{(aq)} \to 2\,\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}$  · (c) single replacement; $\text{Fe(s)} + \text{CuSO}_4\text{(aq)} \to \text{FeSO}_4\text{(aq)} + \text{Cu(s)}$单置换；$\text{Fe(s)} + \text{CuSO}_4\text{(aq)} \to \text{FeSO}_4\text{(aq)} + \text{Cu(s)}$  · (d) combustion; $2\,\text{C}_2\text{H}_6\text{(g)} + 7\,\text{O}_2\text{(g)} \to 4\,\text{CO}_2\text{(g)} + 6\,\text{H}_2\text{O(g)}$燃烧；$2\,\text{C}_2\text{H}_6\text{(g)} + 7\,\text{O}_2\text{(g)} \to 4\,\text{CO}_2\text{(g)} + 6\,\text{H}_2\text{O(g)}$

(a) Magnesium burns in oxygen镁在氧气中燃烧 A1·A1

Type: synthesis (also called combination). Two elements combine to form one compound. General pattern: A + B $\to$ AB.类型：化合（也称合成）。两种单质化合生成一种化合物。通式：A + B $\to$ AB。 $$2\,\text{Mg(s)} + \text{O}_2\text{(g)} \to 2\,\text{MgO(s)}$$ Check: Mg $2=2\,\checkmark$; O $2=2\,\checkmark$. MgO is a white powder (observed as white ash on the magnesium ribbon).验证：Mg $2=2\,\checkmark$；O $2=2\,\checkmark$。MgO 为白色粉末（观察到镁带上有白色灰烬）。

(b) Hydrogen peroxide decomposes过氧化氢分解 A1·A1

Type: decomposition. One compound breaks down into two or more simpler substances. General pattern: AB $\to$ A + B.类型：分解。一种化合物分解为两种或更多种较简单的物质。通式：AB $\to$ A + B。 $$2\,\text{H}_2\text{O}_2\text{(aq)} \to 2\,\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}$$ Check: H $4=4\,\checkmark$; O $4=2+2=4\,\checkmark$. Gas bubbles of O$_2$ are the observable evidence.验证：H $4=4\,\checkmark$；O $4=2+2=4\,\checkmark$。O$_2$ 气泡为可观察证据。

(c) Iron in copper(II) sulfate solution铁在硫酸铜溶液中 A1·A1

Type: single replacement. Fe is above Cu in the activity series, so Fe displaces Cu$^{2+}$. General pattern: A + BC $\to$ AC + B.类型：单置换。Fe 在活动性顺序中位于 Cu 上方，故 Fe 能置换 Cu$^{2+}$。通式：A + BC $\to$ AC + B。 $$\text{Fe(s)} + \text{CuSO}_4\text{(aq)} \to \text{FeSO}_4\text{(aq)} + \text{Cu(s)}$$ Check: Fe $1=1\,\checkmark$; Cu $1=1\,\checkmark$; S $1=1\,\checkmark$; O $4=4\,\checkmark$. Note: FeSO$_4$ is iron(II) sulfate because Fe$^{2+}$ is the ion that forms when Fe displaces Cu$^{2+}$ (each Fe loses 2 electrons, each Cu$^{2+}$ gains 2).验证：Fe $1=1\,\checkmark$；Cu $1=1\,\checkmark$；S $1=1\,\checkmark$；O $4=4\,\checkmark$。注意：FeSO$_4$ 为硫酸亚铁，因为 Fe 置换 Cu$^{2+}$ 时形成 Fe$^{2+}$（每个 Fe 失去 2 个电子，每个 Cu$^{2+}$ 得到 2 个）。

(d) Ethane burns completely in excess oxygen乙烷在过量氧气中完全燃烧 A1·A1

Type: combustion. Hydrocarbon reacts with O$_2$ to produce CO$_2$ and H$_2$O.类型：燃烧。烃与 O$_2$ 反应生成 CO$_2$ 和 H$_2$O。 For $\text{C}_2\text{H}_6$ ($x=2, y=6$): O$_2$ coefficient $= 2 + 6/4 = 3.5$. Multiply through by 2:对 $\text{C}_2\text{H}_6$（$x=2, y=6$）：O$_2$ 系数 $= 2 + 6/4 = 3.5$，全部乘以 2： $$2\,\text{C}_2\text{H}_6\text{(g)} + 7\,\text{O}_2\text{(g)} \to 4\,\text{CO}_2\text{(g)} + 6\,\text{H}_2\text{O(g)}$$ Check: C $4=4\,\checkmark$; H $12=12\,\checkmark$; O $14=8+6=14\,\checkmark$.验证：C $4=4\,\checkmark$；H $12=12\,\checkmark$；O $14=8+6=14\,\checkmark$。
The five reaction types form the backbone of introductory chemistry; every reaction can be classified into one (or occasionally two) of these categories.五种反应类型构成入门化学的骨架；每个反应均可归入其中一种（偶尔两种）类别。 The five types covered in this unit are: (1) synthesis (A + B $\to$ AB), (2) decomposition (AB $\to$ A + B), (3) single replacement (A + BC $\to$ AC + B), (4) double replacement (AB + CD $\to$ AD + CB), (5) combustion (hydrocarbon + O$_2$ $\to$ CO$_2$ + H$_2$O). An exam strategy: always read the reactants first. If there is only one reactant, it must be decomposition. If the reactants are an element and a compound, consider single replacement. If both are compounds, consider double replacement. If one reactant is O$_2$ and the other is a hydrocarbon, it is combustion. If both reactants are elements, it is synthesis.本单元涵盖的五种类型：(1) 化合（A + B $\to$ AB），(2) 分解（AB $\to$ A + B），(3) 单置换（A + BC $\to$ AC + B），(4) 双置换（AB + CD $\to$ AD + CB），(5) 燃烧（烃 + O$_2$ $\to$ CO$_2$ + H$_2$O）。应试策略：先看反应物。若只有一种反应物，必为分解；若反应物为一种单质加一种化合物，考虑单置换；若两种均为化合物，考虑双置换；若一种反应物为 O$_2$ 另一种为烃，则为燃烧；若两种均为单质，则为化合。
Q11MEDIUM中 🇨🇦 ON安 🇨🇦 AB阿 AP-feeder FRQAP 衔接简答题 §6 + §7 Precipitation and net ionic沉淀与净离子 · SCH3U C3.2 [9 marks][9 分]

Water-treatment facility: sodium carbonate added to hard water containing dissolved calcium chloride.水处理厂：向含溶解氯化钙的硬水中加入碳酸钠溶液。

Answer:答案：  (a) yes; calcium carbonate, CaCO$_3$(s)是；碳酸钙 CaCO$_3$(s)  · (b) $\text{Na}_2\text{CO}_3\text{(aq)} + \text{CaCl}_2\text{(aq)} \to \text{CaCO}_3\text{(s)} + 2\,\text{NaCl(aq)}$  · (c) $2\,\text{Na}^+\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} + \text{Ca}^{2+}\text{(aq)} + 2\,\text{Cl}^-\text{(aq)} \to \text{CaCO}_3\text{(s)} + 2\,\text{Na}^+\text{(aq)} + 2\,\text{Cl}^-\text{(aq)}$  · (d) $\text{Ca}^{2+}\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} \to \text{CaCO}_3\text{(s)}$  · (e) Ca$^{2+}$ ions are removed from solution as insoluble CaCO$_3$, eliminating the hardness.Ca$^{2+}$ 离子以不溶性 CaCO$_3$ 形式从溶液中去除，从而消除水的硬度。

(a) Predict precipitate using solubility rules利用溶解度规则预测沉淀 M1·A1

Possible exchange products: CaCO$_3$ and NaCl. NaCl is soluble (all sodium and chloride salts are soluble). CaCO$_3$ is insoluble (carbonates are insoluble except for alkali metal and ammonium carbonates; calcium is not an alkali metal). A precipitate forms: calcium carbonate, CaCO$_3$(s), a white solid.可能的交换产物：CaCO$_3$ 和 NaCl。NaCl 可溶（所有钠盐和氯化物可溶）。CaCO$_3$ 不溶（碳酸盐不溶，碱金属和铵盐的碳酸盐除外；钙不是碱金属）。生成沉淀：碳酸钙 CaCO$_3$(s)，白色固体。

(b) Balanced molecular equation with state symbols含状态符号的配平分子方程式 M1·A1

$$\text{Na}_2\text{CO}_3\text{(aq)} + \text{CaCl}_2\text{(aq)} \to \text{CaCO}_3\text{(s)} + 2\,\text{NaCl(aq)}$$ Check: Na $2=2\,\checkmark$; C $1=1\,\checkmark$; O $3=3\,\checkmark$; Ca $1=1\,\checkmark$; Cl $2=2\,\checkmark$.验证：Na $2=2\,\checkmark$；C $1=1\,\checkmark$；O $3=3\,\checkmark$；Ca $1=1\,\checkmark$；Cl $2=2\,\checkmark$。

(c) Complete ionic equation完全离子方程式 M1·A1

$$2\,\text{Na}^+\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} + \text{Ca}^{2+}\text{(aq)} + 2\,\text{Cl}^-\text{(aq)} \to \text{CaCO}_3\text{(s)} + 2\,\text{Na}^+\text{(aq)} + 2\,\text{Cl}^-\text{(aq)}$$

(d) Net ionic equation净离子方程式 A1

Spectator ions: Na$^+$ and Cl$^-$. Cancel them:旁观离子：Na$^+$ 和 Cl$^-$，消去： $$\text{Ca}^{2+}\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} \to \text{CaCO}_3\text{(s)}$$

(e) How precipitation achieves water softening沉淀反应如何实现软水 M1·A1

By precipitating Ca$^{2+}$ as insoluble CaCO$_3$(s), the carbonate ion removes the dissolved calcium ions that cause hardness; the CaCO$_3$ is then filtered off, leaving softened water.碳酸根离子将 Ca$^{2+}$ 以不溶性 CaCO$_3$(s) 形式沉淀析出，从而去除导致水硬度的溶解钙离子；过滤除去 CaCO$_3$ 后，得到软化水。
Water hardness is caused by dissolved Ca$^{2+}$ and Mg$^{2+}$ ions; precipitation with carbonate is a simple, scalable softening method.水的硬度由溶解的 Ca$^{2+}$ 和 Mg$^{2+}$ 离子引起；碳酸盐沉淀法是一种简单且可扩展的软化方法。 Hard water forms scale (CaCO$_3$ and Mg(OH)$_2$ deposits) in pipes and boilers, reducing heat transfer efficiency. The lime-soda process adds Ca(OH)$_2$ (lime) and Na$_2$CO$_3$ (soda ash) to precipitate both Ca$^{2+}$ and Mg$^{2+}$. Modern municipal plants often use ion-exchange resins instead, but the underlying chemistry is the same net ionic equation: $\text{Ca}^{2+}\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} \to \text{CaCO}_3\text{(s)}$. This question integrates sections 6 and 7 of the unit because writing net ionic equations is the tool that reveals the essential removal chemistry, regardless of the counter-ions chosen.硬水在管道和锅炉中形成水垢（CaCO$_3$ 和 Mg(OH)$_2$ 沉积），降低传热效率。石灰-纯碱法加入 Ca(OH)$_2$（石灰）和 Na$_2$CO$_3$（纯碱）以同时沉淀 Ca$^{2+}$ 和 Mg$^{2+}$。现代市政水厂通常改用离子交换树脂，但底层化学仍是同一净离子方程式：$\text{Ca}^{2+}\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} \to \text{CaCO}_3\text{(s)}$。本题综合了本单元第 6 和第 7 节，因为净离子方程式正是揭示核心去除化学的工具，与选用何种抗衡离子无关。
Q12HARD难 🇺🇸 US美 🇨🇦 AB阿 AP-feeder FRQAP 衔接简答题 §1-7 Cross-topic investigation跨题型综合探究 · HS-PS1-2 · Chem 20 D GO3 [9 marks][9 分]

Three experiments: answer all parts for each.三个实验：回答每个实验的所有小题。

Answer:答案：  (a) $2\,\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\,\text{H}_2\text{O(g)}$; both synthesis and combustion既是化合反应也是燃烧反应  · (b) molecular: $\text{CuCl}_2\text{(aq)} + 2\,\text{NaOH(aq)} \to \text{Cu(OH)}_2\text{(s)} + 2\,\text{NaCl(aq)}$; net: Cu$^{2+}$(aq) + 2OH$^-$(aq)分子式：$\text{CuCl}_2\text{(aq)} + 2\,\text{NaOH(aq)} \to \text{Cu(OH)}_2\text{(s)} + 2\,\text{NaCl(aq)}$；净离子：Cu$^{2+}$(aq) + 2OH$^-$(aq) $\to$ Cu(OH)$_2$(s)  · (c) NR; Ag is below Cu in the activity seriesNR；Ag 在活动性顺序中排于 Cu 下方

(a) Experiment 1: hydrogen burning in limited oxygen实验 1：氢气在有限氧气中燃烧 M1·A1·A1

$$2\,\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\,\text{H}_2\text{O(g)}$$ Check: H $4=4\,\checkmark$; O $2=2\,\checkmark$.验证：H $4=4\,\checkmark$；O $2=2\,\checkmark$。

Classification: This reaction qualifies as both synthesis and combustion. It is a synthesis reaction because two elements (H$_2$ and O$_2$) combine to form one compound (H$_2$O); it satisfies the pattern A + B $\to$ AB. It is simultaneously a combustion reaction because hydrogen is a fuel burning in oxygen to produce an oxide. Both classifications are correct; the question rewards students who recognise the overlap. On most high-school marking schemes, either classification earns the mark; stating both is ideal.分类：该反应既是化合反应，也是燃烧反应。它是化合反应，因为两种单质（H$_2$ 和 O$_2$）化合生成一种化合物（H$_2$O），满足通式 A + B $\to$ AB。它同时也是燃烧反应，因为氢气作为燃料在氧气中燃烧生成氧化物。两种分类均正确；本题奖励能识别两类重叠的学生。在大多数高中评分方案中，任一分类均可得分；两者兼答最佳。

(b) Experiment 2: CuCl$_2$(aq) + NaOH(aq)实验 2：CuCl$_2$(aq) + NaOH(aq) M1·A1·M1·A1

Possible products: Cu(OH)$_2$ and NaCl. NaCl is soluble. Cu(OH)$_2$ is insoluble (most hydroxides are insoluble; only alkali metal and ammonium hydroxides are soluble); it is the blue precipitate observed.可能的产物：Cu(OH)$_2$ 和 NaCl。NaCl 可溶。Cu(OH)$_2$ 不溶（大多数氢氧化物不溶；仅碱金属和铵的氢氧化物可溶）；即所观察到的蓝色沉淀。

Molecular equation:分子方程式：

$$\text{CuCl}_2\text{(aq)} + 2\,\text{NaOH(aq)} \to \text{Cu(OH)}_2\text{(s)} + 2\,\text{NaCl(aq)}$$ Check: Cu $1=1\,\checkmark$; Cl $2=2\,\checkmark$; Na $2=2\,\checkmark$; O $2=2\,\checkmark$; H $2=2\,\checkmark$.验证：Cu $1=1\,\checkmark$；Cl $2=2\,\checkmark$；Na $2=2\,\checkmark$；O $2=2\,\checkmark$；H $2=2\,\checkmark$。

Complete ionic equation:完全离子方程式：

$$\text{Cu}^{2+}\text{(aq)} + 2\,\text{Cl}^-\text{(aq)} + 2\,\text{Na}^+\text{(aq)} + 2\,\text{OH}^-\text{(aq)} \to \text{Cu(OH)}_2\text{(s)} + 2\,\text{Na}^+\text{(aq)} + 2\,\text{Cl}^-\text{(aq)}$$

Net ionic equation (cancel Na$^+$ and Cl$^-$ spectators):净离子方程式（消去旁观离子 Na$^+$ 和 Cl$^-$）：

$$\text{Cu}^{2+}\text{(aq)} + 2\,\text{OH}^-\text{(aq)} \to \text{Cu(OH)}_2\text{(s)}$$

(c) Experiment 3: Ag(s) in Cu(NO$_3$)$_2$(aq)实验 3：Ag(s) 放入 Cu(NO$_3$)$_2$(aq) M1·A1

Standard activity series order (most to least reactive, relevant excerpt): ... Fe, Cu, Ag, Au. Copper (Cu) is above silver (Ag), meaning Cu is more reactive than Ag. For Ag to displace Cu$^{2+}$, Ag would need to be above Cu in the series. Since Ag is below Cu, Ag cannot reduce Cu$^{2+}$ to Cu metal. No reaction occurs. NR.标准活动性顺序（由强到弱，相关片段）：... Fe、Cu、Ag、Au。铜（Cu）排在银（Ag）上方，即 Cu 比 Ag 活动性更强。若 Ag 要置换 Cu$^{2+}$，Ag 需在活动性顺序中排于 Cu 上方。由于 Ag 排在 Cu 下方，Ag 无法将 Cu$^{2+}$ 还原为 Cu 金属。不发生反应。NR。
Cross-topic questions test whether you can apply multiple frameworks in sequence within a single problem.跨主题题目考查能否在同一问题中依次运用多个框架。 Experiment 1 tests classification nuance (synthesis vs combustion overlap). Experiment 2 tests the full three-equation chain for double replacement with a precipitate, plus solubility-rule reasoning to identify Cu(OH)$_2$. Experiment 3 tests activity-series reasoning with a potential single replacement, but requires recognising that the ordering Cu > Ag (not Ag > Cu) means no reaction. A common misconception is to confuse the relative positions of Cu and Ag: remember that both are below H$_2$ in the series (they do not dissolve in HCl), with Cu above Ag. Together the three experiments review all the major skills of this unit.实验 1 考查分类的细微差别（化合与燃烧的重叠）。实验 2 考查双置换沉淀反应的完整三方程链，以及用溶解度规则识别 Cu(OH)$_2$。实验 3 考查活动性顺序分析的潜在单置换，但需认识到 Cu > Ag（而非 Ag > Cu）意味着无反应。常见错误是混淆 Cu 和 Ag 的相对位置：记住两者均排在活动性顺序中 H$_2$ 的下方（不溶于盐酸），其中 Cu 在 Ag 上方。三个实验合在一起，复习了本单元所有主要技能。