Tema 4: Cambios químicos
AQA · GCSE · Chemistry Tema 4 9:51 Narración en inglés · Subtítulos en inglés + 中文 quemados en pantalla
Capítulos
Transcripción
A copper wire and magnesium ribbon behave differently in dilute hydrochloric acid.
铜丝和镁带在稀盐酸中表现不同，电解则用电流推动反应。
Electrolysis can force a change using a current. Begin with particles and conditions, rather than applying one reaction rule to every mixture.
先看粒子和条件，不能把一个规则套用于所有混合物。
Higher Tier electron explanations and calculations are labelled separately.
电子解释与计算的高阶要求分别标明。
The arrow points upwards towards potassium.
箭头向上指向钾，表示活动性增强。
Learn the specified eight metals in order.
记住所列八种金属的顺序。
Carbon predicts whether carbon reduction can extract a metal from its oxide; hydrogen helps with acid reactions and aqueous cathode products. Neither reference substance is a metal.
碳用于判断碳还原，氢用于酸反应和水溶液阴极产物；两者都不是金属。
Use room-temperature water. Sodium may melt into a ball, potassium may burn lilac, lithium floats and fizzes, and calcium can give a cloudy mixture.
使用室温水：钠可能熔成小球，钾可能出现淡紫火焰，锂浮起冒泡，钙可使混合物浑浊。
Do not add steam reactions to this source section.
本节不要求水蒸气反应。
Compare clean samples and matching conditions.
比较时保持样品清洁、条件一致。
More reactive metal atoms tend to form positive ions more readily.
金属越活泼，越易形成阳离子。
The acid–metal scope specifies magnesium, zinc and iron with hydrochloric or sulfuric acid.
本节酸与金属要求是镁、锌、铁与盐酸或硫酸反应。
Metals such as sodium are too vigorous for this ordinary classroom comparison.
钠等过于剧烈，不适合普通课堂比较。
Protect eyes and keep collected hydrogen away from flames except in a controlled small test.
注意护目并避免氢气遇火，只有受控微量检验例外。
Use each successful or unsuccessful displacement to make a comparison. Do not guess the identity of X.
把每个置换或不置换结果变成相对顺序，不能猜X的身份。
Pause for an independent answer.
先独立作答，再看答案。
X replaces Y, so X is more reactive than Y; Z replaces X, so Z is more reactive than X.
X置换Y，Z置换X，得到Z大于X大于Y。
The copper and magnesium tests bracket X between them.
另两项检验说明镁大于X大于铜。
You cannot locate Y or Z relative to every named metal from these results alone.
不能据此确定Y或Z相对所有已知金属的位置。
Identify the substances changing, rather than claiming carbon dioxide is reduced.
指出发生变化的反应物，不能说二氧化碳被还原。
Pause for an independent answer.
先独立作答，再看答案。
Iron oxide becomes iron by losing oxygen. Carbon becomes carbon dioxide by gaining it. Gold may occur native.
氧化铁失氧成为铁，被还原；碳得氧成为二氧化碳，被氧化。
Metals above carbon need another extraction method such as electrolysis; use supplied process details for evaluation.
金可能以单质存在。 碳以上金属需其他提炼方法，如电解；评价时使用题给过程信息。
Carbon dioxide emissions are an environmental disadvantage, not proof that every alternative is impact-free.
二氧化碳排放是环境缺点，但不代表其他方法完全无影响。
Balance each half-equation with electrons and name the ion that gains them.
用电子配平两个半方程，并说清哪种离子得电子。
Pause for an independent answer.
先独立作答，再看答案。
The zinc atom loses two electrons. The copper two-plus ion gains them; say copper ions are reduced. Two electrons cancel when the half-equations are added.
锌原子失去两个电子，铜离子得到电子，要说铜离子被还原。
In magnesium with hydrochloric acid, magnesium is oxidised and hydrogen ions are reduced; chloride is a spectator.
相加时两个电子抵消。 镁与盐酸反应中镁被氧化、氢离子被还原，氯离子不变。
Balance ion charges before the whole equation. Compare the gas with a metal–acid reaction.
先用离子电荷写盐的化学式，再配平方程，比较碳酸盐与金属反应产生的气体。
Pause for an independent answer.
先独立作答，再看答案。
Nitric acid provides nitrate; calcium supplies the positive ion.
硝酸提供硝酸根，钙提供阳离子。
The bracket keeps two complete nitrate ions.
括号表示两个完整硝酸根。
Acids with bases produce salt and water, carbonates also give carbon dioxide, while the specified metal reactions give hydrogen. These are distinct product rules.
酸与碱生成盐和水，与碳酸盐还生成二氧化碳；所列金属与酸生成氢气，不能混用产物规则。
Add copper oxide to warmed dilute sulfuric acid until some solid remains, then filter.
向温热稀硫酸加入氧化铜至有固体剩余，再过滤。
The excess proves the acid has reacted and the filter removes it.
过量固体说明酸已反应完，过滤将其除去。
Concentrate the filtrate gently and cool; separate, wash and dry the crystals.
滤液温和浓缩、冷却结晶，再分离、洗涤、干燥。
This method review does not replace supervised laboratory work.
书面方法不能代替实际实验。
Use the particles and solubility to explain why each proposed step fails.
利用粒子与溶解性解释每步为何错误，再给出改正。
Pause for an independent answer.
先独立作答，再看答案。
The three errors are connected. Using insufficient oxide leaves acid, and a paper filter cannot remove that dissolved acid. Correct the reaction stage first.
三个错误相互关联：氧化铜不足使酸残留，滤纸不能除去溶解的酸，应先改正反应阶段。
Gentle evaporation followed by cooling gives crystals; boiling dry is not the chosen method.
温和蒸发后冷却得到晶体，不应煮干。
Use little cold water to reduce dissolving the crystals.
少量冷水洗涤以减少晶体溶失。
Use a small sample and compare universal-indicator colour with its chart.
用少量样品和通用指示剂色卡估测pH。
A narrow-range titration indicator is for an endpoint rather than the full pH scale.
滴定用窄范围指示剂检测终点，不能测整个pH范围。
Alkalis contain hydroxide ions in water. Neutralisation makes water from hydrogen and hydroxide ions; dilution of an acid brings pH towards seven, not beyond it into an alkali.
碱溶液含氢氧根，中和时与氢离子生成水；酸稀释后趋近中性，不会因此变碱。
Rinse the pipette with alkali and the burette with acid. Clear the burette jet and remove the filling funnel.
移液管用碱润洗，滴定管用酸润洗，排尽尖嘴气泡并移走漏斗。
Use a suitable indicator, swirl over a white tile and finish dropwise. Read at eye level, recording initial and final values.
加合适指示剂，在白瓷板上摇匀，近终点逐滴加入。
Both tiers learn the reacting-volume method; concentration calculations are the Higher extension.
平视并记录初末读数。 两个等级都学体积测量，高阶再算浓度。
Explain each action through mixing, endpoint control or reacting amount.
用混合、控制终点和反应物质的量解释操作。
Pause for an independent answer.
先独立作答，再看答案。
The white tile improves visibility of the indicator change.
白瓷板使颜色变化易见。
A rough trial locates the endpoint; careful repeats provide concordant titres under a stated tolerance. Do not include a rough result automatically in the mean.
粗滴定位终点，精确重复按给定误差限选择相符体积，粗滴不能自动计入平均。
Washing the walls with distilled water is different from rinsing the pipette with water before measuring alkali.
蒸馏水冲洗锥形瓶壁，与取碱前用水润洗移液管不同。
Use sulfuric acid plus two sodium hydroxide. Show volume conversion and the one-to-two ratio.
硫酸与氢氧化钠一比二反应，写出体积换算和摩尔比例。
Pause for an independent answer.
先独立作答，再看答案。
Convert twenty cubic centimetres to zero point zero two zero zero cubic decimetres. Known concentration times volume gives the alkali amount.
20立方厘米换成0.0200立方分米，已知浓度乘体积得到碱0.00200摩尔。
Sulfuric acid reacts with twice as many moles of sodium hydroxide, so divide that amount by two.
硫酸与两倍摩尔数的氢氧化钠反应，因此除以二得酸0.00100摩尔。
Divide acid amount by acid volume, not alkali volume.
酸摩尔数除以酸体积，不能除以碱体积。
Multiply molar concentration by molar mass for grams per cubic decimetre. Check the units cancel.
摩尔浓度乘摩尔质量得到克每立方分米，检查单位约分。
Equal reacting volumes do not in general imply equal concentrations; the balanced ratio matters.
反应体积相等一般不代表浓度相等，要看方程式比例。
Decide what the labels alone tell you, then use the measured whole-number pH difference.
先判断标签本身能说明什么，再用测得整数pH之差比较。
Pause for an independent answer.
先独立作答，再看答案。
The names P and Q alone do not determine their hydrogen-ion concentrations.
仅凭P和Q名称不能确定氢离子浓度。
At a fixed concentration the stronger acid has lower pH. A three-unit measured pH difference gives ten cubed, or a thousand.
同浓度下较强酸pH更低，测得相差三单位对应一千倍。
Neutral water still contains ions; neutrality is a balance of hydrogen and hydroxide ions.
中性水仍含离子，中性表示氢离子与氢氧根的平衡。
The solid lattice has ions but cannot conduct by ion movement because their positions are fixed.
固体晶格虽有离子，但位置固定，不能靠离子移动导电。
Melting or dissolving allows ions to move. At an electrode ions gain or lose electrons and elements are produced. Do not draw electrons travelling through the electrolyte as its charge carriers.
熔化或溶解使离子可移动；离子在电极得失电子形成单质，不能把电解质中载流粒子画成电子。
Aluminium is too reactive for carbon reduction.
铝太活泼，不能用碳还原。
Dissolving its oxide in molten cryolite reduces the heating demand compared with melting the pure oxide. The process still uses substantial electrical energy. Carbon reacts with produced oxygen, making carbon dioxide and consuming the positive electrode.
氧化铝溶于熔融冰晶石，较熔化纯氧化物节省加热能量，但仍耗大量电。
This drawing does not describe an aqueous aluminium salt.
碳与氧气反应生成二氧化碳，消耗正极。 图不表示铝盐水溶液。
List the ions available, including the water contribution in each aqueous solution.
列出可用离子，水溶液中还要考虑水的贡献。
Pause for an independent answer.
先独立作答，再看答案。
These predictions use the specified GCSE rules with inert electrodes.
按本考纲惰性电极规则预测。
A molten salt has no water. In sodium solutions hydrogen is discharged instead of sodium because sodium is above hydrogen. Copper is below hydrogen and deposits.
熔融盐无水，钠水溶液因钠在氢以上而析氢；铜在氢以下而沉积。
Halide gives halogen at the anode; otherwise oxygen forms in this scope.
阳极有卤素离子得卤素，否则本范围产生氧。
Develop a hypothesis before the investigation and specify evidence that could contradict it.
实验前提出假设，并明确什么证据可能不支持它。
Pause for an independent answer.
先独立作答，再看答案。
Use separate labelled solutions and clean electrodes between them.
溶液分开标记，换溶液要清洗电极。
Test small collected gas quantities under supervision; never smell a gas.
监督下用微量气体检验，不得闻气体。
Hydrogen gives a squeaky pop and chlorine bleaches damp litmus.
氢气有爆鸣声，氯气使湿石蕊褪色。
Compare results with the original hypothesis, recognising that no visible bubble is not proof of no reaction. Written predictions do not replace practical participation.
对照原假设评价，没看到气泡不等于没反应，书面预测不能代替实验。
Use electrons on the product side for oxidation and the reactant side for reduction.
氧化把电子放右侧，还原放左侧，最后检查原子和电荷。
Pause for an independent answer.
先独立作答，再看答案。
Zinc ions gain electrons at the cathode; chloride ions lose them at the anode.
锌离子在阴极得电子，氯离子在阳极失电子。
For aqueous oxygen formation balance oxygen and hydrogen first, then add four product electrons to match charge.
水溶液析氧先配平氧氢，再加四个产物电子配平电荷。
The aluminium oxide half-equation uses oxide ions, not hydroxide ions; do not mix the molten and aqueous cases.
氧化铝熔融体系用氧离子，不能与水溶液氢氧根混淆。
Close the answers and attempt the unfamiliar tasks in sheets four point one to four point three. Explain a new metal ordering, repair a salt method, solve a non-one-to-one titration and test an electrolysis hypothesis.
合上答案，独立完成4.1至4.3的新情境：推断金属顺序、改正制盐方法、非一比一滴定计算及电解假设检验。
Respect Higher Tier boundaries and keep actual practical work separate from written review.
保留高阶界限，区分实际实验与书面复习。