Sujet 7 : Chimie organique
AQA · GCSE · Chemistry Sujet 7 11:01 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
Choose the process from what happens to the molecules.
根据分子的变化选择过程。
Heating alone does not tell us which process is taking place.
仅凭加热不能判断是分馏、裂解还是聚合。
We will connect representations, conditions and observations before choosing a reaction.
先结合结构、条件和观察，再判断反应。
Crude oil is finite on a human timescale.
原油在人的时间尺度上是有限资源。
Each compound has its own definite formula, but the mixture contains many compounds.
每种化合物有确定的分子式，但混合物含多种化合物。
Ethanol contains oxygen and is not a hydrocarbon.
乙醇含氧，不是碳氢化合物。
Recognise methane, ethane, propane and butane in molecular and displayed formulae.
识别甲烷、乙烷、丙烷、丁烷的分子式和全展开结构式。
The displayed structures show all atoms and bonds. Each carbon must still have four bonds.
每个碳形成四个键，每个氢形成一个键。
Pause and read every element in each formula. Then apply the alkane relation only to the stated open-chain alkane.
暂停，逐一检查元素，再把烷烃通式用于题目指定的开链烷烃。
The general formula gives one member of the alkane family. It does not give a molecular formula for the whole mixed sample or locate its carbon arrangement.
通式给出烷烃家族成员的组成，不能给整个混合物一个分子式，也不能确定碳的连接方式。
Heated vapours rise and cool.
受热蒸气上升并冷却。
Fractions contain several hydrocarbons with similar boiling points and carbon numbers.
每个馏分含沸点和碳数相近的多种烃。
The lowest-boiling gases can leave at the top.
最低沸点的气体可从塔顶离开。
Distillation leaves carbon chains intact.
分馏不切断碳链。
Viscosity measures resistance to flow.
黏度表示流动阻力。
Do not use boiling point, viscosity and flammability as synonyms.
沸点、黏度和可燃性不是同义词。
A fraction can be useful as a fuel or as chemical feedstock depending on the stated purpose.
根据用途，馏分可作燃料，也可作化工原料。
Pause and check both requirements.
暂停，同时检查两个要求。
Decide whether a flow-time comparison alone establishes an exact viscosity ratio.
判断流动时间比能否直接当作精确黏度比。
The data support the decision for this burner and test. They do not prove that the smaller molecule is always the best fuel or establish all properties of commercial mixtures.
这些数据支持本燃烧器和本测试条件下的选择，不能证明小分子总是最佳燃料，也不能确定混合燃料的所有性质。
With sufficient oxygen, carbon and hydrogen are oxidised to carbon dioxide and water.
氧充足时，碳和氢被氧化为二氧化碳和水。
Change coefficients to balance atoms; changing a subscript changes the substance.
配平时改变系数，不能改变代表物质组成的下标。
Pause to count atoms on both sides.
暂停，核对两边原子数。
Remember that oxygen occurs in both carbon dioxide and water.
二氧化碳和水中都含氧。
The fractional oxygen coefficient is acceptable, but multiplying every coefficient by two gives whole numbers.
分数氧气系数可以使用；所有系数乘二得到整数。
The molecular formula of butane stays unchanged.
丁烷的分子式保持不变。
Count both elements in a supplied cracking equation.
裂解方程要核对碳和氢。
Products include alkanes and alkenes, but a real cracking process can form a mixture rather than one compulsory pair.
产物包括烷烃和烯烃，实际过程可形成混合物，并非只能形成固定的一对产物。
High demand for smaller fuels and chemical feedstocks explains its value.
小分子燃料和化工原料的需求说明其价值。
Ethene, propene, butene and pentene have the stated open-chain single-double-bond formula CnH2n. Compared with the same-carbon alkane, there are two fewer hydrogens.
乙烯、丙烯、丁烯、戊烯对应一个双键的开链通式CnH2n，比同碳数烷烃少两个氢。
The molecular formula alone does not locate the double bond.
分子式本身不能确定双键位置。
Change the double bond to a single bond and attach atoms to both carbons.
把双键改为单键，向原双键的两个碳添加原子。
Retain the carbon skeleton and original hydrogens.
保留原碳骨架和氢。
Chlorine and bromine addition are described at room temperature; identical iodine speed is not promised.
氯、溴加成可在室温描述，但不能保证碘的速率相同。
The top product adds hydrogen. The middle product adds H and OH in the stated orientation. The bottom adds two bromine atoms without removing hydrogens.
上图加氢，中图按指定位置加入H和OH，下图加两个溴而不移除氢。
Do not apply the primary-alcohol oxidation rule to every hydration product.
不能把伯醇氧化规律用于所有水合产物。
Pause to distinguish the observed colour from the interpretation. Include the alkane control under the stated conditions.
暂停，区分观察和解释，并说明指定条件下烷烃对照的现象。
The test supports unsaturation in the comparison.
该对照检验支持不饱和性。
Alkenes often burn smokily in air, but with sufficient oxygen their complete combustion still produces carbon dioxide and water.
烯烃在空气中常有烟焰，但氧充足时仍完全燃烧生成二氧化碳和水。
Hydrogen addition and combustion are different reactions.
加氢与燃烧是不同反应。
Recognise methanol, ethanol, propanol and butanol from names and formulae.
识别甲醇、乙醇、丙醇和丁醇。
The position of OH matters for oxidation, so keep the stated structures with the explanation.
OH位置影响氧化行为，解释必须保留指定结构。
Balanced alcohol equations are required here for combustion only.
本节醇反应只要求配平燃烧方程。
Count the oxygen already present in ethanol.
核对乙醇本身的氧。
Do not claim every alcohol has unlimited water solubility or every OH-position form oxidises to the same acid.
不能说所有醇都无限溶于水，或不同OH位置都氧化为相同酸。
Fermentation uses sugar and yeast, unlike industrial ethene hydration.
发酵使用糖和酵母，与工业乙烯水合不同。
A gas-producing vessel must not be tightly sealed.
产气容器不能完全密封。
Written preparation is separate from a supervised practical and its recorded observations.
书面准备不能代替有监督的实际操作与原始记录。
Pause and separate the control, pressure hazard and measurement limitation.
暂停，分别说明对照、压力风险与测量局限。
The collected gas is not a direct assay of ethanol purity.
收集气体不是乙醇纯度的直接测定。
Keep sugar concentration, temperature, volume and observation time matched.
控制糖浓度、温度、体积和观察时间。
The comparison supports yeast-associated gas formation, not pure ethanol production or proof that every bubble had the same origin.
比较支持与酵母有关的产气，不能证明得到纯乙醇，也不能证明每个气泡来源相同。
Learn methanoic, ethanoic, propanoic and butanoic acids.
学习甲酸、乙酸、丙酸、丁酸。
The diagrams show carbon double-bonded to one oxygen and single-bonded to OH. Count that carbon when identifying the member.
羧基碳与一个氧形成双键、与OH形成单键；判断碳数时要计入这个碳。
Describe the reactions without inventing extra balanced-equation recall or additional ester-name lists.
描述反应，不额外要求酸反应配平方程或其他酯名称。
Recognising ethyl ethanoate is both-tier Separate Chemistry content.
识别乙酸乙酯是化学两层次内容。
Teacher-approved warming and reagents are needed for a real preparation.
实际制备需教师批准的试剂和加热方式。
Compare like concentrations and conditions before inferring strength from pH.
比较强弱前要匹配浓度与条件。
A weak acid can be concentrated, and a strong acid can be dilute.
弱酸可以浓，强酸可以稀。
This explanation is the Higher Tier part of the carboxylic-acid section.
羧酸的这种电离解释属于高阶部分。
Open the double bond, keep the attached atoms, and draw single continuation bonds through both brackets with n outside.
打开双键，保留所有侧基，在两侧括号外画延续单键并写n。
The repeat retains all monomer atoms. No water is released in addition polymerisation.
重复单元保留所有单体原子，加成聚合不脱水。
Pause to remove only the continuation bonds and restore C=C between the two backbone carbons. Retain chlorine and all hydrogens.
暂停，只移除延续键，在两个主链碳之间恢复C = C，保留氯和所有氢。
The supplied monomer is shown completely.
图中给出完整单体，含C = C、三个氢和一个氯。
Its atom count is unchanged in each repeat, so this is addition polymerisation without water loss.
每个重复单元保留这些原子，所以加成聚合不脱水。
A reversed page orientation with the same connectivity is equivalent.
连接关系相同的反向画法等价。
Each monomer needs two functional groups to join at both ends.
每个单体需两个官能团以便两端连接。
At each actual joining event, water forms.
每次实际连接形成水。
The ideal repeating interior omits finite-chain end groups, so use actual links when counting a finite chain.
理想重复单元省略有限链端基，所以有限链的水量要数实际连接。
Glycine condenses through peptide links.
甘氨酸经肽键缩合。
The diagram shows a repeating interior, not the full formula of a finite molecule.
图示是重复链内部，不是有限分子的完整分子式。
Different amino acids can occur in one protein chain; protein monomer recognition remains both-tier.
同一蛋白质链可含不同氨基酸；蛋白质单体识别仍属两层次内容。
Pause to count links between molecules, not merely repeat labels.
暂停，数分子之间的实际连接，不能只数重复标签。
A ring or branched structure would need separate counting.
环状或支化结构需要另行计数。
The relative mass calculation follows conservation of mass in the specified finite chain.
相对质量计算遵循指定有限链中的质量守恒。
The number of waters is tied to actual joining events and remaining end groups.
水的数目由实际连接与剩余端基决定。
This is a flattened ladder model before twisting.
这是扭转前的梯形模型。
Each block represents a whole nucleotide and is labelled by its base.
每个方块代表一个完整核苷酸，只用碱基作标签。
There are four nucleotide types. DNA encodes genetic instructions.
核苷酸有四种，DNA编码遗传信息。
Sharing a monomer does not make two polymers identical.
单体相同不表示聚合物相同。
Use connectivity and evidence to select the method, and keep Higher Tier condensation explanations separate from both-tier natural-polymer recognition.
根据连接关系和证据选方法，把高阶缩聚解释与两层次天然聚合物识别分开。