Sujet 10 : Utilisation des ressources
AQA · GCSE · Chemistry Sujet 10 13:31 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
Begin with the job a material or process must do.
先明确材料或过程需要完成的任务。
A cheap raw material can require expensive treatment.
便宜的原料可能需要昂贵处理。
Practical performance needs a supervised record beyond written questions.
实际操作能力还需课堂监督记录。
Natural resources provide warmth, shelter, food and transport.
自然资源提供保暖、住房、食物和交通。
Synthetic fibres can supplement or replace natural fibres, but their environmental effects still need evaluation.
合成纤维可以补充或替代天然纤维，但仍需评估环境影响。
Pause and distinguish the resource category from the actual management of its stock.
暂停，区分资源类别与实际储量管理。
Use the same mass units.
计算时使用相同质量单位。
Renewable is not a promise of an unlimited supply.
可再生不等于无限供应。
Sustainable development meets present needs without compromising the ability of future generations to meet theirs.
可持续发展满足当代需要，同时不损害后代满足自身需要的能力。
The ratio describes scale, not sustainability.
比例描述规模，并不说明可持续性。
To judge sustainability also consider extraction rate, replenishment and environmental consequences.
还需考虑开采速度、补充速度与环境后果。
Follow the fresh-water route.
沿淡水处理路线看。
Chlorine, ozone or ultraviolet can sterilise.
氯、臭氧或紫外线可用于消毒。
Potable means safe to drink, not a sample consisting only of water molecules.
饮用水指可安全饮用，不是只含水分子的样品。
A normal filter does not remove dissolved salt.
普通过滤不能除去溶解盐。
Choose a treatment using source quality, available energy and required final quality.
应根据原水质量、可用能源和最终质量要求选择方法。
Use consistent volumes and safe heating. Repeat heating and cooling to constant mass when appropriate.
保持取样体积一致并安全加热，适当时反复加热冷却至恒重。
A residue measurement alone cannot certify microbial safety.
仅测残渣不能证明微生物安全。
Pause and subtract the empty mass before dividing by the sample volume.
暂停，先扣除空容器质量，再除以样品体积。
Convert cubic centimetres to cubic decimetres.
把立方厘米换成立方分米。
The measured concentration has a specified method and boundary.
测得浓度具有特定方法和范围。
Additional evidence is needed to decide whether water is safe to drink.
判断是否可饮用，还需其他证据。
Screen large objects and remove grit before sedimentation.
筛去大物体、去除砂砾后进行沉降。
Settled sludge undergoes anaerobic digestion; liquid effluent receives aerobic biological treatment.
沉降污泥进行厌氧消化；液体出水进行好氧生物处理。
Industrial chemicals may need extra processes.
工业化学物可能需要额外处理。
Pause and trace both arrows from sedimentation.
暂停，沿沉降后的两条箭头分别追踪。
Treatment for environmental release is not automatically treatment to drinking-water quality.
达到环境排放要求并不自动等于达到饮用水标准。
Suitable ground water usually needs less treatment than waste water under comparable conditions.
在可比条件下，适当的地下水通常比废水需要更少处理。
Contamination can change the comparison, so avoid an absolute ranking for every source.
但污染状况可改变比较，不能对所有水源作绝对排名。
Plants absorb metal compounds; harvest and burn them to obtain ash containing compounds.
植物吸收金属化合物，收割并焚烧后，灰中仍含化合物。
Bacteria produce a leachate containing metal compounds.
细菌产生含金属化合物的浸出液。
Neither route directly gives a piece of pure copper.
两种路线都不会直接得到一块纯铜。
Pause and use the reactivity series for the displacement.
暂停，利用金属活动性顺序判断置换。
Compare supplied evidence about rock movement, time, yield, energy and waste.
根据题给资料比较搬运岩石、时间、产量、能源与废物。
Count one iron, one copper and one sulfate group on each side.
两侧各有一个铁、一个铜及一个硫酸根。
Biological does not mean immediate or impact-free, and electrolysis also needs energy.
生物方法不等于瞬间完成或没有影响，电解同样耗能。
Quantifiable data still carry uncertainty.
可量化数据仍有不确定性。
Combining different pollutant effects into one score involves value judgements; an LCA is not purely objective.
把不同污染影响合成一个分数需要价值判断，生命周期评估并非完全客观。
Pause and compare twelve n with thirty plus n.
暂停，比较12n与30+n。
The answer must be a whole number of uses, and this dataset concerns energy only.
使用次数应取整数，而且本组数据只涉及能源。
At two tasks the reusable total is thirty-two rather than twenty-four units.
使用两次时，可重复袋为32单位，单次袋为24单位。
Check the real reuse rate and report missing water, litter or pollutant information.
应核查实际重复次数，并说明缺少的用水、垃圾或污染资料。
Reduce use where appropriate, but compare complete routes.
适当时减少使用，但应比较完整路线。
Collection, sorting, washing, transport and melting still require resources.
收集、分类、清洗、运输和熔化仍需资源。
Required final properties determine separation needs.
分离要求取决于最终性能。
Pause and use fifty as the denominator.
暂停，百分比的分母应为50。
Ask whether collection and transport were included.
核查是否计入收集与运输。
An available recycling route is different from an actual recycled product.
可采用回收路线，不等于实际已经回收。
Contamination, sorting and the local collection system affect the result.
污染、分类与当地收集系统都会影响结果。
Compare equivalent nails for the same exposure time and temperature.
在相同时间和温度下比较相同铁钉。
Drying agent removes water; boiling removes dissolved oxygen and oil limits its return.
干燥剂除水，煮沸去除溶解氧，油层限制氧重新进入。
Keep the nail submerged in the oxygen-limited water.
缺氧对照中的铁钉应完全浸没。
Pause and distinguish a barrier from sacrificial protection.
暂停，区分屏障保护与牺牲保护。
State the relative reactivity and the condition that suitable zinc contact remains.
说明相对活动性，并指出锌需要保持适当接触。
A less reactive coating cannot give the same sacrificial protection.
较不活泼的涂层不能提供相同牺牲保护。
Aluminium has an adherent protective oxide layer, which is evidence of reaction rather than evidence that it never reacts.
铝表面有附着氧化膜，这说明它发生了反应，并非永不反应。
Different-sized atoms can hinder sliding of metal layers.
不同大小的原子可阻碍金属层滑动。
Match the composition and properties to the job; do not say every alloy is superior in every way.
应把成分和性能与用途匹配，不能说所有合金在各方面都更好。
Pause and use the fraction eighteen over twenty-four.
暂停，使用18除以24。
Silver, copper or zinc can be alloyed with gold to improve hardness and wear.
金可与银、铜或锌形成合金，以提高硬度和耐磨性。
The mass fraction is not eighteen percent.
质量分数并非百分之十八。
Use supplied data when selecting an unfamiliar alloy rather than guessing from a trade name.
选择陌生合金时应使用题给数据，不应根据商品名称猜测。
Hardness, heat resistance and brittleness can favour different jobs.
硬度、耐热性与脆性适用于不同用途。
Bricks and pottery begin as shaped wet clay; the heating stage changes the material.
砖和陶器由湿黏土成型，加热阶段改变材料。
Do not infer identical density from an identical monomer.
单体相同并不意味着密度相同。
Production conditions affect chain arrangement and therefore properties.
生产条件影响链排列，进而影响性能。
Heating a thermosoftening polymer allows chains to move as intermolecular attractions are overcome.
加热热塑性聚合物可克服分子间吸引，使链移动。
A thermosetting network has covalent cross-links and does not melt; strong heating can decompose it.
热固性网络具有共价交联，不会熔化；强烈加热可能使其分解。
Pause and use both density and mechanical behaviour.
暂停，同时考虑密度与力学表现。
Identify the resin matrix and fibre reinforcement of B.
指出B中树脂基体和纤维增强体。
The strength number alone does not answer the impact question.
仅靠强度数值不能回答抗冲击问题。
C has the highest stated strength but brittleness is a concern.
C的题给强度最高，但脆性是问题。
The provisional choice must be tested against the required service; carbon fibre alone does not identify both composite components.
初步选择仍需根据用途检验；仅说碳纤维不能指出复合材料的两个组分。
Nitrogen comes from air; hydrogen can come from natural gas and steam.
氮来自空气，氢可由天然气和蒸汽制得。
Cooling removes liquid ammonia while unreacted nitrogen and hydrogen return.
冷却分离液氨，未反应氮氢循环回用。
These process steps are required at both tiers.
这些过程步骤两种难度均需掌握。
Pause and apply the coefficient ratio one to three to two.
暂停，使用1比3比2的系数比例。
Separate per-pass conversion from the theoretical maximum and repeated recycling.
区分单次转化、理论最大值与多次循环。
The remaining gases can be recycled.
剩余气体可循环回用。
Recycling improves feedstock use but does not prove zero waste or zero energy consumption.
循环改善原料利用，但不证明零废物或零能耗。
Choose a compromise using rate and equilibrium evidence, including supplied graphs.
应结合速率、平衡证据和题给图表选择折中条件。
Higher temperature can increase rate while reducing equilibrium ammonia yield.
升温可提高速率，同时降低平衡氨产率。
Nitrogen, phosphorus and potassium are different nutrients.
氮、磷、钾是不同营养元素。
Ammonia neutralises nitric acid to make ammonium nitrate; phosphoric acid is not a potassium source.
氨与硝酸中和形成硝酸铵，磷酸并不是钾来源。
Pause and distinguish calcium nitrate and phosphoric acid from the two superphosphate products.
暂停，区分硝酸钙与磷酸以及两种过磷酸盐产品。
Note which product also contains calcium sulfate.
注意哪种还含硫酸钙。
The phosphate fertiliser in single and triple superphosphate is calcium dihydrogen phosphate.
普通过磷酸钙和重过磷酸钙中的磷肥成分是磷酸二氢钙。
Do not label every acid treatment simply superphosphate.
不能把所有酸处理产物都简单称为过磷酸钙。
Pause and separate a reaction method from the quality of the final processed solid.
暂停，区分反应方法与后续加工所得固体的质量。
Use supplied data rather than assuming industry always wins.
使用题给数据，不应假设工业一定更好。
Industry can integrate ammonia and acid manufacture and recover heat, but also needs equipment and emissions control.
工业可整合氨和酸的制造并回收热量，但也需要设备和排放控制。
A laboratory product is not inherently impure.
实验室产品并非天生不纯。
Pause and give a chemical or quantitative reason for each decision, plus a condition or missing datum.
暂停，为每个判断给出化学或定量理由，并说明条件或缺失资料。
Then compare scratched paint and scratched zinc on iron.
再比较铁表面划伤的漆层与锌层。
A sound answer chooses a process for the stated job and limits its claim to the evidence.
可靠答案应根据任务选择过程，并把结论限制在证据范围内。
Written answers support knowledge; practical performance remains separately recorded.
书面答案证明知识，实际操作仍需另行记录。