Sujet 3 : Chimie quantitative
AQA · GCSE · Chemistry Sujet 3 9:40 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
Quantitative chemistry connects measured mass and volume to particles.
定量化学把质量、体积与粒子联系起来。
Begin with the balanced equation and choose a relation that fits the conditions. Keep units visible and name the particles counted.
先配平方程式，再选择适用关系，保留单位并指出计数粒子。
Mole calculations are Higher Tier; mass concentration remains available at both tiers.
摩尔计算属于高阶要求，质量浓度适用于两个等级。
Balance magnesium reacting with oxygen, changing coefficients only. Then compare coefficient-weighted relative masses.
配平镁与氧气反应，只改系数，再比较乘以系数后的相对质量总和。
Pause and attempt before the answer.
先独立作答，再看答案。
Two magnesium atoms and one oxygen molecule form two magnesium oxide formula units. Both sides contain two atoms of each element. The weighted mass totals are eighty on each side.
两个镁原子与一个氧分子生成两个氧化镁式量单位。
Adding twenty-four and thirty-two and comparing with forty ignores coefficients.
两边各有两个镁和氧原子，乘系数后的质量总和均为80。
Subscripts identify a substance and must not be changed to balance.
不能忽略系数，也不能改下标来配平。
A bracket multiplier applies to everything inside the bracket.
括号后的下标作用于括号内全部原子。
Ammonium sulfate contains two nitrogens and eight hydrogens. Relative formula mass has no unit.
硫酸铵含两个氮、八个氢，相对式量132，无单位。
Sulfur contributes thirty-two out of one hundred thirty-two, giving twenty-four point two percent by mass, not by atom count.
硫的质量占32/132，即24.2%，不能用原子个数比例代替。
The plotted curves illustrate possible trends, not universal experimental values.
曲线示意可能趋势，不是通用实验数据。
Metal mass rises when oxygen joins the weighed sample. A carbonate residue loses mass as carbon dioxide leaves. The total system still conserves atoms and mass.
氧气进入称量样品时金属增重，二氧化碳离开时碳酸盐残渣减重。
Explain the system boundary; do not seal a heated gas-producing reaction.
完整系统仍守恒。 应说明系统边界，不得密封加热产气反应。
Use all three valid mass readings. Calculate the mean and range, then identify a source of error that repeated readings might miss.
用三个有效质量读数计算平均值、极差，并指出重复读数可能无法发现的误差。
Pause and attempt before the answer.
先独立作答，再看答案。
Plot the three readings at equally spaced positions on a mass number line.
在质量数轴上标出三个等距读数。 平均2.42克，极差0.04克。
Their mean is two point four two grams and range zero point zero four grams.
若明确采用半极差约定，可写2.42±0.02克。
Under a stated half-range convention, report plus or minus zero point zero two grams. A calibration error could shift all three readings together. Do not discard a reading just because it is far away; record a justified exclusion.
校准误差可能使全部读数同向偏移；排除读数必须有记录和理由。
The mole counts atoms, molecules, ions, electrons or stated formula units. Always name what is counted.
摩尔可以计数原子、分子、离子、电子或指定式量单位，必须说清对象。
One mole of oxygen molecules contains twice as many oxygen atoms.
一摩尔氧分子含两摩尔氧原子。
Molar mass in grams per mole is numerically equal to relative formula mass, but the latter has no unit.
摩尔质量单位为克每摩尔，数值等于无单位的相对式量。
Divide grams by grams per mole to obtain moles.
质量除以摩尔质量得到摩尔数。
Find the molar mass, calculate the amount and multiply by the Avogadro constant.
先求摩尔质量，再算物质的量，最后乘阿伏伽德罗常数。
Pause and attempt before the answer.
先独立作答，再看答案。
Eight grams divided by forty grams per mole gives zero point two zero moles. Multiplying by the number of formula units per mole gives the count.
8克除以40克每摩尔得到0.20摩尔，乘每摩尔式量单位数得到1.20×10的23次方个式量单位。
This is not the count of all atoms: each sodium hydroxide formula unit represents one sodium, one oxygen and one hydrogen.
这不是全部原子数，每个式量单位含一个钠、一个氧和一个氢。
Use the equation ratio after converting the carbonate mass to moles.
先把碳酸钙质量换成摩尔数，再使用方程式比例。
Pause and attempt before the answer.
先独立作答，再看答案。
The one-to-one ratio relates moles, not equal masses.
一比一对应摩尔数，不是质量相等。
Carbon dioxide is lighter per mole than calcium carbonate.
二氧化碳每摩尔比碳酸钙轻。
For a reverse calculation start from product mass, divide by its molar mass, apply the reverse equation ratio and multiply by reactant molar mass.
反算时从产物质量出发，除以摩尔质量、逆向使用比例，再乘反应物摩尔质量。
Convert each substance mass using its actual formula. Oxygen reacts as oxygen molecules, not isolated atoms.
按各物质的实际化学式换算，氧气反应物是氧分子，不是孤立氧原子。
Pause and attempt before the answer.
先独立作答，再看答案。
The smallest amount is zero point one zero moles. Dividing all three amounts gives two to one to two.
最小物质的量为0.10摩尔，相除得到2:1:2。
If fractions remain, multiply the whole ratio by a common number.
若仍有分数，要将整个比例同乘适当整数。
Verify every element and the mass totals rather than treating the arithmetic as sufficient proof.
最后检查每种元素和总质量，不能只信算术。
Compare amount divided by coefficient for each reactant, then use the limiting amount.
把各反应物摩尔数除以系数比较，再用限量反应物计算。
Pause and attempt before the answer.
先独立作答，再看答案。
Magnesium has the smaller amount per coefficient and is used up first.
镁的摩尔数除以系数更小，先耗尽。
The oxygen used is half the magnesium amount, so half the original oxygen remains.
氧气消耗量是镁的一半，因此剩一半氧气。
Check total mass: eleven point two grams before equals eight point zero plus three point two afterward.
反应前11.2克等于反应后8.0加3.2克。
Exactly stoichiometric amounts can use up both reactants together.
恰好按化学计量比混合时，两者可同时耗尽。
Concentration is solute mass per final solution volume.
质量浓度是溶质质量除以最终溶液体积。
Convert cubic centimetres to cubic decimetres by dividing by a thousand. For a smaller portion multiply concentration by its volume.
立方厘米除以1000换成立方分米；求小份溶质质量用浓度乘体积。
At fixed solute mass, increasing the final volume lowers concentration; at fixed volume, more solute raises it. The explanation of these relationships is Higher Tier.
固定溶质质量时增加体积会降低浓度，固定体积时增加溶质会提高浓度；关系解释属高阶。
Compare masses of the same desired product, then distinguish recovery losses from atom economy.
比较同一目标产物的实际和理论质量，并把回收损失与原子经济区分开。
Pause and attempt before the answer.
先独立作答，再看答案。
The denominator is the maximum theoretical product, not the starting reactant mass.
分母是理论最大产物质量，不是起始反应物质量。
Incomplete reversible reaction, recovery losses and side reactions can lower isolated yield.
可逆反应不完全、分离损失和副反应都可能降低分离产率。
A value above one hundred percent calls for checking water, contamination and measurements. Higher Tier may require calculating the theoretical product first from a balanced equation.
超过100%应检查水分、杂质和测量；高阶题可能先要求按方程式求理论产量。
Include coefficients on both sides.
两边都要乘系数。
Atom economy comes from the equation rather than the recovered sample.
原子经济由反应方程式决定，不取决于回收到的样品。
Pause and attempt before the answer.
先独立作答，再看答案。
Both water molecules count as useful product.
两份水都属于目标产物。
Using only eighteen in the numerator falsely gives fifty percent.
分子只用18会误得50%。
An experiment with this equation may still recover less than its theoretical water because atom economy and percentage yield measure different things.
即使原子经济100%，实验回收量也可能低于理论，因为原子经济和产率衡量不同方面。
Route A has better atom utilisation, while route B has higher recovery relative to its own theoretical maximum and a potentially useful by-product.
A的原子利用率更高，B相对于各自理论量的回收率更高，并有可能有用的副产物。
A judgement must consider supplied conditions and missing cost or energy data.
判断需结合条件和缺失的成本、能耗数据。
The percentages alone do not establish product mass per batch because theoretical batch amounts are not supplied.
未给每批理论产量，仅凭百分数不能判断哪种每批产物更多。
A concentration of zero point two zero moles per cubic decimetre states amount per volume, not the amount in every beaker.
0.20摩尔每立方分米表示单位体积物质的量，不是每个烧杯都含0.20摩尔。
For sodium hydroxide, multiplying by forty grams per mole gives eight grams per cubic decimetre. Units cancel to establish the conversion.
氢氧化钠乘40克每摩尔得到8.0克每立方分米，单位约分验证换算。
Start from the known alkali amount. The acid-to-alkali ratio is one to two.
先求已知碱的物质的量，酸碱摩尔比为一比二。
Pause and attempt before the answer.
先独立作答，再看答案。
The acid has half as many moles because each acid formula unit reacts with two hydroxide units.
每份酸与两份碱反应，因此酸摩尔数为碱的一半。
Divide by the acid volume, not the alkali volume.
除以酸体积，不能除以碱体积。
Multiplying the calculated acid amount by two recovers the known alkali amount.
将算出的酸摩尔数乘二应回到已知碱摩尔数。
Equal acid and alkali moles occur only for a one-to-one equation.
只有一比一方程式才有酸碱摩尔数相等。
This specification uses twenty-four cubic decimetres per mole at room temperature and pressure. It is a condition-dependent approximation. Equal gas amounts occupy equal volumes under matching conditions.
本考纲室温20摄氏度、一大气压下用24立方分米每摩尔，是依赖条件的近似。
Do not use the gas volume relation for a solid or condensed liquid merely because it appears in a balanced equation.
同温同压下等摩尔气体体积相同。 不能仅因固体或液体出现在方程式中就套用气体体积关系。
Convert mass to amount, then use room-condition molar gas volume.
先把质量换成物质的量，再用室温摩尔气体体积。
Pause and attempt before the answer.
先独立作答，再看答案。
The amount is zero point one zero zero moles. Multiplying by twenty-four gives two point four zero cubic decimetres, or two thousand four hundred cubic centimetres.
物质的量0.100摩尔，乘24得2.40立方分米，即2400立方厘米。
In reverse, divide a cubic-decimetre volume by twenty-four and multiply by molar mass to find grams.
反算质量时，立方分米体积除以24，再乘摩尔质量。
Compare volume divided by coefficient for gases measured alike. Assume complete reaction for the theoretical result.
对同条件气体，用体积除以系数比较；理论结果假定完全反应。
Pause and attempt before the answer.
先独立作答，再看答案。
The sixty cubic centimetres of nitrogen would require one hundred eighty of hydrogen, but only one hundred fifty are present. Hydrogen limits the theoretical product. Ten cubic centimetres nitrogen remain.
60立方厘米氮气需180立方厘米氢气，实际只有150，因此氢气限制理论产量，生成100立方厘米氨并余10立方厘米氮。
A real reversible reaction can give a smaller ammonia yield than this complete-reaction maximum.
实际可逆反应产量可能低于完全反应理论值。
Close the worked answers. Explain an apparent mass change by naming the gas and system boundary.
合上答案，指出气体和系统边界以解释表观质量变化。
Solve a new mass ratio, limiting mixture and non-one-to-one concentration calculation. Explain why yield and atom economy differ. For gas volumes state temperature and pressure.
独立完成新的质量比例、限量混合物和非一比一浓度计算，区分产率与原子经济。
Use sheets three point one to three point five, including the unfamiliar transfer tasks.
气体体积题说明温压，用练习3.1至3.5进行迁移练习。