יחידה 1: מבנה, קשר והקדמה לאורגני
Pearson Edexcel · International A-Level · Chemistry נושא 1 17:44 קריאת קול באנגלית · תרגום אנגלי + סינית שרוף בתוך הסרטון
פרקים
תרגום טקסט
Use particle models to explain properties, then balanced equations to calculate amounts.
使用粒子模型解释性质，再用配平方程式计算物质的量。
Prerequisites: algebra, powers of ten and conservation of atoms and charge.
先修知识：代数、十的幂次以及原子和电荷守恒。
Convert known quantities to moles before applying a coefficient ratio.
在应用系数比之前，先将已知量转换为摩尔数。
Mass, solution and gas data use different equations and unit systems.
质量、溶液和气体数据使用不同的方程和单位系统。
Magnesium plus two hydrochloric acid yields magnesium chloride and hydrogen gas, with acid in excess.
镁与两分子盐酸反应生成氯化镁和氢气，酸过量。
The amount of magnesium equals mass divided by molar mass, which is 0.480 grams divided by 24.0 grams per mole, giving 0.0200 moles.
镁的物质的量等于质量除以摩尔质量，即 0.480 g 除以 24.0 g mol⁻¹，得到 0.0200 mol。
A one to one ratio gives the amount of hydrogen gas as 0.0200 moles.
1:1 的比例给出氢气的物质的量为 0.0200 mol。
The volume equals amount times molar volume, which is 0.0200 moles times 24 cubic decimeters per mole, resulting in 0.48 cubic decimeters.
体积等于物质的量乘以摩尔体积，即 0.0200 mol × 24 dm³ mol⁻¹ = 0.48 dm³。
How many atoms are in 0.100 moles of methane, C H 4, molecules?
0.100 摩尔甲烷 CH4 分子中含有多少个原子？
Use L equals 6.02 times 10 to the 23 per mole.
取阿伏伽德罗常数 L 等于 6.02 乘以 10 的 23 次方每摩尔。
Pause and justify your choice before the next slide.
请先暂停，在下一张幻灯片出现前说明你的选择理由。
Each CH4 molecule has one C and four H atoms.
每个CH4分子包含1个C和4个H原子。
N sub atoms equals 5 n L, which is 5 times 0.100 mole times 6.02 times 10 to the 23 per mole, which equals 3.01 times 10 to the 23.
N下标atoms等于5nL，即5乘以0.100摩尔乘以6.02乘以10的23次方每摩尔，结果为3.01乘以10的23次方。
Option B.
选项B。
Option A counts molecules, and option C counts only hydrogen atoms.
选项A计算的是分子数，而选项C仅计算氢原子数。
Divide each element mass by its atomic molar mass; form the simplest whole-number ratio.
将每种元素的质量除以其原子摩尔质量，得到最简整数比。
Multiply every empirical subscript by measured molecular mass / empirical mass.
用实测分子质量除以实验式质量，再乘以每个实验式的下标。
Balanced equations conserve atoms and charge; coefficients change, substance formulae do not.
配平的化学方程式遵循原子和电荷守恒；系数可以改变，但物质的化学式不能变。
Percentage yield equals the actual product mass divided by the theoretical product mass, times 100 percent.
产率等于实际产物质量除以理论产物质量，再乘以100%。
Atom economy equals the balanced desired-product mass divided by all the balanced product masses, times 100 percent.
原子经济性等于配平后目标产物质量除以所有配平产物总质量，再乘以100%。
Include coefficients.
需包含化学计量系数。
A high atom economy does not remove practical losses.
高原子经济性并不能消除实际损失。
Known limiting amount, collect gas, record volume, temperature and pressure.
已知限量反应物，收集气体，记录体积、温度和压强。
A leak lowers collected volume, so inferred molar gas volume is too small.
泄漏会降低收集到的气体体积，因此推算出的摩尔气体体积偏小。
For a formula by oxidation, heat to constant mass and avoid losing solid.
通过氧化法确定化学式时，需加热至恒重，并避免固体损失。
Mass spectrometry forms positive gas ions, accelerates them and separates them by m over z, then detects abundance.
质谱法生成带正电的气态离子，将其加速，并按质荷比分离，然后检测丰度。
A doubly charged mass-40 ion gives m over z equal to 20.
一个带两个电荷、质量数为 40 的离子，质荷比为 20。
Lower m over z bends more in magnetic separation.
在磁场分离中，质荷比较低的离子偏转更大。
For chlorine isotope proportions of 3 to 1, the chlorine molecule peaks at 70, 72 and 74 have heights in the ratio 9 to 6 to 1.
氯同位素比例为三比一时，氯分子在 70、72 和 74 处的峰高之比为九比六比一。
An s orbital is spherical; a p orbital has two lobes.
S轨道呈球形。 P轨道有两个瓣。
Each orbital holds at most two opposite-spin electrons.
每个轨道最多容纳两个自旋相反的电子。
s, p and d subshells contain 1, 3 and 5 orbitals: capacities 2, 6 and 10.
S、P和D亚层分别包含1、3和5个轨道，其容量分别为2、6和10。
Fill equal-energy orbitals singly first.
填充能量相等的轨道时，先单电子占据。
Cr is 3d⁵4s¹; Cu is 3d¹⁰4s¹.
铬的电子排布为 3d5 4s1。 铜的电子排布为 3d10 4s1。
Remove 4s first for their ions.
形成离子时，先移除4s电子。
Period 3 generally rises as nuclear charge increases with similar inner shielding.
第三周期总体随核电荷增加而升高，内层屏蔽效应相似。
Al loses a higher-energy 3p electron; S loses an electron from a paired 3p orbital.
铝失去一个较高能量的3p电子；硫从一对配对的3p轨道中失去一个电子。
Three successive ionisation energies are close; the fourth is much larger.
三个连续的电离能数值相近，第四个则大得多。
What follows from the jump alone?
仅凭这一跃变能得出什么结论？
Pause and justify your choice before the next slide.
请先暂停，在下一张幻灯片出现前论证你的选择。
Option A. Three outer electrons were removed before an inner-shell electron.
选项 A：在移除内层电子之前，先移除了三个外层电子。
This supports main group 3, also numbered 13.
这支持主族第 3 族，也称第 13 族。
The limited list does not count all occupied shells.
该有限列表并未计入所有已占据的电子层。
Recognising familiar numerical values is additional evidence, not the jump itself.
识别熟悉的数值是额外证据，而非跃迁本身。
Each orbital holds at most two electrons with opposite spins.
每个轨道最多容纳两个自旋相反的电子。
Fill the three equal-energy p orbitals singly before pairing.
在配对之前，先分别填入三个能量相等的p轨道中。
Pause and classify all four configurations.
请先暂停，对全部四种构型进行分类。
Do not decide the block from the helium column alone.
不要仅凭氦元素所在的列来判断能块。
Explain the subshell receiving the differentiating electron.
请说明接收不同电子的亚层。
Potassium is s-block, scandium d-block and arsenic p-block.
钾是s区，钪是d区，砷是p区。
Helium has a full first shell but its electrons are in an s subshell.
氦具有充满的第一电子层，但其电子位于s亚层中。
Its noble-gas chemical behaviour explains its column; it does not give helium a p electron.
其稀有气体的化学性质解释了它所在的族；这并不赋予氦一个p电子。
Pause to plot every supplied value.
请先暂停，绘制所有提供的数值。
Keep atomic number on the horizontal axis.
将原子序数置于横轴上。
Compare first ionisation of different atoms, not successive ionisations of one atom.
比较不同原子的第一电离能，而非同一原子的逐级电离能。
The logarithmic plot must preserve the order of the values.
对数图必须保持这些数值的相对顺序。
Compare your plot with all thirty-six points.
请将你的图表与全部36个数据点进行对比。
Helium, neon, argon and krypton mark noble-gas peaks.
氦、氖、氩和氪标志着稀有气体峰值。
Moving from helium to lithium, neon to sodium or argon to potassium adds an occupied outer shell.
从氦到锂、从氖到钠或从氩到钾，都会增加一个被占据的外层电子壳。
The d-block does not show a perfectly smooth rise.
d区并未呈现出完全平滑的上升趋势。
The logarithmic plot uses the same data.
对数图使用相同的数据。
It compresses large numerical values and makes ratios easier to compare.
它压缩了较大的数值，使比率更容易比较。
A change of one in the base-ten logarithm represents a factor of ten in energy.
以10为底的对数变化一个单位，代表能量变化一个数量级。
It does not remove the beryllium-to-boron or nitrogen-to-oxygen decreases.
但这并不能消除铍到硼或氮到氧的下降趋势。
Pause and name the particles and attractions in each solid.
请先暂停，说出每种固体中的微粒及作用力。
Explain what is overcome when nitrogen melts, and what remains intact.
解释氮熔化时被克服的是什么，以及什么保持完整。
Use the distinction between a giant covalent network and separate molecules.
请运用巨型共价网络与独立分子之间的区别进行说明。
Lithium and beryllium have metallic structures.
锂和铍具有金属结构。
Boron and carbon form giant covalent networks.
硼和碳形成巨型共价网络。
Nitrogen, oxygen and fluorine form molecular solids, while neon is monatomic.
氮、氧和氟形成分子固体，而氖是单原子分子。
Melting nitrogen overcomes intermolecular attractions, not the triple bond inside each molecule.
熔化氮时克服的是分子间作用力，而非每个分子内部的三键。
Ionic lattice: strong net attraction; fixed ions cannot carry current in the solid.
离子晶格具有强净吸引力。 固态中固定的离子无法导电。
Molten/dissolved ions can move.
熔融或溶解状态的离子可以移动。
Metals conduct through delocalised electrons.
金属通过离域电子导电。
Small molecular substances usually boil by overcoming intermolecular attractions.
小分子物质通常通过克服分子间作用力而沸腾。
NH₄⁺ and CH₄: four bonding regions, tetrahedral, 109.5°.
NH4+和CH4：四个成键区域，四面体构型，键角109.5度。
NH₃: three bonds and one lone pair, pyramidal, about 107°.
NH3：三个成键电子对和一个孤电子对，三角锥形，键角约107度。
H₂O: two bonds and two lone pairs, bent, about 104.5°.
H2O：两个成键电子对和两个孤电子对，V形（或弯曲形），键角约104.5度。
Lone pairs repel more strongly; the angle reduction is not a fixed amount per pair.
孤电子对的排斥作用更强；键角的减小量并非每个孤电子对固定减少相同的度数。
Shared pairs identify covalent bonds.
共用电子对表示共价键。
Lone pairs must also be shown.
孤对电子也必须标出。
These flat diagrams do not show the 3D shapes.
这些平面示意图无法展示三维空间构型。
NH₃ has a lone pair; BF₃ has six electrons around B.
NH3含有一对孤对电子。 BF3中B原子周围有六个电子。
Electronegativity differences create bond dipoles.
电负性差异产生键偶极矩。
CO₂ has polar bonds whose dipoles cancel; water is bent and its dipoles do not cancel.
CO₂具有极性键，其偶极矩相互抵消。 水分子呈弯曲构型，其偶极矩无法抵消。
A small highly charged cation polarises an anion; larger anions distort more easily.
小而高电荷的阳离子会使阴离子极化；较大的阴离子更容易发生变形。
Each ion has a full outer shell.
每个离子都具有完整的价电子层。
Show Na⁺ and Cl⁻ in brackets with their charges.
将Na⁺和Cl⁻用括号括起并标出电荷。
The gained chloride electron comes from sodium; sodium’s displayed outer shell is now its filled inner shell.
氯离子获得的电子来自钠原子。 钠原子显示的价电子层现在变成了其充满的内层电子壳。
Carbon dioxide has sixteen outer electrons.
二氧化碳有16个价电子。
Each C–O double bond has two shared pairs; each oxygen has two lone pairs.
每个C-O双键包含两对共用电子；每个氧原子有两对孤对电子。
The flat electron diagram does not itself show molecular shape.
平面的电子点图本身并不显示分子形状。
Nitrogen has five outer electrons per atom, so the molecule has ten.
每个氮原子有五个外层电子，因此该分子共有十个电子。
Three pairs are shared between the atoms, giving a triple bond.
三个电子对在原子间共享，形成三键。
Each atom also retains one lone pair.
每个原子还保留一对孤对电子。
Count six shared electrons and four lone-pair electrons.
统计六个共享电子和四个孤对电子。
A line formula shows the bonds but hides the two lone pairs.
线形公式显示化学键，但隐藏了两对孤对电子。
NH₄⁺ has eight outer electrons.
NH4+ 有八个外层电子。
Nitrogen donates its lone pair to a hydrogen ion to form one coordinate bond.
氮将其孤对电子提供给氢离子，形成一个配位键。
There is no lone pair remaining on nitrogen, and the ion is tetrahedral.
氮上不再有孤对电子剩余，该离子呈四面体构型。
Each bridging chlorine supplies a pair to the other aluminium.
每个桥接氯原子向另一个铝原子提供一对电子。
Terminal chlorines have three lone pairs; bridging chlorines have two.
端基氯原子有三对孤对电子；桥接氯原子有两对。
Count all forty-eight outer electrons before accepting the diagram.
在确认该结构图之前，请数清所有四十八个外层电子。
Every skeletal end/corner is C; add implied H to give carbon four bonds.
每个骨架末端或拐角都是碳原子。 添加隐含的氢原子，使每个碳原子形成四个键。
Acyclic alkanes: CₙH₂ₙ₊₂.
对于无环烷烃，分子式为C下标nH下标2n加2。
Single-ring cycloalkanes: CₙH₂ₙ. Both are saturated.
对于单环环烷烃，分子式为C下标nH下标2n。
Structural isomers share a molecular formula but differ in atom connectivity.
两者均为饱和化合物。 结构异构体具有相同的分子式，但原子连接方式不同。
Initiation: Cl two yields two Cl dot in UV light. Homolytic cleavage gives one electron to each fragment.
引发：Cl₂在紫外光下生成2个Cl·；均裂使每个碎片各获得一个电子。
Propagation: Cl dot plus CH four yields HCl plus CH three dot. CH three dot plus Cl two yields CH three Cl plus Cl dot.
链增长：Cl· + CH₄ → HCl + CH₃·；CH₃· + Cl₂ → CH₃Cl + Cl·。
Termination: Two CH three dot yields C two H six.
链终止：2个CH₃·结合生成C₂H₆。
Further substitution gives mixtures.
进一步取代会生成混合物。
Cracking makes smaller hydrocarbons including alkenes; balance atoms in proposed products.
裂解生成较小的烃，包括烯烃。 配平拟议产物中的原子。
CO reduces oxygen transport; nitrogen/sulfur oxides can contribute to acid pollution.
一氧化碳会降低氧气运输能力。 氮氧化物和硫氧化物可能导致酸污染。
Bioethanol growth can absorb CO₂, but processing and transport still emit.
生物乙醇的生长可以吸收二氧化碳，但其加工和运输过程仍会排放。
Hydrogen production also matters.
氢气的生产同样重要。
Pentane has three constitutional isomers.
戊烷有三种构造异构体。
Start with a five-carbon chain, then a four-carbon chain with a methyl group, then a three-carbon chain with two methyl groups.
首先画一条五个碳的直链，接着画一条四个碳的主链并带一个甲基取代基，最后画一条三个碳的主链并带两个甲基取代基。
Reversing the numbering does not create another connectivity.
反转编号顺序不会产生新的连接方式。
Hexane, 2-methylpentane and 3-methylpentane are distinct connectivities.
己烷、二甲基戊烷和三甲基戊烷具有不同的连接方式。
2,2-dimethylbutane and 2,3-dimethylbutane complete the five hexane structures.
2,2-二甲基丁烷和2,3-二甲基丁烷构成了全部五种己烷结构。
Keep total carbon count and one saturated ring fixed while changing ring size and substituents.
保持总碳原子数和单个饱和环不变，同时改变环的大小和取代基。
Begin with a six-carbon ring, then a five-carbon ring plus one carbon, then a four-carbon ring plus two.
先从六元环开始，接着是五元环加一个碳原子，最后是四元环加两个碳原子。
On a four-carbon ring, the two extra carbons can occupy positions 1,1 or 1,2 or 1,3.
在四碳环上，两个额外的碳原子可以占据1,1位、1,2位或1,3位。
A three-carbon ring can carry propyl, propan-2-yl, or ethyl and methyl on one ring carbon.
一个三碳环可以在同一个环碳原子上连接丙基、异丙基，或乙基与甲基。
Finish with ethyl/methyl on different carbons and the two distinct trimethyl connectivities.
最后考虑乙基和甲基位于不同碳原子上的情况，以及两种不同的三甲基连接方式。
UV initiation splits the Cl–Cl bond homolytically: one electron goes to each atom.
紫外光引发使Cl–Cl键发生均裂，每个原子各获得一个电子。
C=C has one sigma and one pi bond.
碳碳双键包含一个σ键和一个π键。
Rotation would destroy pi overlap.
旋转会破坏π键的重叠。
E/Z needs two different groups on each double-bond carbon; rank the groups separately.
E/Z构型要求每个双键碳原子上连接两个不同的基团，需分别对基团进行优先级排序。
A curly arrow begins at the pi bond and ends at H; another sends the H–Br bonding pair to Br.
一条弯曲箭头从π键开始，终止于H。 另一条将H–Br成键电子对推向Br。
Bromide supplies a lone pair to the carbocation, forming C–Br.
溴离子向碳正离子提供孤对电子，形成C–Br键。
Hydrogen with a nickel catalyst adds hydrogen; steam with an acid catalyst adds water; bromine adds across the carbon–carbon double bond.
氢气在镍催化下加氢；水蒸气在酸催化下加水；溴加成到碳碳双键上。
Cold dilute acidified manganate seven gives a diol.
冷的稀酸化高锰酸钾生成二醇。
Bromine decolourisation supports unsaturation, not a unique identity.
溴水褪色可支持不饱和性，但并非唯一鉴定依据。
Addition polymerisation opens the carbon–carbon double bond, retains side groups and loses no small molecule.
加成聚合打开碳碳双键，保留侧基，且不脱除小分子。
The pi pair attacks the delta-positive bromine; the bromine–bromine bond pair moves to the delta-negative bromine.
π电子对进攻带部分正电荷的溴原子，Br–Br键的电子对转移到带部分负电荷的溴原子上。
Bromide then supplies a pair to the positive carbon.
随后，溴离子提供一对电子与带正电的碳原子结合。
Open C=C into the polymer backbone, keep the methyl side group, and extend bonds through the brackets.
将C等于C打开，形成聚合物主链，保留甲基侧基，并穿过括号延伸化学键。
The pi bond supplies a pair to hydrogen.
π键提供一对电子给氢原子。
The hydrogen-bromine bond supplies its pair to bromine.
氢溴键提供其电子对给溴原子。
Bromide then donates a lone pair to the positive carbon of the ethyl carbocation.
随后，溴离子向乙基碳正离子的带正电碳原子提供孤对电子。
Either ethene carbon can receive the proton, but both routes give the same connectivity because the carbons are equivalent.
乙烯的任一碳原子均可接受质子，但由于两个碳原子等价，两条路径均得到相同的连接方式。
A 0.480 gram sample produces 0.240 cubic decimetres of gas in a 1 to 1 reaction.
0.480 克样品在一比一反应中产生 0.240 立方分米气体。
Use a molar gas volume of 24.0 cubic decimetres per mole.
取气体摩尔体积为每摩尔 24.0 立方分米。
A leak makes the gas volume too low.
泄漏导致气体体积偏低。
What happens to the inferred sample molar mass?
推断出的样品摩尔质量会如何变化？
Pause and justify your choice before the next slide.
请先暂停，在下一张幻灯片出现前说明你的选择理由。
n equals V divided by V sub m, which is 0.240 dm cubed per mole divided by 24.0 dm cubed per mole, equaling 0.0100 mol.
n 等于 V 除以 Vm，即 0.240 dm³/mol 除以 24.0 dm³/mol，等于 0.0100 mol。
M equals m divided by n, which is 0.480 g divided by 0.0100 mol, equaling 48.0 g per mol.
M 等于 m 除以 n，即 0.480 g 除以 0.0100 mol，等于 48.0 g/mol。
Option A: a leak underestimates n; dividing fixed mass by too small n overestimates M.
选项 A：泄漏会导致 n 的估计值偏小；用固定质量除以过小的 n 会使 M 的估计值偏大。