Tema 2: Enlace, estructura y propiedades de la materia
AQA · GCSE · Chemistry Tema 2 13:15 Narración en inglés · Subtítulos en inglés + 中文 quemados en pantalla
Capítulos
Transcripción
A melting point or conductivity is a property of a bulk sample.
熔点和导电性属于宏观样品的性质。
Explain it by identifying the structure and particles, then the relevant attraction or moving charge carrier.
先确定结构和粒子，再指出相应的引力或能移动的载流粒子，最后联系实验性质。
A single property may support more than one interpretation; use the full pattern of evidence.
单一性质可能支持多种解释，应结合全部证据。
An ionic bond is attraction between opposite ions; electron transfer is how those ions form.
离子键是异号离子之间的引力，电子转移是形成这些离子的过程。
A covalent pair is shared and attracted to both nuclei.
共价电子对被两个原子核共同吸引。
In a metal, electrons extend through the structure and attract the positive ions.
金属中电子在整个结构中离域，并吸引阳离子。
The small drawings are models, not complete giant structures.
这些小图是模型，不是完整的巨型结构。
Mg forms Mg²⁺ and chlorine forms Cl⁻. Which neutral formula follows?
镁形成Mg²⁺，氯形成Cl⁻。
Pause and attempt this independently.
哪一个化学式满足电中性？
Give a reason before revealing the answer.
先独立作答，并说明理由，再看下一页。
Choose MgCl2.
选择MgCl₂。
Magnesium begins with electron arrangement two, eight, two and becomes two, eight. Each chlorine begins two, eight, seven and gains one electron.
镁原子的电子排布由2,8,2变为2,8；每个氯原子由2,8,7得一个电子。
Two chloride ions are needed for one magnesium ion.
一个镁离子需要两个氯离子。
The formula is the simplest charge-balanced ratio, not a separate molecule in the lattice.
化学式表示最简电荷平衡比，不表示晶格中的独立分子。
Check brackets and charges, the one-to-two ion ratio and all eight outer electrons around each chloride.
检查方括号、电荷、一比二的离子比例，以及每个氯离子周围的八个外层电子。
The diagram omits the remaining filled inner shells of magnesium. Magnesium two-plus still has ten electrons.
图中省略了镁剩余的满内层，Mg²⁺仍有十个电子。
Dots and crosses indicate origin only; they are not different kinds of electron.
点和叉只表示来源，不是两类不同的电子。
If the stated representative region has six magnesium ions and twelve chloride ions, reduce the ratio to one to two.
若给出的代表区域有六个镁离子和十二个氯离子，比例约成一比二。
A lattice continues beyond the drawing.
晶格在图外继续延伸。
Do not call its pictured twelve chlorides a discrete molecule, and do not infer a bulk ratio from an arbitrary edge fragment without the stated sharing information.
不要把画出的十二个氯离子当成独立分子，也不要忽略边界共享信息，从任意边缘片段直接推断整体比例。
One hydrogen contributes one outer electron; oxygen contributes six, nitrogen five, carbon four and chlorine seven.
每个氢提供一个外层电子，氧提供六个，氮五个，碳四个，氯七个。
Use one mark type for each atom source.
用不同符号标明电子来源。
One single bond contains a pair of electrons.
一个单键含一对电子。
After bonding, check each hydrogen has two electrons around it and each other atom in these named molecules has eight.
成键后检查每个氢周围有两个电子，其余这些指定分子的原子周围有八个电子。
Two hydrogen atoms each supply one electron. Sharing gives each a filled first shell.
两个氢各提供一个电子，共用后各自第一层充满。
A line between H atoms represents the pair, not one electron.
氢之间的一条线代表一对电子，不是一个电子。
Count fourteen outer electrons: two in the bond and twelve in six lone pairs.
共十四个外层电子：键中两个，六对孤电子对中十二个。
Missing the lone pairs gives an incomplete electron drawing even when the single bond is correct.
即使单键正确，漏掉孤电子对仍是未完成的电子图。
Each oxygen supplies six outer electrons.
每个氧提供六个外层电子。
Four electrons form the double bond and eight remain in lone pairs. The molecule contains twelve outer electrons in total.
四个电子形成双键，八个保留为孤电子对，分子共十二个外层电子。
Each nitrogen supplies five outer electrons.
每个氮提供五个外层电子。
The triple bond uses six electrons; two lone pairs use the other four.
三键使用六个电子，两对孤电子对使用其余四个。
The line formula N triple bond N hides those lone pairs.
N≡N线式省略了孤电子对。
The shared pair contains one hydrogen electron and one chlorine electron.
共用对由氢和氯各提供一个电子。
Hydrogen has two around it; chlorine has eight.
氢周围有两个，氯周围有八个。
Count eight outer electrons overall, not eight around hydrogen.
总外层电子为八个，不是氢周围有八个。
Oxygen supplies six and the two hydrogens supply two more.
氧提供六个，两个氢再提供两个。
Two bonds use four electrons; two lone pairs use the remaining four.
两个键用四个电子，两对孤电子对用另外四个。
This flat drawing shows electron allocation, not the actual bond angle.
平面图表示电子分配，不表示真实键角。
Five nitrogen electrons plus three hydrogen electrons give eight.
氮的五个加三个氢的三个，共八个电子。
The three bonds use six; the lone pair uses two.
三个键用六个，孤电子对用两个。
The diagram is not a claim that the real molecule is flat.
此图不意味着真实分子是平面的。
Carbon supplies four electrons and the four hydrogens supply four. All eight enter shared pairs.
碳提供四个电子，四个氢各提供一个，八个全部形成共用对。
The four lines in a displayed formula each stand for two electrons.
结构式的每条线代表两个电子。
An N≡N line formula omits which feature?
N≡N线式省略了什么？
Pause and attempt this independently. Give a reason before revealing the answer.
先独立作答，并说明理由，再看下一页。
The answer is the two lone pairs.
答案是两对孤电子对。
The line formula clearly shows connectivity and bond order but not all electrons.
线式清楚表示连接和键级，却不表示所有电子。
A ball-and-stick model helps show arrangement but rods and gaps are symbolic.
球棍模型有助于观察排列，但杆和间隙是符号。
A small two-dimensional lattice omits neighbours above and below.
小型二维晶格省略上下邻居。
State what is useful as well as what is omitted.
评价模型时既指出用途，也指出遗漏。
X melts at −15 °C and boils at 65 °C at one pressure.
同一气压下，X熔点为−15°C，沸点为65°C。
Its states at −30, 20 and 90 °C are: Pause and attempt this independently.
在−30、20、90°C分别是什么状态？
Give a reason before revealing the answer.
先独立作答，并说明理由，再看下一页。
The answer is solid, liquid, gas.
答案为固、液、气。
Heating changes energy and particle motion; atoms do not become larger pieces of gas.
加热改变能量和粒子运动，原子不会变成更大的气体碎块。
Melting allows particles to move past one another while remaining close.
熔化后粒子仍相近，但可以相互移动；沸腾时距离增大得多。
Boiling separates them much further. Cooling reverses these transfers.
冷却使能量向外转移，过程反向进行。
The comparison uses transition data at the stated pressure.
判断依赖指定气压下的转变数据。
This model-limitation discussion is Higher Tier.
本页模型局限属于高阶内容。
Without attractions, hard spheres alone cannot explain energy needed to separate particles or different boiling temperatures.
没有引力的硬球模型不能解释分离粒子所需能量，也不能解释不同沸点。
Keep useful information such as particle-number conservation, but add an attraction account where needed.
保留粒子数守恒等有用信息，并在需要时补充引力。
Bulk hardness and melting point describe many interacting particles.
宏观硬度和熔点描述大量相互作用的粒子。
Aqueous means dissolved in water, not simply liquid.
aq表示溶于水，不只是液态。
The magnesium reaction includes a solid metal, an acid solution, a dissolved salt and a hydrogen gas product.
镁的反应含固体金属、酸溶液、溶解盐和氢气产物。
Read observations and conditions before assigning states; formula letters alone do not determine them.
先看条件和观察，再写状态符号；不能只凭化学式字母判断。
A charged particle is not enough; it must move to carry current through the sample.
有电荷还不够，粒子必须能移动才能传导电流。
Solid sodium chloride has fixed ions and does not conduct. Molten sodium chloride has moving ions and conducts.
固体氯化钠的离子固定，不能导电；熔融时离子能移动，可以导电。
Dissolved ions also move; dissolution is not the same process as melting.
溶液中的离子也能移动，但溶解不是熔化。
Explain the stated sample rather than assuming all salts dissolve readily.
依据题设样品解释，不要假定所有盐都易溶。
Water vapour remains water molecules.
水蒸气中仍是水分子。
Boiling overcomes attractions between molecules, not the oxygen-hydrogen bonds inside each one.
沸腾克服分子之间的引力，不打断分子内部的氧氢键。
Comparable larger molecules tend to have stronger intermolecular attractions and higher boiling points. Molecule size is a trend, not an exact universal prediction; shape and other interactions can matter.
可比较的较大分子通常分子间力更强、沸点更高，但分子大小只是趋势，形状和其他相互作用也会影响结果。
In a polymer, strong covalent bonds run along each chain and relatively strong intermolecular attractions act between the large molecules.
聚合物每条链内有强共价键，大分子之间有较强分子间力。
A repeat unit has continuation bonds crossing brackets.
重复单元的延伸键穿过括号。
In silica or diamond, covalent bonds form a continuous network.
二氧化硅和金刚石则构成连续共价网络。
Softening a polymer is not equivalent to breaking every covalent bond in a giant network.
聚合物软化不等于打断巨型网络中的所有共价键。
An alloy is harder than its pure metal and still conducts.
合金比纯金属硬，但仍能导电。
Which explanation fits?
哪项解释符合？
Pause and attempt this independently. Give a reason before revealing the answer.
先独立作答，并说明理由，再看下一页。
Different-sized atoms distort the regular metal layers, making sliding harder.
不同大小的原子扰乱金属规则层状排列，使滑动更困难。
Mobile delocalised electrons remain, so charge can flow.
离域电子仍能移动，所以仍可导电。
Increasing hardness does not imply fixing every electron. Those electrons also transfer thermal energy through a metal.
硬度增加不意味着所有电子都固定；这些电子也传递热能。
The left projection shows diamond; the right compares silica.
左图是金刚石，右图对比二氧化硅。
Diamond is hard because a rigid network resists deformation. It has no mobile electrons for conduction.
金刚石的刚性网络抵抗变形，因此硬；没有可移动电子，所以不导电。
The flat projection does not show tetrahedral bond angles and is only part of the giant structure.
平面投影不表示四面体键角，也只是巨型结构的一部分。
Three covalent neighbours leave one delocalised electron per carbon. This explains conduction.
每个碳有三个共价邻居，留下一个离域电子，解释导电。
Weaker attractions between layers allow sliding and explain lubrication.
层间较弱引力允许层滑动，解释润滑。
Do not say there are no forces at all between layers; there are no covalent bonds between them.
不要说层间完全没有作用力，应说没有共价键。
Graphene is one atom thick.
石墨烯只有一个原子厚。
Strong connections make it useful in composites; delocalised electrons allow conduction for electronics.
强连接适用于复合材料，离域电子适用于电子器件导电。
A cropped drawing has edge atoms, but the interior pattern gives three neighbours per carbon.
截取图存在边缘原子，内部每个碳的重复模式有三个邻居。
A fullerene cage is a finite molecule rather than an endless diamond network.
富勒烯笼是有限分子，不是无限延伸的金刚石网络。
Fullerenes contain carbon rings; nanotubes are cylindrical forms.
它含碳环，纳米管是圆筒形式。
Their shapes and properties support uses in materials, electronics and medicine.
这些形状和性质支持材料、电子和医学应用。
The tube on this slide shows shape only, not every carbon bond.
图中管只表示形状，不表示全部碳键。
Nanoparticles are one to one hundred nanometres. Fine particles span one hundred to two thousand five hundred, and coarse particles to ten thousand.
纳米颗粒为1至100纳米，细颗粒为100至2500纳米，粗颗粒延伸至10000纳米。
A twenty-nanometre particle spans about two hundred typical atomic diameters; this is not an atom count.
20纳米约跨越200个典型原子直径，但这不是原子数量。
The smallest nano structures may contain of the order of hundreds of atoms, not every particle in the whole range.
最小纳米结构可能含数百个原子，不是整个范围内每颗都只有数百个。
A cube side decreases from 100 nm to 10 nm. Its surface-area/volume quotient: Pause and attempt this independently.
模型立方体边长从100纳米减为10纳米，表面积除以体积的商如何变化？
Give a reason before revealing the answer.
先独立作答，并说明理由，再看下一页。
Choose the tenfold rise.
选择增大十倍。
For side one hundred nanometres, area divided by volume is zero point zero six inverse nanometres; for ten nanometres it is zero point six.
边长100纳米时商为0.06每纳米，10纳米时为0.6每纳米。
These cubes are schematic, not to scale.
图为示意，并非按比例绘制。
The quotient has an inverse-length unit, because area and volume have different dimensions.
面积和体积量纲不同，所以这个商有长度倒数单位。
Do not confuse the area of one particle with area per amount of material.
不要混淆单个粒子的面积和每单位材料的面积。
Splitting a fixed total volume into smaller particles increases exposed area.
把固定总体积分成更小粒子会增加暴露面积。
For a catalyst, this can give more surface contacts with reactants per unit mass and reduce the amount needed. Activity also depends on the particular material and conditions.
催化剂每单位质量可与反应物接触的表面更多，因此可能减少用量；活性也取决于材料及条件。
Equal masses remove pollutant in 12 minutes as nano material and 40 minutes in bulk.
等质量材料去除污染物，纳米形式用12分钟，块状形式用40分钟。
Does this establish safety in a skin product?
这能证明其用于皮肤产品安全吗？
Pause and attempt this independently. Give a reason before revealing the answer.
先独立作答，并说明理由，再看下一页。
The test supports faster pollutant removal under its conditions. It does not prove skin safety or performance for a different purpose.
该试验支持指定条件下更快去除污染物，不能证明皮肤安全或另一用途的性能。
Compare the intended benefit with material-specific exposure, health and environmental evidence.
把预期收益与具体材料的暴露、健康、环境证据比较。
Small size alone establishes neither harm nor safety.
小尺寸本身既不证明有害，也不证明安全。
Recommend further evidence matched to the uncertainty, rather than an unsupported yes or no.
提出与不确定性相匹配的进一步证据。
Close the reference.
合上参考资料。
Draw each required electron diagram and count its electrons.
画出所有指定电子图并计数。
Explain high melting point using the relevant forces, and conduction using charged particles that can move.
用相关作用力解释高熔点，用能移动的带电粒子解释导电。
Separate graphite layer sliding from electron movement.
把石墨层滑动与电子移动区分开。
Finish by evaluating a new nanoparticle claim, stating the evidence and what remains unknown.
最后评价新的纳米颗粒论断，说明证据及未知之处。
Use sheets two point one to two point four for independent practice.
用练习2.1至2.4进行独立练习。