chemistry-subject test2 w. solutions
应用化学专业英语翻译(第二版)
Unit10 Nomenclature of Hydrocarbons碳氢化合物的命名Alkanes烷烃理想的，每一种化合物都应该由一个明确描述它的结构的名称，并且通过这一名称能够画出它的结构式。
为了这一目的，全世界的化学家接受了世界纯粹与应用化学会（IUPAC）建立的一系列规则。
这个系统就是IUPAC系统，或称为日内瓦系统，因为IUPAC的第一次会议是在瑞士日内瓦召开的。
不含支链的烷烃的IUPAC命名包括两部分（1）表明链中碳原子数目的前缀；（2）后缀－ane，表明化合物是烷烃。
用于表示1至20个碳原子的前缀见表10.1表10.1中前4个前缀是由IUPAC选择的，因为它们早已在有机化学中确定了。
实际上，它们甚至早在它们成为规则之下的结构理论的暗示之前，它们的地位就确定了。
例如，在丁酸中出现的前缀but－，一种表示在白脱脂中存在的四个碳原子的化合物（拉丁语butyrum白脱（黄油））。
表示5个或更多碳原子的词根来源于希腊或拉丁词根。
含取代基的烷烃的IUPAC名称由母体名称和取代基名称组成，母体名称代表化合物的最长碳链，取代基名称代表连接在主链上的基团。
来源于烷烃的取代基称为烷基。
字母R－被广泛用来表示烷烃的存在.烷烃的命名是去掉原烷基名称中的－ane加上后缀－yl。
例如，烷基CH3CH2－称为乙基。
CH3－CH3乙烷（原碳氢化合物）CH3CH2－乙基（一个烷基）下面是IUPAC的烷烃命名规则：1. 饱和碳氢化合物称为烷烃。
2. 对有支链的碳氢化合物，最长的碳链作为主链，IUPAC命名按此主链命名。
3. 连接在主链上的基团称为取代基。
每一取代基有一名称和一数字.这一数字表示取代基连接在主链上的碳原子的位置。
4. 如果有多于一个的相同取代基，要给出表示支链位置的每个数字。
而且，表示支链数目的数字由前缀di－，tri－，tetra－，penta－等表示。
5. 如果有一个取代基，主链碳原子编号从靠近支链的一端开始，使支链位号最小。
iGCSE chemistry Section 2 lesson 5
The iGCSE Chemistry course
Content
Section 1 Principles of Chemistry Section 2 Chemistry of the Elements Section 3 Organic Chemistry Section 4 Physical Chemistry Section 5 Chemistry in Society
Lesson 5
g) Tests for ions and gases
g) Tests for ions and gases
2.37 describe tests for the cations: i Li+, Na+, K+, Ca2+ using flame tests ii NH4+, using sodium hydroxide solution and identifying the ammonia evolved iii Cu2+, Fe2+ and Fe3+, using sodium hydroxide solution
Eg. Calcium forms Ca2+
Sodium ions, Na+, and Magnesium ions, Mg2+, are examples of cations
Group 3 metals form trivalent cations
Eg. Aluminium forms Al3+
What’s a cation (and now we mention it, what’s an
剑桥雅思阅读5test2翻译及答案
剑桥雅思阅读5test2翻译及答案剑桥雅思阅读5原文(test2)1You should spend about 20 minutes on Questions 1-13, which are based on Reading Passage 1 below.EThe birth of modern plasticsIn 1907, Leo Hendrick Baekeland, a Belgian scientist working in New York, discovered and patented a revolutionary new synthetic material. His invention, which he named ‘Bakelite,’ was of enormous technological importance, and effectively launched the modernplastics industry.The term ‘plastic’ comes from the Greek plassein, meaning ‘to mould’. Some plastics are derived from natural sources, some are semi-synthetic (the result of chemical action on a natural substance), and some are entirely synthetic, that is, chemically engineered from the constituents of coal or oil. Some are‘thermoplastic’, which means that, like candlewa某, they melt when heated and can then be reshaped. Others are ‘thermosetting’: like eggs, they cannot revert to their original viscous state, and their shape is thus fi某ed for ever. Bakelite had the distinction of being the first totally synthetic thermosetting plastic.The history of today’s plastics begins with the discovery of a series of semi-synthetic thermoplastic materials in the mid-nineteenth century. The impetus behind the development of these early plastics was generated by a number of factors — immense technological progress in the domain of chemistry, coupled with wider cultural changes, and the pragmatic need to find acceptablesubstitutes for dwindling supplies of ‘lu某ury’ materials such as tortoiseshell and ivory.Baekeland’s interest in plastics began in 1885 when, as a young chemistry student in Belgium, he embarked on research into phenolic resins, the group of sticky substances produced when phenol (carbolic acid) combines with an aldehyde (a volatile fluid similar to alcohol). He soon abandoned the subject, however, only returning toit some years later. By 1905 he was a wealthy New Yorker, having recently made his fortune with the invention of a new photographic paper. While Baekeland had been busily amassing dollars, some advances had been made in the development of plastics. The years 1899 and 1900 had seen the patenting of the first semi-synthetic thermosetting material that could be manufactured on an industrial scale. In purely scientific terms, Baekeland’s major contribution to the field is not so much the actual discovery of the material to which he gave his name, but rather the method by which a reaction between phenol and formaldehyde could be controlled, thus making possible its preparation on a commercial basis. On 13 July 1907, Baekeland took out his famous patent describing this preparation, the essential features of which are still in use today.The original patent outlined a three-stage process, in which phenol and formaldehyde (from wood or coal) were initially combined under vacuum inside a large egg-shaped kettle. The result was a resin known as Novalak which became soluble and malleable when heated. The resin was allowed to cool in shallow trays until it hardened, and then broken up and ground into powder. Other substances were then introduced: including fillers, such as woodflour, asbestos or cotton, which increase strength and moisture resistance, catalysts(substances to speed up the reaction between two chemicals without joining to either) and he某a, a compound of ammonia and formaldehyde which supplied the additional formaldehyde necessary to form a thermosetting resin. This resin was then left to cool and harden, and ground up a second time. The resulting granular powder was raw Bakelite, ready to be made into a vast range of manufactured objects. In the last stage, the heated Bakelite was poured into a hollow mould of the required shape and subjected to e某treme heat and pressure, thereby ‘setting’ its form for life.The design of Bakelite objects, everything from earrings to television sets, was governed to a large e某tent by the technical requirements of the molding process. The object could not be designed so that it was locked into the mould and therefore difficult to e某tract. A common general rule was that objects should taper towards the deepest part of the mould, and if necessary the product was molded in separate pieces. Moulds had to be carefully designed sothat the molten Bakelite would flow evenly and completely into the mould. Sharp corners proved impractical and were thus avoided, giving rise to the smooth, ‘streamlined’ style pop ular in the 1930s. The thickness of the walls of the mould was also crucial: thick walls took longer to cool and harden, a factor which had to be considered by the designer in order to make the most efficient use of machines.Baekeland’s invention, al though treated with disdain in its early years, went on to enjoy an unparalleled popularity which lasted throughout the first half of the twentieth century. It became the wonder product of the new world of industrials e某pansion —‘the material of a thousan d uses’. Being both non-porous and heat-resistant, Bakelite kitchen goods were promoted as being germ-freeand sterilisable. Electrical manufacturers seized on its insulating properties, and consumers everywhere relished its dazzling array of shades, delighted that they were now, at last, no longer restricted to the wood tones and drab browns of the preplastic era. It then fell from favour again during the 1950s, and was despised and destroyed in vast quantities. Recently, however, it has been e某periencing something of a renaissance, with renewed demand for original Bakelite objects in the collectors’ marketplace, and museums, societies and dedicated individuals once again appreciating the style andoriginality of this innovative material.Questions 1-3Complete the summary.Choose ONE WORD ONLY from the passage for each answer.Write your answers in bo某es 1-3 on your answer sheet.Some plastics behave in a similar way to 1……… in that they melt under heat and can be moulded into new forms. Bakelite was unique because it was the first material to be both entirely 2……… in origin, and thermosetting.There were several reasons for the research into plastics in the nineteenth century, among them the great advances that had been made in the field of 3…………and the search for alternatives to natural resources like ivory.Questions 4-8Complete the flow-chart.Choose ONE WORD ONLY from the passage for each answer.Write your answers in bo某es 4-8 on your answer sheet.The Production of Bakelite图片6Questions 9 and 10Choose TWO letters A-E.Write your answers in bo某es 9 and 10 on your answer sheet.NB Your answers may be given in either order.Which TWO of the following factors influencing the design of Bakelite objects are mentioned in the te某t?A the function which the object would serveB the ease with which the resin could fill the mouldC the facility with which the object could be removed from the mouldD the limitations of the materials used to manufacture the mouldE the fashionable styles of the periodQuestions 11-13Do the following statements agree with the information given in Reading Passage 1?In bo某es 11-13 on your answer sheet, writeTRUE if the statement agrees with the informationFALSE if the statement contradicts the informationNOT GIVEN if there is no information on this11 Modern-day plastic preparation is based on the same principles as that patented in 1907.12 Bakelite was immediately welcomed as a practical and versatile material.13 Bakelite was only available in a limited range of colours.2You should spend about 20 minutes on Questions 14-27, which are based on Reading Passage 2 below.What’s so funny?John McCrone reviews recent research on humorThe joke comes over the headphones: ‘Which side of a dog has the most hair? The left.’ No, not funny. Try again. ‘Which side of a dog has the most hair? The outside.’ Hah! T he punchline is silly yet fitting, tempting a smile, even a laugh. Laughter has always struck people as deeply mysterious, perhaps pointless. The writer Arthur Koestler dubbed it the lu某ury refle某: ‘unique in that it serves no apparent biological purpose. ’Theories about humour have an ancient pedigree. Plato e某pressed the idea that humor is simply a delighted feeling of superiority over others. Kant and Freud felt that joke-telling relies on building up a psychic tension which is safely punctured by the ludicrousness of the punchline. But most modern humor theorists have settled on some version of Aristotle’s belief that jokes are based on a reaction to or resolution of incongruity, when the punchline is either a nonsense or, though appearing silly, has a clever second meaning.Graeme Ritchie, a computational linguist in Edinburgh, studies the linguistic structure of jokes in order to understand not only humor but language understanding and reasoning in machines. He says that while there is no single format for jokes, many revolve around a sudden and surprising conceptual shift. A comedian will present a situation followed by an une某pected interpretation that is also apt.So even if a punchline sounds silly, the listener can see thereis a cle ver semantic fit and that sudden mental ‘Aha!’ is the buzz that makes us laugh. Viewed from this angle, humor is just a form of creative insight, a sudden leap to a new perspective.However, there is another type of laughter, the laughter ofsocial appeasement and it is important to understand this too. Playis a crucial part of development in most young mammals. Rats produce ultrasonic squeaks to prevent their scuffles turning nasty. Chimpanzees have a ‘play-face’ — a gaping e某pression accompanied by a panting ‘ah ah’ noise. In humans, these signals have mutated into smiles and laughs. Researchers believe social situations, rather than cognitive events such as jokes, trigger these instinctual markers of play or appeasement. People laugh on fairground rides or when tickled to flag a play situation, whether they feel amused or not.Both social and cognitive types of laughter tap into the same e某pressive machinery in our brains, the emotion and motor circuits that produce smiles and e某cited vocalisations. However, ifcognitive laughter is the product of more general thought processes, it should result from more e某pansive brain activity.Psychologist Vinod Goel investigated humour using the new technique of ‘single event’ functional magnetic resona nce imaging (fMRI). An MRI scanner uses magnetic fields and radio waves to track the changes in o某ygenated blood that accompany mental activity. Until recently, MRI scanners needed several minutes of activity and so could not be used to track rapid thought processes such as comprehending a joke. New developments now allow half-second‘snapshots’ of all sorts of reasoning and problem-solving activities.Although Goel felt being inside a brain scanner was hardly the ideal place for appreciating a joke, he found evidence that understanding a joke involves a widespread mental shift. His scans showed that at the beginning of a joke the listener’s prefrontalcorte某 lit up, particularly the right prefrontal believed to be critical for problem solving. But there was also activity in the temporal lobes at the side of the head (consistent with attempts to rouse stored knowledge) and in many other brain areas. Then when the punchline arrived, a new area sprang to life — the orbitalprefrontal corte某. This patch of brain tucked behind the orbits of the eyes is associated with evaluating information.Making a rapid emotional assessment of the events of the momentis an e某tremely demanding job for the brain, animal or human. Energy and arousal levels may need to be retuned in the blink of an eye. These abrupt changes will produce either positive or negative feelings. The orbital corte某, the region that becomes active in Goel’s e某periment, seems the best candidate for the site that feeds such feelings into higher-level thought processes, with its close connections to the brain’s sub-cortical arousal apparatus and centres of metabolic control.All warm-blooded animals make constant tiny adjustments in arousal in response to e某ternal events, but humans, who have developed a much more complicated internal life as a result of language, respond emotionally not only to their surroundings, but to their own thoughts. Whenever a sought-for answer snaps into place, there is a shudder of pleased recognition. Creative discovery being pleasurable, humans have learned to find ways of milking this natural response. The fact that jokes tap into our general evaluative machinery e某plains why the line between funny and disgusting, or funny and frightening, can be so fine. Whether a joke gives pleasure or pain depends on a person’s outlook.Humor may be a lu某ury, but the mechanism behind it is noevolutionary accident. As Peter Derks, a psychologist at William and Mary College in Virginia, says: ‘I like to think of humour as the distorted mirror of the mind. It’s creative, perceptual, analytical and lingual. If we can figure out how the mind processes humor, then we’ll have a pretty good handle on how it works in general.’Questions 14-20Do the following statements agree with the information given in Reading Passage 2?In bo某es 14-20 on your answer sheet, writeTRUE if the statement agrees with the informationFALSE if the statement contradicts the informationNOT GIVEN if there is no information on this14 Arthur Koestler considered laughter biologically important in several ways.15 Plato believed humour to be a sign of above-average intelligence.16 Kant believed that a successful joke involves the controlled release of nervous energy.17 Current thinking on humour has largely ignored Aristotle’s view on the subject.18 Graeme Ritchie’s work links jokes to artificial intelligence.19 Most comedians use personal situations as a source of humour.20 Chimpanzees make particular noises when they are playing.Questions 21-23The diagram below shows the areas of the brain activated by jokes.Label the diagram.Choose NO MORE THAN TWO WORDS from the passage for each answer.Write your answers in bo某es 21-23 on your answer sheet.Questions 24-27Complete each sentence with the correct ending A-G below.Write the correct letter A-G in bo某es 24-27 on your answer sheet.24 One of the brain’s most difficult tasks is to25 Because of the language they have developed, humans26 Individual responses to humour27 Peter Derks believes that humourA react to their own thoughts.B helped create language in humans.C respond instantly to whatever is happening.D may provide valuable information about the operation of the brain.E cope with difficult situations.F relate to a person’s subjective views.G led our ancestors to smile and then laugh.3You should spend about 20 minutes on Questions 28-40, which are based on Reading Passage 3 below.The Birth of Scientific EnglishWorld science is dominated today by a small number of languages, including Japanese, German and French, but it is English which is probably the most popular global language of science. This is notjust because of the importance of English-speaking countries such as the USA in scientific research; the scientists of many non-English-speaking countries find that they need to write their research papers in English to reach a wide international audience. Given theprominence of scientific English today, it may seem surprising that no one really knew how to write science in English before the 17th century. Before that, Latin was regarded as the lingua franca1 for European intellectuals.The European Renaissance (c. 14th-16th century) is sometimes called the ‘revival of learning’, a time of renewed interest in the ‘lost knowledge’ of classical times. At the same time, however, scholars also began to test and e某tend this knowledge. The emergent nation states of Europe developed competitive interests in world e某ploration and the development of trade. Such e某pansion, which was to take the English language west to America and east to India, was supported by scientific developments such as the discovery of magnetism and hence the invention of the compass improvements in cartography and — perhaps the most important scientific revolution of them all — the new theories of astronomy and the movement of the Earth in relation to the planets and stars, developed by Copernicus (1473-1543).England was one of the first countries where scientists adopted and publicised Copernican ideas with enthusiasm. Some of these scholars, including two with interests in language — John Wallis and John Wilkins — helped found the Royal Society in 1660 in order to promote empirical scientific research.Across Europe similar academies and societies arose, creating new national traditions of science. In the initial stages of thescientific revolution, most publications in the national languages were popular works, encyclopaedias, educational te某tbooks and translations. Original science was not done in English until the second half of the 17th century. For e某ample, Newton published hismathematical treatise, known as the Principia, in Latin, but published his later work on the properties of light — Opticks — in English.There were several reasons why original science continued to be written in Latin. The first was simply a matter of audience. Latin was suitable for an international audience of scholars, whereas English reached a socially wider, but more local, audience. Hence, popular science was written in English.A second reason for writing in Latin may, perversely, have been a concern for secrecy. Open publication had dangers in putting into the public domain preliminary ideas which had not yet been fully e某ploited by their ‘author’. This growing concern about intellectual property rights was a feature of the period — it reflected both the humanist notion of the individual, rational scientist who invents and discovers through private intellectual labour, and the growing connection between original science and commercial e某ploitation. There was something of a social distinction b etween ‘scholars and gentlemen’ who understood Latin, and men of trade who lacked a classical education. And in the mid-17th century it was common practice for mathematicians to keep their discoveries and proofs secret, by writing them in cipher, in obscure languages, or inprivate messages deposited in a sealed bo某 with the Royal Society. Some scientists might have felt more comfortable with Latin precisely because its audience, though international, was socially restricted. Doctors clung the most keenly t o Latin as an ‘insider language’.A third reason why the writing of original science in English was delayed may have been to do with the linguistic inadequacy of English in the early modern period. English was not well equipped to dealwith scientific argument. First it lacked the necessary technical vocabulary. Second, it lacked the grammatical resources required to represent the world in an objective and impersonal way, and to discuss the relations, such as cause and effect, that might hold between comple某 and hypothetical entities.Fortunately, several members of the Royal Society possessed an interest in Language and became engaged in various linguistic projects. Although a proposal in 1664 to establish a committee for improving the English lan guage came to little, the society’s members did a great deal to foster the publication of science in English and to encourage the development of a suitable writing style. Many members of the Royal Society also published monographs in English. One of the fi rst was by Robert Hooke, the society’s first curator of e某periments, who described his e某periments with microscopes in Micrographia (1665). This work is largely narrative in style, based on a transcript of oral demonstrations and lectures.In 1665 a new scientific journal, Philosophical Transactions, was inaugurated. Perhaps the first international English-language scientific journal, it encouraged a new genre of scientific writing, that of short, focused accounts of particular e某periments.The 17th century was thus a formative period in the establishment of scientific English. In the following century much of this momentum was lost as German established itself as the leading European language of science. It is estimated that by the end of the 18th century 401 German scientific journals had been established as opposed to 96 in France and 50 in England. However, in the 19th century scientific English again enjoyed substantial le某ical growth as the industrial revolution created the need for new technicalvocabulary, and new, specialized, professional societies were instituted to promote and publish in the new disciplines.lingua franca: a language which is used for communication between groups of people who speak different languagesQuestions 28-34Complete the summary.Choose NO MORE THAN TWO WORDS from the passage for each answer. Write your answers in bo某es 28-34 on your answer sheet.In Europe, modern science emerged at the same time as the nation state. At first, the scientific language of choice remained 28…………… . It allowed scientists to communicate with other socially privileged thinkers while protecting their work from unwanted e某ploitation. Sometimes the desire to protect ideas seems to have been stronger than the desire to communicate them,particularly in the case of mathematicians and 29…………… . In Britain, moreover, scientists worried that English had neither the 30…………… nor the 31………… to e某press their ideas. This situation only changed after 1660 when scientists associated with the 32………… set about developing English. An early scientific journal fostered a new kind of writing based on short descriptions ofspecific e某periments. Although English was then overtaken by 33……… , it developed again in the 19th century as a direct result of the 34……………….Questions 35-37Do the following statements agree with the information given in Reading Passage 3?In bo某es 35-37 on your answer sheet, writeTRUE if the statement agrees with the informationFALSE if the statement contradicts the informationNOT GIVEN if there is no information on this35 There was strong competition between scientists in Renaissance Europe.36 The most important scientific development of the Renaissance period was the discovery of magnetism.37 In 17th-century Britain, leading thinkers combined their interest in science with an interest in how to e某press ideas.Questions 38-40Complete the table.Choose NO MORE THAN TWO WORDS from the passage for each answer. Write your answers in bo某es 38-40 on your answer sheet.Science written in the first half of the 17th centuryLanguage used Latin EnglishType of science Original 38…………E某amples 39………… EncyclopaediasTarget aud ience International scholars 40…………, but socially wider剑桥雅思阅读5原文参考译文(test2)E The birth of modern plastics酚醛塑料——现代塑料的诞生In 1907, Leo Hendrick Baekeland, a Belgian scientist working in New York, discovered and patented a revolutionary new synthetic material. His invention, which he named ‘Bakelite,’ was of enormous technological importance, and effectively launched the modernplastics industry.1907年，比利时科学家Leo Hendrick Baekeland在纽约工作时发现了一种全新的合成材料，并申请了专利。
A level chemistry unit 2 2014.6 ,answer
Mark Scheme (Results) Summer 2014IAL Chemistry (WCH02/01)Edexcel and BTEC QualificationsEdexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at .Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert email service helpful. /contactusPearson: helping people progress, everywhereOur aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: /ukSummer 2014Publications Code IA038350*All the material in this publication is copyright© Pearson Education Ltd 2014General Marking Guidance∙All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark thelast.∙Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather thanpenalised for omissions.∙Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie.∙There is no ceiling on achievement. All marks on the mark scheme should be used appropriately.∙All the marks on the mark scheme are designed to be awarded.Examiners should always award full marks if deserved, i.e. if theanswer matches the mark scheme. Examiners should also beprepared to award zero marks if the candidate’s response is notworthy of credit according to the mark scheme.∙Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may belimited.∙When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must beconsulted.∙Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.∙Mark schemes will indicate within the table where, and which strands of QWC, are being assessed. The strands are as follows:i) ensure that text is legible and that spelling, punctuation andgrammar are accurate so that meaning is clearii) select and use a form and style of writing appropriate to purposeand to complex subject matteriii) organise information clearly and coherently, using specialistvocabulary when appropriateUsing the Mark SchemeExaminers should look for qualities to reward rather than faults topenalise. This does NOT mean giving credit for incorrect or inadequateanswers, but it does mean allowing candidates to be rewarded foranswers showing correct application of principles and knowledge.Examiners should therefore read carefully and consider every response:even if it is not what is expected it may be worthy of credit.The mark scheme gives examiners:∙ an idea of the types of response expected∙ how individual marks are to be awarded∙ the total mark for each question∙ examples of responses that should NOT receive credit./ means that the responses are alternatives and either answer shouldreceive full credit. ( ) means that a phrase/word is not essential for the award of the mark,but helps the examiner to get the sense of the expected answer.Phrases/words in bold indicate that the meaning of the phrase or theactual word is essential to the answer.ecf/TE/cq (error carried forward) means that a wrong answer given in anearlier part of a question is used correctly in answer to a later part of thesame question.Candidates must make their meaning clear to the examiner to gain themark. Make sure that the answer makes sense. Do not give credit forcorrect words/phrases which are put together in a meaningless manner.Answers must be in the correct context.Quality of Written CommunicationQuestions which involve the writing of continuous prose will expectcandidates to:∙ write legibly, with accurate use of spelling, grammar and punctuation inorder to make the meaning clear∙ select and use a form and style of writing appropriate to purpose and tocomplex subject matter∙ organise information clearly and coherently, using specialist vocabularywhen appropriate.Full marks will be awarded if the candidate has demonstrated the aboveabilities.Questions where QWC is likely to be particularly important are indicated(QWC) in the mark scheme, but this does not preclude others.Section A (multiple choice)Correct Answer Reject MarkQuestionNumber1 1(a) BCorrect Answer Reject MarkQuestionNumber1(b) A1 Correct Answer Reject MarkQuestionNumber1 2 BQuestionCorrect Answer Reject MarkNumber1 3 CCorrect Answer Reject MarkQuestionNumber4 (a) D1 Correct Answer Reject MarkQuestionNumber1 4 (b) BCorrect Answer Reject MarkQuestionNumber1 5 ACorrect Answer Reject MarkQuestionNumber1 6 CCorrect Answer Reject MarkQuestionNumber1 7 ACorrect Answer Reject MarkQuestionNumber1 8 BCorrect Answer Reject MarkQuestionNumber9 D1 Correct Answer Reject MarkQuestionNumber1 10 CQuestionCorrect Answer Reject MarkNumber1 11 (a) DCorrect Answer Reject MarkQuestionNumber1 11 (b) CCorrect Answer Reject MarkQuestionNumber1 12 CQuestionCorrect Answer Reject MarkNumber13 B1 Correct Answer Reject MarkQuestionNumber1 14 AQuestionCorrect Answer Reject MarkNumber15 B1 Correct Answer Reject MarkQuestionNumber16 D1 Correct Answer Reject MarkQuestionNumber17 C1TOTAL FOR SECTION A = 20 MARKSSection BQuestionNumberAcceptable Answers Reject Mark18 (a) NaCl + H2SO4→ HCl + NaHSO4ALLOWMultiplesHNaSO42NaCl + H2SO4→ 2HCl + Na2SO4IGNOREstate symbols even if incorrectCOMMENTALLOWCapitals or lower case in formulae1QuestionNumberAcceptable Answers Reject Mark18 (b) Ammonia (gas) / NH3Allow Ammonia solution/ NH3(aq) (1)White smoke/solidALLOWwhite cloud /Dense white fumes (1)The observation mark is consequentialon use of ammonia.If name and formula are given, bothmust be correct. AmmoniumIncorrect identification of white smokeMisty fumes / steamy fumes/ white gas/ white ppt2QuestionNumberAcceptable Answers Reject Mark18 (c) White ppt/solidALLOWwhite crystals (1)IGNOREidentification of white solid, even ifwrong(ppt/solid) dissolves (in excess)/(colourless) solution formsALLOW(ppt/solid) disappears/ soluble(1)IGNOREclear solution(c.NH3) dissolves AgBr (as well asAgCl) (1)Just “white”Cream pptother colours of solutionDissolvesbromide ions/ bromineJust “Only AgCl dissolves indilute NH3”c.NH3 dissolves other things3TOTAL FOR Q18 = 6 MARKSNumber19(a)(i)OR7x and 5 • around the bromine. (1)Total of 8 electrons round each oxygenOne octet MUST INCLUDE the electronrepresented by * (1)ALLOWx for oxygen and • for bromine if clearElectrons in bonds to be shown in rows egxx •• or x•x• between the relevant atoms;non-bonded electrons not in pairs..All dots or all crosses then max 1Two dative covalent bonds by the bromineto the oxygens then max 1 (loses first mark)IGNOREcircles round outer shells of atoms2QuestionNumberAcceptable Answers Reject Mark19(a)(ii) There are vacant (3)d orbitals / Theyare using (3)d orbitalsALLOWSub-shells for orbitalsUse of D for d2dp/ f orbitalsShell for sub-shell1Number19 (b)(i) (n=8.35 ÷ 167 = ) 0.05(00) (mol) (1)Ignore any units even if incorrect.(c= 0.05 ÷ 0.25 =) 0.2(00) (mol dm-3)TE on incorrect number of moles in firstmark (1)Correct answer without working scores (2)If final units are given they must be correct. ALLOW1sfmol /dm3 OR M mol /dm-32QuestionNumberAcceptable Answers Reject Mark19(b)(ii)(0.0025 x 6 = ) 0.015 (mol) (1)(0.015 x 166 = 2.49 (g))TE from first mark (1)2.6 ≤ value ≤ 5.0 (g)TE for third mark as long as a calculationhas been done for second mark. Valuesshould be at least 0.1 g above calculatedvalue and less than double calculated value.(1)ALLOW1sf for suitable mass3QuestionNumberAcceptable Answers Reject Mark19(b)(iii) (0.001 x 2 =) 0.002/ 2 x 10-3 (mol)(1)(V = 0.002 ÷ 0.1x1000 =) 20 (cm3 )ALLOW0.02 dm3/ 0.020 dm3(1)If units are not in cm3 they must bestatedTE from incorrect number of molCorrect answer without workingscores (2)0.020.02 dm-32QuestionNumberAcceptable Answers Reject Mark19(b)(iv)Mass of KBrO3 (1)Second mark depends on correctchoice in first.Percentage error/ uncertainty large with a small massOR Mass is only to 1sf(1) IGNOREcalculation, even if incorrect Just“Mass is only to2 decimalplaces” /“mass is only0.07g”/“mass is not accurate”2TOTAL FOR Q19 = 12 MARKSQuestionNumberAcceptable Answers Reject Mark 20 *(a) These marks are independentThe outer electrons are further from the nucleus / the electron being removed is further from the nucleus/ larger atomic radius (in calcium)ALLOWCa has one more shell/ more shells (of electrons) (1)More shielding (in calcium) (1)ORReverse argument for magnesiumALLOWDiscussion based on trend going down group without specifying Mg and Ca IGNORErepulsion between shells Larger ionic radius (in Ca)Just “Calcium is larger”Reference to molecules, delocalised electronsJust “Ca has more energy levels”Two more shellsAny reference to polarising powerof ions2QuestionAcceptable Answers Reject MarkNumber320 (b) Electrons are promoted/ jump /become excited to higher energylevel (1)Electron(s) return/ fall back to lowerenergy levelALLOWto ground state (1)Release of (visible ) light (energy)upon return / energy is released invisible spectrumALLOWrelease of photons upon return (1)Acceptable Answers Reject MarkQuestionNumber1 20 (c)(i) CaO + 2HNO3→ Ca(NO3)2 + H2OIgnore state symbols even if incorrectQuestionNumberAcceptable Answers Reject Mark20(c)(ii) Observation mark:(Calcium nitrate) produces a brown/red-brown gasALLOWNO2 for gasFumes for gasOR(Potassium nitrate) does not producea brown gasIGNOREOxygen is given off / Gas given offrelights a glowing splint(1)Second mark (can also be anobservation):(Only calcium nitrate) produces theoxideOR(Only potassium nitrate) producesthe nitriteORcalcium nitrate is less stable to heatORpotassium nitrate decomposes at ahigher temperature/takes longer toproduce oxygen (1)ALLOW“Calcium nitrate produces a whitesolid and potassium nitrateproduces a yellow solid”as an alternative for either markNOTEReject comparisons with one correctand one incorrect statement (thisapplies to both marks)Flame coloursReference toother incorrectproducts.2Number20 (d)(i) Hydrogen (gas) / H2If name and formula are given bothmust be correct1QuestionNumberAcceptable Answers Reject Mark20(d)(ii) White ppt/white solid/goes milky/goescloudy/ white suspension (1)Ca(OH)2 + CO2→ CaCO3 + H2O(1)ALLOWAlternative answerWhite precipitate forms which dissolveswith excess carbon dioxide (1)Ca(OH)2 + 2CO2→ Ca(HCO3)2(1)Whitesolution /anysolutionproduced2QuestionNumberAcceptable Answers Reject Mark20(d)(iii) (One of): Sr(OH)2/Ba(OH)2/Ra(OH)2OR(One of):Strontium/Barium/RadiumhydroxideIf name and formula given then bothmust be correctSrOH/ BaOH/RaOHJust Sr/ Ba/ RaMg(OH)2/MgOH/magnesiumhydroxide/Be(OH)2/BeOH/berylliumhydroxide1Number20 (e)(i) White ppt/solidALLOWWhite crystals (1)(BaSO4 is insoluble but) MgSO4 is(very) soluble / MgSO4 gives acolourless solution/ MgSO4 gives noprecipitateALLOWBaSO4 does not dissolveTE on first mark if it stated that aprecipitate formed even if colour iswrong/ missing(1)White ppt ofBaCl2 / MgCl2Extraobservations eg effervescence Magnesium is soluble / bariumis insolubleA precipitate of magnesiumsulfate formsand thendissolvesJust “MgSO4 is more soluble /less insoluble” Reference to solubility of chloridesThere would beno reaction2QuestionNumberAcceptable Answers Reject Mark20(e)(ii) Barium sulfate is not absorbed/ isinsolubleIGNOREComments on X-raysBarium sulfate is not digestedBarium sulfate is unreactive/ doesnot react with stomach acidsReferences to toxicity.Just ‘Barium’ 1QuestionNumberAcceptable Answers Reject Mark20 (f) First mark:(Increase) concentration of HCl(1)Second markMore particles/ moles of (HCl) in thesame volumeOR more (frequent/ successful )collisionsAllow second mark only if factor isconcentration (1)-------------------------------Any two from three of the followingfor third and fourth marks:Reduce particle size / use powder(instead of lumps)/ use finely divided(solid) (1)(Increases) surface area (1)more (frequent/ successful ) collisions(1)ALLOWReverse arguments Increase concentration of CaCO3 /HCl and CaCO3 /reactantsIncrease kinetic energy ofparticlesIncrease kinetic energy ofparticles4QuestionAcceptable Answers Reject MarkNumber1 20(g) Pressure only affects gaseous reactions/there are no gaseous reactants (orproducts) /there is no significant volumechange/ liquids are incompressibleALLOWpressure doesn’t affect solids/ solutionsNote: there are many possible correctways of expressing the idea thatpressure only affects rate of reactionsinvolving gases.IGNORENumber of moles in reaction doesn’tchangeTOTAL FOR Q20 = 20 MARKSTOTAL FOR SECTION B = 38 MARKSSection CQuestionNumberAcceptable Answers Reject Mark 21 (a) Primary (1)Part of the molecule whichdetermines how it will react / atom or group responsible for its reactions / group where chemical reactions occur/ part of the molecule responsible for its (chemical) propertiesALLOWThe part of the molecule which reacts / Group responsible for its characteristics (1)IGNOREGroup which determines how the molecule behaves Molecule responsible for reactions2QuestionNumberAcceptable Answers Reject Mark21 (b) C20H30OCorrect number of carbons (1)Rest of formula correct (stand alonemark, even if C incorrect) (1)Note: C20H29OH scores first markonlyIgnore working (structural formula)if shown as long as a molecularformula is given Juststructuralformula2QuestionNumberAcceptable Answers Reject Mark21(c)(i) Reflux apparatus produces carboxylic/ retinoicacidORcompletely oxidizes the alcohol(1)Convert to distillationALLOWuse condenser in horizontal position/description of distillation/ sketch of distillationapparatus (1)Oxidizing agent should be limiting/not inexcess/remove aldehyde as it is formed/ removebefore further oxidation (1)ALLOWUse excess alcohol‘Product’ for ‘aldehyde’Oxidizes to aketoneFractionaldistillationJust ‘thecollection ofaldehyde’3QuestionNumberAcceptable Answers Reject Mark21(c)(ii) Cr2O72-(aq)+14H+(aq)+ 6e-→2Cr3+(aq)+7H2O(l)+6 +3Orange GreenOne mark for the correct numbers of hydrogens(1)One mark for the correct numbers of chromiumsand electrons (1)One mark for each oxidation number with sign.If sign is missing penalise once onlyALLOW6+ , 3+ (2)One mark for both colours (1)Any othercolour withorange/Green-blue5Number21(c)(iii) (Retinal) (strong) absorption at1740-1720 (due to C=O bond)OR(Retinal) (weak) absorption at2900–2820/ 2775–2700 (due to C-Hbond)ALLOWWavenumber/ peak/ stretch for“absorption” (1)No absorption at 3750–3200/absorption at 3750-3200 shows notall retinol converted (1)Ignore comments on absorptions at3300-2500Absorption at1725-17001700-16802Number21(c)(iv)Any one of the following:ALLOW the following circles in retin ol Any additional area circled Circlesincluding anyC atom other than those ofthe doublebond circledon the mark scheme1Number21(c)(v) Round the carbon there arethree areas with electrons / 3 regionsof electron density/ 3 areas ofelectron densityALLOWThree bond pairs IF answer says thatdouble bond can be treated as onebond (1)Electron pairsrepel/ go to maximum separation/goto minimum repulsion (1)ALLOWBonds repelThe answer must clearly refer toelectrons/ bonds/ bonding pairs atsome point to score these marks.Trigonal planarALLOWTriangular planar (1)Round thecarbon there are3 bondsC with a lonepairatoms repelmaximumrepulsion/minimumseparation3QuestionNumberAcceptable Answers Reject Mark21 (d)Accept any orientation of =O and –OH andlength of bonds.Allow the OH displayed COOH addedto finalsingle bond OOH added1Number21 (e) Observation and precaution marks aredependent on correct reagent.EITHERReagentPCl5 / phosphorus((V)) chloride /phosphorus pentachlorideALLOWPhosphoric(V) chloride (1)ObservationSteamy/misty/white fumes (1)IGNORETests on steamy fumes eg litmusPrecautionUse of fume cupboard (1)IGNOREneed for safety goggles and lab coats. Incorrect reasons given for use of fume cupboard.Need for dry equipmentUse of gloves OR ALLOWReagentSodium/ Na (1) ObservationFizzing/Bubbles (1)IGNOREsodium dissolvesPrecautionHandle with gloves/tweezers (1) IGNOREnaked flamesneed for dry equipmentneed for safety goggles and lab coats. Whitesmoke/solid Dense white fumesGas mask3TOTAL FOR SECTION C (Question 21) = 22 MARKSTOTAL FOR PAPER = 80 MARKSAppendix A:Question 19ai: Additional Guidance.Pearson Education Limited. Registered company number 872828 with its registered office at Edinburgh Gate, Harlow, Essex CM20 2JE。
美国麻省理工学院化学工程专业介绍.doc
美国麻省理工学院化学工程专业介绍学校名称：美国麻省理工学院(剑桥)MassachusettsInstituteofTechnology(Cambridge)所在位置：美国，77MassachusettsAvenueCambridge,MA02139-4307(617)253-1000创建时间：1861 QS排名：2 USNEWS排名：5 学费：40732录取率：0.1申请美国留学的学生越来越多，那么美国麻省理工学院的化学工程专业怎么样呢?这是很多移民人士比较感兴趣的问题。
和出国移民网一起来看看吧!下面是小编整理的相关资讯，欢迎阅读。
美国麻省理工学院化学工程专业简介麻省理工学院的化学工程专业历史起源于19世纪晚期。
1888年，麻省理工学院化学教授LewisM.Norton创造了化学工程课程。
1891年，麻省理工学院化学工程学院成为第一个授予化学工程学士学位的地方。
1898年弗兰克·索普教授出版的工业化学概况，被认为是化学工程的第一本教科书。
1907年，麻省理工学院成为第一所授予化学工程专业博士学位的学校。
从那时起，麻省理工学院化学工程系领导全美化学工程的发展。
麻省理工学院化学工程专业基本申请要求学费$46,400申请截止日期12月1日TOEFL要求100GPA要求无可申请学期秋季IELTS要求7.0GRE要求要求GRESubject自愿麻省理工学院化学工程专业课程介绍麻省理工学院化学工程专业理科硕士学科的学生需要完成30学分的课程，学生可以选择主题方向或者是报告方向。
在主题方向的选择，要求该专业的学生需要完成24个学分的授课课程和6个学分的主题，其主题要求基于学生所直接研究表现。
在报告方向的选择，要求该专业的学生需要完成27个学分的授课课程和2个学分的报告，该报告要求根据文献回顾或设计表现的项目。
麻省理工学院化学工程专业前置课程·化学工程中的数字方法(NumericalMethodsinChemicalEngineering10.34)·化学工程热力学(ChemicalEngineeringThermodynamics10.40)·运输现象分析(AnalysisofTransportPhenomena10.50)·化学反应工程(ChemicalReactorEngineering10.65)麻省理工学院化学工程专业设置化学工程系有以下研究领域，分别是：·热力学与分子计算(ThermodynamicsandMolecularComputations)·催化剂与反应工程(CatalysisandReaction Engineering)·系统设计与工程(Systems DesignandEngineering)·运输过程(Transport Processes)·生物工程(Biological Engineering)·材料(Materials)·聚合物(Polymers)·表面与结构(SurfacesandStructures)·能源与环境工程(EnergyandEnvironmentalEngineering)麻省理工学院化学工程专业就业情况CollegeFactual的研究数据显示：麻省理工学院化学工程专业毕业生的平均年薪起薪为80,000美元，在职业生涯中期的薪资大约年薪为111,000美元。
电子教案与课件：《化工专业英语》 unit2
With qualitative logic we cannot identify what "too high" means — we need quantitative logic to do that.
用定性逻辑，我们不能确定什么是“太Βιβλιοθήκη ” ，这部分我们需要用定量逻辑来做。
For example, a quantitative statement about the river might state that “when the mass of organic matter entering a certain river equals x kilograms per day, the amount of oxygen in the stream is y.”
我们将社会科学和自然科学区分开来，前 者研究人及其作为家庭、部落、社区、种 族和国家而生活在一起的方式，后者研究 自然和物理世界。
Natural science includes such diverse disciplines as biology, chemistry, geology, physics, and environmental science.
当然，环境科学的研究重点是自然环境。
By this, we mean the atmosphere, the land, the water and their inhabitants as differentiated from the built environment.
由此，我们认为大气、土地、水和居住环 境这些跟建筑环境还是有区别的。
分类和泛化的存在，最重要的是它使科学 成为有组织的知识体系。
化学选择性必修二实验计划
化学选择性必修二实验计划英文回答：Chemistry Elective II Experiment Plan.Experiment 1: Determination of the Activation Energy of a Chemical Reaction.Objective: To determine the activation energy of a chemical reaction using the Arrhenius equation.Materials: Reactants, catalyst, thermometer, stopwatch, graduated cylinder, pipettes.Procedure:Prepare a series of solutions with different concentrations of the reactants.Add a catalyst to each solution.Measure the temperature of each solution and start the stopwatch.Monitor the temperature and record the time taken for the reaction to complete.Plot a graph of ln(rate) vs. 1/T and determine the slope of the line.Use the slope of the line to calculate the activation energy (Ea) using the Arrhenius equation:ln(rate) = -Ea/RT + ln(A)。
Experiment 2: Synthesis and Characterization of a Coordination Complex.Objective: To synthesize and characterize a coordination complex using spectroscopic techniques.Materials: Metal salt, ligand, solvent, spectrophotometer, IR spectrophotometer.Procedure:Prepare a solution of the metal salt and the ligand in a solvent.Heat the solution to reflux and allow the reaction to proceed for a specified time.Filter the reaction mixture and isolate the coordination complex.Characterize the coordination complex using UV-Vis spectrophotometry and IR spectroscopy.Determine the stoichiometry and geometry of the complex.Experiment 3: Determination of the Equilibrium Constant of a Chemical Reaction.Objective: To determine the equilibrium constant of achemical reaction using spectrophotometry.Materials: Reactants, solvent, spectrophotometer, cuvettes.Procedure:Prepare a series of solutions with different concentrations of the reactants at equilibrium.Measure the absorbance of each solution at a specific wavelength.Plot a graph of absorbance vs. concentration and determine the slope of the line.Use the slope of the line to calculate the equilibrium constant (K) using the Beer-Lambert law: A = εbc.中文回答：化学选择性必修二实验计划。
认知诊断中的 Q 矩阵简述
261区域治理MARKET作者简介：邱 瑾，生于1996年，江西师范大学研究生，研究方向为心理统计与测量。
认知诊断中的Q 矩阵简述江西师范大学心理学院 邱瑾摘要：作为新一代的心理测量理论，在教育测量领域认知诊断受到越来越多的青睐。
它不仅能在细粒度的层面上，评估个体在学科能力上的知识掌握状态，而且可以根据评估结果，给教师提供个体接下来的发展学习的道路。
在运用认知诊断理论作认知诊断评估的过程中，Q矩阵起着至关重要的作用。
关键词：认知诊断；Q矩阵；知识状态中图出版号：B842.1文献标识码：A文章编号：2096-4595（2020）28-0261-0001认知诊断是一项通过评估被试者的认知技能或特定知识（属性）集合的掌握情况的技术。
正如Tasuka 所说，它至少包含“Q 矩阵理论”和“诊断分类”两大部分。
在目前的认知诊断的领域中，有相当多的研究正把热点集中于Q 矩阵的修正和估计研究上。
这一研究热点不仅仅只存在于Q 矩阵修正方法开发研究、Q 矩阵估计方法开发研究，国内同样还存在着对Q 矩阵相关研究的综述上。
例如，王晓军，丁树良和罗芬（2019）作了Q 矩阵的综述，存在层级属性关系的Q 矩阵的算法、作用，刘永和涂冬波（2015）讨论的是根据被试反应估计Q 矩阵的6种方法。
一、认知诊断的目的在认知诊断的过程中，Q 矩阵是必不可少的一部分。
在认识Q 矩阵的概念和作用之前，我们应该了解认知诊断的目的。
Leighton 和Gierl(2007)指出，认知诊断致力于评估受测者的特定知识结构和加工技能的掌握情况，它是基于个人的认知过程上的诊断。
在认知过程中加工技能也可称之为属性，认知诊断评估最后会给出受测者在多个属性上的属性掌握状态用以表示被试的知识状态。
关键的问题是，如何通过被试作答项目的反应来判断被试的属性掌握状态？重点在于Q 矩阵的建立。
Q 矩阵理论主要是确定测验项目所测的间接观察的认知属性，并把它转化为可直接监测的理想反应模式，将被试在“黑箱子”里的认知状态与在项目上直观的作答反应联系起来，进而为了解并推测被试的认知状态提供基础。
basic solution化学
basic solution化学英文回答：What is a basic solution?A basic solution is a solution that has a pH greater than 7.0. This means that the solution contains more hydroxide ions (OH-) than hydrogen ions (H+). Basic solutions are also known as alkaline solutions.What are the properties of basic solutions?Basic solutions have a number of characteristic properties, including:They are slippery to the touch.They taste bitter.They turn red litmus paper blue.They react with acids to form salts and water.What are the uses of basic solutions?Basic solutions are used in a variety of applications, including:Cleaning agents: Basic solutions are often used as cleaning agents because they can dissolve grease and dirt.Deodorants: Basic solutions can be used as deodorants because they can neutralize acids that cause body odor.Medicines: Basic solutions are used in a number of medicines, such as antacids and cough syrups.How can you make a basic solution?There are a number of ways to make a basic solution, including:Dissolving a base in water: The most common way to make a basic solution is to dissolve a base in water. Some common bases include sodium hydroxide (NaOH), potassium hydroxide (KOH), and calcium hydroxide (Ca(OH)2).Mixing an acid and a base: You can also make a basic solution by mixing an acid and a base. When an acid and a base react, they form a salt and water. The salt will be basic if the base is stronger than the acid.Safety precautions.Basic solutions can be harmful if they are not handled properly. It is important to wear gloves and eye protection when working with basic solutions. Basic solutions should also be stored in a safe place away from children and pets.中文回答：什么是碱性溶液？碱性溶液是pH值大于7.0的溶液。
官方SAT2化学真题解析Subj...
官方SAT2化学真题解析Subj...Chemistry SAT Subject Tests?Answer Explanations to Practice Questions from Getting Ready for the SAT Subject TestsVisit the SAT? student website toget more practice and study tips forthe Subject Test in Chemistry.SAT? Subject Test in ChemistryIn this document, you will find detailed answer explanations to all of the chemistry practice questions from Getting Ready for the SAT Subject Tests. By reviewing these answer explanations, you can familiarize yourself with the types of questions on the test and learn your strengths and weaknesses. The estimated difficulty level is based on a scale of 1 to 5, with 1 the easiest and 5 the most difficult.To find out more about the SAT Subject Tests, visit us at /doc/b17639eaaeaad1f346933fac.html.1. Difficulty: 2Choice (C) is correct. To answer this question, you mustrecognize which of the choices above are acid solutions.Only choices (A) and (C) satisfy this requirement. Choice(B) refers to a neutral salt solution, choice (D) is a solutionof an alcohol, and choice (E) is a basic solution. Choices(A) and (C) are both acidic solutions, but choice (A) is astrong acid that is completely ionized in aqueous solution, whereas choice (C) is only partially ionized in aqueoussolution. Since the concentrations of all the solutions are the same, you do not need to consider this factor. The hydrogen ion concentration of a 0.1 molar acetic acid solution isconsiderably smaller than 0.1 molar. The hydrogen ionconcentration in choice (A) is equal to 0.1 molar. Thus, the correct choice is choice (C), a weakly acidic solution.2. Difficulty: 3Choice (E) is correct. To answer this question, you needto understand the pH scale, which is a measure of thehydrogen ion concentration in solution and is defined aspH = -log [H+]. The higher the pH, the lower the hydrogen ion concentration and the more basic the solution. Among the choices given above, choice (E) is the most basic solution.3. Difficulty: 4Choice (A) is correct. To answer this question, you needto know that acids react with bases to form salts andwater. Since the question refers to equal volumes of eachsolution, assume 1 liter of each solution is available.Barium hydroxide solution is a strong base (i.e., it iscompletely ionized in water), and 1 liter of 0.05 M Ba(OH)2 provides 0.1 mole of OH- ions in solution. When 1 liter of this solution is added to 1 liter of either 0.1 M NaCl,0.1 M CH3OH or 0.1 M KOH, no reactions occur and theresulting solutions remain basic; that is, the pH will begreater than 7 in each case. When 0.1 mole of OH- ionsreacts with 0.1 mole of acetic acid, the resulting solutionwill also be basic and have a pH greater than 7 becauseacetic acid is a weak acid (i.e., it is incompletely ionized inwater). The acetic acid reacts with the OH- ions as follows: HC2H3O2 + OH- C2H3O2- + H2OThe acetate ion produced is a strong base, which hydrolyzes in water to yield a solution containing more OH- ions than H+ ions. When 1 liter of 0.05 M Ba(OH)2 reacts with1 liter of 0.1 M HCl, there is a reaction between 0.1 mole ofOH- ions and 0.1 mole of H+ to form 0.1 mole of H2O. The resulting solution contains Ba2+ ions and Cl- ions and equal concentrations of OH- and H+ ions. The solution formed is neutral, and the pH is 7; therefore choice (A) is the correct answer.4. Difficulty: 3Choice (A) is correct. The chemical formula of sodiumacetate is NaC2H3O2. The sodium ion is Na+, and theacetate ion is C2H3O2-. The Na+ ion has the form X+.5. Difficulty: 1Choice (C) is correct. The chemical formula of aluminumoxide is Al2O3. The aluminum ion is Al3+, and the oxide ion is O2-. The Al3+ ion has the form X3+.6. Difficulty: 2Choice (A) is correct. The chemical formula of potassiumphosphate is K3PO4. The potassium ion is K+, and thephosphate ion is PO43-. The K+ ion has the form X+, andthe phosphate ion has the form XO43- ; therefore the correct answer is choice (A).7. Difficulty: 3Choice (C) is correct. The highest principal quantum number in this electron configuration is 3 (valence electrons 3s2 3p4), so the element is in the third period of the periodic table. The atom has two s and four p valence electrons, so it is in theoxygen group. The atom is therefore an atom of sulfur.8. Difficulty: 3Choice (A) is correct. The Ca2+ ion is a Ca atom that haslost two electrons. Ca has an atomic number of 20 and has20 electrons, so Ca2+ must have 18 electrons. The neutralatom with 18 electrons (and 18 protons) is Ar. Note thatAr is two elements before Ca in the periodic table.9. Difficulty: 2Choice (E) is correct. The element with the lowest atomicnumber that has any electrons in an f orbital is cerium(atomic number 58), which has the electron configuration[Xe] 4f 1 5d1 6s2. All elements with an atomic numberequal to or greater than 58 have ground-state atoms withelectrons in f orbitals. Uranium has an atomic number of92; therefore, the correct answer is (E).10. Difficulty: 2Choice (A) is correct. Argon is a noble gas, and noble gases are relatively unreactive.101. Difficulty: 3Both statements are true, and statement II gives the reason that statement I is correct. The correct answer is therefore true, true, correct explanation.102. Difficulty: 3It is true that diamond melts at a very high temperature(over 3,500°C). Substances with ionic bonding do have high melting points, but the bonding in diamond is network covalent, not ionic. (Note that unless both statements aretrue, it is not necessary to determine whether the secondstatement is a correct explanation of the first statement.) 103. Difficulty: 3Ionization energy depends on effective nuclear chargeand the distance of the electron from the nucleus. Outerelectrons are partially shielded from the nucleus by innerelectrons, so the effective nuclear charge is about the same for atoms of elements in the same group of the periodictable. Potassium (K) and lithium (Li) are in the samegroup of the periodic table, but the lithium atom has fewer occupied shells and a smaller atomic radius. The outermost electron in Li is in a shell that is close to the nucleus, and the outermost electron in K is in a shell that is relatively far from the nucleus. Therefore, there is less attraction between the outermost electron in a K atom and its nucleus thanbetween the outermost electron in a Li atom and its nucleus, and it is easier to remove an electron from a K atom, resulting in a lower first ionization energy for K than for Li.It is true that K (atomic number 19) has more protons inits nucleus than does Li (atomic number 3), but this is notthe reason for the relative ionization energies. (Note, forexample, that Ca has more protons in its nucleus than does K, but Ca has a higher first ionization energy than does K.) The correct answer is true, true, not a correct explanation. 104. Difficulty: 4In the activity series, Zn is relatively easy to oxidize to Zn2+, while Cu is relatively inactive. Therefore, if Zn metal were placed in a solution containing Cu2+ ions, Zn would beoxidized to Zn2+ ions, and Cu2+ would be reduced to Cumetal. Both statements are true, and statement II gives the reason that statement I is correct. The correct answer istherefore true, true, correct explanation.105. Difficulty: 5When a system at equilibrium is disturbed, the systemwill shift in a way that partially offsets the effect of thedisturbance (LeChatelier’s principle). Thus, when HC2H3O2 is added to the equilibrium mixture, the equilibriumwill shift to the right. This increases the concentration ofC2H3O2- and H3O+. Statement I is therefore false. Changes in the concentrations of reactants and products do not affect thevalue of the equilibrium constant; if the temperature does not change, the value of the equilibrium constantdoes not change. Statement II is therefore false. The correct answer is false, false.11. Difficulty: 3Choice (C) is correct. This is a question that concernsthe concentration of a diluted solution. One way to solvethe problem is through the use of ratios. In this question,a solution of nitric acid is diluted 10-fold; therefore, theconcentration of the solution will decrease by a factor of 10, that is, from 0.100 molar to 0.010 molar. Alternatively, you could calculate the number of moles of H+ ions present and divide this value by 0.50 liter: (0.100 × 0.050)/0.5 = M of the diluted solution.12. Difficulty: 2Choice (D) is correct. This question pertains to thebalancing of chemical equations. To answer this questioncorrectly, you need to recognize that both mass and charge must be conserved in any chemical equation. With this in mind, the chemical equation is correctly written as2 Cu2+(aq) + 4 I-(aq) → 2 CuI(s) + I2(s)The coefficient for I-(aq) is 4.13. Difficulty: 4Choice (C) is correct. This is a laboratory-orientedquestion pertaining to the measurement of gas pressures. It demands higher-level analytical skills that involve drawing conclusions from results obtained in an experiment. To answer this question correctly, you must first understandthat, in an open type of manometer, the air exerts pressure on the column of liquid in the open side of the U-tube, andthe gas being studied exerts pressure on the other side of the U-tube. It is clear, then, that statement I is true sincethe data given show that the manometer is open-ended,and its right side is exposed to the atmosphere. Statement II is also true because the level of liquid mercury is higherin the right side, which is exposed to the atmosphere, than in the left side, which is exposed to the gas. Thus the gaspressure is greater than atmospheric pressure. StatementIII is not correct because the pressure of the gas in thebulb, expressed in millimeters of mercury, is equal to thedifference in height, h, of the two mercury levels, plus theatmospheric pressure. Thus only statements I and II aretrue, and the correct answer is choice (C).14. Difficulty: 3Choice (B) is correct. This is a question on states ofmatter. Y ou must convert the description of the physicalphenomenon given in the question to graphical form.When a liquid is cooled slowly, its temperature will decrease with time. Thus the first portion of a graph depicting this phenomenon must show a decrease when temperature isplotted against time. When a pure liquid substance reaches its fusion (melting) point, continued cooling will release heat with time as the substance solidifies. During thisperiod, there is no drop in temperature. After the substance has completely solidified, further cooling will cause an additional drop in temperature. The only graph shown that accurately depicts the events described is choice (B), which is the correct answer.15. Difficulty: 3Choice (A) is correct. This is a question on chemicalbonding that requires you to apply the principles ofmolecular bonding. Each of the molecules given is correctly paired with the term describing its molecular geometry except choice (A). The geometry of PF3 is not trigonalplanar but trigonal pyramidal because this geometrycorresponds to a maximum possible separation of theelectron pairs around the central atom, phosphorus, andtherefore yields the most stable configuration; the centralatom of the molecule is surrounded by the three singlebonds and one unshared electron pair. Thus, the correctanswer is choice (A). Note that this is the type of questionthat asks you to identify the one solution to the problemthat is inappropriate.16. Difficulty: 2Choice (C) is correct. This question tests your ability tobalance chemical equations. The stoichiometry of thecorrectly balanced equation indicates that 2 moles of SO2(g) are needed to react completely with 1 mole of O2(g) to form2 moles of SO3.17. Difficulty: 3Choice (B) is correct. This is a question on stoichiometrythat tests the important skill of scientific reasoning basedon experimental evidence. The question indicates that 100 percent of the composition of the compound analyzed can be accounted for with the elements hydrogen and carbon.Thus, this compound is a hydrocarbon and choice (A)is a correct statement. It is not the correct answer to thequestion, however, because you can deduce more specific conclusions about this compound from the informationgiven. The relative percentage composition providesevidence that the atomic ratio of carbon to hydrogenin the compound must be 86/12 : 14/1.0, or about 1 : 2.Therefore, you can conclude that the empirical formula for the compound is CH2. Thus choice (B) is a better answer than choice (A). Since you do not know the total numberof moles of the compound used for analysis, you cannotcalculate the molar mass or derive the molecular formulafor this compound. Therefore choices (C) and (D) cannotbe determined from the information given and thus are not correct answers to the question. It is known, however, that a substance with an empirical formula of CH2 cannot have a triple bond. Therefore choice (E) is incorrect.18. Difficulty: 2Choice (D) is correct. The molar mass of C3H8 is 44.0 g/mol.According to the equation, four moles of water are formed for each mole of propane combusted. The molar massof water is 18.0 g/mol, so the mass of water formed is18.0 g × 4 = 72.0 g.19. Difficulty: 1Choice (D) is correct. Each mole of KHSO4 contains fourmoles of oxygen atoms, so 0.50 mol of KHSO4 containstwo moles of oxygen atoms. One mole is 6.0 × 1023, and2 (6.0 × 1023) = 12 × 1023 = 1.2 × 1024.20. Difficulty: 3Choice (E) is correct. Air contains mostly N2 (molar mass28 g/mol) and O2 (molar mass 32 g/mol). The molar mass ofcarbon dioxide is 44 g/mol, so carbon dioxide is more dense than air. The other statements are true and are not correct choices.21. Difficulty: 3Choice (D) is correct. Going down a group of the periodictable, atomic radius and atomic mass increase. Becauseeffective nuclear charge is about the same within a groupand radius increases going down a group, ionization energy decreases going down a group.22. Difficulty: 4Choice (E) is correct. In aqueous solution, HI can donate aproton and form I-, NH4+ can donate a proton to form NH3, HCO3- can donate a proton to form CO32-, and H2S can donate a proton to form HS-. NH3 does not donate a proton in aqueous solution to form NH2-; therefore, choice (E) is the correct answer.2011 The College Board. College Board, achieve more, SAT and the acorn logo are registered trademarks of the College Board. SAT Subject Tests is a trademark owned by the College Board.。
化学学科介绍英语作文
化学学科介绍英语作文Chemistry: An Introduction.Chemistry, the study of matter and its properties, is a fundamental science that has shaped our understanding ofthe natural world and played a pivotal role intechnological advancements. From the materials we use in everyday life to the medicines that heal us, chemistry underpins countless aspects of modern society.Branches of Chemistry.Chemistry encompasses a diverse array of subfields,each focused on specific aspects of matter and its behavior:Inorganic Chemistry: Explores elements and compounds that do not contain carbon atoms, such as metals, salts,and semiconductors.Organic Chemistry: Concerned with carbon-basedcompounds, including those found in living organisms and synthetic materials.Analytical Chemistry: Develops methods for identifying and quantifying substances in various samples.Physical Chemistry: Examines the physical properties of matter, such as its structure, energy, and reactivity.Biochemistry: Integrates chemistry and biology to understand the chemical processes that occur in living systems.Key Concepts in Chemistry.At the heart of chemistry lie several fundamental concepts that govern the behavior of matter:Atoms: The basic building blocks of all matter, consisting of a nucleus surrounded by electrons.Elements: Pure substances composed of only one type ofatom.Molecules: Combinations of two or more atoms that form distinct chemical entities.Chemical Bonding: The forces that hold atoms together in molecules and compounds.Chemical Reactions: Processes that involve the rearrangement of atoms to form new substances.Applications of Chemistry.Chemistry has a profound impact on various fields and industries, including:Materials Science: Development of new materials with enhanced properties, such as strength, flexibility, and conductivity.Medicine: Creation of drugs, vaccines, and diagnostic tools to treat and prevent diseases.Energy: Research on renewable energy sources, batteries, and fuel cells.Environmental Science: Monitoring and mitigating pollution, studying climate change, and finding sustainable solutions.Food Science: Ensuring food safety, developing new food products, and addressing global food security.Importance of Chemistry.Chemistry is essential for understanding the world around us and addressing global challenges. It enables us to:Innovate: Develop new technologies and products to improve our lives and solve problems.Solve Healthcare Challenges: Find cures for diseases, design targeted treatments, and improve patient outcomes.Promote Sustainability: Find sustainable solutions to reduce environmental impact and preserve resources.Foster Innovation: Inspire future generations of scientists and engineers to explore the frontiers of chemistry.Conclusion.Chemistry is a dynamic and interdisciplinary sciencethat plays a crucial role in shaping our world. Its applications extend far beyond the laboratory, touching countless aspects of our lives. By understanding the principles of chemistry, we gain a deeper appreciation for the natural world and the potential for scientific advancements to improve society. As we venture into the future, chemistry will undoubtedly continue to shape our understanding of matter and its transformative applications.。
我的理想是科学家研究化学作文
我的理想是科学家研究化学作文英文回答：As a scientist, my ideal career would be to conduct research in the field of chemistry. Chemistry is a fascinating subject that explores the composition, structure, properties, and reactions of matter. I am passionate about delving into the intricacies of chemical compounds and their interactions, and I believe that by contributing to the advancement of chemical knowledge, I can make a meaningful impact on the world.In order to pursue a career as a chemist, I would need to obtain a strong foundation in chemistry through academic study and hands-on laboratory experience. This would involve earning a bachelor's degree in chemistry or a related field, and potentially pursuing further education at the graduate level. I would also seek out opportunities to participate in research projects and internships to gain practical skills and insight into the scientific process.In addition to academic and practical training, I believe that effective communication and collaboration are essential skills for a successful scientist. The ability to convey complex scientific concepts to a diverse audience and work effectively within a team are crucial for advancing research and making meaningful contributions to the field of chemistry.Overall, my ideal career as a scientist would involve conducting innovative research, collaborating with other experts in the field, and ultimately contributing to the development of new knowledge and technologies in the realm of chemistry.中文回答：作为一名科学家，我的理想职业是在化学领域进行研究。
课程思政教学体系中化学类课程案例式教学的研讨
2021| 02教学研究当代化工研究n s Modern Chemical R esearch-i-J J课程思政教学休系中化学类课程案例式教学的研讨*韩永生1王若愚1车平2(1.中国科学院大学化学工程学院北京1000492.北京科技大学化学与生物工程学院北京100083)摘要：化学类课程在理工科院校中占有重要的地位，它的实践性和社会应用性很强。
本文通过不同高校在化学教学中的案例式教学实例来说明化学类课程中自然融入课程思政元素，课程思政融入到化学类课程教学中需要“自上而下”来实现，需要从以下几方面努力：合理 的课程群设置、加强基层教学组织建设和发挥教师在课堂教学中积极性、主动性和创造性。
关健词：课程思政；教学设计；课程设置中图分类号：G64; 06 文献标识码：ADiscussion on Case Teaching of Chemistry Courses in Ideological and PoliticalTeaching SystemHan Yongsheng1,Wang Ruoyu1,Che Ping2(1.School of Chemical Engineering,University of Chinese Academy of Sciences,Beijing, 1000492.School of Chemistry and Biological Engineering,Beijing University of Science&Technology,Beijing, 100083)Abstract: Chemistry courses occupies an important position in science & engineering colleges and comprehensive universities since it has strong p racticality and industrial applications. This article uses case-based teaching examples o f c hemistry in different universities to illustrate that chemistry courses naturally incorporate ideological a nd p olitical curriculum. It n eeds a ''top-down'1d esign to realize the integration o f i deological and political curriculum into the teaching o fchemistry courses. The f ollowing issues are significant: reasonable curriculum group setting, strengthening the construction of b asic teaching organization, and g iving f ull attention to teachers' enthusiasm, initiative and creativity in teaching.Key words i ideological and p olitical education% instructional design% curriculum design课程思政，是当前高等教育课程改革的热点和发展方 向，思政课是落实立德树人根本任务的关键课程，而其它课 程中全面贯彻课程思政也是传播知识，传授美德，让社会主 义核心价值观的种子在学生们心中生根发芽的重要途经。
多模态过程的全自动离线模态识别方法
第42卷第1期自动化学报Vol.42,No.1 2016年1月ACTA AUTOMATICA SINICA January,2016多模态过程的全自动离线模态识别方法张淑美1王福利1,2谭帅3王姝1,2摘要多模态是复杂工业生产过程的普遍特性.不同模态具有不同的过程特性,需要建立不同的模型,因此离线建模数据的模态划分与识别是整个多模态过程建模的关键问题之一.目前,常用的聚类算法需要对其结果进行人工分析和后续处理,无法真正实现多模态过程的全自动模态识别.因此,本文提出一种全自动的多模态过程离线模态识别方法.首先通过宽度为H的大切割窗口对数据进行切割,利用改进的K-means聚类算法对窗口单元进行聚类;根据聚类结果,对稳定模态淹没现象进行处理,得到模态的初步划分结果;最终,利用小滑动窗口L,对稳定模态及过渡模态交接区域进行细划分,准确定位稳定模态与过渡模态的分割点.算法实现了多模态过程的全自动离线识别,并给出合理有效的识别结果.仿真分析表明此方法能够实现模态的自动识别,且识别结果准确.关键词模态识别,多模态过程,过渡模态,稳定模态,全自动引用格式张淑美,王福利,谭帅,王姝.多模态过程的全自动离线模态识别方法.自动化学报,2016,42(1):60−80DOI10.16383/j.aas.2016.c150048A Fully Automatic Offline Mode Identification Method for Multi-mode ProcessesZHANG Shu-Mei1WANG Fu-Li1,2TAN Shuai3WANG Shu1,2Abstract Multimode is a general characteristic of complex industrial processes.Different modes have different process characteristics,and different models should be established.Therefore,offline mode identification is one of the critical problems for multimode processes modelling.Presently,the commonly used clustering methods cannot realize offline mode identification of multimode processes automatically because human analysis and further processing are needed to gain thefinal identification result.A fully automatic offline mode identification method is proposed in the paper.First, the data is divided into a series of data segments by a cutting window with the designated width H.The improved K-means clustering method is used to assign the segments into different clusters.According to the clustering result,the missing stable modes are dealt with to obtain the preliminary mode identification results.Finally,the regions between the stable modes and transitional modes are further analyzed by a small moving window L to determine the accurate boundaries between different modes.The mode identification of multimode processes can be realized automatically by the method for a reasonable and effective identification result.Feasibility and practical value of the method are evaluated by case study.Key words Mode identification,multi-mode process,transition mode,stable mode,fully automaticCitation Zhang Shu-Mei,Wang Fu-Li,Tan Shuai,Wang Shu.A fully automatic offline mode identification method for multi-mode processes.Acta Automatica Sinica,2016,42(1):60−80收稿日期2015-03-04录用日期2015-08-31Manuscript received March4,2015;accepted August31,2015国家自然科学基金(61533007,61374146,61403072),流程工业综合自动化国家重点实验室基础科研业务费(2013ZCX02-04),中央高校基本科研专项资金(N140404020),华东理工大学探索研究专项基金(22A 201514050)资助Supported by National Natural Science Foundation of China (61533007,61374146,61403072),State Key Laboratory of Syn-thetical Automation for Process Industries Fundamental Re-search Funds(2013ZCX02-04),the Fundamental Research Funds for the Central Universities(N140404020),and the Fundamen-tal Research Funds for East China University of Science and Technology(22A201514050)本文责任编委胡昌华Recommended by Associate Editor HU Chang-Hua1.东北大学信息科学与工程学院沈阳1108192.流程工业综合自动化国家重点实验室(东北大学)沈阳1108193.化工过程先进控制和优化技术教育部重点实验室(华东理工大学)上海2002371.College of Information Science and Engineering,Northeast-ern University,Shenyang1108192.State Key Laboratory of Synthetical Automation for Process Industries(Northeastern随着社会的进步,工业生产更加倚重于可以生产多种产品的高效过程,过程生产方案的变动或者产品类型的改变会导致生产过程出现具有不同过程特性的多种模态,多模态也成为复杂工业生产过程中普遍存在的一种特性.因此,多模态过程成为研究热点之一.多模态生产过程包含多个稳定生产模态,不同的稳定模态之间存在不同的过渡模态[1].不同模态的过程特性差别很大,变量相关性也有着显著的差异,故无法用一个模型对多个不同的模态特性同时进行描述.为了解决多模态特性对过程监测、故障诊断等问题的影响,Yu等[2−3]提出了基于高斯混合University),Shenyang110819 3.Key Laboratory of Advanced Control and Optimization for Chemical Processes of Ministry of Education(East China University of Science and Technology), Shanghai2002371期张淑美等:多模态过程的全自动离线模态识别方法61模型(Gaussian mixture model,GMM)的多模态处理方法.在此基础上,Xie等[4]提出了滑动GMM 的方法,解决多模态过程中单个模态的动态问题. Ge等[5−6]利用混合概率PCR的思想,巧妙解决了多模态过程的质量回归问题.这些基于贝叶斯思想的方法将所有模态放在一起建立一个大的混合模型,无需进行模态识别.但此类方法只考虑了稳定模态,并没有考虑过渡模态的影响.如果将这些方法直接应用到含有过渡模态的过程中,则会导致过渡模态描述信息不完全,在监测的过程中可能给出错误的结论.但是过渡模态是多模态过程的一大特征,在许多过程中是无法忽略的.因此,对于多模态过程,尤其含有过渡模态的多模态过程,无论是过程优化、过程监测、故障诊断或者过程状态评价等等,都需要对差异明显的多个模态采用不同的方法分别进行分析.为了解决该问题,Ng等[7]提出了一种基于重叠PCA的联合监测方法,但是该方法无法分辨稳定模态与过渡模态.Lu等[8]和Zhao等[9]提出了多模型的监测思想并将基于该思想的一系列监测方法[10−13]成功应用到了多阶段批次过程和多模态连续过程中.Zhang等[14−15]有效提取所有模态之间的公共信息,将每个模态分成公共子空间和独立子空间,提高了监测效果.这些多模型方法能够有效解决过渡模态所带来的影响,但是在应用时首先需要对离线数据进行模态识别和划分,然后才能针对不同模态建立不同模型.也就是说,离线建模数据的模态划分与识别是实现多模态过程建模、优化、监测以及状态评价的基础和首要前提,是研究多模态过程的关键问题之一.离线建模数据的模态划分与识别就是将一组正常操作数据按照其生产模态的不同进行类别划分,同时辨别出生产数据对应的模态信息(某一稳定模态或某过渡模态)[1],对于简单的有模态指示变量的生产过程,可以根据已知的模态指示变量准确地判断出建模数据所对应的运行模态.但大多工业生产过程相对复杂,通常没有模态指示变量,因此需要提出一种合理有效的方法来实现过程模态的自动区分与识别.目前,已经有许多聚类算法应用到模态识别中,例如K-means聚类算法[16]、减聚类算法[17]、模糊C 均值聚类算法[18−19]等.但这些算法在处理多模态过程时存在以下问题:1)一个较完整的多模态过程的操作数据,尤其一些连续生产的大工业过程,包含了多个稳定模态以及稳定模态间的过渡模态,样本容量大、复杂度高.如果数据不加处理,直接应用聚类算法,则计算量大,且容易给出错误的聚类结果.2)这些聚类算法只能将数据分成不同的类,无法给出具体的模态信息.尤其是当多模态过程含有过渡模态时,聚成的子类属于哪个模态,这些子类之间有什么关系,这些信息都无法根据聚类结果直接获取.3)过程数据经常会出现一些奇异点,这些奇异点对模态的划分有着很大的影响,因此需要对奇异点现象进行处理.4)由于过渡模态具有一定的动特性,多模态过程数据复杂,聚类算法的参数选取比较困难.不准确的参数有可能导致某一稳定模态被划分到不同子类中,从而出现稳定模态被淹没的现象.因此,这些聚类算法只适用于多模态过程的初始分类,要得到具体的模态信息,需要对其聚类结果进行进一步分析.目前,大多数实际过程都是对聚类结果进行简单人为处理,而且没有考虑到初始聚类结果不理想的情况,因此当前的一些处理算法存在两大主要问题:1)不能实现完全自动的模态识别;2)给出错误识别结果的可能性很大.为了解决上述问题,本文提出一种系统的全自动离线模态识别方法.首先,选取长度为H的切割窗口对离线建模数据进行切割,将各窗口作为分析的基本单元,提取每个窗口内数据的均值,通过改进的K-means聚类算法[16],对窗口均值向量进行聚类;针对由于聚类算法参数不准确等原因造成的稳定模态被埋没的现象进行处理,实现稳定模态和过渡模态的初步划分;在初步确定的模态基础上,选定一个小滑动窗口L,对稳定模态及过渡模态交接区域进行进一步细划分,准确定位稳定模态与过渡模态的分割点,实现多模态建模数据的模态识别.1多模态过程的离线模态识别多模态过程中,稳定模态是生产过程运行状态比较稳定的过程,变量基本在某一种状态下稳定运行,因此时间窗口内的变量均值在稳定模态期间变化不大.相比于稳定模态,过渡模态是从一种稳定运行模式向另一种稳定运行模式渐近式量变的动态过程,出现了较多的动态变量,变量的均值随时间方向也呈现出一种动态的渐变趋势.根据多模态过程的这一特性,本文对多模态过程进行离线模态识别.假设一个多模态生产过程,其建模数据表示为最常用的二维矩阵形式ˆX∈R N×J,其中N和J分别表示过程的采样个数以及过程变量个数.首先对数据进行无量纲化处理[20],即将不同变量的方差归一,从而使每一个变量在数据模型中都具有同等的权重,消除了过程变量测量值的量程差异对聚类结果的影响,无量纲处理后的数据记为X.1.1模态初步确定首先,寻找过程数据中的稳定模态和过渡模态,实现过程模态的初步确定.62自动化学报42卷步骤1.窗口切割选取长度为H的切割窗口对二维数据矩阵X沿着采样方向进行分割,即将矩阵X转置后沿横轴方向分割.设N可表示为N=K×H+d(0≤d<H).这样N个采样数据被分割为K个窗口,每个切割窗口按着采样时间方向顺次排列.第k,k=1,2,···,K个窗口的数据表示为X Tk∈R J×H.求取这些二维矩阵的均值向量,记为x k,这些均值向量表征了窗口内变量的相关信息,成为模态划分的基本单元.步骤2.聚类分析下面利用改进的K-means聚类算法对均值向量进行聚类.该聚类算法的具体步骤简述如下:算法的输入是均值向量集合x1,x2,···,x K,以及两个子类中心的最小距离阈值θ.算法的输出是子类数量C s t.子类中心w1,w2,···,w Cst,以及每个窗口属于不同子类的隶属关系u(k):x k→1,2,···,C st.变量i,k及c分别是算法中迭代次数、分类模式以及聚类中心的索引.1)从K个被分类单元中,任意选择C0个单元向量作为初始聚类中心W i,c(c=1,2,···,C0).如果C0个初始聚类中心选取的过于集中或者个数较少,有可能造成局部最优,导致最终形成的子类比真正的子类要少,给出错误结果.因此,一般常用方法是从被分类单元向量中均匀抽取C0个模式,建议C0=K/3∼K/2.2)若两个子类中心的距离dist(w i,c1,w i,c2)小于预定的阈值θ,剔除其中一个聚类中心,更新子类数量,记为C i+1.3)计算每个单元向量x k(k=1,2,···,K)到所有聚类中心的距离,若x k和第c∗类的中心的距离最小,则将x k的隶属关系定义为u(k)=c∗.4)根据单元向量的隶属关系重新计算新的聚类中心dist(w i,c,c=1,2,···,C st).5)如果算法满足收敛条件则结束,否则返回2),进行下一次迭代计算.收敛条件有:两次迭代中的聚类中心距离的变化小于一个很小的阈值 ，或者每个子类中各单元向量x k到子类中心的距离平方和以及子类之间的距离平方和达到最小.上述分类算法最终可以将整个多模态过程分成C st子类.步骤3.时段划分将窗口数据沿着时间轴展开,时间上连续且属于同一个子类的窗口单元划分到同一个时段.假设整个过程被初步划分为M0个时段,各个时段按照时间方向有序排列.第m个子时段记为ˆX m ,m=1,2,···,M0.ˆX m属于第c个子类,即u(ˆX m)=c,其中c=1,2,···,C st.第m个子时段时间长度记为ˆS m.由于聚类算法的输入是按照时间顺序排列的窗口均值向量,因此按照时间顺序,将时间上相邻且属于同一子类的窗口单元划分到同一个时段中,这样就使得那些具有相似过程状态的连续时间窗口被包含在同一个时段中,时段内部也呈现出时间的连续性.也就是说,如果过程变量在某段时间内保持一致,这段时间上的均值向量将会被聚类算法划分在同一个时段中.当过程变量的状态发生改变时,对应的均值窗口被分到不同的时段.步骤4.稳定模态确定根据经验知识定义稳定模态最短运行时间S min,将时段长度大于稳定模态最短运行时间S min的时段定义为稳定模态,且隶属于同一子类的稳定模态定义为同一种稳定模态.如果两个相邻的稳定模态之间所有时段的运行长度之和大于S min,说明在两个稳定模态之间仍然含有稳定模态.这是由于多模态过程既含有稳定模态,又含有动态特性明显的过渡模态,数据特性复杂.聚类算法在聚类时由于参数设定不恰当或者陷入局部最优等问题,可能使得某稳定模态被聚成比较多的子时段,导致该模态在初步识别时无法满足稳定模态的识别条件,造成稳定模态识别上的缺失,这里将这种现象定义为稳定模态的淹没现象.我们需要对该现象进行分析处理.假设现在已确定有A和B两种稳定模态,其中稳定模态A和B相邻,且A和B之间所有时段的长度之和大于S min,则做以下处理:1)在稳定模态A和模态B之间的所有子时段中寻找运行时间最长的一段,作为中间新稳定模态的基准时段.如果运行时间最长的时段不只一个,则选择最靠近中间位置的时段.这是由于两个稳定模态之间一定含有过渡模态,且稳定模态运行时间远远大于过渡模态运行时间,所以若稳定模态A和稳定模态B之间含有稳定模态,则该模态应该位于靠近模态A和B之间的中间位置.假设第k个时段是该稳定模态的基准时段,记为ˆX base=ˆX k,其长度ˆS base=ˆS k.与ˆX base相邻的两个时段分别定义为ˆX k1和ˆX k2,其中k1=k−1,k2=k+1.2)以ˆX base为基准,比较其左右两个相邻的时段与ˆX base的相似度,即计算γ(ˆX base,ˆX k1)以及γ(ˆX base,ˆX k2).时段ˆX1和ˆX2的相似度定义为γ1,2=γˆX1,ˆX2=exp−X1−X22J(1)其中,X1和X2分别为时段ˆX1和ˆX2的均值向量.3)如果两个时段相似度相等,则将两个时段同1期张淑美等:多模态过程的全自动离线模态识别方法63时加入到ˆX base中,否则选择相似度高的时间段加入到ˆX base中,组成新的ˆX base.ˆX base =[ˆX baseˆX k2],γbase,k1<γbase,k2 [ˆX k1ˆXbaseˆXk2],γbase,k1=γbase,k2 [ˆX k1ˆXbase],γbase,k1>γbase,k2(2)ˆS base =ˆSbase+ˆS k2,γbase,k1<γbase,k2ˆSk1+ˆS base+ˆS k2,γbase,k1=γbase,k2ˆSk1+ˆS base,γbase,k1>γbase,k2(3)4)如果时段满足终止条件则认为稳定模态已经确定,否则返回2),进行下一次迭代计算.终止条件为:迭代中最高相似度小于阈值σ,而且ˆX base的时间长度ˆS base大于稳定模态最短运行时间S min.5)计算新确定稳定模态与所有已知的稳定模态的中心距离d inew= X new−X i ,其中i=A,B, X new和X i分别为新确定稳定模态和第i个已知稳定模态的样本均值.如果最小的距离小于预定的阈值θ,则将该稳定模态可定义为相应的稳定模态(例如,d Anew =min(i Anew)且小于阈值θ,则将新确定稳定模态定义为A稳定模态),否则说明该稳定模态是没有出现过的新模态,定义为C稳定模态.6)计算新确定稳定模态与相邻的前一稳定模态A之间所有时段的运行长度S A new.判断S A new是否大于S min.如果S A new<S min,认为新确定模态与模态A之间不存在其他稳定模态.否则,新稳定模态与模态A之间仍然存在稳定模态,按照新稳定模态的确定方法,继续寻找稳定模态,即重复1)∼5)所描述的提取过程,直到找出新确定模态与模态A 之间所有的稳定模态.7)对新确定稳定模态与相邻稳定模态B之间的时段进行处理,处理方法与6)类似,直到找出新确定模态与模态B之间所有的稳定模态.除了考虑两个相邻稳定模态之间是否存在新稳定模态,仍有两种情况需要考虑:1)过程起始状态为非稳定模态时,分析过程起点即第1个时刻到第一个稳定模态起点之间的数据是否含有新稳定模态;2)过程终止状态为非稳定模态时,分析最后一个稳定模态的终止时刻到过程终点之间是否含有新稳定模态.分析过程与两个稳定模态(A和B为例)之间的分析过程类似.经过上述处理,多模态过程的所有稳定模态都已确定,而且稳定模态之间的长度均小于稳定模态最短运行时间S min.步骤5.处理两个相邻稳定模态之间的时段对于两个相邻的稳定模态,有以下两种情况:1)相邻的两个稳定模态分别为两种不同的模态A和B,则认为两个稳定模态之间的所有子时段属于稳定模态A到模态B的过渡模态.2)相邻两个稳定模态为同一种模态A,则将两个稳定模态之间的所有子时段定义为稳定模态A的扰动过程.这种现象的发生是由于实际过程在正常运行的时候,出现较长时间的扰动,使得一些变量发生变化,经过控制系统的调节,过程再次达到稳定时的新状态与扰动前的状态相同,因此扰动前后的两种模态可以看作同一种模态.通过步骤1∼5,算法实现了多模态过程中所有的稳定模态以及过渡模态的识别与初步划分.模态的初步划分过程示意图如图1所示.1.2模态准确划分模态划分的关键是确定当前模态的结束时间,即确定下一种模态的开始时间.上一节实现了多模态过程各模态的初步划分,但是聚类单元为时间长度一定的切割窗口,因此模态的初步划分只能确定模态开始转变的起始窗口,却无法得知窗口内模态开始发生变化的具体时刻,因此需要对稳定模态与过渡模态相邻的窗口进行更深层次的分析.如图2所示,假设从第1个窗口到第k1−1个窗口为稳定模态A,从k1个窗口开始进入过渡模态.此时可能有两种情况:1)过渡模态的开始时间发生在第k1个窗口的前半段.那么第k1−1个窗口完全属于稳定模态,第k1个窗口数据以过渡模态为主,其均值向量到稳定模态A中心距离偏大,因此没有聚到稳定模态A所在的类中.2)过渡模态的开始时间发生在第k1−1个窗口的后半段.稳定模态的过程数据占据第k1−1个窗口的大部分,因此其均值向量离稳定模态A所在的类中心距离最近,故将此窗口归到稳定模态中.第k1个窗口内数据完全属于过渡模态,到稳定模态类中心的距离很大,因此将其分到了过渡模态起始子时段中.基于以上分析,为了准确判断稳定模态的结束时间,需要从第k1−1个窗口开始用较短的滑动窗口L重新分析判断,滑动步长为h.如图2所示,通过比较每个滑动窗口与稳定模态A的相似度来确定过渡模态的起始时间,定义滑动窗口与稳定模态的相似度,如式(4)所示:γx t,X A=exp−x t−X A2J(4)其中, x t为第t个小滑动窗口的均值,X A为稳定模态A的变量均值.相似度定量指示了各窗口与稳定模态的相似度.当多模态过程从稳定模态逐渐进入过渡状态,其变量64自动化学报42卷图1多模态过程的模态初步识别过程Fig.1Preliminary mode identification for multimode processes1期张淑美等:多模态过程的全自动离线模态识别方法65图2多模态准确识别过程(以AC 过渡模态起始时刻为例)Fig.2Exact mode identification for multimode processes (i.e.beginning of the transitional mode AC )66自动化学报42卷越来越偏离前一稳定模态所在的稳定状态,相应地,与稳定模态的相似度也逐渐减少.本文引入可以调节的相似度阈值α作为边界参数,按照如下的准则分析模态识别问题:如果t1是第一个满足以下条件的窗口:从第t1个小窗口开始,连续有r个窗口都满足γt<α,即满足时,认为过程从t1个小窗口进入到过渡模态.将第t1个小窗口的起始位置作为过渡模态的开始,则此过渡模态起始时刻为(k1−2)×H+(t1−1)×h+1.γt1−1≥α,γt<α,t=t1,t1+1,···,t1+r−1(5)过渡模态终止时间的确定与起始时刻的确定类似,不同的是考虑过渡模态与下一个稳定模态C的相似度,而且当多模态过程从过渡模态逐渐进入稳定模态时,滑动窗口与稳定模态C的相似度逐渐增大.AC过渡模态到第k2−1个窗口结束,从第k2个窗口进入下一稳定模态C.此时需要对从第k2−1个窗口开始用较短的滑动窗口L进行分析.通过比较每个滑动窗口与稳定模态C的相似度来确定过渡模态的终止时间.如果t2是第一个满足以下条件的窗口:从第t2个小窗口开始,连续有r个窗口都满足γt≥α,认为过程从t2个小窗口进入到稳定模态.将第t2个小窗口的起始位置记为下一稳定模态的开始,则过渡模态终止时刻为(k2−2)×H+(t2−1)×h+1.γt2−1<α,γt≥α,t=t2,t2+1,···,t2+r−1(6)由于规定窗口长度H不得大于最短过渡时间,因此,一般不会出现过渡模态被淹没的现象.即使出现这一情况,也不会影响最终的识别结果.例如,由于窗口长度H取值过大,导致初步识别时某一过渡模态被淹没,即两个稳定模态直接相连,中间没有过渡模态.这种情况下,说明被淹没的过渡模态一定位于两个稳定模态相连接的两个H窗口内.根据本节提出的准确划分方法,对这两个相邻的H窗口进行分析处理,便可确定过渡模态的起始时间和终止时间.1.3重要参数分析在模态识别过程中,涉及到几个重要参数的选取,影响模态的辨识结果.1)大切割窗口H一个比较完整的多模态过程操作数据,包含了不同的稳定模态以及稳定模态间具有动特性的过渡模态,样本容量大,复杂度高.如果不经过任何处理,直接对其进行聚类,奇异点等异常数据的存在以及过渡模态的动特性会大大影响聚类效果.因此,大切割窗口的设定一方面可以快速处理大量的建模数据,降低复杂度,另一方面可以有效减少异常数据对聚类的影响.H的选取会直接影响模态的初步确定.首先, H不得大于最短过渡时间长度,否则可能造成过渡模态的淹没现象.H越大,对噪声、奇异点的冗余能力越强.但同时,各模态起始终止时间的初步确定范围越宽.H越小,各模态起始终止时间的初步确定更加准确,但噪声奇异点等随机扰动的影响越大,聚类复杂度越高.在实际应用中,可以根据实际过程选取恰当的切割窗口.例如,在实际过程中,离线数据的样本数很大,数据质量较差,可选取较大的H.如果离线数据样本数较少,或者过渡时间较短,便选取较小的H.2)小滑动窗口L为了更加精确地确定各模态起始终止时间,需要对稳定模态与过渡模态相交接的两个窗口进行分析,此段数据长度为2H.因此需要一个小的滑动窗口L来进行分析.L越大,去噪能力越强,但过渡模态起始、终止时刻的确定误差有可能越大.L越小,噪声或奇异点等随机扰动导致错误结果的可能性越大.滑动步长h一般不能取得太大,否则会造成漏点现象.h越大,滑动速度越快,结果误差可能越大. h越小,滑动速度越慢,但结果越精确.在小滑动窗口L和滑动步长h的实际选取过程中,需要根据实际情况,选取合适的窗口长度,保证识别的有效准确.例如,H较小时,h完全可以取一个较小的值,甚至取为1.H很大时,考虑到算法的运算速度,h 可以取一个较大的值,但是如果过程对识别精度要求非常高,h取值也不能太大.3)相似度阈值α在模态识别的过程中,阈值α的选取会直接影响到模态识别精度和监测效果.α越大,要求窗口数据与稳定模态相似度越高,稳定模态的数据离稳定模态中心越近,但过高的阈值会导致原本属于稳定模态的数据也划分到过渡模态中,在增大过渡模态范围的同时丢失了部分稳定模态信息.α太小,会将过渡模态数据划分到稳定模态中,从而无法准确定位模态的起始终止位置.因此,针对不同的模态,结合过程数据提供的信息,定义一个弹性相似度阈值α.以图2为例,在初步划分时,从第一个窗口到第k1−1个窗口为稳定模态A,从k1个窗口开始进入过渡模态.AB过渡模态到第k2−1个窗口结束,从第k2个窗口进入下一稳定模态B.经过分析知道,过渡模态确切起始位置应该在。
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选择性必修二化学课本读后感英文回答：Chemistry Elective II is an advanced chemistry textbook designed for high school students who have a strong foundation in the subject. The book covers a wide range of topics, including the periodic table, chemical bonding, thermodynamics, kinetics, and electrochemistry. It is written in a clear and concise style, and it is well-organized and easy to follow.One of the strengths of Chemistry Elective II is its emphasis on problem-solving. Each chapter includes numerous practice problems and exercises, which allow students to apply their knowledge and skills to real-world situations. The book also contains a wealth of supplemental resources, such as worked-out examples, review questions, and self-assessment quizzes. These resources help students to master the material and to prepare for exams.Overall, Chemistry Elective II is an excellent resource for high school students who are interested in pursuing a career in chemistry or a related field. The book provides a comprehensive overview of the subject, and it is written in a way that is both accessible and engaging. I highly recommend this book to any student who is serious about learning chemistry.中文回答：选择性必修二化学是我高二化学的教科书，它的难度相较于必修一来说有一定提升。
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MCAT Subject TestsDear Future Doctor,The following Subject Test and explanations contains questions not in test format and should be used to practice and to assess your mastery of the foundation content necessary for success on the MCAT. Simply memorizing facts is not sufficient to achieve high scores; however, an incomplete understanding of basic science knowledge will limit your ability to think critically. Think of building your content knowledge as learning the vocabulary and practicing MCAT-like questions as actually speaking.All rights are reserved pursuant to the copyright laws and the contract clause in your enrollment agreement and as printed below. Misdemeanor and felony infractions can severely limit your ability to be accepted to a medical program and a conviction can result in the removal of a medical license. We offer this material for your practice in your own home as a courtesy and privilege. Practice today so that you can perform on test day; this material was designed to give you every advantage on the MCAT and we wish you the best of luck in your preparation.Sincerely,Albert ChenExecutive Director, Pre-Health Research and DevelopmentKaplan Test Prep© 2003 Kaplan, Inc.All rights reserved. No part of this book may be reproduced in any form, by Photostat, microfilm, xerography or any other means, or incorporated into any information retrieval system, electronic or mechanical without the written permission of Kaplan, Inc. This book may not be duplicated, distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement._________________________________________________________G E N E R A L C H E M I S T R Y S U B J E C T T E S T2 General Chemistry Subject Test 21. Which one of the following reactions will beaccompanied by an increase in entropy?A. Na(s) + H2O(l) ∅ NaOH(aq) + H2(g)B. I2(g) ∅ I2(s)C. H2SO4(aq) + Ba(OH)2(aq) ∅ BaSO4(s) +H2O(l)D. H2(g) + 1/2O2(g) ∅ H2O(l)E. None of the above2. Which one of the following has the largest ionicradius?A. Na+B. K+C. Mg++D. Al3+E. Cl–3. The heat of combustion of gaseous ammonia,NH3(g), is 81 kcal/mole. How much heat is released in the reaction of 34 grams of ammonia with excessoxygen?A. 40.5 kcalB. 60.3 kcalC. 75.8 kcalD. 81 kcalE. 162 kcal4. What is the percent composition by mass of Al inAl2(SO4)3?A. 7.9 %B. 31.6 %C. 15.8 %D. 12.7 %E. 22.3 % 5. Compound _H of formation(kcal/mole, 25 °C, 1 atm) H2O(l) –68.4H2O(g) –57.8C2H6(g) –20.2C2H4(g) +12.5C2H2(g) +54.2CO(g) –26.4 CO2(g) –94.1Based on the values contained in the table, what is theheat of combustion of one mole of ethylene at 298Kand 1 atm pressure?A. 316.1 kcalB. 12.5 kcalC. –291.1 kcalD. –316.1 kcalE. –337.3 kcal6. The reaction below is carried out at constanttemperature in a 2.0 liter vessel.N2 + O2∅ 2NOAt equilibrium, the vessel is found to contain 8 molesof NO, 2 moles of N2 and 4 moles of O2. What is thevalue of K eq for this reaction under these conditions?A. 1B. 8C. 0.25D. 2E. 0.57. I Aslow→fast→fast→BII B + C BC* (activated complex)III BC* D + EGiven the above series of reactions leading to theformation of products D and E, which is the rate-determining step?A. IB. IIC. IIID. I and IIE. II and IIIK A P L A N________________________________________________________________________________________1G E N E R A L C H E M I S T R Y S U B J E C T T E S T2 ________________________________________________________________2 ________________________________________________________________________________________K A P L A N_________________________________________________________G E N E R A L C H E M I S T R Y S U B J E C T T E S T 2 8.According to VSEPR Theory, the H-Te-H bond angle in H 2Te should be closest to A. 90°. B. 120°. C. 109.5°. D. 149.5°. E. 180°.9.A 300 mL flask containing nitrogen at a pressure of 100 torr and a 200 mL flask containing oxygen at a pressure of 200 torr are connected such that the gases are allowed to fill the combined volume of the two flasks. What is the partial pressure of nitrogen in the combined volume? A. 60 torr B. 80 torr C. 100 torr D. 150 torr E. 200 torr10. Which of the following statements is true with regard to the diagram below?U W X Y Z Reaction CoordinateFreeK A P L A N ________________________________________________________________________________________3A. Going from U to W requires less energy than goingform X to Y.B. Going from U to X occurs more readily than goingfrom X to Z.C. Although in the forward direction the reaction X toZ will occur more rapidly than from U to X, in the reverse reaction Z to X will occur more slowly than X to U.D. The reaction U to X is exothermic while thereaction from X to Z would be endothermic. E. The overall reaction U to Z is endothermic.11. Which of the elements below is the leastelectronegative?A. FB. ClC. BrD. KE. C12. Which one of the electronic configurations belowrepresents the atom with the highest ionization energy?A. 1s 22s 22p 63s 1B. 1s 22s 22p 63s 23p 64s 23d 104p 65s 14d 10C. 1s 22s 22p 63s 23p 64s 23d 104p 65s 24d 105p 5D. 1s 22s 22p 63s 23p 63s 23p 5E. 1s 22s 22p 63s 23p 613. Electron density studies have revealed that X and Yhave an equal number of electrons. Which of the following could be X and Y respectively?A. Ca + and KB. H + and HeC. Cl and FD. O – and S +E None of the above 14.1st Ionization 2nd Ionization Energy (eV) Energy (eV)U 5.6 5.6V1.72.9 X 1.1 13.6 Y 12.4 2.8 Z2.9 1.7From the information given in the table above, which of the following is most probably a Group IA metal? A. U B. V C. X D. Y E. ZG E N E R A L C H E M I S T R Y S U B J E C T T E S T 2 ________________________________________________________________4 ________________________________________________________________________________________K A P L A N15. 2 HCl + Na 2CO 3 ∅ 2 NaCl + H 2O + CO 2(g )(Atomic weights: H = 1, Cl = 35, Na = 23, O=16, C=12)36 g of HCl are mixed with excess Na 2CO 3. If the reaction is carried out at STP, how many liters of CO 2 are formed? A. 5.60 B. 11.2 C. 16.8 D. 22.4 E. 44.816. Which of the following elements would be the mostreactive with fluorine?A. ArB. TiC. ClD. LiE. IQuestions 17 -18 refer to the following kinetic rates:T rial [A]o (M) [B]o (M) rate (M/min) 1 1 x 10–2 3 x 10–3 2 x 10–3 2 1 x 10–2 6 x 10–3 8 x 10–3 3 2 x 10–2 1.2 x 10–23.2 x 10–217. What is the reaction order with respect to A?A. 1B. 1.5C. 2D. 3E. 018. What would be the initial rate of the reaction in afourth trial if [A]o = 4 x 10–2 M and [B]o = 3 x 10–3 M?A. 1 x 10–3 M/minB. 2 x 10–3 M/minC. 8 x 10–3 M/minD. 8 x 10 –2 M/minE. Cannot be determined from the information given19. A spontaneous reaction is always accompanied byA. a negative change in enthalpy.B. a positive change in enthalpy.C. a negative change in enthalpy and a positivechange in entropy.D. a positive change in enthalpy and a negativechange in entropy.E. a negative change in Gibbs free energy.20. Combustion of a 50 gram sample of an unknownhydrocarbon yields 132 grams of CO 2 and 126 grams of H 2O. Approximately how many grams of carbon were contained in the original sample?A. 14 gB. 28 gC. 36 gD. 63 gE. 66 g21. Under which of the following sets of conditions (temperature and pressure) will the ideal gas law be least likely to apply?A. 250 K, 4 atmB. 250 °C, 400 atmC. 25 °C, 4 torrD. 25 K, 400 torrE. 25 K, 400 atm22. The following reaction is first order with respect to Wand first order with respect to X. If the concentration of W and X are doubled what happens to the rate of the reaction?W + X ∅ Y + Z A. It is doubled. B. It is quadrupled. C. It is squared.D. It is increased but it cannot be determined how much from the information given.E. It remains the same.23. If atom A has a valence of 7 and atom B has avalence of 2, which compound consisting of atoms A and B would be the most likely to form?A. BAB. B 2AC. BA 2D. B 2A 2E. B 3A 2_________________________________________________________G E N E R A L C H E M I S T R Y S U B J E C T T E S T224. In moving down column II A of the Periodic TableA. electron affinity increases.B. ionization energy increases.C. electronegativity remains constant.D. the oxidation state remains constant.E. the shielding effect decreases.25. Which of the following choices lists elements, fromleft to right, with decreasing atomic radii?A. Li, Be, B, C, NB. Be, Mg, Ca, Sr, BaC. Li, Mg, Sc, Zn, TaD. Au, Ag, CuE. Two of the above26. Under which of the following conditions will areaction always be in thermodynamic equilibrium?A. ∆ H∆ S> 1B. ∆ H∆ S< 1C. ∆ H∆ S= 1D. ∆ HT ∆S= 1E. ∆ H∆ S< 027. An ideal gas is contained in a one-liter vessel at 350K. What additional information is sufficient todetermine the pressure of the gas?A. the molecular weight of the gasB. the number of gas moleculesC. the molecular weight and the mass of the gasD. B or CE. A, B, or C 28. An ideal gas is placed in a 10 liter container that isfree to expand. If the temperature (K) is doubled andthe pressure is changed from 60 mmHg to 20 mmHg,what is the new volume in liters?A. 2060x 12x 10B. 2060x 2 x 10C. 6020x 12D. 6020x 1( 2 x 10)E. 6020x 2 x 1029. The concentration of which of the followingsubstances would NOT decrease as a result of anincrease in the temperature?X2 3AB + P _H = -10 kcal/moleA. AB onlyB. both AB and PC. X2Y onlyD. both X2Y and BDE. Cannot be determined without a value for thechange in entropy30. Boron found in nature has an atomic weight of10.811 and is made up of the isotopes 10B (mass10.013 amu) and 11B (mass 11.0093 amu). Whatpercentage of naturally occurring boron is made up of 10B and 11B respectively?A. 30; 70B. 25; 75C. 20; 80D. 15; 85E. 10; 90STOP! END OF TEST.K A P L A N________________________________________________________________________________________5G E N E R A L C H E M I S T R Y S U B J E C T T E S T2 ________________________________________________________________THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE.6 ________________________________________________________________________________________K A P L A N_________________________________________________________G E N E R A L C H E M I S T R Y S U B J E C T T E S T 2K A P L A N ________________________________________________________________________________________7GENERAL CHEMISTRY SUBJECT TEST 2ANSWER KEY1. A2. E3. E4. C5. E6. B7. A 8. C 9. A 10. C 11. D 12. E13. A 14. C 15. B 16. D 17. E 18. B19. E 20. C 21. E 22. B 23. C 24. D25. E 26. D 27. D 28. E 29. D 30. CG E N E R A L C H E M I S T R Y S U B J E C T T E S T 2 ________________________________________________________________8 ________________________________________________________________________________________K A P L A NEXPLANATIONS1. AQualitative thermodynamics questions can be answered by applying the definitions of the various state functions. This question asks about entropy, which can be defined as the energetic equivalent of randomness, or disorder. The question thus translates into “Which of the following will have an increase in disorder as the reaction proceeds from left to right?” Recall that the general order of increasing entropy goes: solid<liquid<ions in solution<gas. Choice A is thus correct since it shows one mole of a highly ordered, crystalline solid, metallic sodium, and one mole of a somewhat well-ordered, hydrogen-bonding liquid, water, transforming into two moles of aqueous ions, Na + and OH -, and one mole of gas, H 2. The relatively well-ordered reactants are replaced by highly disordered products as the reaction proceeds, and a net increase in disorder, or entropy, is the result. As for the wrong choices, choice B shows the deposition of gaseous iodine. This represents a large decrease in entropy as the random motion of the gas phase molecules is replaced by the highly ordered crystal structure of the solid phase. Choice C shows a reaction between aqueous sulfuric acid and aqueous barium hydroxide; this reaction is simultaneously a precipitation, as the barium ions combine with the sulfate ions to form the ionic solid, barium sulfate, and a neutralization, as the aqueous acid reacts with the aqueous base to form water.Precipitation takes aqueous ions to ionic solid and thus represents a decrease in entropy. Neutralization takes aqueous ions to molecules of liquid and is accompanied by a net decrease in disorder, therefore C is accompanied by a decrease in entropy. Choice D shows the formation of liquid water or, alternatively, the combustion of gaseous hydrogen. By either description, this reaction proceeds with a large decrease in entropy as the mole and a half of mixed gas phase molecules transform into one mole of pure liquid.2. EIn general, a cation is smaller than a neutral atom of the same element since the removal of an electron or electrons to form the cation leaves fewer electrons to shield one another from the attractive force of the nucleus and to repel one another. The remaining electrons on the cation are drawn closer to the nucleus, resulting in a reduction in size. By this same reasoning, it follows that the greater the positive charge, the smaller the ion. Conversely, anions are generally larger than the corresponding neutral atoms due to an increased number of electrons. A size trend is also observed in ions of equal charge: as one moves down a group, or column of the Periodic Table, the atomic and ionic radii increase due to the larger size of the highest occupied orbital. Therefore sodium atoms are smaller than potassium atoms, and Na + ions are smaller than K + ions. If we apply these facts and place the ions into a partial order by size, we have:K + > Na + > Mg 2+ > Al 3+Thus choices A, C, and D can be eliminated, and we must now compare the remaining species, K + and Cl -. When potassium, atomic number 19, gives up an electron to form the cation K +, it is left with 18 electrons and is thus isoelectronic to argon, atomic number 18. Chlorine, atomic number 17, takes on an electron to form Cl -, and thus it likewise becomes isoelectronic to argon. Each species is isoelectronic to the other, and the relative radii will be affected most by the attractive force exerted on the outer electrons by the positive charge of the nucleus. It follows that the more protons the nucleus contains, i.e., the larger the atomic number is, the smaller the ionic radius will be. The final comparison is thus Cl - > K +, verifying choice E as the credited choice.3.EThis question combines some basic stoichiometry with thermodynamic values for a reaction. We are given the ∆H comb per mole of ammonia and the mass of ammonia in grams. Applying dimensional analysis, we must convert the mass of ammonia into moles, then multiply by kcal/mol to find the total number of kcal. Ammonia has a formula weight of 14 + 3(1) = 17 g/mol, therefore: (34 g)/(17 g/mol) = 2 mol (2 mol).(81 kcal/mol) = 162 kcal.Note that it is not necessary to come up with a balanced equation for the combustion reaction to do this problem._________________________________________________________G E N E R A L C H E M I S T R Y S U B J E C T T E S T 2K A P L A N ________________________________________________________________________________________94. CThe percent by mass of an element in a compound is calculated by multiplying the atomic mass of that element by the number of atoms of that element in one formula of the compound, then dividing by the total formula weight of the compound and, finally, converting the result to a percentage. In this example there are two atoms of aluminum (atomic weight 27) and three sulfate ions (formula weight = 32 + 4(16) = 96) per formula of aluminum sulfate, so the calculation goes:2(27)/(2(27) + 3(96)) = 54/342.which rounds off and converts to 15.8%. Calculation can, as usual, be avoided by noticing that 54/342 will beapproximately equal to 50/350, or roughly 15%. Choice A is the result of inserting one aluminum atomic weight into the top of the calculation, while choices B, D, and E are the results of other random errors in calculation.5.EThis question simultaneously tests the basic definition of combustion and thermodynamic calculations involving heats of formation, along with a bit of basic organic nomenclature. To calculate thermodynamic values for a reaction or process from given heats of formation, we must set up a correctly balanced equation for the process and then subtract the sum of the heats of formation, ∆H f , of the reactants from the sum of the heats of formation of the products. To do so for this problem, we must first identify ethylene as the C 2H 4(g ) entry in the table and then write a balanced equation for its combustion:C 2H 4(g ) + 3 O 2(g ) ∅ 2 CO 2(g ) + 2 H 2O (l )Adding the values for the ∆H f (products) = 2 ∆H f (CO 2(g )) + 2 ∆H f (H 2O (l )) and subtracting the sum of the ∆H f (reactants) = ∆H f (C 2H 4(g )) + 3 ∆H f (O 2(g )), the result is :∆H comb (C 2H 4(g )) = ∆H f (products) - ∆H f (reactants) =[2 ∆H f (CO 2(g )) + 2 ∆H f (H 2O (l ))]- [∆H f (C 2H 4(g )) + 3 ∆H f (O 2(g )) = 2(-94) + 2(-68.4) -(12.5) -3(0) = -337.3 kcal/mol(Recall that the heat of formation of any standard state element, like O 2(g ) here, is zero.) As for the wrong choices, notice first that choices A and B could have been immediately eliminated sincecombustion, by nature, is exothermic; a correct answer to this question must therefore be negative. Choice D results from using the heat of formation of gaseous water rather than that of the liquid water that should be expected at 25o C, and choice C is the result obtained by adding 12.5 (when it should have been subtracted) in addition to using the value for gaseous rather than liquid water.6.BWe can calculate the equilibrium constant for this reaction by setting up the reaction quotient in terms of theconcentrations of the reactants and products, then solving based on the numbers given in the problem. Since we are told that the vessel is at equilibrium the reaction quotient is the equilibrium constant. For the given reaction, the reaction quotient is:Q c = K eq = [NO]2[N 2][O 2]Since the reaction takes place in a 2.0 liter container, we can convert the given numbers of moles into moles per literby dividing each by two, then plug the results into the expression for Q c : [NO] = 8 mol/2 L = 4 mol/L [N 2] = 2 mol/2 L = 1 mol/L [O 2] = 4 mol/2 L = 2 mol/l∴ K eq = [NO]2[N2][O2] =42(1)(2) = 87. AThis question can be answered based on fundamental definitions; the rate-determining step is simply the slowest step in a proposed mechanism. Of the three steps shown, only step I is labeled as 'slow' and must therefore be thebottleneck in the overall process. As such, it must be the rate-determining step.8. CTellurium, Te, is a Group VIA element, found below oxygen and sulfur in the Periodic Table. It should thus have six valence electrons available for bonding. If the central Te atom uses two of its valence electrons to form sigma bonds to the two hydrogens in H2Te, it will still have two pairs of nonbonding electrons remaining. According to VSEPR theory H2Te fits the general formula AB2U2 and will have tetrahedral electronic geometry and angular molecular geometry, with bond angles of 109.5_. VSEPR theory makes the same prediction for H2O and H2S, and similar predictions for other tetrahedral electronic species such as methane and ammonia.(Note: In reality, the bond angles generally vary from those predicted by the theory because of differences in repulsion between bonded and nonbonded electron pairs and because of orientational differences in the orbitals involved in the bonding.)9. AThis question about gas volumes and pressures can best be answered by applying Dalton's law of partial pressures, which states that the pressure exerted by any one gas in a mixture of ideal gases is the same as it would be if the gas were to occupy the same volume alone. If we focus on the nitrogen we can approach this problem as if the volume increases from the initial 300 mL to the combined 500 mL and apply Boyle's law: P1V1 = P2V2 (at constant temperature ). Since an increase in volume will lead to a decrease in pressure, the final pressure must be less than 100 torr. Thus we canautomatically eliminate choices C, D, and E before completing the calculation as follows:P2 = P1V1/V2 = (100 torr)(300 mL)/(500 mL) =3/5(100 torr) = 60 torr.10. CThis question is probably best approached by examining the answer choices in search of the one true statement required. Choice A is false; it appears from the reaction coordinate diagram provided that W and Y are at approximately the same level while U is lower in energy than is X. It will thus take more energy to go from the lower energy state U to the W/Y level than it will to go from the higher X energy level to that of W/Y. In other words, the activation energy forthe reaction U ∅ X is larger than the activation energy for the reaction X ∅ Z. Choice B is essentially equivalent to choice A, since a reaction "occurs more readily" if and only if its activation energy is smaller. Choice C is true. In theforward direction, X ∅ Z has a smaller energy of activation (and is therefore a faster reaction) than does U ∅ X.Meanwhile, the free energy of activation for Z ∅ X, that is, the energy required to go from Z to Y, is larger than the energy required to move from X to W. Therefore, in the reverse direction, the reaction Z ∅ X will be slower than the reaction X ∅ U. Choices D and E are both wrong for two reasons. First, the diagram shown is a free energy diagram (∆G on the vertical axis) rather than an enthalpy diagram (∆H on the vertical axis); as such we cannot directly determinechanges in enthalpy to which the terms exothermic and endothermic apply. What we can determine from a free energy diagram is whether a given reaction is spontaneous or not. Secondly, even if we did attempt to infer that enthalpy and free energy were following parallel trends, the reaction U ∅ X is an uphill move on the diagram and would thus beendothermic (making choice D wrong) while the reaction U ∅ Z results in a net decrease in energy and would thus be exothermic (disqualifying choice E).11. DElectronegativity increases toward the upper right-hand corner of the Periodic Table, with fluorine, F, the most electronegative element and cesium, Cs, the least. Electronegativity is a property of atoms in bonds, i.e., it is actually a molecular property, therefore rare gases like helium, He, which do not form compounds, are not included in theelectronegativity scale. (Heavier elements like francium, Fr, are not included because their half-lives are so short that their compounds can not be isolated.) The question can therefore be translated as "Which of the following is closest to the lower left-hand corner of the Periodic Table?" Choice D, potassium, lies far to the left of the other entries and as such is the least electronegative.12. EIonization energy is defined as the amount of energy required to remove an electron from a gas phase atom or ion, and increases toward the upper right-hand corner of the Periodic Table, with helium, He, having the largest such energy requirement of any neutral atom. This question can therefore be reworded as "Which of the following configurations corresponds to the element which is closest to the upper right-hand corner of the Periodic Table?" We can count the total number of electrons and, given that these are atoms rather than ions, thereby translate the answer choices into theircorresponding elemental symbols:A. total number of electrons = atomic number = 11 _ NaB. total number of electrons = atomic number = 47 _ AgC. total number of electrons = atomic number = 53 _ ID. total number of electrons = atomic number = 17 _ ClE. total number of electrons = atomic number = 18 _ ArWe can then see that choice E, argon, lies furthest to the upper right and therefore has the greatest ionization energy.It is not necessary to convert the electron configurations into the atoms which they represent if one notices that choice E is the only one with all its subshells filled and would therefore be the hardest to remove an electron from.13. ATwo species with the same number of electrons are said to be isoelectronic; this question can thus be reworded to read "Which of the following is an isoelectronic pair?" Calcium has atomic number 20. This means that a neutral atom of calcium has twenty protons in its nucleus and twenty electrons around its nucleus. Positive charges result from theremoval of outer electrons, thus to acquire a charge of positive one, calcium would need to lose one electron from its outermost, or 4s, orbital, leaving Ca+ with a total of 19 electrons. Potassium, meanwhile, has atomic number 19 and thus also has 19 electrons as a neutral atom. These two species are then seen to be isoelectronic, verifying choice A as the correct choice. The species in each of the other choices do not have the same number of electrons; H+ has no electrons while He has two, Cl has 17 while F has 9, and O- has 9 electrons while S+ has 15. These choices, along with choice E, can thus be eliminated.14. CIonization energy is defined as the amount of energy required to remove an electron from a given species. Ionization energy is usually expressed in energy per particle (as it is here) or in energy per mole, and energy, in turn, is usuallyexpressed in electron volts (eV), joules, or kilojoules. The first ionization energy of an element is the energy required to remove an electron from a neutral atom of that element, while the second ionization energy is the energy required toremove a second electron, i.e., to remove an electron from the +1 cation. A particularly useful piece of logic/knowledgehere is that it will be more difficult to remove an electron from a positively-charged species than it will be to remove an electron from a neutral version of the same species. In other words, the second ionization energy of an element is always greater than its first ionization energy. Based on this fact alone, choices A, D, and E can be eliminated. To choosebetween the remaining two choices, we must reason out that a small value for ionization energy corresponds to relative ease of removal of the electron. A Group IA metal atom will lose its first electron with relative ease, but after that it will possess an electronic configuration similar to that of a noble gas. It will therefore be difficult to remove another electron, implying a high second ionization energy. Looking at the numbers in the table, we see that both remaining choices have relatively small values for first ionization energy, but that X has a much higher second ionization energy. It follows that element V is most likely a Group IIA element (e.g., Mg, Ca) while element X is most likely a member of Group IA (e.g., Na, K). Choice C, as indicated above, is thus the best choice.15. BAccording to the balanced equation, the complete reaction of two moles of HCl should result in the formation of one mole of CO2. We are given that 36 grams, or one mole, of HCl reacts; there should thus be half a mole of CO2 formed.Assuming that the gas formed behaves ideally, and recalling that one mole of any ideal gas occupies 22.4 liters at STP, this half a mole of CO2 should occupy a volume of 11.2 liters as stated in choice B.16. DReading from left to right across the Periodic Table, metallic character decreases as electronegativity increases. The most reactive combinations of elements are those with the greatest differences in electronegativity, and this difference can be most easily predicted by observing the distance between the elements on the Periodic Table. This question can thus be translated to read: "Which of the following lies furthest from F on the Periodic Table?" Lithium, in Group IA, is therefore the correct answer. Titanium, choice B, would be the second most reactive of the options offered, while choices C and E, chlorine and iodine, are Group VIIA elements like fluorine, and thus would be only slightly reactive with it. Choice A, argon, is an inert gas and would not be reactive at all.17. EThe rate law for any reaction, A + B ∅ C, is given by:rate = k[A]x[B]ywhere k is the rate constant at the temperature at which the reaction is carried out, [A] and [B] are the concentrations of the two reactants, and x and y are the reaction orders with respect to A and B respectively. This question asks us to find x from the experimental rate data provided. To simplify the tabulated rate data, we can rewrite the table as small whole number multiples of the smallest entry in each column:line # [A]o [B]o rate1 1 1121 2 42 4 163Looking at lines 1 and 2, where the concentration of reactant A is held constant, we see that a doubling in the concentration of B results in a quadrupling of the rate, therefore the reaction is second order with respect to B. We can now look at lines 2 and 3 (or 1 and 3) to determine the required reaction order. From line 2 to line 3, the concentration ofB is doubled; the reaction rate should thus quadruple due to the influence of the concentration increase of second orderreactant B. We can divide out this quadrupling effect and create a hypothetical line 4:。