湖南省益阳市箴言中学2015届高三化学上学期第二次模拟考试新人教版

益阳市箴言中学2015届高三第二次模拟考试英语本试题卷分四个部分，包括听力理解、语言知识运用、阅读和书面表达。

时量120分钟。

满分150分。

PartⅠListening Comprehension(30 marks)Section A （22.5 marks）Directions: In this section, you will hear six conversations between two speakers. For each conversation, there are several questions and each question is followed by three choices marked A, B and C. Listen carefully and then choose the best answer for each question. You will hear each conversation TWICE.Conversation 11.Where is the park?A. In the east end of the town.B. In the west end of the town.C. At the end of the road.2. How does the man suggest going there ?A. By car.B. By bike.C. On foot.Conversation 23. What do we learn about the watch?A. It‟s a gold watch.B. It‟s a diamond watch.C. I t‟s a second-hand watch.4. How much is it?A.＄150.B.＄115.C.＄1,500.Conversation 35. Where did David go probably?A. To a party.B. To his workplaceC. To a motorcycle repair shop.6. What did the man find by the phone?A. Da vid‟s phone book.B. David‟s text bookC. David‟s address book.Conversation 47.What season is it most likely now?A. Autumn.B. Summer.C. Spring.8. Why did the woman come here?A . To attend a wedding. B. To teach skating. C. To visit John.9.Where does the man work now?A. At the radio station.B. At Bank of America.C. In Chicago.Conversation 510.How much time did the man have to impress the woman?A . Eight seconds. B. Fifty seconds. C. Three minutes.11.Why was the interview so short?A. The woman can know a person just by a glance .B. The woman had to meet many people in a day.C. The woman has an important meeting later.12.What showed that the man was careful?A. Cleaning the feet at the door.B. Taking off the cap when coming.C. Clean hair and fingernails.Conversation 613. What can we learn from the conversation?A. The woman likes the story of Morning Star.B. The man has seen Laura Harper‟s other movies before Roses.C. The woman thinks Hugh Wayne will win the Best Actor.14. Which of the following is true?A. Laura Harper has won the Beat Actress more than once.B. Took Off has already won Tony Baker many prizes.C. Julia Adam is noticed for her acting in Roses.15. What does the woman mean by saying …We‟re on the same page about this‟?A. The two speakers both read the book Roses before it was made into a movie.B. The two speakers are both reading news about Roses on the Internet.C. The two speakers both agree who will win the Best Actress.Section BDirections: In this section, you will hear a mini-talk. Listen carefully and then fill in the numbered blanks with the information you’ve got. Fill in each blank with NO MORE THAN 3 WORDS.You will hear the mini-talk TWICEPAER TWO: LANGUAGE KNOWLEDGE （45 points）Section A（15 points）Directions: Beneath each of the following sentences there are 4 choices marked A, B, C and D. Choose one answer that best completes the sentence.21.Half of the world‟s population is under the age of 25 and when they are n ot involved in thedecision-making process, they are not aware of ______ is taking place.A. whatB. whomC. whichD. whose22.Although we haven‟t got the accurate statistics, it is estimated that _____ 300 factories in thisprovince closed down during the economic crisis.A. partlyB. brieflyC. approximatelyD. relatively23.The use of several senses gives the brain more connections and associations, making it easier______ information later, which assists memory and learning.A. findB. findingC. foundD. to find24.I have learned a lot about Asian customs, ______ in the small village for three years in the early1990s.A. livedB. to liveC. having livedD. to have lived25.Dad decided to build a small tool room with a lock, ______ he would keep his best tools so my brother couldn‟t reach them.A. whyB. whereC. whoD. which26.My parents always ask me what I am doing at school and ________ I am prepared for the collegeentrance examination.A. thatB. whatC. whichD. whether27.No sooner ________ on the train than it began to pull out of the station.A. had we gotB. we had gotC. got weD. we got28.He seems to be quite relaxed, his hands ________ behind his back.A. crossingB. crossedC. to crossD. cross29.Looking back through their files, I found that they ________ this faithfully for more than three years.A. has been doingB. was doingC. had been doneD. had been doing30.According to the report, the US invested its money abroad at high rates of return, ________ the rest of the world invested in the US at low rates of return.A. whileB. whenC. sinceD. as31.By the time the traffic ______, we will have run out of gas.A. clearsB. clearedC. will clearD. has been clearing32._____ to drive after drinking, some drivers are still trying their luck, which is dangerous.A. Not remindedB. Reminded notC. Reminding notD. Not having reminded33.Experts suggest that people should _____ some hobbies, which may help reduce pressure anddepression.A. take upB. put upC. bring upD. set up34.—I hear Prof. Huang will give us a lecture this weekend.—Well, I would rather he _____ tomorrow.A. cameB. will comeC. had comeD. would come35.As for the plan some agree while others don‟t. I‟m one of _____ opposed to it.A.whomB. thoseC. those whoD. who isSection B（18 points）Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in blanks with a word or phrase that best fits the context.With his leg lame and his teeth uneven, a boy thought of himself as the most unfortunate child in the world. He seldom played with his classmates, and when asked to answer questions, he always 36 his head without a word.One spring, his father brought home some saplings (树苗). 37 of his children would planta sapling and he promised, “Whoever grows his sapling best shall get a gift.”The boy certainly wanted to get the gift.But seeing his brothers and sisters watering the trees, he hit upon a(n) 38 : he hoped the tree he planted would die soon. So watering it once or twice, he never 39 it.Several days later, he was surprised to find it didn't die, but grew so many fresh 40 . Compared with those of his brothers and sisters, his appeared greener. His father kept his 41 ，bought the boy a gift and said he would become an outstanding 42 after growing up.From then on, the boy slowly became optimistic and confident. One evening, he suddenly43 his biology teacher once said that plants generally grow at night. Why not go to see the tree?When he came to the courtyard, he found his father working near the tree! Instantly he44 ：Father had been secretly watering his tree! He returned to his room, tears 45 in his eyes.Decades passed. The boy didn't become a botanist. 46 , he became the U．S. President. His name was Franklin Roosevelt.47 is the best nourishment (滋养品) of life. Even though it is just a bucket of water, it can make the tree of life grow well!36．A. held B．raised C．lowered D．covered37．A. Both B．None C..One D．Each38．A. word B．promise C．idea D．way39．A. appealed to B．attended to C．adapted to D．turned to40．A. roots B．leaves C．branches D．seeds41．A. word B．balance C．agreement D．opinion42．A. teacher B．gardener C．president D．botanist43．A. believed B．recalled C．repeated D．knew44．A. remembered B．understood C．wondered D．admitted45．A. welling B．falling C．dropping D．crying46．A. Therefore B．Besides C．Moreover D．Instead47．A. Love B．Water C．Disability D．FatherSection C (12 points)Directions: Complete the following passage by filling in each blank with one word that best fits the context.Dear boys and girls,I feel honored to speak here on behalf of all your teachers .I‟m really sorry that it will be only several months 48 you graduate from high school .We have spent an unforgettable time in the past three years, during 49 I have been greatly impressed by your good qualities such as honesty, diligence, co-operation and so on.50 in my teaching career have I met such wonderful students 51 you.We have an endless journey in pursuit of knowledge. Graduation 52 marks a stage of one‟s education .Set a high goal and you‟ll achieve more in your future life.I do believe that all your efforts will eventually pay 53 .I as well as your other teachers 54 praying that every one of you will be admitted to 55 ideal college.Yours,English teacherPART THREE: READING COMPREHENSION（30 points）Directions: Read the following pass three ages. Each passage is followed by several questions or unfinished statements. For each of them there are 4 choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage.A.Schouten University is a highly respected online education provider offering officially-recognized degree programs across the globe. Our mission is to provide students with a high-quality mixed learning MBA program and personal coaching to help them to become world-class managers.MBA with Unique Skills & Personal Development ProgramStudents will benefit from 30 years of soft skills training experience from Schouten & Nielsen, the leading soft skills training institute in Europe. To become an effective manager, theory alone is not enough, and being able to put theory into practice is more important today than ever before. With the Schouten University MBA program you will have the unique opportunity to develop such soft skills as influencing skills, leadership skills, communication skills and motivational skills.International Network + 2 Residential Study weeks in Cambridge & ShanghaiAt Schouten University, students, tutors and coaches meet each other online. But there is also the opportunity for students to study face to face by way of the two Residential(居住的)Study weeks. One week will be held in Shanghai (China) and the other week at Cambridge (UK). Shanghai is the commercial and industrial center of the world’s leading economy. Cambridge is home to some of Europe’s oldest and best universities. Residential weeks will be held several times a year. Therefore, you can choose when to participate, depending on the progress of your study. In theresidential weeks cooperation between students will be an important element. Students will work together on real company cases. During the residential week students will be supervised by a professional trainer.*Residentials are an obligatory (必须的) part of the program and the costs of the residential weeks are covered by your tuition.*The fee includes the study weeks’ housing, food and participation but not the costs of travel.Maximum FlexibilitySchouten University introduces a new way of learning: Study 2. 0. Students will study online through my and interact via rich social media. It allows busy working professionals like you to study a world class MBA degree anytime, anywhere at any pace.Great SupportAccess to the University’s virtual learning environment (VLE) 24/7, and your study coach, tutors are all here to give you strong personal support to ensure your success.56. Most graduates of Schouten University will work in the field of ________.A. educationB. militaryC. medicineD. business57. According to the advertisement, which of the following DOESN’T belong to soft skills?A. Leadership skills.B. Communication skills.C. Driving skills.D. Motivational skills.58. The time to participate the residential weeks is ________.A. flexibleB. fixedC. unnecessaryD. inconvenient59. If you attend the two Residential Study weeks, you will have to pay for ________ besides your tuition.A. housingB. participation feesC. mealsD. travel costs60. The advertisement is probably intended for ________.A. full-time college studentsB. working businessmenC. studying teenagersD. retired managersBOne of our biggest fears nowadays is that our kids might some day get lost in a “sea of technology” rather than experiencing the natural world. TV and computer games are leading to a serious disconnect between kids and the great outdoors, which will change the wild places of the world, its creatures and human health for the worse, unless adults get working on child's play.Each of us has a place in nature we go sometimes, even if it was torn down. We cannot be the last generation to have that place. At this rate, kids who miss the sense of wondering outdoors will not grow up to be protectors of natural landscapes. “If the decline in parks use continues across North America, who will defend parks against encroachment (蚕食)?” asks Richard Louv, author of Lost Child in the Woods.Without having a nature experience, kids, can turn out just fine, but they are missing out a huge enrichment of their lives. That applies to everything from their physical health and mental health, to stress levels, creativity and cognitive (认知的) skills. Experts predict modern kids will have poorer health than their parents—and they say a lack of outside play is surely part of it. Research suggests that kids do better academically in schools with a nature component and that playing in nature fosters (培养) leadership by the smartest, not by the toughest. Even a tiny outdoor experience can create wonder in a child. The three-year-old turning over his first rock realizes he is not alone in the world.A clump of trees on the roadside can be the whole universe in his eyes. We really need to value that more.Kids are not to blame. They are over-protected and frightened. It is dangerous out there from time to time, but repetitive stress from computers is replacing breaking an arm as a childhood rite of passage (仪式).Everyone, from developers to schools and outdoorsy citizens, should help regain for our kidssome of the freedom and joy of exploring, taking friendship in fields and woods that cement (增强) love, respect and need for landscape. As parents, we should devote some of our energies to taking our kids into nature. This could yet be our greatest cause.61. The main idea of Paragraph 2 is that ________.A. kids missing the sense of wondering outdoorsB. parks are in danger of being gradually encroachedC. Richard Louv is the author of Lost Child in the WoodsD. children are expected to develop into protectors of nature62. According to the passage, children without experiencing nature will ________.A. keep a high sense of wonderB. be over-protected by their parentsC. be less healthy both physically and mentallyD. change wild places and creatures for the better63. According to the author, children's breaking an arm is ________.A. the fault on the part of their parentsB. the natural experience in their growing upC. the result of their own carelessness in playD. the effect of their repetitive stress from computers64. What does “sea of technology” mean in the first paragraph?A. The technology of TV and computer games.B. The technology of food.C. The technology of sea food.D. The technology of catching animals in the sea.65. In writing this passage, the author mainly intends to ________.A. blame children for getting lost in computer gamesB. encourage children to protect parks from encroachmentC. show his concern about children's lack of experience in natureD. inspire children to keep the sense of wonder about things aroundCIn a recent study, researchers have estimated the energy required to produce bottled water, including the energy required to manufacture plastic, make the plastic into bottles, process the water, fill and seal the bottles, transport the bottles, and chill the bottles for use. Combining all the energy input totals for treatment and distribution, researchers found that producing bottled water requires between 5.6 and 10.2 million joules(焦耳) of energy per liter, depending on transportation factors. That's up to 2, 000 times the energy required to produce tap water.To break down the energy requirements, researchers found that producing the plastic bottles and transporting the bottles greatly dominated(控制) the energy input. Although some companies are experimenting with producing lightweight bottles or using recycled materials, the researchers calculated that the manufacturing cost is about 4 million joules of energy per plastic bottle weighing 38 grams, and the cap weighing 2 grams.“Our previous work had suggested that bottled water production was an energy­ intensive process, but we were surprised to see the results,” the researchers said.Transportation costs vary depending on the distance and mode of transport, and both factors depend on the type of bottled water. There are two main kinds of bottled water in the US. “spring water”, which comes from an underground spring, and “purified water”, which is city tap water that has received further treatment. For purified water distributed locally by truck within Los Angels, the total transportation energy is about 1.4 million joules per liter. In the second situation, spring water shipped from Fiji (such as Fiji Spring Water) across the Pacific to Los Angels, and then delivered locally by truck, requires about 4 million joules per liter for transportation. For the spring water, the transportation energy equaled (in the case of Fiji) the energy required to produce the bottle.With this data, the researchers hope that future studies will have the ability to make specific estimates for different situations, and possibly find ways to cut energy costs.66. From the first paragraph we can know that the energy required to ______.A. produce bottled water includes the energy for producing the bottlesB. distribute bottled water includes the energy required to drink itC. distribute bottled water is much smaller than the energy for treating itD. produce bottled water is much smaller than the energy for producing tap water67. To reduce the energy required to produce bottled water, some companies are trying ______.A. to fill tap water into the plastic bottles directlyB. to replace the plastic bottles with metal containersC. to produce lightweight bottles with recycled materialsD. to use bicycles to transport and deliver bottled water locally68. What's Paragraph 4 mainly talking about?A. Treatment of bottled water needs no energy input at all.B. Bottled water production is an energy- intensive process.C. Transportation of bottled water takes up the most energy input.D. Production of the plastic bottles takes up the most energy input.69. We can infer from the passage that ______.A. there are two main kinds of bottled water in the USB. world consumption of bottled water has been increasingC. transportation costs have something to do with the mode of transportD. bottled water produced in Spring is more expensive than purified water70. What's the main purpose of the passage?A. To help consumers themselves make more environmentally sustainable choices.B. To advise the government to take severe measures to stop producing bottled water.C. To urge the consumers to drink tap water instead of bottled water for the sake of the earth.D. To try the best to find ways to produce bottled water cheaply and quickly for the companies.PART FOUR: WRITING（45 points）Section A (10 points)Directions: Read the following passage. Complete the diagram/ Fill in the numbered blanks by using the in formation in the passage. Write NO MORE THAN 3 WORDS for each answer. Write your answers on your answer sheet.Would you work for nothing? Non-paid internships(实习生) have been common in the US for some time now. But these days, they are becoming more and more popular in other countries, too. Is this a good thing?For young people, there are clear benefits to doing an internship. Firstly, interns get a valuable insight into an industry they might be interested in working for. Also, having done an internship looks good on your CV, which is very important in today‟s competitive job market. A lucky few may be hired at the en d of the internship, so it‟s a good way of getting a foot in the door. For other interns, there is even extra welfare: deals on clothes if they‟re working in the fashion in industry, free trips if they‟re working for a travel agency…and so on.There are obvious benefits to companies, too. They can help select employees who are better. “If you put two young people to work as interns for twelve weeks, you‟ll soon see who is good.” said one manager. Besides, they can save money. “If I use interns, I don‟t have to pay another salary, or worry about paying pensions or giving people overtime pay.” said another.So, what do interns do? Most tasks are fairly routine. These may include photocopying, filing documents, or writing up notes from meetings. At times, interns may be given jobs with more responsibility, such as making sales calls, writing newsletters or updating websites.However, for some interns the experience can become unpleasant. “I worked for a theatre company for six weeks and spent the whole time photocopying scripts and making the tea and coffee. It is dull.” said one intern. On the other hand, non-paid internships violate the labor law in some countries, where workers must be paid the minimum wage and paid for overtime.Internship can provide benefits to both sides, but these unpaid workers are also easy to take advantage of!Title 71.____________I. Facts:▲72. __________ in the US.▲More and more popular in other countries.II. 73.___________:▲For young people●Getting a v aluable insight into 74.___________.●Looking good on their CV.● 75.__________ after the internship.▲For others●Extra welfare such as deals on clothes and 76.__________.▲ 77.__________.●Helping select 78.__________.●Saving money.III .Duties:▲79.___________ tasks.▲ More responsible jobs.IV. Problems:▲Doing 80.__________.▲Violating the labor law in some countries.Section B (10 points)Directions: Read the following passage. Answer the questions according to the information given in the passage and required word limit. Write your answers on your answer sheet.Pollution is a growing problem that is facing the international community, but even the actions that we take can make a difference. Littering is creating environmental problems all over the world and is often created by careless actions. If we guard our actions against careless littering we can ensure a better world for everyone.Littering is a severe problem in many countries around the world. For example, the African city of Kampala is suffe ring from the consequences of littering. For many years the city‟s poor didn‟t have a proper way to dispose of trash, so garbage filled the streets. Over the years much of the plastic garbage would slowly be pushed into the ground. This plastic litter does n‟t allow rain water to seep （渗透）into the ground and has caused flooding. In many of Kampala‟s poor neighborhoods whenever it rains people‟s homes become flooded with water. This water is often dirty and has allowed diseases to spread. This flooding has become so bad that the Ugandan government has banned plastic bags in the entire country to stop the littering.Trash often finds its way into the water. Today the world‟s oceans are believed to have millions of tons of trash floating around in them. This creates a problem for wildlife because they may accidentally eat the trash and die from it. Scientists have found on very remote islands baby birds that have died because their stomachs were full of trash. Fish often get tangled in plastic and will also die. Our actions in our own countries affect the most remote parts of the planet.We can all be part of the solution to stop this pollution, just by preventing careless littering. If you need to throw something away, find a trash can. If you see a friend littering tell them not to. By making this small change we can create a better world for all.81. What is the cause of littering? (No more than 3words)_____________________________________________________________________82. How does plastic litter cause flooding? (No more than 11words)_____________________________________________________________________83. If you need to throw something away, what should you do? (No more than 5words)_____________________________________________________________________84. What‟s the topic of this pas sage? (No more than 5words)_____________________________________________________________________Section C (25 points)Directions: Write an English composition according to the instructions given below in Chinese.据《中国青年报》报道，浙江省高中生从新学期开始必须选修一门新课程，如修马桶、做凳子、换灯泡等日常生活实用技术。

益阳市箴言中学2014—2015学年高二3月月考理科化学试题时量： 90分钟总分： 100 分可能用到的相对原子质量：H-1 C-12 O-16 S-32 Fe-56 Ni-59 Cu-64一、选择题（每题只有一个选项是正确的．每题3分，共54分）1、下列属于放热反应的是A、浓硫酸的稀释B、铝热反应C、氢气还原氧化铜D、氢氧化钡晶体与氯化铵晶体混合2、已知：CH4(g)+2O2(g)=CO2(g)+2H2O(g)ΔH＝- Q1；2H2(g)+O2(g)=2H2O(g) ΔH＝- Q2； H2O(g)=H2O(l) ΔH＝- Q3常温下，取体积比为4：1的甲烷和H2的混合气体112L（标准状况下），经完全燃烧后恢复到常温，则放出的热量为A、4Q1+0.5Q2B、4Q1+Q2+10Q3C、4Q1+2Q2D、4Q1+0.5Q2+9Q33、钢铁发生吸氧腐蚀时，负极上发生的电极反应是A、2H+＋2e－＝H2B、Fe2+＋2e－＝FeC、2H2O＋O2＋4e－＝4OH－D、Fe3+＋e－＝Fe2+4、某溶液中由水电离产生的c(H+)＝1×10－10mol/L，该溶液的溶质不可能是A、KOHB、H2SO4C、HClO4D、MgSO45、常温常压下，在带有相同质量活塞的容积相等的甲、乙两容器里，分别充有二氧化氮和空气，现分别进行下列两上实验：（N2O4⇌2NO2△H＞0）（a）将两容器置于沸水中加热（b）在活塞上都加2kg的砝码在以上两情况下，甲和乙容器的体积大小的比较，正确的是A、（a）甲＞乙，（b）甲＞乙B、（a）甲＞乙，（b）甲=乙C、（a）甲＜乙，（b）甲＞乙D、（a）甲＞乙，（b）甲＜乙第5题图第6题图第7题图6、根据上图，可判断出下列离子方程式中错误的是A、2Ag+（aq）+Cd（s）=2Ag（s）+Cd 2+（aq）B、Co2+（aq）+Cd（s）=Co（s）+Cd2+（aq）C、2Ag（s）+Cd2+（aq）=2Ag+（aq）+ Cd（s）D、2Ag +（aq）+Co（s）=2Ag（s）+Co2+（aq）7、如上图，图Ⅰ表示10 mL量筒中液面的位置，A与B、B与C刻度间相差1 mL，图II表示50mL滴定管中液面的位置，D与E刻度间相差1 mL。

益阳市箴言中学2015届高三第二次模拟考试化 学可能用到的相对原子质量：H--1 N--14 O--16 Na--23 Al--27 Mg--24 Cl--35.5S--32 Cu--64 Fe--56 Ba--137 一、选择题（每小题仅一个正确答案，18×3分） 1.下列应用不涉及氧化还原反应的是A Na 2O 2用作呼吸面具的供氧剂B 工业上电解熔融状态Al 2O 3制备AlC 工业上利用合成氨实现人工固氮D 实验室用NH 4Cl 和Ca(OH)2制备NH 32. W 克下列物质电火花引燃后的产物通过足量的Na 2O 2后，固体增重大于W 克的是 A.CO 与H 2的混合气体 B.CH 3CH 2OH C.HCOOH D.CH 3COOH 3．足量下列物质与等质量的铝反应，放出氢气且消耗溶质物质的量最少的是 A ．氢氧化钠溶液 B ．稀硫酸 C ．盐酸 D ．稀硝酸4．已知X 、Y 中含有相同的元素，Z 、W 中也含有相同的元素，根据反应X+H 2OY+ H 2 ； Z + H 2O W + O 2 （方程式均未配平），可推断X 、Y 中及Z 、W 中相同元素的化合 的高低顺序为A 、X ＞Y 、Z ＞WB 、X ＜Y 、Z ＜WC 、X ＞Y 、Z ＜WD 、X ＜Y 、Z ＞W5．向含大量Na ＋、Cl －的溶液中通入足量的NH 3后，该溶液中还可能大量存在的离子组是A ．K ＋、Br －、CO 32－B ．Al 3＋、H ＋、MnO 4－C ．NH 4＋、Fe 3＋、SO 42－D ．Ag ＋、Cu 2＋、NO 3－6．科学家指出，食用虾类等水生甲壳类动物的同时服用维生素C 容易中毒，这是因为对 人体无害的＋5价砷类物质在维生素C 作用下，能够转化为有毒的＋3价砷类化合物。

下列说法不．正确的是 A ．维生素C 具有还原性 B ．上述过程中砷元素发生还原反应C ．上述过程中＋5价砷类物质作氧化剂D ．＋5价砷转化为＋3价砷时，失去电子7.设NA为阿伏伽德罗常数的值。

益阳市箴言中学2015届高三第二次模拟考试生物总分100分时量90分钟一、选择题（每题2分，共60分，每题只有一个正确答案，请将答案序号填在答卷相应位置）1.生物大分子通常都有一定的分子结构规律，即是由一定的基本结构单位，按一定的排列顺序和连接方式形成的多聚体，下列表述正确的是A．若该图为一段肽链的结构模式图，则1表示肽键，2表示中心碳原子，3的种类有20种B．若该图为一段RNA的结构模式图，则1表示核糖，2表示磷酸基团，3的种类有4种C．若该图为一段单链DNA的结构模式图，则1表示磷酸基团，2表示脱氧核糖，3的种类有4种D．若该图表示多糖的结构模式图，淀粉、纤维素和糖原是相同的2.有关下图中某蛋白质的叙述，正确的是A．形成该蛋白质时共脱掉126个水分子 B. 该蛋白质含有两条肽链C．该蛋白质的R基中共含16个氨基 D．该蛋白质共有111个肽键3.下列有关生命活动的描述，正确的是①酶不一定都能与双缩脲试剂发生反应而呈现紫色②温度超过80℃，所有酶的活性都因空间结构改变而失活③磷脂，核酸和ATP都含有C、H、O、N、P五种元素④突触前膜释放肾上腺素的过程体现膜的流动性这一功能特性⑤细胞中携带遗传信息的物质不一定是遗传物质⑥噬菌体的遗传物质彻底水解后能得到4种脱氧核苷酸⑦成熟mRNA的碱基数与多肽链中的氨基酸数之比为3∶1⑧麦芽糖和蔗糖水解的产物中都有葡萄糖⑨着丝点数不一定等于染色体数，有核糖体的不一定是真核细胞⑩丙酮酸彻底氧化分解产生CO2的过程中没有O2参与反应A. ①③⑤⑧⑩B. ①②③⑤⑩C. ①④⑤⑥⑩D. ②④⑥⑦⑨⑩4.2009年2月9日，加拿大麦吉尔大学卫生中心研究所杰纳斯•雷克研究小组认为，肿瘤细胞能释放一种叫微泡的“气泡”让肿瘤与血管内皮细胞进行交流，并改变这些内皮细胞的行为。

这些微泡在离开肿瘤组织时携带一种特殊的癌症蛋白。

当微泡与内皮细胞融合，它们所携带的这些癌症蛋白就会触发促进新血管异常形成的机制。

益阳市箴言中学2015届高三第二次模拟考试数学（文科）试题满分：150分 时量：120分钟 第Ⅰ卷（选择题，共50分）一、选择题（本大题包括10小题，每小题5分，共50分，每小题给出的四个选项中，只．有一项．．．是符合题目要求的，请将正确选项填涂在答题卡上） 1．已知集合M ＝｛x|x ＜3}，N ＝｛x|0862<+-x x ｝，则M ∩N ＝（ ）. A ．∅ B ．｛x|0＜x ＜3} C ．｛x|1＜x ＜3} D ．｛x|2＜x ＜3} 2．复数31ii--等于（ ）. A ．i 21+ B.12i - C.2i + D.2i - 3．在△ABC中，60,A a b =︒==，则B 等于（ ）. A ．45°或135°B ．135°C ．45°D ．以上答案都不对4．条件甲“a ＞1”是条件乙“a ＞a ”成立的（ ）. A ．既不充分也不必要条件 B ．充要条件 C ．充分不必要条件 D ．必要不充分条件5．已知向量b a ,的夹角为︒60，且,1||,2||==b a则向量a 与b a 2+的夹角为（ ）.A. ︒150B. ︒120C. ︒60D. ︒30 6．定义在R 上的函数f(x)满足f(x)= ⎩⎨⎧>---≤-0),2()1(0),4(log 2x x f x f x x ，则f （3）的值为( ).A.-1B. -2C.1D. 27、已知函数()()()f x x a x b =--（其中a b >）的图象如下面右图所示，则函()x g x a b =+ 的图象是( ).A B C D 8．已知函数，x sin ）3x 2（cos f(x)2++=π则 ）x （f 是 （ ）. A ．周期为2π的奇函数B ．周期为2π的偶函数C ．周期为π的奇函数D ．周期为π的非奇非偶函数f (x )9．已知f x x x f x f x a f b f ()ln ()()'()()'()=>==0712，的导数是，若，，c f ='()13，则a 、b 、c的大小关系是（ ）.A. c<b<aB. a<b<cC. b<c<aD. b<a<c10．设定义在R 上的函数()f x 满足以下两个条件：（1）对,()()0x R f x f x ∀∈+-=都有立；（2）当20,(2)'()0x x x f x <+≥时. 则下列不等式关系中正确的是（ ）. A ．(1)(0)f f -≤B ．(1)(2)f f ≥C ．(2)(3)f f -≤-D ．(2)(0)f f ≥第Ⅱ卷 非选择题 （共100分）二、填空题 （本大题共5个小题，每小题5分，共25分）11.若a ＞0，b ＞0，且函数224)(23+--=bx ax x x f 在x ＝1处有极值，则ab 的最大值等 于 .12、点O 在ABC ∆内部且满足022 =++OC OB OA ，则ABC ∆的面积与凹四边形ABOC的面积之比为 . 13.设函数f(x)=x-1x,对任意x [1,∈+∞），f(mx)+mf(x)<0恒成立，则实数m 的取值范 围是________. 14、已知,2)4tan(=+πx 则x 2tan 的值为__________.15、对于函数()2cos ([0,])f x x x π=-∈与函数21()ln 2g x x x =+有下列命题： ①函数()f x 的图像关于2x π=对称；②函数()g x 有且只有一个零点；③函数()f x 和函数()g x 图像上存在平行的切线；④若函数()f x 在点P 处的切线平行于函数()g x 在点Q 处的切线，则直线PQ 的斜率为1.2π-其中正确的命题是 。

湖南省益阳市箴言中学2015届高三上学期第二次月考化学试卷一、选择题（共18小题，每小题3分，满分54分）1．下列应用不涉及氧化还原反应的是（）A．N a2O2用作呼吸面具的供氧剂B．工业上电解熔融状态的Al2O3制备AlC．工业上利用合成氨实现人工固氮D．实验室用NH4Cl和Ca（OH）2制备NH3考点：氧化还原反应．.专题：氧化还原反应专题．分析：氧化还原反应的本质特征是反应前后元素化合价的发生变化；依据元素化合价变化分析判断；解答：解：A、Na2O2用作呼吸面具的供氧剂，过氧化钠和二氧化碳反应生成碳酸钠和氧气，发生了氧化还原反应，故A不符合；B、工业上电解熔融状态的Al2O3制备Al，是电解氧化铝发生氧化还原反应，故B不符合；C、工业上利用合成氨实现人工固氮，是单质气体单质和氢气化合生成氨气，发生了氧化还原反应，故C不符合；D、NH4Cl和Ca（OH）2制备NH3是复分解反应，故D符合；故选D．点评：本题考查了氧化还原反应的概念应用，本质特征化合价变化的判断，较简单．2．（3分）W克下列物质电火花引燃后的产物通过足量的Na2O2后，固体增重大于W克的是（）A．C O与H2的混合气体B．C H3CH2OHC．H COOH D．C H3COOH考点：化学方程式的有关计算．.专题：计算题．分析：H2、CO在整个过程中发生的化学方程式为：2H2+O22H2O、2Na2O2+2H2O=4NaOH+O2↑；2CO+O22CO2、2Na2O2+2CO2=2Na2 CO3+O2，由反应方程式可知，过氧化钠增加的质量即为H2、CO的质量．因此只要是CO或H2或它们的混合气体或化学组成符合（CO）m•（H2）n，固体增重等于W克，不符合此CO）m•（H2）n形式的，多O的增重的质量小于ag，多C的增重的质量大于ag，满足题意．解答：解：A．H2、CO在整个过程中发生的化学方程式为：2H2+O22H2O、2Na2O2+2H2O=4NaOH+O2↑；2CO+O22CO2、2Na2O2+2CO2=2Na2CO3+O2，由反应方程式可知，过氧化钠增加的质量即为H2、CO的质量，所以固体增重等于W克，故A 错误；B．乙醇CH3CH2OH可表示为（CO）1•（H2）3•C，固体增重大于W克，故B正确；C．甲酸HCOOH，可以写成CO1（H2）•O，固体质量增加小于W克，故C错误；D．CH3COOH可表示为（CO）2•（H2）2，固体增重等于W克，故D错误；故选B．点评：本题考查化学方程式的有关计算，注意根据方程式利用差量法进行解答，侧重考查学生分析解决问题的能力，题目难度中等．3．）足量下列物质与等质量的铝反应，放出氢气且消耗溶质物质的量最少的是（）A．氢氧化钠溶液B．稀硫酸C．盐酸D．稀硝酸考点：铝的化学性质．.专题：元素及其化合物．分析：首先硝酸与金属铝反应不生成氢气，根据生成物的化学式：Na[Al（OH）4]、Al2（SO4）3、AlCl3，通过物料守恒可直接判断出等量的铝消耗NaOH物质的量最少．解答：解：设Al为1mol，A．铝与氢氧化钠溶液反应生成Na[Al（OH）4]，1molAl消耗1molNaOH；B．铝与稀硫酸反应生成Al2（SO4）3，1mol铝消耗1.5mol硫酸；C．铝与盐酸反应生成AlCl3，1mol铝消耗3mol盐酸；D．硝酸与金属铝反应不生成氢气，综合以上分析可知放出氢气且消耗溶质物质的量最少的是氢氧化钠溶液，故选A．点评：本题考查铝的性质，侧重于物质的量的判断，解答时可根据生成物的化学式，从质量守恒的角度分析，易错点为D，注意硝酸与铝反应不生成氢气．4．已知X、Y中含有相同的元素，Z、W中也含有相同的元素，根据反应X+H2O→Y+H2；Z+H2O→W+O2 （方程式均未配平），可推断X、Y中及Z、W中相同元素的化合价的高低顺序为（）A．X＞Y、Z＞W B．X＜Y、Z＜W C．X＞Y、Z＜W D．X＜Y、Z＞W考点：氧化还原反应．.专题：氧化还原反应专题．分析：X+H2O→Y+H2↑中，H元素的化合价降低，则X、Y中相同元素的化合价升高；Z+H2O→W+O2↑中O元素的化合价升高，则Z、W中相同元素的化合价降低，据此分析解答．解答：解：氧化还原反应中，某元素的化合价升高（降低），则一定存在元素化合价的降低（升高），X+H2O→Y+H2↑中，H元素的化合价降低，则X、Y中相同元素的化合价升高，即化合价为Y＞X；Z+H2O→W+O2↑中O元素的化合价升高，则Z、W中相同元素的化合价降低，即化合价为Z＞W，故选D．点评：本题考查了氧化还原反应，根据得失电子确定元素化合价升降，注意反应中各元素化合价变化，题目难度不大．5．（3分）向存在大量Na+、Cl﹣的溶液中通入足量的NH3后，该溶液中还可能大量存在的离子组是（）D．A g+、Cu2+、NO3﹣A．K+、Br﹣、CO32﹣B．A l3+、H+、MnO4﹣ C．N H4+、Fe3+、SO42﹣考点：离子共存问题．.专题：离子反应专题．分析：向存在大量Na+、Cl﹣的溶液中通入足量的NH3后，溶液呈碱性，该溶液中还可能大量存在的离子与Cl﹣、OH﹣不反应．解答：解：A．在碱性条件下离子之间不发生任何反应，可大量共存，故A正确；B．Al3+、H+与OH﹣不能大量共存，故B错误；C．Fe3+与OH﹣不能大量共存，故C错误；D．Ag+、Cu2+与OH﹣不能大量共存，故D错误．故选A．点评：本题考查离子共存问题，题目难度不大，注意加入氨水后溶液呈碱性，与OH﹣反应的离子不能大量共存．6．（3分）科学家指出，食用虾类等水生甲壳类动物的同时服用维生素C容易中毒，这是因为对人体无害的+5价砷类物质在维生素C的作用下，能够转化为有毒的+3价砷类化合物．下列说法不正确的是（）A．维生素C具有还原性B．上述过程中砷元素发生还原反应C．上述过程中+5价砷类物质作氧化剂D．+5价砷转化为+3价砷时，失去电子考点：氧化还原反应．.专题：压轴题；氧化还原反应专题．分析：根据+5价砷转化为有毒的+3价砷，则+5价砷类物质在反应中作氧化剂，发生还原反应，而这一转化是维生素C参加反应的导致的，则维生素C具有还原性，在氧化还原反应中得电子化合价降低，失去电子化合价升高．解答：解：A、+5价砷类物质在维生素C的作用下，能够转化为有毒的+3价砷类化合物，则维生素C具有还原性，故A正确；B、砷元素的化合价降低，则发生还原反应，故B正确；C、反应中含化合价降低元素的物质为氧化剂，则+5价砷类物质作氧化剂，故C正确；D、化合价降低，元素得到电子，故D错误；故选D．点评：本题考查氧化还原反应，明确信息的利用及元素化合价的分析是解答本题的关键，并注意氧化还原反应中的相关概念来解答．7．（3分）设NA为阿伏伽德罗常数的值．下列说正确的是（）A．高温下，0.2molFe与足量水蒸气反应，生成的H2分子数目为0.3NAB．标准状况下，22.4L盐酸含有NA个HCl分子C．氢氧燃料电池正极消耗22.4L（标准状况）气体时，电路中通过的电子数目为2NA D．5NH4NO3═2HNO3+4N2↑+9H2O反应中，生成28g N2时，转移的电子数目为3.75NA考点：阿伏加德罗常数．.专题：阿伏加德罗常数和阿伏加德罗定律．分析：A．铁与水蒸气反应生成四氧化三铁，产物中铁元素的平均化合价为+价，根据电子守恒计算出生成氢气的物质的量；B．标准状况下，盐酸不是气体，不能使用标况下的气体摩尔体积计算盐酸的物质的量；C．氢氧燃料电池中正极为氧气，标况下22.4L氧气的物质的量为1mol，消耗1mol 氧气转移了4mol电子；D．根据化合价变化判断生成28g氮气转移的电子式．解答：解：A．铁与水蒸气反应生成的是四氧化三铁，0.2mol铁完全反应转移的电子为：0.2mol×=mol，根据电子守恒，生成氢气的物质的量为：=mol，生成的H2分子数目为NA，故A错误；B．标况下盐酸不是气体，不能使用标况下的气体摩尔体积计算22.4L盐酸的物质的量，故B错误；C．氢氧燃料电池中负极为氢气、正极为氧气，则正极消耗1mol氧气时转移了4mol 电子，电路中通过的电子数目为4NA，故C错误；D．5NH4NO3═2HNO3+4N2↑+9H2O反应中，铵根离子中﹣3价N原子被氧化成氮气，化合价升高了3价，消耗5mol铵根离子会生成4mol氮气，转移的电子的物质的量为：3×5mol=15mol，则生成1mol氮气转移的电子为：15mol×=3.75mol，转移的电子数目为3.75NA，故D正确；故选D．点评：本题考查阿伏加德罗常数的应用，题目难度中等，注意明确标况下气体摩尔体积的使用条件，准确弄清分子、原子、原子核内质子中子及核外电子的构成关系；选项D为易错点，注意该反应中化合价变化情况．8．（3分）若以w1和w2分别表示浓度为a mol/L和b mol/L氨水的质量分数，且知2a=b，则下列判断正确的是（氨水的密度比纯水的小）（）A．2w1=w2 B．2w2=w1 C．w2＞2w1 D．w1＜w2＜2w1考点：物质的量浓度的相关计算．.专题：物质的量浓度和溶解度专题．分析：根据c=表示氨水的物质的量浓度，结合氨水的浓度越大密度越小，进行判断．解答：解：设质量分数w1的氨水密度为ρ1g/mL，质量分数w2的氨水的为ρ2g/mL，质量分数w1的氨水的物质量浓度为a=mol/L，质量分数w2的氨水的物质量浓度为b=mol/L，由于2a=b，所以2×mol/L=mol/L，故2ρ1w1=ρ2w2，氨水的浓度越大密度越小，所以ρ1＞ρ2，故w2＞2w1，故选C．点评：本题考查物质的量浓度的有关计算，难度中等，关键理解物质的量浓度与质量分数之间的关系，注意氨水、酒精的浓度越大、密度越小．9．（3分）下列指定反应的离子方程式正确的是（）A．C u溶于稀硝酸HNO3：Cu+2H++NO3﹣=Cu2++NO2↑+H2OB．（NH4）2Fe（SO4）2溶液与过量NaOH溶液反应制Fe（OH）2：Fe2++2OH﹣=Fe（OH）2↓C．向NaAlO2溶液中通入过量CO2制Al（OH）3：CO2+AlO2﹣+2H2O=Al（OH）3↓+HCO3﹣D．用CH3COOH溶解CaCO3：CaCO3+2H+=Ca2++H2O+CO2↑考点：离子方程式的书写．.专题：离子反应专题．分析：A．Cu和稀硝酸反应生成硝酸铜、NO和水；B．二者反应生成氢氧化亚铁、一水合氨；C．二者反应生成氢氧化铝和碳酸氢钠；D．弱电解质写化学式．解答：解：A．Cu和稀硝酸反应生成硝酸铜、NO和水，离子方程式为3Cu+8H++2NO3﹣=3Cu2++2NO↑+4H2O，故A错误；B．二者反应生成氢氧化亚铁、一水合氨，离子方程式为Fe2++2NH4++4OH﹣=Fe（OH）2↓+2NH3．H2O，故B错误；C．二者反应生成氢氧化铝和碳酸氢钠，离子方程式为CO2+AlO2﹣+2H2O=Al（OH）3↓+HCO3﹣，故C正确；D．弱电解质写化学式，离子方程式为CaCO3+2CH3COOH=Ca2++2CH3COO﹣+H2O+CO2↑，故D错误；故选C．点评：本题考查离子方程式正误判断，明确物质的性质及离子方程式书写规则是解本题关键，单质、氧化物、弱电解质、气体、沉淀、络合物都写化学式，注意要结合原子守恒、电荷守恒及转移电子守恒，题目难度不大，易错选项是B，注意B中离子反应先后顺序及反应物的量，为易错点．10．下列依据相关实验得出的结论正确的是（）A．向某溶液中加入稀盐酸，产生的气体通入澄清石灰水，石灰水变浑浊，该溶液一定是碳酸盐溶液B．用铂丝蘸取少量某溶液进行焰色反应，火焰呈黄色，该溶液一定是钠盐溶液C．将某气体通入溴水中，溴水颜色褪去，该气体一定是乙烯D．向某溶液中滴加KSCN 溶液，溶液不变色，滴加氯水后溶液显红色，该溶液中一定含Fe2+考点：常见阳离子的检验；常见气体的检验；常见阴离子的检验．.专题：物质检验鉴别题．分析：A、二氧化硫通入澄清石灰水也会变浑浊；B、钾元素的焰色反应若不用钴玻璃也会发出黄色火焰；C、溴水中溴单质是强氧化剂，通入的气体只要是还原性气体都可以使溴水褪色；D、依据亚铁离子的检验方法分析判断．解答：解：A、向某溶液中加入稀盐酸，产生的气体通入澄清石灰水，石灰水变浑浊，该溶液可能是碳酸盐溶液或碳酸氢盐溶液或亚硫酸盐溶液，故A错误；B、用铂丝蘸取少量某溶液进行焰色反应，火焰呈黄色，钾元素的焰色反应若不用钴玻璃也会发出黄色火焰，该溶液不一定是含钠元素的溶液，故B错误；C、将某气体通入溴水中，溴水颜色褪去，通入的气体只要是还原性气体或能发生加成的有机气体都可以使溴水褪色，如二氧化硫、硫化氢等，该气体不一定是乙烯，故C错误；D、向某溶液中滴加KSCN 溶液，溶液不变色，说明不含铁离子，滴加氯水后溶液显红色，氯气氧化亚铁离子为铁离子，遇到硫氰酸钾溶液生成血红色溶液证明该溶液中一定含Fe2+，故D正确；故选D．点评：本题考查了常见离子检验方法和现象判断，主要是碳酸盐、亚硫酸盐、溴单质、亚铁离子等物质性质的分析应用，焰色反应的实验方法应用，题目难度中等．11．（3分）我省盛产矿盐（主要成分是NaCl，还含有SO42﹣等其他可溶性杂质的离子）．下列有关说法正确的是（）A．有矿盐生成食盐，除去SO42﹣最合适的实际是Ba（NO3）2B．工业上通过电解氯化钠溶液制备金属钠和氯气C．用酚酞试液可鉴别饱和食盐水和饱和纯碱溶液D．室温下，AgCl在水中的溶解度小于在食盐中的溶解度考点：物质的分离、提纯的基本方法选择与应用；盐类水解的应用；难溶电解质的溶解平衡及沉淀转化的本质；金属冶炼的一般原理；物质的分离、提纯和除杂．.专题：离子反应专题；电离平衡与溶液的pH专题．分析：A．除杂不能引入新杂质；B．电解氯化钠溶液，生成NaOH、氢气、氯气；C．纯碱溶液显碱性，酚酞的变色范围为8﹣10；D．存在AgCl（s）⇌Ag+（aq）+Cl﹣（aq），结合平衡移动分析．解答：解：A．除杂不能引入新杂质，则除去SO42﹣最合适的实际是BaCl2，故A错误；B．电解氯化钠溶液，生成NaOH、氢气、氯气，则电解熔融NaCl制备金属钠和氯气，故B错误；C．纯碱溶液显碱性，酚酞的变色范围为8﹣10，则加酚酞变红的为碳酸钠，无色的为NaCl，可鉴别，故C正确；D．存在AgCl（s）⇌Ag+（aq）+Cl﹣（aq），在NaCl溶液中抑制溶解，则溶解度小，则室温下，AgCl在水中的溶解度大于在食盐中的溶解度，故D错误；故选C．点评：本题考查较综合，涉及混合物的分离提纯、物质的鉴别、电解原理、溶解平衡移动等，综合性较强，注重高频考点的考查，题目难度不大．12．下列实验方案中，不能测定Na2CO3和NaHCO3混合物中Na2CO3质量分数（）A．取a克混合物充分加热，减重b克B．取a克混合物与足量稀盐酸充分反应，加热、蒸干、灼烧，得b克固体C．取a克混合物与足量稀硫酸充分反应，逸出气体用碱石灰吸收，增重b克D．取a克混合物与足量Ba（OH）2溶液充分反应，过滤、洗涤、烘干，得b克固体考点：钠的重要化合物．.专题：压轴题．分析：实验方案是否可行，关键看根据测量数据能否计算出结果．A、此方案利用碳酸氢钠的不稳定性，利用差量法即可计算质量分数；B、根据钠守恒，可列方程组求解；C、C项应先把水蒸气排除才合理；D、根据质量关系，可列方程组求解．解答：解：A、在Na2CO3和NaHCO3中，加热能分解的只有NaHCO3，故A项成立；B、反应后加热、蒸干、灼烧得到的固体产物是NaCl，Na2CO3和NaHCO3转化为NaCl时的固体质量变化不同，由钠元素守恒和质量关系，可列方程组计算，故B项成立；C、C项中碱石灰可以同时吸收CO2和水蒸气，则无法计算，故C项错误；D、Na2CO3和NaHCO3转化为BaCO3时的固体质量变化不同，利用质量关系来计算，故D项成立．故选：C．点评：本题看似是实验设计，实际上是从定性和定量两个角度考察碳酸钠和碳酸氢钠性质的不同．13．（3分）对相同状况下的12C18O和14N2两种气体，下列说法正确的是（）A．若质量相等，则质子数相等B．若原子数相等，则中子数相等C．若分子数相等，则体积相等D．若体积相等，则密度相等考点：阿伏加德罗定律及推论；质量数与质子数、中子数之间的相互关系．.专题：原子组成与结构专题．分析：本题需要根据阿伏伽德罗定律及其推论来解答，要针对4个选项中相同的量来分别计算其要求的其他量是否相等．解答：解：对于 12C18O和14N2，它们的摩尔质量分别为（12+18）g/mol和（14+14）g/mol，即30g/mol和28g/mol，二者所含质子数分别为6+8=14和7+7=14，所含中子数分别为（12﹣6）+（18﹣8）=16和（14﹣7）+（14﹣7）=14．A、设质量均为1g，则它们所含质子的物质的量之比为：=14：15，因为微粒数之比等于物质的量之比，所以其质子数之比也是14：15，故A错误．B、二者均为双原子分子，原子数相等时，二者的物质的量相等，其中子数之比为16：14=8：7，即原子数相等时，中子数不相等，故B错误．C、相同条件时，分子数之比等于物质的量之比等于体积之比，则分子数相等，其体积必然相等，故C正确．D、相同条件下，密度之比等于摩尔质量之比，其密度之比为30：28=15：14，即体积相等时，其密度不相等，故D错误．故选C．点评：解答本题要熟练记忆阿伏伽德罗定律及其推论，并能把握各自的前提条件，才能灵活运用，学习新课时本类题会成为难点，因此要知难而进，才能突破难点．14．下列叙述Ⅰ和Ⅱ均正确并且有因果关系的是（）选项叙述Ⅰ叙述ⅡA NH4Cl为强酸弱碱盐用加热法除去NaCl中的NH4ClB Fe3+具有氧化性用KSCN溶液可以鉴别Fe3+C 溶解度：CaCO3＜Ca（HCO3）2 溶解度：Na2CO3＜NaHCO3D SiO2可与HF反应氢氟酸不能保存在玻璃瓶中A．A B．B C．C D．D考点：铵盐；硅和二氧化硅；二价Fe离子和三价Fe离子的检验．.专题：氮族元素；碳族元素；几种重要的金属及其化合物．分析：A．氯化铵是强酸弱碱盐，且不稳定；B．铁离子具有氧化性，能氧化还原性物质，铁离子和硫氰根离子反应生成血红色硫氰化铁溶液；C．碳酸钙的溶解度小于碳酸氢钙，但碳酸钠的溶解度大于碳酸氢钠；D．二氧化硅是酸性氧化物，但能和氢氟酸反应．解答：解：A．氯化铵是强酸弱碱盐，其水溶液呈酸性，氯化铵不稳定，加热易分解，两者没有因果关系，故A错误；B．铁离子具有氧化性，能氧化还原性物质，铁离子和硫氰根离子反应生成血红色硫氰化铁溶液，可以用硫氰化钾溶液检验铁离子，两者没有因果关系，故B错误；C．碳酸钙的溶解度小于碳酸氢钙，但碳酸钠的溶解度大于碳酸氢钠，二者没有因果关系，故C错误；D．二氧化硅能和氢氟酸反应，玻璃中含有二氧化硅，所以玻璃瓶不能保存氢氟酸，二者有因果关系，故D正确；故选D．点评：本题考查较综合，明确二氧化硅的性质，铁离子的检验是高考热点，应熟练掌握．15．（3分）X、Y、Z、W有如图所示的转化关系，则X、W可能是①C、O2 ②AlCl3、NaOH ③Fe、HNO3 ④S、O2（）A．①②③B．①②C．③④D．①②③④考点：无机物的推断；常见金属元素的单质及其化合物的综合应用．.专题：推断题．分析：将选项中的物质带入验证即可，图象中X、Y都能与W反应．解答：解：①X为C、W为O 2时，Y为CO，Z为CO 2，且C与O2完全反应即生成CO 2，故正确；②若X为AlCl 3、W为NaOH，则Y为Al（OH）3，Z为NaAlO 2，过量NaOH与AlCl 3反应生成NaAlO 2，故正确；③若X为Fe、W为HNO3，Fe过量时生成Fe（NO 3）2，继续与HNO3反应生成Fe（NO3）3，Fe在过量HNO 3中生成Fe（NO 3）3，故正确；④若X为S、W为O2，S与氧气反应生成二氧化硫，在催化剂条件下二氧化硫与氧气反应生成三氧化硫，但硫单质不能与氧气反应生成直接三氧化硫，故错误．点评：本题考查元素化合物中的连续反应，思维容量较大，需要对物质间的转换关系及转换条件非常熟悉，否则易出错．16．（3分）某氯化镁溶液的密度为1.18g•cm﹣3，其中镁离子的质量分数为5.1%，则300mL 该溶液中Cl﹣的物质的量约等于（）A．0.37 mol B．0.63 mol C．0.74 mol D．1.50 mol考点：物质的量浓度的相关计算．.专题：物质的量浓度和溶解度专题．分析：根据m=ρV计算氯化镁溶液的质量，根据m（（Mg2+）=m（溶液）×ω（（Mg2+）计算镁离子质量，再根据n=计算镁离子的物质的量，根据电荷守恒氯化镁溶液中n（Cl ﹣）=2n（Mg2+）．解答：解：氯化镁溶液的质量为：1.18g•cm﹣3×300mL=354g，镁离子的质量为：354g×5.1%=3.54×5.1g，镁离子的物质的量为：=0.75mol，根据电荷守恒氯化镁溶液中n（Cl﹣）=2n（Mg2+）=2×0.75mol=1.5mol，故选D．点评：考查物质的量浓度、质量分数、常用化学计量的有关计算，比较基础，注意公式的灵活运用．17．（3分）（2012•江苏）下列物质的转化在给定条件下能实现的是（）①Al2O3NaAlO2（aq）Al（OH）3②S SO3H2SO4③饱和NaCl（aq）NaHCO3Na2CO3④Fe2O3FeCl3（aq）无水FeCl3⑤MgCl2（aq）Mg（OH）2MgO．A．①③⑤B．②③④C．②④⑤D．①④⑤考点：镁、铝的重要化合物；二氧化硫的化学性质；铁的氧化物和氢氧化物．.专题：元素及其化合物．分析：①氧化铝与氢氧化钠反应生成偏铝酸钠，偏铝酸钠溶液通入二氧化碳，生成氢氧化铝．②硫燃烧生成二氧化硫．③在饱和食盐水中通入氨气，形成饱和氨盐水，再向其中通入二氧化碳，在溶液中就有了大量的钠离子、铵根离子、氯离子和碳酸氢根离子，其中NaHCO3溶解度最小，所以析出NaHCO3，加热NaHCO3分解生成碳酸钠．④氧化铁与盐酸反应生成氯化铁，Fe3+水解，加热蒸发得不到无水FeCl3．⑤氯化镁与石灰乳转化为更难溶的氢氧化镁，氢氧化镁不稳定，加热分解生成氧化镁．解答：解：①氧化铝与氢氧化钠反应生成偏铝酸钠，偏铝酸钠溶液通入二氧化碳，发生反应2NaAlO2+CO2+3H2O=2Al（OH）3↓+Na2CO3，生成氢氧化铝，故①正确；②硫燃烧生成二氧化硫，不能生成三氧化硫，故②错误；③在饱和食盐水中通入氨气，形成饱和氨盐水，再向其中通入二氧化碳，在溶液中就有了大量的钠离子、铵根离子、氯离子和碳酸氢根离子，其中NaHCO3溶解度最小，析出NaHCO3，加热NaHCO3分解生成碳酸钠，故③正确；④氧化铁与盐酸反应生成氯化铁，Fe3+水解Fe3++3H2O⇌2Fe（OH）3+HCl，加热蒸发HCl挥发，平衡向右移动，得不到无水FeCl3，故④错误；⑤氯化镁与石灰乳转化为更难溶的氢氧化镁，氢氧化镁不稳定，加热分解生成氧化镁，故⑤正确．故①③⑤正确．故选：A．点评：考查元素化合性质、侯德榜制碱法、盐类水解等，难度中等，注意侯德榜制碱法要先通氨气，后通二氧化碳，以便获得高浓度离子溶液．18．（3分）1.52g 铜镁合金完全溶解于50mL 密度为1.40g/mL、质量分数为63%的浓硝酸中，得到NO2和N2O4的混合气体1120mL（标准状况），向反应后的溶液中加入1.0mol/L NaOH 溶液，当金属离子全部沉淀时，得到2.54g沉淀．下列说法不正确的是（）A．该合金中铜与镁的物质的量之比是2：1B．得到2.54 g沉淀时，加入NaOH溶液的体积是600 mLC．N O2和N2O4的混合气体中，NO2的体积分数是80%D．该浓硝酸中HNO3的物质的量浓度是14.0 mol/L考点：化学方程式的有关计算；有关混合物反应的计算．.专题：计算题．分析：A．金属离子全部沉淀时，得到2.54g沉淀为氢氧化铜、氢氧化镁，故沉淀中氢氧根的质量为2.54g﹣1.52g=1.02g，根据n=计算氢氧根的物质的量，根据电荷守恒可知，金属提供的电子物质的量等于氢氧根的物质的量，令铜、镁合金中Cu、Mg的物质的量分别为xmol、ymol，根据提供的电子物质的量与二者质量之和列方程计算x、y的值，据此解答；B．根据钠离子守恒可知，氢氧化钠的物质的量等于反应后溶液中硝酸钠的物质的量，根据氮元素守恒计算硝酸钠的物质的量，再根据V=计算需要氢氧化钠溶液的体积；C．根据n=计算NO2和N2O4混合气体的物质的量，令二氧化氮的物质的量为amol，根据电子转移列放出计算，进而计算二氧化氮的体积分数；D．根据c=计算该浓硝酸的物质的量浓度．解答：解：A．金属离子全部沉淀时，得到2.54g沉淀为氢氧化铜、氢氧化镁，故沉淀中氢氧根的质量为 2.54g﹣1.52g=1.02g，氢氧根的物质的量为=0.06mol，根据电荷守恒可知，金属提供的电子物质的量等于氢氧根的物质的量，令铜、镁合金中Cu、Mg的物质的量分别为xmol、ymol，则：，解得，故合金中铜与镁的物质的量之比是0.02mol：0.01mol=2：1，故A正确；B．根据钠离子守恒可知，氢氧化钠的物质的量等于反应后溶液中硝酸钠的物质的量，根据氮元素守恒可知，硝酸钠的物质的量为0.05L×14mol/L﹣0.04mol﹣（0.05﹣0.04）×2=0.64mol，故需要氢氧化钠溶液的体积为=0.64L=640mL，故B错误；C．NO2和N2O4混合气体的物质的量为=0.05mol，令二氧化氮的物质的量为amol，则四氧化二氮的物质的量为（0.05﹣a）mol，根据电子转移守恒可知，a×1+（0.05﹣a）×2×1=0.06，解得a=0.04，故NO2的体积分数是×100%=80%，故C正确；D．该浓硝酸密度为 1.40g/mL、质量分数为63%，故该浓硝酸的物质的量浓度为mol/L=14mol/L，故D正确；故选B．点评：本题考查混合物的有关计算，题目难度中等，理解反应发生的过程是关键，是对学生综合能力的考查，注意根据守恒思想进行的解答．二、非选择题（共四道小题共计46分）19．（12分）通常情况下，微粒A和B为分子，C和E为阳离子，D为阴离子，它们都含有10个电子；B溶于A后所得的物质可电离出C和D；A、B、E三种微粒反应后可得C和一种白色沉淀．请回答：（1）用化学符号表示下列4种微粒：A：H2O ；B：NH3 ；C：NH4+ ；D：OH﹣．（2）写出A、B、E三种微粒反应的离子方程式：Al3++3NH3+3H2O═Al（OH）3↓+3NH4+或Mg2++2NH3+2H2O═Mg（OH）2↓+2NH4+．考点：无机物的推断；原子核外电子排布．.专题：推断题．分析：常见的10电子分子有H2O、NH3、CH4、HF等，常见的10电子阳离子有Na+、Mg2+、Al3+、NH4+、H3O+，常见的10电子阴离子有F﹣、OH﹣，根据“B溶于A后所得的物质可电离出C和D”，可推出A为H2O、B为NH3、C为NH4+、D为OH﹣，再根据A、B、E反应后可得C和一种白色沉淀，可推出E为Mg2+或Al3+，从而得出答案．解答：解：常见的10电子分子有H2O、NH3、CH4、HF等，常见的10电子阳离子有Na+、Mg2+、Al3+、NH4+、H3O+，常见的10电子阴离子有F﹣、OH﹣，根据“B溶于A后所得的物质可电离出C和D”，可推出A为H2O、B为NH3、C为NH4+、D为OH﹣，再根据A、B、E反应后可得C和一种白色沉淀，可推出E为Mg2+或Al3+，（1）由上述分析可知，A为H2O；B为NH3；C为NH4+；D为OH﹣；故答案为：H2O；NH3；NH4+；OH﹣；（2）A为H2O、B为NH3，E为Mg2+或Al3+，A、B、E三种微粒反应的离子方程式为：Al3++3NH3+3H2O═Al（OH）3↓+3NH4+或Mg2++2NH3+2H2O═Mg（OH）2↓+2NH4+；故答案为：Al3++3NH3+3H2O═Al（OH）3↓+3NH4+或Mg2++2NH3+2H2O═Mg（OH）2↓+2NH4+．点评：考查物质（微粒）的推断、离子反应等，难度中等，掌握常见的10电子微粒及其性质是推断的关键，B溶于A后所得的物质可电离出C和D，是突破口．20．（10分）硫酸钠﹣过氧化氢加合物（xNa2SO4•yH2O2•zH2O）的组成可通过下列实验测定：①准确称取1.7700g样品，配制成100.00mL溶液A．②准确量25.00mL溶液A，加入盐酸酸化的BaCl2溶液至沉淀完全，过滤、洗涤、干燥至恒重，得到白色固体0.5825g．③准确量取25.00mL溶液A，加适量稀硫酸酸化后，用0.020 0mol•L﹣1 KMnO4溶液滴定至终点，消耗KMnO4溶液25.00mL．H2O2与KMnO4反应的离子方程式如下：2MnO4﹣+5H2O2+6H+═2Mn2++8H2O+5O2↑（1）已知室温下BaSO4的Ksp=1.1×10﹣10，欲使溶液中c（SO42﹣）≤1.0×10﹣6mol•L ﹣1，应保持溶液中c（Ba2+）≥_ 1.1×10﹣4 mol•L﹣1．（2）上述滴定若不加稀硫酸酸化，MnO4﹣被还原为MnO2，其离子方程式为2MnO4﹣+3H2O2═2MnO2+3O2↑+2OH﹣+2H2O ．（3）通过上述实验事实，该样品的组成为（写化学式）2NaSO4•H2O2•2H2O．（4）上述实验③KMnO4溶液滴定至终点的现象是）溶液由无色变紫红色¡£怎样判断滴定达到了终点当最后一滴KMnO4溶液滴下时，溶液突然由无色变紫红色，且30秒内不褪色．考点：化学方程式的有关计算．.专题：计算题．分析：（1）根据沉淀溶解平衡常数Ksp=c（SO42﹣）•c（Ba2+ ）来计算；（2）不加稀硫酸酸化，MnO4﹣被还原为MnO2，双氧水被氧化生成氧气，根据电荷守恒可知有氢氧根离子生成，再根据原子守恒判断是否有水生成，配平书写；（3）生成的白色固体0.5825g为硫酸钡，根据n=计算硫酸钡的物质的量，根据硫。

益阳市箴言中学2014—2015学年高一10月月考 化 学 时量：60分钟 满分100分 相对原子质量：H 1 C 12 O 16 Na 23 N 14 S 32 Fe 56 Cl 35.5 第I卷（选择题 共60分） 选择题部分共15小题，每小题给出四个选项中只有一个选项正确，每小题4分，共60分。

1．如果你家里的食用花生油混有水份，你将采用下列何种方法分离( )B．蒸馏C．分液D．萃取 2．下列实验操作中，不合理的是( ) ① ② ③ A．洗涤沉淀时(如图①)，向漏斗中加适量水，搅拌并滤干 B．用CCl4提取碘水中的碘，选③ C．蒸馏时蒸馏烧瓶中液体的体积不能超过容积的2/3，液体也不能蒸干 D．粗盐提纯，选①和② ( )( )( ) B．CH4C．CO2 D．SO2 6．下列实验操作中，主要不是从安全因素考虑的是( ) A．点燃氢气前一定要检验氢气的纯度 B．未使用完的白磷要随时收集起来，并与空气隔绝 C．用氢气还原氧化铜时，要先通一会儿氢气，再加热氧化铜 D．酒精灯不用时，必须盖上灯帽 7．鉴别SO42－时所选用的试剂及顺序最合理的是( )用NA表示阿伏加德罗常数的值，下列叙述中正确的是( ) A．1 mol·L－1 NaCl溶液中含有NA个Na＋ B．25 ℃，101 kPa64 g SO2中含有的原子数为3NA C．在常温常压下.4 L Cl2含有的分子数为 NA D．标准状况下11.2 L CCl4含有的分子数为0.5NA Na2CO3俗名纯碱，下面对纯碱采用不同分类法进行的分类中正确的是( ) A．Na2CO3是碱 B．Na2CO3是酸式盐 C．Na2CO3是难溶性盐 D．Na2CO3是钠盐 Na2SO4溶液，正确的方法是( )Na2SO4 溶于100ml水中 ② 将32.2g Na2SO4·10H2O溶于少量水中，再用水稀释至100 ml ③ 将20 ml 5.0 mol/L Na2SO4溶液用水稀释至100 ml A．①②B．②③C． ③D．①②③ 11．医学上对血液中毒最常用的净化手段是血液透析。

益阳市箴言中学2015—2016学年高二10月月考 化学试题 时量：90分钟 满分：100分 一、选择题（本题包括17个小题，每小题3分，共51分，每小题只有一个答案符合题意） 1．在下列反应中，反应物的总能量低于生成物的总能量的是 A．2H2＋O22H2O B．CaO＋CO2=CaCO3 aCO3CaO＋CO2D．C2H5OH＋3O2 2CO2＋3H2O 2．在 xA(气)＋yB(气) zC(气)＋wD(气) 的可逆反应中，经1分钟后A减少a mol/L，B减少a/3 mol/， C增加2a/3 mol/L， D增加a mol/L， 则x、y、z、w的比例关系是 A．3 : 1 : 2 : 3 B．1 : 3 : 3 : 1 C．3 : 2 : 1 : 3 D．1 : 2 : 3 : 2 3．反应A(g)+3B(g) 2C(g)+2D(g)，在不同情况下测得反应速率，其中反应速率最快的是 A．υ(D)=0.4 mol / ?L·s? B．υ(C)=0.5 mol / ?L·s? C．υ(B)=0.6 mol / ?L·s? D．υ(A)=0.15 mol / ?L·s? 4．下列条件的改变，一定能加快化学反应速率的是 A．增大压强 B．升高温度C．增大反应物的量 D．减小生成物的浓度 2NH3的正逆反应速率可用各反应物或生成物浓度的变化来表示。

下列关系中能说明反应已达到平衡状态的是 A．υ正(N2)=υ逆(NH3) B．3υ正(N2)=υ正(H2)C．υ正(N2)=3υ逆(H2) D．2υ正(H2)=3υ逆(NH3) 6．能增加反应物分子中活化分子的百分数的是 A．使用催化剂 B．增大固体表面积 C．增大压强 D．增加浓度 7．某化学反应其△H=—122 kJ/mol，?S=231 J/(mol·K)，则此反应在下列哪种情况下可自发进行 A．在任何温度下都不能自发进行 B．在任何温度下都能自发进行 C．仅在高温下自发进行 D．仅在低温下自发进 8. 对于3Fe(s)＋4H2O(g) Fe3O4(s)＋4H2(g)，反应的·L－1,则化学平衡常数为 A．1/54 B．1/6.75 C．1/27 D．1 9．在密闭容器中，2(g)+3H2(g) 2NH3（g） ΔH＜0，达到甲平衡。

益阳市箴言中学2015届高三第二次模拟考试数学〔理科〕考试时间：150分钟 总分：150分一、选择题〔每题5分〕1.设集合{}12A x R x =∈-<，{}2,x B y R y x R =∈=∈，如此AB ＝〔 〕A ．∅B ．[)0 3，C ．()0 3，D ．()1 3-，2.集合{}05≤-=a x x A ，{}06>-=b x x B ,N b a ∈,，且{}2,3,4A B N ⋂⋂=，如此整数对()b a ,的个数为( )A.20B. 25C. 30D. 423.设函数()ln(1)f x x x =+- ，记(1),(3),c (7)a f b f f ===如此 〔 〕 A.c a b << B.a b c << C.c b a << D.b c a <<4.设f′〔x 〕是函数f 〔x 〕的导函数，将y=f 〔x 〕和y=f′〔x 〕的图象画在同一个直角坐标系中，不可能正确的答案是〔 〕 A ．B ．C ．D ．5.直线y=4x 与曲线y=x 3在第一象限内围成的封闭图形的面积为〔 〕 A ． 2B ． 4C ． 2D ． 46.设函数()f x 的定义域为D ，如果对于任意的1x D ∈，存在唯一的2x D ∈，使得12()()2f x f x C +=成立〔其中C 为常数〕，如此称函数()y f x =在D 上的均值为C ，现在给出如下4个函数：①3y x =②4sin y x =③lg y x =④2x y =，如此在其定义域上的均值为 2的所有函数是下面的〔〕A. ①②B. ③④C. ①③④D. ①③7.设f 〔x 〕=|lnx|，假设函数g 〔x 〕=f 〔x 〕﹣ax 在区间〔0，3]上有三个零点，如此实数a 的取值范围是〔 〕 A ．〔0，〕B ． 〔，e 〕C ． 〔0，]D ． [，〕8.设函数)(x f 的定义域为实数集R ，且)()1()2(x f x f x f -+=+，假设2)4(-=f ，如此函数1)2011(2)(++=xx e f e x g 的最小值是 A.1B.3C.3lnD.2ln9.如图，正△ABC 的中心位于点G 〔0，1〕，A 〔0，2〕，动点P 从A 点出发沿△ABC 的边界按逆时针方向运动，设旋转的角度∠AGP=x〔0≤x≤2π〕，向量在=〔1，0〕方向的射影为y 〔O 为坐标原点〕，如此y关于x 的函数y=f 〔x 〕的图象是〔 〕10.函数()M f x 的定义域为实数集R ,满足()1,0,M x M f x x M ∈⎧=⎨∉⎩(M 是R 的非空真子集),在R 上有两个非空真子集,A B ,且A B =∅,如此()()()()11A B A B f x F x f x f x +=++的值域为A ．20,3⎛⎤ ⎥⎝⎦B ．{}1C ．12,,123⎧⎫⎨⎬⎩⎭D ．1,13⎡⎤⎢⎥⎣⎦二、填空题〔每题5分〕 11.命题p ：“∀x ∈R ，∃m ∈R,4x－2x ＋1＋m ＝0〞，且命题非p 是假命题，如此实数m 的取值范围为________．12.假设函数f 〔x 〕在定义域D 内某区间I 上是增函数，且在I 上是减函数，如此称y=f 〔x 〕在I上是“弱增函数〞．函数h 〔x 〕=x 2﹣〔b ﹣1〕x+b 在〔0，1]上是“弱增函数〞，如此实数b 的值为________． 13.函数2ln )(bx x a x f -=图象上一点))2(,2(f P 处的切线为22ln 23++-=x y ，假设方程0)(=+m x f 在区间],1[e e内有两个不等实根，如此实数m 的取值范围是14.定义在R 上的奇函数()f x 满足(4)()f x f x +=，且在[]2,0上的解析式为()⎩⎨⎧≤<≤≤-=21,sin 10),1(x x x x x x f π，如此_______641429=⎪⎭⎫⎝⎛+⎪⎭⎫⎝⎛f f 15.函数()()f x g x ''、分别是二次函数()f x 和三次函数()g x 的导函数，它们在同一坐标系下的图象如下列图，设函数()()()h x f x g x =-，如此(1),(0),(1)h h h -的大小关系为三、解答题16〔此题12分〕.在中，角所对的边分别为，，〔1〕求的大小；〔2〕假设，求的取值范围.17〔此题12分〕.如图，四棱锥P ABCD -的底面ABCD 是正方形，侧棱PD ⊥底面ABCD ，PD DC =，E 是PC 的中点．〔I 〕证明：PA //平面BDE ；〔II 〕求二面角B DE C --的平面角的余弦值；〔Ⅲ〕在棱PB 上是否存在点F ，使PB ⊥平面DEF ？证明你的结论．18〔此题12分〕.设2()f x x x =+,用)(n g 表示()f x 当[,1](*)x n n n N ∈+∈时的函数值中整数值的个数.(1)求)(n g 的表达式.(2)设32*23()()n n n a n N g n +=∈,求2121(1)nk n k k S a -==-∑. (3)设12(),2n n n n g n b T b b b ==+++,假设)(Z l l T n ∈<,求l 的最小值.19〔此题13分〕．经销商用一辆J 型卡车将某种水果从果园运送(满载)到相距400 km 的水果批发市场．据测算，J 型卡车满载行驶时，每100 km 所消耗的燃油量u(单位：L)与速度v(单位：km/h)，的关系近似地满足u ＝除燃油费外，人工工资、车损等其他费用平均每小时300元．燃油价格为每升(L)7.5元．(1)设运送这车水果的费用为y(元)(不计返程费用)，将y 表示成速度v 的函数关系式； (2)卡车该以怎样的速度行驶，才能使运送这车水果的费用最少？20〔此题13分〕.抛物线24y x =的焦点为F 2，点F 1与F 2关于坐标原点对称，以F 1,F 2为焦点的椭圆C 过点2⎛ ⎝⎭. 〔Ⅰ〕求椭圆C 的标准方程；〔Ⅱ〕设点T )0,2(，过点F 2作直线l 与椭圆C 交于A,B 两点，且22F A F B λ=，假设[]2,1,TA TB λ∈--+求的取值范围.21〔此题13分〕.函数()ln f x x a x =-，1(), (R).ag x a x +=-∈〔1〕假设1a =，求函数()f x 的极值；〔2〕设函数()()()h x f x g x =-，求函数()h x 的单调区间； 〔3〕假设在[]1,e 〔e 2.718...=〕上存在一点0x ，使得0()f x <0()g x 成立，求a 的取值范围.数学〔理科〕答案1.C2.C3.B4.D5.D6.D7.D8.B9.c 10.B假设A x ∈，如此1)(,0)(,1)(===x f x f x f B A B A ，1)(=x F ；假设B x ∈，如此,0)(=x f A 1)(,1)(,1)(===x F x f x f B A B ；假设B x A x ∉∉，，如此0)(=x f A ，0)(=x f B ，.1)(,0)(==x F x f B A 应当选B.11.m 1≤12.1 13.]12,1(2e +14.51615.)1()1()0(-<<h h h16.解：〔1〕由条件结合正弦定理有：，从而有：，.〔2〕由正弦定理得：，，，即：.17.解：法一：〔I 〕以D 为坐标原点，分别以DA 、DC 、DP 所在直线为x 轴、y 轴、z 轴建立空间直角坐标系，设2PD DC ==，如此(2,0,0)A ，(0,0,2)P ，(0,1,1)E ，(2,2,0)B)0,2,2(),1,1,0(),2,0,2(==-=DB DE PA设1(,,)n x y z =是平面BDE 的一个法向量，如此由1100n DE n DB ⎧⋅=⎪⎨⋅=⎪⎩，得0220y z x y +=⎧⎨+=⎩ 取1y =-，得1(1,1,1)n =-．∵1220PA n ⋅=-=，1,//PA n PA BDE PA BDE ∴⊥⊄∴，又平面平面〔II 〕由〔Ⅰ〕知1(1,1,1)n =-是平面BDE 的一个法向量，又2(2,0,0)n DA ==是平面DEC 的一个法向量．设二面角B DE C --的平面角为θ，由图可知12,n n θ=<>∴12121223cos cos ,3||||32n n n n n n θ⋅=<>===⋅⨯．故二面角B DE C --的余弦值为33．〔Ⅲ〕∵)1,1,0(),2,2,2(=-=DE PB ∴0220,.PB DE PB DE =+-=∴⊥假设棱PB 上存在点F ，使PB ⊥平面DEF ，设)10(<<=λλPB PF ， 如此(2,2,2)PF λλλ=-，(2,2,22)DF DP PF λλλ=+=-由0PF DF •=得22442(22)0λλλλ+--= ∴PBPF 31)1,0(31=∈=，此时λ即在棱PB 上存在点F ，13PF PB=，使得PB ⊥平面DEF ．法二：〔I 〕连接AC ，AC 交BD 于O ，连接OE ．在PAC ∆中，OE 为中位线，∴OE //PAPA BDE ⊄又平面,∴PA //平面BDE ．〔II 〕PD ⊥底面ABCD ,∴平面PDC ⊥底面ABCD ，CD 为交线,BC ⊥CD∴平面BCE ⊥平面PDC ，PC 为交线, PD =DC ，E 是PC 的中点∴DE ⊥PC∴DE ⊥平面PBC ，∴DE ⊥BE ∴BEC ∠即为二面角B DE C --的平面角．设PD DC a ==，在Rt BCE ∆中,263,,,cos 223CE a BC a BE a BEC ===∴∠=故二面角B DE C --的余弦值为33．〔Ⅲ〕由〔II 〕可知DE ⊥平面PBC ，所以DE ⊥PB ，所以在平面PDE 内过D 作DF ⊥PB ，连EF ，如此PB ⊥平面DEF ．在Rt PDB ∆中,PD a =，2BD a =，3PB a =，33PF a =．所以在棱PB 上存在点F ，13PF PB=，使得PB ⊥平面DEF ．18.解对,函数在单增,值域为,故.(2),故=-n(2n+1)(3)由得,且两式相减,得于是故假设且,如此的最小值是7.19.所以当v ＝100时，y 取得最小值．答当卡车以100 km/h 的速度驶时，运送这车水果的费用最少．(16分) 20.〔Ⅰ〕设椭圆的半焦距为c ，由题意得1=c ，设椭圆C 的标准方程为)0(12222>>=+b a b y a x ，略21.解：〔Ⅰ〕()f x 的定义域为(0,)+∞，当1a =时，()ln f x x x =-，11()1x f x x x -'=-= ，所以()f x 在1x =处取得极小值1.〔Ⅱ〕1()ln a h x x a x x +=+-， x (0,1) 1(1,)+∞ ()f x ' — 0+ ()f x 极小22221(1)(1)[(1)]()1a a x ax a x x a h x x x x x +--++-+'=--==①当10a +>时，即1a >-时，在(0,1)a +上()0h x '<，在(1,)a ++∞上()0h x '>，所以()h x 在(0,1)a +上单调递减，在(1,)a ++∞上单调递增；②当10a +≤，即1a ≤-时，在(0,)+∞上()0h x '>，所以，函数()h x 在(0,)+∞上单调递增.〔III 〕在[]1,e 上存在一点0x ，使得0()f x <0()g x 成立，即 在[]1,e 上存在一点0x ，使得0()0h x <，即 函数1()ln a h x x a x x +=+-在[]1,e 上的最小值小于零. 由〔Ⅱ〕可知①即1e a +≥，即e 1a ≥-时， ()h x 在[]1,e 上单调递减，所以()h x 的最小值为(e)h ，由1(e)e 0e a h a +=+-<可得2e 1e 1a +>-， 因为2e 1e 1e 1+>--，所以2e 1e 1a +>-； ②当11a +≤，即0a ≤时， ()h x 在[]1,e 上单调递增，所以()h x 最小值为(1)h ，由(1)110h a =++<可得2a <-；③当11e a <+<，即0e 1a <<-时， 可得()h x 最小值为(1)h a +，因为0ln(1)1a <+<，所以，0ln(1)a a a <+<故(1)2ln(1)2h a a a a +=+-+>此时，(1)0h a +<不成立.综上讨论可得所求a 的范围是：2e 1e 1a +>-或2a <-.。