【精准解析】广东省深圳市第二高级中学2019-2020学年高一下学期第四学段考试化学试题
广东省深圳市第二高级中学2019_2020学年高一历史下学期第四学段考试试题含解析
试题分析：本题主要考查的是获取和解读材料信息的能力。从材料中可以得出的时间信息是1400—1820年，根据所学知识可知，铁制农具等生产工具的改进是春秋战国时期的，不属于这个时期，故①不符合，排除。所以答案选D。
考点：古代中国的经济·农业的主要耕作方式和土地制度·明清时期农业的发展
9. 白种人主要分布在欧洲，但是到了近代，在北美洲和大洋洲白种人成为当地的主要居民，造成这种现象的主要原因是（ ）
15.斯塔夫里阿诺斯在《全球通史》中记载：“（1840年）素有印度的曼彻斯特（英国纺织工业中心）之称的达卡，已从一个非常富庶的城市衰败为异常贫穷的小镇……这种衰败不仅发生在达卡，而且发生在其他所有地区。”“这种衰败”在中国的表现是
A. 19世纪四五十年代，英国对华鸦片贸易十分猖獗
B. 19世纪四五十年代，东南沿海妇女纺织半数停工
A. 土著居民的减少B. 欧洲的殖民扩张
C. 种族之间通婚D. 大量战争移民
【答案】B
【解析】
试题分析：随着新航路的开辟和早期殖民扩张，欧洲国家积极对北美洲和大洋洲进行殖民扩张，大量白人涌入这些地区，而当地土著居民受到迫害，如美洲印第安人几乎遭到种族灭绝，因此选B.CD与史实不符，A不是主要原因。
考点：欧洲的殖民扩张
4.官营手工业长期占据中国古代手工业的主导地位，其产品
A. 全部销往海外
B. 占据市场大部分份额
C. 质量普遍不高
D. 主要满足官府需要
【答案】D
【解析】
本题主要考查的是对知识的再现再认能力。根据所学知识可知，官营手工业的产品不进入市场，主要满足官府的需要，故D项正确，AB项错误；官营手工业的产品质量比较好，技术水平高，故C项错误；
5.《新唐书》记载：“唐制……日暮鼓八百声而门闭。五更二点鼓自内发，诸街鼓承振，坊市门皆起。”陆游《老学庵笔记》载：“京都街鼓今尚废，后生读唐诗文及街鼓者，往往茫然不能知。”唐宋街鼓这一变化从一个侧面反映了宋代
2019-2020学年深圳市第二高级中学高三英语第四次联考试卷及答案解析
2019-2020学年深圳市第二高级中学高三英语第四次联考试卷及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AIn September, something terrible happened on the west coast ofTasmania,Australia. As many as 380 pilot whales became stranded(搁浅) in shallow water there and later died. This might have beenAustralia’s largest stranding event on record, the BBC reported.But this large amount of whale stranding is not uncommon. For centuries, it has happened all over the world and has puzzled scientists. Scientists say the cause is often unknown. But they have offered many different explanations.Some say the whales chase small fish for food and end up in shallow water because they are not paying attention to where they are going.Others think the stranding has something to do with Earth’s geomagnetic field (地磁场). They say that a geomagnetic compass in whales’ brains controls their position. Unusual changes in Earth’s magnetic field can affect the whales’ compasses and send them in the wrong direction.Another explanation suggests that stranding is caused by the close relationships that whales have. Pilot whales travel in large groups. One lead whale might mistakenly lead the whole group to shallow water. “And if one gets into trouble, the others will not leave,” said Sheryl Gibney, a leading biologist fromNew Zealand. “Some will come in and try to help, they get trapped on the beach, then more will come.”The whales are trapped by mistake or out of sympathy(同情). Once they get stranded, they will likely die. According to the National Oceanic and Atmospheric Administration of theUS, less than 10 percent of all stranded whales survive.1. What do we know about the stranding inAustralia?A. It happened on theeast coast ofAustralia.B. It caused the death of over 300 pilot whales.C. It is commonly seen in September each year.D. It was the largest stranding event in the world.2. According to Gibney, the pilot whales are the animals that________.A. are kind to each otherB. are easy to lose directionC. are too huge to float in the seaD.are silly to follow the leading whale3. What is the story mainly about?A. How human activity has affected whales.B. What might cause whale strandings.C. How whales find their direction while traveling.D. What scientists are doing to save stranded whales.BAs an eco-minimalist, Su Yige has maintained an environmentally friendly and sustainable lifestyle for the past three years while studying in Canada. She takes her own bag when she goes grocery shopping and uses second-hand items as often as possible. She avoids almost all paper-related products unless she has to use a public toilet while away from home.Diligence and thrift are time-honored traditional Chinese virtues. Su's family is a good example of this, according to the native of Weihai, Shandong province. Although they have little formal “green” knowledge, her parents lead a very environmentally friendly life.For example, the family has used the same hair dryer for more than a decade, and Su remembers many of her mother's clothes from as far back as kindergarten. “As long as something can still be used, my mother will not replace it with a new item,” she said.“I frequently asked my father to bring the plastic bag back home after dumping our waste in the trash bin. He was unhappy, and argued that instead of making that request, I should go downstairs to dump the waste myself” she said. In another move, her father criticized her for doing too much shopping online. Eventually, they both made steps toward becoming better environmentalists. Her father brings the bag back for reuse and she has only bought two pieces of clothing online in the past six months.Back in Canada, Su is looking forward to finding a job related to sustainable development in China after she graduates as a computer science major in the summer.4. Which of the following best describes Su Yige?A. Conservative.B. Nostalgic.C. Economical.D. Productive.5. What can be found about Su's mother according to the third paragraph?A. She has a lot of formal green knowledge.B. She regards using the same items as a lifestyle.C. She always wears old clothes due to lack of money.D. She will not replace the old items until they are out of style.6. What can be inferred about Su and her father?A. Both of them like to criticize each other.B. Su's father is particular about her clothes.C. The relationship between them is very tense.D. They urge each other to become more environmentally friendly.7. What can we learn about the author soon after she graduates in Canada?A. She will stay there to look for a job.B. She will put effort into computer science.C. She will devote herself to her motherland's future.D. She will come back to China to stay with her parents.CIn the summer of 2016, I gave a talk at a small conference in northernVirginia. I began by admitting that I’d never had a social-media account; I then outlined arguments for why other peopleshould consider removing social media from their lives. The event organizers uploaded the video of my talk to YouTube. Then it was shared repeatedly on Facebook and Instagram and, eventually, viewed more than five million times. I was both pleased and annoyed by the fact that my anti-social-media talk had found such a large audience on social media.I think of this event as typical of the love-hate relationships many of us have with Facebook, Instagram, and other social-media platforms. On the one hand, we’ve grown cautious about the so-called attention economy, which, in the name of corporate(公司的) profits, destroys social life gradually and offends privacy. But we also benefit from social media and hesitate to break away from it completely. Not long ago, Imet a partner at a large law firm in Washington, D.C., who told me that she keeps Instagram on her phone because she misses her kids when she travels; looking through pictures of them makes her feel better.In recent months, some of the biggest social-media companies, Facebook and Twitter, in particular, have promised various reforms. In March, Mark Zuckerberg announced a plan to move his platform toward private communication protected by end-to-end encryption(端对端加密); later that month, he put forward the establishment of a third-party group to set standards for acceptable content.All of these approaches assume that the reformation of social media will be a complex, lengthy, and gradual process. But not everyone sees it that way. Alongside these official responses, a loose collective of developers that calls itself the IndieWeb has been creating another alternative. They are developing their own social-media platforms, which they say will preserve what’s good about social media while getting rid of what’s bad. They hope to rebuild social media according to principles that are less corporate and more humane(人道的).8. Why did the author feel annoyed when his video was spread online?A. His video caused many arguments.B. His video was shared without his permission.C. His talk was opposed by a large amount of people.D. His video’s popularity on social media is against his talk.9. Why does the author mention the story of his partner in paragraph 2?A. To prove that social media has some benefits.B. To advise people to break away from social media.C. To tell the negative effects social media may produce.D. To describe people’s complicated relationships with social media.10. What is the purpose of the reform made by some social-media companies?A. To attract more users.B. To improve network environment.C. To make more profits.D. To provide more convenientservice.11. What does the IndieWeb intend to do?A. Develop new social-media platforms.B. Remove social media from people’s lives.C. Improve the existing social-media principles.D. Help social-media companies to make reformation.DDogs are often referred to as “man's best friend”. But MacKenzie, a four-pound Chihuahua (奇瓦瓦狗), who was named winner of the 2020 American Hero Dog competition, is making the world a better place for humans and animals alike. Often called the “Oscars for dogs”, the award recognizes dogs who make great contributions to society.This year's competition attracted over 400 competitors from across the country. While all were impressive, it was tiny MacKenzie who won the judges' hearts. Born at a rescue shelter in Hilton, New York, in 2013, she had acleft palate (腭裂) that required her to be tube fed for the first year of her life. A life-saving operation, performed in 2014, gave her the ability to eat and drink independently, enabling the tiny dog to focus on doing what she loved most: taking care of others.The seven-year-old Chihuahua is now gainfully “employed” by the Mid Foundation, a Rochester, New York-based non-profit organization that shelters and cares for animals born with disabilities. MacKenzie's official job is “to provide love and care for baby rescue animals born with birth defects”. The Chihuahua is good at her job and hasnurturedmany different species-from puppies to kittens to turkeys, squirrels, birds and even a goat. She acts as their mother and teaches them how to socialize, play, and have good manners.In addition to her role as an animal caretaker, MacKenzie also has the important job of greeting the foundation's volunteers and friends. The incredible dog, who has lost her ability to bark, also visits area schools to help children understand physical disabilities in both animals and people. Her heart-warming and inspiring story makes MacKenzie worthy of America's top dog honor!12. What made MacKenzie American Hero Dog?A. Being man's best friend.B. Her struggle with disabilities.C. Rescuing animals with disabilities.D. Her contributions to a better world.13. What can we infer about MacKenzie from Paragraph 2?A. Her growth path was not easy.B. She was deserted by her owner.C. She was operated on at two years old.D. She still needs taking care of by others.14. What does the underlined word “nurtured” in Paragraph 3 mean?A. Trained.B. Comforted.C. Tended.D. Abused.15. Which can be a suitable title for the text?A. MacKenzie—The Most Hard-working DogB. MacKenzie—America's “Most Heroic Dog”C. Chihuahua—Inspiration of Positive EnergyD. Chihuahua—Appeal for Animals' Protection第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
8 平抛与斜面结合(含答案)
【平抛与斜面结合】1、如图所示，在坡度一定的斜面顶点以大小相同的初速v同时水平向左与水平向右抛出两个小球A 和B，两侧斜坡的倾角分别为30°和60°，小球均落在坡面上，若不计空气阻力，则A和B两小球的运动时间之比为（）A．3：4 B．4：3 C．1：3 D．3：1【来源】广东省深圳市第二高级中学2019-2020学年高一（下）第四学段物理试题【答案】C【解析】ABCD．A球在空中做平抛运动，落在斜面上时，有解得同理对B有由此解得ABD错误C正确。
故选C。
2、如图所示，从倾角为的斜面顶点A将一小球以初速度水平抛出，小球经过t落在斜面上B点，重力加速度为g，斜面足够长，不计空气阻力，则下列说法正确的有（）A ．从A 到B 的运动时间为B ．AB 的长度为C ．初速度变为2倍，抛出到落在斜面的时间t 变为2倍D ．改变初速度大小，小球落在斜面上时速度与斜面的夹角大小不变【来源】河北省唐山市开滦一中2019-2020学年高一（下）期末物理试题【答案】ACD【解析】AC.根据位移偏角的正切值小球在空中飞行的时间由于位移的偏角不变，若初速度变为2倍，则小球抛出到落在斜面的时间t 变为2倍，故AC 正确；B.小球的水平位移 0022tan v x v t g θ==所以AB 的长度为故B 错误；D.做平抛运动的小球，速度偏角的正切值是位移偏角正切值的2倍，而小球落在斜面上位移的偏角不变，则位移偏角正切值不变，速度偏角的正切值也不变，所以改变初速度大小，小球落在斜面上时速度与斜面的夹角大小不变，故D正确。
故选ACD。
3、如图所示，从斜面上的A点以速度水平抛出一个物体，飞行一段时间后，落到斜面上的B点；若仍从A点抛出物体，抛出速度为，不计空气阻力，下列说法正确的是（）A. 物体的飞行时间不变B. 物体的位移变为原来的C. 物体落到斜面上的速度变为原来的D. 物体落到斜面上时速度方向不变【答案】D【解析】【解答】根据可知，当初速度减半时，飞行的时间减半，A不符合题意；根据x=v0t可知，物体的水平位移变为原来的1/4，竖直位移也变为原来的1/4，则物体的位移变为原来的1/4，B不符合题意；水平初速度减半时，根据v y=gt可知，落到斜面上的竖直速度变为原来的一半，可知物体落到斜面上的速度变为原来的1/2，C不符合题意；根据为定值，则物体落到斜面上时速度方向不变，D符合题意；故答案为：D.4、如图，从斜面上的A点以速度v0水平抛出一个物体，飞行一段时间后，落到斜面上的B点，已知AB = 75 m， = 37°，不计空气阻力，下列说法正确的是（）A．物体的位移大小为60 m B．物体飞行的时间为6 sC．物体的初速度v0大小为20 m/s D．物体在B点的速度大小为30 m/s【来源】江西省南昌市八一中学2019-2020学年高一（下）期中物理试题【答案】C【解析】A ．物体的位移等于首末位置的距离，大小为75 m ，故A 错误；B ．平抛运动的竖直位移根据得，物体飞行的时间 2245s 3s 10h t g ⨯===故B 错误；C ．物体的初速度故C 正确；D ．物体落到B 点的竖直分速度根据平行四边形定则知，物体落在B 点的速率故D 错误。
广东省深圳市第二高级中学2019_2020学年高一地理下学期第四学段考试试题含解析.doc
广东省深圳市第二高级中学2019-2020学年高一地理下学期第四学段考试试题（含解析）考试时间：75分钟；命题人：注意事项：1．答题前填写好自己的姓名、班级、考号等信息2．请将答案正确填写在答题卡上第I卷（选择题)一、单选题（每题2分，共70分）产蛋崖是贵州某县特有的地质奇观，泥岩构成的崖壁上镶嵌着石灰岩构成的卵形“石蛋”，“石蛋”30年左右才脱落一个。
完成下面小题。
1. 崖壁产下石蛋的地质作用是A. 风化B. 风蚀C. 水蚀D. 搬运2. 产蛋崖岩石类型为A. 崖壁-变质岩，石蛋-沉积岩B. 崖壁-沉积岩，石蛋-沉积岩C. 崖壁-变质岩，石蛋-侵入岩D. 崖壁-沉积岩，石蛋-变质岩【答案】1. A 2. B【解析】【1题详解】泥岩构成的崖壁上镶嵌着石灰岩构成的卵形“石蛋”，崖壁岩石在温度变化、阳光、水、生物作用下会发生破坏作用，会逐渐剥离脱落，则“石蛋”会松动脱落，这属于风化作用，A正确；贵州多山地、气候湿润，风力较小，因此崖壁产下石蛋的地质作用不是风蚀作用，B错误；崖壁上没有经常性的水流，因此崖壁产下石蛋的地质作用不是水蚀作用或搬运作用，C、D错误。
故选A。
【2题详解】材料信息表明，产蛋崖上的崖壁由泥岩构成，应为颗粒很小的泥沙沉积而形成，应属于沉积岩，排除A 、C ；材料材料信息表明，镶嵌在崖壁上“石蛋”是由石灰岩构成，石灰岩是上温暖广阔的海洋环境的海水沉积和生物沉积作用形成，应属于沉积岩，B 符合题意，排除D 。
故选B 。
【点睛】岩石的成因类型有岩浆岩、变质岩、沉积岩等。
（1）岩浆岩：岩浆侵入或喷出冷凝而成。
如花岗岩、玄武岩、流纹岩、安山岩。
（2）沉积岩：外力作用下形成。
如砾岩、砂岩、页岩、石灰岩等。
（3）变质岩：变质作用下形成。
片麻岩、大理岩、石英岩、板岩等属于变质岩。
下图是在我国某地区拍摄的一幅景观图，读图完成下面小题。
3. 形成该地貌的外力作用是（ ） A. 流水侵蚀B. 流水堆积C. 风力侵蚀D. 风力堆积4. 该地貌主要分布于我国（ ） A. 东南丘陵B. 华北平原C. 塔里木盆地D. 黄土高原5. 据图分析，该地近期的风向是（ ） A. 西北风B. 东南风C. 西南风D. 东北风【答案】3. D 4. C 5. A 【解析】 【3题详解】风力堆积形成沙丘和沙垅（风力作用一般在干旱半干旱地区表现显著），图中地表形态是沙丘，是风力的沉积作用形成的，故D 正确； 流水侵蚀作用常常形成沟谷（v 型）、瀑布和峡谷，故A 不符合题意； 流水堆积作用常常在山前形成冲积扇或洪积扇、在河流中下游形成冲积平原和河口三角洲，，故B 不符合题意；风力侵蚀作用形成风蚀蘑菇和风蚀洼地，故C 不符合题意；故选D 。
广东省深圳市第二高级中学2021-2022高一英语下学期第四学段考试试题
广东省深圳市第二高级中学2021-2022高一英语下学期第四学段考试试题时间：120分钟满分：150分命题人：第I卷I. 听力 (共两节，满分20分)第一节 (共 5 小题; 每小题1分, 满分5分)听下面 5 段对话，每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项,听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What size jacket is the man wearing now?A. Small.B. Medium.C. Large.2. What did the man buy for his new flat?A. A chair.B. A bed.C. A table.3. Who is the woman probably?A. A receptionist.B. A director.C. A sales manager.4. What will the woman doA. Reserve the book.B. Buy a book.C. Borrow the book.5. What are the speakers talking about?A. A project.B. A program.C. Food.第二节（共15小题; 每小题1分, 满分15分）听下面5段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项，听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟; 听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料, 回答第6、7题。
6. Where will the man go on Tuesday?A. To Edinburgh.B. To Paris.C. To London.7. When will the man interview the main speaker?A. On Tuesday morning.B. On Tuesday afternoon.C. On Wednesday afternoon 听第7段材料，回答第8至9题。
2019-2020学年广东省深圳市第二高级中学高一下学期第四学段考试化学试题(解析版)
广东省深圳市第二高级中学2019-2020学年高一下学期第四学段考试试题可能用到的相对原子质量：H 1 C 12 N 14 O 16 S 32 Fe 56 Mn 55 Cu 64 一、单项选择题1.最近日本科学家确认世界还存在着另一种“分子足球”60N ，它与60C 分子结构相似，60N 在高温和撞击后会释放出巨大能量，下列对60N 的说法不正确．．．的是（ ） A. 60N 高温和撞击后应该生成2N B. 60N 和2N 都是氮的同素异形体 C. 60N 中氮原子之间由共价键结合D. 60N 和14N 都是氮的同位素『答案』D 『解析』『详解』A ．由信息N 60在高温和撞击后会释放出巨大能量，转化为稳定物质氮气，则N 60高温和撞击后应该生成N 2，故A 正确；B ．N 60和N 2都是单质，且均只含N 元素，则均为氮的同素异形体，故B 正确；C ．N 60中只有一种非金属元素N ，氮原子之间由共价键结合，故C 正确；D ．N 60为单质，14N 为原子，二者不是同位素关系，故D 错误； 故选D 。
2.下列说法正确的是（ ）A. 双原子分子中的共价键一定是非极性键B. 任何原子的原子核都是由质子和中子构成的C. 非极性键可能存在于离子化合物中D. 质子数相同的微粒，它们的化学性质基本相同『答案』C 『解析』『详解』A. 同种元素形成双原子分子中的共价键，一定是非极性键，而不同种元素形成的双原子分子如HCl 中的共价键，是极性键，故A 错误； B. H 原子中没有中子，故B 错误；C. 非极性键可能存在于离子化合物中如Na 2O 2，故C 正确；D. H 2O 和HF 质子数相同，但化学性质不同，故D 错误；故答案为C。
3.下列有关化学用语表示正确的是( )A. 甲烷的电子式B. 乙烷的结构简式：C2H6C. CH4分子的球棍模型：CD. 核内有8个中子的碳原子：86『答案』A『解析』『详解』A．甲烷为共价化合物，C最外层形成8个电子，电子式为，故A正确；B．乙烷的分子式为C2H6，其结构简式为CH3CH3，故B错误；C．为比例模型，甲烷分子的球棍模型为，故C错误；C，D．碳原子核内有6个质子，则核内有8个中子的碳原子的质量数为14，其核素表示为146故D错误；故答案为A。
2023-2024学年广东省深圳市第二高级中学高一下学期第四学段考试化学试卷
2023-2024学年广东省深圳市第二高级中学高一下学期第四学段考试化学试卷1.下列有关材料的说法，不正确的是A．中国深海载人潜水器“蛟龙号”的发动机使用的氮化硅陶瓷属于新型无机非金属材料B．华为首款5G手机搭载了智能7nm制程SOC“麒麟980”手机芯片的主要成分是二氧化硅C．光伏电池是以硅为基底的硅太阳能电池，将光能直接转化为电能D．2022年北京冬奥会展示的智能服饰含有石墨烯，石墨烯与金刚石互为同素异形体2.化学与生活、未来密切相关，下列有关叙述错误的是A．在家用燃煤中适量的生石灰能有效减少二氧化硫的排放量B．利用清洁能源如太阳能、潮汐能、风能来发电C．采用纳米二氧化钛光触媒技术，将汽车尾气中的NO 和CO 转化为无害气体D．大量排放SO 2或CO 2都会导致酸雨的形成3.已知碘化氢分解吸热，分以下两步完成：2HI(g)→H2(g)+2I•(g)；2I•(g)→I2(g)，下列图像最符合上述反应历程的是A．B．C．D．4.普通水泥在固化过程中自由水分子减少并产生Ca(OH)2，溶液呈碱性。
根据这一物理化学特点，科学家发明了电动势法测水泥初凝时间。
此法的原理如图所示，反应的总方程式为：2Cu+Ag2O=Cu2O+2Ag，下列有关说法正确的是（）A．装置中电流方向由Cu经导线到Ag 2 OB．测量原理示意图中，Ag 2 O为负极C．负极的电极反应式为：2Cu+2OH – -2e – =Cu 2 O+H 2 OD．电池工作时，OH –向正极移动5.代表阿伏加德罗常数，下列说法不正确的是A．一定条件下，将和混合反应生成，转移的电子总数为B．中含有的硅氧键数为C．常温下，氯气含有个电子D．28gN 2和组成的混合气体中含有的原子数为6.下列离子方程式书写正确的是A．足量通入少量石灰水中：B．溶液与溶液共热：C．与稀硝酸反应：D．硫酸氢钠溶液与少量反应：7.水合肼(N2H4·H2O)又称水合联氨，是重要的氢能源稳定剂，具有较强的还原性。
2019-2020学年深圳市第二高级中学高三英语第四次联考试题及答案解析
2019-2020学年深圳市第二高级中学高三英语第四次联考试题及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AEast Yorkshire has typical unpredictable British weather. So here are some ideas to keep everybody happy when the weather is not the most ideal.William's Den, North CaveThe outdoor and indoor areas are suitable for children of all ages to have fun.There are nests to explore, rope bridges to cross, a tree-house and a slide. The attached Kitchen provides fresh food made from locally sourced ingredients serving a selection of treats.East Riding Leisure CentresKnown for a fun learner pool alongside an incredible fun zone with two slides as well, it is perfect for kids to find their feet in the water, have fun and explore. Its 6 climbing walls offer a different challenge on each. This place is suitable for anyone over the age of 4 and you can refuel at cafe with fresh food, snacks and cakes.Sewerby Hall and GardensWhen the weather’s not sure, take cover in the Hall and learn how life was in the early 1900’s for the residents and workers of the house. Then explore the zoo and meet the pigs, parrots and penguins! Kids of all ages are welcome.Withernsea LighthouseThere’s no limitation to the age of kids to climb Withernsea Lighthouse, which is 144 steps to the top, with full views of the East Yorkshire Coast at the top of it. Enjoy the museum on the ground floor and learn what life is like working and living in a lighthouse. The souvenir shop provides attractive gifts for visitors at a fair price.1.Which one is unsuitable for kids of all ages?A.William’s Den, North Cave.B.East Riding Leisure Centres.C.Sewerby Hall and Gardens.D.Withernsea Lighthouse.2.Where can kids enjoy food?A.In William’s Den, North Cave and Sewerby Hall and Gardens.B.In East Riding Leisure Centres and Withernsea LighthouseC.In William’s Den, North Cave and East RidingLeisure Centres.D.In Sewerby Hall and Gardens and Withernsea Lighthouse.3.Where does this passage probably come from?A.A geography textbook.B.A science report.C.A finance magazine.D.A travel brochure.BMark Twain,the famous American writer,was once traveling in France.He went by trainto Dijon.He was very tired and wanted to sleep.He therefore asked the conductor to wake him up when the train came to Dijon.But first he explained he was a very heavy sleeper,“I may possibly protest(抗议)loudly when you try to wake me up,” he said to the conductor.“But don’t take any notice of what I say.Just put me off the train anyway.”Then Mark Twain went to ter,when he woke up it was night time and the train had reached Paris already.He realized at once that the conductor had forgotten to wake him up at Dijon.He was so angry that he ran to the conductor and began to shout at him.“I have never been so angry in my life,” Mark Twain said.The conductor looked at him calmly(平静地).“You are not half so angry as the American whom I put off the train atDijon,” he said.4. Mark Twain knew that he was a heavy sleeper,so ________.A. he protested loudly to the conductorB. he did not sleep before he arrived inDijonC. he told the conductor to wake him up no matter how loudly he might protestD. he slept lightly that time5. The conductor didn’t wake up Mark Twain atDijonbecause ________.A. he didn’t take Mark Twain’s words seriouslyB. he forgot Mark Twain’s words when the train came toDijonC. he did not want to bear his protestD. he mistook another American traveler for Mark Twain6. The American whom the conductor put off the train ________.A. did not want to get off atDijonB. wanted to get off atParisC. wanted to get off atDijonD. did not want to get off atParis7. Which of the following is TRUE?A. The conductor didn’t take Mark Twain’s words seriously.B. The conductor did take Mark Twain’s words seriously.C. The conductor was a heavy sleeper.D. Mark Twain must get off atParis.CTrue friend from a distant landInternational friends can come in many shapes and sizes. However they arrive on the scene, they are always appreciated, especially in times ofturbulence. Michael Lindsay (1909 — 94) is a case in point.After graduating from the University of Oxford in 1931, Lindsay arrived in China in 1937 on a ship from Canada intending merely to teach economics at Beiping Yenching University, the current Peking University. He traveled with Norman Bethune the Canadian doctor who dedicated his life to medicine in China during the War of Resistance Against Japanese Aggression (1931 — 45).Lindsay left Beijing at the end of 1941, and went to the Communist-led Jinchaji base(晋察冀边区)together with his wife Li Xiaoli, who was a Shanxi local and a former student of his.“Lindsay was inspired by the hard work local Chinesepeople made in fighting against Japanese aggression. After learning that Chinese armies on the frontlines were suffering supply shortages, Lindsay was determined to lend a hand," says Yan Juhai, a retired official from Shanxi's Lyuliang city. "In the Jinchaji base, NieRongzhen( 1899 — 1992), the base's top commander, appointed him to work as the communications expert, where he became a full-time radio technician contributing greatly in communication technology innovation and training," says Yan.In 1944 Lindsay and his wife arrived in Yan'an in Shaanxi province and helped fix its telecommunication equipment. They stayed for 18 months in Yan'an, and after the victory in the War of Resistance Against Japanese Aggression, they decided to return to the United Kingdom to visit their parents.8. Which word in the following can best replace the underlined word?A. peaceB. warC. stressD. sorrow9. Why did Lindsay go to the Communist-led Jinchaji base according to Yan Juhai?A. Because his wife was a Shanxi local.B. Because he was warmly invited by NieRongzhen.C. Because he was touched by Chinese people's effortsD. Because he was more interested in communications.10. What can we learn from the article?A. Lindsay and his wife stayed in China for over 8 years.B. Lindsay's wife was also skilled in communication technology.C. Lindsay was appointed as the communications consultant in Shanxi province.D. Lindsay taught economics in the Jinchaji base besides his work as a radio expert.11. Where is the text probably from?A. a guidebookB. a novelC. a diaryD. a newspaperDLight pollution is a significant but overlooked driver of the rapid decline of insect populations, according to the most comprehensive review of the scientific evidence to date.Artificial light at night can affect every aspect of insects' lives, the researchers said. "We strongly believe artificial light at night — in combination with habitat loss, chemical pollution.invasive (入侵的) species, and climate change — is driving insect declines, " the scientists concluded after assessing more than 150 studies.Insect population collapses have been reported around the world, and the first global scientific review published in February,said widespread declines threatened to cause a "catastrophic collapse of nature's ecosystems".There are thought to be millions of insect species, most still unknown to science, and about half are active at night. Those active in the day may also be disturbed by light at night when they are at rest.The most familiar impact of light pollution is moths (飞蛾) flapping around a bulb, mistaking it for the moon. Some insects use the polarisation of light to find the water they need to breed, as light waves line up after reflecting from a smooth surface. But artificial light can scupper (使泡汤) this. Insects areimportant prey (猎物) for many species, but light pollution can tip the balance in favour of the predator if it traps insects around lights. Such increases in predation risk were likely to cause the rapid extinction of affected species, the researchers said.The researchers said most human-caused threats to insects have analogues in nature, such as climate change and invasive species. But light pollution is particularly hard for insects to deal with.However, unlike other drivers of decline, light pollution is ly easy to prevent. Simply turning off lights that are not needed is the most obvious action, he said, while making lights motion-activated also cuts light pollution.Shading lights so only the area needed is lit up is important. It is the same with avoiding blue-white lights, which interfere with daily rhythms. LED lights also offer hope as they can be easily tuned to avoid harmful colours and flicker rates.12. What is discussed in the passage?A. Causes of declining insect populations.B. Consequences of insect population collapses.C. Light pollution: the key bringer of insect declines.D. Insect declines: the driver of the collapsed ecosystem.13. What is the 5th paragraph mainly about?A. How light travels in space.B. How light helps insects find food.C. How the food chain is interrelated.D. How light pollution affects insects.14. What does the underlined word"analogues"in Paragraph 6probably mean?A. Selective things.B. Similar things.C. Variations.D. Limitations.15. What is the purpose of the last paragraph?A. To offer solutions.B. To give examples.C. To make comparisons.D. To present arguments.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020年广东省深圳市第二高级中学高一数学理下学期期末试题含解析
2020年广东省深圳市第二高级中学高一数学理下学期期末试题含解析一、选择题：本大题共10小题，每小题5分，共50分。
在每小题给出的四个选项中，只有是一个符合题目要求的1. 函数y=lnx﹣6+2x的零点为x0，x0∈（）A．（1，2）B．（2，3）C．（3，4）D．（5，6）参考答案：B【考点】二分法求方程的近似解．【专题】计算题；函数思想；综合法；函数的性质及应用．【分析】分别求出f（2）和f（3）并判断符号，再由函数的单调性判断出函数唯一零点所在的区间．【解答】解：∵f（2）=ln2﹣2＜0，f（3）=ln3＞0，∴f（x）=lnx+2x﹣6的存在零点x0∈（2，3）．∵f（x）=lnx+2x﹣6在定义域（0，+∞）上单调递增，∴f（x）=lnx+2x﹣6的存在唯一的零点x0∈（2，3）．故选：B．【点评】本题主要考查函数零点存在性的判断方法的应用，要判断个数需要判断函数的单调性，属于基础题．2. 已知角θ的终边与单位圆的一个交点为,则的值是 ( )A. B. C. D.参考答案：D略3. 函数y=cos2x+sinx的值域为（）A．[﹣1，1] B．[1，] C．[﹣1，] D．[0，1]参考答案：C【考点】34：函数的值域．【分析】令sinx=t∈[﹣1，1]，可得函数y=cos2x+sinx=1﹣t2+t=﹣+=f（t），t∈[﹣1，1]，再利用二次函数的单调性即可得出值域．【解答】解：令sinx=t∈[﹣1，1]，则函数y=cos2x+sinx=1﹣t2+t=﹣+=f（t），t∈[﹣1，1]，f（t）max=，又f（﹣1）=﹣1，f（1）=1，可得f（t）min=f（﹣1）=﹣1．∴f（t）∈．故选：C．4. 设集合,,则下列结论正确的是[ ]A. B. C. D.都不对参考答案：B5. （3分）已知x，y为正实数，则（）A．2lgx+lgy=2lgx+2lgy B．2lg（x+y）=2lgx?2lgyC．2lgx?lgy=2lgx+2lgy D．2lg（xy）=2lgx?2lgy参考答案：D考点：有理数指数幂的化简求值；对数的运算性质．专题：函数的性质及应用．分析：直接利用指数与对数的运算性质，判断选项即可．解答：解：因为a s+t=a s?a t，lg（xy）=lgx+lgy（x，y为正实数），所以2lg（xy）=2lgx+lgy=2lgx?2lgy，满足上述两个公式，故选D．点评：本题考查指数与对数的运算性质，基本知识的考查．6. 已知则（）A. B. C. D.参考答案：D略7. 已知y=f(x)是奇函数，且满足，当时，，则y=f(x)在(1,2)内是（）（A）单调增函数，且（B）单调减函数，且（C）单调增函数，且（D）单调减函数，且参考答案：A8. 一枚质地均匀的硬币，如果连续抛掷1000次，那么第999次出现正面朝上的概率是A． B． C． D．参考答案：D略9. 已知，，，，那么（）A、B、C、D、参考答案：D10. 全集U={x∈Z|0＜x≤8}，M={1，3，5，7}，N={5，6，7}，则?U（M∪N）=( )A．{5，7} B．{2，4} C．{2，4，8} D．{1，3，5，6，7}参考答案：C【考点】交、并、补集的混合运算．【专题】集合思想；分析法；集合．【分析】由集合M，N求出M并N，然后求出全集U，则?U（M∪N）可求．【解答】解：由全集U={x∈Z|0＜x≤8}={1，2，3，4，5，6，7，8}，M={1，3，5，7}，N={5，6，7}，得M∪N={1，3，5，7}∪{5，6，7}={1，3，5，6，7}，则?U（M∪N）={2，4，8}．故选：C．【点评】本题考查了交、并、补集的混合运算，是基础题．二、填空题:本大题共7小题,每小题4分,共28分11. 给出以下四个结论：①若函数的定义域为[1,2]，则函数的定义域是[4,8]；②函数（其中，且）的图象过定点(1,0)；③当时，幂函数的图象是一条直线；④若，则的取值范围是；⑤若函数在区间(－∞,1]上单调递减，则的取值范围是[1,+∞).其中所有正确结论的序号是.参考答案：①④⑤12. 已知，函数，若正实数，满足，则、的大小关系是.参考答案：>略13. 一个车间为了规定工时定额，需要确定加工零件所花费的时间，为此进行了5次试验，收集数据如表：由表中数据，求得线性回归方程=0.65x+，根据回归方程，预测加工70个零件所花费的时间为分钟．参考答案：102【考点】BQ：回归分析的初步应用．【分析】根据表中所给的数据，求出横标和纵标的平均数，得到样本中心点，进而得到线性回归方程，再令x=70，即可得出结论．【解答】解：由题意， =（10+20+30+40+50）=30， =（64+69+75+82+90）=76，∴回归直线过样本中心点（30，76），代入线性回归方程，可得a=56.5，∴x=70时，y=0.65×70+56.5=102．故答案为：102．【点评】本题考查线性相关及回归方程的应用，解题的关键是得到样本中心点，为基础题．14. 如果是奇函数，则= .参考答案：－2略15. 已知的三个内角,,所对的边分别为，，，，.若，且，则角＝参考答案：16. 设，，求和：__________．参考答案：见解析当时，，当时，，当，且时，∴．17. 过点（1，3）且与直线x+2y﹣1=0平行的直线方程是．参考答案：x+2y﹣7=0【考点】直线的一般式方程与直线的平行关系．【专题】计算题；规律型；方程思想；直线与圆．【分析】求出直线的斜率，然后求解直线方程．【解答】解：与直线x+2y﹣1=0平行的直线的斜率为：，由点斜式方程可得：y﹣3=﹣（x﹣1），化简可得x+2y﹣7=0．故答案为：x+2y﹣7=0．【点评】本题考查直线方程的求法，考查计算能力．三、解答题：本大题共5小题，共72分。
广东省深圳市第二高级中学2019-2020学年高一数学下学期第四学段考试试题[含答案]
广东省深圳市第二高级中学2019-2020学年高一数学下学期第四学段考试试题时间：120分钟 满分：150分注意事项：1．本试题分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。
2．答题前，考生务必将自己的班级、姓名、准考证号填写在答题卡相应的位置。
3．全部答案在答题卡上完成，答在本试题卷上无效。
考试结束后，将答题卡交回。
第Ⅰ卷一．选择题：本大题共12小题，每小题5分．在每小题给出的四个选项中，有且只有一项是符合题目要求的．1．已知平面向量(1,2)a =-r ，(1,0)b =r ，则向量3a b +=r r （ ）A ．(2,6)--B ． ()2,6-C ．(2,6)D ．(2,6)-2．设向量()2,a m =r ，()3,1b =-r ，若()2a a b ⊥-r r r ，则实数m =（ ） A ．2或－4B ．2C ．14-或12D ．－4 3．在△ABC 中，3c =，45B =︒，60C =︒，则b =（ ）A ．22B ．32C ．322D ．24．如下图的矩形长为5，宽为2，在矩形内随机地撒300颗黄豆，数得落在阴影部分的黄豆数为138颗，则我们可以估计出阴影部分的面积为（ ）A ．235B ．2350C ．10D ．不能估计5．下列图形中，不是．．三棱柱展开图的是（ ） A ． B ． C ． D ．6．若a ，b 是异面直线，b ，c 是异面直线，则直线a ，c 的位置关系为（ ）A ．相交、平行或异面B ．相交或平行C ．异面D ．平行或异面7．关于某设备的使用年限x （单位：年）和所支出的维修费用y （单位：万元）有如下统计数据表：根据上表可得回归直线方程$$1.23y x a=+，据此估计，该设备使用年限为10年时所支出的维修费用约是（ ）A. 12.08万元B. 12.28万元C. 12.38万元D. 12.58万元8．设非零向量a r ，b r 满足a b a b +=-r r r r ，则（ ）A ．a b r r ⊥B ．a b =r rC ．a r //b rD ．a b >r r9．在△ABC 中，如果sin :sin :sin 2:3:4A B C =，那么cos C =（ ）A ．23B ．23-C ．13-D ．14- 10．设l 是直线，α，β是两个不同的平面，下列命题正确的是（ ）A ．若//l α，l β//，则//αβB ．若//l α，l β⊥，则αβ⊥C ．若αβ⊥，l α⊥，则l β⊥D ．若αβ⊥，//l α，则l β⊥11．在△ABC 中，AD 为BC 边上的中线，E 为AD 的中点，则向量EB =u u u v（ ） A ．3144AB AC -u u u v u u u v B ．1344AB AC -u u u v u u u v C ．3144+AB AC u u u v u u u v D ．1344+AB AC u u u v u u u v 12．设ABC ∆的内角A ，B ，C 的对边分别是a ，b ，c ．已知2cos 0b a C -=，sin 3sin()A A C =+，则2bc a=（ ）A B ．9 C ．23 D 第Ⅱ卷二．填空题：本大题共4小题，每小题5分．13．已知向量(,4),(3,2)a m b ==-r r ，且a b r r ∥，则m =________.14．已知圆柱的轴截面为正方形，且圆柱的体积为54π，则该圆柱的侧面积为________.15．若向量,2,()a b a b a b a ==-⊥v v v v v v v 满足，则向量a v 与b r 的夹角等于________．16．在四面体ABCD 中，AD AC BC BD ===，AB CD ==球O 是四面体ABCD 的外接球，过点A 作球O 的截面，若最大的截面面积为9π，则四面体ABCD 的体积是____.三．解答题：解答应写出文字说明，证明过程或演算步骤．17. （本小题满分10分）在某中学举行的物理知识竞赛中，将三个年级参赛学生的成绩在进行整理后分成5组，绘制出如图所示的频率分布直方图，图中从左到右依次为第一、第二、第三、第四、第五小组.已知第三小组的频数是15.（1）求成绩在50-70分的频率；（2）求这三个年级参赛学生的总人数；（3）求成绩在80-100分的学生人数.18．（本小题满分12分）某班共有学生45人，其中女生18人，现用分层抽样的方法，从男、女学生中各抽取若干学生进行演讲比赛，有关数据见下表（单位：人） 性别学生人数 抽取人数 女生18 y 男生 x3 （1）求和y ；（2）若从抽取的学生中再选2人做专题演讲，求这2人都是男生的概率．19．（本小题满分12分）已知ABC ∆的内角A,B,C 所对的边分别为a ,b ,c ,且32,cos 5a B ==. （1）若4b =，求sin A 的值；（2）若4ABC S ∆=,求b ,c 的值.20．（本小题满分12分）如图，四棱锥P ABCD -中，底面ABCD 是正方形，PA ⊥底面ABCD .（1）求证：BD ⊥平面PAC ；（2）若2,23AB PA ==，求点A 到平面PCD 的距离.21．（本小题满分12分）在△ABC 中，内角A ，B ，C 所对的边分别为a ，b ，c ，且222b a c ac =+-，（1）求角B 的大小；（2）若a =c =2，求△ABC 的面积；（3）求sin A ＋sin C 的取值范围.22.（本小题满分12分）如图，已知四棱锥P ABCD -的底面ABCD 是平行四边形，PA ⊥平面ABCD ，M 是AD 的中点，N 是PC 的中点．（1）求证：//MN PAB 平面；（2）若平面PMC PAD ⊥平面，求证：CM AD ⊥.高一数学第四学段考试参考答案1-12 BADACA CADBAD 13. -6 14. 36π 15. 450 16. 32317.（1）成绩在50-70分的频率为：0.03100.04100.7⨯+⨯=.（2）第三小组的频率为：0.015100.15⨯=.这三个年级参赛学生的总人数（总数=频数/频率）为：15/0.15100=（人）（3）成绩在80-100分的频率为：0.01100.005100.15⨯+⨯=则成绩在80-100分的人数为：1000.1515⨯=（人）.18.解：（1）由题意可得，451827x =-=，又31827y =，所以2y =； （2）记从女生中抽取的2人为1a ，2a ，从男生中抽取的3人为1b ，2b ，3b ， 则从抽取的5人中再选2人做专题演讲的基本事件有()12,a a ，()11,a b ，()12,a b ，()13,a b ，()21,a b ，()22,a b ，()23,a b ，()12,b b ，()13,b b ，()23,b b 共10种．设选中的2人都是男生的事件为A ，则A 包含的基本事件有()12,b b ，()13,b b ，()23,b b 共3种．因此()310P A =． 故2人都是男生的概率为310． 19.(1)∵3cos 05B =>,且0B π<<，∴4sin 5B ==， 由正弦定理得sin sin a b A B=， ∴42sin 25sin 45a B A b⨯===； (2)∵1sin 42ABC S ac B ∆==， ∴142c 425⨯⨯⨯=， ∴5c =， 由余弦定理得2222232cos 25225175b ac ac B =+-=+-⨯⨯⨯=，∴b =20.（1）因为底面ABCD 是正方形，所以AC BD ⊥,因为PA ⊥底面ABCD ，所以PA BD ⊥,又因为PA AC A =I ，所以BD ⊥平面PAC .（2）设点A 到平面PCD 的距离为h因为PA ⊥底面ABCD ，所以PA CD ⊥，又AD CD ⊥,PA AD A ⋂=，所以CD ⊥平面PAD ，所以CD PD ⊥,由已知得4PD === 所以三角形PCD 的面积为：11S 4242=⨯⨯=， 所以11433A PCD V S h h -== 依题PA 为三棱锥P ACD -的高，所以三棱锥P ACD -的体积为：11122332P ACD ACD V PA S -∆=⋅=⨯⨯⨯=又因为P ACD A PCD V V --=，所以43h =h =所以点A 到平面PCD 21.（Ⅰ）由.2222a c b cosB ac+-=，得12cosB =， 所以3B π＝；（Ⅱ）由（Ⅰ）得1602ABC S acsin =︒V =（Ⅲ）由题意得23sinA sinC sinA sin A π⎛⎫+=+- ⎪⎝⎭ 322sinA =+6A π⎛⎫=+ ⎪⎝⎭. 因为0＜A ＜23π，所以3336sin Aπ⎛⎫<+≤⎪⎝⎭.故所求的取值范围是332⎛⎤⎥⎝，.22.（1）取PB的中点E，连接EA，EN，在△PBC中，EN//BC且，又，AD//BC，AD＝BC所以EN//AM，，EN＝AM.所以四边形ENMA是平行四边形，所以MN//AE. 又，，所以MN//平面PAB. （2）过点A作PM的垂线，垂足为H，因为平面PMC⊥平面PAD，平面PMC∩平面PAD＝PM，AH⊥PM，所以AH⊥平面PMC，又所以AH⊥CM.因为PA⊥平面ABCD，所以PA⊥CM.因为PA∩AH＝A，所以CM⊥平面PAD.又所以CM⊥AD.。
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2019-2020学年度第四学段考试高一化学试卷可能用到的相对原子质量：H 1 C 12 N 14 O 16 S 32 Fe 56 Mn 55 Cu 64 一、单项选择题1.最近日本科学家确认世界还存在着另一种“分子足球”60N ，它与60C 分子结构相似，60N 在高温和撞击后会释放出巨大能量，下列对60N 的说法不正确．．．的是（ ） A. 60N 高温和撞击后应该生成2N B. 60N 和2N 都是氮的同素异形体 C. 60N 中氮原子之间由共价键结合 D. 60N 和14N 都是氮的同位素【答案】D 【解析】【详解】A ．由信息N 60在高温和撞击后会释放出巨大能量，转化为稳定物质氮气，则N 60高温和撞击后应该生成N 2，故A 正确；B ．N 60和N 2都是单质，且均只含N 元素，则均为氮的同素异形体，故B 正确；C ．N 60中只有一种非金属元素N ，氮原子之间由共价键结合，故C 正确；D ．N 60为单质，14N 为原子，二者不是同位素关系，故D 错误； 故选D 。
2.下列说法正确的是（ ）A. 双原子分子中的共价键一定是非极性键B. 任何原子的原子核都是由质子和中子构成的C. 非极性键可能存在于离子化合物中D. 质子数相同的微粒，它们的化学性质基本相同 【答案】C 【解析】【详解】A. 同种元素形成双原子分子中的共价键，一定是非极性键，而不同种元素形成的双原子分子如HCl 中的共价键，是极性键，故A 错误； B. H 原子中没有中子，故B 错误；C. 非极性键可能存在于离子化合物中如Na 2O 2，故C 正确；D. H 2O 和HF 质子数相同，但化学性质不同，故D 错误；故答案为C。
3.下列有关化学用语表示正确的是( )A. 甲烷的电子式B. 乙烷的结构简式：C2H6C. CH4分子的球棍模型：CD. 核内有8个中子的碳原子：86【答案】A【解析】【详解】A．甲烷为共价化合物，C最外层形成8个电子，电子式为，故A正确；B．乙烷的分子式为C2H6，其结构简式为CH3CH3，故B错误；C．为比例模型，甲烷分子的球棍模型为，故C错误；C，D．碳原子核内有6个质子，则核内有8个中子的碳原子的质量数为14，其核素表示为146故D错误；故答案为A。
4.A、B是两种前20号元素，A比B的核电荷数少5个，B原子M层比L层电子少2个，则A、B分别是( )。
A. Li、NaB. Na、SC. Na、KD. K、Br 【答案】B【解析】【详解】B原子的M层比L层少2个电子，则其M层电子数=8-2=6，故B为S元素，A比B的核电荷数少5个，应为11号元素，为Na元素。
故答案为B。
5.如图所示是部分短周期元素原子半径与原子序数的关系，则下列说法正确的是（）A. Z 、N 两种元素形成的化合物是离子化合物B. 通过M 、N 含氧酸的酸性强弱，可以比较两者的非金属性C. Z 的氧化物能分别溶解于Y 的氢氧化物和N 的氢化物的水溶液中D. 由X 与M 两种元素组成的化合物能与水反应 【答案】C 【解析】 【分析】同周期主族元素自左而右原子半径减小，同主族自上而下原子半径增大，故前7种元素处于第二周期，后7种元素处于第三周期，结合原子序数可知，X 为O 元素，Y 为Na 元素，Z 为Al 元素，M 为Si 元素，N 为Cl 元素，以此解答该题。
【详解】A ．Z 、N 两种元素形成的化合物为氯化铝，为共价化合物，故A 错误；B ．比较非金属性强弱关系时需要比较最高价含氧酸的酸性，含氧酸的描述不准确，故B 错误；C ．Z 的氧化物为氧化铝，是两性氧化物，能与Y 的氢氧化物氢氧化钠反应生成偏铝酸钠，能与N 的氢化物水溶液盐酸反应生成氯化铝，故C 正确；D ．由X 与M 两种元素组成的化合物SiO 2不与水反应，故D 错误； 故答案为C 。
6.下列叙述错误的是（ ）A. 13C 和14C 属于同一种元素，它们互为同位素 B. 6Li 和7Li 的电子数相等，中子数也相等 C. 14C 和14N 质量数相等，中子数不相等 D. 1mol23592U的中子数比1mol23892U的中子数少3N A 个【答案】B 【解析】A 、二者均为碳元素的原子，互为同位素，A 正确；B 、二者的中子数分别为3、4，B 错误；C 、二者的中子数分别为8、7，C 正确；D 、由N=A -Z 可知1 mol 23592U 的中子数比1 mol 23892U 的中子数少3N A 个,D 正确。
答案选B 。
7.在常温常压下断裂1 mol 指定共价键时吸收的能量如下表：有关CH 4与Cl 2反应的下列说法正确的是 A. 1mol 甲烷生成CCl 4最多消耗2mol 氯气 B. 反应只生成1molCH 3Cl 时放热109.6kJC. 反应的有机产物常温下都是液态D. 该反应为放热反应，无需提供能量即可实现 【答案】B 【解析】【详解】A ．甲烷中的四个氢原子均可以被Cl 原子取代，同时生成HCl ，1 mol 甲烷中有4mol 氢原子，和氯气光照充分反应，最多消耗4mol 氯气，故A 错误；B ．反应只生成1molCH 3Cl 时，断裂1molC-H 键和1molCl-Cl 键，吸收(413.4+243)kJ 能量，生成1molC-Cl 键和1molH-Cl 键，释放(330+436)kJ 能量，所以放出的能量为(766-656.4)kJ=109.6kJ ，故B 正确；C ．CH 3Cl 在常温下为气体，故C 错误；D ．该反应需要在光照条件下进行，需要提供能量，故D 错误； 故答案为B 。
8.在一个密闭容器中发生反应：2SO 2(g)+O 2(g)2SO 3(g)。
已知反应过程中的某一时间SO 2、O 2和SO 3的浓度分别为：0.2mol/L 、0.1mol/L 和0.2mol/L 。
当反应达到平衡时可能存在的数据正确的是 A. c (O 2)＝0.2mol/L B. c (SO 2)＝0.25mol/L C. c (SO 3)＝0.4mol/L D. c (SO 2)＝c (SO 3)＝0.15mol/L【答案】B 【解析】可逆反应有一定的限度，反应物不能完全转化为生成物，反应过程中的某一时间SO2、O2和SO3的浓度分别为：0.2mol/L、0.1mol/L和0.2mol/L，若极限转化，则SO2、O2和SO3的浓度分别为：0.4mol/L、0.2mol/L和0mol/L或0mol/L、0mol/L和0.4mol/L。
【详解】A．极限转化时c(O2)＝0.2mol/L，但可逆反应不能完全转化，所以c(O2)<0.2mol/L，故A错误；B．根据分析可知0＜c(SO2)＜0.4mol∙L-1 ，故B正确；C．极限转化时c(SO3)＝0.4mol/L，但可逆反应不能完全转化，所以c(SO3)<0.4mol/L，故C错误；D．根据硫元素守恒，应有c(SO2)+c(SO3)=0.4mol•L-1，所以不可能出现c(SO2)＝c(SO3)＝0.15mol/L，故D错误；故答案为B。
9.对于可逆反应 A(g)＋3B(s) 2C(g)＋2D(g)，在不同条件下的化学反应速率如下，其中表示的反应速率最快的是A. v(A)＝0.5 mol·L－1·min－1B. v(B)＝1.2 mol·L－1·s－1C. v(C)＝0.1 mol·L－1·s－1D. v(D)＝0.4 mol·L－1·min－1【答案】C【解析】【详解】A．v(A)=0.5mol/(L•min)；B．B为固体，不能用B表示反应速率；C．v(A)=12v(C)=0.05mol/(L•s)=3mol/(L•min)；D．v(A)=12v(D)=0.2mol/(L•min)；根据分析可知，反应速率最快的为C，故选C。
10.分子式为410C H的烃的一氯代物有（）。
A. 3种B. 4种C. 5种D. 6种【答案】B【详解】分子式为410C H 的烃有两种同分异构体：3223CH CH CH CH (正丁烷)和(异丁烷)，正丁烷和异丁烷各有2种不同化学环境的氢原子，故410C H 的一氯取代物有4种，故选B 。
11.下列说法正确的是（ ）A. CHCl 3只有一种结构，说明甲烷是以碳原子为中心的正四面体结构B. 甲烷是造成“光化学烟雾”的气体C. 标准状况下，11.2LCCl 4中含有C —Cl 键的数目为2N AD. 在点燃甲烷气体之前必须检验其纯度 【答案】D 【解析】【详解】A ．若甲烷为平面正方形结构，CHCl 3也只一种结构，所以不能说明甲烷是正四面体结构，CH 2Cl 2只有一种结构可以说明，故A 错误；B ．氮氧化物是造成“光化学烟雾”的气体，而不是甲烷，故B 错误；C ．标况下四氯化碳不是气体，11.2L 四氯化碳的物质的量不是0.5mol ，故C 错误；D ．甲烷为易燃气体，若不纯，点燃时容易爆炸，所以在点燃甲烷气体之前必须检验其纯度，故D 正确； 故答案为D 。
12.利用含碳化合物合成燃料是解决能源危机的重要方法，已知CO(g)＋2H2(g)CH 3OH(g)反应过程中的能量变化情况如图所示，曲线Ⅰ和曲线Ⅱ分别表示不使用催化剂和使用催化剂的两种情况。
下列判断正确的是（ ）A. 生成1mol CH 3OH(g)时，该反应吸收的热量为91 kJ 。
B. 加入催化剂可以减少反应最终放出的热量C. 1mol CO(g)与2molH 2(g)的总能量大于1mol CH 3OH(g)的总能量D. 如果该反应生成CH 3OH (l)，则放出的能量会减少 【答案】C 【解析】 【分析】催化剂可以降低反应活化能，所以曲线I 为不使用催化剂的能量变化，曲线II 为使用催化剂的能量变化。
【详解】A ．该反应的反应物能量高于生成物能量，为放热反应，根据图中数据可知，生成1mol CH 3OH(g)时，该反应释放的能量为91 kJ ，故A 错误； B ．催化剂只改变活化能，不影响反应的热效应，故B 错误；C ．据图可知，反应物的能量高于生成物，即1mol CO(g)与2molH 2(g)的总能量大于1mol CH 3OH(g)的总能量，故C 正确；D ．CH 3OH 由气态变为液体需要放出能量，所以如果生成CH 3OH (l)，则放出的能量会增多，故D 错误； 故答案为C 。
13.有a 、b 、c 、d 四个金属电极，有关的实验装置及部分实验现象如下：实验装置部分实验现象 a 极质量减小，b 极质量增大b 极有气体产生，c 极无变化d 极溶解，c 极有气体产生 电流从a 极流向d 极由此可判断这四种金属的活动性顺序是( ) A. a>b>c>d B. b>c>d>a C. d>a>b>c D. a>b>d>c【答案】C 【解析】【详解】装置一是原电池，a 极质量减小，说明a 极金属易失去电子形成离子，故a 极金属比b 极金属活泼；装置二没有形成原电池，可知b 比c 活泼，且c 位于金属活动性顺序表中氢后面；装置三和四均形成原电池，易知d比c活泼，d比a活泼。