河南省周口市沈丘县县直高中2014-2015学年高一上学期第一次月考化学试卷
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2022年河南省周口市沈丘县第一中学高一化学月考试题含解析一、单选题(本大题共15个小题,每小题4分。
在每小题给出的四个选项中,只有一项符合题目要求,共60分。
)1. 五种短周期元素A、B、C、D、E的原子序数依次增大,A和C同族,B和D同族,C离子和B离子具有相同的电子层结构。
A和B、D、E均能形成共价型化合物。
A和B形成的气态化合物在水中呈碱性,C和E形成化合物CE。
下列说法不正确的是()A. B离子的半径大于C离子的半径B. 将C的单质放入CuSO4溶液中会有蓝色沉淀产生C. E的最高价氧化物对应水化物的化学式为HEO4D. A和B、D、E均能形成共价化合物,其中稳定性最弱的是A、E形成的化合物参考答案:D2. 黄铁矿(FeS2)中-1价硫元素在酸性条件下可以与ClO3-发生反应,其离子方程式如下_____ClO3-+_____ FeS2+_____H+= _____ClO2↑+ _____Fe3++ _____ SO42- + _____H2O(未配平):下列说法正确的是A. ClO3-是氧化剂,Fe3+是还原产物B. ClO3-的氧化性强于Fe3+C. 生成lmol SO42-时,转移7mol 电子D. 向反应后的溶液加KSCN 溶液,可观察到有红色沉淀产生参考答案:B 点睛:明确解题方法和思路、理解概念抓实质,解题应用靠特征,即从氧化还原反应的实质——电子转移,去分析理解有关的概念,而在实际解题过程中,应从分析元素化合价变化这一氧化还原反应的特征入手。
本题中需要注意的是反应中失去电子的元素是两种,即S和Fe,尤其是选项C中电子转移计算时容易忽视亚铁离子也失去电子,为易错点。
3. 将15 mL 2 mol/L Na2CO3溶液逐滴加入到40 mL 0.5 mol/L MCl n盐溶液中,恰好将溶液中的M n+离子完全沉淀为碳酸盐,则MCl n中n值是()A.4 B.3 C.2 D.1参考答案:B略4. 下列叙述中正确的是A.虽然Cu能导电,但Cu是非电解质B.BaSO4虽然难溶于水,但BaSO4是电解质C.SO3溶于水能导电,所以SO3是电解质D.NaNO3溶液能导电,故NaNO3溶液是电解质参考答案:B略5. 氯元素在自然界有35Cl 和37Cl 两种同位素,在计算式34.969×75.77%+36.966×24.23% =35.453中A 、75.77%表示35Cl 的质量分数B 、24.23%表示35Cl 的丰度C 、35. 453表示氯元素的相对原子质量D 、36.966表示37Cl 的质量数 参考答案:C略6. 下列各组物质,不能按右图(“→”表示反应一步完成)关系转化的是选项 a b c A S SO 3 H 2SO 4 B SiO 2 Na 2SiO 3 H 2SiO 3 C AlCl 3 NaAlO 2 Al(OH)3 DCuCuOCu(OH)2AD7. 下列操作中,完全正确的一组是( )①用试管夹夹持试管时,试管夹从试管底部往上套,夹在试管的中上部②给盛有液体的体积超过1/3容积的试管加热 ③把鼻孔靠近容器口去闻气体的气味④将试管平放,用纸槽往试管里送入固体粉末后,然后竖立试管⑤取用放在细口瓶中的液体时,取下瓶塞倒放在桌面上,倾倒液体时,瓶上的标签对着地面 ⑥将烧瓶放在桌上,用力塞紧塞子⑦用坩埚钳夹取加热后的蒸发皿 ⑧将滴管垂直伸进试管内滴加液体⑨稀释浓硫酸时,把水迅速倒入盛有浓硫酸的量筒中⑩检验装置的气密性时,把导管的一端浸入水中,用手捂住容器的外壁或用酒精灯微热A .①④⑤⑦⑩B .①④⑦⑩C .①④⑤⑦⑧⑩D .④⑤⑦⑧⑩参考答案:B略8. 下列物质中,能用作漂白剂的是( )A .H 2SO 4(浓)B .NaOHC .NaClOD .CaCl 2参考答案:C解:A .浓硫酸具有强氧化性,但不能做漂白剂,通常作吸水剂和脱水剂,故A 不选; B .氢氧化钠具有强氧化性,但是没有漂白性,不作漂白剂,故B 不选; C .次氯酸钠在水中会生成次氯酸,次氯酸具有漂白性,常用作漂白剂,故C 选; D .氯化钙不具有漂白剂,实验室常用作干燥剂,故D 不选。
县直高中高一英语上学期第一次月考第I卷(三部分,共85分)第一部分:阅读理解(共20小题;每小题2分,满分40分)阅读下列短文,从每题所给的选项中,选出最佳选项。
ALao She wrote Teahouse in 1957. The play shows the audience life in China between 1898 and 1945. It takes place in a teahouse in old Beijing and it tells us the story of Wang Life and his customers.It asks us to see the teahouse as the centre of the neighborhood. Finally, it says goodbye old Beijing and its people.The story starts in 1898 during the Qing Dynasty. It continues in 1916, and finally, it brings the audience to the end of the Anti-Japanese War in 1945.After the war, Wang loses the teahouse and he dies.Lao She was born in Beijing in 1899. His parents sent him to the Teahouse’s School in Beijing and he learned to teach. Form 1924 to 1929 he taught Chinese to the English in London. He wrote many plays, novels and short stories. He was named a “People’s Artist” and a “Great Master of Language”. He was one of the greatest Chinese writers of the 20th century.In Lao She’s Teahouse today, waiters brin g tea to the customers and sell them delicious Chinese food. If you like Beijing Opera, folk music, acrobatics or magic shows, you can enjoy them in the teahouse. Lao She’s Teahouse gives a wonderful welcome to everyone from China and from all over the world.1. What’s the main idea of Paragraph 4?A. Lao She.B. Lao She’s Teahouse.C. The story of Teahouse.D. The play Teahouse.2. How many years of life in China is shown in the play Teahouse?A. 57 years.B. 45 years.C. 98 years.D. 47 years.3. Lao She was all of the following except __________.A. a boss of TeahouseB. a people’s ArtistC. a Great Master of LanguageD. one of the greatest Chinese writers of the 20th century4. Who we re welcome to Lao She’s Teahouse today?A. Only Chinese customers.B. Only foreign customers.C. Every customer from home and abroad .D. Only Beijing opera audience.BPeople all over the world celebrate the New Year. However, not all countries celebrate in the same way, and in some countries, the New Year doesn’t begin on the same date every year. It can depend on the seasons, the moon or the sun.In many countries, the New Year begins on 1st January, but people start celebrating on 31st December, New Year’s Eve. In New York many people go to celebrate in Time Square. While they’re waiting for the New Year, they listen to music, sing traditional songs and have fun. Just before 12 o’clock, everyone counts down from 10: 10, 9, 8 …As soon as it’s 12 o’clock, everyone shouts very loudly, “Happy New Year!” and they say, "Out with the old year and in with the new!" Suddenly there are fireworks and special drinks. Sometimes people sing an old song called Auld Lang Syne.New Year’s Day is often a family day. Some people relax at home. Some families get together for a special meal with grandparents, aunts, uncles and cousins. When the weather is fine, many families go out for a walk.On New Year’s Day, many people make resolutio ns for the New Year. They write down a list of things, such as “I will help out more at home. I will work harder at school.” or “I won’t spend so much time playing video games.” when they have made their list, they read it to their family or friends and promise to follow their resolutions.5. Where do many people in New York spend the New Year’s Eve?A. In Times Square.B. At home.C. Outside their homes.D. In their grandparents’ home.6. What do peo ple do just before 12 o’clock on New Year’s Eve in New York?A. Shout loudly.B. Listen to music.C. Sing traditional songs.D. Count down from 10: 10, 9, 8, …7. The underlined phrase “ out with” in the passage means__________.A. Goodbye toB. Hello toC. WelcomeD. Evening to8. Why do people read their list of resolutions to their family or friends on New Year’s Day?A. Because they want to tell their family or friends their resolutions.B. Because they are going to play video games.C. Because they will help out more at school.D. Because they promise to carry out the plans.CFor the first time, Zhang Shengyu, a student from Laiyang Middle School in Shandong, didn’t ask her mother to buy her a new schoolbag or any new pens for the beginning of the school term. “My old schoolbag can still be used, so there is no need to buy a new one,” Zhang said, “I’ve decided to be thrifty from now on.”As the c enter of the family’s attention, many of the one-child generation (一代人) do not know about the hardships of life. They sometimes have no idea of the value of things. Many students would throw away pens and erasers that haven’t been used up and buy new ones. Some like to compete with each other to see whose clothes and school things are more expensive.On its website in January, the Ministry of Education called on students to form good habits such as working hard, being helpful and thrifty, to avoid wasting food and to pay more attentionto study and others.Many schools consider teaching the value of money to be an important lesson and encourage their students to be thrifty.Students at No. 44 Middle School of Shijiazhuang, Hebei planted many fruit trees in their school. Through taking care of the trees, the students learned how to value things.At Tie Yi Middle School in Xi’an, students organized a group to collect waste and reuse them. “What we students do are small things, but I think with everyone’s work, we can make a difference,” said Li Dongdong, 16, a student f rom the group.9. The underlined word “thrifty” in Paragraph 1 means ______.A. giving money to the poorB. spending lots of moneyC. spending no money at allD. using money carefully and wisely10. In January, the Ministry of Education called on students to do the followings EXCEPT _________.A. working hardB. being thriftyC. helping othersD. protecting environment11. We can conclude that Li Dongdong ____________.A. took care of treesB. planted fruit treesC. collected wasteD. didn’t ask his mother for a new school bag12. What is the best title of the passage?A. School activitiesB. Waste not, want notC. Make a differenceD. The problems of one-child generationDHaving a teenager at home can be a real headache. Many teenagers smoke, drink and even fight. They often do something dangerous.Earlier studies have shown that teenagers are more likely (可能) to make irrational decisions than people from any other age group, including children and adults. Is it that teenagers are too young to tell right from wrong? Not really. So what’s the reason?According to Laurence, a teacher from a US university, the reason is that teenagers care a lot about how their peers (同龄人) view them—that is “peer influence”.As children enter their teenage years, they spend more time with their friends and classmates, and also they care more about what they think of them. This makes teenagers make decisions without thinking about the costs.In a test, a group of teenagers were asked to play a video driving game. When they played with their friends watching around them, they took more chances and drove more carelessly because that would increase their possibility of winning. But when they played alone, they drove more safely. Why do peers have such a big influence on teens’ behavior?As Laurence sees it, a teenager’s brain is like a car with a good accelerator (油门) but a weak brake (刹车). The “accelerator” is fully developed by teenage years while the “brake” is still not fully developed. When teenagers are watched by their peers, they usually push hard on the accelerator. With their weak brake, it is likely that they are going to end up in an accident. But the good news is that a violent (暴力的)teenager doesn’t necessa rily become a violent adult. About two-thirds to three-quarters of violent youth grow out of it. “They get more self-controlled.”13. Why are teenagers likely to make wrong decisions according to the text?A. They like difficult tasks.B. They don’t care about costs.C. They are too young to tell right from wrong.D. They care a lot about what their peers think of them.14. How does Laurence explain the influence of peer pressure on teens’ behavior?A. By comparing it with how a car works.B. By comparing it with an accident.C. By showing research findings.D. By giving examples.15. What can we conclude from the text?A. Teenagers are easy to have traffic accidents.B. Peer influence is bad on teenagers.C. Most of the teenagers become less violent when they grow up.D. Teenagers shouldn’t spend all the free time with their friends.ETaking good notes is a time-saving skill that will help you to become a better student in several ways, 16 Second, your notes are excellent materials to refer to (参考)when you are studying for a test. Third, note-taking offers variety to your study time and helps you to hold your interest.You will want to take notes during classroom discussions and while reading a textbook or doing research for a report.__17_____Whenever or however you take notes, keep in mind that note-taking is a selective process. 18The following methods may work best for you.●Read the text quickly to find the main facts and ideas in it.●Carefully read the text and watch for words that can show main points and supporting facts.● Write your notes in your own words.●____19__● Note any questions or ideas you may have about what was said or written.As you take notes, you may want to use your own shorthand(速记). When you do, be sure that you understand your symbols(符号) and that you use them all the time. 20A. Use words, not complete sentences.B. There are three practical note-taking methods.C. You must write your notes on separate paper.D. Otherwise, you may not be able to read your notes later.E. you will also want to develop your own method for taking notes.F. That means you must first decide what is important enough to include in your notes.G. First, the simple act of writing something down makes it easier for you to understand and remember it.第二节:完形填空(共20小题;每小题1.5分,满分30分)阅读下面短文,从短文后各题所给的四个选项(A、B、C和D)中,选出可以填入空白处的最佳选项。
沈丘县县直高中2014-2015学年第一学期高一必修一第一章化学测试卷(人教版)时间:90分钟,分数:100分原子量:H—1 O---16 K---39 Cl---35.5 N—14 S—32 C--12 一,选择题(3x20=60分)1.下列操作或发生事故时的处理方法正确的是( )A.过滤时,用玻璃棒搅动漏斗内液体,使之过滤完全B.燃着的酒精灯打翻失火,应立即用水浇灭C.配制一定浓度的NaCl溶液时,定容振荡摇匀后,发现液面低于刻度线,再加水到刻度线D.浓硫酸不慎溅到皮肤上,应立即用大量水冲洗,最后再涂上3%~5%的小苏打溶液2.下列说法中正确的是( )A.KOH的摩尔质量是56 gB.1 mol H2O的质量是18 g/molC.Cl2的摩尔质量是35.5 g/molD.1 mol N2的质量是28 g3.绿色植物是空气天然的“净化器”,研究发现,1万平方米柳杉每月可以吸收160 kg SO2,则1万平方米柳杉每月吸收的SO2的物质的量为( )A.164 kg B.2.5 molC.2 500 mol D.2 500 g/mol4.下列实验操作正确的是( )5.下列各组物质中分子数相同的是( )A.2 L CO和2 L CO2B.9 g H2O和标准状况下11.2 L CO2C.标准状况下1 mol O2和22.4 L H2OD.0.2 mol H2和4.48 L HCl气体6.用N A代表阿伏加德罗常数的值,下列说法正确的是( )A.0.5 mol锌粒与足量盐酸反应产生11.2 L H2B.标准状况下,11.2 L CCl4所含分子数为0.5N AC.0.1 mol CH4所含氢原子数为0.4N AD.常温常压下,28 g N2中所含原子个数为N A7.在相同的温度和压强下,4个容器中分别装有4种气体。
已知各容器中的气体和容器的容积分别是a.CO2,100 mL;b.O2,200 mL;c.N2,400 mL;d.CH4,600 mL,则4个容器中气体的质量由大到小的顺序是( )A.a>b>c>d B.b>a>d>cC.c>d>b>a D.d>c>a>b8.为除去某物质中所含的杂质,所选用的试剂或操作方法正确的是( )A.①②③ B.②③④C.①③④D.①②④9.下列叙述中正确的是()A.摩尔是物质的量的单位,每摩尔物质含有6.02×1023个分子B.1 mol氧的质量为16 gC.0.5 mol He约含有6.02×1023个电子D.2H既可表示2个氢原子又可表示2m o l氢分子10.在0.5 mol Na2SO4中,含有Na+数约是()A.6.02×1023B.3.01×1023 C.0.5 D.111.相等物质的量的CO和CO2的下列比较中正确的是()①所含的分子数目之比为1:1 ②所含的O原子数目之比为1:2③所含的原子总数目之比为2:3 ④所含的C原子数目之比为1:1 ⑤所含的电子数目之比为7:11A.①②B.②③C.④⑤D.①②③④⑤12.现有下列四种因素:①温度和压强②所含微粒数③微粒本身大小④微粒间的距离,其中对气体物质体积有显著影响的是() A.①②③B.①③④C.①②④D.①②③④13.只给出表中甲和乙对应的量,不能用来求物质的量的是()14.的原子数相等,则这两个容器的体积之比是()A.2:1 B.1:2 C.2:3 D.1:315.下列说法中正确的是()A.32 g O2占有的体积约为22.4 LB.22.4 L N2含阿伏加德罗常数个氮分子C.在标准状况下,22.4 L水的质量约为18 gD.22 g二氧化碳与标准状况下11.2 L HCl含有相同的分子数16.下列叙述正确的是()A.同温同压下,相同体积的物质,其物质的量一定相等B.任何条件下,等物质的量的甲烷和一氧化碳所含的分子数一定相等C.1 L一氧化碳气体一定比1 L氧气的质量小D.相同条件下的一氧化碳气体和氮气,若物质的量相等则体积相等,但质量不相等17.有一种气体的质量是14.2 g,体积是4.48 L(标准状况),该气体的摩尔质量是()A.28.4 B.28.4 g·mol-1 C.71 D.71 g·mol-118.实验室使用的浓盐酸的溶质质量分数为36.5%,密度为1.19 g·cm-3,则物质的量浓度是()A.11.9 B.11.9 mol·L-1 C.0.012 mol·L-1D.无法计算19.实验室需用2mol·L-1氯化钠溶液450 mL,配置时应选用容量瓶的规格和称取氯化钠的质量分别是()A.450 mL,52.7 g B.500 mL,58.5 gC.1000 mL,117 g D.任意规格,111.2 g20.配制一定物质的量浓度的某溶液,下列情况会使配制结果偏低的是()A.未冷却即转移、定容B.未洗烧杯及玻璃棒C.定容时俯视刻度线观察液面D.容量瓶中原有少量蒸馏水二,填空题:(每空1分,共40分)21.实验室欲用NaOH固体配制1.0 mol/L的NaOH溶液480 mL:(1)配制时,必须使用的玻璃仪器有________、________、________、________。
河南省沈丘县县直高级中学2013-2014学年高一化学上学期期中试题新人教版一、选择题(本题包括20小题,每小题3分,共60分。
每题只有一个选项符合或最符合题意)1、下列反应一定属于氧化还原反应的是()A.化合反应 B.分解反应C.置换反应 D.复分解反应2、下列实验操作均要用玻璃棒,其中玻璃棒的作用及其目的相同的是()①过滤②蒸发③溶解④向容量瓶转移液体A.①和② B.①和③C.③和④ D.①和④3、、下列电离方程式中,一定错误的是()A. Na2CO3 === Na2++CO32- B.NaHSO4 ==== Na+ +H+ +SO42-C. H2SO4 ==== 2H+ +SO42- D.KClO3 ===== K+ +ClO3 一4、用N A表示阿伏德罗常数,下列叙述正确的是()A.标准状况下,22.4LH2O含有的分子数为1 N AB.常温常压下,1.06g Na2CO3含有的Na+离子数为0.02 N AC.通常状况下,1 N A个CO2分子占有的体积为22.4LD.在0.5mol·/L的Mg SO4溶液中,Mg2+数目为0.5 N A5、在物质分类中,前者包括后者的是()A.氧化物、化合物 B.化合物、电解质C.溶液、胶体 D.溶液、分散系6、在某无色透明的酸性溶液中,能大量共存的离子组是()A.Na+ 、K+、SO42-、HCO3-B.Cu2+、Ba2+、NO3-、SO42-C.Na+、 K+、Cl-、 NO3-D.Fe3+、K+、SO42-、Cl-7、能正确表示下列化学反应的离子方程式的是()A.氢氧化钡溶液与硫酸的反应 OH-+H+= H2OB.澄清的石灰水与稀盐酸反应 Ca(OH)2 + 2H+ = Ca2+ + 2H2OC.铜片插入硝酸银溶液中 Cu + Ag+ = Cu2+ + AgD.碳酸钙溶于稀盐酸中 CaCO3+2H+=Ca2++H2O+CO2↑8.下列叙述正确的是()A.1 mol H2O的质量为18g/molB.CH4的摩尔质量为16gC.3.01×1023个SO2分子的质量为32gD.标准状况下,气体的摩尔体积是22.4L9、小明体检的血液化验单中,葡萄糖为 5.9×10-3mol/L。
河南省周口市沈丘县实验中学高一化学月考试卷含解析一、单选题(本大题共15个小题,每小题4分。
在每小题给出的四个选项中,只有一项符合题目要求,共60分。
)1. 不能用来表示其结构的微粒是A.Ne B.F- C.Al3+ D.S2-参考答案:D略2. 下列物质中属于电解质的是A. CO2B. K2SO4C.石墨D. NaOH溶液参考答案:B略3. 已知X元素的原子核外电子数为n,X2-离子和Y3+离子的核外电子排布相同,则Y原子的质子数为()A.n+1 B.n+2C.n+4 D.n+5参考答案:D解析Y3+的电子数为n+2,则Y的质子数=Y的电子数=n+5。
4. 在无色强酸性溶液中,下列各组离子能够大量共存的是A. Na+、Ba2+、NO3-、SO42-B. NH4+、HCO3-、Cl-、K+C. Cl -、K+、NO3-、NH4+D. Al3+、Cu2+、Cl-、SO42-参考答案:C 【分析】强酸性溶液,含大量的H+,根据离子之间不能结合生成水、气体、沉淀,不能发生氧化还原反应等,则离子能大量共存,并结合离子的颜色来解答。
【详解】A. Ba2+与SO42-会反应生成沉淀,不能大量共存,A项错误;B. 酸性条件下,HCO3?不能大量共存,发生复分解反应生成二氧化碳和水,B项错误;C. 酸性条件下该组离子不会发生反应,离子共存,C项正确;D. Cu2+为蓝色,不符合溶液无色的要求,D项错误;答案选C。
【点睛】解答离子共存问题时需要注意题设中的限定条件,如无色透明,则常见的有颜色的离子如Cu2+、Fe3+、Fe2+、MnO4-、Cr2O72-、CrO42-等不符合题意,本题D项便可直接排除;还有一些限定条件如:常温下与Al反应生成氢气的溶液时,该溶液可能为酸溶液,也可能为碱溶液。
解题时只要多加留意,细心严谨,便可快速选出正确答案。
5. 下列每组物质发生状态变化所克服的粒子间的相互作用属于同种类型的是A.食盐和蔗糖熔化B.钠和硫熔化C.碘和干冰升华D.二氧化硅和氧化钠熔化参考答案:C6. 以“物质的量”为中心的计算是化学计算的基础,下列与“物质的量”相关的计算正确的是A.现有等物质的量的CO和CO2,这两种气体中氧原子的质量之比为1∶2B.5.6 g CO和22.4 L CO2中含有的碳原子数一定相等C.将20 g NaOH固体溶解在500 mL水中所得溶液的物质的量浓度为1 mol·L-1D.a g Cl2中有b个氯原子,则阿伏加德罗常数N A的数值可表示为71b/a参考答案:AA. 设CO和CO2的物质的量均为1mol,这两种气体中氧原子的质量之比为1×1×16∶1×2×16=1:2,A 正确;B.没有给定气体的存在状态,无法确定22.4 L CO2的物质的量,也就无法比较两种气体中含有的碳原子数是否相等,B错误;C. 20 g NaOH固体物质的量为0.5mol,溶剂水的体积为500 mL,溶液的体积不等于500 mL,所以所得溶液的物质的量浓度不是1 mol·L-1,C错误;D. 根据公式:n=m/M=N/N A可知,N A×2×a/71=b,所以N A=71b/2a,D错误;综上所述,本题选A。
沈丘县高级中学2018-2019学年高二上学期第一次月考试卷化学班级__________ 座号_____ 姓名__________ 分数__________一、选择题的是()1.【2017届浙江省绍兴市高三上学期适应性考试】下列说法不正确...A.硬脂酸甘油酯在NaOH溶液中水解完全后,加入饱和食盐水,下层析出硬脂酸钠B.天然高分子化合物淀粉或纤维素最终的水解产物都是葡萄糖C.鸡蛋清的溶液中加入福尔马林溶液,鸡蛋清凝聚,蛋白质变性D.不同种类的氨基酸能以不同的数目和顺序彼此结合,形成更复杂的多肽化合物2.织物漂白剂亚氯酸钠(NaClO 2)在溶液中可生成ClO2、HClO2、ClO、Cl-等,其中HClO2和ClO2都具有漂白作用,但ClO2是有毒气体。
25 ℃时,各组分含量随pH变化情况如图所示(Cl-没有画出)。
下列说法错误的是A.25 ℃时,HClO2的电离平衡常数的数值K a=10-6B.使用该漂白剂的最佳pH为3.0C.25 ℃时,等浓度的HClO2溶液和NaClO2溶液等体积混合后,混合溶液中:c(HClO2)+2c(H+)=c(ClO)+2c(OH-)D.该温度下的NaClO 2溶液中c(Na+)>c(ClO)>c(OH-)>c(H+)3.下列物质仅能在水溶液中导电的是A.HC1 B.NaHSO3C.NaOH D.CH3CH2OH4.下列叙述正确的是A.25℃时,用醋酸溶液滴加等浓度NaOH溶液至pH=7,V(CH3COOH)>V(NaOH)NH)B.常温下pH=2 的盐酸和pH=12 的氨水等体积混合后c(Cl-)>c(+4C.CH3COONH4溶液显中性,所以CH3COONH4对水的电离不影响D.常温下pH=12氨水稀释10倍后,溶液的pH=115.室温下,用相同物质的量浓度的HCl溶液,分别滴定物质的量浓度均为0.1mol·L-1的三种碱(AOH、BOH和DOH)溶液,滴定的曲线如图所示,下列判断正确的是A.滴定时,随着盐酸溶液的滴入,水电离出的c(H+)始终增大B.滴定至P点时,溶液中:c(Cl-)>c(B+)>c(BOH)>c(OH-)>c(H+)C.pH=7时,三种溶液中c(Cl-)相等D.当中和百分数达100%时,将三种溶液混合后:c(AOH)+c(BOH)+c(DOH)=c(H+)-c(OH-)6.【2017届湖北省恩施一中高三上学期开学考试】两种气态烃的混合气1 L,完全燃烧后,生成2 LCO2和2 L水蒸气(相同状况),则该混合烃的组合及其体积比是()A.CH4和C3H6,体积比2∶1 B.C3H8和C2H4,体积比2∶1C.C2H6和C2H2,体积比1∶1 D.C2H6和C2H2,体积比2∶17.下列说法正确的是A.碘单质的升华过程中,只需克服分子间作用力B.NH4Cl属于离子化合物,该物质中只存在离子键C.在N2、CO2和SiO2物质中,都存在共价键,它们都是由分子构成D.金刚石和足球烯(C60)均为原子晶体8.下列说法正确的是A.瘦肉精学名盐酸克伦特罗,结构如右图,可以发生取代、加成、水解、酯化、消去反应B.C3H7OH与C4H9OH分子间脱水,最多可生成6种醚C.1mol最多可与含4mol NaOH的水溶液完全反应D.某油脂的化学式为,它既属于酯类又属于高分子化合物9.关于下列各图的叙述,正确的是A .甲表示H 2与O 2发生反应过程中的能量变化,则H 2的燃烧热为483.6 kJ·mol -1B .乙表示恒温恒容条件下发生的可逆反应2NO 2(g )N 2O 4(g )中,各物质的浓度与其消耗速率之间的关系,其中交点A 对应的状态为化学平衡状态C .丙表示A 、B 两物质的溶解度随温度变化情况,将A 、B 饱和溶液分别由t 1 ℃升温至t 2 ℃时,溶质的质量分数B =AD .丁表示常温下,稀释HA 、HB 两种酸的稀溶液时,溶液pH 随加水量的变化,则同浓度的NaA 溶液的pH 大于NaB 溶液10.混合动力汽车(HEV )中使用了镍氢电池,其工作原理如图所示:其中M 为储氢合金,MH 为吸附了氢原子的储氢合金,KOH 溶液作电解液。
河南省周口市沈丘县县直高级中学2014-2015学年高二上学期第一次月考物理试题考试时间:90分钟,分值 :100分一、选择题(本大题共12个小题,每小题4分,共48分) 1.关于元电荷下列说法正确的是( )A .元电荷实际上是指电子和质子本身B .所有带电体的电荷量一定等于元电荷的整数倍C .元电荷的值通常取作e = 1.60×10-19CD .电荷量e 的数值最早是由美国科学家密立根用实验测得的2.关于点电荷以下说法正确的是( )A .足够小的电荷就是点电荷B .一个电子,不论在任何情况下都可视为点电荷C .点电荷是一种理想化的模型D .一个带电体能否看成点电荷,不是看他尺寸的绝对值,而是看它的形状和尺寸对相互作用力的影响能否忽略不计 3.关于电场线的以下说法中,正确的是( )A .电场线上每一点的切线方向都跟电荷在该点的受力方向相同B .沿电场线的方向,电场强度越来越小C .电场线越密的地方同一检验电荷受的电场力就越大D .顺着电场线移动电荷,电荷受电场力大小一定不变4.一带电粒子射入固定在O 点的点电荷的电场中,粒子轨迹如图虚线abc 所示,图中实线 是同心圆弧,表示电场的等势面,不计重力,可以判断( )A .粒子受到静电排斥力的作用B .粒子速度a b v v >C .粒子动能kc ka E E =D .粒子电势能pc pbE E > 5.关于等势面的说法正确的是 ( )A .电荷在等势面上移动时不受电场力的作用,所以说电场力不做功B .等势面上各点的场强相等C .等势面一定跟电场线垂直D .两个等势面永不相交6.一个带正电的质点,电荷量q =2.0×10-9C ,在静电场中由a 点移动到b 点。
在这过程中 除电场力外,其它力做的功为6.0×10-5J ,质点的动能增加了8.0×10-5J ,则a 、b 两 间的电势差U AB 为( ) A .1×104 V B .-1×104 VC .4×104 VD .-7×104 V7.对公式d E U ⋅=的理解,下列说法正确的是( )A .在相同的距离上的两点,电势差大的其场强也必定大B .匀强电场中,沿着电场线方向相等的距离上的电势降落必定相等C .此公式适用于所有电场D .公式中的d 是通过两点的等势面间的距离8.某电容器上标有“1.5μF ,9V ”,则该电容器( )A .所带电荷量不能超过1.5×10-6CB .所带电荷量不能超过1.35×10-5CC .所加的电压不应超过9VD .该电容器的击穿电压为9V9.由电场强度的定义式qFE =可知,在电场中的同一点 ( )A .电场强度E 跟F 成正比,跟q 成反比B .无论点电荷所带的电量如何变化,qF始终不变 C .电场中某点的场强为零,则在该点的电荷受到的电场力一定为零D .一个带电的小球在P 点受到的电场力为零,则P 点的场强一定为零10.如图所示,电场中有A 、B 两点,则下列说法中正确的是( )A .电势B A ϕϕ>,场强B A E E > B .电势B A ϕϕ>,场强B A E E <C .将q +电荷从A 点移到B 点电场力做了正功D .将q -电荷分别放在A 、B 两点时具有的电势能PB PAE E >11.将平行板电容器两极板之间的距离、电压、电场强度大小和极板所带的电荷量分别用d 、U 、E 和Q表示.下列说法正确的是( )A .保持U 不变,将d 变为原来的两倍,则E 变为原来的一半B .保持E 不变,将d 变为原来的一半,则U 变为原来的两倍C .保持d 不变,将Q 变为原来的两倍,则U 变为原来的一半D .保持d 不变,将Q 变为原来的一半,则E 变为原来的一半12.如图所示,实线为不知方向的三条电场线,从电场中M 点以相同速度飞出a 、b 两个带电粒子,运动轨迹如图中虚线所示,则( )A .a 一定带正电,b 一定带负电B .a 的速度将减小,b 的速度将增加C .a 的加速度将减小,b 的加速度将增加D .两个粒子的电势能一个增加一个减小第Ⅱ卷(非选择题,共52分)二、填空题(每空2分,共22分。
县直高中2020年上期高二第一次月考试题一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1、△8,45C=75=ABC a A ==o o中,,则b () A. B、C)412.锐角△ABC 的面积为33,a =4,b=3,则角C 的大小为( )A .75°B .60°C .45°D .30°3、△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ,已知2220b c a bc +-+=,则角A 等于( )A.π3B. π6 C 、23π D . 23π或3π 4、△ABC 中,sin :sin :sin 4:7:8A B C =则△ABC 一定为( )A .锐角三角形B 、直角三角形C 、钝角三角形D 、等腰三角形()()()()()()11357921121121121 --()----n nn n a A n B n C n D n +=-+-5、数列, , , , ,的一个通项公式 、 、、 、L L6、数列n a 中,112a = ,121n n a a +=- 则5a =( )A 、25-B 、65C 、67D 、10-7、在△ABC中,若AB C =150°,则它的外接圆的面积为( )A 、πB 、2πC 、 3πD 、4π8、等差数列n a 中,5671,a a a ++= 则有39a a +=( )A 、 2B 、23 C 、13D 、19.△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ,,若10a =,B =45°,b =7,则△ABC ( )A .无解B .仅有一解C .仅有两解D .无法判断10.一船向正北航行,看见正西方向有相距10海里的两个灯塔恰好与它在一条直线上,继续航行半小时后,看见一灯塔在船的南偏西60°方向上,另一灯塔在船的南偏西75°方向上,则这艘船的速度是每小时( )A .5海里B .53海里C .10海里D .103海里11.△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ,若b 2=ac ,且c =2a ,则cos B 等于( )A 、14B C . 34D12、在△ABC 中,sin 2A ≤sin 2B +sin 2C -sin B sin C ,则A 的取值范围是( )A .(0,π6]B .[π6,π)C .[π3,π)D .(0,π3]二、填空题。
县直高中高一第一次月考生物试题一选择题1下列属于生物致癌因子的是()A.病毒引起细胞癌变B.紫外线引起皮肤癌C.防腐剂引起消化道癌变D.吸烟引起肺癌2下列细胞可能发生癌变的是()A.游离组织形态改变B.核增大,染色质固缩C.膜透性增大,运输功能降低D.酶活性降低3 近十几年,癌症患者的发病率和死亡率上升的原因是()①人口老龄化②吸烟③不健康的生活方式④缺乏体育锻炼A.①②B.②C.①②③④D.③④4 应用工程技术诊断疾病的过程中必须使用基因探针才能达到检测疾病的目的。
这里的基因探针是指()A.用于检测疾病的医疗器械B.用放射性同位素或荧光分子等标记的DNA分子C.合成α-球蛋白的DNAD.合成苯丙氨酸羟化酶的DNA片段5 基因治疗是指()A把健康的外源基因导入有基因缺陷的细胞中,达到治疗疾病的目的[学#科#XB.对有基因缺陷的细胞进行修复,从而使其恢复正常,达到治疗疾病的目的C.运用人工诱变的方法,使有基因缺陷的细胞发生基因突变,恢复正常D.运用基因工程技术,把有缺陷的基因切除,达到基因治疗的目的6 下列是研究生物学问题的方法,其中正确的步骤是()①推论②结论③发现问题④实验⑤观察⑥提出假设A.③⑥①④⑤②B.③⑥④⑤①②C.⑤③①⑥④②D.⑤③⑥①④②7 下列有关观察的说法,不正确的是()A.观察必须通过感官收集信息B.有时需要一些辅助工具C.亲自观察的结果一定正确D.观察到的结果必须如实地记载8 人们为了得到更加准确的实验数据,以反映出生物学现象所渗透的规律性本质,通常对测量及其结果的处理要求是()A.随机测量、取中间值B.随机测量、取平均值C.测量多次、取平均值D.测量多次、取中间值9下列关于癌症病因的叙述,错误的是()A.长期接触放射性物质,如电离辐射、X射线等B.长期接触一些有毒的化学物质,如砷、苯等C.由遗传因素决定,不受其他条件影响D.由引起癌变的病毒引起10.下列甲、乙两表分别是生物体及人体组织、器官的含水量。
河南省周口市沈丘县县直高级中学2014—2015学年度上学期第一次月考高一英语试题命题人:贾纪奎第I卷(三部分,共85分)第一部分:阅读理解(共20小题;每小题2分,满分40分)阅读下列短文,从每题所给的选项中,选出最佳选项。
ALao She wrote Teahouse in 1957. The play shows the audience life in China between 1898 and1945. It takes place in a teahouse in old Beijing and it tells us the story of Wang Life and his customers.It asks us to see the teahouse as the centre of the neighborhood. Finally, it says goodbye old Beijing and its people.The story starts in 1898 during the Qing Dynasty. It continues in 1916, and finally, it brings the audience to the end of the Anti-Japanese War in 1945.After the war, Wang loses the teahouse and he dies.Lao She was born in Beijing in 1899. His parents sent him to the Teahouse’s School in Beijing and he learned to teach. Form 1924 to 1929 he taught Chinese to the English in London. He wrote many plays, novels and short stories. He was named a “People’s Artist” and a “Great Master of Language”. He was one of the greatest Chinese writers of the 20th century.In La o She’s Teahouse today, waiters bring tea to the customers and sell them delicious Chinese food. If you like Beijing Opera, folk music, acrobatics or magic shows, you can enjoy them in the teahouse. Lao She’s Teahouse gives a wonderful welcome to everyone from China and from all over the world.1. What’s the main idea of Paragraph 4?A. Lao She.B. Lao She’s Teahouse.C. The story of Teahouse.D. The play Teahouse.2. How many years of life in China is shown in the play Teahouse?A. 57 years.B. 45 years.C. 98 years.D. 47 years.3. Lao She was all of the following except __________.A. a boss of TeahouseB. a people’s ArtistC. a Great Master of LanguageD. one of the greatest Chinese writers of the 20th century4. Who were welcome to Lao She’s Teahouse today?A. Only Chinese customers.B. Only foreign customers.C. Every customer from home and abroad .D. Only Beijing opera audience.BPeople all over the world celebrate the New Year. However, not all countries celebrate in the same way, and in some countries, the New Year doesn’t begin on the same date every year. It can depend on the seasons, the moon or the sun.In many countries, the New Year begins on 1st January, but people start celebrating on 31stDecember, New Year’s Eve. In New York many people go to celebrate in Time Square. While they’re waiting for the New Year, they listen to music, sing traditional songs and have fun. Just before 12 o’clock, everyone counts down from 10: 10, 9, 8 … As soon as it’s 12 o’clock, everyone shouts very loudly, “Happy New Year!” and they say, "Out with the old year and in with the new!" Suddenly there are fireworks and special drinks. Sometimes people sing an old song called Auld Lang Syne.New Year’s Day is often a family day. Some people relax at home. Some families get together fora special meal with grandparents, aunts, uncles and cousins. When the weather is fine, many families go out for a walk.On New Ye ar’s Day, many people make resolutions for the New Year. They write down a list of things, such as “I will help out more at home. I will work harder at school.” or “I won’t spend so much time playing video games.” when they have made their list, they read it to their family or friends and promise to follow their resolutions.5. Where do many people in New York spend the New Year’s Eve?A. In Times Square.B. At home.C. Outside their homes.D. In their grandparents’ home.6. What do people do just before 12 o’clock on New Year’s Eve in New York?A. Shout loudly.B. Listen to music.C. Sing traditional songs.D. Count down from 10: 10, 9, 8, …7. The underlined phrase “ out with” in the passage means__________.A. Goodbye toB. Hello toC. WelcomeD. Evening to8. Why do people read their list of resolutions to their family or friends on New Year’s Day?A. Because they want to tell their family or friends their resolutions.B. Because they are going to play video games.C. Because they will help out more at school.D. Because they promise to carry out the plans.CFor the first time, Zhang Shengyu, a student from Laiyang Middle School in Shandong, didn’t ask her mother to buy her a new schoolbag or any new pens for the beginning of the school term. “My old schoolbag can still be used, so there is no need to buy a new one,” Zhang said, “I’ve decided t o be thrifty from now on.”As the center of the family’s attention, many of the one-child generation (一代人) do not know about the hardships of life. They sometimes have no idea of the value of things. Many students would throw away pens and erasers that haven’t been used up and buy new ones. Some like to compete with each other to see whose clothes and school things are more expensive.On its website in January, the Ministry of Education called on students to form good habits such as working hard, being helpful and thrifty, to avoid wasting food and to pay more attention to study and others.Many schools consider teaching the value of money to be an important lesson and encourage their students to be thrifty.Students at No. 44 Middle School of Shijiazhuang, Hebei planted many fruit trees in their school. Through taking care of the trees, the students learned how to value things.At Tie Yi Middle School in Xi’an, students organized a group to collect waste and reuse them. “What we students do are small things, but I think with everyone’s work, we can make a difference,” said Li Dongdong, 16, a student from the group.9. The underlined word “thrifty” in Paragraph 1 means ______.A. giving money to the poorB. spending lots of moneyC. spending no money at allD. using money carefully and wisely10. In January, the Ministry of Education called on students to do the followings EXCEPT _________.A. working hardB. being thriftyC. helping othersD. protecting environment11. We can conclude that Li Dongdong ____________.A. took care of treesB. planted fruit treesC. collected wasteD. didn’t ask his mother for a new school bag12. What is the best title of the passage?A. School activitiesB. Waste not, want notC. Make a differenceD. The problems of one-child generationDHaving a teenager at home can be a real headache. Many teenagers smoke, drink and even fight. They often do something dangerous.Earlier studies have shown that teenagers are more likely (可能) to make irrational decisions than people from any other age group, including children and adults. Is it that teenagers are too young to tell right from wrong? Not really. So what’s the reason?According to Laurence, a teacher from a US university, the reason is that teenagers care a lot about how their peers (同龄人) view them—that is “peer influence”.As children enter their teenage years, they spend more time with their friends and classmates, and also they care more about what they think of them. This makes teenagers make decisions without thinking about the costs.In a test, a group of teenagers were asked to play a video driving game. When they played with their friends watching around them, they took more chances and drove more carelessly because that would increase their possibility of winning. But when they played alone, they drove more safely.Why do peers have such a big influence on teens’ behavior?As Laurence sees it, a teenager’s brain is like a car with a good accelerator (油门) but a weak brake (刹车). The “accelerator” is fully developed by teenage years while the “brake” is still not fully developed. When teenagers are watched by their peers, they usually push hard on the accelerator. With their weak brake, it is likely that they are going to end up in an accident.But the good news is that a violent (暴力的)teenager doesn’t necessa rily become a violent adult. About two-thirds to three-quarters of violent youth grow out of it. “They get more self-controlled.”13. Why are teenagers likely to make wrong decisions according to the text?A. They like difficult tasks.B. They don’t care about costs.C. They are too young to tell right from wrong.D. They care a lot about what their peers think of them.14. How does Laurence explain the influence of peer pressure on teens’ behavior?A. By comparing it with how a car works.B. By comparing it with an accident.C. By showing research findings.D. By giving examples.15. What can we conclude from the text?A. Teenagers are easy to have traffic accidents.B. Peer influence is bad on teenagers.C. Most of the teenagers become less violent when they grow up.D. Teenagers shouldn’t spend all the free time with their friends.ETaking good notes is a time-saving skill that will help you to become a better student in several ways, 16 Second, your notes are excellent materials to refer to (参考)when you are studying for a test. Third, note-taking offers variety to your study time and helps you to hold your interest.You will want to take notes during classroom discussions and while reading a textbook or doing research for a report.__17_____Whenever or however you take notes, keep in mind that note-taking is a selective process. 18The following methods may work best for you.●Read the text quickly to find the main facts and ideas in it.●Carefully read the text and watch for words that can show main points and supporting facts.● Write your notes in your own words.●____19__● Note any questions or ideas you may have about what was said or written.As you take notes, you may want to use your own shorthand(速记). When you do, be sure that you understand your symbols(符号) and that you use them all the time. 20A. Use words, not complete sentences.B. There are three practical note-taking methods.C. You must write your notes on separate paper.D. Otherwise, you may not be able to read your notes later.E. you will also want to develop your own method for taking notes.F. That means you must first decide what is important enough to include in your notes.G. First, the simple act of writing something down makes it easier for you to understand and remember it. 第二节:完形填空(共20小题;每小题1.5分,满分30分)阅读下面短文,从短文后各题所给的四个选项(A、B、C和D)中,选出可以填入空白处的最佳选项。
河南省周口市沈丘县县直高中2014-2015学年高一上学期第一次月考化学试卷一.选择题(3x20=60分)1.(3分)下列操作或发生事故时的处理方法正确的是()A.过滤时,用玻璃棒搅动漏斗内液体,使之过滤完全B.燃着的酒精灯打翻失火,应立即用水浇灭C.配制一定浓度的NaCl溶液时,定容振荡摇匀后,发现液面低于刻度线,再加水到刻度线D.浓硫酸不慎溅到皮肤上,应立即用大量水冲洗,最后再涂上3%~5%的小苏打溶液2.(3分)下列说法中正确的是()A.K OH的摩尔质量是56 gB.1 mol H2O的质量是18 g/molC.C l2的摩尔质量是35.5 g/molD.1 mol N2的质量是28 g3.(3分)绿色植物是空气天然的净化器,1万平方米柳杉每月可以吸收160kg SO2,则1万平方米柳杉每月吸收的SO2的物质的量为()A.164 kg B.2.5 mol C.2 500 mol D.2 500 g/mol4.(3分)下列实验操作正确的是()A.检查容量瓶是否漏水B.给液体加热C.称量D.过滤5.(3分)下列各组物质中分子数相同的是()A.2 L CO和2 L CO2B.9 g H2O和标准状况下11.2 L CO2C.标准状况下1 mol O2和22.4 L H2OD.0.2 mol H2和4.48 L HCl气体6.(3分)用N A代表阿伏加德罗常数的值,下列说法正确的是()A.0.5 mol锌粒与足量盐酸反应产生11.2 L H2B.标准状况下,11.2 L CCl4所含分子数为0.5N AC.0.1 mol CH4所含氢原子数为0.4N AD.常温常压下,28 g N2中所含原子个数为N A7.(3分)在相同的温度和压强下,4个容器中分别装有4种气体.已知各容器中的气体和容器的容积分别是a.CO2,100mL;b.O2,200mL;c.N2,400mL;d.CH4,600mL,则4个容器中气体的质量由大到小的顺序是()A.a>b>c>d B.b>a>d>c C.c>d>b>a D.d>c>a>b8.(3分)为除去某物质所含的杂质,所选用的试剂或操作方法正确的是()序号物质杂质除杂质用选用的试剂或操作方法① KNO3溶液KOH 滴入稀HNO3同时用PH计测定至溶液呈中性② FeSO4溶液C uSO4加入过量铁粉并过滤③ H2CO2依次通过盛有NaOH溶液和浓硫酸的洗气瓶④ NaNO3CaCO3加稀盐酸溶解、过滤、蒸发、结晶A.①②③B.②③④C.①③④D.①②③④9.(3分)下列叙述中正确的是()A.摩尔是物质的量的单位,每摩尔物质含有6.02×1023个分子B.1mol氧的质量为16gC.0.5molHe约含有6.02×1023个电子D.2H既可表示2个氢原子又可表示2mol氢原子10.(3分)在0.5mol Na2SO4中,含有的Na+数约是()A.3.01×1023B.6.02×1023C.0.5 D.111.(3分)物质的量相等的CO和CO2相比较,下列有关叙述中正确的是()①它们所含的分子数目之比为1:1②它们所含的氧原子数目之比为1:2③它们所含的原子总数目之比为2:3④它们所含的C原子数目之比为1:1⑤它们所含的电子数目之比为7:11.A.①和④B.②和③C.①②③④D.①②③④⑤12.(3分)现有下列四种因素:①温度和压强②所含微粒数③微粒本身大小④微粒间的距离,其中对气体物质体积有显著影响的是()A.①②③B.①③④C.①②④D.①②③④13.(3分)只给出表中甲和乙对应的量,不能用来求物质的量的是()甲乙A 物质的微粒数阿伏加德罗常数B 标准状况下气体摩尔体积标准状况下气体体积C 固体体积固体密度D 非标准状况下物质的质量物质的摩尔质量A.A B.B C.C D.D14.(3分)在同温同压下,A容器的氯气和B容器的氨气中,若它们所含的原子数相等,则这两个容器体积之比是()A.2:1 B.1:2 C.2:3 D.1:315.(3分)下列说法中正确的是()A.32 g O2占有的体积约为22.4 LB.22.4 L N2含阿伏加德罗常数个氮分子C.在标准状况下,22.4 L水的质量约为18 gD.22 g二氧化碳与标准状况下11.2 L HCl约含有相同的分子数16.(3分)下列叙述正确的是()A.同温同压下,相同体积的物质,其物质的量一定相等B.任何条件下,等物质的量的甲烷和一氧化碳所含的分子数一定相等C.1 L一氧化碳气体一定比1 L氧气的质量小D.相同条件下的一氧化碳气体和氮气,若物质的量相等则体积相等,但质量不相等17.(3分)有一种气体的质量是14.2g,体积是4.48L(标准状况),该气体的摩尔质量是()A.28.4 B.28.4 g•mol﹣1C.71 D.71 g•mol﹣118.(3分)实验室使用的浓盐酸的溶质质量分数为36.5%,密度为1.19g•cm3,则物质的量浓度是()A.11.9 B.11.9mol•L﹣1C.0.012mol•L﹣1D.无法计算19.(3分)实验室需用2mol•L﹣1氯化钠溶液450mL,配置时应选用容量瓶的规格和称取氯化钠的质量分别是()A.450 mL,52.7 g B.500 mL,58.5 gC.1000 mL,117 g D.任意规格,111.2 g20.(3分)配制一定物质的量浓度的某溶液,下列情况会使配制结果偏低的是()A.未冷却即转移、定容B.未洗涤烧杯及玻璃棒C.定容时俯视刻度线D.容量瓶中原有少量蒸馏水二.填空题:(每空1分,共40分)21.(5分)实验室欲用NaOH固体配制1.0mol/L的NaOH溶液480mL:(1)配制时,必须使用的玻璃仪器有.(2)要完成本实验该同学应称出NaOHg.(3)某同学欲称量NaOH的质量,他先用托盘天平称量烧杯的质量,天平平衡后的状态如图.烧杯的实际质量为g(4)使用容量瓶前必须进行的一步操作是.(5)在配制过程中,其他操作都是正确的,下列操作会引起误差偏高的是.①没有洗涤烧杯和玻璃棒②转移溶液时不慎有少量洒到容量瓶外面③容量瓶不干燥,含有少量蒸馏水④定容时俯视刻度线⑤未冷却到室温就将溶液转移到容量瓶并定容⑥定容后塞上瓶塞反复摇匀,静置后,液面低于刻度线,再加水至刻度线.22.(2分)24.5g H2SO4的物质的量是mol,1.50molNa2CO3的质量是克.23.(4分)2.5molBa(OH)2中含有个Ba2+,含有个OH﹣;5molNa2SO4溶于水电离出Na+个;0.6mol的Al2(SO4)3中含Al3+ mol.24.(4分)0.5mol O3分子数约是个、原子数约是个、质子数约是个.25.(3分)从1L1mol/LNaOH溶液中取出100mL,则这100mLNaOH溶液的物质的量浓度是mol/L,含NaOH的质量g.26.(5分)某同学将标准状况下44.8L的HCl气体溶于水,配成500mL的溶液.(1)根据公式,可知该HCl气体的物质的量为.(2)根据公式,可知所配成的盐酸溶液的物质的量浓度为.(3)100mL该盐酸溶液中含HCl的物质的量为.27.(5分)现有NaC1、Na2SO4和NaNO3的混合溶液,选择适当的试剂将其转化为相应的沉淀或固体,从而实验C1﹣、和的相互分离,实验过程如下:请回答下列问题:(1)写出上述实验过程中所用试剂的名称:试剂1为,试剂2为,试剂4为(2)加入过量试剂3的目的是.(3)在加入试剂4后,获得晶体D的实验操作④的名称是.28.(6分)某化学课外活动小组从实验室取出硫酸试剂,试剂瓶上标签的部分内容如图1所示.该小组欲配制450mL 0.2mol•L﹣1的稀硫酸溶液:(1)所需试剂瓶内硫酸溶液的体积为mL.(2)所用量筒和容量瓶的规格为和.(3)浓硫酸稀释时的操作是.(4)下列为该小组的操作,可能导致配制溶液浓度偏高的是.A.移液前未冷却至室温B.定容时加水多了,用滴管吸出C.定容时俯视刻度线(5)图2是该小组转移溶液的示意图,图中的错误是.29.(6分)(1)0.5mol CH4的质量为g,在标准状况下的体积为L;含个氢原子.(2)将标准状况下22.4L的HCl配成200mL的溶液,所得溶液的物质的量浓度为;配制300mL 0.5mol/L氯化钠溶液,需要1.5mol/L的氯化钠溶液的体积为mL.河南省周口市沈丘县县直高中2014-2015学年高一上学期第一次月考化学试卷参考答案与试题解析一.选择题(3x20=60分)1.(3分)下列操作或发生事故时的处理方法正确的是()A.过滤时,用玻璃棒搅动漏斗内液体,使之过滤完全B.燃着的酒精灯打翻失火,应立即用水浇灭C.配制一定浓度的NaCl溶液时,定容振荡摇匀后,发现液面低于刻度线,再加水到刻度线D.浓硫酸不慎溅到皮肤上,应立即用大量水冲洗,最后再涂上3%~5%的小苏打溶液考点:化学实验安全及事故处理.分析:A、用玻璃棒搅拌过滤器中的液体可能将滤纸弄破,使实验失败;B、酒精与水互溶,应用抹布扑灭燃着的酒精灯;C、定容振荡摇匀后,发现液面低于刻度线,不需再加水,再加水到刻度线,导致所配浓度偏低;D、浓硫酸具有腐蚀性,不慎溅到皮肤上,应立即用大量水冲洗,最后再涂上3%~5%的小苏打溶液.解答:解:A、用玻璃棒搅拌过滤器中的液体可能将滤纸弄破,使实验失败,故A错误;B、酒精与水互溶,应用抹布扑灭燃着的酒精灯,故B错误;C、定容振荡摇匀后,发现液面低于刻度线,不需再加水,再加水到刻度线,导致所配浓度偏低,故C错误;D、浓硫酸具有腐蚀性,不慎溅到皮肤上,应立即用大量水冲洗,最后再涂上3%~5%的小苏打溶液,故D正确;故选D.点评:本题主要考查基本实验操作等相关内容,掌握各自的操作注意事项是解题的关键.2.(3分)下列说法中正确的是()A.K OH的摩尔质量是56 gB.1 mol H2O的质量是18 g/molC.C l2的摩尔质量是35.5 g/molD.1 mol N2的质量是28 g考点:摩尔质量.分析:一种物质的摩尔质量在数值上等于该物质的相对分子质量或相对原子质量,单位为g/mol.据此解答.解答:解:A、摩尔质量的单位为g/mol,故A错误;B、质量的单位为g,故B错误;C、氯气的相对分子质量为71,故其摩尔质量应为71g/mol,故C错误;D、氮气的摩尔质量为28g/mol,1mol氮气的质量m=nM=1mol×28g/mol=28g,故D正确.故选D.点评:本题考查了摩尔质量的数值和单位,应注意的是摩尔质量的单位是g/mol,而质量的单位为g.3.(3分)绿色植物是空气天然的净化器,1万平方米柳杉每月可以吸收160kg SO2,则1万平方米柳杉每月吸收的SO2的物质的量为()A.164 kg B.2.5 mol C.2 500 mol D.2 500 g/mol考点:物质的量的相关计算.分析:根据n=计算SO2的物质的量.解答:解:1万平方米柳杉每月可以吸收160kg SO2,则1万平方米柳杉每月吸收的SO2的物质的量为n===2500mol,故选C.点评:本题考查二氧化硫的污染及物质的量的计算,题目难度不大,注意把握物质的量与质量等物理量的计算公式的运用.4.(3分)下列实验操作正确的是()A.检查容量瓶是否漏水B.给液体加热C.称量D.过滤考点:化学实验方案的评价.分析:A.盖上塞子,上下颠掉摇匀;B.试管中液体不能超过容积的,且试管夹应在试管口的处;C.NaOH易潮解,具有腐蚀性;D.漏斗下端应紧靠烧杯内壁.解答:解:A.盖上塞子,上下颠掉摇匀,图中操作合理,故A正确;B.试管中液体不能超过容积的,且试管夹应在试管口的处,图中加热操作不合理,故B错误;C.NaOH易潮解,具有腐蚀性,应在烧杯中称量,故C错误;D.漏斗下端应紧靠烧杯内壁,图中操作不合理,故D错误;故选A.点评:本题考查化学实验方案的评价,为高频考点,涉及实验基本操作和实验基本技能等,把握实验基本操作和图中装置的作用为解答的关键,注意实验的评价性分析,题目难度不大.5.(3分)下列各组物质中分子数相同的是()A.2 L CO和2 L CO2B.9 g H2O和标准状况下11.2 L CO2C.标准状况下1 mol O2和22.4 L H2OD.0.2 mol H2和4.48 L HCl气体考点:物质的量的相关计算.分析:在标准状况下,气体的Vm=22.4L/mol,根据气体含有的分子数N=nN A=N A=N A结合物质存在的条件、状态解答该题.解答:解:A.相同条件下,相同体积的任何气体具有相同的分子数,2 L CO和2 L CO2两种气体存在的外界条件不一定相同,则体积相同时物质的量不一定相同,故A错误;B.n(H2O)==0.5mol,n(CO2)==0.5mol,二者的物质的量相等,则分子数相同,故B正确;C.水为液态,22.4L时物质的量不是1mol,故C错误;D.氯化氢存在的条件未知,不一定为0.2mol,故D错误;故选B.点评:本题主要考查气体摩尔体积的相关计算,可以根据所学知识来回答,注意公式的使用范围,难度不大.6.(3分)用N A代表阿伏加德罗常数的值,下列说法正确的是()A.0.5 mol锌粒与足量盐酸反应产生11.2 L H2B.标准状况下,11.2 L CCl4所含分子数为0.5N AC.0.1 mol CH4所含氢原子数为0.4N AD.常温常压下,28 g N2中所含原子个数为N A考点:阿伏加德罗常数.分析:A、依据化学方程式定量关系计算生成氢气0.5mol,温度压强不知不能计算气体体积;B、标准状况四氯化碳不是气体;C、依据甲烷含有氢原子数计算;D、依据n=计算物质的量,结合分子式计算原子数;解答:解:A、依据化学方程式定量关系计算生成氢气0.5mol,温度压强不知不能计算气体体积,故A错误;B、标准状况四氯化碳不是气体,11.2 L CCl4物质的量不是0.5mol,故B错误;C、依据甲烷含有氢原子数计算,0.1 mol CH4所含氢原子数为0.4N A,故C正确;D、依据n=计算物质的量=1mol,结合分子式计算原子数为2N A,故D错误;故选C.点评:本题考查了阿伏伽德罗常数的分析应用,主要是气体摩尔体积条件分析,物质的量计算微粒数的方法和注意问题,掌握基础是关键,题目较简单.7.(3分)在相同的温度和压强下,4个容器中分别装有4种气体.已知各容器中的气体和容器的容积分别是a.CO2,100mL;b.O2,200mL;c.N2,400mL;d.CH4,600mL,则4个容器中气体的质量由大到小的顺序是()A.a>b>c>d B.b>a>d>c C.c>d>b>a D.d>c>a>b考点:阿伏加德罗定律及推论;物质的量的相关计算.专题:阿伏加德罗常数和阿伏加德罗定律.分析:根据n=,计算各物质的物质的量,根据m=nM计算各气体的质量,然后进行计较.解答:解:设在该温度和压强下,气体的气体摩尔体积为V m,a容器内气体的质量为:m(CO2)==g;b容器内气体的质量为:m(O2)==g;c容器内气体的质量为:m(N2)==g;d容器内气体的质量为:m(CH4)==g,则4个容器中气体的质量由大到小的顺序是c>d>b>a.故选C.点评:本题考查物质的量的相关计算,题目难度不大,注意物质的量与质量、体积有关计算公式的运用.8.(3分)为除去某物质所含的杂质,所选用的试剂或操作方法正确的是()序号物质杂质除杂质用选用的试剂或操作方法① KNO3溶液KOH 滴入稀HNO3同时用PH计测定至溶液呈中性② FeSO4溶液C uSO4加入过量铁粉并过滤③ H2CO2依次通过盛有NaOH溶液和浓硫酸的洗气瓶④ NaNO3CaCO3加稀盐酸溶解、过滤、蒸发、结晶A.①②③B.②③④C.①③④D.①②③④考点:物质的分离、提纯和除杂.专题:实验评价题.分析:①KOH和硝酸至中性恰好生成硝酸钾和水;②过量Fe,与硫酸铜反应生成硫酸亚铁和Cu;③二氧化碳与碱反应,而氢气不能;④加盐酸溶解会引入杂质.解答:解:①KOH和硝酸至中性恰好生成硝酸钾和水,能除去杂质KOH;②过量Fe,与硫酸铜反应生成硫酸亚铁和Cu,然后过滤得到硫酸亚铁,除去杂质硫酸铜;③二氧化碳与碱反应,而氢气不能,然后利用浓硫酸干燥得到纯净的氢气,除去杂质二氧化碳;④加盐酸溶解会引入杂质氯离子,应选择加硝酸;显然①②③都符合除杂原则,不引入新杂质,而④中加入盐酸,引进了Cl﹣,不符合题意,故选A.点评:本题考查混合物的分离、提纯,明确物质的性质及性质差异选择分离方法即可解答,注意除杂的原则是去除杂质,但不引进新的杂质,更不能使提纯物质减少,题目难度不大.9.(3分)下列叙述中正确的是()A.摩尔是物质的量的单位,每摩尔物质含有6.02×1023个分子B.1mol氧的质量为16gC.0.5molHe约含有6.02×1023个电子D.2H既可表示2个氢原子又可表示2mol氢原子考点: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、物质并不都是由分子构成.物质由分子、原子或离子构成.B、使用物质的量必须指明具体的物质或化学式.C、稀有气体是单原子分子,每个氦原子含有2个电子,根据N=nN A计算氦原子数目.D、2H表示2个氢原子,不能表示2mol氢原子.解答:解:A、有的物质由分子构成如氧气等,有的物质由原子构成如二氧化硅等,有的物质由离子构成如氯化钠等,物质并不都是由分子构成,故A错误;B、1mol氧指代不明确,不清楚是指氧气还是氧原子,故B错误;C、稀有气体是单原子分子,0.5molHe约含有电子数为0.5mol×2×6.02×1023mol﹣1=6.02×1023,故C正确;D、2H表示2个氢原子,不能表示2mol氢原子,故D错误.故选:C.点评:考查学生对物质的量与化学用语理解、物质构成及常用化学计量计算,难度不大,注意使用物质的量必须指明具体的物质或化学式.10.(3分)在0.5mol Na2SO4中,含有的Na+数约是()A.3.01×1023B.6.02×1023C.0.5 D.1考点:阿伏加德罗常数;电解质在水溶液中的电离.分析:根据化学式为Na2SO4,利用硫酸钠的物质的量来确定钠离子的物质的量,再利用N=n×N A来计算钠离子的数目.解答:解:0.5molNa2SO4中钠离子的物质的量为0.5mol×2=1mol,则由N=n×N A可知,则钠离子的个数为1mol×N A mol﹣1=N A≈6.02×1023,故选B点评:有关阿伏伽德罗常数的计算,题目在注重对计算关系的考查的同时,又隐含对物质状态、物质结构、氧化还原反应、电离、水解知识的考查.11.(3分)物质的量相等的CO和CO2相比较,下列有关叙述中正确的是()①它们所含的分子数目之比为1:1②它们所含的氧原子数目之比为1:2③它们所含的原子总数目之比为2:3④它们所含的C原子数目之比为1:1⑤它们所含的电子数目之比为7:11.A.①和④B.②和③C.①②③④D.①②③④⑤考点:阿伏加德罗定律及推论.专题:计算题;阿伏加德罗常数和阿伏加德罗定律.分析:根据N=nN A计算微粒数目.①物质的量相同,分子数相同;②1个CO和CO2分子中所含氧原子数分别为1个、2个;③1个CO和CO2分子中所含原子数分别为2个、3个;④1个CO和CO2分子中所含碳原子数都是1个;⑤CO和CO2分子中所含电子数分别为14个、22个.解答:解:①由于N=nN A,所以物质的量相同,分子数相同,故①正确;②CO和CO2分子中所含氧原子数分别为1个、2个,相等物质的量的CO和CO2所含氧原子的物质的量之比为1:2,由于N=nN A,所含氧原子数之比为1:2,故②正确;③CO和CO2分子中所含原子数分别为2个、3个,相等物质的量的CO和CO2所含原子的物质的量之比为2:3,由于N=nN A,所含原子数之比为2:3,故③正确;④CO和CO2分子中所含碳原子数都是1个,相等物质的量的CO和CO2所含碳原子的物质的量之比为1:1,由于N=nN A,所含碳原子数之比为1:1,故④正确;⑤CO和CO2分子中所含电子数分别为14个、22个,相等物质的量的CO和CO2所含电子的物质的量之比为14:22=7:11,由于N=nN A,所含电子数之比为7:11,故⑤正确.故选D.点评:本题考查常用化学计量的有关计算、阿伏伽德罗定律及推论,题目难度不大,注意有关公式的利用与影响气体体积的因素.12.(3分)现有下列四种因素:①温度和压强②所含微粒数③微粒本身大小④微粒间的距离,其中对气体物质体积有显著影响的是()A.①②③B.①③④C.①②④D.①②③④考点:阿伏加德罗定律及推论.分析:影响体积的因素有微粒的大小,微粒之间的距离以及所含微粒数等因素,对于气体来说,微粒本身大小远小于微粒间的距离,条件不同,距离不同.解答:解:对于气体来说,温度和压强不同,气体分子之间的距离不同,而微粒本身大小远小于微粒间的距离,则微粒本身大小可忽略不计,微粒数目越多,则体积越大,所以影响气体体积的因素主要有:温度和压强、所含微粒数以及微粒间的距离,故选C.点评:本题考查影响气体体积的因素,题目难度不大,注意相关基础知识的积累,把握好课本相关知识,是解答该题的关键.13.(3分)只给出表中甲和乙对应的量,不能用来求物质的量的是()甲乙A 物质的微粒数阿伏加德罗常数B 标准状况下气体摩尔体积标准状况下气体体积C 固体体积固体密度D 非标准状况下物质的质量物质的摩尔质量A.A B.B C.C D.D考点:物质的量的相关计算.分析:A.根据N=nN A进行判断;B.标准状况下V=nVm,据此判断;C.只能求出固体的质量,无法求解其物质的量;D.根据n=进行判断.解答:解:A.已知物质的微粒数和阿伏加德罗常数,根据N=nN A可知,n=可以计算出物质的量,故A正确;B.已知标准状况下的气体摩尔体积和标准状况下的气体的体积,根据V=nVm可知,n=可以计算出物质的量,故B正确;C.已知固体的体积和固体的密度,根据m=ρV可以计算固体的质量,不能求出物质的量,故C错误;D.已知物质的质量和物质的摩尔质量,根据n=可以计算出物质的量,故D正确.故ABD可以计算出物质的量,C不能求出物质的量.故选:C.点评:本题考查物质的量有关计算,涉及对常用化学计量的计算公式的理解,难度较小,注意基础知识的学习掌握.14.(3分)在同温同压下,A容器的氯气和B容器的氨气中,若它们所含的原子数相等,则这两个容器体积之比是()A.2:1 B.1:2 C.2:3 D.1:3考点:阿伏加德罗定律及推论.专题:阿伏加德罗常数和阿伏加德罗定律.分析:同温同压下,气体摩尔体积相等,A、B容器中原子个数相等,则A、B容器中氯气和氨气的物质的量之比为2:1,根据V=nV m判断其体积关系.解答:解:同温同压下,气体摩尔体积相等,A、B容器中原子个数相等,则A、B容器中氯气和氨气的物质的量之比为2:1,根据V=nV m知,相同条件下气体体积之比等于其物质的量之比,所以这两个容器体积之比为2:1,故选A.点评:本题考查了阿伏伽德罗定律及其推论,明确各个物理量之间的关系即可解答,题目难度不大.15.(3分)下列说法中正确的是()A.32 g O2占有的体积约为22.4 LB.22.4 L N2含阿伏加德罗常数个氮分子C.在标准状况下,22.4 L水的质量约为18 gD.22 g二氧化碳与标准状况下11.2 L HCl约含有相同的分子数考点:物质的量的相关计算.专题:计算题.分析:A.氧气所处状态下的气体摩尔体积不一定是22.4L/mol;B.氮气不一定处于标况下;C.标况下,水为液态,22.4L水的质量远远大于18g;D.根据n==计算气体物质的量进行比较.解答:解:A.32 g O2的物质的量为1mol,但氧气所处状态下的气体摩尔体积不一定是22.4L/mol,其占有的体积不一定是22.4L,故A错误;B.氮气不一定处于标况下,22.4 L N2不一定为1mol,则不一定含有阿伏加德罗常数个氮分子,故B错误;C.标况下,水为液态,密度约是1g/ml,22.4L水的质量远远大于18g,故C错误;D.22 g二氧化碳物质的量为=0.5mol,标准状况下11.2 L HCl物质的量为=0.5mol,二者物质的量相等,则含有相同的分子数,故D正确,故选D.点评:本题考查物质的量有关计算、气体摩尔体积,比较基础,注意气体摩尔体积的使用条件与对象.16.(3分)下列叙述正确的是()A.同温同压下,相同体积的物质,其物质的量一定相等B.任何条件下,等物质的量的甲烷和一氧化碳所含的分子数一定相等C.1 L一氧化碳气体一定比1 L氧气的质量小D.相同条件下的一氧化碳气体和氮气,若物质的量相等则体积相等,但质量不相等考点:物质的量与其浓度和气体摩尔体积的综合应用.分析:A.同温同压下具有相同的气体摩尔体积,必须是相同体积的气体才能具有相同的物质的量;B.根据N=nN A可知,物质的量相等时,则一定含有相同的分子数;C.没有告诉在相同条件下,无法判断二者的物质的量大小;D.一氧化碳和氮气的摩尔质量相同,物质的量相等则质量相等.解答:解:A.同温同压下,相同体积的气体,其物质的量一定相等,没有指出研究对象为气体,故A错误;B.甲烷和一氧化碳的物质的量相等,根据N=nN A可知,二者一定含有相同的分子数,故B 正确;C.如果在相同条件下,1L一氧化碳气体和1L氧气具有相同的物质的量,则1L一氧化碳气体一定比1L氧气的质量小,没有告诉体积,无法比较二者的质量大小,故C错误;D.相同条件下等物质的量一氧化碳和氮气具有相同的体积,由于二者的摩尔质量都是28g/mol,则等物质的量的二者具有相同的质量,故D错误;故选B.点评:本题考查了有关物质的量的计算与判断,题目难度中等,注意掌握物质的量与摩尔质量、气体摩尔体积、阿伏伽德罗常数等物理量之间的转化关系,明确气体摩尔体积的影响因素及使用条件.17.(3分)有一种气体的质量是14.2g,体积是4.48L(标准状况),该气体的摩尔质量是()A.28.4 B.28.4 g•mol﹣1C.71 D.71 g•mol﹣1考点:摩尔质量;物质的量的相关计算.专题:计算题.分析:根据M==计算该气体的摩尔质量.解答:解:M====71g/mol,故选D.点评:本题考查了摩尔质量的计算,难度不大,注意基本公式的掌握.18.(3分)实验室使用的浓盐酸的溶质质量分数为36.5%,密度为1.19g•cm3,则物质的量浓度是()A.11.9 B.11.9mol•L﹣1C.0.012mol•L﹣1D.无法计算考点:物质的量浓度的相关计算;溶液中溶质的质量分数及相关计算.专题:物质的量浓度和溶解度专题.分析:根据c=计算该浓盐酸的物质的量浓度.解答:解:盐酸的溶质质量分数为36.5%,密度为1.19g•cm3,则该浓盐酸的物质的量浓度为mol/L=11.9mol/L.故选:B.。