2015-2016新乡市高二下期末试卷答案

导读：　2015-2016新乡市高二下期末试卷答案(共5篇)新乡市2015-2016学年高二下期终结性评价测试答案高二参考答案一、现代文阅读(9分，每小题3分)1 C 解析 “发展到鼎盛时期”错，原文只说“蔚成风气”。2 B 解析 原因表述错误，由第三段可知，原因主要是南方秀丽的山水可以使他们从对现实的迷惘、懑闷中解脱出来。3 A解析 “爱竹源于他们的性...

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2015-2016新乡市高二下期末试卷答案(一)
新乡市2015-2016学年高二下期终结性评价测试答案

高二参考答案

一、现代文阅读(9分，每小题3分)

1.C 解析 “发展到鼎盛时期”错，原文只说“蔚成风气”。

2.B 解析 原因表述错误，由第三段可知，原因主要是南方秀丽的山水可以使他们从对现实的迷惘、懑闷中解脱出来。

3. A解析 “爱竹源于他们的性格”错，应源于竹子的特点与“历史上某些„„产生深刻的影响”。

二、古代诗文阅读(36分)

(一)文言文阅读(19分)

4.A

5.B (将“权臣或辅政大臣”改为皇后、皇太后或太皇太后等女性统治者)

6.C (“杀人之夫而夺其妻”的是郭千户而不是刘福)

7.(1) (杨惟中)宣扬国威，发布政令，让他们都登记户口设置属官，(然后)才返回，皇帝于是有重用他的意图。(“俾”“籍”“用”各1分，句意通顺2分)(5分)

(2) (杨惟中)邀请儒士赵复、王粹等在书院讲学，他于是通晓圣贤之学，慷慨激昂地想用圣贤之道助益国家。( “延”

“慨然”“济”各1分，句意通顺2分)(5分)

参考译文：

杨惟中字彦诚，是弘州人。金朝末年，以孤儿的身份侍奉太宗，懂得读书，有胆略，太宗器重他。他二十岁的时候，奉命出使西域三十多个国家，宣扬国威，发布政令，让他们都登记户口设置属官，然后才返回，皇帝于是有重用他的意图。

皇子阔出征伐宋朝，命 令杨惟中在军中行中书省事。攻克宋朝枣阳、光化等军，攻克光、随、郢、复等州，以及襄阳、德安府，共获得名士数十人。收集程颢、程颐的著作送往燕都，建立 宋朝大儒周敦颐的祠堂，建立太极书院，邀请儒士赵复、王粹等在书院讲学，杨惟中于是通晓圣贤之学，慨然想以圣道助益国家。杨惟中任中书令，太宗去世，太后 临朝听政，杨惟中一人以宰相的身份负担天下大事。

定宗即位，平阳道断事 官斜彻横行不法，皇帝诏令杨惟中代皇帝前去宣扬政令，安抚百姓，杨惟中审讯之后将其处死。金朝灭亡后，金朝将领武仙在邓州溃败，余党逃散于太原、真定之 间，占据了大明川，使用金朝开兴年

号，人数达到数万，抢掠数千里。皇帝诏令会集各道的兵力进讨，没有战胜。杨惟中执符节前往劝谕，其首领投降，其余党全部 平定了。

宪宗即位，世祖以太弟 的身份镇守金莲川，有权设立衙署和任命官员。于是在汴梁设立河南道经略司，奏请任命杨惟中为经略使，让他在唐、邓、申、裕、 嵩、汝、息、亳、颍等州屯田。刚灭金时，由监河桥万户刘福为河南道总管，刘福贪婪残酷，欺压残害金朝的遗民二十多年。杨惟中到任后，召刘福来听训，刘福称 病不来。杨惟中在座位上放了大木棒，又召他，刘福不得已，有数千人护卫着来见杨惟中，杨惟中便手握大棒将他打倒。几天后刘福死了，河南由此清明安定。杨惟 中改任陕右四川宣抚使。当时各部队的将领横行奢侈，侵害百姓，有个叫郭千户的尤其厉害，杀人之夫而抢夺其妻，杨惟中将其处死示众，关中于是秩序安定。杨惟 中对人说：“我并非喜欢杀戮，国家没有法纪，致使这种贼人残害百姓，百姓们无处控告，就算我不想除掉他们，行吗?”

己未年，世祖统率东路军，奏请任命杨惟中为江淮京湖南北路宣抚使，让他建立行御史台，因在世祖之先启

行，宣布朝廷的恩泽信誉，蒙古、汉军的各部队将领都听从他的节制。部队返回时，杨惟中在蔡州去世，时年五十五岁。

(二)古代诗歌阅读(11分)

9.（1）从内容上看，是作者远离官场后的生活写照，定下了全词的思想感情基调。（2分）从结构上看，“出红尘”是全词的“诗眼”，统领全词。因为“出红尘”，才有了下文所描写的悠闲自在、超然旷达的生活情景。（3分）

（2）词的下片通过晚来垂钓时月映江面、水天一色、孤鸿明灭等景物的描写，展现的是一幅澄澈清雅、闲适宁静的月夜垂钓图（3分），表现了词人对红尘官场的轻蔑和不愿同流合污的情怀，寄寓了作者坚持纯洁高尚的操守，向往自由宁静生活的人生追求。（3分）

（三）名篇名句默写（6分）

10.（1）地崩山摧士死 然后天梯石栈相钩连

（2）苟全性命于乱世 不求闻达于诸侯

（3）艰难苦恨繁霜鬓 潦倒新停浊酒杯

三、文学类文本阅读（25分）

11．（1） C E （每个3分，全对5分）A项“他们厌恶战

争，想尽快结束”，于文无依据；B项，强加因果；D项只揭示了部分原因，人们的沉默，更多的是人们对杰夫卡夫斯基的举动的不解。

（2）①照应题目，以美好的“歌声”，表达对美好生活的向往；②渲染宁静而压抑的氛围；③表现人物内心世界的复杂情绪，以情感人，引发人们对战争的思考；④推动故事情节的发展。（写对3个满分，酌情考虑）

（3）①他热爱音乐，如果不是战争，现在一定是一名钢琴师；②他满怀爱心，他喜欢听孩子们在教堂唱歌的感觉，男孩绊倒时他毫不犹豫地去搀扶；③他勇于自我牺牲，甘愿为保护反抗入侵的“孩子”而接受军法处置。

（4）答案：示例一：我认为杰夫卡夫斯基是一个合格的士兵。①从资历上看，杰夫卡夫斯基是一个“二战”老兵，德军入侵苏联，他成为一名保家卫国的战士，为了祖国利益而战。②从信仰上看，杰夫卡夫斯基对“伟大的莫斯科”有着无比信任，体现了其军人的坚定信仰。③从军事素质看，杰夫卡夫斯基训练有素，当男孩拿出手枪的那一刻，他习惯性地下手并保住了自己的性命。（言之有理，酌情给分）

2015-2016新乡市高二下期末试卷答案(二)
2015-2016学年河南省新乡市高二下学期期末考试语文试题(扫描版)

2015-2016新乡市高二下期末试卷答案(三)
2015-2016学年高二下学期期末统考英语试题带答案(精品)

2015—2016学年度高二级第二学期期末试题（卷）

英 语

本试卷满分共150分，分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。

第Ⅰ卷

注意事项：

1. 答第Ⅰ卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

2. 将答案写在答题卡上对应题目的答案标号下。不能答在本试卷上，否则无效。 第一部分 阅读理解（共两节，满分60分）

第一节 （共25小题；每小题2分，满分50分）

阅读下列短文，从每题所给的四个选项（A、B、C、和D）中，选出最佳选项，并在答题卡上填该项或涂黑。

A

A city child‘s summer is spent in the street in front of his home, and all through the long summer vacations I sat on the edge of the street and watched enviously the other boys on the block play baseball. I was never asked to take part even when one team had a member missing—not out of special cruelty, but because they took it for granted I would be no good at it. They were right, of course.

I would never forget the wonderful evening when something changed. The baseball ended about eight or eight thirty when it grew dark. Then it was the custom of the boys to retire to a little stoop(门廊) that stuck out from the candy store on the corner and that somehow had become theirs. No grownup ever sat there or attempted to. There the boys would sit, mostly talking about the games played during the day and of the game to be played tomorrow. Then long silences would fall and the boys would wander off one by one. It was just after one of those long silences that my life as an outsider changed. I can no longer remember which boy it was that summer evening who broke the silence with a question: but whoever he was, I nod to him gratefully now. ―What‘s in those books you‘re always reading?‖ he asked casually. ―Stories,‖ I answered. ―What kind?‖ asked somebody else without much interest.

Nor do I know what drove me to behave as I did, for usually I just sat there in silence, glad enough to be allowed to remain among them; but instead of answering his question, I told them for two hours the story I was reading at the moment. The book was Sister Carrie. They listened bug-eyed and breathless. I must have told it well, but I think there was another and deeper reason that made them to keep an audience. Listening to a tale being told in the dark is one of the most

ancient of man‘s entertainments, but I was offering them as well, without being aware of doing it, a new and exciting experience.

The books they themselves read were the Rover Boys or Tom Swift or G.A.Henty. I had read them too, but at thirteen I had long since left them behind. Since I was much alone I had become an enthusiastic reader and I had gone through the books-for-boys series. In those days there was no reading material between children‘s and grownups‘ books or I could find none. I had gone right from Tome Swift and His Flying Machine to Theodore Dreiser and Sister Carrie. Dreiser had hit my young mind, and they listened to me tell the story with some of the wonder that I had had in reading it.

The next night and many nights thereafter, a kind of unspoken ritual (仪式) took place. As it grew dark, I would take my place in the center of the stoop and begin the evening‘s tale. Some nights, in order to taste my victory more completely, I cheated. I would stop at the most exciting part of a story by Jack London or Bret Harte, and without warning tell them that that was as far as I had gone in the book and it would have to be continued the following evening. It was not true, of course; but I had to make certain of my new-found power and position. I enjoyed the long summer evenings until school began in the fall. Other words of mine have been listened to by larger and more fashionable audiences, but for that tough and athletic one that sat close on the stoop outside the candy store, I have an unreasoning love that will last forever.

1. Watching the boys playing baseball, the writer must have felt ________.

A. bitter and lonely B. special and different

C. pleased and excited D. disturbed and annoyed

2. The writer feels grateful even now to the boy who asked the question because the boy ________.

A. invited him to join in their game

B. liked the book that he was reading

C. broke the long silence of that summer evening

D. offered him an opportunity that changed his life

3. According to Paragraph 3, story-telling was popular among the boys basically because ________.

A. the story was from a children‘s book B. listening to tales was an age-old practice

C. the boys had few entertainments after dark

D. the boys didn‘t read books by themselves

4.The boys were attracted to Sister Carrie because ________.

A. it was written by Theodore Dreiser

B. it was specifically targeted at boys

C. it gave them a deeper feeling of pleasure

D. it talked about the wonders of the world 5. Sometimes the writer stopped at the most exciting part of a story to _______.

A. play a mean trick on the boys

B. experience more joy of achievement

C. add his own imagination to the story

D. help the boys understand the story better

6. What is the message conveyed in the story?

A. One can find his position in life in his own way.

B. Friendship is built upon respect for each other.

C. Reading is more important than playing games.

D. Adult habits are developed from childhood.

B

In 1978, I was 18 and was working as a nurse in a small town about 270 km away from Sydney, Australia. I was looking forward to having five days off from duty. Unfortunately, the only one train a day back to my home in Sydney had already left. So I thought I‘d hitch a ride (搭便车).

I waited by the side of the highway for three hours but no one stopped for me. Finally, a man walked over and introduced himself as Gordon. He said that although he couldn‘t give me a lift, I should come back to his house for lunch. He noticed me standing for hours in the November heat and thought I must be hungry. I was doubtful as a young girl but he assured （使…放心）me I was safe, and he also offered to help me find a lift home afterwards. When we arrived at his house, he made us sandwiches. After lunch, he helped me find a lift home.

Twenty-five years later, in 2003, while I was driving to a nearby town one day, I saw an elderly man standing in the glaring heat, trying to hitch a ride. I thought it was another chance to repay someone for the favour I‘d been given decades earlier. I pulled over and picked him up. I made him comfortable on the back seat and offered him some water.

After a few moments of small talk, the man said to me, ―You haven‘t changed a bit, even your red hair is still the same.‖ I couldn‘t remember where I‘d met him. He then told me he was the man who had given me lunch and helped me find a lift all those years ago. It was Gordon.

A. her work delayed her trip to Sydney

B. she was going home for her holidays

C. the town was far away from Sydney

D. she missed the only train back home

8. Which of the following did Gordon do according to Paragraph 2?

A. He helped the girl find a ride.

B. He gave the girl a ride back home.

C. He bought sandwiches for the girl.

D. He watched the girl for three hours.

9. The reason why the author offered a lift to the elderly man was that A. she realized he was Gordon

B. she had known him for decades

C. she was going to the nearby town

D. she wanted to repay the favour she once got

10. What does the author want to tell the readers through the story?

A. Giving sometimes produces nice results.

B. Those who give rides will be rapid.

C. Good manners bring about happiness.

D. People should offer free rides to others.

C

What makes a gift special? Is it the price you see on the gift receipt? Or is it the look on the recipient's face when they receive it that determines the true value? What gift is worth the most? This Christmas I was debating what to give my father. My dad is a hard person to buy for because he never wants anything. I pulled out my phone to read a text message from my mom saying that we were leaving for Christmas shopping for him when I came across a message on my phone that I had locked. The message was from my father. My eyes fell on a photo of a flower taken in Wyoming. and underneath a poem by William Blake. The flower, a lone dandelion standing against the bright blue sky, inspired me. My dad had been reciting those words to me since I was a kid. That may even be the reason why I love writing. I decided that those words would be my gift to my father.

I called back. I told my mom to go without me and that I already created my gift. I sent the photo of the cream-colored flower to my computer and typed the poem on top of it. As I was arranging the details another poem came to mind. The poem was written by Edgar Allan Poe; my dad recited it as much as he did the other. I typed that out as well and searched online for a background to the words of it. The poem was focused around dreaming, and after searching I found the perfect picture. The image was painted with blues and greens and purples, twisting together to create the theme and wonder of a dream. As I watched both poems passing through the

printer, the white paper coloring with words that shaped my childhood. I felt that this was a gift that my father would truly appreciate.

Christmas soon arrived. The minute I saw the look on my dad's face as he unwrapped those swirling black letters carefully placed in a cheap frame, I knew I had given the perfect gift.

11. The idea for a special gift began to form when the author was______.

A. doing shopping

C. reading a message B. having a debate D. leaving for Wyoming

12. The author's inspiration for the gift came from_____.

A. a photo of a flower

C. a call from the mother B. a story about a kid D. a text about Christmas

13. The underlined word "it" in Paragraph 3 refers to a poem by_____.

A. the father

B. the author D. Edgar Allan Poe C. William Blake

14. The author made the gift by_____.

A. searching for the poems online

C. painting the letters in three colors B. drawing the background by hand D. matching the words with pictures

15. What is the main purpose of the passage?

A. To show how to design images for gifts.

B. To suggest making gifts from one's heart.

C. To explain how computers help create gifts.

D. To describe the gifts the author has received

D

Planning a visit to the UK? Here we help with ways to cut your costs.

AVOID BIG EVENTS Big sporting events, concerts and exhibitions can increase the cost of accommodation and make it harder to find a room. A standard double room at the Thistle Brighton on the final Friday of the Brighton Comedy Festival (19 Oct.) cost £169.15 at Booking.com. A week later, the same room cost £118.15.

If you can be flexible and want to know dates to avoid—or you‘re looking for a big event to pass your time—check out sites such as Whatsonwhen.com, which allow you to search for events in the UK by city, date and category.

STAYAWAY FROM THE STATION If traveling to your destination by train, you may want to find a good base close to the station, but you could end up paying more for the sake of convenience at the start of your holiday.

Don‘t be too choosy about the part of town you stay in. Booking two months in advance, the cheapest room at Travelodge‘s Central Euston hotel in London for Saturday 22 September was

2015-2016新乡市高二下期末试卷答案(四)
2015-2016学年高二下学期期末考试数学(文)试题带答案

2016年春季学期期考试题

高二数学(文科)

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. 复数zi(i1)（i为虚数单位）的共轭复数是（ ）.

A．1i B．1i C．1i D．1i

2. 命题“对任意xR,都有x20”的否定为( ).

A．对任意xR,使得x20 B．不存在xR,使得x20

22C．存在x0R, 都有x00 D．存在x0R, 都有x00

3．“(2x1)x0”是“x0”的( ).

A．充分不必要条件

C．充分必要条件 B．必要不充分条件 D．既不充分也不必要条件

4. 设z是复数, 则下列命题中的假命题是( ).

A．若z20, 则z是实数

C．若z是虚数, 则z20 B．若z20, 则z是虚数 D．若z是纯虚数, 则z20

x2y2

1上的一点P到椭圆一个焦点的距离为3，则P到另一个焦点的距离5. 椭圆2516

为( ).

A．2

C．5 B．3 D．7

6. 若f(x)x2，则f(x)在x＝1处的导数为( ).

A．2x

x2y2

1的右焦点为(3,0)，则该双曲线的离心率等于( ). 7．已知双曲线2a5 B．2 C．3 D．4

A

．34 B

． C． D． 23144

8. 曲线y＝x3－2x＋4在点(1,3)处的切线的倾斜角为( ).

A．30°

C．60° B．45° D．120°

9. 设抛物线y2＝8x上一点P到y轴的距离是4，则点P到该抛物线焦点的距离是( ).

A．6 B．7 C．8 D．12

x2y2

10．若双曲线22【2015-2016新乡市高二下期末试卷答案】

1则其渐近线方程为( ). ab

A．y2x B．y2x 1C．yx 2D

．yx 11．已知函数y＝2x3＋ax2＋36x－24在x＝2处有极值，则该函数的一个递增区间是( ).

A．(2,3) B．(3，＋∞)

D．(－∞，3) C．(2，＋∞)

x2y2→⊥12．已知F1，F2是椭圆Ca＋b＝1(a>b>0)的两个焦点，P为椭圆C上一点，且PF1

→.若△PFF的面积为9，则b＝( )． PF212

A. 1 B. 2 C. 3 D. 4

二、填空题：(本大题共4小题，每小题5分，共20分.)

13．设复数z＝1＋2i(i是虚数单位)，则|z|＝________.

14．曲线yx2在点(1,1)处的切线方程为_______.

x2

y21上，顶点A是椭圆的一个焦点，且椭圆15. 已知△ABC的顶点B、C在椭圆4

的另外一个焦点在BC边上，则△ABC的周长是________．

x2

y21，过左焦点F作倾斜角为的直线交椭圆于A、B两点，则16. 已知椭圆：69

弦AB的长为__________.

三、解答题：(本大题共6小题，满分70分．解答须写出文字说明，证明过程或演算步骤)

17．（本小题满分10分）

计算: (1) (1i)(1i)(12i)2；

3－2i2－31－i(2) ； 2＋i

18. （本小题满分12分）

x22设F1和F2是双曲线4y＝1的两个焦点，点P在双曲线右支上，且满足∠F1PF2

＝90°，求△F1PF2的面积为S.

19．（本小题满分12分）

3已知直线x＋y－1＝0与椭圆x2＋by2＝4b的取值范围．

20.（本小题满分12分）

设x＝－2，x＝4是函数f(x)＝x3＋ax2＋bx的两个极值点．

(1) 求常数a，b；

(2) 判断x＝－2，x＝4是函数f(x)的极大值点还是极小值点，并说明理由．

21. （本小题满分12分）

1已知某工厂生产x件产品的成本为C＝25 000＋200x＋40x2 (元)．

(1) 要使平均成本最低应生产多少件产品？ ．．．．

(2) 若产品以每件500元出售，要使利润最大，应生产多少件产品？

22. （本小题满分12分）

x22已知椭圆C1：4y＝1，椭圆C2以C1的长轴为短轴，且与C1有相同的离心率．

(1)求椭圆C2的方程；

→＝2OA→，求直线AB的方(2)设O为坐标原点，点A，B分别在椭圆C1和C2上，OB

程．

2015-2016新乡市高二下期末试卷答案(五)
2015-2016学年河南省新乡市第二中学分校高二下期6月月考化学试卷(解析版)

河南省新乡市第二中学分校2015-2016学年高二下期6月月考化

学试卷（解析版）

1．下列物质的制备方法正确的是（ ）

A．氯化铝溶液和硫化钠溶液反应制备Al2S3

B.用镁粉和空气反应制备Mg3N2【2015-2016新乡市高二下期末试卷答案】

C．用铝粉和MnO2制得锰

D.用电解熔融氯化铝的方法制得单质铝

【答案】C

【解析】

试题分析：A、硫化钠和氯化铝溶液反应，硫离子和铝离子发生双水解反应生成氢氧化铝沉淀和硫化氢，产物，不选A；B、镁粉在空气中和氧气也反应，会生成氧化镁，错误，不选B；

C、铝热法可冶炼锰，正确，选C；D、氧化铝属于分子晶体，熔融状态下不导电，工业上电解熔融氧化铝的方法是冶炼铝，错误，不选D。

考点： 制备实验方案的设计，镁和铝的重要化合物

2．下列物质含有两种官能团的是( )

A.CH2Cl2 B.CH2=CHCl C.C2H5OH D.C3H8

【答案】B

【解析】A中含有2个氯原子，只有一种官能团；B中含有碳碳双键和Cl原子，有2种官能团；C中只有羟基，D没有官能团。

3．已知一定温度和压强下，合成氨反应：N2（g）+3H2（g）

2NH3（g）；△H=-92．0KJ·mol-1，将1mol N2和3mol H2充入一密闭容器中，保持恒温恒压，在催化剂存在时进行反应，达到平衡时，测得N2的转化率为α1，放出热量为Q1。若在相同条件下，起始时在该容器中充入2mol NH3，反应达到平衡时测得NH3的转化率为α2 ，吸收热量为Q2。下列说法错误的是（ ）

A．Q1 + Q2 = 92．0KJ B．α1 + α2 = 1

C．Q1 + Q2 = 184．0KJ D．α1/α2 = Q1/ Q2

【答案】C

【解析】

4．现代建筑的门窗框架常用电解加工成的古铜色合金制造。取合金样品进行如下实验（每一步试剂均过量）：

由此可以推知该合金的组成可能为

A．Al、Mg(OH)2、Si、S、Cu

C．Al2O3、SiO2、Fe、C、Cu B．Al、Fe、Mg、Si、Cu D．Al、Si、C、CuO、Cu

【答案】B

【解析】合金里每种金属都具有本身的化学性质。能溶于NaOH生成气体的有Al、Si，弄溶于盐酸生成气体的有Fe和Mg。

5．某元素原子3p能级上有一个空轨道，则该元素为( )

A．Na B．Mg C．Al D．Si

【答案】D

【解析】某元素原子3p能级上有一个空轨道，为3s23p2是硅

6．我国第五套人民币中的一元硬币材料为钢芯镀镍，硬币制作时钢芯应该做

A．阳极 B．阴极 C．正极 D．负极

【答案】B

【解析】

正确答案:B

电镀时，待镀金属放在阴极，镀层金属放在阳极。

7

【答案】B

【解析】【2015-2016新乡市高二下期末试卷答案】

试题分析：A、将水蒸气通过灼热的铁粉，粉末黑色，现象错误，故A错误；B、溶液显紫红色，说明生成I2，则证明氧化性：Fe3+＞I2，故B正确；C、Cu与硝酸反应生成硝酸铜，溶液变蓝，同时生成NO和水，该反应不属于置换反应，故C错误；D、向AgNO3溶液中滴加过量氨水，先生成AgOH后被过量的氨水溶解生成络离子，则Ag+与NH3∙H2O不能大量共存，故D错误；故选B。

【考点定位】考查化学实验方案的评价

【名师点晴】本题考查化学实验方案的评价，为高频考点，把握物质的性质及发生的反应为解答的关键，涉及钠的化合物与酸反应、银氨溶液的配制、硝酸的性质及氧化还原反应等，综合性较强。对几个实验方案的正确与错误、严密与不严密、准确与不准确做出判断．要考虑是否完全合理、有无干扰现象、经济上是否合算和对环境有无污染等。

-18．用已准确称量过的氯化钠固体配制2.00mol·L的NaCl溶液0.2L，需要用到的仪器是：

①坩埚②分液漏斗③容量瓶④烧杯⑤胶头滴管⑥烧瓶⑦玻璃棒⑧托盘天平⑨药匙

A．③④⑤⑦⑨ B．①②⑤⑥⑧ C．③④⑤⑦ D．③④⑤⑥

【答案】C

【解析】

试题分析：准确配制一定体积、一定浓度的溶液，在已准确称量后还需要用烧杯溶解，用玻璃棒搅拌、引流，用胶头滴管定容，用容量瓶盛装溶液。选项为：C。

考点：考查准确配制一定体积、一定物质的量浓度的溶液需要用到的仪器等相关知识。

9．下列说法正确的是( )

A．O2、O2互为同位素

B．H2O、D2O、T2O的化学性质不同 1618

C．和是两种不同结构

D．白磷与红磷互为同素异形体

【答案】D

【解析】

试题分析：A．O、O互为同位素，而16O2、18O2是由同位素原子形成的单质，化学性质1618

相同，是同一物质，错误；B．H2O、D2O、T2O是由H元素的三种同位素原子形成的水，它们

3的化学性质相同，错误；C．由于饱和碳原子是sp杂化，甲烷是正四面体结构，所以其二

溴代物是同一物质，结构相同，错误；D．白磷与红磷是磷元素的不同性质的单质，二者互为同素异形体，正确。

考点：考查有机物的概念的应用正误判断的知识。

10．在恒温恒压下，向密闭容器中充入4molSO2和2molO2，发生如下反应：2 SO2（g）+O2（g）

2SO3（g）△H＜0 。2min后，反应达到平衡，生成SO3为1．4mol,同时放出热量Q kJ，则下列分析正确的是

A．在该条件下，反应前后的压强之比为6 : 5．3

B．若反应开始时容器体积为2L，则v（SO3）＝0．35mol/（L·min）

C．若把“恒温恒压下”改为“恒压绝热条件下”反应，平衡后n（SO3）＜1．4mol

D．若把“恒温恒压下”改为“恒温恒容下”反应，达平衡时放出热量大于Q kJ

【答案】C

【解析】

试题分析：A．由于反应是在恒温恒压下，在该条件下，反应前后的压强相等，A错误；B．反应达到平衡时， 生成SO3为1．4mol，则反应消耗SO2和O2分别是1．4mol、0．7mol；平衡时气体的物质的量分别是SO2：2．6mol，O2：1．3mol，SO3:1．4mol，平衡时总物质的量是

5．3mol,开始时气体的物质的量是6mol,所以平衡时的容器的容积是：（5．3÷6）L, 则v（SO3）＝1．4mol ÷（5．3÷6）L ÷2min＝0．79mol/（L•min），B错误；C．由于该反应是放热反应，所以若把“恒温恒压下”改为“恒压绝热条件下”反应，则升高温度，根据平衡移动原理，化学平衡向吸热反应方向即向逆反应方向移动，所以达到平衡时产生的SO3的物质的量必然小于1．4mol，C正确；D．由于该反应的正反应是气体体积减小的反应，随着反应的进行，容器内的气体压强减小，所以若把“恒温恒压下”改为“恒温恒容下”反应，则反应后容器内气体压强减小，所以平衡逆向移动，达平衡时放出热量小于Q kJ，D错误，答案选C。

考点：考查可逆反应的化学反应速率、化学平衡的移动的知识

11．下列离子能大量共存的是

【答案】C

【解析】

+2-3+-试题分析：A、H与SiO3反应生成H2SiO3沉淀，不能大量共存，错误；B、Fe与SCN反应生

成红色Fe（SCN）3，不能大量共存，错误；C、离子之间都不反应，可以大量共存，正确；D、

H与AlO2反应生成氢氧化铝沉淀，不能大量共存，错误，答案选C。

考点：考查离子大量共存的判断

12．一氯代物有2种，二氯代物有4种的烃是

A．丙烷 B．2-甲基丙烷 C．丙烯 D．苯

【答案】A

【解析】

试题分析：丙烷分子中有2种H原子，其一氯代物有2种，为CH3CH2CH2Cl，CH3CH（Cl）CH3，CH3CH2CH2Cl中含有3种H原子，再次一氯取代有3种产物，CH3CH（Cl）CH3中含有2种H原子，再次一氯取代有2种产物，其中有两种为同一物质，故丙烷的二氯代物有4种，选项A满足题意；2-甲基丙烷分子中都有2种H原子，其一氯代物有2种，为（CH3）（CH3）2CHCH2Cl，

（CH3）2CHCH2Cl中含有3种H原子，再次一氯取代有3种产物，（CH3）3CCl中含有13CCl，

种H原子，再次一氯取代有1种产物，其中有两种为同一物质，故2-甲基丙烷的二氯代物有3种，选项B不正确；丙烯分子有三种环境的H原子，一氯代物有3种，选项C不正确；苯分子中只有1种H原子，其一氯代物只有1种，故D不符合；

考点：有机物取代产物的判断 涉及等效氢的判断

13．下列说法正确的是：

－1－1A．把100mL3mol·L的H2SO4跟100mL水混合，硫酸的物质的量浓度变为1.5 mol·L

－1－1-B．把200mL3mol·L的BaCl2溶液跟100mL3mol·L的KCl溶液混合后，溶液中的c(Cl)

－1仍然是3mol·L

C．把100 g 20%的NaCl溶液跟100 g H2O混合后，NaCl溶液的质量分数是10%

D．把100 mL 20%的NaOH溶液跟100 mL H2O混合后，NaOH溶液的质量分数是10%

【答案】C

【解析】 试题分析：A、稀释前后溶质硫酸的物质的量不变，混合溶液体积小于200ml，所以稀释后

-1-1硫酸的物质的量浓度大于为1.5mol•L，故A错误；B、3mol•L的BaCl2溶液中氯离子浓度

-1为6mol/L，3mol•L的KCl溶液中氯离子浓度为3molL，氢离子浓度介于3mol/L～6mol/L，

若忽略体积变化，混合后氯离子浓度约为=5mol/L，故B错误；C、混合后溶液质量为200g，溶质氯化钠的质量不变为100g×20%=20g，所以混合后氢氧化钠溶液质量分数为20/200×100%=10%，故C正确；D、水的密度小于氢氧化钠溶液密度，混合后溶液的质量小于2倍的100mL20%的氢氧化钠溶液质量，所以混合后的氢氧化钠溶液的质量分数大于10%，故D错误．故选：C．

考点： 溶液的稀释。

14．在三个密闭容器中分别充入Ne、H2、O2三种气体，当它们的温度和密度都相同时，这三种气体的压强（p）从小到大的是

A．p (Ne) < p (H2) < p (O2) B．p (O2) < p (Ne) < p (H2)

C．p (H2) < p (O2) < p (Ne) D．p (H2) < p (Ne) < p (O2)

【答案】B

【解析】

试题分析：由于气体的密度相同，则相同体积内物质的质量相等，而m=n·M,气体的摩尔质量越大，则单位体积内气体的物质的量就越小，在相同体积内气体的物质的量越小，容器内气体的压强就越小。相等分子质量：O2>Ne>H2；所以气体的物质的量n(H2)>n(Ne)>n(O2).所+-

以这三种气体的压强（p）从小到大的是p(O2)<p(Ne)<p(H2),因此选项是B。

考点：考查容器内气体的密度与气体的相对分子质量和气体的物质的量的关系的知识。

15．天津是我国研发和生产锂离子电池的重要基地。锂离子电池正极材料是含锂的二氧化钴

＋(LiCoO2)，充电时LiCoO2中Li被氧化，Li迁移并以原子形式嵌入电池负极材料碳(C6)中，

以LiC6表示。电池反应为CoO2＋LiC6

－LiCoO2＋C6，下列说法正确的是 ＋A．充电时，电池的阴极反应为LiC6－e===Li＋C6

＋－B．放电时，电池的正极反应为CoO2＋Li＋e===LiCoO2

C．羧酸、醇等含活泼氢的有机物可用作锂离子电池的电解质

D．锂离子电池的比能量(单位质量释放的能量)低

【答案】B

【解析】

＋－试题分析：充电相当于电解池，阴极得到电子，电极反应式为Li＋C6＋e===LiC6，A不正

确。放电相当于原电池，正极得到电子，所以B正确。羧酸、醇等含活泼氢的有机物可与金属锂反应，不能用作电解质，C不正确。金属锂的摩尔质量小，电池的比能量高。所以答案选B。

考点：二次电池

【名师点睛】“加减法”书写电极反应式

(1)先确定原电池的正负极，列出正负极上的反应物质，并标出相同数目电子的得失。

(2)根据氧化还原反应原理写出电极反应式

①负极反应

负极失去电子发生氧化反应。注意负极反应生成的阳离子与电解质溶液中的阴离子是否共存。若不共存，则该电解质溶液中的阴离子应写入负极反应式。

②正极反应

正极得到电子发生还原反应。当正极上的反应物质是O2时：若电解质溶液为中性或碱性，

－－－则水必须写入正极反应式中，O2生成OH，写为O2＋2H2O＋4e===4OH；若电解质溶液为酸

＋＋－性，则H必须写入正极反应式中，O2生成水，写为O2＋4H＋4e===2H2O。

(3)写出电池总反应方程式

结合电子守恒将正负极电极反应式相加即得到电池总反应方程式。

(4)若已知电池反应的总反应式，可先写出较易书写的一极的电极反应式，然后在电子守恒的基础上，总反应式减去较易写出的一极的电极反应式，即得到较难写出的另一极的电极反应式。

16．第三周期元素A、B、C，其原子序数依次增大，已知A的焰色为黄色，C的某种氧化物是形成酸雨的主要原因，且这三种元素的最高价氧化物的水化物有一种具有两性，且他们两两之间都能反应生成盐和水。

（1）填出元素符号A __________B __________C__________

2－ （2） C 离子结构示意图为

（3）A和C形成的化合物的电子式为 ，属于 化合物。

【答案】（1）Na、Al 、S ； （2） ；（3）

、离子化合物

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