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初一化学上学期考试665

初一化学上学期考试665
初一化学上学期考试665

(E)属第一代磺酰脲类口服降糖药

(2) 符合Glibenclamide的描述是

(A)含磺酰脲结构,具有酸性,可溶于氢氧化钠溶液

(B)结构中脲部分不稳定,在酸性溶液中受热易水解

(C)属第二代磺酰脲类口服降糖药

原文地址:2011年高考试题——上海化学(有答案)作者:黄山化学

2011年普通高等学校招生全国统一考试(上海卷)

化学

本试卷分为满分150分,考试时间120分钟。

相对原子质量:H-l C-12 N-14 O-16 F-19 Na-23 Mg-24 Si-28 S-32 Cu-64 I-127。第I卷(共66分)

一、选择题(本题共10分,每小题2分,只有一个正确选项,答案涂写在答题卡上。)1.下列离子在稳定人体血液的pH中起作用的是

A.Na+ B.HCO3-C.Fe2+ D.Cl-

2.从光合作用的反应原理6CO2+6H2O C6H12O6+6O2可知碳是农作物生长的必需元素之一。关于用二氧化碳生产肥料的评价正确的是

A.缓解地球温室效应,抑制农作物营养平衡

B.加剧地球温室效应,促进农作物营养平衡

C.缓解地球温室效应,促进农作物营养平衡

D.加剧地球温室效应,抑制农作物营养平衡

3.据报道,科学家开发出了利用太阳能分解水的新型催化剂。下列有关水分解过程的能量变化示意图正确的是

4.下列有关化学用语能确定为丙烯的是

5.高铁酸钾( K2FeO4)是一种新型的自来水处理剂,它的性质和作用是

A.有强氧化性,可消毒杀菌,还原产物能吸附水中杂质

B.有强还原性,可消毒杀菌,氧化产物能吸附水中杂质

C.有强氧化性,能吸附水中杂质,还原产物能消毒杀菌

D.有强还原性,能吸附水中杂质,氧化产物能消毒杀菌

二、选择题(本题共36分,每小题3分,只有一个正确选项,答案涂写在答题卡上。)6.浓硫酸有许多重要的性质,在与含有永分的蔗糖作用过程中不能显示的性质是

A.酸性B.脱水性C.强氧化性D.吸水性

7.下列溶液中通入SO2一定不会产生沉淀的是

A.Ba(OH)2 B.Ba(NO3)2 C.Na2S D.BaCl2

8.高炉炼铁过程中既被氧化又被还原的元素是

A.铁B.氮C.氧D.碳

9.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

10.草酸晶体(H2C2O4·2H2O) 100℃开始失水,101.5℃熔化,150℃左右分解产生H2O、CO和CO2。用加热草酸晶体的方法获取某些气体,应该选择的气体发生装置是(图中加热装置已略去)

11.根据碘与氢气反应的热化学方程式

(i) I2(g)+ H2(g)2HI(g)+ 9.48 kJ (ii) I2(S)+ H2(g) 2HI(g) - 26.48 kJ

下列判断正确的是

A.254g I2(g)中通入2gH2(g),反应放热9.48 kJ

B.1 mol固态碘与1 mol气态碘所含的能量相差17.00 kJ

C.反应(i)的产物比反应(ii)的产物稳定

D.反应(ii)的反应物总能量比反应(i)的反应物总能量低

12.甲醛与亚硫酸氢钠的反应方程式为HCHO+NaHSO3 HO-CH2-SO3Na,反应产物俗称“吊白块”。关于“吊白块”的叙述正确的是

A.易溶于水,可用于食品加工B.易溶于水,工业上用作防腐剂

C.难溶于水,不能用于食品加工D.难溶于水,可以用作防腐剂

13.某溶液中可能含有Na+、NH4+、Ba2+、SO42-、I-、S2-。分别取样:①用pH计测试,溶液显弱酸性;②加氯水和淀粉无明显现象。为确定该溶液的组成,还需检验的离子是A.Na+ B.SO42-C.Ba2+ D.NH4+

14.某物质的结构为,关于该物质的叙述正确的是

A.一定条件下与氢气反应可以生成硬脂酸甘油酯

B.一定条件下与氢气反应可以生成软脂酸甘油酯

C.与氢氧化钠溶液混合加热能得到肥皂的主要成分

D.与其互为同分异构且完全水解后产物相同的油脂有三种

15.β—月桂烯的结构如右图所示,一分子该物质与两分子溴发生加成反应的产物(只考虑位置异构)理论上最多有

A.2种B.3种C.4种D.6种

16.用电解法提取氯化铜废液中的铜,方案正确的是

A.用铜片连接电源的正极,另一电极用铂片B.用碳棒连接电源的正极,另一电极用铜片

C.用氢氧化钠溶液吸收阴极产物D.用带火星的木条检验阳极产物

17.120 mL含有0.20 mol碳酸钠的溶液和200 mL盐酸,不管将前者滴加入后者,还是将后者滴加入前者,都有气体产生,但最终生成的气体体积不同,则盐酸的浓度合理的是A.2.0mol/L B.1.5 mol/L C.0.18 mol/L D.0.24mol/L

三、选择题(本题共20分,每小题4分,每小题有一个或两个正确选项。只有一个正确选项的,多选不给分;有两个正确选项,选对一个给2分,选错一个,该小题不给分,答案涂写在答题卡上。)

18.氧化还原反应中,水的作用可以是氧化剂、还原剂、既是氧化剂又是还原剂、既非氧化剂又非还原剂等。下列反应与Br2+SO2+2H2O=H2SO4+2HBr相比较,水的作用不相同的是A.2Na2O2+2H2O=4NaOH+O2↑B.4Fe(OH)2+O2+2H2O=4Fe(OH)3

C.2F2+2H2O=4HF+O2 D.2Al+2NaOH+2H2O=2NaAlO2+3H2↑19.常温下用pH为3的某酸溶液分别与pH都为11的氨水、氢氧化钠溶液等体积混合得到

a、b两种溶液,关于这两种溶液酸碱性的描述正确的是

A.b不可能显碱性B.a可能显酸性或碱性

C.a不可能显酸性D.b可能显碱性或酸性

20.过氧化钠可作为氧气的来源。常温常压下二氧化碳和过氧化钠反应后,若固体质量增加了28 g,反应中有关物质韵物理量正确的是(NA表示阿伏加德罗常数)

二氧化碳

碳酸钠

转移的电子

A

1mol

NA

B

22.4L

1mol

C

106 g

1mol

D

106g

2NA

21.在复盐NH4Fe(SO4)2溶液中逐滴加入Ba(OH)2溶液,可能发生的反应的离子方程式是A.Fe2++SO42-+Ba2++2OH-=BaSO4↓+Fe(OH)2↓

B.NH4++Fe3++ 2SO42-+ 2Ba2++ 4OH-=2BaSO4↓+ Fe(OH)3↓+ NH3·H2O

C.2Fe3++ 3SO42-+ 3Ba2++6OH-=3BaSO4↓+ 2Fe(OH)3↓

D.3NH4++ Fe3++3SO42-+ 3Ba2++ 6OH-=3BaSO4↓+Fe(OH)3↓+3NH3·H2O

22.物质的量为0.10 mol的镁条在只含有CO2和O2混合气体的容器中燃烧(产物不含碳酸镁),反应后容器内固体物质的质量不可能为

A.3.2g B.4.0g C.4.2g D.4.6g

第II卷(共84分)

四、(本题共24分)

23.工业上制取冰晶石(Na3AlF6)的化学方程式如下:

2Al(OH)3+ 12HF+ 3Na2CO3=2Na3AlF6+ 3CO2↑+ 9H2O

根据题意完成下列填空:

(1)在上述反应的反应物和生成物中,属于非极性分子的电子式,属于弱酸的电离方程式。

(2)反应物中有两种元素在元素周期表中位置相邻,下列能判断它们的金属性或非金属性强弱的是

(选填编号)。

a.气态氢化物的稳定性b.最高价氧化物对应水化物的酸性

c.单质与氢气反应的难易d.单质与同浓度酸发生反应的快慢

(3)反应物中某些元素处于同一周期。它们最高价氧化物对应的水化物之间发生反应的离子方程式为。

(4)Na2CO3俗称纯碱,属于晶体。工业上制取纯碱的原料是。

24.雄黄(AS4S4)和雌黄(As2S3)是提取砷的主要矿物原料,二者在自然界中共生。根据题意完成下列填空:

(1)As2S3和SnCl2在盐酸中反应转化为As4S4和SnCl4并放出H2S气体。若As2S3和SnCl2正好完全反应,As2S3和SnCl2的物质的量之比为。

(2)上述反应中的氧化剂是,反应产生的气体可用吸收。(3)As2S3和HNO3有如下反应:As2S3+ 10H++ 10NO3—=2H3AsO4+ 3S+10NO2↑+ 2H2O 若生成2mol H3AsO4,则反应中转移电子的物质的量为。若将该反应设计成一原电池,则NO2应该在(填“正极”或“负极”)附近逸出。

(4)若反应产物NO2与11.2L O2(标准状况)混合后用水吸收全部转化成浓HNO3,然后与过量的碳反应,所产生的CO2的量(选填编号)。

a.小于0.5 mol b.等于0.5 mol c.大于0.5mol d.无法确定

25.自然界的矿物、岩石的成因和变化受到许多条件的影响。地壳内每加深1km,压强增大约25000~30000 kPa。在地壳内SiO2和HF存在以下平衡:SiO2(s) +4HF(g) SiF4(g)+ 2H2O(g)+148.9 kJ

根据题意完成下列填空:

(1)在地壳深处容易有气体逸出,在地壳浅处容易有沉积。

(2)如果上述反应的平衡常数K值变大,该反应(选填编号)。

a.一定向正反应方向移动b.在平衡移动时正反应速率先增大后减小

c.一定向逆反应方向移动d.在平衡移动时逆反应速率先减小后增大

(3)如果上述反应在体积不变的密闭容器中发生,当反应达到平衡时,(选填编号)。

a.2v正(HF)=v逆(H2O) b.v(H2O)=2v(SiF4)

c.SiO2的质量保持不变d.反应物不再转化为生成物

(4)若反应的容器容积为2.0L,反应时间8.0 min,容器内气体的密度增大了0.12 g/L,在这段时间内HF的平均反应速率为。

26.实验室制取少量溴乙烷的装置如右图所示。根据题意完成下列填空:

(1)圆底烧瓶中加入的反应物是溴化钠、和1:1的硫酸。配制体积比1:1的硫酸所用的定量仪器为(选填编号)。

a.天平b.量筒c.容量瓶d.滴定管

(2)写出加热时烧瓶中发生的主要反应的化学方程式。

(3)将生成物导入盛有冰水混合物的试管A中,冰水混合物的作用是。

试管A中的物质分为三层(如图所示),产物在第层。

(4)试管A中除了产物和水之外,还可能存在、(写出化学式)。

(5)用浓的硫酸进行实验,若试管A中获得的有机物呈棕黄色,除去其中杂质的正确方法是(选填编号)。

a.蒸馏b.氢氧化钠溶液洗涤

c.用四氯化碳萃取d.用亚硫酸钠溶液洗涤

若试管B中的酸性高锰酸钾溶液褪色,使之褪色的物质的名称是。(6)实验员老师建议把上述装置中的仪器连接部分都改成标准玻璃接口,其原因是:。

27.CuSO4·5H2O是铜的重要化合物,有着广泛的应用。以下是CuSO4·5H2O的实验室制备流程图。

根据题意完成下列填空:

(1)向含铜粉的稀硫酸中滴加浓硝酸,在铜粉溶解时可以观察到的实验现象:、

(2)如果铜粉、硫酸及硝酸都比较纯净,则制得的CuSO4·5H2O中可能存在的杂质是,除去这种杂质的实验操作称为。

(3)已知:CuSO4+2NaOH=Cu(OH)2↓+ Na2SO4

称取0.1000 g提纯后的CuSO4·5H2O试样于锥形瓶中,加入0.1000 mol/L氢氧化钠溶液28.00 mL,反应完全后,过量的氢氧化钠用0.1000 mol/L盐酸滴定至终点,耗用盐酸20.16 mL,则0.1000 g该试样中含CuSO4·5H2O g。

(4)上述滴定中,滴定管在注入盐酸之前,先用蒸馏水洗净,再用。

在滴定中,准确读数应该是滴定管上蓝线所对应的刻度。

(5)如果采用重量法测定CuSO4·5H2O的含量,完成下列步骤:

①②加水溶解③加氯化钡溶液,沉淀④过滤(其余步骤省略)

在过滤前,需要检验是否沉淀完全,其操作是

(6)如果1.040 g提纯后的试样中含CuSO4·5H2O的准确值为1.015 g,而实验测定结果是l.000 g 测定的相对误差为。

六、(本题共20分)

28.异丙苯(),是一种重要的有机化工原料。

根据题意完成下列填空:

(1)由苯与2-丙醇反应制备异丙苯属于反应;由异丙苯制备对溴异丙苯的反应试剂和反应条件为。

(2)异丙苯有多种同分异构体,其中一溴代物最少的芳香烃的名称是。(3)α-甲基苯乙烯()是生产耐热型ABS树脂的一种单体,工业上由异丙苯催化脱氢得到。写出由异丙苯制取该单体的另一种方法(用化学反应方程式表示)。

(4)耐热型ABS树脂由丙烯腈(CH2=CHCN)、1,3-丁二烯和α-甲基苯乙烯共聚生成,写出该树脂的结构简式(不考虑单体比例)。

29.化合物M是一种治疗心脏病药物的中间体,以A为原料的工业合成路线如下图所示。已知:RONa+ R’X→ROR’+ NaX

根据题意完成下列填空:

(1)写出反应类型。反应①反应②

(2)写出结构简式。A C

(3)写出的邻位异构体分子内脱水产物香豆素的结构简式。

(4)由C生成D的另一个反应物是,反应条件是。

(5)写出由D生成M的化学反应方程式。

(6)A也是制备环己醇( )的原料,写出检验A已完全转化为环己醇的方法。

七、(本题共16分)

30.氨和联氨(N2H4)是氮的两种常见化合物,在科学技术和生产中有重要的应用。

根据题意完成下列计算:

(1)联氨用亚硝酸氧化生成氮的另一种氢化物,该氢化物的相对分子质量为43.0,其中氮原子的质量分数为0.977,计算确定该氢化物的分子式。

该氢化物受撞击则完全分解为氮气和氢气。4.30g该氢化物受撞击后产生的气体在标准状况下的体积为L。

(2)联氨和四氧化二氮可用作火箭推进剂,联氨是燃料,四氧化二氮作氧化剂,反应产物是氮气和水。

由联氨和四氧化二氮组成的火箭推进剂完全反应生成72.0kg水,计算推进剂中联氨的质量。

(3)氨的水溶液可用于吸收NO与NO2混合气体,反应方程式为

6NO+ 4NH3=5N2+6H2O

6NO2+ 8NH3=7N2+12H2O

NO与NO2混合气体180 mol被8.90×103g氨水(质量分数0.300)完全吸收,产生156mol 氮气。吸收后氨水密度为0.980 g/cm3。

计算:①该混合气体中NO与NO2的体积比。

②吸收后氨水的物质的量浓度(答案保留1位小数)。

(4)氨和二氧化碳反应可生成尿素CO(NH2)2。尿素在一定条件下会失去氨而缩合,如两分子尿素失去一分子氨形成二聚物:

已知常压下120 mol CO(NH2)2在熔融状态发生缩合反应,失去80mol NH3,生成二聚物(C2H5N3O2)和三聚物。测得缩合产物中二聚物的物质的量分数为0.60,推算缩合产物中各缩合物的物质的量之比。

10.《周易》64卦,其实,讲的就是人生所可能面临的64种()。(分数:10分)

标准答案:B

学员答案:A

A.情景

B.情境

初一上英语期末考试题

北京市西城区2014-2015学年上学期初中七年级期末考试 英语试卷 试卷卷面共90分,口试10分,合计100分。 笔试时间共100分钟。 听力理解(共20分) 一、听对话。选出与对话内容相符的图片。每段对话读两遍。(共4分,每小题1分) 二、听句子。选择恰当的答语。每个句子读两遍。(共3分,每小题1分) 5. A. Beef. B. Oranges. C. Coffee. 6. A. Oh. Who is your grandma? B. Wow! Can I cook it? C. Really? How does it work? 7. A. I see. Here you are.

B. Yes. it’s a good idea to watch,TV. C. Oh. thank God. We’ll have a blue sky. 三、听对话。选出最佳答案。每段对话读两遍。(共6分,每小题1分) 请听一段对话,完成第8至第9小题。 8. Where are they going? A. Paris. B. London. C. New York. 9. When are they going there? A. In January. B. In February. C. In March. 请听一段对话,完成第10至第11小题。 10. When can the students have breakfast at school? A. At half past six. B. At six. C. At ten past six. 11. How many rules have they talked about? A. One. B. Two. C. Three. 请听一段对话,完成第12至第13小题。 12. What can Steven do? A. Play football. B. Drive a car. C. Cook. 13. What can we know about from the game? A. A country. B. A friend. C. A school. 四、听短文,根据短文内容记录关键信息。短文读两遍。(共4分,每小题1分)Time to get up At 14. __________a. m. Things to do for her job Work in a newspaper 15. __________and write something for most of the day. Time to get up At 5:45 a. m. Things to do for his job Walk about 12 miles. stop at about 90 buildings and 17. __________about 500 letters a day.

初一入学考试英语试卷

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