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初一化学上学期考试1023

初一化学上学期考试1023
初一化学上学期考试1023

17、稳定化合物、不稳定化合物、固熔体体系相图分析、应用 A B C

18、分配定律与萃取计算 A B

五、化学平衡

考试内容(考点)考试目标1、化学反应吉布斯函数变ΔrGm A B

自己收集整理的

错误在所难免

仅供参考交流

如有错误

请指正!谢谢

2008年高考(全国Ⅱ)理综化学部分试题(WORD)

6.2008年北京奥运会的"祥云"火炬所用的燃料的主要成分是丙烷

下列有关丙烷的叙述中不正确的是

A.分子中碳原子不在一条直线上

B.光照下能够发生取代反应

C.比丁烷更易液化

D.是石油分馏的一种产品

7.实验室现有3种酸碱指示剂

基PH变色范围如下

甲基橙:3.1~4.4 石蕊:5.0~8.0 酚酞:8.2~10.0

用0.1000mol/L NaOH溶液滴定未知浓度的CH3COOH溶液

反应恰好完全时

下列叙述中正确的是

A. 溶液呈中性

可选用甲基橙或酚酞作指示剂

B. 溶液呈中性

只能选用石蕊作指示剂

C. 溶液呈碱性

可选用甲基橙或酚酞作指示剂

D. 溶液呈碱性

只能选用石蕊作指示剂

8.对于IVA族元素

下列叙述中不正确的是

A. SiO2和CO2中

Si和O、C和O之间都共价键

B. Si、C、Ge的最外层电子数都是4

层电子数都是8

C. SiO2和CO2中都是酸性氧化物

在一定条件下都能和氧化钙反应

D. 该族元素的主要化合价是+4和+2

9.取浓度相等的NaOH和HCl溶液

以3∶2体积比相混和

所得溶液的PH等于12

则原溶液的浓度为

A.0.01mol/L

B. 0.017mol/L

C. 0.05mol/L

D. 0.50mol/L

10.右图为直流电源电解稀Na2SO4水溶液的装置

通电后在石墨电极a和b附近分别滴加一滴石蕊试液

下列实验现象中正确的是

A. 逸出气体的体积a电极的小于b电极的

B. 一电极逸出无味气体

另一电极逸出刺激性气味气体

C. a电极附近呈红色

b电极附近出现蓝色

D. a电极附近呈蓝色

b电极附近出现红色

11.某元素的一种同位素X的原子质量数为A

含N个中子

它与1H原子组成的HmX分子

在ag HmX分子中含质子的物质的量是

A.(A-N+m)mol

B.(A-N)mol

C. ( A-N)mol

D. (A-N+m)mol

12.(NH4)2SO4在高温下分解

产物是SO2、H2O、N2和NH3

在该反应的化学方程式中

化学计量数由小到大的产物分子依次是

A. SO2、H2O、N2、NH3

B. N2、SO2、H2O 、NH3

C. N2、SO2、NH3、H2O

D. H2O、NH3、SO2、N2

13.在相同温度和压强下

对反应CO2(g)+H2(g)=CO(g)+H2O (g)进行甲、乙、丙、丁四组实验, 实验起始时放入容器内各组分的物质的量见下表

物质的量

CO2

H2

CO

H2O

a mol

a mol

0 mol

0 mol

2a mol

a mol

0 mol

0 mol

0 mol

0 mol

a mol

a mol

a mol

0 mol

a mol

a mol

上述四种情况达到平衡后

n(CO)的大小顺序是

A.乙=丁>丙=甲B乙>丁>甲>丙C.丁>乙>丙=甲D.丁>丙>乙>甲

26.红磷P(S)和Cl2发生反应生成KCL3和PCL5

反应过程和能量关系如图所示(图中的

△H表示生成1mol产物的数据)

根据上图回答下列问题

(1)P和Cl2反应生成PCl3的热化学方程式;

(2)PCl5分解生成PCl3和Cl2的热化学方程式;

上述分解反是一个可逆反应

温度T1时

在密闭容器中加入0.8mol PCl5

反应达到平衡时还剩余0.6mol PCl5

其分解率α1等于;若反应温度由T1升高到T2

平衡时PCl5分解率α2

α1 α2 (填"大于"

"小于"或"等于");

(3)工业上制备PCl5通常分两步进行

先将P和Cl2反应生成中间产物PCl3

然后降温

再和Cl2反应生成PCl5

原因是;

(4)P和Cl2分两步反应生成1mol PCl3的△H1= ;P和Cl2一步反应生成1mol PCl5的△H2 △H1(填"大于"

"小于"或"等于");

(5)P Cl5与足量水反应

最终生成两种酸

其化学方程式是

27.(15分)

Q、R、X、Y、Z为前20号元素中的五种

Q的低价氧化物与X单质分子的电子总数相等

R与Q同族

Y和Z的离子与Ar原子的电子结构相同且Y原子序数小于Z

.

(1)Q的最高价氧化物

其固态属于晶体

俗名叫:

(2)R的氢化物分子的空间构型是

属于分子(填"极性"或"非极性");它与X形成的化合物可作为一种重要的陶瓷材料其化学式为;

(3)X的常见氢化物的空间构型是

它的另一氢化物X2H4是火箭燃料的成分

其电子式是;

(4)Q分别与Y、Z形成的共价化合物的化学式是和;Q与Y形成的化合物的电子式为

属于分子(填"极性"或"非极性")

28.(13分)

某钠盐溶液可能含有阴离子NO3-、CO32-、SO32-、SO42-、Cl-、Br-、I-、为鉴别这些离子

分别取少量溶液进行以下实验:

①测得混合液呈碱性;

②加HCl后

生成无色无味气体

该气体能使饱和石灰水溶液变浑浊;

③加CCl4后

滴加少量氯水

振荡后

CCl4后层未变色;

④加BaCl2溶液产生白色沉淀

分离

在沉淀中加入足量盐酸

沉淀不能完全溶解;

⑤加HNO3酸化后

再加过量AgNO3

溶液中析出白色沉淀

(1)分析上述5个实验

写出每一步实验鉴定离子的结论与理由

实验①;

实验②;

实验③;

实验④;

实验⑤;

(2)上述5个实验不能确定是否存在的离子是

29.(17分)

A、B、C、D、E、F、G、H、I、J均为有机化合物

根据以下框图

回答问题:

(1)B和C均为有支链`的有机化合物

B的结构式为;C在浓硫酸作用下加热反应只能生成一种烯烃D D的结构简式为

(2)G能发生银镜反应

也能使溴的四氯化碳溶液褪色

则G的结构简式为

(3)写出⑤的化学反应方程式

⑨的化学反应方程式

(4)①的反应类型

④的反应类型

⑦的反应类型

(5)与H具有相同官能团的H的同分异构体的结构简式为

参考答案

CDBCDACA

26.答案:(1)2P(s)+3Cl2(g)=2PCl3(g) △H=-612kJ/mol

(2)PCl5(g)=PCl3(g)+Cl2(g) △H=+93kJ/mol 25% 小于

(3)因为PCl5分解反应是吸热反应

温度太高

不利于PCl5的生成

(4)-306kJ/mol 大于

(5)PCl5+4H2O=H3PO4+5HCl

27.答案:(1)分子

干冰

(2)正四面体

非极性Si3N4

(3)三角锥形

28.答案(1)①说明含有CO32-、因为只有它水解显碱性

②不含有SO32-、因SO2有刺激性气味

③不含有Br-、I-

因两者均能与氯水反应后溶解于CCl4显色

④说明含有SO42-

因BaSO4溶液于盐酸

⑤说明含有Cl-

因AgNO3与Cl-反应生成的AgCl不溶于稀HNO3

(2)NO3-

29.答案:(1)

(2)

(3)

(4)水解

取代

氧化

(5)CH3CH=CHCOOH CH2=CHCH2COOH

??

??

??

??

2、溶胶制备、净化,胶团结构式表示 A B

3、溶胶的动力性质(布朗运动、沉降平衡) A B

4、溶胶的光学性质(丁达尔现象、光的散射) A B

5、胶团的双电层结构,电泳、电渗、ξ电势 A B C

6、憎液溶胶的聚沉 A B

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人教版初一英语期末考试试题以及标准答案

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