最新物质构成的奥秘测试题及答案.docx

最新物质构成的奥秘测试题及答案

一、物质构成的奥秘选择题

1．以下事实，不能说明分子间有间隔的是（）

A．海绵能吸水B．用打气筒能将气体压缩

C．物体有热胀冷缩的现象D．酒精和水混合后，总体积小于两者的体积之和

【答案】 A

【解析】

【详解】

A.海绵能吸水，是因为海绵中有无数细小孔隙，当置于水中时，水会充满这些孔隙，不能

说明微粒之间有间隔，此选项错误；B. 用打气筒能将气体压缩，是通过加压减小分子间的

间隔，能说明分子之间有间隔，此选项正确； C. 物体热胀冷缩，是因为分子间的间隔随温度的变化而变化的结果，能说明分子间有间隙，此选项正确； D. 酒精和水混合后，总体积小于两者的体积之和，是因为酒精和水都是由分子构成的，分子之间有间隔，当二者混合

时，一部分分子填充了另一种分子之间的间隔，因此总体积会减少，能说明分子之间有间

隔，此选项正确。故选A。

【点睛】

本题主要考查了分子的特征，会用分子的知识解释日常生活中的现象，反过来，通过日常

生活中的现象会总结分子的特征。

2．如图，是两种气体发生反应的微观示意图，其中相同的球代表同种原子，相关说法正

确的是( )

A．分子在化学变化中不可分

B．反应后生成了两种新的化合物

C．原子在化学反应中可分

D．化学反应前后原子的种类不变

【答案】 D

【解析】

试题分析：由图可知，反应前后原子的种类不变，生成物只有一种，是化合物；

考点：化学反应的微观过程

3．富硒大米中的硒元素具有防癌和抗癌作用。右图是硒元素在元素周期表中的信息及硒原

子的原子结构示意图。下列有关说法错误的是（）。

A．图示中x = 18B．硒的相对原子质量是78.96

C．硒属于非金属元素D．硒原子在反应中易失去 6 个电子

【答案】 D

【解析】

A、2+8+x+6=34， x=18，正确；

B、由图可知，硒的相对原子质量是78.96，正确；

C、硒字的偏旁是石字，故是固态非金属元素，正确；

D、硒原子在反应中易得到 6 个电子形成8 电子稳定结构，错误。故选D。

4．生活中废弃的铝质易拉罐、破铁锅、废铜线等可归为一类加以回收，它们属于（）A．化合物

【答案】 B

B．金属或合金C．氧化物D．非金属

【解析】

铝质易拉罐、破铁锅、废铜线都属于废旧金属材料，主要成分是金属单质或合金，与有机

物氧化物非金属无任何关系，应属于金属或合金．

故选 B

点评：此题是对废旧金属的考查，属于废旧金属回收的知识，属基础性知识考查题．

5．如图所示是甲、乙粒子的结构示意图，下列有关说法正确的是（）

A．甲粒子是阳离子B．乙粒子是原子

C．两种粒子属于同种元素

【答案】 C

D．甲粒子的化学性质比乙粒子稳定

【解析】

【分析】

【详解】

A、甲粒子的质子数是8，核外电子数是8，属于原子．故 A 说法不正确；

B、乙粒子的质子数是8，核外电子数是10，属于阴离子．故 B 说法不正确；

C、由于甲、乙的核内的质子数相同，属于属于同种元素．故 C 说法正确；

D、由上述分析可知，甲粒子的最外层电子数是6，在反应中已得到两个电子，乙粒子是带

两个单位的负电荷阴离子．故 D 说法不正确．

故选 C．

6．芯片是内含集成电路的硅片，下图是硅元素在元素周期表中的相关信息，下列说法正

确的是

A．硅元素属于金属元素

B．硅元素的原子序数为14

C．硅元素的相对原子质量为28.09g

D．硅元素是地壳中含量最高的元素

【答案】 B

【解析】

A、硅元素属于非金属元素，故 A 错误；

B、硅元素的原子序数为 14，故 B 正确；

C、硅元素的相对原子质量为 28.09，相对原子质量没有单位，故 C 错误；

D、氧元素是地壳中含量

最高的元素，其次是硅元素，故 D 错误。

点睛∶原子序数等于原子核内质子数等于核外电子数，元素分为金属元素，非金属元素和

稀有气体元素。

7．用“”和“”代表两种不同的单质分子，它们在一定条件下能发生化学反应，反应

前后的微观示意图如下所示，下列说法正确的是（）

A．该反应是化合反应B．该反应有 2 种生成物

C．每个生成物分子由 3 个原子构成D．参加反应的“”和“”分子的个数比是 2:1

【答案】 A

【解析】

【详解】

A.根据模拟模型可知，反应前有两种分子，反应后生成了一种分子，即该反应是两种物质

发生化学反应生成一种物质，属于化合反应，此选项正确； B. 反应后出现了一种新分子，即该反应有 1 种生成物，此选项错误； C. 根据模拟模型可知，每个生成物分子由 4 个原子构成，此选项错误； D. 根据模拟模型可知，反应后有 1 个剩余，所以参加反应的

“”和“”分子的个数比是6： 2=3:1，此选项错误。故选A。

8．如图是两种元素的原子结构示意图，据图分析，下列判断错误的是

A．甲元素属于金属元素

B．甲和乙两种元素的化学性质相似

C．甲和乙形成化合物的化学式为MgCl2

D．乙元素的原子在化学反应中易得电子

【答案】 B

【解析】

【分析】

【详解】

A、X 是 12 号元素，是镁元素，故是金属元素，故正确；

B、 X、 Y 两种元素的最外层电子数不同，故化学性质不相似，故错误；

C、 X 是镁元素， Y 是氯元素，二者形成的化合物是2，

正确；

MgCl

D、 Y 元素的原子最外层的电子数大于7 ，在化学反应中易得电子，正确。故选B。

9．科学家已研究出高能微粒N5+，则N5+的结构可能是( )

A． N5+是由 5 个氮原子构成的单质B．N5+中只有质子，没有中子和电子C．每个N5+中含有35 个质子和34 个电子D． N5+中含有35 个质子和35 个电子【答案】C

【解析】

【分析】

【详解】

A、高能微粒N5+带一个单位的正电荷，属阳离子，选项 A 错误；

B、一个氮原子有7 个质子，那么N5中含有 35 个质子，带一个单位的正电荷，说明失去了一个电子，那么有34 个电子，选项 B 错误；

C、一个氮原子有7 个质子，那么N5+中含有 35 个质子，带一个单位的正电荷，说明失去了一个电子，那么有34 个电子，选项C正确；

D、一个氮原子有 7 个质子，那么 N5+中含有 35 个质子，带一个单位的正电荷，说明失去了一个电子，那么有 34 个电子，选项 D 错误。故选 C。

10．对下列事实的解释错误的是 ()

事实解释

A湿衣服晾干分子在不停运动

B铁轨接头处留有一定的缝隙原子之间有间隔

固体中离子不能自由移动，溶液中离子能自C氯化钠固体不导电，溶液导电

由移动

D氢氧化钠溶液显碱性溶液中含有氢氧原子

A． A B． B C． C D． D

【答案】 D

【解析】

事实解释

A湿衣服晾干分子在不停运动，向空气中扩散，故 A 正确；

B 铁轨接头处留有一定的缝

原子之间有间隔，防止热胀冷缩，故 B 正确；隙

C 氯化钠固体不导电，溶液固体中离子不能自由移动，溶液中离子能自由移动，导电故 C正确；

D氢氧化钠溶液显碱性溶液中含有氢氧离子，不是原子，故 D 错误。

点睛∶ 分子基本性质⑴分子质量非常小、分子的体积很小。⑵ 、分子并不是静止不动的，

它总是在不断的运动。⑶ 、分子间有一定的间隙。⑷同种物质的分子，性质相同；不同种

物质的分子，性质不相同。酸碱盐溶液之所以能够导电是因为溶液中有自由移动的离子。11．根据下表提供的信息，下列判断错误的是

第一周期

第二周期

第三周期

A．元素所在的周期数等于其原子的电子层数

B．原子序数与元素原子核电荷数在数值上相同

C．第二、三周期元素的原子从左至右最外层电子数逐渐增多

D．金属元素的原子，其最外层电子数一般少于 4 个，在化学反应中易得到电子，趋向达

到相对稳定结构

【答案】D

【解析】

试题分析：A、元素所在的周期数等于其原子的电子层数，正确，B、原子序数与元素原子

核电荷数在数值上相同，正确，C、第二、三周期元素的原子从左至右最外层电子数逐渐增

多，正确， D、金属元素的原子，其最外层电子数一般少于 4 个，在化学反应中易失去电子，趋向达到相对稳定结构，而不是易得到电子，错误，故选D

考点：元素周期表中的规律，核外电子的排布

12．根据如图 R2-的离子结构示意图，推测m 的值为

A．10B． 16C． 18D． 20

【答案】 B

【解析】

【分析】

【详解】

由题意可知，图中是 R2-的离子结构示意图，则 R 原子的核外电子数是 2+8+8-2=16 ，原子中，

质子数 =核外电子数，即 m 的值为 16。

选 B。

【点睛】

原子结构示意图中，圆圈内的数字表示质子数，外面的弧线表示电子层，弧线上的数字表

示该电子层上的电子数。质子数大于核外电子数的是阳离子；小于核外电子数的是阴离子；

等于核外电子数的是原子。

原子由原子核和核外电子构成，原子核由质子和中子构成，原子中，质子数=核电荷数

外电子数 =原子序数。

=核

13．下列关于分子、原子、离子的说法正确的是（）

A．分子是保持物质性质的一种粒子

B．化学反应中任何离子都不能再分

C．物体有热胀冷缩现象，主要是因为物体中粒子大小随温度的改变而改变。

D．分子、原子、离子都可以直接构成物质

【答案】 D

【解析】

【详解】

A、由分子构成的物质，分子是保持其化学性质的最小粒子，故说法错误；

B、离子包括带电的原子或原子团，由原子形成的离子化学变化中不会再分，而由原子团

形成的离子化学变化中可能再分，比如碳酸钙高温分解中，碳酸根离子分开了，反应生成

了二氧化碳，故说法错误；

C、物体有热胀冷缩现象，主要是因为物体中的粒子间隔随温度的改变而改变，故说法错

误；

D、分子、原子、离子都可以直接构成物质，故选项正确。故选 D 。

14．下列关于分子、原子、离子的叙述，正确的是

A．分子是保持物质性质的最小粒子B．只有带电的原子才叫离子

C．分子、原子、离子都可以直接构成物质D．原子是变化中最小的粒子

【答案】 C

【解析】

【分析】

【详解】

A、分子是保持物质化学性质的最小粒子，错误；

B、带电的原子或原子团叫离子，错误；

C、分子、原子、离子都可以直接构成物质，如水是由分子构成的，铁是由原子构成的，氯

化钠是由离子构成的，正确；

D、原子是化学变化中的最小粒子，错误。

故选 C。

15．下列说法正确的是（）

A．所有原子的原子核都是由质子和中子构成

B．在同一种物质中同种元素的化合价可能不相同

C．由同一种元素组成的物质一定是单质，不可能是化合物

D．分子、原子都是不带电的粒子，所以不带电的粒子一定是分子或原子

【答案】 B

【解析】

试题分析： A、氢原子核内只有一个质子，没有中子，故 A 错；

B、在硝酸铵中铵根中的氮元素显﹣ 3 价，硝酸根中的氮元素显+5 价，所以在同一种物质中同种元素的化合价可能不相同，故 B 正确；

C、由同种元素组成的纯净物是单质，由同一种元素组成的物质不一定是纯净物，例如氧

气和臭氧在一块属于混合物，故 C 错；

D、不带电的粒子不一定是分子或原子，可能是中子，故 D 错．

故选 B．

考点：原子的定义与构成；单质和化合物的概念；常见元素与常见原子团的化合价．

点评：解答本题关键是要知道原子核内一定有质子，不一定有中子；由同种元素组成的纯

净物是单质；中子不带电，熟悉在同一种物质中同种元素的化合价可能不相同．

16．科学家已研究出一种高能微粒N5+，关于它的说法正确的是

A． N5+是由氮元素组成的单质B．每个 N5+中含有 35 个质子和34 个电子C． N5+中只有质子，没有中子和电子D．每个 N5+中含有 35 个质子和36 个电子【答案】 B

【解析】

【分析】

【详解】

A、 N5+是一种高能微粒，不属于单质，选项 A 不正确；

B、每个 N5+中含有 35 个质子， N5+中的电子总数为五个氮原子的电子总数之和再减去一个

电子，为 35-1=34 ，选项 B 正确；

C、 N5+只是失去了一个电子，中子数没有发生改变，选项 C 不正确；

D、 N5+中的电子总数为五个氮原子的电子总数之和再减去一个电子，为35-1=34 ，选项 D 不正确。故选B。

17．如图是元素周期表的一部分，以下表述正确的是

A．氟原子核内的中子数为9

B．氟、硫、氯都是非金属元素

C．氯原子的相对原子质量是35.45g

D．氟和氯位于元素周期表同一周期

【答案】 B

【解析】

【分析】

【详解】

A、氟原子的核内中子数不是9，质子数是9，该选项说法不正确；

B、氟、硫、氯三种元素都属于非金属元素，该选项说法正确；

C、氯原子的相对原子质量是35.45，单位不是g，该选项说法不正确；

D、氟元素和氯元素核外电子层数不同，位于元素周期表的不同周期，该选项说法不

正确。

故选 B。

18．下列化学家中，发现元素周期律并制出元素周期表的是

A．道尔顿B．拉瓦锡

C．侯德榜D．门捷列夫

【答案】 D

【解析】

A、道尔顿在化学学科中的主要成就，是首先发现了电子，提出了原子论．B拉瓦锡在化学学科中的主要成就，是首次利用天平为化学研究的工具进行定量实验，并测定了空气的组

成． C. 侯德榜发明了连续生产纯碱与氯化铵的联合制碱新工艺，以及碳化法合成氨流程制碳

酸氢铵化肥新工艺； D. 门捷列夫在化学学科中的主要成就，是发现了元素周期律，并首次编

制了元素周期表，选 D

19．下列结构示意图中，属于阳离子的是（）

A．B．C．D．

【答案】 A

【解析】

试题分析：质子数小于核外电子数的，因电子数大于质子数，所以微粒带负电．其中 B 质子数小于电子数是阳离子．CD 质子数等于电子数表示原子． A 中质子数小于电子数，表示

阴离子．

考点：原子结构示意图与离子结构示意图

点评：非金属元素最外层电子数一般大于 4 个，所以在发生化学变化时，易得电子，带负电，

形成阴离子．

20．如图 A 是钾元素在元素周期表中的部分信息，图 B 是钾原子的结构示意图，下列判断

不正确的是

A．钾的相对原子质量是39.10B．钾原子在化学反应中容易失去电子

C．钾元素在元素周期表中排在第四周期D．钾元素和钾离子的电子层数相同

【答案】 D

【解析】

【分析】

【详解】

A、根据元素周期表一格提供的信息可知，钾的相对原子质量是39.10，正确；

B、钾元素的原子最外层电子数是1，在化学反应中容易失去电子，正确；

C、钾元素核外电子层数是4，在元素周期表中应该排在第四周期，正确；

D、钾元素和钾离子的电子层数不相同，钾原子核外电子层数是4，钾离子核外电子层数是3，错误。故选D。

英语补考试卷及答案

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