2015北京西城区初三化学二模试题及答案

北京市西城区2015年初三二模试卷

化 学 2015.6

考生须知

1.本试卷共8页,共两部分,38道小题,满分80分。考试时间100分钟。 2.在试卷和答题卡上准确填写学校名称、姓名和准考证号。 3.答案一律填写在答题卡上,在试卷上作答无效。 4.考试结束,将试卷和答题卡一并交回。

可能用到的相对原子质量

H 1 C 12 O 16 Na 23 Al 27 S 32 Fe 56

第一部分 选择题

(每小题只有一个选项符合题意) 1.地壳中含量最多的元素是

A .氧

B .硅

C .铝

D .铁 2.下列过程中发生化学变化的是

A .铁矿石炼铁

B .干冰制冷

C . 海水晒盐

D .风能发电 3.下列属于氧化物的是 A .O 2 B .SO 2 C .KCl D .KOH 4.下列二氧化碳的用途中,不正确...的是 A .供给呼吸 B .用作气体肥料 C .生产碳酸饮料 D .用于灭火 5.下列符号表示1个氢分子的是

A .H

B .H +

C .H 2

D .2H

6.公共场所的环境要靠我们一起来维护。下列图标中表示“禁止吸烟”的是

7.尿素【CO(NH 2) 2】是一种高效化肥,尿素属于

A .磷肥

B .氮肥

C .钾肥

D .复合肥 8.下列物质放入水中,能形成无色溶液的是 A .豆浆 B .冰块 C .硫酸铜 D .蔗糖 9.下列物质的化学式中,书写正确的是

A .氧化镁MgO 2

B .氯化铁FeCl 2

C .硝酸银 AgNO 3

D .氢氧化铜CuOH 10.下列物质露置于空气中,一段时间后,质量增大的是 A .碳酸钙 B .氯化钠

C .浓盐酸

D .氢氧化钠 11.用右图净水器处理浑浊的河水,下列叙述正确的是 A .能消灭水中的细菌 B .能得到纯净物 C .能减少河水的异味 D .能除去钙镁化合物 12.下列做法正确的是

A .氧气用于切割金属

B .氢氧化钠用于中和酸性土壤

C .电器着火用水浇灭

D .焚烧废弃塑料解决“白色污染” 13.下列物品所使用的主要材料,属于天然有机高分子材料的是

A

B C D

A.塑料玩具 B.棉线帽子 C.陶瓷碗 D.尼龙提包14.下列操作能鉴别空气、氧气和二氧化碳 3 瓶气体的是

A.闻气味 B.将集气瓶倒扣在水中

C.观察颜色 D.将燃着的木条伸入集气瓶中

15.清凉油具有散热、醒脑、提神的功效,其主要成分为薄荷脑(化学式为C10H20O)。

下列有关薄荷脑的说法不正确

...的是

A.薄荷脑是3种元素组成的有机物B.1个薄荷脑分子中含有31个原子

C.薄荷脑中碳元素的质量分数最大 D.薄荷脑中碳、氢元素的质量比为1︰2 16.下列实验操作正确的是

A.加热液体 B.稀释浓硫酸 C.检验气密性 D.熄灭酒精灯

17. 20℃时,取甲、乙、丙、丁四种物质各10 g,分别加入到四个盛有50 g水的烧杯中,充分溶解,情况如下表(温度保持20℃),下列说法正确的是

物质甲乙丙丁未溶解固体的质量/g 4.2 2 0 9.2

A.所得溶液一定都是饱和溶液

B.溶质质量分数最大的是丁溶液

C.所得四种溶液的质量关系为:丙>乙>甲>丁

D.20℃时四种物质溶解度的关系为:丁>甲>乙>丙

18.甲和乙在一定条件下恰好完全反应生成丙和丁。结合微观示意图分析,下列结论正确的是

A.甲的化学式为H2O

B.反应中甲、乙、丁的分子个数比为1︰2︰1

C.若32 g乙参加反应,则生成18 g丁

D.丁中氧元素的质量一定等于参加反应的乙的质量

19.下列实验方案能达到实验目的的是

选项实验目的实验方案

A 区分氢氧化钠溶液和氢氧化钙溶液取样,加入无色酚酞溶液

B 检验氯化钙溶液中含有稀盐酸

取样,加入适量的碳酸钙

C 除去氯化铜溶液中的氯化锌加入过量的锌,过滤

D 证明氢氧化钠已部分变质取样,加入少量的稀盐酸

20.下列4个坐标图分别表示4个实验过程中某些量的变化,其中正确的是

甲乙丙丁

海水

分离膜

淡水

A .向盐酸中加入水

B .10 mL 5%的H 2O 2

溶液分解

C .将饱和石灰水升温

D .向一定量的稀盐酸

中加入铁粉

第二部分 非选择题

〖生活现象解释〗

21.(2分)化学与人体健康息息相关。

(1)“铁强化酱油”中的铁元素可预防 (填“贫血症”或“佝偻病”)。 (2)食用下列食品,通常不会..

对人体健康造成危害的是 (填序号)。 A. 霉变的大米、花生 B. 甲醛溶液浸泡的水产品 C. 牛奶经发酵后得到的酸奶 22.(2分)金属在生产、生活中广泛应用。

(1)铜丝用作导线,利用了铜的延展性和 性。

(2)不能用铁制容器盛放硫酸铜溶液的原因是 (用化学方程式表示)。

23.(3分)下图为生活中一些物质的pH 。依据图示回答: (1)柠檬汁呈 性。 (2)碱性最强的物质是_____。

(3)蚊虫叮咬后,皮肤因蚊虫分泌出的蚁酸而痛痒, 可帮助减轻痛痒的物质是 (答1种即可)。 24. (2分)水是生命活动中不可缺少的物质。 (1)生活中将硬水转化成软水的方法是 。 (2)膜分离技术可用于淡化海水(如右图所示), 通过膜的主要微粒是 。

25.(2分)日常生活中,食品保鲜的方法很多。

(1)山梨酸钾是常用的防腐剂,它可由山梨酸和氢氧化钾发生中和反应制得。山梨酸钾属于 (填

“酸”、“碱”或“盐”)。

(2)生活中常用隔绝空气的方法防腐,例如 (举1例即可)。 26.(3分)“××糕点”主要营养成分如右表所示。请根据该表回答: (1)所含营养素中属于无机盐的是 。 (2)表中未列出的一类营养素是 。 (3)能为人体提供能量的营养素是 。 27.(3分)钠元素对人类生命和生活具有重要意义。 (1)食盐是生活中最常用的 。

(2)小苏打可治疗胃酸过多症,该反应的化学方程式是 。 (3)固体叠氮化钠(NaN 3)被广泛应用于汽车安全气囊。撞击后30毫秒内,NaN 3迅速分解生成两种单

质,该反应的化学方程式是 。

28.(3分)下列是生活中常用的清洗剂。

营养素 每100 g 含有 蛋白质 4.8 g 糖 类 40 g ××糕点

油 脂 20 g 碳酸钙 400 mg

维生素 1.8 mg

名称

84消毒液

洁厕灵

污渍爆炸盐 产品示例

有效成分 次氯酸钠 盐酸 过碳酸钠

(1)次氯酸钠(NaClO )中氯元素的化合价为 。 (2)洁厕灵在使用时禁止与皮肤接触,其原因是 。

(3)“污渍爆炸盐”溶于水后会生成Na 2CO 3和H 2O 2,与洁厕灵混合时产生的气体可能是 。 〖科普阅读理解〗

29.(5分)阅读下面科普短文。

【资料1】融雪剂按组成分为有机融雪剂和无机融雪剂。 有机融雪剂的主要成分为醋酸钾,一般用于机场等重要场所。

无机融雪剂以氯化钠、氯化钙等为主要成分,其融雪原理与相同条件下,不同浓度的食盐水的凝固点有关(见下表)。

溶剂质量(g) 100 100 100 100 100 溶质质量(g ) 0 3.6 7.5 11.1 29.9 凝固点(℃)

-2.3

-4.2

-6.3

-21

【资料2】钛(Ti )是一种银白色的金属。它具有良好的耐高温、耐低温、抗酸碱以及高强度、低密度的特点,广泛用于航空工业;它无毒且具有优良的生物相容性,是非常理想的医用金属材料,可用来制造人造骨等。

金属钛不仅能在空气中燃烧,也能在二氧化碳或氮气中燃烧。在稀有气体和高温条件下,用四氯化钛和镁发生置换反应可制备金属钛。 依据资料,回答下列问题:

(1)若飞机场使用融雪剂,应选择融雪剂的类别是 。

(2)从无机融雪剂的融雪原理得出:100 g 溶剂中,溶质质量和凝固点的关系是 。 (3)钛金属广泛应用于航空工业的原因是 (答出2条即可)。 (4)金属钛着火时 (填“能”或“不能”)用二氧化碳灭火。 (5)四氯化钛制备金属钛的化学方程式是 。

〖生产实际分析〗

北京市从2010年开始试点实施垃圾分类,依据对生活垃圾进行分类处理与综合利用的部分流程,回答30题和31题。

30.(3分)【资料】焚烧垃圾产生的烟气含有害的气体,如HCl 、SO 2等

(1)可利用厨余垃圾得到 。

(2)1 t 塑料瓶可回炼600 kg 的无铅汽油和柴油。塑料瓶属于垃圾分类中的 类。

(3)石灰浆的作用是吸收CO2和。

31.(3分)【资料】①

②绿矾(FeSO4·7H2O)是一种浅绿色晶体,加热至70℃以上易分解

(1)I中主要发生3个反应,其中生成气体的化学方程式是。

(2)II中向FeSO4溶液里加入铁屑,依次经过:加热浓缩、过滤、降温结晶、(填操作名称),得到绿矾。其中铁屑的作用是。

〖物质组成和变化分析〗

32.(5分)A~F为初中常见物质,它们由氢、碳、氧、氯、钠、钙中的2~3种元素组成。

(1)气体A、B由两种相同元素组成且可以相互转化。A常用于冶炼金属,它的化学式为。若B 在高温条件下转化成A的反应是化合反应,该反应的化学方程式是。

(2)固体C和固体D含相同金属元素。用它们的混合物完成如下图所示实验。

已知:Na2CO3 + CaCl2 = CaCO3↓+ 2NaCl

回答下列问题:

① E的化学式是。

②加入F的溶液产生气体的化学方程式是。

③最终得到的溶液中的溶质是。

〖基本实验〗

33.(5分)根据下图回答问题。

(1)实验室加热高锰酸钾制取氧气的化学方程式是,所选用的发生装置是______(填序号,下同),所选用的收集装置是。

(2)实验室制取二氧化碳的化学方程式是,用装置C收集二氧化碳气体,验满

的方法是。

34.(2分)“去除粗盐中难溶性杂质”的实验操作如下图所示:

A.溶解B.蒸发 C.过滤

(1)正确的操作顺序是(填序号)。

(2)操作B中,用玻璃棒不断搅拌的目的是。

35.(3分)根据下列实验回答问题。

A B C

(1)A中观察到的实验现象是。

(2)B中水面下白磷不燃烧的原因是。

(3)C中观察到烧杯内壁有无色液滴,迅速将烧杯倒转过来,倒入澄清石灰水,石灰水变浑浊。若该气体为纯净物,其可能是。

〖实验原理分析〗

36.(5分)化学小组同学利用下图装置进行相关实验。

(1)若甲中放入块状大理石,乙中放入澄清石灰水,打开b,

将分液漏斗中过量的盐酸注入,关闭a,此时乙中反应

的化学方程式是,一段时间后,关闭b,观察到

乙中液体变澄清,产生此现象的原因是。

(2)若甲中放入氢氧化钠固体,乙中放入水,关闭b,将分液

漏斗中的盐酸注入,关闭a。观察到甲中导管内液面上升至

一定高度,一段时间后液面下降且最终高于烧瓶中的液面。

甲中反应的化学方程式是。同学们解释“甲中导管内液面上升”

的原因有氢氧化钠与盐酸反应放出热量、、等。

〖科学探究〗

37.(5分)“侯氏制碱法”是我国近代著名科学家侯德榜发明的一种连续生产纯碱与氯化铵的联合制碱工艺,同学们查阅相关资料并进行模拟实验。

【查阅资料】

①生产原理:NaCl(饱和溶液)+ NH3 + CO2 + H2O = NaHCO3(固体)↓+ NH4Cl

②主要生产流程如下图所示:

③氨气(NH3)是一种有刺激性气味的气体,极易溶于水,其水溶液显碱性。

【理论分析】

(1)依据资料①可知相同条件下,NaHCO3的溶解度 NH4Cl的溶解度(填“>”或“<”)。

(2)NaHCO3受热分解制得纯碱及两种常见的氧化物,该反应的化学方程式是。

【模拟实验】验证侯氏制碱法能获得纯碱

步骤实验装置实验操作及现象解释与结论

1

从a管通入NH3,一段

时间后,从b管通入

CO2,溶液中有固体析

(1)蘸有稀硫酸的棉花的

作用是

(2)通NH3的a管若伸到

液面下,会造成的后

果是

铜片

80℃

a

分液漏斗

b

甲乙

蘸有稀硫酸

饱和食盐水

的棉花

2

加热,导管口有气泡,澄清石灰水变浑浊

步骤1析出的固体是碳酸氢钠

3

滴加足量稀盐酸,有气泡产生,固体逐渐消失

步骤2加热后得到的固体是纯碱

【实验反思】同学们认为步骤3的结论存在不足。依据题中信息,设计实验确定步骤2加热后得到的固

体成分,其方案是 。 〖实际应用定量分析〗

38.(4分)工业上常用NaOH 溶液处理铝土矿(主要成分为A12O 3),反应的化学方程式为:

Al 2O 3+2NaOH=2NaAlO 2+H 2O (杂质不参与反应也不溶于水)。若用160 kg NaOH 配制的溶液恰好能处理340 kg 铝土矿,最终得到1004 kg NaAlO 2 溶液。

计算:(1)该铝土矿中A12O 3的质量分数。(2)加入的NaOH 溶液中水的质量。

足量稀盐酸

少许步骤2

步骤1 析出的固体

澄清

石灰水

加热后的固体

北京市西城区2015年初三二模

化学试卷参考答案及评分标准 2015.6

第一部分 选择题(每小题只有一个选项符合题意,共20个小题,每小题1分,共20分。) 题号 l 2 3 4 5 6 7 8 9 10 答案 A A B A C A B D C D 题号 1l 12 13 14 15 16 17 18 19 20 答案

C

A

B

D

D

C

C

D

B

B

说明:除特别注明外,以下每空1分。其他合理答案均可给分。 第二部分 非选择题(共18个小题,共60分。) 〖生活现象解释〗 21.(2分)

(1)贫血症 (2)C 22.(2分)

(1) 导电 (2)Fe + CuSO 4 FeSO 4 + Cu 23.(3分)

(1)酸 (2)厨房清洁剂 (3)肥皂(或肥皂水) 24.(2分)

(1)煮沸 (2)水分子 25.(2分)

(1)盐 ( 2 ) 真空包装 26.(3分)

(1)CaCO 3 ( 2 ) 水 (3)糖类、油脂和蛋白质 27.(3分)

(1)调味品 ( 2 ) NaHCO 3 + HCl NaCl + H 2O + CO 2↑

(3)2NaN 3 2Na + 3N 2↑ 28.(3分)

(1)+1 ( 2 ) 盐酸具有腐蚀性 (3)CO 2和O 2 〖科普阅读理解〗 29.(5分)

(1)有机融雪剂 ( 2 ) 溶质的质量越大,溶液的凝固点越低 (3)耐低温、低密度 (4)不能 (5)TiCl 4 + 2Mg 2MgCl 2 + Ti

〖生产实际分析〗 30.(3分)

(1)肥料 ( 2 ) 可回收物 (3)SO 2和HCl 31.(3分)

(1)Fe + H 2SO 4 FeSO 4 + H 2↑ ( 2 ) 过滤 防止FeSO 4被氧化 〖物质组成和分析〗 32.(5分)

(1)CO ( 2 ) CO 2 + C 2CO

(3)① HCl ②2HCl + Na 2CO 3 2NaCl + CO 2↑ + H 2O ③ NaCl 和Na 2CO 3 〖基础实验〗 33. (5分)

高温

撞击

高温 稀有气体

(1)2KMnO

4 K 2MnO 4 + MnO 2 + O 2↑ A C (或E ) (2)CaCO 3 + 2HCl CaCl 2 + H 2O + CO 2↑ 将燃着的木条放在瓶口,若熄灭,则已满 34. (2分)

(1)A C B (2)防止液体局部过热导致液滴飞溅(或均匀受热) 35. (3分)

(1)乙中溶液由无色变成红色 (2)没有与氧气接触 (3)CH 4 〖实验原理分析〗 36. (5分)

(1)Ca(OH)2 + CO 2 CaCO 3↓+ H 2O

关闭b ,甲中气压增大,过量的稀盐酸被压入乙中,与CaCO 3反应

(2)NaOH + HCl NaCl + H 2O 氢氧化钠固体溶于水放热 注入的盐酸在 烧瓶中占据一部分体积 〖科学探究〗 37.(5分)

【理论分析】(1)< (2)2NaHCO 3 Na 2CO 3 + H 2O + CO 2↑ 【模拟实验】 (1)稀硫酸可吸收..NH ..

3. (2)液体倒吸 【实验反思】取步骤2加热后得到的固体少许于试管中,加热,若固体的质量不再减少,

说明得到的固体为Na 2CO 3(或取步骤2加热后得到的固体少许于试管中, 加热,将产生的气体通入澄清石灰水,若不变浑浊,则得到的固体为Na 2CO 3)

〖实际应用定量分析〗

38.(4分) [解] (1)设铝土矿中的Al 2O 3的质量为x 。 Al 2O 3 + 2NaOH 2NaAlO 2 + H 2O 102 80 x 160 kg

x =204 kg 铝土矿中Al 2O 3的质量分数= ×100% = 60% (2)加入的NaOH 溶液中水的质量= 1004 kg – 204kg – 160kg = 640 kg 答:铝土矿中Al 2O 3的质量分数为60%,加入的NaOH 溶液中水的质量为640 kg 。

= 102

160 kg x 80 ……………………(1分)

……………………(1分) ……(1分) 340 kg

204 kg

…………… (1分)

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