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1、vip 会员免费1.(2018年全国卷I)硫酸亚铁锂(LiFePO4)电池是新能源汽车的动力电池之一。采用湿法冶金工艺回收废旧硫酸亚铁锂电池正极片中的金属,其流程如下:下列叙述错误的是()A.合理处理废旧电池有利于保护环境和资源再利用B.从“正极片”中可回收的金属元素有Al、Fe、Li C.“沉淀”反应的金属离子为Fe3+D.上述流程中可用硫酸钠代替碳酸钠【解析】A、废旧电池中含有重金属,随意排放容易污染环境,因此合理处理废旧电池有利于保护环境和资源再利用,A正确;B、根据流程的转化可知从正极片中可回收的金属元素有Al、Fe、Li,B正确;C、得到含Li、P、Fe的滤液,加入碱液生成氢氧化铁沉
2、淀,因此“沉淀”反应的金属离子是Fe3,C正确;D、硫酸锂能溶于水,因此上述流程中不能用硫酸钠代替碳酸钠,D错误。【答案】D 2.(2018年全国卷I)下列说法错误的是()A.蔗糖、果糖和麦芽糖均为双糖B.酶是一类具有高选择催化性能的蛋白质C.植物油含不饱和脂肪酸酯,能使Br?/CCl4褪色D.淀粉和纤维素水解的最终产物均为葡萄糖【解析】A、果糖不能再发生水解,属于单糖,A错误;B、酶是由活细胞产生的具有催化活性和高度选择性的蛋白质,B正确;C、植物油属于油脂,其中含有碳碳不饱和键,因此能使Br2/CCl4溶液褪色,C正确;D、淀粉和纤维素均是多糖,其水解的最终产物均为葡萄糖,D正确。答案选A
3、。【答案】A 3.(2018年全国卷I)在生成和纯化乙酸乙酯的实验过程中,下列操作未涉及的是()A.A B.B C.C D.Dvip 会员免费【解析】A、反应物均是液体,且需要加热,因此试管口要高于试管底,A正确;【答案】D 4.(2018年全国卷I)NA是阿伏加德罗常数的值,下列说法正确的是()A.16.25 g FeCl3水解形成的Fe(OH)3胶体粒子数为0.1 NAB.22.4 L(标准状况)氨气含有的质子数为18NAC.92.0 g甘油(丙三醇)中含有羟基数为1.0NAD.1.0 mol CH4与 Cl2在光照下反应生成的CH3Cl 分子数为1.0NA【解析】A、16.25g 氯化铁
4、的物质的量是16.25g 162.5g/mol 0.1mol,由于氢氧化铁胶体是分子的集合体,因此水解生成的 Fe(OH)3胶体粒子数小于0.1 NA,A错误;B、标准状况下22.4L 氩气的物质的量是1mol,氩气是一个Ar 原子组成的单质,其中含有的质子数是18 NA,B正确;C、1 分子丙三醇含有3 个羟基,92.0g 丙三醇的物质的量是1mol,其中含有羟基数是3 NA,C错误;D、甲烷与氯气在光照条件下发生取代反应生成的卤代烃不止一种,因此生成的CH3Cl分子数小于1.0 NA,D错误。【答案】B 5.(2018年全国卷I)环之间共用一个碳原子的化合物称为螺环化合物,螺2,2 戊烷(
5、)是最简单的一种。下列关于该化合物的说法错误的是()A.与环戊烯互为同分异构体B.二氯代物超过两种C.所有碳原子均处同一平面D.生成 1 molC5H12至少需要2 molH2【解析】A、螺 2,2 戊烷的分子式为C5H8,环戊烯的分子式也是C5H8,结构不同,互为同分异构体,A正确;B、分子中的 8 个氢原子完全相同,二氯代物中可以取代同一个碳原子上的氢原子,也可以是相邻碳原子上或者不相邻的碳原子上,因此其二氯代物超过两种,B正确;C、由于分子中4 个碳原子均是饱和碳原子,而与饱和碳原子相连的4 个原子一定构成四面体,所以分子中所有碳原子不可能均处在同一平面上,C错误;D、戊烷比螺2,2 戊
6、烷多 4 个氢原子,所以生成 1 molC5H12至少需要 2 molH2,D正确。【答案】C 6.(2018年全国卷I)主族元素W、X、Y、Z的原子序数依次增加,且均不大于20。W、X、Z最外层电子数之和为10;W与 Y同族;W与 Z 形成的化合物可与浓硫酸反应,其生成物可腐蚀玻璃。下列说法正确的是()A.常温常压下X的单质为气态文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2
7、G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6
8、T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T
9、1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7
10、D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10
11、U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P
12、1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:C
13、K7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4vip 会员免费B.Z 的氢化物为离子化合物C.Y 和 Z形成的化合物的水溶液呈碱性D.W 与 Y具有相同的最高化合价【解析】根据以上分析可知W、X、Y、Z 分别是 F、Na、Cl、Ca。则 A、金属钠常温常压下是固态,A错误;B、CaH2中含有离子键,属于离子化合物,B正确;C、Y与 Z 形成的化合物是氯化钠,其水溶液显中性,C错误;D、F 是最活泼的非金属,没有正价,Cl 元素的最高价是+7价,D错误。答案选B。【答案】B 7.(2018年全国卷I)最近我国科学家设计了一种CO2+H2S协同转化装置,实现对天然气中CO2
14、和 H2S的高效去除。示意图如图所示,其中电极分别为ZnO 石墨烯(石墨烯包裹的ZnO)和石墨烯,石墨烯电极区发生反应为:EDTA-Fe2+-e-EDTA-Fe3+2EDTA-Fe3+H2S2H+S+2EDTA-Fe2+该装置工作时,下列叙述错误的是()A.阴极的电极反应:CO2+2H+2e-CO+H2O B.协同转化总反应:CO2+H2SCO+H2O+S C.石墨烯上的电势比ZnO 石墨烯上的低D.若采用 Fe3+/Fe2+取代 EDTA-Fe3+/EDTA-Fe2+,溶液需为酸性B、根据石墨烯电极上发生的电极反应可知+即得到H2S2e2H+S,因此总反应式为CO2+H2SCO+H2O+S,
15、B正确;C、石墨烯电极为阳极,与电源的正极相连,因此石墨烯上的电势比ZnO 石墨烯电极上的高,C错误;D、由于铁离子、亚铁离子均易水解,所以如果采用Fe3/Fe2取代 EDTA-Fe3/EDTA-Fe2,溶液需要酸性,D正确。文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6
16、T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T
17、1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7
18、D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10
19、U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P
20、1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:C
21、K7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4vip 会员免费【答案】C 8.(20
22、18 年全国卷I)醋酸亚铬(CH3COO)2CrH2O 为砖红色晶体,难溶于冷水,易溶于酸,在气体分析中用作氧气吸收剂。一般制备方法是先在封闭体系中利用金属锌作还原剂,将三价铬还原为二价铬;二价铬再与醋酸钠溶液作用即可制得醋酸亚铬。实验装置如图所示,回答下列问题:(1)实验中所用蒸馏水均需经煮沸后迅速冷却,目的是_,仪器 a 的名称是 _。(2)将过量锌粒和氯化铬固体置于c 中,加入少量蒸馏水,按图连接好装置,打开K1、K2,关闭 K3。c 中溶液由绿色逐渐变为亮蓝色,该反应的离子方程式为_。同时 c 中有气体产生,该气体的作用是_。(3)打开 K3,关闭 K1和 K2。c 中亮蓝色溶液流入d
23、,其原因是 _;d 中析出砖红色沉淀,为使沉淀充分析出并分离,需采用的操作是_、_、洗涤、干燥。(4)指出装置d 可能存在的缺点_。【解析】(1)由于醋酸亚铬易被氧化,所以需要尽可能避免与氧气接触,因此实验中所用蒸馏水均需煮沸后迅速冷却,目的是去除水中溶解氧;根据仪器构造可知仪器a 是分液(或滴液)漏斗;%网(2)c 中溶液由绿色逐渐变为亮蓝色,说明Cr3被锌还原为Cr2,反应的离子方程式为Zn+2Cr3Zn2+2Cr2;锌还能与盐酸反应生成氢气,由于装置中含有空气,能氧化Cr2,所以氢气的作用是排除c 中空气;(3)打开 K3,关闭 K1和 K2,由于锌继续与盐酸反应生成氢气,导致 c 中压
24、强增大,所以 c 中亮蓝色溶液能流入d 装置,与醋酸钠反应;根据题干信息可知醋酸亚铬难溶于水冷水,所以为使沉淀充分析出并分离,需要采取的操作是(冰浴)冷却、过滤、洗涤、干燥。(4)由于 d 装置是敞开体系,因此装置的缺点是可能使醋酸亚铬与空气接触被氧化而使产品不纯。【答案】(1).去除水中溶解氧 (2).分液(或滴液)漏斗 (3).Zn+2Cr3Zn2+2Cr2 (4).排除 c 中空气 (5).c中产生 H2使压强大于大气压 (6).(冰浴)冷却 (7).过滤 (8).敞开体系,可能使醋酸亚铬与空气接触9.(2018年全国卷I)焦亚硫酸钠(Na2S2O5)在医药、橡胶、印染、食品等方面应用广
25、泛。回答下列问题:(1)生产 Na2S2O5,通常是由NaHSO3过饱和溶液经结晶脱水制得。写出该过程的化学方程式_。(2)利用烟道气中的SO2生产 Na2S2O5的工艺为:文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7
26、D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10
27、U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P
28、1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:C
29、K7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO
30、10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG
31、6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4vip 会员免费pH=4.1 时,中为 _溶液(写化学式)。工艺中加入Na2CO3固体、并
32、再次充入SO2的目的是 _。(3)制备 Na2S2O5也可采用三室膜电解技术,装置如图所示,其中SO2碱吸收液中含有NaHSO3和 Na2SO3。阳极的电极反应式为 _。电解后,_室的 NaHSO3浓度增加。将该室溶液进行结晶脱水,可得到Na2S2O5。(4)Na2S2O5可用作食品的抗氧化剂。在测定某葡萄酒中Na2S2O5残留量时,取 50.00 mL葡萄酒样品,用 0.01000 molL-1的碘标准液滴定至终点,消耗10.00 mL。滴定反应的离子方程式为_,该样品中Na2S2O5的残留量为_gL-1(以 SO2计)。【解析】(1)亚硫酸氢钠过饱和溶液脱水生成焦亚硫酸钠,根据原子守恒可知
33、反应的方程式为2NaHSO3Na2S2O5+H2O;(2)碳酸钠饱和溶液吸收SO2后的溶液显酸性,说明生成物是酸式盐,即中为NaHSO3;要制备焦亚硫酸钠,需要制备亚硫酸氢钠过饱和溶液,因此工艺中加入碳酸钠固体、并再次充入二氧化硫的目的是得到 NaHSO3过饱和溶液;(3)阳极发生失去电子的氧化反应,阳极区是稀硫酸,氢氧根放电,则电极反应式为2H2O4e4H+O2。阳极区氢离子增大,通过阳离子交换膜进入a 室与亚硫酸钠结合生成亚硫酸钠。阴极是氢离子放电,氢氧根浓度增大,与亚硫酸氢钠反应生成亚硫酸钠,所以电解后a 室中亚硫酸氢钠的浓度增大。(4)单质碘具有氧化性,能把焦亚硫酸钠氧化为硫酸钠,反应
34、的方程式为S2O52+2I2+3H2O 2SO42+4I+6H;消耗碘的物质的量是0.0001mol,所以焦亚硫酸钠的残留量(以SO2计)是。【答案】(1).2NaHSO3Na2S2O5+H2O (2).NaHSO3 (3).得到 NaHSO3过饱和溶液 (4).2H2O 4e4H+O2 (5).a (6).S2O52+2I2+3H2O 2SO42+4I+6H (7).0.128 10.(2018年全国卷I)采用 N2O5为硝化剂是一种新型的绿色硝化技术,在含能材料、医药等工业中得到广泛应用。回答下列问题(1)1840 年 Devil用干燥的氯气通过干燥的硝酸银,得到N2O5。该反应的氧化产物
35、是一种气体,其分子式为_。(2)F.Daniels等曾利用测压法在刚性反应器中研究了25时 N2O5(g)分解反应:文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1
36、H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D
37、2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U
38、6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1
39、T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK
40、7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO1
41、0U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4vip 会员免费其中 NO2二聚为 N2O4的反应可以迅速达到平衡。体系的总压强p随时间t的变化如下表所示(t=时,N2O4(g)完全分解):t/m
42、in 0 40 80 160 260 1300 1700 p/kPa 35.8 40.3 42.5.45.9 49.2 61.2 62.3 63.1 已知:2N2O5(g)2N2O5(g)+O2(g)H1=-4.4 kJ mol-12NO2(g)N2O4(g)H 2=-55.3 kJ mol-1则反应 N2O5(g)2NO2(g)+O2(g)的 H=_ kJmol-1。研究表明,N2O5(g)分解的反应速率。t=62 min 时,测得体系中pO2=2.9 kPa,则此时的=_ kPa,v=_kPa min-1。若提高反应温度至35,则 N2O5(g)完全分解后体系压强p(35)_63.1 kP
43、a(填“大于”“等于”或“小于”),原因是 _。25时 N2O4(g)2NO2(g)反应的平衡常数Kp=_kPa(Kp为以分压表示的平衡常数,计算结果保留1 位小数)。(3)对于反应2N2O5(g)4NO2(g)+O2(g),R.A.Ogg 提出如下反应历程:第一步 N2O5NO2+NO3快速平衡第二步 NO2+NO3NO+NO2+O2慢反应第三步 NO+NO32NO2快反应其中可近似认为第二步反应不影响第一步的平衡。下列表述正确的是_(填标号)。Av(第一步的逆反应)v(第二步反应)B反应的中间产物只有NO3C第二步中NO2与 NO3的碰撞仅部分有效D第三步反应活化能较高【解析】(1)氯气在
44、反应中得到电子作氧化剂,硝酸银中只有氧元素化合价会升高,所以氧化产物是氧气,分子式为O2;(2)已知:、2N2O5(g)2N2O4(g)+O2(g)H1 4.4kJ/mol、2NO2(g)N2O4(g)H2 55.3kJ/mol 根据盖斯定律可知2即得到N2O5(g)2NO2(g)+1/2O2(g)H1 53.1kJ/mol;根据方程式可知氧气与消耗五氧化二氮的物质的量之比是1:2,又因为压强之比是物质的量之比,所以消耗五氧化二氮减少的压强是2.9kPa 25.8kPa,则此时五氧化二氮的压强是35.8kPa 5.8kPa 30.0kPa,因此此时反应速率v文档编码:CK7D2G7F5Y7 H
45、O10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 Z
46、G6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编
47、码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7
48、 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3
49、 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文
50、档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5Y7 HO10U6T6I3G3 ZG6P1T1H4E4文档编码:CK7D2G7F5