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    emergency,and treat burns.As a student reporter,I was impressed by the organization and importance of this event.Firesafety is crucial for everyone,and this event helped raise awareness among students andprovided them with practical skills that could save their lives in a real emergency.Overall,thisevent was informative and rewarding.I hope that our school continues to organize events likethis in the future.部分解析阅读理解第一节A篇主题语境:人与自然一一自然生态本文是应用文。文章介绍了一个由澳大利亚博物馆专家创建的鲨鱼展。21.B。理解具体信息。根据第一段中的visitors will..explore the contents of a shark's stomach可知,在鲨鱼展,参观者可以探索鲨鱼胃里的东西。22.C。理解具体信息。根据Ticket prices部分中的Family(admits4)2A+2C/1A+3C:S76可知,题干中的一对夫妇和他们两个上小学的孩子符合家庭票(包含两个大人、两个小孩),由此可知答案。23.C。理解具体信息。根据You'll see:部分中的The fastest shark一a life--size model of theshortfin mako shark,which can reach speeds of up to7okm/hour可知,速度最快的鲨鱼是theshortfin mako shark。B篇主题语境:人与社会一一社会本文是记叙文。在意大利的一个小镇上,有一座由Bruno Ferrin纯手工打造的游乐园。24.B。理解具体信息。根据第一段中的I could open a food stand in the woods和第四段中的He started work at 4 a.m.and finished at noon.So he looked for part-time work.He and his wife,Marisa,started a food stand in the woods on Montello可知,Ferrin最初打算是找一个兼职工作,所以决定在那片森林里开个食品摊。25.B。理解具体信息。根据第二段中的This ride,like all the37 attractions in the amusementpark,operates on human effort and the laws of physics可知,Ai Pioppi的游乐设施都是靠人力和物理定律运行的。26.C。推断。根据倒数第二段中的Encouraged by the success,he made more rides和最后一Ferrin's favorite time is what he still spends in his workshop,working old steel tubes intomore pieces for his amusement park."What I love about my life is that I am truly free,"he says.“I work on what I want when I want.”可知,Ferrin沉浸在自己的工作室,自由工作,手工打造Ai Pioppi的游乐设施,由此可知,他是一个有创造力、自由奔放的人。27.A。理解具体信息。根据最后一段中的And Ferrin still keeps the amusement park free.Overthe past 50 years friends have suggested the introduction of a fee.He resists."I want people tocome here,have a bite,enjoy the fresh air and play,”he says.可知,Ferrin的游乐园Ai Pioppi将继续免费让游客游玩。C篇主题语境:人与社会一一科学与技术本文是说明文。软件更新搭配微小芯片可使智能手机成为完整的射频识别(FID)读取器。28.D。推断。根据第二段中的The phone becomes capable of identifying objects based on

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    玉冠高三第一轮夏周测卷教净②实验能够证实Fc+可与SC反应,则c中应观察到的现象是】(3)①由图可知F+与氣气反应生成F+,F与鹰反应又生成F,整个过褪没有消耗F+。学须作用过程记(⑤)实验中若未加适量的水进行稀释,则无法通过现象得出结论。推测加水稀释的目的可他有衡中,FC的作用是催化作用。后续实验颜色的变化易于观察;降低c(F+),②梅1与Fe+反应生成Fe2,F+得电子是氧化剂,酶1是还原利,故具有还原性的降是1。【解题分析(2)用过量NH HCO溶液与FeSO,溶液反应得到FcCO,同时反应产生(NH),SO,、HzO,CO,反应的离子方程式为Fe+2HCO一FeCO,十H,O十CO2个③该过程总反应的化学方程式是C,Ha0,十60.省化iC02+6H,O.(3)通过对实验进行对比可知,实验b排除了加水稀释对沉淀溶解衡移动的影响。【答案】(1)Fc3+(2分)(4)②Fe与SCN反应,降低了溶液中c(SCN).使衡Fe-3SCN一Fe(SCN)3逆向移动,溶液s(2)1.可行(2分)[Fc(SCN),]双小,因而应该观察到的实验现泉是溶液红色变浅。1,①将亚铁离子氧化成铁离子(2分)(5)加水稀驿的目的可能有使后续实验颜色的变化易于观察:降低c(Fe+),使衡Fc+3SCN一F(SCN②红色(2分)逆向移功,溶液中c[Fe(SCN,]减小,因而溶液红色变浅.③高(2分)【答案1)还原(3分)(3)①催化作用(2分②晦1(2分)(2)Fe2-+2HCO万-FC0¥+H2O+CO24(3分)(3)排除FCO溶解产生的F+对尖脸的影响(3分)③C,H:0,+50,雀化剂-6C0+6H0(3分)(4)①1ml0.5mol·L1H,S0,溶液(或其他合理答柴)(3分》13.(17分)“绿色化学”在推动社会可持续发展中发挥着重要作用。某科研团队设计了一种熔盐液相②溶液红色变泼(3分)氧化法制备高价铬盐的新工艺,该工艺不消耗除铬铁可、氢氧化钠和空气以外的其他原料,不产生(5)使衡Fe+3SCN一Fe(SCN):边向移动(3分}废弃物,实现了Cr一Fe一Al-Mg的深度利用和Na*的内循环。工艺流程如下:12.(17分)铁是人体必需的微量元素,铁摄人不足可能引起缺铁性贫血气体AI.黑木耳中含有比较丰富的铁元素,某研学小组进行实验测定黑木耳中铁元素的含量。NazCr2O2O2(g+熔融Na0HH00(1)铁元素的分离:在坩埚中高温灼烧黑木耳,使之完全灰化。用足量的酸充分溶解,过滤,滤液中溶液4+工序③溶液+物质Vo介稳铁元素的存在形式是Fe+、Fe(CrO2),NaOH溶液(循环(含AL,0,高温连(2)铁元素含量测定:研学小组提出如下测量方案。MgO)续氧化工序①过滤态相分离过量气体A滤渣】Al(OH)(s)+i.沉淀法:向(I)滤液中加人足量的NaOH溶液,过滤、洗涤沉淀、加热烘干、灼烧、称量。请溶液工序@物质Y无伍评价该测量方案是否可行。(填“可行”或“不可行”)。过量气体A+H,0g一→工序@的溶液ⅱ.比色法:流程如下。0L滤液,0含F心溶液K、溶液国剥透光率数据处理速-物质-这解Mec0,付☒Meo溶液固体Ⅲ①加入H:O的目的是混合气体V请回答下列问题:②溶液a的颜色是」③溶液颜色越深,光的透过能力越差,即透光率越小,含铁量越(填“高”或低”)。(])高温连续氧化工序中被氧化的元素是〔填元素符号)。(2)工序①的名称为。滤渣I的主要成分是Ⅱ.在肺部,血红蛋白中的亚铁血红素与O2结合,把O,送到各个组织器官。(填化学式)。(3)工序③中发生反应的离子方程式为(3)已知葡萄糖的分子式是CHO。铁元素参与人体内的呼吸作用的示意图如下图(部分中间(4)热解工序产生的混合气体最适宜返回工序(填“①”、“②”、“③”或“④")参与内循环。产物已略去)(5)工序④溶液中的铝元素恰好完全转化为沉淀的pH为。(通常认为溶液中离子浓度H,0小于105mol·L'时沉淀完全;Al(OH):十OH=A1(OH)5K=10°,Km=10-4;KAI(OH)3]-10-33}【答案(1)Cr,Fe(2分)(2)水浸(或溶解)(2分):FeO、MgO3分)(3)2Na+2CCg+2C02+HL0Cr:C号+2N3HC03¥(3分)①呼吸作用过程中,Fer的作用是(4)②(3分)(5)8.374分)②具有还原性的酶是(填“醇1”或“薛2”)。③该过程总反应的化学方程式是【解题分析】(1)在坩塥中商温均浇黑木耳,使之完全灰化,黑木耳中的铁元素会转化为铁的氧化物,因此得到的物质中一定全有铁的氧化物,铁的氧化物用酸充分没泡溶解,过滤,端液中铁元素的存在形式是F、F€。(2)「,由沉度法测定原理可知,流测量方案可行,当加入足量的NOH溶液,溶液中F~和F均生成沉淀,且F(OH)1在空气中易放黛化生成F(OH)1,溶液中的铁元素最终完全转化为FeO,称量,计算、得出结论。丽.①加入H0的目的是将亚铁离子氧化成铁离子,②向各P~的淳液中加入KSCN常液变红色,数a游液的颜色是红色③考泼颜色怒深,泛羽F的浓度总大·光的透过能力总差,即透光率越小.合铁量越高【24G32C(高奢1化学-R-必考-0GA-Y】2322{24G3ZC(新高考)·化学-R-必考-QGA-YI扫码使用◆》夸克扫描王

  • 2024届衡水金卷先享题 分科综合卷 全国乙卷 英语(一)1试题

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