山东省齐鲁名校教科研协作体19所名校2020┄2021届高三第二次调研考试化学试题Word版 含解析
解析山东省齐鲁名校协作体2021年高三上学期第二次联考物理试卷
２0２0┄20２1学年山东省齐鲁名校协作体高三（上）第二次联考物理试卷一、本题共１４小题，每小题4分，共56分．在每小题给出的四个选项中,有的小题只有一个选项正确，有的小题有多个选项正确，全选对的得4分;选对但不全的得２分；有选错或不答的得０分.１.下列说法正确的是（)A．牛顿在得出力不是维持物体运动的原因这一结论的过程中运用了理想实验的方法B.在“探究弹性势能的表达式”的活动中,为计算弹簧弹力所做功，把拉伸弹簧的过程分为很多小段，拉力在每小段可以认为是恒力,用各小段做功的代数和代表弹力在整个过程所做的功，物理学中把这种研究方法叫做“微元法”，那么由加速度的定义a=,当△t非常小的时候，就可以表示物体在t时刻的瞬时加速度，上述论断就运用了“微元法”Ｃ．用比值法定义物理量是物理学中一种重要的物理科学方法，公式a=就运用了比值定义法Ｄ．万有引力可以理解为任何有质量的物体都要在其周围空间产生一个引力场，而一个有质量的物体在其它有质量的物体所产生的引力场中都要受到该引力场的引力（即万有引力）作用,这情况可以与电场相类比，那么在地球的引力场中的重力加速度就可以与电场中的电场强度相类比2．下列说法正确的是( )A．物体做匀加速曲线运动时所受的力应该是均匀增加的B.空军跳伞时降落伞刚刚张开的瞬间伞绳对人的拉力稍稍大于人对伞绳的拉力C.静摩擦力有时可能阻碍物体的相对运动D．力是改变物体运动状态的原因3.一物体受到两个外力的作用,沿某方向做匀速直线运动.若将其中一个力的方向旋转９0°，保持这个力的大小和另一个力不变,则物体可能做（)A.匀速直线运动B．匀加速直线运动C．匀速圆周运动ﻩD.轨迹为抛物线的曲线运动４．如图所示,一根轻质弹簧的上端拴接在天花板上,下端拴接在一个物块上,物块的下端是地面，整个装置处于静止状态,请问：物块的受力情况可能有几种情况？( )A.2 Ｂ.３ﻩC.4 Ｄ．55．根据前人研究的成果,火星和地球沿着各自的椭圆轨道绕太阳运行,根据开普勒行星运动定律以及能量方面的相关知识可以判定下属选项正确的是：(已知行星引力势能的表达式为ＥP=﹣,其中Ｍ和m分别为太阳和行星的质量,ｒ为太阳和行星之间的距离），（）A．太阳位于火星运行轨道的中心B.当它们由各自的近日点运动到远日点时,它们（与太阳组成的系统)的引力势能都要增大Ｃ.当它们由各自的近日点运动到远日点时,动能都要减小,所以它们(与太阳组成的系统)的机械能也要减小D.如果采用理想化模型法将两个行星的运动轨迹看成正圆的话,它们运行时各自的向心加速度一定相同６.一钢球从某高度自由下落在一放在水平地面的弹簧上,从钢球与弹簧接触到压缩到最短的过程中，弹簧的弹力F、钢球的加速度ａ、重力所做的功ＷG以及小球的机械能E与弹簧压缩量ｘ的变化图线如图(不考虑空气阻力），选小球与弹簧开始接触点为原点，建立图示坐标系，并规定向下为正方向，则下述选项中的图象符合实际的是（）A．ﻩB.C．D.７.如图所示，在光滑绝缘水平面上，两个带等量负电的点电荷分别固定在Ａ、B两点,O为AB连线的中点,MN为AB的垂直平分线．在MＮ之间的C点由静止释放一个带正电的小球(可视为质点)，若不计空气阻力,则( ）Ａ．C点的场强大于Ｏ点的场强，Ｃ点的电势高于O点的电势B．小球从C点运动至距离该点最远位置的过程中，其所经过各点所受的电场力先减小后增大C.小球从C点运动至距离该点最远位置的过程中，其电势能先减小后增大D．若在小球运动过程中，两个点电荷所带电荷量同时等量地缓慢增大,则小球往复运动过程中的最大速度将逐渐减小8.如图所示电路中,电源电动势为E,内阻为r,Ｒ１和R3均为定值电阻，R T为热敏电阻（温度越高，电阻越低）．当环境温度较低时合上电键S,当环境的温度逐渐升高时，若三个电表A1、A2和V的示数分别用I１、Ｉ2和U表示.则各个电表示数的变化情况是（）A．Ｉ1增大,I2不变，U增大ﻩB．I1减小,I2增大,Ｕ减小C．I１增大，I2减小，U增大Ｄ.I１减小,Ｉ2不变,U减小9．如图(甲）所示,一根粗绳ＡB的长度为l,其质量均匀分布，在水平外力F的作用下,沿水平面做匀加速直线运动.绳上距A端x处的张力T与x的关系如图（乙）所示．下列说法中正确的是（)A．粗绳可能受到摩擦力作用ﻩB．粗绳一定不受摩擦力作用C．可以求出粗绳的质量ﻩD.可以求出粗绳运动的加速度10．如图所示,倾角为37°的斜面长１0m，一块质量为m=2ｋg的小物块静止放置于斜面底端．小物块在平行于斜面的48N的推力作用下，沿斜面向上运动，小物块和斜面间的动摩擦因数μ=0．5.当物块向上运动了9m而到达B时才撤去外推力,则关于小物块的运动下列说法正确的是( ）A.小物块先做匀加速直线运动,后做匀减速直线运动，运动到斜面的顶端速度刚好为零Ｂ.在外推力F作用在物块上的过程中,推力F做的功要大于物体克服重力和摩擦力所做的功C．从小物块开始运动到最后飞出斜面到最终落回地面的整个过程中,小物块所受外推力做的功和滑动摩擦力做的功的代数和等于物块机械能的变化量D.物块落地的动能为3５2J11．带有同种电荷的各种带电粒子(不计重力)沿垂直电场方向入射到平行带电金属板之间的电场中，并都能从另一侧射出．以下说法正确的是( )A．若粒子的带电量和初动能相同,则离开电场时它们的偏向角φ相同Ｂ.若质量不同的带电粒子由静止开始经相同电场加速后进入该偏转电场，则离开电场时它们的偏向角φ相同C．若带电粒子由静止开始经相同电场加速后进入该偏转电场，离开电场时其偏移量y与粒子电荷量成正比Ｄ.若带电粒子以相同的初速度进入该偏转电场,离开电场时其偏移量ｙ与粒子的荷质比成正比１2.有一沿x轴分布的电场，设定+x方向为正方向.其电场强度E随ｘ变化的图象如图所示．下列说法正确的是（)A.O点的电势最低B.ｘ2点的电势最高C．正电荷从ｘ１移动到ｘ３克服电场力做的功等于其电势能的减小量D.x1与x3的场强相同13.如图是一个多用电表的简化电路图.S为单刀多掷开关，通过操作开关，接线柱O可以接通1,也可以接通2、3、４、5或６.下列说法正确的是（)A.当开关Ｓ分别接1或2时,测量的是电流，其中S接2时量程较小B．当开关S分别接３或4时，测量的是电阻,其中A是黑表笔C．当开关S分别接5或6时，测量的是电压,其中B是红表笔D．当开关S分别接５和６时,测量的是电压，其中S接6时量程较大１4．20１3年６月20日，女航天员王亚平在“天宫一号”目标飞行器里成功进行了我国首次太空授课.授课中的一个实验展示了失重状态下液滴的表面张力引起的效应．在视频中可观察到漂浮的液滴处于周期性的“脉动”中．假设液滴处于完全失重状态,液滴的上述“脉动”可视为液滴形状的周期性的微小变化(振动),如图所示．已知液滴振动的频率表达式为f=k，其中k为一个无单位的比例系数，r为液滴半径,ρ为液体密度；σ(其单位为Ｎ/m）为液体表面张力系数，它与液体表面自由能的增加量△Ｅ(其单位为J)和液体表面面积的增加量△S有关,则在下列关于σ、△E和△S关系的表达式中，可能正确的是（)A．σ=△E×△SﻩB．σ=ﻩC.σ＝ D.σ=二、本题2小题，共27分，其中第1５小题1６分；第16小题１１分.把答案填到题中横线上或按要求作图.15．利用如图1所示的装置做探究弹力和弹簧伸长的关系的实验．所用钩码每只的质量皆为3０ｇ．实验中,先测出不挂钩码时弹簧的长度，再将5个钩码逐个加挂在弹簧下端，稳定后测出相应的弹簧总长度,将数据填在表中.(弹力始终未超过弹性限度,取g=9.8m/ｓ２)记录数据组1２3456钩码总质量(g）0３060901２0150弹簧总长(cm)18.0021．3324.6027.9331．234.56弹力大小(N)(１）上表记录数据中有一个不符合规范，它是第组中的数据(填写数据的名称)，应记作 .（２）根据实验数据将对应的弹力大小计算出来并填入表内相应的空格内（保留3位有效数字).（3）在图２坐标纸中作出弹簧弹力大小F跟弹簧总长度L之间的关系图线．（4)根据图线求得该弹簧的劲度k= N/m．(保留3位有效数字)（5）写出弹簧中弹力的大小F跟弹簧总长度L之间的函数关系的解析式，注意:前5问不要求考虑弹簧自重对实验结果的影响.(6)有的同学认为在上述实验中，如果将弹簧水平放置来测量自然长度，就会消除因为弹簧的自重而引起的长度的改变量,你认为这种观点是否正确（填写:正确或者不正确).记录数据组12３4５６钩码总质量(g）０３06０９0120150弹簧总长（ｃｍ）18．0021.３324.602７.93３1.234．5６(7）有的同学认为如果将图象的横坐标改成弹簧长度相对于自然长度的改变量,这样才能直观的得出胡克定律所要表达的规律.[必修一课本中胡克定律的内容:弹簧发生弹性形变时，弹力的大小跟弹簧伸长（或者缩短）的长度成正比．］正由于考虑到弹簧自重可能会对实验结果产生影响,该同学准确测得弹簧水平放置时的自然长度为16.89cm.根据以上信息可以算得弹簧的质量为g.（求解本问时可以将弹簧的自重全部看成位于最低点，结果保留两位有效数字）.(8)请选用以上所有实验数据和结果，根据需要在下方给定表格中添加左数第一列中的数据名称,并填写相关数据，然后在图3中做出弹簧中的力随着弹簧后来的长度相对于其自然长度的改变量（△x）的变化而变化的关系(9）有的同学认为弹簧竖直放置时，其自重会使第(4）问中测得的劲度系数不准确，请你帮该同学定量分析一下他的观点是否正确？．16.李华同学的爸爸是个电气工程师,李华经常看到爸爸在用如图1所示的多用电表进行一些测量.在高中物理课堂上学习了多用电表的用法之后，爸爸给他出了一道题目，让他通过测量找到如图2所示的二极管的负极．（1)李华同学做了如下两步具体的操作：第一，将多用电表选择开关旋转到电阻挡的×1档，经过之后,他把红表笔接在二极管的短管脚上,把黑表笔接在二极管的长管脚上,发现二极管发出了耀眼的白光，如图3所示；然后他将两表笔的位置互换以后,发现二极管不发光．这说明二极管的负极是(填写“长管脚”或者“短管脚”)所连接的一极．(2）李华同学的好奇心一下子就被激发起来了.他琢磨了一下,然后又依次用电阻挡的×1档，×10档,×1０0档,×1K档分别进行了二极管导通状态的准确的测量,他发现二极管发光的亮度越来越(填写“大”或者“小”),请帮助他分析一下具体的原因．(3）爸爸说,欧姆表内的电池用过了一段时间,电动势稍微减小，内电阻可能增大很多倍．自己可以设计实验测量一下电动势和内电阻.李华立刻准备设计实验方案进行相关测量，现在备有如下器材可供选择:A．干电池１节 B.滑动变阻器(０～20Ω) C.滑动变阻器(0~1kΩ）D．电压表(０~3V） E.电流表(0~0.6A) F．电流表（0~3A)G.开关、导线若干①其中滑动变阻器应选,电流表应选.（只填器材前的序号)②为了最大限度的减小实验误差,请在图4虚线框中画出该实验最合理的电路图．③某同学根据实验数据画出的U﹣I图象如图５所示．由图象可得电池的电动势为Ｖ，内电阻为Ω.三、本题共3小题，共27分．解答应写出必要的文字说明、方程式和重要的演算步骤，只写出最后答案的不能得分.有数值计算的题,答案必须明确写出数值和单位.１７．高速公路上甲乙两车在同一车道上同向行驶,甲车在前,乙车在后,速度均为v０=1０8k ｍ/h，距离s0＝10０ｍ，t＝０时刻甲车遇紧急情况后，甲乙两车的加速度随时间变化如图所示(图1为甲，图2为乙），取运动方向为正方向.(1）t=3s时甲车速度是多少？此时甲乙两车间距是多少?(2)通过计算说明两车在0~9s内会不会相撞？(3）上一问中如果相撞,则求出相撞时间;如果不相撞,则求第９s末两车的间距.（４）分别作出两车的速度时间图线．１8．(２０15秋•山东月考）１0个相同的扁木块一个挨一个的放在水平地面上．每个木块的质量m=0.40ｋｇ,其长度为Ｌ=0.50m.木块原来都静止,它们与地面间的静、动摩擦因数都是μ1=0.１０．在左边第一块木块的左端点放一块Ｍ=1.0０ｋg的小铅块,它与木块间的静、动摩擦因数都为μ2＝0.2０.现突然给铅块一个向右的初速度v０=4.3m/ｓ，使其在木块上滑行,试问：(1)当铅块刚刚滑至哪一块木块上时，下面的木块开始滑动？(2）当铅块和下方的几个木块开始同时运动的瞬间，此时铅块的瞬时速度大小为多少？(３)铁块最终落在地上还是停在哪一块木块上？(设铅块的线度与L相比可忽略．）19．如图甲所示，在边界MN左侧存在斜方向的匀强电场E1,在ＭN的右侧有竖直向上、场强大小为E2=0.4N/C的匀强电场,还有垂直纸面向内的匀强磁场B（图甲中未画出）和水平向右的匀强电场Ｅ3（图甲中未画出）,B和E3随时间变化的情况如图乙所示，P1P2为距M Ｎ边界2.28ｍ的竖直墙壁，现有一带正电微粒质量为４×１0﹣７kg,电量为1×10﹣5C,从左侧电场中距MN边界m的A处无初速释放后，沿直线以1m／s速度垂直MＮ边界进入右侧场区,设此时刻ｔ=0s，取ｇ=１0m/s2.求:(1)MN左侧匀强电场的电场强度Ｅ1(siｎ37°=０.６）;（2)带电微粒在MN右侧场区中运动了1.５s时的速度；(3）带电微粒在ＭN右侧场区中运动多长时间与墙壁碰撞?(≈0.19)2020┄２０21学年山东省齐鲁名校协作体高三（上)第二次联考物理试卷参考答案与试题解析一、本题共14小题，每小题4分，共５6分.在每小题给出的四个选项中,有的小题只有一个选项正确，有的小题有多个选项正确,全选对的得4分；选对但不全的得2分;有选错或不答的得０分.1．下列说法正确的是( ）Ａ.牛顿在得出力不是维持物体运动的原因这一结论的过程中运用了理想实验的方法Ｂ．在“探究弹性势能的表达式”的活动中,为计算弹簧弹力所做功,把拉伸弹簧的过程分为很多小段，拉力在每小段可以认为是恒力，用各小段做功的代数和代表弹力在整个过程所做的功，物理学中把这种研究方法叫做“微元法”,那么由加速度的定义a=,当△t非常小的时候，就可以表示物体在t时刻的瞬时加速度，上述论断就运用了“微元法”C.用比值法定义物理量是物理学中一种重要的物理科学方法，公式a＝就运用了比值定义法Ｄ.万有引力可以理解为任何有质量的物体都要在其周围空间产生一个引力场，而一个有质量的物体在其它有质量的物体所产生的引力场中都要受到该引力场的引力（即万有引力）作用，这情况可以与电场相类比，那么在地球的引力场中的重力加速度就可以与电场中的电场强度相类比【分析】根据物理学史和常用的物理方法解答，记住著名物理学家的主要贡献和常用的方法即可解答.【解答】解:A、伽利略通过理想斜面实验，得出力不是维持物体运动原因的结论,力是改变物体运动状态的原因，故A错误；B、由加速度的定义a=，当△ｔ非常小的时候,就可以表示物体在ｔ时刻的瞬时加速度，上述论断运用了“极限思维法”，故B错误;C、公式ａ=是牛顿第二定律的表达式，不是运用了比值定义法，故C错误;D、将万有引力可以理解为任何有质量的物体都要在其周围空间产生一个引力场，与电场相类比,那么在地球的引力场中的重力加速度就可以与电场中的电场强度相类比,故D正确;故选：D2．下列说法正确的是（）Ａ.物体做匀加速曲线运动时所受的力应该是均匀增加的B.空军跳伞时降落伞刚刚张开的瞬间伞绳对人的拉力稍稍大于人对伞绳的拉力Ｃ.静摩擦力有时可能阻碍物体的相对运动D.力是改变物体运动状态的原因【分析】匀加速曲线运动的加速度不变，合力不变．静摩擦力有时阻碍物体的相对运动趋势.力是改变物体运动状态的原因．根据牛顿第一定律、第三定律分析.【解答】解：A、物体做匀加速曲线运动时加速度不变，由牛顿第二定律知物体所受的力应该是不变的；故A错误．B、伞绳对人的拉力与人对伞绳的拉力是一对作用力和反作用力，由牛顿第三定律知这两个力大小应该相等；故B错误.Ｃ、静摩擦力只能阻碍物体的相对运动趋势；故C错误.D、根据牛顿第一定律可知，力是改变物体运动状态的原因，故D正确.故选:D3.一物体受到两个外力的作用,沿某方向做匀速直线运动.若将其中一个力的方向旋转９0°，保持这个力的大小和另一个力不变,则物体可能做（)Ａ.匀速直线运动 B.匀加速直线运动C.匀速圆周运动D.轨迹为抛物线的曲线运动【分析】物体原来处于平衡状态，一个力的方向旋转９0°后,将另一个力与旋转后的力进行合成,根据合力的方向与速度方向的关系，判断物体的运动情况．【解答】解:物体原来处于平衡状态,一个力的方向旋转90°后，旋转后的力与另一个力进行合成；A、由于一定存在加速度，因此不可能匀速直线运动,故Ａ错误；B、若合力与速度同向,物体做匀加速直线运动，若合力与速度反向,物体做匀减速直线运动,故Ｂ正确；C、由于匀速圆周运动合力始终指向圆心,是个变力,所以不可能做匀速圆周运动，故C错误；D、若合力与速度垂直，物体做类平抛运动,故D正确;故选BD．４.如图所示，一根轻质弹簧的上端拴接在天花板上，下端拴接在一个物块上，物块的下端是地面，整个装置处于静止状态，请问：物块的受力情况可能有几种情况?（)Ａ．２ﻩB．3ﻩＣ.4 D．5【分析】物体处于静止状态，受力平衡,合力为零,分弹簧处于原长、压缩、拉伸三个状态对物体受力分析即可.【解答】解：弹簧可能处于压缩状态和伸长状态或者是自然长度的状态．如果弹簧处于压缩状态,则物块的受力情况如图２所示;如果弹簧处于拉伸状态,则我们还要分析弹簧拉伸的程度,如果此时弹簧的拉力恰好与物块的重力相等，则物块的受力情况如图3所示；如果弹簧中的拉力小于物块的重力的话,则物块的受力分析图如图4所示;如果弹簧处于自然长度,则物块只受重力和地面的弹力两个力的作用，其受力情况如图5所示.综上所述，本题有四种答案.故C正确.故选：C5．根据前人研究的成果，火星和地球沿着各自的椭圆轨道绕太阳运行,根据开普勒行星运动定律以及能量方面的相关知识可以判定下属选项正确的是:（已知行星引力势能的表达式为E P=﹣，其中M和ｍ分别为太阳和行星的质量,ｒ为太阳和行星之间的距离)，（) A.太阳位于火星运行轨道的中心B．当它们由各自的近日点运动到远日点时，它们(与太阳组成的系统）的引力势能都要增大C．当它们由各自的近日点运动到远日点时,动能都要减小,所以它们（与太阳组成的系统）的机械能也要减小D．如果采用理想化模型法将两个行星的运动轨迹看成正圆的话，它们运行时各自的向心加速度一定相同【分析】根据开普勒第一定律判断太阳的位置,根据判断引力势能的变化,两行星只受到万有引力作用,所以在运动过程中机械能守恒,加速度是矢量．【解答】解:A、根据开普勒第一定律可知，太阳应该位于焦点上,故Ａ错误;B、根据可知,当它们由各自的近日点运动到远日点时,它们(与太阳组成的系统)的引力势能都要增大，故B正确;Ｃ、由于两行星只受到万有引力作用,所以在运动过程中机械能守恒，故C错误;D、它们运行时各自的向心加速度大小相等,方向不同,故D错误.故选:Ｂ６．一钢球从某高度自由下落在一放在水平地面的弹簧上,从钢球与弹簧接触到压缩到最短的过程中，弹簧的弹力F、钢球的加速度a、重力所做的功W G以及小球的机械能E与弹簧压缩量ｘ的变化图线如图（不考虑空气阻力）,选小球与弹簧开始接触点为原点,建立图示坐标系,并规定向下为正方向，则下述选项中的图象符合实际的是（)A．Ｂ.ﻩC.ﻩD．【分析】开始小球做自由落体运动，与弹簧接触后，弹力逐渐增大，开始小于重力,再等于重力，后大于重力,写出各个物理量与x的表达式，从而判断图象是否正确.【解答】解：A、由于向下为正方向，而弹簧中的弹力方向向上,所以选项A中的图线应该向下，故A错误;B、小球接触弹簧上端后受到两个力作用:向下的重力和向上的弹力．在接触后的前一阶段，重力大于弹力，合力向下,而弹力F＝kｘ，则加速度a＝,故B正确;Ｃ、根据重力做功的计算式ＷG＝mgｘ可知C正确;D、小球和弹簧整体的机械能守恒,小球的机械能不守恒,故D错误．故选:BＣ7．如图所示，在光滑绝缘水平面上，两个带等量负电的点电荷分别固定在A、Ｂ两点，O 为AB连线的中点,MN为AＢ的垂直平分线.在MN之间的C点由静止释放一个带正电的小球(可视为质点),若不计空气阻力,则( ）A．C点的场强大于O点的场强，C点的电势高于O点的电势B.小球从C点运动至距离该点最远位置的过程中,其所经过各点所受的电场力先减小后增大Ｃ．小球从C点运动至距离该点最远位置的过程中,其电势能先减小后增大D．若在小球运动过程中,两个点电荷所带电荷量同时等量地缓慢增大，则小球往复运动过程中的最大速度将逐渐减小【分析】本题要根据等量同种点电荷电场线的分布情况，抓住对称性，分析试探电荷的受力情况,分析其运动情况，电场力做功情况,判断电势能的变化情况.【解答】解：A、根据等量同种点电荷电场线的分布情况和电场线的疏密和方向可知,C点的场强大于Ｏ点的场强,C点的电势高于O点的电势.故A正确;B、根据等量同种点电荷电场线的分布情况和电场线的疏密可知,从Ｍ到Ｎ点场强的变化有两种可能,或者先增大再减小再增大再减小,或者先减小再增大;故B错误；C、小球从Ｃ点运动至距离该点最远位置的过程中,由于开始时受到的电场力的方向向下,所以电场力先做正功后做负功，结合电场力做的功与电势差的变化之间的定量关系可知其电势能先减小后增大,故C正确.D、若在小球运动过程中，两个点电荷所带电荷量同时等量地缓慢增大,则带电小球在各处受到的电场力增大，所以小球往复运动过程中的最大速度将增大，故D错误.故选：AC8．如图所示电路中,电源电动势为E,内阻为r，R1和R３均为定值电阻，RＴ为热敏电阻（温度越高,电阻越低)．当环境温度较低时合上电键S,当环境的温度逐渐升高时，若三个电表A1、Ａ2和V的示数分别用I1、Ｉ2和U表示．则各个电表示数的变化情况是( )A.I1增大，I２不变，U增大ﻩＢ.Ｉ1减小,I2增大，U减小Ｃ．I1增大,I2减小，U增大ﻩD.Ｉ1减小,Ｉ2不变，U减小【分析】本题首先要理清电路,确定电压表测得什么电压,电流表测得什么电流,抓住电动势和内阻不变,再采用局部→整体→局部的方法，利用闭合电路欧姆定律进行分析，先从环境的温度逐渐升高，热敏电阻RＴ电阻减小入手分析.【解答】解：由图知电压表测量路端电压，电流表A1测量流过Ｒ１的电流,电流表Ａ2测量流过R T的电流；。
2020-2021学年济南齐鲁学校高三生物第二次联考试题及答案
2020-2021学年济南齐鲁学校高三生物第二次联考试题及答案一、选择题：本题共15小题，每小题2分，共30分。
每小题只有一个选项符合题目要求。
1. 物质甲作为抑制剂能与蔗糖酶结合或分离，从而改变蔗糖酶的活性。
在适宜温度、pH等条件下，某同学将蔗糖酶和物质甲的混合液均分为若干份，分别加入到不同浓度的等量蔗糖溶液中，发现蔗糖的水解速率随蔗糖溶液浓度的升高而增大。
下列分析与该实验不相符的是（）A.蔗糖溶液浓度的升高能导致物质甲与蔗糖酶的分离B.物质甲与蔗糖酶的结合使酶降低活化能的能力降低C.物质甲与蔗糖酶的结合能改变酶的高效性和专一性D.反应体系中没有甲时，蔗糖溶液的浓度不会改变酶活性2. 正确选择实验材料和方法是得出正确结论的前提，下列有关实验材料或方法的选择，正确的是（）A. 利用废旧物品制作的真核细胞模型——概念模型B. 人鼠细胞融合实验研究生物膜的流动性——荧光标记法C. 提取并研究细胞膜的化学成分——公鸡的成熟红细胞D. 观察叶绿体和线粒体——人口腔上皮细胞3. 如图表示溶酶体的产生和“消化”衰老线粒体的过程，有关说法错误的是（）A. 囊泡融合的过程体现了膜的流动性B.a、b、c、d膜结构的主要成分是磷脂和蛋白质C.d是衰老的线粒体，不能为生命活动供能D.e起源于高尔基体，b来自内质网4. 如图表示某植物的非绿色器官在氧浓度为a、b、c、d时，CO2释放量和O2吸收量的变化（呼吸底物均为葡萄糖）。
下列相关叙述正确的是：( )A. 氧浓度为a是最适于储藏该植物器官B. 氧浓度为b时，无氧呼吸消耗葡萄糖的量是有氧呼吸的4倍C. 氧浓度为c时，无氧呼吸最弱D. 氧浓度为d时，产生CO2的场所只有线粒体5. 下列关于实验操作步骤的叙述中，正确的是（）A. 用于鉴定还原糖的斐林试剂甲液和乙液，可直接用于蛋白质的鉴定B. 用显微镜观察活细胞中的线粒体时，可用健那绿进行染色C. 鉴定蛋白质时需水浴加热D. 观察菠菜细胞叶绿体时，需经盐酸处理使细胞死亡6. 如图是生物圈中碳循环示意图，下列相关分析错误的是（）A. 生物圈通过碳循环实现碳元素的自给自足B.A是消费者，C是生产者，碳在各成分间以二氧化碳的形式传递C. 对E过度开发利用会打破生物圈中碳循环的平衡D. 碳循环过程需要能量驱动，同时又是能量的载体7. 有一种“生物活性绷带”的原理是先采集一些细胞标本，让其在特殊膜片上增殖，5~7天后，将膜片敷到患者伤口上，膜片会将细胞逐渐“释放”到伤口处，并促进新生皮肤层生长，达到愈合伤口的目的。
2020-2021学年山东省高考二模考试数学试题（文）及答案解析
2020-2021学年⼭东省⾼考⼆模考试数学试题（⽂）及答案解析⼭东省⾼三下学期⼆模考试⾼三数学（⽂科）试题第Ⅰ卷（共50分）⼀、选择题：本⼤题共10个⼩题,每⼩题5分,共50分.在每⼩题给出的四个选项中，只有⼀项是符合题⽬要求的.1.设全集U R =，集合{}2|20M x x x =+->，11|()22x N x -?=≥，则()U M N =I e（） A ．[]2,0-B ．[]2,1-C ．[]0,1D ．[]0,22.若复数(1)(3)mi i ++（i 是虚数单位，m R ∈）是纯虚数，则复数31m ii+-的模等于（） A ．1 B ．2 C ．3 D ．43.已知平⾯向量a r 和b r 的夹⾓为60?，(2,0)a =r ，||1b =r ，则|2|a b +=r r（）A ．20B ．12C ．D ．4.已知3cos 5α=，cos()10αβ-=，且02πβα<<<，那么β=（）A ．12πB ．6π C ．4π D ．3π 5.设3log 6a =，4log 8b =，5log 10c =，则（） A ．a b c >>B ．b c a >>C ．a c b >>D ．b a c >>6.某产品的⼴告费⽤x 万元与销售额y 万元的统计数据如表：根据上表可得回归⽅程9.4y x a =+，据此模型预测，⼴告费⽤为6万元时的销售额为（）万元 A ．63.6B ．65.5C ．72D ．67.77.下列说法正确的是（）A ．命题“x R ?∈，使得210x x ++<”的否定是：“x R ?∈，210x x ++>”B ．命题“若2320x x -+=，则1x =或2x =”的否命题是：“若2320x x -+=，则1x ≠或2x ≠”C ．直线1l ：210ax y ++=，2l ：220x ay ++=，12//ll 的充要条件是12a = D ．命题“若x y =，则sin sin x y =”的逆否命题是真命题8.已知双曲线22221x y a b-=（a >，0b >）的两条渐进线与抛物线24y x =的准线分别交于A ，B两点，O 为坐标原点，若AOB S ?=e =（）A ．32B ．2C ．2 D9.已知某空间⼏何体的三视图如图所⽰，则该⼏何体的体积为（）A ．403B ．343C ．4210+D ．436 10.已知函数|ln |,0,()(2),2,x x e f x f e x e x e <≤?=?-<f x b -=+（b R ∈）的四个实根从⼩到⼤依次为1x ，2x ，3x ，4x ，对于满⾜条件的任意⼀组实根，下列判断中⼀定成⽴的是（） A ．122x x += B ．2234(21)e x x e <<-C ．340(2)(2)1e x e x <--<D ．2121x x e <<第Ⅱ卷（共100分）⼆、填空题（每题5分，满分25分，将答案填在答题纸上）11.已知函数221,1,()log (1),1,x x f x x x ?-≤=?->?则7(())3f f = ．12.在长为5的线段AB 上任取⼀点P ，以AP 为边长作等边三⾓形，3和3的概率为．13.设x，y满⾜约束条件360,20,0,0,x yx yx y--≤-+≥≥≥则22x y+的最⼤值为．14.执⾏如图所⽰的程序框图，则输出的结果是．15.若对任意的x D∈，均有()()()g x f x h x≤≤成⽴，则称函数()f x为函数()g x到函数()h x在区间D上的“任性函数”．已知函数()f x kx=，2()2g x x x=-，()(1)(ln1)h x x x=++，且()f x 是()g x到()h x在区间[]1,e上的“任性函数”，则实数k的取值范围是．三、解答题（本⼤题共6⼩题，共75分.解答应写出⽂字说明、证明过程或演算步骤.）16.某⾷品⼚为了检查甲、⼄两条⾃动包装流⽔线的⽣产情况，随机在这两条流⽔线上各抽取40件产品作为样本，并称出它们的重量（单位：克），重量值落在[495,510)内的产品为合格品，否则为不合格品，统计结果如表：（Ⅰ）求甲流⽔线样本合格的频率；（Ⅱ）从⼄流⽔线上重量值落在[]505,515内的产品中任取2个产品，求这2件产品中恰好只有⼀件合格的概率．17.已知函数()4sin cos()33f x x x π=++，0,6x π??∈. （Ⅰ）求函数()f x 的值域；（Ⅱ）已知锐⾓ABC ?的两边长a ，b 分别为函数()f x 的最⼩值与最⼤值，且ABC ?的外接圆半径为32，求ABC ?的⾯积． 18.如图，在四棱锥S ABCD -中，四边形ABCD 为矩形，E 为SA 的中点，2SB =，3BC =，13SC =．（Ⅰ）求证：//SC 平⾯BDE ；（Ⅱ）求证：平⾯ABCD ⊥平⾯SAB ．19.已知等⽐数列{}n a 的前n 项和为n S ，且163n n S a +=+（a N +∈）．（Ⅰ）求a 的值及数列{}n a 的通项公式；（Ⅱ）设122233(1)(221)(log 2)(log 1)n n n n n n b a a --++=++，求{}n b 的前n 项和n T ． 20.已知椭圆C ：22221(0)x y a b a b +=>>经过点，左右焦点分别为1F 、2F ，圆222x y +=与直线0x y b ++=相交所得弦长为2．（Ⅰ）求椭圆C 的标准⽅程；（Ⅱ）设Q 是椭圆C 上不在x 轴上的⼀个动点，Q 为坐标原点，过点2F 作OQ 的平⾏线交椭圆C 于M 、N 两个不同的点，求||||MN OQ 的取值范围． 21.已知函数21()2ln (2)2f x x a x a x =-+-，a R ∈．（Ⅰ）当1a =-时，求函数()f x 的极值；（Ⅱ）当0a <时，讨论函数()f x 单调性；（Ⅲ）是否存在实数a ，对任意的m ，(0,)n ∈+∞，且m n ≠，有()()f m f n a m n->-恒成⽴？若存在，求出a 的取值范围；若不存在，说明理由．⾼三数学（⽂科）试题答案⼀、选择题1-5:ACDCA 6-10:BDDBB⼆、填空题13 12.2513.52 14.8 15.[]2,2e - 三、解答题16.解：（Ⅰ）由表知甲流⽔线样本中合格品数为814830++=，故甲流⽔线样本中合格品的频率为300.7540=．（Ⅱ）⼄流⽔线上重量值落在[]505,515内的合格产品件数为0.025404??=，不合格产品件数为0.015402??=．设合格产品的编号为a ，b ，c ，d ，不合格产品的编号为e ，f ．抽取2件产品的基本事件空间为{(,)a b Ω=，(,)a c ，(,)a d ，(,)a e ，(,)a f ，(,)b c ，(,)b d ，(,)b e ，(,)b f ，(,)c d ，(,)c e ，(,)c f ，(,)d e ，(,)d f ，}(,)e f 共15个．⽤A 表⽰“２件产品恰好只有⼀件合格”这⼀基本事件，则{(,)A a e =，(,)a f ，(,)b e ，(,)b f ，(,)c e ，(,)c f ，(,)d e ，}(,)d f 共8个，故所求概率815P =． 17.解：（Ⅰ）1()4sin (cos )22f x x x x =?-+22sin cos x x x =-+sin 22x x =2sin(2)3x π=+，∵06x π≤≤，∴22333ππ≤+≤，sin(2)13x π≤+≤，∴函数()f x的值域为2??．（Ⅱ）依题意a =2b =，ABC ?的外接圆半径4r =，sin 2a A r ===，sin 232b B r ===，cos 3A =，1cos 3B =，sin sin()sin cos cos sin 3C A B A B A B =+=+=，∴11sin 2223ABC S ab C ?==?=． 18.证明：（Ⅰ）连接AC 交BD 于F ，则F 为AC 中点，连接EF ，∵E 为SA 的中点，F 为AC 中点，∴//EF SC ，⼜EF ?⾯BDE ，SC ?⾯BDE ，∴//SC 平⾯BDE ．（Ⅱ）∵2SB =，3BC =，13SC =，∴222SB BC SC +=，∴BC SB ⊥，⼜四边形ABCD 为矩形，∴BC AB ⊥，⼜AB 、SB 在平⾯SAB 内且相交，∴BC ⊥平⾯SAB ，⼜BC ?平⾯ABCD ，∴平⾯ABCD ⊥平⾯SAB ．19.解：（Ⅰ）∵等⽐数列{}n a 满⾜163n n S a +=+（a N +∈），1n =时，169a a =+；2n ≥时，1166()3(3)23n n n n n n a S S a a +-=-=+-+=?.∴13n n a -=，1n =时也成⽴，∴169a ?=+，解得3a =-，∴13n n a -=.（Ⅱ）122233(1)(221)(log 2)(log 1)n n n n n n b a a --++=++1222(1)(221)(1)n n n n n --++=+12211(1)(1)n n n -??=-+??+?? .当n 为奇数时，22222221111111()()11223(1)(1)n T n n n ??=+-++++=+??++??…；当n 为偶数时，n T =22222221111111()()11223(1)(1)n n n ??+-++-+=-??++??…. 综上，1211(1)(1)n n T n -=+-+． 20.解：（Ⅰ）由已知可得：圆⼼到直线0x y b ++=的距离为11=，所以b =，⼜椭圆C经过点，所以221413a b+=，得到a = 所以椭圆C 的标准⽅程为22132x y +=．（Ⅱ）设00(,)Q x y ，11(,)M x y ，22(,)N x y ，OQ 的⽅程为x my =，则MN 的⽅程为1x my =+.由22,1,32x my x y =+=??得222226,236,23m x m y m ?=??+??=?+?即22022026,236.23m x m y m ?=??+?=+所以0||||OQ y ==由221,1,32x my x y =++=??，得22(23)440m y my ++-=，所以122423m y y m +=-+，122423y y m =-+，12||||MN y y =-====所以||||MNOQ====，因为2 11m+≥，所以21011m<≤+，即212231m<+≤+，即213221m≤<++，所以||23||MNOQ≤<，即||||MNOQ的取值范围为[,2) 3．21.解：（Ⅰ）当1 a=-时，21()2ln32f x x x x=+-，2232(1)(2)x x x xf x xx x x-+--=+-==．当01x<<或2x>时，'()0f x>，()f x单调递增；当12x<<时，'()f x<，()f x单调递减，所以1x=时，5()(1)2f x f==-极⼤值；2x=时，()(2)2ln24 f x f==-极⼩值．（Ⅱ）当0a<时，2'()(2)ax=-+-2(2)2x a x ax+--=(2)()x x ax-+=，①当2a->，即2a<-时，由'()0f x>可得02x<<或x a>-，此时()f x单调递增；由'()0 f x<可得2x a<<-，此时()f x单调递减；②当2a-=，即2a=-时，'()0f x≥在(0,)+∞上恒成⽴，此时()f x单调递增；③当2a-<，即20a-<<时，由'()0f x>可得0x ax>，此时()f x单调递增；由'()0f x<可得2a x-<<，此时()f x单调递减．综上：当2a <-时，()f x 增区间为(0,2)，(,)a -+∞，减区间为(2,)a -；当2a =-时，()f x 增区间为(0,)+∞，⽆减区间；当20a -<<时，()f x 增区间为(0,)a -，(2,)+∞，减区间为(,2)a -．（Ⅲ）假设存在实数a ，对任意的m ，(0,)n ∈+∞，且m n ≠，有()()1f m f n a m ->-恒成⽴，不妨设0m n >>，则由()()1f m f n a m ->-恒成⽴可得：()()f m am f n an ->-恒成⽴，令()()g x f x ax =-，则()g x 在(0,)+∞上单调递增，所以'()0g x ≥恒成⽴，即'()0f x a -≥恒成⽴，∴2(2)0ax a a x-+--≥，即2220x x a x --≥恒成⽴，⼜0x >，∴2220x x a --≥在0x >时恒成⽴，∴2min11(2)22a x x ??≤-=-??，∴当12a ≤-时，对任意的m ，(0,)n ∈+∞，且m n ≠，有()()1f m f n a m ->-恒成⽴.。
山东省齐鲁名校协作体2020┄2021届高三上学期第二次调研联考 英语试题含解析
齐鲁名校教科研协作体山东省部分重点中学2021届高三第二次联考英语试题命题学校：临沂一中命题人：白爱华卢蕊注意事项：1本试卷分第I卷（选择题）和第II卷（非选择题）两部分。
2答题前，考生务必将自己的姓名、准考证号填写在本试卷相应的位置。
3.全部答案在答题卡上完成，答在本试卷上无效。
第I卷改编第一节（每小题1.5分，满分7.5分）听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What will the woman do this weekend？A. Go swimming.B. Camp in the mountain.C. Go hiking.2. Who catches the fish？A. Jason.B. The man.C. The woman.3. How many people will go to the park？A. 2.B. 3.C. 4.4. What is going to be cleaned？A. The toilets.B. The bedrooms.C. The living room.5. What will the man buy？A. Nothing.B. A computer.C. A cellphone.第二节（每小题1.5分，满分22.5分）听下面5段对话或独白。
每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听下面一段对话，回答第6和第7两个小题。
6. What's the man's father？A. A scientist.B. A businessman.C. A travel agent.7. What does the man probably like most now？A. Science.B. Business.C. Travel.听下面一段对话，回答第8和第9两个小题。
山东省齐鲁名校新高考语文二模试卷及答案解析
山东省齐鲁名校新高考语文二模试卷注意事项：1．答卷前，考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。
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将条形码粘贴在答题卡右上角"条形码粘贴处"。
2．作答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案。
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考试结束后，请将本试卷和答题卡一并交回。
1、阅读下面的文字,完成下面小题。
人们常说眼见为实,但是却有那么一类图形,能够骗过你的眼睛,当你发现真相后。
中科院发表研究论文,揭开了复杂的视觉错觉的。
研究人员探索了真实光流运动向错觉光流运动转化的脑神经生理机制。
研究人员发现了错觉信息转化机制,能够帮助人们理解视觉信息是怎么在脑区之间的传递过程以及从局部到整体的视觉信息整合。
视觉错觉,对于大脑来说却是一种真实的感知觉,它反映的是人视网膜物理输入和大脑视皮层感知之间的不一致,是人类大脑通过复杂的脑区之间的相互作用和海量神经计算而产生的。
人们注视某个图片中心的黑点,( ),然而事实上圆环并没有任何物理转动。
这种整体运动视觉错觉感知的强弱,与构成图形的局部细节密切相关。
虽然错觉广为人知,但它在大脑中是如何产生的脑神经编码机制,有待全部解开。
视觉错觉现象是好是坏?科学家认为,由于视觉错觉图形在自然界条件下极少存在,对人们的影响,因此这种现象并没有好坏的属性,它只是视觉系统基本运作方式的本能体现。
1．依次填入文中横线上的成语,全都恰当的一项是A．茅塞顿开冰山一角莫衷一是不足挂齿B．茅塞顿开九牛一毛莫衷一是微乎其微C．如梦初醒九牛一毛错综复杂不足挂齿D．如梦初醒冰山一角错综复杂微乎其微2．文中画横线的句子有语病,下列修改最恰当的一项是A．研究人员发现了错觉信息转化机制,能够帮助人们理解视觉信息是怎么在脑区之间传递的,以及是怎么从局部到整体进行视觉信息整合的。
2020-2021学年济南齐鲁学校高三英语第二次联考试卷及答案解析
2020-2021学年济南齐鲁学校高三英语第二次联考试卷及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AMy Biggest ChallengeAs a writer on an adventure sports magazine, I’dalways fought shy of doing the adventurous stuff myself, preferring instead to observe the experts from a safe distance and relay their experiences to readers in the form of written language. Thus, when I was challenged to take part in a mountain climb in aid of raising money for charity-and to write about it afterwards-I was unwilling, to say the least.I was lucky enough to have a brilliant climbing coach called Keith, who put me through my paces after my daily work. He gave me knowledge about everything from the importance of building muscle groups to how to avoid tiredness through nutrition. It quickly became apparent that the mechanics of climbing were more complex than I had imagined. There was the equipment and techniques I’d never even heard of, all of which would come in handy on the snow-capped peak I’d be climbing.Aware of the challenge, Keith made a detailed action plan and I forced myself to stick to it, doing a daily workout at the gym and going on hikes with a heavy pack. I perfected my technique on the climbing wall and even went to climb the mountains to get vital experience. My self-belief increased alongside my muscle power and I became confident about finishing the climb.All too soon I was on a plane to my destination. On that day, when I looked up at the mountain, I thought of abandoning it. But then I remembered all the hard work I’d done and how disappointed Keith would be if I gave up at the last minute-not to mention letting down the charity and the sense of failure I’d experience myself. With a deep breath I gathered my equipment and headed out into the sunshine to meet the rest of the group.And as I sit here now, tapping away on my laptop, I’m amazed at the details in which I can recall every second of the climb: the burning muscles, the tiredness, the minor problems along the way. Could I have been better prepared? Possibly. Would I be back for another go? Thankfully not. The feeling of being excited when I stood on top of the world is a never-to-be-repeated experience but one I will enjoy forever nevertheless.1. At the beginning of the activity, the author revealed his ______.A. disappointment in the coming adventure.B. expectation of writing about his experience.C. lack of enthusiasm for the challenge he’d been offered.D. curiosity about taking part in the mountain climb for charity.2. What did the author realise during his climb training?A. The knowledge about climbing was really confusing.B. The equipment was the key factor to reach the peak.C. Climbing was much more complicated than expected.D. Hard training was far more important than making plans.3. How did the author feel after he succeeded in climbing the mountain?A. He was relieved that he wouldn’t have to do it again.B. He was well satisfied that he had done his best for it.C. He was surprised that he had managed to complete it.D.He was regretful that it wasn’t as smooth as imagined.BThere will be more plastic than fish in the world’s oceans by 2050. That is what a new report from the World Economic Forum and Ellen MacArthur Foundation warns. If the current trend continues, the report said, oceans will contain one ton of plastic for every three tons of fish in 2025. By 2050, plastics will weigh more than fish. The problem is that each year at least 8 million tons of plastics end up in oceans around the world. This is the same as dumping the contents of one garbage truck into the ocean every minute.Not all plastic ends up in the ocean because someone throws a plastic bottle into the water. Plastic containers and other trash thrown onto streets and sidewalks often are swept into oceans. Unlike other types of trash in the ocean, the plastic never bio-degrades. There is a way to slow the amount of plastics going into the oceans — people can recycle more. Currently only about 14 percent of plastics are recycled. Research in Europe shows as much as 53 percent of plastic could be recycled using available technology. The report says that another solution is using less plastic for packaging products. But that is not likely to happen.“Given plastic packaging’s many benefits, both the possibility and desirability of an across-the-board dramatic reduction in the volume of plastic packaging used is clearly low, ”the report said. But the authors note reducing the use of plastics should be tried“where possible. ”For decades, scientists warned that plastics are killing fish. Research shows that fish are dying from choking after eating plastics. Another cause of death is that plastics cause“intestinal blockage and starvation, ”the environmental group said.4. Why is the garbage truck mentioned in Paragraph 1 ?A. To explain how plastics end up in the ocean.B. To warn people against the ocean pollution.C. To clarify the seriousness of the problem.D. To point out some details of the report.5. Why is plastic more dangerous than other types of trash?A. There is too much of it.B. It poisons the ocean water.C. It is from different sources.D. It is hard to break down.6. What is the author’s attitude to the suggested solutions?A. Pessimistic.B. Curious.C. Unconcerned.D. Terrified.7. What can we infer from the text?A. People are not aware of the problem.B. Recycling is limited by lack of technology.C. Plastic packaging has become part of our life.D. Fish in the ocean will be replaced by plastics.CThe idea of growing food in a desert would make most people laugh but this is quickly becoming a reality. There are currently two desert farms in the world where quality vegetables are being planted cheaply and easily.Sundrop Farms, based in South Australia, uses experimental greenhouses to grow tomatoes, peppers and cucumbers. The biggest challenge of growing food in a desert, obviously, is the lack of available water. The researchers at Sundrop Farms have gotten aroundthis problemby using the sun to desalinate (淡化) sea water. It can also be used to control the temperature of the greenhouses.Without depending on limited resources such as land and fresh water Sundrop Farms has made farming a practice. This can increase the world’s food supplies. Another benefit ofthis kind of farming is that it can be done anywhere, thus reducing the costs of transporting food to distant locations. Yet another benefit is that it reduces the need for pesticides (杀虫剂).Another experimental desert farm is the Sahara Forest Project, which began in Qatar in December 2012. Greenhouses in the farm are cooled by saltwater. Solar power and other technologies are used together to help make vegetation (植被) grow in the desert environment. As deserts have expanded over recent years around the world due to global warming, this project could solve the problem.The result form the Qatar project were better than expected and in June of 2014, Jordan agreed to hostanother one. This will be much bigger than the Qatar project and the project members will have even more opportunities to test their experiments on a much larger scale. It is not clear yet that desert farming resents the future of farming but these projects have shown some success in the field.8. What does “this problem” in paragraph 2 refer to?A. Sea water is bard to purify.B. The desert is short of water.C. The temperature is high in the desert.D. Desert farms aren’t fit to plant vegetables.9. What is one of the characters of desert farming?A. It needs more pesticides.B. It saves delivery costs.C. It has a location limit.D. It solves food waste problems10. What can we know about the Sabara Forest Project?A. It lives up to expectationsB. It can help produce more foodC. It is started to prevent global warmingD. It uses technology to produce saltwater11. What can be inferred about desert farming from the last paragraph?A. It still has problems to solve.B. It represents the future of farming.C. Its early success has aroused interest in it.D. Its aim is to create more job opportunities.DYou’re in a crowd of people who are all asking for the same thing. How do you make your voice heard above the rest? Be different. Don’t shout. Lisa, 25, was waiting to board a plane flying fromLondontoAustriafor Christmaswhen the flight was cancelled.“There were about a hundred of us unable to leave,” she says. “Everyone else was shouting at the airport staff. Instead of joining in, I walked up to the man behind the ticket desk very quietly and said, ‘This must be so awful for you! I don’t know how you deal with these situations—it’s not even your fault. I could never handle it as well as you are.’ Without my even asking, he found me a seat on another airline with an upgrade to first class. He was happy to do a favor forsomeone who was appreciative instead of unfriendliness.”Flattery (恭维) is an essential element of the sweet-talk strategy. “It’s human psychology that stroking a person’s ego (自我) with a few well-directed praises makes them want to prove you right,” says apsychologist. “Tell someone they’re pretty and they’ll instantly fix their hair; praise their sense of humor and they’ll tell a joke.”You need help and there’s ly no reason that the person will want to lend a hand. Allison, 26. a lawyer, realized she’d made a huge mistake on a batch of documents. “The only way I could fix the problem was to get the help of a colleague who I knew didn’t like me,” she said.Allison then went to the woman’s office and explained her problem. “As I was saying to the boss the other day you’re the only person who would know how to handle a situation like this, what would you suggest I do?” “Feeling pumped up (鼓励), she set about helping me and we finished the job on time, and she was happy to help.” Allison said.12. Whatwould have happened at the airport according to paragraph 1?A. The departure hall was filled with noise.B. Someone screamed just lo be different.C. The passengers waited on board patiently.D. The airport stuff were rude to the passengers.13. Why did the man put Lisa on another airline?A. He admired Lisa’s beauty.B. He appreciated her attitude.C. He was ready to help others.D. He was blamed for the cancellation.14. What is the third paragraph mainly about?A. The potential benefits of ego.B. The strategy to start small talk.C. The great importance of flattery.D. The value of humor in daily life.15. What can we learn about Allison’s colleague?A. She was a popular lawyer.B. She was always ready to help others.C. She always got praise from Allison.D. She did a great favor for Allison eventually.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020届山东省齐鲁名校高三第二次调研考试化学试题
2020届山东省齐鲁名校高三第二次调研考试化学试题★祝考试顺利★注意事项：1、考试范围：高考范围。
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可能用到的相对原子质量：H 1 C 12 N 14 O 16 Mg 24 Cl 35.5 Ca 40 Cu 64Ⅰ卷一、选择题（每题3分，只有1个正确答案）1．德国著名行业杂志《应用化学》上刊登文章介绍：某中德联合研究小组设计制造了一种“水瓶”，用富勒烯(C60)的球形笼子作“瓶体”，一种磷酸盐作“瓶盖”，恰好可将一个水分子关在里面。
下列说法正确的是()A．水、双氧水、水玻璃都是纯净物B．石墨和C60互称为同位素C．磷酸钙是可溶性强电解质D．一定条件下石墨转化为C60是化学变化2.下列选项说法不正确的是()A．25 ℃时，0.1 mol·L－1 CH3COOH溶液的pH>1，证明CH3COOH是弱电解质B．BaSO4的水溶液不易导电，故BaSO4是弱电解质C．25 ℃时，0.1 mol·L－1的硫化氢溶液比等浓度的硫化钠溶液的导电能力弱D．醋酸、一水合氨、水都是弱电解质3．设阿伏加德罗常数的值为N A。
2020-2021学年济南齐鲁学校高三生物第二次联考试卷及参考答案
2020-2021学年济南齐鲁学校高三生物第二次联考试卷及参考答案一、选择题：本题共15小题，每小题2分，共30分。
每小题只有一个选项符合题目要求。
1. 某种H﹢-ATPase是一种位于膜上的载体蛋白，具有A TP水解酶活性，能够利用水解ATP释放的能量逆浓度梯度跨膜转运H﹢。
①将某植物气孔的保卫细胞悬浮在一定pH的溶液中（假设细胞内的pH高于细胞外），置于暗中一段时间后，溶液的pH不变。
①再将含有保卫细胞的该溶液分成两组，一组照射蓝光后溶液的pH明显降低；另一组先在溶液中加入H﹢-ATPase的抑制剂（抑制ATP水解），再用蓝光照射，溶液的pH不变。
根据上述实验结果，下列推测不合理的是A.H﹢-ATPase位于保卫细胞质膜上，蓝光能够引起细胞内的H﹢转运到细胞外B.蓝光通过保卫细胞质膜上的H﹢-ATPase发挥作用导致H﹢逆浓度梯度跨膜运输C.H﹢-ATPase逆浓度梯度跨膜转运H﹢所需的能量可由蓝光直接提供D.溶液中的H﹢不能通过自由扩散的方式透过细胞质膜进入保卫细胞2. 交感神经和副交感神经是神经系统的重要组成部分，下列有关它们的叙述正确的是（）A.它们包括传入神经和传出神经B.它们都属于中枢神经系统中的自主神经系统C.它们通常共同调节同一器官，且作用通常相反D.交感神经使胃肠蠕动加强，副交感神经使胃肠蠕动减弱3. 下列选项中不符合含量关系“c=a+b，且a>b”的是（）A.a非必需氨基酸种类、b必需氨基酸种类、c人体蛋白质的氨基酸种类B.a各细胞器的膜面积、b细胞核的膜面积、c生物膜系统的膜面积C.a线粒体的内膜面积、b线粒体的外膜面积、c线粒体膜面积D.a叶肉细胞的自由水、b叶肉细胞的结合水、c叶肉细胞总含水量4. 关于下图所示实验的叙述中，错误的是()A.甲装置是实验组，乙装置是对照组B.H2O2溶液的浓度属于无关变量C.乙试管产生气泡的速率比甲试管快D.两支试管中H2O2溶液的体积应相同5. 下列关于生物组织中还原糖、蛋白质的鉴定，叙述正确的是（）A. 可用斐林试剂甲液和乙液、蒸馏水来鉴定葡萄糖和蛋白质B. 在组织样液中加入斐林试剂后不产生砖红色沉淀说明没有还原糖C. 实验结束时，要将剩余的斐林试剂装入棕色瓶，以便长期保存备用D. 变性后的蛋白质不能与双缩脲试剂发生紫色反应6. 赫尔希和蔡斯的T2噬菌体侵染大肠杆菌实验证实了DNA是遗传物质，下列关于该实验的叙述正确的是（）A. 实验需分别用含32P和35S的培养基培养噬菌体B. 搅拌目的是使大肠杆菌破裂，释放出子代噬菌体C.35S标记噬菌体的组别，搅拌不充分可致沉淀物的放射性增强D.32P标记噬菌体的组别，放射性同位素主要分布在上清液中7. 下列关于神经调节的叙述，错误的是（）A.突触前膜释放的神经递质只能作用于神经细胞B.条件反射的建立需要大脑皮层的参与，可以消退C.大脑感知膝盖下方的叩击后可以有意识抑制膝跳反射D.碰到烛火时的缩手可能会快过大脑对火烧感觉的感知8. 植物在白天“醒来”晚上“入睡”的现象与叶绿体活动有关，而叶绿体活动受TRXL2酶和2CP酶的共同影响。
山东省齐鲁名校教科研协作体2020┄2021届高考化学模拟试卷Word版 含解析
山东省齐鲁名校教科研协作体2021年高考化学模拟试卷一、选择题（每小题5分，共42分）1．（5分）2021年全国多个城市出现了严重的雾霾天气，十面“霾”伏，自强不“吸”等网络语流行，（）A．目前加碘食盐中主要添加的是KIB．光化学烟雾与大量排放碳氢化合物和氮氧化合物有关C．淀粉、纤维素和聚乙烯都属于天然高分子化合物D．光纤通信使用的光缆的主要成分和太阳能电池使用的材料都是SiO22．（5分）某学习兴趣小组讨论辨析以下说法，其中说法正确的是（）①通过化学变化可以实现16O与18O间的相互转化；②向Fe（OH）3胶体中逐渐滴入过量的稀硫酸，能发生“先沉淀后溶解”的现象；③向无色溶液中滴加盐酸生成能使澄清石灰水变浑浊的无色气体，证明溶液一定含CO32﹣或者HCO3﹣；④Na、S、C分别跟氧气反应均能得到两种不同的氧化物；⑤等物质的量的Na2O、Na2O2阴离子数之比为1：2；⑥加入铝粉后产生大量氢气的溶液中，NH4+、Na+、NO3﹣、CI﹣可以大量共存．A．②⑥B．②④⑤C．②D．②④⑥3．（5分）X、Y、Z、W是短周期元素，原子序数依次递增，Y、Z、W原子的最外层电子数之和为14，X与Z位于同一主族，Y元素单质既能与盐酸反应也能与NaOH溶液反应，Z原子的最外层电子数是次外层电子数的一半，下列说法正确的是（）A．气态氢化物的稳定性：X＜Z＜WB．X、Z、W氧化物对应水化物酸性的强弱顺序为Z＜X＜WC．室温下，含Y元素的盐形成的水溶液其pH＜7D．Z元素的单质和氧化物均既能与NaOH溶液反应也能与某种酸反应4．（5分）有机物M的结构简式如图所示，下列说法正确的是（）A．分子式为C17H18O3B．M中有4个饱和碳原子C．M可以使溴水和酸性高锰酸钾溶液褪色，但它们的反应类型不同D．1molM常温下最多可以与6molH2反应5．（5分）某温度下在容积为2L的密闭容器中，发生反应2X（g）+Y（g）═2W（g）△H＜0，当充入2molX和1molY，经20s达到平衡时生成了0.4molW．下列说法正确的是（）①升高温度，W的体积分数减小，△H增大；②以Y的浓度变化表示的反应速率为0.01mol（L•s）；③在其它条件不变的情况下，增加1molX，则X和Y的转化率均提高；④增大压强，平衡向正反应方向移动，该反应的平衡常数增大；⑤再向容器中通入2molX和1molY，达到平衡时，X、Y的体积分数均减小．A．①⑤B．⑤C．②③④D．④⑤6．（5分）关于下列各实验装置的叙述中，正确的是（）A．图①可用于制取少量NH3或用MnO2和浓盐酸制取Cl2B．可用从a处加水的方法检验图②装置的气密性C．实验室可用图③的方法收集Cl2或NH3D．利用图④装置制硫酸和氢氧化钠，其中c为阴离子交换膜、b为阳离子交换膜7．（5分）下列各种说法中，不正确的是（）A．物质的量浓度均为0.1mol•L﹣1的①Na2CO3溶液、②CH3COONa溶液、③NaOH溶液、水的电离程度的大小顺序是①＞②＞③B．将100mLpH=a的盐酸与100mLpH=b的Ba（OH）2溶液混合后恰好中和，则a+b=13C．向10mL浓度为0.1mol•L﹣1的CH3COOH溶液中滴加相同浓度的氨水，在滴加过程中（NH4+）/c （NH3•H2O）先增大再减小D．等物质的量的二元弱酸H2X与其钾盐K2X﹣的混合液中：c（K+）=c（H2X）+c （HX﹣）+c（X2﹣）二、非选择题（本大题有小题，每空分，共分）8．（16分）随着大气污染的日趋严重，“低碳减排”备受关注，研究NO2、SO2、CO 等大气污染气体的处理具有重要意义．（1）图1是在101kP a，298k条件下1molNO2和1molCO反应生成1molCO2和1molNO过程中能量变化示意图．已知：N2（g）+O2（g）=2NO（g）△H=+179.5kJ/mol2NO（g）+O2（g）═2NO2（g）△H=﹣112.3kJ/mol请写出NO（g）与CO反应生成无污染气体的化学方程式．（2）将0.20molNO和0.10molCO充入一个容积恒定为1L的密闭容器中发生如图所示的反应，反应达到平衡时，生成了0.02molCO2．①下列说法正确的是．（填序号）a．容器内的压强不发生变化说明该反应达到平衡b．当向容器中加再充入0.20molNO时，平衡向正反应方向移动，K不变c．升高温度后，K值减小，NO的转化率减小d．向该容器内充入He气，压强增大，反应速率增大②反应达到平衡时，若保存温度不变，此时再向容器中充入CO2、NO各0.060mol，平衡将移动（填“正向”、“逆向”或“不”）（3）直接排放煤燃烧产生的烟气会引起严重的环境问题．①煤燃烧产生的烟气含氮的氧化物，用CH4催化还原NO x可以消除氮氧化物的污染．例如：CH4+2NO2=N2+CO2+2H2O，当生成14gN2时，转移电子数目是②将煤燃烧产生的二氧化碳回收利用，可达到低碳排放的目的．如图2是通过人工光合作用，以CO2（g）和H2O（g）为原料制备HCOOH和O2的原理示意图．催化剂b表面发生的电极反应式为．③25℃时，部分物质的电离平衡常数如图所示：化学式HCOOH H2CO3HCIO电离平衡常数 1.77×10﹣4K i1=4.3×10﹣7K i2=5.6×10﹣11 3.0×10﹣10请回答下列问题：同浓度的HCOO﹣、HCO3﹣、CO32﹣、ClO﹣结合H+的能力有强到弱的顺序是，物质的量浓度均为0.1mol•L﹣1的下列四种物质的溶液：a、Na2CO3；b、NaClO；c、HCOONa；d、NaHCO3，pH由小到大的顺序是（填编号）9．（17分）錳、铬、钴、镍虽不是中学阶段常见的金属元素，但在工业生产中有着重要作用．Ⅰ．（1）自然界Cr主要以+3价和+6价存在．+6价的Cr能引起细胞的突变，可以用亚硫酸钠将其还原为+3价的铬．完成并配平下列离子方程式：Cr2O72﹣+SO32﹣+=Cr3++SO42﹣+H2OⅡ．某化学研究性学习小组拟从废旧干电池中回收二氧化锰制取碳酸锰．①将干电池剖切、分选得到黑色混合物（主要成分为MnO2）洗涤、过滤、烘干．②将上述固体按固液体积比2：9加入浓盐酸、加热，反应完全后过滤、浓缩．③向上述溶液中加入Na2CO3溶液，边加边搅拌，再过滤即可得到碳酸锰．（2）在第②步中，将上述固体与浓盐酸混合的目的是，反应的离子反应方程式：（3）过程②中，浓缩需要使用酒精灯、三脚架，还需要的仪器有Ⅲ．（4）NiSO4•6H2O是一种绿色易溶于水的晶体，广泛用于镀镍、生产电池等，可由电镀废渣（除含镍外，还含有：Cu、Zn、Fe、Cr等杂质）为原料获得．操作步骤如下：①向滤液Ⅰ中加入FeS是为了除去Cu2+、Zn2+等杂质，除去Cu2+的离子方程式为．②对滤液Ⅱ先加W，再测pH，W最适宜的试剂是，调pH的目的是．③滤液Ⅲ溶质的主要成分是NiSO4，加Na2CO3过滤后，再加适量稀硫酸溶解又生成NiSO4，这两步操作的目的是．10．（1）某校研究性学习小组同学用如图1所示实验时发现BaCl2溶液中出现白色沉淀，且白色沉淀不溶于盐酸．二氧化硫通入氯化钡溶液中理论上不产生沉淀，实际因受到空气和溶液中氧气的影响而产生沉淀．①写出加热时试管中发生反应的化学方程式．②用离子方程式表示生成该白色沉淀的原因：．（2）为了避免产生沉淀，某化学小组设计了如图2所示的实验装置．请回答下列问题：①检查装置A气密性的方法是；②装置B中反应开始前，应进行的操作是；③装置C中苯的作用是；④装置B中反应结束后，需再打开装置A中活塞，将产生的氢气导入后续装置一段时间，目的是；⑤一同学取实验后的C中溶液少许，滴加一种无色溶液，也产生不溶于盐酸的白色沉淀，滴加的试剂不可能是下列物质中的，A、NaOH溶液B、新制氯水C、H2O2溶液D、酸性KMnO4溶液⑥若D中NaOH溶液恰好完全反应生成NaHSO4溶液，写出该溶液中的平衡方程式，该溶液呈酸性，则溶液中离子浓度大小顺序为．（选做题）（从11、12、13题中任选一题作答，若多选，按所做的第一题得分）【化学---化学与技术】11．（12分）工业上制取硝酸铵的流程图如图，请回答下列问题：（1）在上述工业制硝酸的生产中，B设备的名称是，其中发生反应的化学方程式为．（2）在合成硝酸的吸收塔中通入空气的目的是．（3）生产硝酸的过程中常会产生一些氮的氧化物，可用如下两种方法处理：碱液吸收法：NO+NO2+2NaOH═2NaNO2+H2ONH3还原法：NH3+NO2\frac{\underline{\;催化剂\;}}{\;}N2+H2O①配平上述反应方程式；②以上两种方法中，符合绿色化学的是．（4）某化肥厂用NH3制备NH4NO3．已知：由NH3制NO的产率是96%、NO制HNO3的产率是92%，则制HNO3，所用去的NH3的质量占总耗NH3质量（不考虑其它损耗）的%．【化学--物质结构与性质】12．许多金属及他们的化合物在科学研究和工业生产中具有许多用途，回答下列有关问题（1）下列有关的说法正确的是A、第一电离能大小：S＞P＞SiB、因为晶格能CaO比KCl高，所以KCl比CaO熔点低C、SO2与CO2的化学性质类似，分子结构也都呈直线型，相同条件下SO2的溶解度更大D、分子晶体中，共价键键能越大，该分子晶体的熔沸点越高（2）镍（Ni）可形成多种配合物，且各种配合物有广泛的用途．某镍配合物结构如图所示，分子内含有的作用力有（填序号）．A．氢键B．离子键C．共价键D．金属键E．配位键组成该配合物分子且同属第二周期元素的电负性由大到小的顺序是．甲基（﹣CH3）中C 原子的杂化方式为．（3）铁和铜在生产和生活中有重要应用，基态Fe2+的M层电子排布式为，用晶体的x射线衍射发可以测得阿伏伽德罗常数，对金属铜的测定得到以下结果：晶胞为面心立方最密堆积，边长为361pm（提示：1pm=10﹣10cm，3.613=47.05），又知铜的密度为9.00g•cm﹣3，则铜晶胞的质量是g（保留两位小数）；阿伏加德罗常数为（保留两位小数）．【化学---有机化学基础】13．酚是重要的化式原料，通过下列流程可合成阿司匹林、香料和一些高分子化合物．已知：（1）写出水杨酸A分子中官能团名称．（2）写出反应③的化学方程式．（3）写出反应类型：④，⑦．（4）检验阿司匹林样品中混有水杨酸的试剂是．（5）写出F的结构简式；（6）写出符合下列条件的水杨酸A的同分异构体：．①能发生水解反应；②遇FeCI2溶液呈紫色；③能发生银镜反应．山东省齐鲁名校教科研协作体2021年高考化学模拟试卷参考答案与试题解析一、选择题（每小题5分，共42分）1．（5分）2021年全国多个城市出现了严重的雾霾天气，十面“霾”伏，自强不“吸”等网络语流行，（）A．目前加碘食盐中主要添加的是KIB．光化学烟雾与大量排放碳氢化合物和氮氧化合物有关C．淀粉、纤维素和聚乙烯都属于天然高分子化合物D．光纤通信使用的光缆的主要成分和太阳能电池使用的材料都是SiO2考点：常见的生活环境的污染及治理．专题：化学应用．分析：A．目前加碘食盐中主要添加的是碘酸钾；B．光化学烟雾和氮氧化物、碳氢化合物有关；C．聚乙烯属于合成高分子化合物；D．光导纤维的主要成分为二氧化硅，太阳能电池使用的材料都是Si．解答：解：A．目前加碘食盐中主要添加的是KIO3，故A错误；B．光化学烟雾与大量排放碳氢化合物和氮氧化合物有关，故B正确；C．淀粉、纤维素属于天然高分子化合物，聚乙烯属于合成高分子化合物，故C错误；D．光纤通信使用的光缆的主要成分是晶体SiO2，太阳能电池使用的材料主要是Si，故D 错误；故选：B．点评：本题考查了生活中化学知识，熟悉加碘盐的成分、光化学烟雾的成因、天然高分子化合物的定义、二氧化硅、硅的性质是解题关键，题目难度不大．2．（5分）某学习兴趣小组讨论辨析以下说法，其中说法正确的是（）①通过化学变化可以实现16O与18O间的相互转化；②向Fe（OH）3胶体中逐渐滴入过量的稀硫酸，能发生“先沉淀后溶解”的现象；③向无色溶液中滴加盐酸生成能使澄清石灰水变浑浊的无色气体，证明溶液一定含CO32﹣或者HCO3﹣；④Na、S、C分别跟氧气反应均能得到两种不同的氧化物；⑤等物质的量的Na2O、Na2O2阴离子数之比为1：2；⑥加入铝粉后产生大量氢气的溶液中，NH4+、Na+、NO3﹣、CI﹣可以大量共存．A．②⑥B．②④⑤C．②D．②④⑥考点：核素；胶体的重要性质；离子共存问题；钠的重要化合物；常见阴离子的检验．分析：①化学变化的实质：分子的分裂，原子的重新组合；②根据胶体的性质判断，加入稀硫酸产生聚沉现象，H2SO4与Fe（OH）3反应，所以出现先沉淀后溶解现象；③能使澄清石灰水变浑浊的无色气体可能是二氧化硫；④S在氧气中点燃只能生成二氧化硫；⑤Na2O2和Na2O两种物质中的阴离子分别是过氧根离子和氧离子；⑥能与铝片反应生成氢气的溶液可能呈碱性，也可能呈酸性，分别判断在碱性或酸性条件下离子之间能否发生反应．解答：解：16O与18O互为同位素，二者不需要通过化学变化实现，故①错误，向Fe（OH）3胶体中逐滴滴入过量的稀硫酸，先发生胶体的聚沉，继续加入硫酸，发生酸碱中和反应，Fe（OH）3溶解，故②正确；若溶液中含有CO32﹣、HCO3﹣、SO32﹣、HSO3﹣等离子时，与盐酸反应产生能够使澄清石灰水变浑浊的无色气体，故③错误；S跟氧气反应只生成SO2，故④错误；等物质的量的Na2O、Na2O2阴离子数之比为1：1，故⑤错误，加入铝粉能够生成氢气的溶液可能是强酸性或强碱性溶液，强酸性条件下，NO3﹣具有强氧化性，不能生成氢气，强碱性条件下，NH4+不能大量共存，故⑥错误；故选C．点评：本题主要考查的是同位素性质、胶体的性质、离子共存等知识点，难度不大．3．（5分）X、Y、Z、W是短周期元素，原子序数依次递增，Y、Z、W原子的最外层电子数之和为14，X与Z位于同一主族，Y元素单质既能与盐酸反应也能与NaOH溶液反应，Z原子的最外层电子数是次外层电子数的一半，下列说法正确的是（）A．气态氢化物的稳定性：X＜Z＜WB．X、Z、W氧化物对应水化物酸性的强弱顺序为Z＜X＜WC．室温下，含Y元素的盐形成的水溶液其pH＜7D．Z元素的单质和氧化物均既能与NaOH溶液反应也能与某种酸反应考点：原子结构与元素周期律的关系．专题：元素周期律与元素周期表专题．分析：Y元素的单质既能与盐酸反应也能与NaOH溶液反应，应为Al元素，Z原子的最外层电子数是次外层电子数的一半，应为Si元素，X与Z位于同一主族，则X为C元素，Y、Z、W原子的最外层电子数之和为14，则W的最外层电子数为14﹣3﹣4=7，且原子序数最大，应为Cl元素，结合元素周期律的递变规律判断元素对应的单质、化合物的性质．解答：解：Y元素的单质既能与盐酸反应也能与NaOH溶液反应，应为Al元素，Z原子的最外层电子数是次外层电子数的一半，应为Si元素，X与Z位于同一主族，则X为C 元素，Y、Z、W原子的最外层电子数之和为14，则W的最外层电子数为14﹣3﹣4=7，且原子序数最大，应为Cl元素，A．气态氢化物的稳定性应该是：Z＜X＜W，故A错误；B．只能判断X、Z、W最高价氧化物对应水化物酸性的强弱顺序为Z＜X＜W，故B错误；C．Y是铝元素，含铝元素的盐如AlCl3水溶液显酸性，NaAlO2水溶液显碱性，故C错误；D．硅和二氧化硅既能与NaOH溶液反应也能与氢氟酸反应，故D正确；故选D．点评：本题考查元素的推断及元素化合物的性质，题目难度中等，注意从物质的性质作为推断题的突破口，注意常见既能与盐酸反应也能与NaOH溶液反应的物质，题中从常见元素化合物的性质入手考查，是一道位置、结构、性质有机结合综合考查学生分析能力、推断能力的典型题目，难度较大．4．（5分）有机物M的结构简式如图所示，下列说法正确的是（）A．分子式为C17H18O3B．M中有4个饱和碳原子C．M可以使溴水和酸性高锰酸钾溶液褪色，但它们的反应类型不同D．1molM常温下最多可以与6molH2反应考点：有机物的结构和性质．专题：有机物的化学性质及推断．分析：由结构可知分子式，分子中含碳碳双键、﹣OH、﹣COOC﹣，结合烯烃、醇、酯的性质来解答．解答：解：A．M中有6个饱和碳原子，分子式为C17H20O3，故A错误；B．甲基、﹣CH（CH3）2中的C均为饱和碳原子，且与﹣OH相连环状结构中含2个饱和碳原子，故B错误；C．M含双键，使溴水褪色属于加成反应，使酸性高锰酸钾溶液褪色是氧化反应，故C正确；D．苯环、双键与氢气发生加成反应，则1mol M常温下最多可以与5 mol H2反应，故D错误．故选C．点评：本题考查有机物的结构与性质，为高频考点，把握结构中官能团与性质的关系为解答的关键，侧重醇、烯烃等有机物性质的考查，题目难度不大．5．（5分）某温度下在容积为2L的密闭容器中，发生反应2X（g）+Y（g）═2W（g）△H＜0，当充入2molX和1molY，经20s达到平衡时生成了0.4molW．下列说法正确的是（）①升高温度，W的体积分数减小，△H增大；②以Y的浓度变化表示的反应速率为0.01mol（L•s）；③在其它条件不变的情况下，增加1molX，则X和Y的转化率均提高；④增大压强，平衡向正反应方向移动，该反应的平衡常数增大；⑤再向容器中通入2molX和1molY，达到平衡时，X、Y的体积分数均减小．A．①⑤B．⑤C．②③④D．④⑤考点：化学平衡的影响因素．专题：化学平衡专题．分析：①升高温度，W的体积分数减小，则平衡逆向移动，但△H不变；②用W的浓度变化表示的反应速率，v=；③增加X的量，平衡正向移动，则X的转化率会减小，Y的转化率增大；④增大压强，平衡向气体体积减小的方向移动，即向着正反应方向移动，但反应的平衡常数不变；⑤再向容器中通入 2 mol X 和 1 mol Y，相当是对原平衡加压，平衡正向移动，达到平衡时，X、Y 的转化率均增大，X、Y 的体积分数均减小．解答：解：①升高温度，W的体积分数减小，则平衡逆向移动，但△H不变，因为△H与方程式的写法有关，方程式确定，△H确定，①错误；②用W的浓度变化表示的反应速率v=0.4mol/2L•20s=0.01mol/（L•s），所以用Y表示的反应速率0.005 mol/（L•s），②错误；③增加X的量，平衡正向移动，则X的转化率会减小，Y的转化率增大，③错误；④增大压强，平衡向气体体积减小的方向移动，即向着正反应方向移动，但反应的平衡常数不变，④错误；⑤再向容器中通入 2 mol X 和 1 mol Y，相当是对原平衡加压，平衡正向移动，达到平衡时，X、Y 的转化率均增大，X、Y 的体积分数均减小，⑤正确．选B．点评：本题考查化学反应速率及化学平衡的影响因素，为高频考点，把握压强、浓度对反应速率及平衡的影响即可解答，注重基础知识的考查题目难度不大．6．（5分）关于下列各实验装置的叙述中，正确的是（）A．图①可用于制取少量NH3或用MnO2和浓盐酸制取Cl2B．可用从a处加水的方法检验图②装置的气密性C．实验室可用图③的方法收集Cl2或NH3D．利用图④装置制硫酸和氢氧化钠，其中c为阴离子交换膜、b为阳离子交换膜考点：化学实验方案的评价．专题：实验评价题．分析：A．MnO2和浓盐酸制取Cl2应在加热条件下进行；B．根据压强差原理判断；C．应用向上排空法收集氯气；D．根据阴、阳离子的定向移动判断．解答：解：A．图①可以利用碱石灰（或生石灰）与浓氨水制取氨气，用MnO2和双氧水或Na2O2与水反应制取氧气，不能用MnO2和浓盐酸制取Cl2，故A错误；B．图②关闭止水夹，从a处加水，U型管中产生液面差，静置一段时间，若液面差不变气密性好，若不能形成液面差或液面差发生变化，气密性不好，故B正确；C．氯气密度比空气的大，收集时应从长导管通入，而氨气的密度比空气的小，收集时应从短导管通入，故C错误；D．以Pt电解Na2SO4溶液，实际上是电解水，阳极放电的是OH﹣，该极区H+与阴离子交换膜交换的SO42﹣形成H2SO4溶液，而阴极放电的是H+，该极区的OH﹣，与阳离子交换膜交换的Na+形成NaOH溶液，故b为阴离子交换膜、c为阳离子交换膜，故D错误．故选B．点评：本题综合考查化学实验方案的评价，为高频考点，侧重于学生的分析能力和实验能力的考查，注意把握实验的严密性和可行性的评价，难度中等．7．（5分）下列各种说法中，不正确的是（）A．物质的量浓度均为0.1mol•L﹣1的①Na2CO3溶液、②CH3COONa溶液、③NaOH溶液、水的电离程度的大小顺序是①＞②＞③B．将100mLpH=a的盐酸与100mLpH=b的Ba（OH）2溶液混合后恰好中和，则a+b=13C．向10mL浓度为0.1mol•L﹣1的CH3COOH溶液中滴加相同浓度的氨水，在滴加过程中（NH4+）/c （NH3•H2O）先增大再减小D．等物质的量的二元弱酸H2X与其钾盐K2X﹣的混合液中：c（K+）=c（H2X）+c （HX﹣）+c（X2﹣）考点：盐类水解的应用；离子浓度大小的比较．专题：盐类的水解专题．分析：A．酸碱抑制水的电离，易水解的盐促进水电离，水解程度越大，水的电离程度越大；B．酸碱恰好中和，则盐酸中n（H+）等于氢氧化钡中n（OH﹣）；C．根据一水合氨的电离平衡常数可知，该比值与氢氧根离子成反比，电解氨水的过程中，溶液中氢氧根离子浓度逐渐增大；D．根据物料守恒分析．解答：解：A．物质的量浓度均为0.1mol•L﹣1的①Na2CO3溶液、②CH3COONa溶液、③NaOH溶液，因醋酸的酸性比碳酸的酸性强，N a2CO3水解程度大于CH3COONa，故Na2CO3的pH大于CH3COONa，NaOH是强碱，pH最大，所以水的电离程度的大小顺序是①＞②＞③，故A正确；B．氯化氢和氢氧化钡都是强电解质，酸碱恰好中和，则盐酸中n（H+）等于氢氧化钡中n（OH﹣），溶液混合后恰好中和：10﹣a×0.1=10b﹣14×0.1，化简得a+b=14，故B错误；C．当向CH3COOH溶液中滴加相同浓度的氨水，开始时溶液为CH3COOH和CH3COONH4的混合物，由CH3COONH4的水解常数K h=，随着氨水的加入，c（H+）逐渐减小，Kh不变，则变小，当加氨水至溶液显碱性时，氨水的电离常数K b=，c（OH﹣）与氢离子浓度成反比，随着氨水的滴入，氢氧根离子浓度逐渐增大，电离常数K不变，所以逐渐减小，即始终减小，故C错误；D．等物质的量的二元弱酸H2X与其钾盐K2X的混合溶液中，根据元素守恒可知n（X）=n（K），则有c（K+）=c（H2X）+c（HX﹣）+c（X2﹣），故D正确．故选BC．点评：本题考查溶液中离子的浓度的比较，明确物质的化学式及水解、电离是解答本题的关键，并注意溶液中的溶质及其物质的量的关系、水解及抑制水解的因素等来解答．二、非选择题（本大题有小题，每空分，共分）8．（16分）随着大气污染的日趋严重，“低碳减排”备受关注，研究NO2、SO2、CO 等大气污染气体的处理具有重要意义．（1）图1是在101kP a，298k条件下1molNO2和1molCO反应生成1molCO2和1molNO过程中能量变化示意图．已知：N2（g）+O2（g）=2NO（g）△H=+179.5kJ/mol2NO（g）+O2（g）═2NO2（g）△H=﹣112.3kJ/mol请写出NO（g）与CO反应生成无污染气体的化学方程式2NO（g）+2CO（g）⇌N2（g）+2CO2（g）△H=﹣759.8KJ/mol．（2）将0.20molNO和0.10molCO充入一个容积恒定为1L的密闭容器中发生如图所示的反应，反应达到平衡时，生成了0.02molCO2．①下列说法正确的是abc．（填序号）a．容器内的压强不发生变化说明该反应达到平衡b．当向容器中加再充入0.20molNO时，平衡向正反应方向移动，K不变c．升高温度后，K值减小，NO的转化率减小d．向该容器内充入He气，压强增大，反应速率增大②反应达到平衡时，若保存温度不变，此时再向容器中充入CO2、NO各0.060mol，平衡将逆向移动（填“正向”、“逆向”或“不”）（3）直接排放煤燃烧产生的烟气会引起严重的环境问题．①煤燃烧产生的烟气含氮的氧化物，用CH4催化还原NO x可以消除氮氧化物的污染．例如：CH4+2NO2=N2+CO2+2H2O，当生成14gN2时，转移电子数目是2.408×1024②将煤燃烧产生的二氧化碳回收利用，可达到低碳排放的目的．如图2是通过人工光合作用，以CO2（g）和H2O（g）为原料制备HCOOH和O2的原理示意图．催化剂b表面发生的电极反应式为CO2+2H++2e﹣=HCOOH．③25℃时，部分物质的电离平衡常数如图所示：化学式HCOOH H2CO3HCIO电离平衡常数 1.77×10﹣4K i1=4.3×10﹣7K i2=5.6×10﹣11 3.0×10﹣10请回答下列问题：同浓度的HCOO﹣、HCO3﹣、CO32﹣、ClO﹣结合H+的能力有强到弱的顺序是CO32﹣＞ClO﹣＞HCO3﹣＞HCOO﹣，物质的量浓度均为0.1mol•L﹣1的下列四种物质的溶液：a、Na2CO3；b、NaClO；c、HCOONa；d、NaHCO3，pH由小到大的顺序是c＜d＜b＜a （填编号）考点：热化学方程式；原电池和电解池的工作原理；化学平衡的影响因素；化学平衡状态的判断．专题：基本概念与基本理论．分析：（1）由图1，可得热化学方程式：①．2NO2（g）+4CO（g）=N2（g）+4CO2（g）△H=2×[﹣（368﹣134）KJ/mol]=﹣468KJ/mol；已知：②．N2（g）+O2（g）=2NO（g）△H=+179.5kJ/mol③．2NO（g）+O2（g）=2NO2（g）△H=﹣112.3kJ/mol根据盖斯定律，（③+①﹣②）可得：2NO（g）+2CO（g）⇌N2（g）+2CO2（g），焓变也进行相应的计算；（2）发生反应：2NO（g）+2CO（g）⇌N2（g）+2CO2①a．反应前后气体物质的量减小，随反应进行容器内气体压强减小，容器内压强不变，说明到达平衡；b．增大反应物浓度，平衡正向移动，化学平衡常数只受温度影响；c．升高温度后，K值减小，说明平衡逆向进行；d．向该容器内充入He气，总压增大，气体分压不变，所以反应速率不变；②根据K=计算平衡常数，再计算浓度商Qc，与平衡常数比较判断反应进行方向；（3）①反应中化合价降低的元素只有N元素，其化合价由+4降低为0计算，计算生成氮气的物质的量，进而计算转移电子数目；②由图可知，左室投入水，生成氧气与氢离子，催化剂a表面发生氧化反应，为负极，右室通入二氧化碳，酸性条件下生成HCOOH；③电离平衡常数越大，酸性越强，电离平衡常数越小，其对应酸根离子结合H+能力越强，水解程度越大，碱性越强．解答：解：（1）由图1可得热化学方程式：2NO2（g）+4CO（g）=N2（g）+4CO2（g）△H=2×[﹣（368﹣134）KJ/mol]=﹣468KJ/mol；已知：②．N2（g）+O2（g）=2NO（g）△H=+179.5kJ/mol③．2NO（g）+O2（g）=2NO2（g）△H=﹣112.3kJ/mol根据盖斯定律，（③+①﹣②）可得：2NO（g）+2CO（g）⇌N2（g）+2CO2（g）△H=﹣112.3kJ/mol﹣468kJ/mol﹣112.3=﹣759.8KJ/mol，故答案为：2NO（g）+2CO（g）⇌N2（g）+2CO2（g）△H=﹣759.8KJ/mol；（2）发生反应：2NO（g）+2CO（g）⇌N2（g）+2CO2，①a．正反应为气体物质的量减小的反应，随反应进行，容器内压强减小，当容器内的压强不发生变化，说明该反应达到平衡，故a正确；b．当向容器中加再充入0.20mol NO时，平衡向正反应方向移动，平衡常数只受温度影响，K值不变，故b正确；c．升高温度后，K值减小，说明平衡逆向进行，NO的转化率减小，故c正确；。
2020年济南齐鲁学校高三生物第二次联考试卷及答案
2020年济南齐鲁学校高三生物第二次联考试卷及答案一、选择题：本题共15小题，每小题2分，共30分。
每小题只有一个选项符合题目要求。
1.人体中的P53蛋白是由一种抑癌基因（P53基因）控制合成的蛋白质，可以抑制DNA的复制和细胞分裂，有抗肿瘤的作用，P53蛋白由一条含393个氨基酸组成的肽链加工而成。
下列叙述不正确的是（）A.各种氨基酸之间的不同在于R基的不同B.393个氨基酸脱水缩合成P53蛋白时，相对分子质量减少了7056C.组成P53蛋白的氨基酸最多21种D.构成P53蛋白的肽链呈直线且在同一平面上2.下图为一组有关生长素的对照实验，有关实验的叙述不正确的（）A.甲和乙对照证明光照不影响生长素的分布B.该实验不能证明胚芽鞘尖端产生生长素C.该对照实验证明生长素能促进植物生长D.该实验不能证明感受光刺激的部位在胚芽鞘尖端3.下图为植物组织培养的基本过程，其中构成愈伤组织的细胞为同种类型。
下列叙述错误的是（）A.①②过程中均发生了细胞分化B.①②过程中均发生了细胞分裂C.温度会影响①②过程进行的速度D.图示过程体现了植物的体细胞具有全能性4.PK基因广泛存在于高等动、植物体内，它控制合成的丙酮酸激酶能促进丙酮酸和ATP产生。
下列推断正确的是（）A.PK基因突变一定会导致丙酮酸激酶的分子结构发生变化B. 丙酮酸激酶的活性高低只会影响到细胞的无氧呼吸C.RNA聚合酶读取到突变PK基因上的终止密码时停止转录D. 细胞内PK基因的异常表达会影响某些离子的跨膜运输5.下列有关洋葱根尖分生区细胞有丝分裂的说法，正确的是（）A.分裂前期由细胞两极的中心粒发出星射线形成纺锤体B.显微镜下同一个视野里，处于中期的细胞数目最多C.细胞分裂后期着丝点分裂，DNA和染色体数目加倍D.细胞分裂末期高尔基体的活动显著加强，细胞板扩展形成细胞壁6.溶酶体是细胞的“消化车间”，内部含有多种水解酶。
下列叙述错误的是A. 溶酶体执行功能时会伴随其膜组分的更新B. 溶酶体通过胞吞作用吞噬侵入细胞的病毒或病菌C. 衰老或损伤的细胞器可被溶酶体消化分解D. 溶酶体属于生物膜系统，能合成水解酶执行其功能7.下列关于人体内物质跨膜运输的叙述，正确的是（）A.胰岛素通过被动运输运出胰岛细胞B.钾离子通过协助扩散进人神经元C.葡萄糖进入红细胞不需要ATP提供能量D.吞噬细胞吞噬细菌与膜上载体蛋白有关8.对下图所表示的生物学意义的描述，正确的是A. 若图甲表示雄果蝇精原细胞染色体组成图，体细胞中最多含有四个染色体组B. 对图乙代表的生物测交，其后代中，基因型为AADD的个体的概率为1/4C. 图丙细胞处于有丝分裂后期，染色单体数、DNA数均为8条D. 图丁所示家系中男性患者明显多于女性患者，该病是伴X隐性遗传病9.DNA的碱基或染色体片段都可能存在着互换现象，下列相关叙述错误的是（）A.基因的两条链上相同位置的碱基互换可引起基因突变B.姐妹染色单体相同位置的片段互换可引起基因重组C.非同源染色体之间发生基因的互换可引起染色体结构变异D.减数第一次分裂时染色体互换会引起配子种类的改变10.下列关于实验操作步骤的叙述中，正确的是（）A.用于鉴定还原糖的斐林试剂甲液和乙液，可直接用于蛋白质的鉴定B.鉴定还原糖时，要加入斐林试剂甲液摇匀后，再加入乙液，且必须要现配现用C.用于鉴定蛋白质的双缩脲试剂A液与B液要混合均匀后，再加入含样品的试剂中D.脂肪的鉴定实验中需要用显微镜才能看到被苏丹Ⅲ染成橘黄色的脂肪颗粒11.生物大分子通常都有一定的分子结构规律，即由一定的基本组成单位，按一定的排列顺序和连接方式形成多聚体，下列表述正确的是（）A.若该图为一段肽链的结构模式图，则1表示肽键，2表示中心碳原子，3的种类约有21种B.若该图为一段RNA的结构模式图，则1表示核糖，2表示磷酸基团，3的种类有4种C.若该图为一段单链DNA的结构模式图，则1表示磷酸基团，2表示脱氧核糖，3的种类有4种D.若该图表示多糖结构模式图，淀粉、纤维素和糖原的空间结构是相同的12.2020年10月，为防止疫情出现聚集性爆发，青岛市在5天时间内完成了千万人级别的全员核酸检测，其效率和速度让众多外媒为之折服，也传递出中国抗击疫情的力量和决心。
2020-2021学年济南齐鲁学校高三英语第二次联考试题及答案
2020-2021学年济南齐鲁学校高三英语第二次联考试题及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AOne day when I was 5, my mother blamed me for not finishing my rice and I got angry. I wanted to play outside and not to be made to finish eating my old rice. When angrily opening the screen door (纱门) with my foot, I kicked back about a 12-inch part of the lower left hand corner of the new screen door. But I had no regret, for I was happy to be playing in the backyard with my toys.Today, I know if my child had done what I did, I would have blamed my child, and told him about how expensive this new screen door was, and I would have delivered a spanking (打屁股) for it. However, my parents never said a word. They left the corner of the screen door pushed out, creating an opening, in the defense against unwanted insects.For years, every time I saw that corner of the screen, it would constantly make me think about my mistake. For years, I knew that everyone in my family would see that hole and remember who did it. For years, every time I saw a fly buzzing (嗡嗡) in the kitchen, I would wonder if it came in through the hole that I had created with my angry foot. Iwould wonder if my family members were thinking the same thing, silently blaming me every time a flying insect entered our home, making life more terrible for us all. My parents taught me a valuable lesson, one that a spanking or stern (严厉的) words perhapscould not deliver. Their silent punishment for what I had done delivered a hundred stern messages to me. Above all, it has helped me become a more patient person and not burst out so easily.1. When the author damaged the door, his parents _______.A. gave him a spankingB. left the door unrepairedC. told him how expensive it wasD. blamed him for what he had done2. The experience may cause the author _______.A. not to go against his parents’ willB. to have a better control of himselfC. not to make mistakes in the futureD. to hide his anger away from others3. What is the main idea of this text?A. Parents is the best way to solve problems.B. Parents are the best teachers of their children.C. Adults should ignore their children’s bad behavior.D. Silent punishment may have a better effect on educating people.BEver wondered if dogs can learn new words? Yes, say researchers as they have found that talented dogs may have the ability to grasp new words after hearing them only four times.While previous evidence seems to show that most dogs do not learn words, unless eventually very well trained, a few individuals have shown some extraordinary abilities, according to a study published in the journal Scientific Reports.“We wanted to know under which conditions the gifted dogs may learn novel words” said researcher xuekw Claudia Fugazza from theEötvösLorándUniversityinHungary. For the study, the team involved two gifted dogs, Whisky and Vicky Nina. The team exposed the dogs to the new words in two different conditions.In the exclusion-based task, presented with seven known toys and one new toy, the dogs were able to select the new toy when presented with a new name. Researchers say this proves that dogs can choose by exclusion when faced with a new word, they selected the only toy which did not have a known name.However, this was not the way they would learn the name of the toy. In fact, when they were presented with one more equally new name to test their ability to recognize the toy by its name, the dogs got totally confused and failed.The other condition, the social one, where the dogs played with their owners who pronounced the name of the toy while playing with the dog, proved to be the successful way to learn the name of the toy, even after hearing it only 4 times. “The rapid learning that we observed seems to equal children’s ability to learn many new words at a fast rate around the age of 18 months,” Fugazza says. “But we do not know whether the learning mechanisms(机制) behind this learning are the same for humans and dogs. ”To test whether most dogs would learn words this way, 20 other dogs were tested in the same condition, but none of them showed any evidence of learning the toy names, confirming that the abilityto learn words rapidly in the absence of formal training is very rare and is only present in a few gifted dogs.4. What was the purpose of the study published in Scientific Reports?A. To better train dogs’ ability to learn new words.B. To further confirm previous evidence about dogs.C. To prove extraordinary memory abilities of gifted dogs.D. To explore favorable conditions for gifted dogs’ new-word learning.5. How did the dogs react when exposed to two new names in the first condition?A. Slow to understand.B. Quick to learn.C. At a loss.D. In a panic.6. What was found about dogs’ new-word learning in the social condition?A. Learning through playing applied to most dogs.B. The social condition helped dogs learn new words.CDogs’ new-word learning turned out to be less effective.D. Dogs shared similar learning mechanisms with children.7. Which of the following is the best title for the text?A. Gifted Dogs Can Learn New Words Rapidly.B. Dogs Identify Newly-named Toys by Exclusion.C. Dogs Can Acquire Vocabulary through Tons of Training.D. Gifted Dogs Have Similar Learning Abilities to Humans.CMusic is said to be a universal language. But for Chase Burton, a deaf filmmaker fromTexas, music has always been a totally different experience.“When I was a kid, I’d lie on the floor so I could feel the vibrations (震动，颤动) from my brother’s band rocking out below my body, ” the 33-year-old man said. “That was one of the first times I began building a relationship with music.”In 2016, his ability to experience music changed dramatically, thanks to California-based technology company Not Impossible Labs. It designed a vibrating suit that enables deaf people to “feel” music through their skin. Consisting of a body harness (背带), ankle and wrist belts, the suit translates audio into a range of vibrating pulses that are felt at 24 contact points.Burtonhas been trialing the suit for four years.“The sound hits different parts of your body, ” saidBurton. “Maybe it will strikeme down in my ankles first. And then I’ll start to feel the vibrations in my back. And then I’ll feel some pulsations in my wrist.”The creators want to extends the tactile (触觉) musical experience beyond the deaf community. In 2018, they gave out 150 of the suits at a rock concert inLas Vegaswhere half the audience members were deaf and half were able to hear.Since then, Not Impossible Labs has been working to improve the technology and says it’s ready to go to market soon. Eventually, the creators want the suit to become a consumer product, accessible to all. The company’s talent and business development director, Jordan Richardson, said that the technology could be used inlive sports broadcasts, video games and theme parks.As a writer and director who’s been working to make the movie world more accessible,Burtonhopes that the vibrating suit will be available to his film audiences in the future. He believes the suit canenhanceemotions while watching a movie – for hearing as well as deaf people. “I see the tech as a real opportunity to help people understand that music for movies doesn’t always need to be enjoyed through the ears”.8. Why would Chase Burton lie on the floor when he was a kid?A. To feel some pulsations in his wrist.B. To feel the vibrations from his brother’s band.C. To expand the tactile musical experience.D. To begin building a bond with films.9. What do we know about Not Impossible Labs from the passage?A. It was started by Chase Burton in 2016.B. Its products have been used in live sports broadcasts.C. It is a technology company based inCalifornia.D. Its administrative director is Jordan Richardson.10. Which can replace the underlined word “enhance” in the last paragraph?A. createB. expressC. coverD. strengthen11. What is the best title for the passage?A. Vibrating Suit Allows Deaf People to “Feel” MusicB. Tech Company Provides Free Suits for Deaf PeopleC. Deaf People Enjoy Rock Music with Free SuitD. Movies Need to Be Enjoyed Through the EarsDPoaching and habitat loss have threatened Africa's two species of elephants, taking them closer toward the edge of disappearance, according to a new report released by the International Union for Conservation of Nature(IUCN).Before this update, Africa's elephants were grouped together and were evaluated as vulnerable by the IUCN. This is the first time the two species have been sorted separately. In the past, elephants were mostly considered as either Asian elephants or African elephants. Forest and savanna elephants were typically classified as subspecies of African elephants.The African forest elephant is now listed as critically endangered and the African savanna elephant as endangered. The number of African forest elephants fell by more than 86% over a 31 -year assessment period. The population of African savanna elephants dropped by at least 60% over the last 50 years, according to the IUCN, which tracks the assessment risk of the world's animals. Africa currently has an estimated 415,000 elephants, counting the two species together.Both elephant species experienced significant population decreases because of poaching. Although it peaked in 2011, illegal hunting still happens and continues to threaten elephant populations. African elephants also face continued habitat loss as their land isconvertedfor agriculture or other uses.There is some good conservation news, the IUCN points out. Anti-poaching measures, combined with better land use planning to support better human-wildlife relationships, have helped conservation efforts. Some forest elephant population figures have stabilized in well-managed areas in Gabon and the Republic of Congo and savanna population figures have remained stable or have been growing, particularly in the Kavango-Zambezi Transfrontier Conservation Area in southern Africa.But with constant demand for ivory and increasing human pressures on Africa's wild lands, concern for Africa's elephants is high, and the need to creatively conserve and wisely manage these animals and their habitats is more severe than ever.12. What can be inferred from the new report about African elephants?A. They are divided into three kinds.B. They are dying out.C. Their threat is mainly from poaching.D. Their population has grown in Africa.13. What does the author mainly tell us in Paragraph 3?A. The detailed number of African elephants.B. The similarities of African elephants.C. The different types of African elephants.D. The present situation of African elephants.14. What does the underlined word "converted" in Paragraph 4 probably mean?A. Expanded.B. Protected.C. Transformed.D. Forbidden.15. What's the authors attitude to the present situation of African elephants?A. Hopeless.B. Optimistic.C. Uncertain.D. Worried.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020-2021学年山东省高三上学期第二次调研联考物理试题及答案解析
山东省部分名校高三第二次调研联考注意事项：1．本试卷分第I 卷（选择题）和第II 卷（非选择题）两部分，共110分，考试时间为90分钟.2．请将第I 卷正确答案的序号填到第II 卷相应的答题表中. 3．第II 卷用蓝、黑色钢笔或圆珠笔直接答在试卷上. 4．答卷前将密封线内的项目填写清楚.考试学校：莱芜一中、临沂一中 北镇中学 济南一中 等第I 卷（选择题，共42分）一、本题共14小题，每小题4分，共56分.在每小题给出的四个选项中，有的小题只有一个选项正确，有的小题有多个选项正确，全选对的得4分；选对但不全的得2分；有选错或不答的得0分.（原创）1．下列说法正确的是：（ ）A ．牛顿在得出力不是维持物体运动的原因这一结论的过程中运用了理想实验的方法B ．在“探究弹性势能的表达式”的活动中，为计算弹簧弹力所做功，把拉伸弹簧的过程分为很多小段，拉力在每小段可以认为是恒力，用各小段做功的代数和代表弹力在整个过程所做的功，物理学中把这种研究方法叫做“微元法”，那么由加速度的定义t a ∆∆v=，当t ∆非常小的时候，t∆∆v 就可以表示物体在t 时刻的瞬时加速度，上述论断就运用了“微元法” C ．用比值法定义物理量是物理学中一种重要的物理科学方法，公式mFa =就运用了比值定义法 D ．万有引力可以理解为任何有质量的物体都要在其周围空间产生一个引力场，而一个有质量的物体在其它有质量的物体所产生的引力场中都要受到该引力场的引力（即万有引力）作用，这情况可以与电场相类比，那么在地球的引力场中的重力加速度就可以与电场中的电场强度相类比 【答案】D【解析】选项A 是伽利略；选项B 后面部分为极限法；选项C 的mFa =不是定义式【考点】综合考查物理科学方法（原创）2.下列说法正确的是A.物体做匀加速曲线运动时所受的力应该是均匀增加的B.空军跳伞时降落伞刚刚张开的瞬间伞绳对人的拉力稍稍大于人对伞绳的拉力C.静摩擦力有时可能阻碍物体的相对运动D.力是改变物体运动状态的原因【答案】D【解析】选项A物体做匀加速曲线运动时所受的力应该是不变的；选项B中二力应该相等；选项C静摩擦力只能阻碍物体的相对运动趋势；【考点】牛顿运动定律、摩擦力（原创）3. （多选）一物体受到两个恒定外力的作用，沿某方向做直线运动。
2021届济南齐鲁学校高三英语第二次联考试卷及答案
2021届济南齐鲁学校高三英语第二次联考试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AIf you had the opportunity to live forever, would you take it? Keeping your body alive indefinitely still seems like an impossibility, but some scientists think that digital technology may have the answer: creating a digital copy of your “self” and keeping it “alive” online long after your physical body has ceased to function.In effect, the proposal is to clone a person electronically. Unlike the familiar physical clones — children that have identical features as their parents, but that are completely separate organisms with a separate life — your electronic clone would believe itself to be you. How might this be possible? The first step would be to mapthe brain.How? One plan relies on the development of nanotechnology (纳米技术). Ray Kurzweil — one of the kings of artificial intelligence — predicts that within two or three decades we will have nano transmitters that can be put into the bloodstream. Inthe capillaries (毛细血管) of the brain, they would line up alongside the neurons and detect the details of the cerebral (大脑的) electronic activity. They would be able to send that information to a receiver inside a special helmet, so there would be no need for any wires sticking out of the head.As a further step, Ray Kurzweil also imagines the nano transmitters being able to connect you to a world of virtual reality on the Internet, similar to what was shown in the film “Matrix”. With the nano transmitters in place, by thought alone, you could log on to the Internet and instead of the pictures coming up on your screen, they would play inside your mind. Rather than send your friends e-mails you would agree to meet up on some virtual tropical beach.Some peoplebelieve that they can enjoy life after death. But why wait for that when you could have a shot of nanobots (纳米机器人) and upload your brain onto the Internet and live forever as a virtual surfer?One snag: to exist on the net you will have to have your neural network parked on the computer of a web-hosting company. These companies want real money in real bank accounts every year or they will wipe your bit of the hard disc and sell the space to someone else. With your body six feet underground how will you pay?1. Which of the following statements is TRUE according to the passage?A. Nano transmitters can help map the human brain.B. Electronic clones recreate the original human body.C. Electronic clones may put their physical selves into movies.D. Nano transmitters use a helmet to detect the cerebral activities.2. What is the author’s attitude towards electronic clones?A. Optimistic and careful.B. Interested and unconvinced.C. Excited and confused.D. Assured and critical.3. The author asks “how will you pay?” at the end of the article, because ________.A. you can’t pay to exist on the Internet if you are physically deadB. you can’t pay for hard disc space if you don’t have a bank accountC. you can’t pay for a special service if too many people want to use itD.you can’t pay the web-hosting company if you don’t have a neural networkBI don’t think I can recall a time whenI wasn’t aware of the beauty of the ocean. Growing up inAustralia, I had the good fortune of having the sea at my side. The first time I went toHalfmoonBay,I suddenly had the feeling of not being able to feel the ground with my feet anymore.For my 10th birthday, my sister and I were taken out to theGreat Barrier Reef. There were fish in different color1 s, caves and layers of coral. They made such an impression on me. When I learned that only one percent ofAustralia’sCoral Seawas protected, I was shocked. Australian marine (海洋的) life is particularly important because the reefs have more marine species than any other country on earth. But sadly, only 45% of the world’s reefs are considered healthy.This statistic is depressing, so it’s important for usto do everything to protect them. The hope that theCoral Searemains a complete ecosystem has led me to take action. I’ve become involved with the Protect Our Coral Sea activity, which aims to create the largest marine park in the world. It would serve as a place where the ocean’s species will all have a safe place forever.Together, Angus and I created a little video and we hope it will inspire people to be part of the movement. Angus also shares many beautiful childhood memories of the ocean as a young boy, who grew up sailing, admiring the beauty of the ocean, and trying to find the secrets of ocean species.4. What can we learn about the author from the underlined sentence inPara. 1?A. He seldom went surfing at the sea.B. He forgot his experiences about the ocean.C. He never went back to his hometown.D. He had a wonderful impression ofHalfmoonBay.5. What is Australian marine life like according to the second paragraph?A. It is escaping from theCoral Seagradually.B. It depends on reefs for living greatly.C. It may be faced with danger.D. It is protected better than that in other oceans.6. What’s the purpose of The Protect Our Coral Sea activity?A. It is intended to contribute to a complete ecosystem.B. It is intended to prevent more marine species being endangered.C. It is intended to set up a large nature reserve for reefs.D. It is intended to raise more teenagers’ environmental awareness.7. Why do Angus and the author create a little video?A. To urge more people to take action toprotect the marine species.B. To inspire more people to explore the secret of the ocean.C. To share their childhood experiences about the ocean.D. To bring back to people their memory of ocean species.CI come to theUnited Statesten year ago. I would always say that I was trying to study, but there were always things like work and my kids that would not allow me to start.Now I realized that those were only excuses. What stopped me was that I was afraid to start studying again. I always believed I would learn by myself.One day, however, my son told me that he was sad because his friends would come over and I didn’t understand them because I didn’t speak English. He was also sad because I could never help him with his homework. That same day, I told myself, “Rocio, you have to start believing in yourself and you will see you can make it.”The next day, I went downtown to look for a big banner (横幅) in front of the school which said that they offered classes for adults. I came in to see if I could join, but the classes were closed already. That night I took the kids to the movies, and on the way back, I told them we would take a new route. I ended up getting lost. That’s the way I foundChaffeyCollege. The following Monday, I went to ask for information. They told me that summer school was starting that week.That’s how I started studying English last summer. It is difficult, but I have had great rewards. My daughter had to write a story for school. It was about the female they most admired and why . She wrote that I was the person she most admired because I had started going to College. I will never forget this.8. According to the passage, the author probably is a .A. teacherB. doctorC. fatherD. mother9. What made the author make up her mind to study English?A. What her son said.B. What her daughter said.C. Thinking about herself.D. Thinking about her daughter.10. How did they findChaffeyCollege?A. On the way to the movies.B. They took a new route and got lost.C. Ask a stranger for information.D. According to the banner.11. Which of the following is NOT true?A. The author came to theUnited Statesfrom another country.B. The author had two children at least including one daughter.C. What really changed the author’s life was she believed in herself.D. The author wrote that she was the person her son most admired.DSomeday soon an emoji (表情符号)might really save lives.Hiroyuki Komatsu is a Google engineer who suggested adding a series of new emojis to the standard emoji library. It could help those with food allergies (过敏)understand what they are eating anywhere inthe world. Emojis should cover characters representing major food causing allergies. They make people understand what are used in foods even in foreign countries and safely select meals.Emojis are universal because they are chosen and developed by the Unicode Consortium, a non-profit company that oversees, develops and maintains how text is represented. This is in regards to all software products and standards. It's thanks to the Unicode Standard that when you text a friend six pizza emojis, they’ll see those six pizza pieces on their phone. This is true regardless of whether they use an iPhone or an Android.Because emojis are everywhere and visual(视觉的),they could be helpful for restaurants and food packaging designers. They can communicate whether a product is made with common causing-allergy food. But as Komatsu’s advice argues, many of the most common causing-allergy foods are missing or poorly represented by the present emoji library. For example, there is an emoji for octopus, but nothing for squid. There is a loaf of bread that could symbolize grain, but a picture of wheat could be clearer. The emojis can be more direct when symbolizing foods.It’s not uncommon for the Unicode Consortium to add new emojis to the library: several food-related emojis were put into use last June, including some long-waited food emojis. Apple included support for multiracial emojis in a recent iOS update. An artist even recreated Moby-Dick in emoji characters. Some might be sorry for the continuing death of the written word if Komatsu’s suggestion is accepted, but look on the bright side: if you ever see that happy poop on a box, you’ll know to stay away.12. How will emojis save lives according to the text?A. By showing what the food contains visually.B. By telling the safest places in the world.C. By teaching people how to treat allergies.D. By adding standard emojis about safety.13. What does paragraph 3 mainly tell us?A. Emojis have the same meanings around the world.B. The Unicode Consortium is a non-profit company.C. What emojis represent is different in different places.D. Different mobile operating systems have different emojis.14. What can be the reason for Komatsu’s advice?A. Emojis are easy to mix up.B. Present emojis are not enough.C. Emojis can't interest most users.D. Emojis can't represent foods directly.15. What is the author’s attitude to Komatsu's suggestion?A. Doubtful.B. Worried.C. Supportive.D. Uninterested.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020-2021学年济南齐鲁学校高三英语二模试卷及答案
2020-2021学年济南齐鲁学校高三英语二模试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AFine art fairs（艺术博览会）are the trend of the 21st century, with new art and antique（古玩）fairs and festivals springing up in diverse parts of the world. Here is a list of four noteworthy art fairs.Art Basel, Basel, SwitzerlandThe granddaddy of art fairs, Art Basel, was established in 1970 by a group of local art gallerists and is the biggest contemporary art fair in the world. Art Basel takes place over a 5-day period each June in Basel, Switzerland. The high cost of renting space for gallery owners is offset（抵消）by the huge attendance at the fair. For example in 2010, about 60,000 visitors attended Art Basel.Frieze Art Fair, London“Frieze Art Fair was established in 2003 and is one of the few fairs to focus only on contemporary art and living artists.v"Thefair takes place every October in Regent's Park, London. It features over 170 of the most exciting contemporary art galleries in the world. ”In addition to the fair which began in 2003, the fair owners Matthew Slotover and Amanda Sharp publish Frieze, an international art magazine established in 1991 and devoted to contemporary art.TEFAF Maastricht, the NetherlandsEstablished in 1975 as The Pictura Fine Art Fair, and renamed The European Fine Art Foundation（TEFAF）, Maastricht in 1996, the fair includes 260 of the world's most famous art and antique dealers from 16 countries.The 24th edition of the TEFAF fair held March 18 — 27, 2011 featured 260 dealers exhibiting approximately 30, 000 artworks and antiques with an overall value of $ US 1. 4 billion.ARCO, MadridARCO Madrid was established in 1982 and is one of Europe's leading and popular art fairs. In addition to the exhibiting galleries（in 2011, 197 international art galleries participated）, a seriesof lectures and specially focused exhibitions take place.1.How does Art Basel cover the expense of renting space?A.By selling tickets.B.By selling expensive exhibits.C.By donation from dealers.D.By support from the government.2.The owners of Frieze Art Fair are also in charge of____.A.170 living artists.B.An international art magazine.C.30,000 artworks and antiques.D.A series of lectures.3.Which of the following has the longest history?A.Art BaselB.Frieze Art FairC.TEFAFD.ARCOBThe idea came to him when he least expected it. Alvin Irby was at a barbershop when he saw one of his former students sitting in the shop with a bored look on his face. That’s when Irby realized that by pairing barbershops and books, he might be able to inspire young boys to read.Alvin Irby, a former kindergarten and first-grade teacher, knows how important it is for young children to read. He also knows that young boys in particular often don’t have adult male role models who inspire them to read. “Many young boys may literally never see a man reading in school during the years when they’re learning to read because there are so few male elementary school teachers,” Irby toldMashable.That’s where the barbershops come in. Four years ago, Irby launched Barbershop Books as a way to not just get books into the hands of young boys, but also to create community reading spaces in a place where kids go frequently. Since itsinceptionin 2013, the program has created kid-friendly reading spaces in 50 barbershops in 12 states throughout the United States.Irby isn’t the first person to see the connection between barbershops and books and boys. Hair stylist Courtney Holmes, launched a program a few years ago offering free haircuts to kids as long as they read to him while he cuts their hair.That’s the kind of environment that Irby wants to promote with his program. The reading spaces created by Barbershop Books help to spark an interest in books by showing kids that reading is about more than just spelling and vocabulary skills, it’s about making reading a low-stress activity that can help them relax, laugh and have fun.“Our belief is that if we can create positive reading experiences early and often for young boys, then they will choose to read for fun,” Irby noted, adding, “This is really what Barbershop Books is about, getting young boys to say three words: I’m a reader.”4. What happened to Alvin when he was at a barbershop?A. He found it easy for young people to get bored.B. He offered a barbershop to his former student.C. He thought of a way to encourage young readers.D. He realized the importance of reading for young boys.5. What is the function of Barbershop Books?A. To attract more customers who love films.B. To provide free haircuts to book lovers.C. To show the influence of reading on children.D. To create a reading environment fbr children.6. It can be inferred from the passage that ________.A. reading is a low-stress activity that is relaxingB. Barbershop Books is only suitable for young boysC. Irby attaches great importance to school educationD. Barbershop Books can arouse (引起) young people’s interest in reading7. What does the underlined word “inception” in the 3rdparagraph mean?A. discovery.B. success.C. popularity.D. beginning.CIvrea is a town in the Piedmont province of northern Italy. It is known for its localcarnival（狂欢节）organized in February.The main part of the carnival is the famous Battle of the Oranges (La battaglia delle arance). The Battle includes nine teams who throw oranges at each other during three carnival days – Sunday, Monday and Tuesday.Oranges were not always used in the battle. In the middle ages people used beans. Twice a year the local feudal lord gave a pot of beans to the poor families who, as sign of rebellion, threw them out of their homes. Later beans became part of carnival as sort of “ammunition” for throwing at people passing by.It is still not known why exactly people started using oranges. It is believed that the origin for this tradition is in the mid 19th century. The legend says that local girls, standing on balconies, started to throw some oranges, together with confetti,lupins(白羽扇豆)and flowers, onto the parade carriages. The girls actually wanted to draw attention of boysin the carriages. Boys “answered” by throwing some objects back at girls. Little battle started that way.Battle of the Oranges got strict rules after the World War II. The battles are organized on town's squares. The battles are fought between teams in carriages (symbolizing local the guards of localtyrant(恶霸)) and the teams walking beside those carriages (symbolizing rebellious people of Ivrea).Oranges for the event are brought from the island of Sicily. The oranges used are of low quality, not suitable for humans. About 270,000 kilograms of oranges are used each year.The carnival ends with a silent march on the night of “Fat Tuesday”. The Carnival "general" says goodbye to everyone with the phrase "See you next Fat Thursday at 1 p.m."Special prizes are awarded to three best foot teams, three carriages drawn by two horses and three carriages drawn by four horses. Different elements are judged like for example throwing ability, fair play or decoration of carriages.8. The word "ammunition" (paragraph 3) is closest in meaning to________.A. decorationB. bulletsC. advertisementD. presents9. In Battle of the Oranges, the teams walking beside carriages act as ________.A. the Carnival generalB. the local feudal lordC. the rebellious people of IvreaD. the guards of local tyrant(暴君)10. Which of the following statements about Battle of the Oranges is NOT true?A. In the middle ages, people threw beans out of homes as a sign of rebellion.B. Girls throw oranges towards carriages in the parade to attract the attention of boys inside.C. Tons of oranges are used every year to hold the festival.D. People started to throw oranges because they are of low quality and not suitable to eat.11. What is this passage mainly about?A. A traditional activity in an Italian carnivalB. The origin of Battle of the OrangeC. How people enjoy themselves in the Orange CarnivalD. The rules of activities in carnivals in ItalyDThe AI research arm of Alibaba created a machine learning model that received a higher score on the Stanford Question Answering Dataset than humans. The database consists of more than 100,000 questions to test reading comprehension.In early January this year the Alibaba AI software machine scored 82.44 on the test while humans scored 82.304. Besides, computers and AI have already defeated humans, for example in games such as chess. However, it seemed that language skills were superior in humans as machines find languages hard to master.A large number of call center employees, often in developing countries, may be out of work soon if the AIrobots are cheaper and as effective as human labor. Soon when you phone a company for information the conversation will go like this: “ We are sorry but all our robots are busy right now. We value your call. Please stay on the line until a robot is free to serve you. There are just 12 callers ahead of you.” A robot will serve you some popular tunes while you wait.Si Luo, who is a chief scientist of natural language processing at Alibaba’s AI research group noted that questions such as “What causes rain?” can now be answered with a high degree of accuracy by robots. Si Luo said, “ We believe the foundational technology can be gradually applied to a lot of applications such as customer service, museum tutorials, and online responses to inquiries from patients, freeing human efforts in a new way.”Si Luo’s team is working closely with Ali Xiaomi, a mobile customer service chatbot. Ali Xiaomi can be customized to be used on Alibaba’s platforms such as Taobao and Tmall. The new AI robots could answer consumers’ questions as they did the Stanford questions. The robots would look for the answers from prepared information. However, there are limits to what the system will be able to do. If questions do not have clear-cut answers, or the questions asked are too unclear or ungrammatical, the robot may not be able to deal with them.12. What can we learn about the Alibaba AI software machine?A. It has been tested in some areas.B. It has become popular since January.C. It has offered a special learning style.D. It has made people interested in reading.13. What does the example in paragraph 3 show about the AI robots?A. They should have better language skills.B. They may replace humans in some fields.C. They need to be customized to serve customers.D. They will be widely used in developing countries.14. How does Si Luo feel about the foundational technology?A. Doubtful.B. Worried.C. Curious.D. Confident.15. What can we infer from the lastparagraph about Ali Xiaomi?A. It needs to improve in some ways.B. It is connected with another system.C. It is a platform to show good service.D. It can answer any questions accordingly.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
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山东省19所名校2021年高三第二次调研考试化学试题本试卷分第I卷和第Ⅱ卷两部分，满分100分。
考试用时90分钟。
可能用到的相对原子质量：H—1 C—12 N—14 O—16 Na—23 Al—27S—32 Cu—64 Ba—137第I卷（选择题共48分）注意事项：1．第I卷共16小题。
2．每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑。
如需改动，用橡皮擦干净后，再选涂其他答案标号。
不涂在答题卡上，只答在试卷上不得分。
一、选择题（本题包括16小题，每小题只有一个选项符合题意）1．化学在生产和日常生活中有着重要的应用。
下列说法错误的是（）A．半导体行业中有一句话：“从沙滩到用户”，计算机芯片的材料是二氧化硅B．葡萄糖注射液不能产生丁达尔效应现象，不属于胶体C．淀粉、油脂、蛋白质都能水解，但水解产物不同D．纯碱可用于生产普通玻璃，日常生活中也可用纯碱溶液来除去物品表面的油污【答案】A【命题立意】本题考查化学与生活、生产等知识点。
【解析】计算机芯片的材料是高纯硅，A错误；葡萄糖注射液不属于胶体，不能产生丁达尔效应现象，B正确；淀粉、油脂、蛋白质都能水解，但水解产物不同，C正确；纯碱可用于生产普通玻璃，日常生活中也可用纯碱溶液的碱性来除去物品表面的油污，D正确。
2．下列说法正确的是（）A．玻璃钢是一种复合材料，由基体和增强体组成，基体起骨架作用B．SO2溶于水，其水溶液能导电，说明SO2是电解质C．食醋、纯碱、食盐分别属于酸、碱、盐D ．生铁、不锈钢、青铜都属于合金【答案】D【命题立意】本题考查电解质、合金等概念辨析。
【解析】玻璃钢是一种复合材料，由基体和增强体组成，增强体起骨架作用，A 错误；SO 2溶于水，其水溶液能导电，说明亚硫酸是电解质，B 错误；纯碱属于盐，C 错误；生铁、不锈钢、青铜都属于合金，D 正确。
3．设N A 为阿伏加德罗常数的值，下列叙述正确的是（ ）A ．用石墨作电极电解饱和食盐水时，若阴极得到2N A 个电子，则阳极产生22.4 L 气体B ．2 L 0.05 mol 1L -⋅乙酸溶液中H +数为0.1N AC ．等物质的量的Na 2O 和Na 2O 2中含有的阴离子总数不同D ．常温常压下，Na 2O 2与足量H 2O 反应，共生成0.2 mol O 2，转移电子的数目为0.4N A【答案】D【命题立意】本题考查阿伏伽德罗常数及其应用【解析】阳极产生22.4 L 气体，没有说明是否是标准状况，A 错误；2 L 0.05 mol 1L -⋅乙酸溶液中H +数小于0.1N A ，原因是醋酸是弱电解质，部分电离，B 错误；等物质的量的Na 2O 和Na 2O 2中含有的阴离子总数相同，都含有一个阴离子，C 错误；Na 2O 2与足量H 2O 反应，过氧化钠既是氧化剂又是还原剂，生成0.2 mol O 2，转移电子的数目为0.4N A ，D 正确。
4．在下列溶液中，各组离子一定能够大量共存的是（ ）A ．由水电离的()4110c H +-=⨯mol 1L -⋅的溶液中：Ca 2+、K +、Cl —、HCO 3- B ．在pH=1的无色溶液：Fe 2+、Mg 2+、NO 3-、Cl —C ．含有大量Al 3+的溶液：Na +、C1一、NO 3-、OH —D ．含大量OH —的溶液中：CO 23-、Cl —、F —、K +【答案】D【命题立意】本题考查离子共存问题【解析】由水电离的()4110c H +-=⨯mol 1L -⋅的溶液中溶液呈酸性或者碱性，碳酸氢根不能大量共存，A 错误；在pH=1的无色溶液中Fe 2+、硝酸根发生含有大量氧化还原反应，不能共存，B 错误；Al 3+与氢氧根不能共存，C 错误；含大量OH —的溶液中：CO 23-、Cl —、F —、K +能大量存在，D 正确。
5．下列推断合理的是（ ）A ．一定条件下，使用催化剂能加快反应速率并能提高反应物的平衡转化率B ．蛋白质和油脂都属于高分子化合物，一定条件下都能水解C ．甲烷和氯气反应生成一氯甲烷，与苯和硝酸反应生成硝基苯的反应类型相同D ．将SO 2通入品红溶液，溶液褪色后加热恢复原色；将SO 2通入溴水，溴水褪色后加热也能恢复原色【答案】C【命题立意】本题考查反应速率、蛋白质、油脂、有机反应类型等知识点。
【解析】催化剂不能改变转化率，A 错误；油脂不属于高分子化合物，B 错误；甲烷和氯气发生取代反应，苯和硝酸也发生取代反应，C 正确；二氧化硫与溴水发生氧化还原反应，生成硫酸和氢溴酸，加热不能复原，D 错误。
6．某溶液中含有NH 4+、Mg 2+、Fe 2+、A13+和SO 24-五种离子，若向其中加入过量的Ba （OH ）2溶液，微热并搅拌，再加入过量的氢碘酸，溶液中大量减少的离子有（ ）A ．4种B ．3种C ．2种D ．1种【答案】C【命题立意】本题考查离子共存的有关知识点。
【解析】加入过量的Ba （OH ）2溶液，微热并搅拌，减少的是硫酸根离子和铵根离子，再加入过量的氢碘酸，离子不减少，故选C 。
7．若司机酒后驾车，可通过对其呼出的气体进行检验而查出，所利用的化学反应如下：2CrO 3（红色）+3C 2H 5OH+3H 2SO 4==Cr 2（SO 4）3（绿色）+3CH 3CHO+6H 2O 。
关于该反应，下列叙述不正确的是（ ）A．每1 mol CrO3发生氧化反应，转移3 mol e—B．C2H5OH是还原剂C．CrO3在反应中表现氧化性D．C2H5OH在反应中失去电子【答案】A【命题立意】本题考查氧化还原反应的有关内容。
【解析】CrO3在反应中表现氧化性，得到电子，发生还原反应，1mol得到3mol电子，C2H5OH是还原剂发生氧化反应，A错误。
8．下列叙述正确的是（）A．Na、Al、Fe金属单质在一定条件下与水反应都生成H2和对应的碱B．漂白粉和明矾都常用于自来水的处理，二者的作用原理不相同C．将SO2通入Ca（ClO）2溶液可生成CaSO3沉淀D．向FeCl2溶液中通入Cl2反应的离子方程式为：Fe2++Cl2==Fe3++2C1—【答案】B【命题立意】本题考查元素化合物的有关内容。
【解析】铁和水蒸气反应生成四氧化三铁和氢气，A错误；漂白粉和明矾都常用于自来水的处理，二者的作用原理不相同，漂白粉是利用了强氧化性，明矾是利用电离出来的铝离子水解生成氢氧化铝胶体的吸附性来净化水，B正确；将SO2通入Ca（ClO）2溶液可生成CaSO4沉淀，C错误；向FeCl2溶液中通入Cl2反应的离子方程式为：2Fe2++Cl2==2Fe3++2C1—，D错误。
9．下列与实验相关的叙述中，正确的是（）A．加热NH4C1晶体时，将湿润的红色石蕊试纸靠近试管口，检验NH3的生成B．向某溶液中加入足量盐酸酸化，无现象，再加入BaCl2溶液，有白色沉淀生成，说明原溶液中存在2SO4C．取少量溶液X，向其中加入适量新制氯水，再加几滴KSCN溶液，溶液变红，说明X溶液中一定含有Fe2+D．用饱和碳酸钠溶液除去二氧化碳中的氯化氢【答案】B【命题立意】本题考查元素化合物的知识，物质的检验、实验操作等知识点。
【解析】加热NH4C1晶体时，氯化铵分解产生的氨气、氯化氢遇冷重新生成氯化铵，得不到氨气，A错误；向某溶液中加入足量盐酸酸化，无现象，再加入BaCl2溶液，有白色沉淀生成，说明原溶液中存在2SO ，B正确；取少量溶液X，向其中加入适量新制氯水，再加几4滴KSCN溶液，溶液变红，说明X溶液中可能含有Fe2+，也可能含有Fe3+，C错误；用饱和碳酸氢钠溶液除去二氧化碳中的氯化氢，D错误。
10．下列有关元素的性质及其递变规律正确的是（）A．第ⅦA族元素从上到下，其氢化物的稳定性逐渐增强B．第二周期元素从左到右，最高正价从+1递增到+7C．同周期元素（除0族元素外）从左到右，原子半径逐渐减小D．同周期金属元素的化合价越高，其原子失电子能力越强【答案】C【命题立意】本题考查元素周期表和元素周期律的有关内容【解析】第ⅦA族元素从上到下，其氢化物的稳定性逐渐减弱，A错误；第二周期元素F无正价，B错误；同周期元素（除0族元素外）从左到右，原子半径逐渐减小，C正确；同周期金属元素的化合价越高，其原子失电子能力不一定越强，D错误。
11．下列有关实验原理或操作正确的是（）A．选择合适的试剂，用图1所示装置可分别制取少量CO2、NO和O2B．除去干燥CO2中混有的少量SO2，可将混合气体依次通过盛有足量酸性KmnO4溶液、浓硫酸的洗气瓶C．洗涤沉淀时（如图2），向漏斗中加适量水，搅拌并滤干D．将碘水倒入分液漏斗，加适量乙醇，振荡后静置，可将碘萃取到乙醇中【答案】B【命题立意】本题考查化学实验的基本操作【解析】收集NO不能用排空气法，A错误；酸性KmnO4溶液能氧化二氧化硫，B正确；洗涤沉淀时不能用玻璃棒搅拌，C错误；乙醇与水互溶，不能用作萃取剂，D错误。
12．某兴趣小组设计如图微型实验装置。
实验时，先断开K2，闭合K1，两极均有气泡产生；一段时间后，断开K1，闭合K2，发现电流表A指针偏转。
下列有关描述正确的是（）A．断开K2，闭合K1时，铜电极为阳极B．断开K2，闭合K1时，石墨电极附近溶液变红C．断开K1，闭合K2时，铜电极发生还原反应D．断开K1，闭合K2时，石墨电极作正极【答案】D【命题立意】本题考查原电池及电解池的原理及应用。
【解析】断开K2，闭合K1时，装置为电解池；两极均有气泡生成，说明铜为阴极，铜电极附近氢离子放电，溶液变红；故A、B错误；断开K1，闭合K2时，装置为燃料电池，铜作负极发生氧化反应，石墨作正极，故电流计发生偏转，C错误，D正确。
13．对于反应CO（g）+H2O（g）CO2（g）+H2（g）△H<0，在其他条件不变的情况下，下列说法正确的是（）A．加入催化剂，改变了反应的途径，反应的△H也随之改变B．改变压强，平衡不发生移动，反应放出的热量不变C．升高温度，反应速率加快，正向移动D．平衡时，其他条件不变，升高温度可使该反应的平衡常数增大【答案】B【命题立意】本题考查反应热与化学平衡、平衡常数等知识点。
【解析】加入催化剂不能改变反应的△H ，A 错误；反应前后气体的物质的量之和不变，改变压强，平衡不发生移动，反应放出的热量不变，B 正确；．升高温度，反应速率加快，平衡向吸热反应方向移动，逆向移动，C 错误；平衡时，其他条件不变，升高温度可使该反应的平衡常数减小，D 错误。
14．下列说法正确的是（ ）A ．铅蓄电池在放电过程中，负极质量减小，正极质量增加B ．原子核内有10个中子的氧原子188O C ．铜锌原电池工作时，电子沿外电路从铜电极流向锌电极D ．新制饱和氯水在光照下有气泡产生，其主要成分是氯气【答案】B【命题立意】本题考查铅蓄电池、原电池的原理，原子核、氯水等性质。