浙江省名校新高考研究联盟(Z20联盟)2019届高三第一次联考英语试题(含答案)

0608 Z20名校联盟2021届高三第一次联考英语试题选择题部分第一部分听力（共两节，满分30分）第一节（共5小题：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 time is it now?A.9:35B:9:20.C:9:052.What will the woman have?A.Tea.B.Coffee.k.3.What does the man want to buy?A.A car.B.An apartment.C.A necklace.4.What does the man mean?A.I he woman can try cooking this term.B.Cooking is time-consuming.C.Next term's schedule is lighter than this term’s5.What does the man think of the new drug?A.Ineffectiveeful.C.Successful第二节（共15小题；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，各小题给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料．回答第6、7题。

6.Why won’t the man see The Dark Night?A.He can’t get the ticket.B.He is not interested in it.C.He thinks it too horrible.7.Where will the speakers go to see a movie?A.To the Central TheaterB.To the Red Star CinemaC.To the Sunshine Movie House听第7段材料，回答第8至9题。

浙江省名校新高考研究联盟（Z20）2019届高三第一次联考数学试题一、选择题（本大题共10小题，共40.0分）1.已知集合，，则A. B.C. D. 或【答案】C【解析】解：集合，，．故选：C．先求出集合A，B，由此能求出．本题考查交集的求法，考查交集定义、不等式性质等基础知识，考查运算求解能力，是基础题．2.设复数z满足为虚数单位，则A. B. i C. D. 1【答案】B【解析】解：由，得．故选：B．把已知等式变形，再利用复数代数形式的乘除运算化简得答案．本题考查了复数代数形式的乘除运算，是基础题．3.设函数，则的值为A. B. C. D. 2【答案】C【解析】解：函数，，．故选：C．推导出，从而，由此能求出结果．本题考查函数值的求法，考查函数性质等基础知识，考查运算求解能力，是基础题．4.已知m，n是空间中两条不同的直线，，是两个不同的平面，则下列命题正确的是A. 若，，，，则B. 若，，则C. 若，，则D. 若，，，则【答案】D【解析】解：由m，n是空间中两条不同的直线，，是两个不同的平面，得：在A中，若，，，，则与相交或平行，故A错误；在B中，若，，则n与相交、平行或，故B错误；在C中，若，，则或，故C错误；在D中，若，，，则由面面垂直的判定定理得，故D正确．故选：D．在A中，与相交或平行；在B中，n与相交、平行或；在C中，或；在D中，由面面垂直的判定定理得．本题考查命题真假的判断，考查空间中线线、线面、面面间的位置关系等基础知识，是中档题．5.已知实数x，y满足约束条件，则的最大值为A. 1B. 4C. 2D.【答案】B【解析】解：作出实数x，y满足约束条件对应的平面区域如图：阴影部分由得，平移直线，由图象可知当直线经过点A时，直线的截距最大，此时z最大由解得．代入目标函数得．即目标函数的最大值为4．故选：B．作出不等式对应的平面区域，利用线性规划的知识，通过平移即可求z的最大值．本题主要考查线性规划的应用，利用数形结合是解决线性规划题目的常用方法利用平移确定目标函数取得最优解的条件是解决本题的关键．6.已知双曲线C：，则“”是“双曲线C的焦点在x轴上”的A. 充分不必要条件B. 必要不充分条件C. 充要条件D. 既不充分也不必要条件【答案】A【解析】解：由题意得：双曲线C的焦点在x轴上或，或，或推不出，“”是“双曲线C的焦点在x轴上”的充分不必要条件．故选：A．根据充分条件和必要条件的定义分别进行判断即可．本题主要考查充分条件和必要条件的判断，根据充分条件和必要条件的定义是解决本题的关键．7.函数的图象可能是A. B.C. D.【答案】A【解析】解：函数，可得是奇函数排除C；当时，，图象在x轴的上方，排除D；当时，，排除B；故选：A．根据奇偶性，单调性结合特殊点，即可求解．本题考查了函数图象变换，是基础题．8.已知，是椭圆与的左、右焦点，过左焦点的直线与椭圆交于A，B两点，且满足，，则该椭圆的离心率是A. B. C. D.【答案】B【解析】解：由题意可得：，，可得，，，，，，可得，可得．故选：B．利用已知条件，画出图形，通过三角形的边长关系，求解椭圆的离心率即可．本题考查椭圆的简单性质的应用，考查数形结合以及转化思想的应用．9.已知实数a，b，c，d满足，，则的最小值是A. 10B. 9C.D.【答案】B【解析】解：，，，，当且仅当时，取等号．则，当且仅当时，且，时，的最小值为9，故选：B．利用基本不等式求得，再利用基本不等式求得的最小值．本题主要考查基本不等式的应用，式子的变形，是解题的关键和难点，属于中档题．10.已知三棱锥的所有棱长为是底面内部一个动点包括边界，且M到三个侧面PAB，PBC，PAC的距离，，成单调递增的等差数列，记PM与AB，BC，AC所成的角分别为，，，则下列正确的是A. B. C. D.【答案】D【解析】解：依题意知正四面体的顶点P在底面ABC的射影是正三角形ABC的中心O，由余弦定理可知，，，其中，表示直线MO与AB的夹角，同理可以将，转化，，，其中，表示直线MO与BC的夹角，，，其中，表示直线MO与AC的夹角，由于是公共的，因此题意即比较OM与AB，BC，AC夹角的大小，设M到AB，BC，AC的距离为，，则，其中是正四面体相邻两个面所成角，，所以，，成单调递增的等差数列，然后在中解决问题由于，可知M在如图阴影区域不包括边界从图中可以看出，OM与BC所成角小于OM与AC所成角，所以，故选：D．根据题意分析，将问题转化为：比较OM与AB，BC，AC的大小然后在中可以解决．本题考查了异面直线及其所成角，属难题．二、填空题（本大题共7小题，共36.0分）11.已知随机变量的分布如表所示，则______，______．【答案】【解析】解：随机变量的分布可得，可得，所以．．故答案为：；．利用分布列求解m，求解期望，利用方差公式求解即可．本题主要考查离散型随机变量的分布和数学期望、方差等基础知识，熟记期望、方差的公式是解题的关键．12.某几何体的三视图如图所示，则该几何体的体积为______，表面积为______．【答案】24 60【解析】解：由三视图还原原几何体如图：该几何体为直三棱柱，底面为直角三角形，则其体积为．表面积为．故答案为：24；60．由三视图还原原几何体，可知原几何体为直三棱柱，底面为直角三角形，从而可求几何体的体积和表面积．本题考查由三视图求面积、体积，关键是由三视图还原原几何体，是中档题．13.若的展开式中，的系数为6，则，______，常数项的值为______．【答案】1 15【解析】解：的展开式的通项公式为，令，求得，可得的系数为，．令，求得，可得常数项的值为，故答案为：1；15．在二项展开式的通项公式中，令x的幂指数等于3，求出r的值，即可求得的系数，再根据的系数为6，求得a的值；在二项展开式的通项公式中，令x的幂指数等于0，求出r的值，即可求得常数项的值．本题主要考查二项式定理的应用，二项展开式的通项公式，二项式系数的性质，属于基础题．14.在中，角A，B，C所对的边分别为a，b，c，，且外接圆半径为，则______，若，则的面积为______．【答案】3【解析】解：，且外接圆半径R为，由正弦定理，可得：，，由余弦定理，可得：，解得：，．故答案为：3，．由已知利用正弦定理可求a的值，进而根据余弦定理可求bc的值，根据三角形面积公式即可计算得解．本题主要考查了正弦定理，余弦定理，三角形面积公式在解三角形中的应用，属于基础题．15.沿着一条笔直的公路有9根电线杆，现要移除2根，且被移除的电线杆之间至少还有2根电线杆被保留，则不同的移除方法有______种【答案】21【解析】解：把6根电线杆放好，7为空，选择两个放入需要移除的电线杆，这样这两根需要移除的电线杆中间至少有一根，然后再把余下一根放到这两根中间去，所以有，故答案为：21．把6根电线杆放好，7为空选择两个放入需要移除的电线杆，这样这两根需要移除的电线杆中间至少有一根，然后再把余下一根放到这两根中间去，问题得以解决．本题考查了排列组合在实际生活中的应用，属于中档题．16.已知向量，满足，，则的取值范围为______．【答案】【解析】解，，又，，，又，设为向量，的夹角，，又，，，，故答案为：先由，，得然后由三角函数的有界性，得然后计算即可本题考查了向量的数量积及三角函数的有界性，属难度为中档题17.设函数，当时，记的最大值为，则的最小值为______．【答案】【解析】解：由去绝对值可得在的最大值为，，，中之一，由题意可得，，，，上面四个式子相加可得即有，可得的最小值为．故答案为：．由题意可得在的最大值为，，，中之一，可得四个不等式，相加，再由绝对值不等式的性质，即可得到所求最小值．本题考查函数的最值求法，注意运用函数取最值的情况，以及绝对值不等式的性质，考查运算能力和推理能力，属于中档题．三、解答题（本大题共5小题，共74.0分）18.已知函数Ⅰ求的最小正周期及单调递增区间；Ⅱ求在区间上的最大值．【答案】解：Ⅰ．的最小正周期，令，，得，，的单调递增区间为，；Ⅱ时，，，在区间上的最大值为3．【解析】Ⅰ利用三角函数的诱导公式化简，由周期公式计算得的最小正周期，由，可解得函数的单调增区间；Ⅱ由x的范围求出的范围，进一步求出的范围，则答案可求．本题考查正弦函数的周期性及单调性，考查了正弦函数的值域，属于基础题．19.如图，在四棱锥中，平面ABCD，，，，Q为棱PC上的一点，且．Ⅰ证明：平面平面ABCD；Ⅱ求直线QD与平面PBC所成角的正弦值．【答案】证明：Ⅰ连结AC，BD，交于点O，则由∽，得，，，平面ABCD，平面ABCD，又平面QBD，平面平面ABCD．解：Ⅱ过D作平面PBC的垂线，垂足为H，则即为直线QD与平面PBC所成角，设为，设，，，即，解得，，直线QD与平面PBC所成角的正弦值．【解析】Ⅰ连结AC，BD，交于点O，推导出，平面ABCD，由此能证明平面平面ABCD．Ⅱ过D作平面PBC的垂线，垂足为H，则即为直线QD与平面PBC所成角，设为，设，由，求出，由此能求出直线QD与平面PBC所成角的正弦值．本题考查面垂直的证明，考查线面角的正弦值的求法，考查空间中线线、线面、面面间的位置关系等基础知识，考查运算求解能力，考查数形结合思想，是中档题．20.已知数列的前n项和为，且满足且Ⅰ当，时，求数列的前n项和：Ⅱ若是等比数列，证明：．【答案】解：Ⅰ当，时，，前n项和；Ⅱ证明：可得，时，，由是等比数列，可得，且，即，，，则，则，．【解析】Ⅰ当，时，，运用分组求和方法，结合等差数列和等比数列的求和公式，计算可得所求和；Ⅱ运用等比数列的通项公式，可得a，b的值，进而得到，运用裂项相消求和和不等式的性质，即可得证．本题考查数列的求和方法：分组求和和裂项相消求和，考查等差数列和等比数列的求和公式的运用，考查化简运算能力，属于中档题．21.已知抛物线的焦点为F，点，且．Ⅰ求抛物线方程；Ⅱ设A，B是抛物线上的两点，当F为的垂心时，求直线AB的方程．【答案】解：Ⅰ，解得：，所以C：；Ⅱ由，设，，因为F是的垂心，所以，有，故，所以设AB：与C：联立得，令，有，由韦达定理，，，因为F是的垂心，所以，即同理，得，所以，解得，又因为，所以AB：．【解析】Ⅰ由两点间距离公式列式，求得即可；Ⅱ根据垂心性质得AB的斜率，可设出AB的方程，与抛物线联立，利用韦达定理，列式可得．本题考查了直线与抛物线的综合属难题．22.设，已知函数，．Ⅰ若恒成立，求a的范围：Ⅱ证明：存在实数a使得有唯一零点．【答案】解：Ⅰ，，，恒成立，，解得，又当时，，在单调递增，，综上所述；证明：Ⅱ设的零点为，有，则，令，则，，在上存在零点，设为，取，则，，，设的零点为，则在上递增，在上递减，函数存在两个零点，，函数在，上递减，在上递增，函数存在唯一的零点，综上所述存在，符合题意．【解析】Ⅰ先求导，根据导数和函数的单调性的关系可得当时，，Ⅱ设的零点为，有，则，构造函数，再求导，设在上存在零点，设为，取，代入到中，根据导数和函数最值的关系，即可求出．本题考查导数知识的运用，函数的单调性，函数零点的问题，解题的关键是正确求导，合理构造，属于难题．。

浙江省名校新高考研究联盟（Z20联盟）2019届第一次联考语文试题考生须知：1.本卷满分150分，考试时间120分钟。

2.答题前务必将自己的姓名，准考证号用黑色字迹的签字笔或钢笔分别填写在试题卷和答题纸规定的地方。

3.答题时，请按照答题纸上“注意事项”的要求，在答题纸相应的位置上规范答题，在本试卷纸上答题一律无效。

4.考试结束后，只需上交答题卷。

一、语言文字运用（共20分）1.下列各句中，没有错别字且加点字的注音全都正确的一项是（3分）A.场（cháng）院四周都岑（cén）寂了下来，只有苇塘里蛙声一片：老书记望着院子里刚收获的玉米，感慨地说：“真是堆积如山哪！”B.南曲柔缓婉转的曲调，和北方遒劲（jìn）朴实的声调不能互相调节，只好改弦更（gng）张，但无论南曲或北曲，都并未超出杂剧的范畴。

C.众多大牌导演在蜇伏了一年以后，在市场的召（zhào）唤下纷纷“出山”争霸天下，使得今年的华语片市场格外热闹；影片（pin）各具特色，不会混淆。

D.这是一幅饱蘸（zhàn）着大师才情的作品，无论谁见到，都会感到运筹帷幄的豪迈席卷而来，可以说它至今依然是我国这一领域的翘（qiào）楚。

阅读下面的文字，完成2、3题。

（5分）中华山河表里形胜，自然的奇迹似乎是顽皮的山鹿率性奔跃之后留下的斑驳蹄印。

两千余座石窟、十万余尊佛像从北魏开始启程，一路逶迤走到唐的时空廊坊。

一代复一代的工匠勒绳搭架在山崖钎凿锤打，劳作的汗渍在衣衫上形成片片云图，他们湿漉漉的发髻几近浸透了半个洛阳城。

［甲］诸佛、菩萨、罗汉、力士、夜叉、飞天……，或立或坐或卧，姿态万千，神情自若，既悲又悯，俯视世间芸芸众生的喜怒哀乐。

［乙］佛的目光要把这一切包涵，无论你是谁，从哪里来，在智慧与光明的化身面前，多舛的肉身都俨然一粒尘埃。

［丙］光明使所有的物象透明空盈，苦思冥想日日追求的重如泰山的名利权势，在此且放下——因为，你在佛的面前永远是孩子。

浙江省名校新高考研究联盟（Z20）2019届高三第一次联考数学试题一：选择题。

1.已知集合，，则A. B.C. D. 或【答案】C【解析】【分析】利用一元二次不等式的解法化简集合,再由交集的定义求解即可.【详解】集合，，．故选C．【点睛】研究集合问题，一定要抓住元素，看元素应满足的属性.研究两集合的关系时，关键是将两集合的关系转化为元素间的关系，本题实质求满足属于集合且属于集合的元素的集合.2.设复数满足为虚数单位，则A. B. i C. D. 1【答案】B【解析】【分析】把已知等式变形，利用复数的除法运算法则：分子、分母同乘以分母的共轭复数，化简复数，从而可得结果. 【详解】由，得．故选B．【点睛】本题考查了复数代数形式的乘除运算，是基础题．复数是高考中的必考知识，主要考查复数的概念及复数的运算．要注意对实部、虚部的理解，掌握纯虚数、共轭复数、复数的模这些重要概念，复数的运算主要考查除法运算，通过分母实数化转化为复数的乘法，运算时特别要注意多项式相乘后的化简，防止简单问题出错，造成不必要的失分.3.设函数，则的值为A. B. C. D. 2【答案】C【解析】【分析】由分段函数，先求=ln2，然后根据判断范围再由分段函数另一段求出值【详解】，=ln2，ln2，即=【点睛】本题主要考察分段函数求函数值，这类题目，需要判断自变量所在范围，然后带入相应的解析式解答即可4.已知是空间中两条不同的直线，，是两个不同的平面，则下列命题正确的是A. 若，，，，则B. 若，，则C. 若，，则D. 若，，，则【答案】D【解析】【分析】利用与相交或平行判断；根据与相交、平行或判断；根据或判断；由面面垂直的判定定理得．【详解】由，是空间中两条不同的直线，，是两个不同的平面，得：若，，，，则与相交或平行，故错误；若，，则与相交、平行或，故错误；若，，则或，故错误；若，，，则由面面垂直的判定定理得，故正确．故选D．【点睛】本题考查命题真假的判断，考查空间中线线、线面、面面间的位置关系等基础知识，是中档题．空间直线、平面平行或垂直等位置关系命题的真假判断，除了利用定理、公理、推理判断外，还常采用画图（尤其是画长方体）、现实实物判断法（如墙角、桌面等）、排除筛选法等；另外，若原命题不太容易判断真假，可以考虑它的逆否命题，判断它的逆否命题真假，原命题与逆否命题等价.5.已知实数满足约束条件，则的最大值为A. 1B. 4C. 2D.【答案】B【解析】【分析】由约束条件作出可行域，化目标函数为直线方程的斜截式，数形结合得到最优解，联立方程组求得最优解的坐标，把最优解的坐标代入目标函数得结论.【详解】作出实数满足约束条件对应的平面区域如图阴影部分由得，平移直线，由图象可知当直线经过点时，直线的截距最大，此时最大由解得．代入目标函数得．即目标函数的最大值为4．故选B．【点睛】本题主要考查线性规划中，利用可行域求目标函数的最值，属于简单题.求目标函数最值的一般步骤是“一画、二移、三求”：（1）作出可行域（一定要注意是实线还是虚线）；（2）找到目标函数对应的最优解对应点（在可行域内平移变形后的目标函数，最先通过或最后通过的顶点就是最优解）；（3）将最优解坐标代入目标函数求出最值.6.已知双曲线：，则“”是“双曲线的焦点在轴上”的A. 充分不必要条件B. 必要不充分条件C. 充要条件D. 既不充分也不必要条件【答案】A【解析】【分析】根据充分条件和必要条件的定义，结合总表示焦点在轴上判断即可．【详解】双曲线的焦点在轴上或，或，或推不出，“”是“双曲线的焦点在轴上”的充分不必要条件．故选A．【点睛】判断充分条件与必要条件应注意：首先弄清条件和结论分别是什么，然后直接依据定义、定理、性质尝试.对于带有否定性的命题或比较难判断的命题，除借助集合思想化抽象为直观外，还可利用原命题和逆否命题、逆命题和否命题的等价性，转化为判断它的等价命题；对于范围问题也可以转化为包含关系来处理.7.函数的图象可能是A.B.C.D.【答案】A【解析】【分析】利用排除法,由是奇函数排除；排除；排除；从而可得结果. 【详解】因为，可得是奇函数排除；当时，，点在轴的上方，排除；当时，，排除；故选A．【点睛】本题通过对多个图象的选择考查函数的图象与性质，属于中档题.这类题型也是近年高考常见的命题方向，该题型的特点是综合性较强、考查知识点较多，但是并不是无路可循.解答这类题型可以从多方面入手，根据函数的定义域、值域、单调性、奇偶性、特殊点以及时函数图象的变化趋势，利用排除法，将不合题意的选项一一排除.8.已知，是椭圆与的左、右焦点，过左焦点的直线与椭圆交于，两点，且满足，，则该椭圆的离心率是A. B. C. D.【答案】B【解析】【分析】由，，利用椭圆的定义，求得，，，可得，，由二倍角公式列方程可得结果.【详解】由题意可得：，，可得，，，，，，,可得，可得．故选B．【点睛】本题考查椭圆的简单性质的应用以及椭圆的离心，属于难题．离心率的求解在圆锥曲线的考查中是一个重点也是难点，一般求离心率有以下几种情况：①直接求出，从而求出;②构造的齐次式，求出;③采用离心率的定义以及圆锥曲线的定义来求解．9.已知实数，满足，，则的最小值是A. 10B. 9C.D.【答案】B【解析】【分析】利用基本不等式求得，则，展开后再利用基本不等式可求得的最小值．【详解】，，，，当且仅当时，取等号．则，当且仅当时，且，时，的最小值为9，故选B．【点睛】本题主要考查利用基本不等式求最值，属于难题.利用基本不等式求最值时，一定要正确理解和掌握“一正，二定，三相等”的内涵：一正是，首先要判断参数是否为正；二定是，其次要看和或积是否为定值（和定积最大，积定和最小）；三相等是，最后一定要验证等号能否成立（主要注意两点，一是相等时参数是否在定义域内，二是多次用或时等号能否同时成立）.10.已知三棱锥的所有棱长为是底面内部一个动点包括边界，且到三个侧面，，的距离，，成单调递增的等差数列，记与，，所成的角分别为，，，则下列正确的是A. B. C. D.【答案】D【解析】【分析】利用公式将问题转化为：比较与，，夹角的大小，然后判断到，，的距离，在中确定所在区域，利用数形结合可以解决．【详解】依题意知正四面体的顶点在底面的射影是正三角形的中心，则，，其中，表示直线、的夹角，，，其中，表示直线、的夹角，，，其中，表示直线的夹角，由于是公共的，因此题意即比较与，，夹角的大小，设到，，的距离为，，则，其中是正四面体相邻两个面所成角，所以，，成单调递增的等差数列，然后在中解决问题由于，结合角平分线性质可知在如图阴影区域不包括边界从图中可以看出，、所成角小于所成角，所以，故选D．【点睛】本题考查了异面直线及其所成角，以及公式的应用，考查了转化思想与数形结合思想的应用，属于难题．若直线与其在平面内的射影所成的角为，平面内任意直线与、成的角为，则.二：填空题。

浙江省名校新高考研究联盟（Z20联盟）2019届第一次联考英语参考答案听力：1-5ABCBA6-10BBABB11-15CACBB16-20BCAAC阅读：21-23BAA24-26DCA27-30BADA31-35BFGDC完型：36-40CBADA41-45BDACC46-50BDBAD51-55CABDC语法填空：anized 61.satisfied 57.was invited62.encouragement58.for59.a63.and64.who60.readily65.yourself应用文：One possible version:Dear Sir/Madam,I’m planning to sign up for the writing course for IELTS tests you offer this summer,and now writing to ask you for some basic information.First,I’d like to know when the course will begin and how long it will last,so that I can get things prepared in advance.Besides,I’m curious about the class size,which will directly decide how much practice and attention I can get during the class.By the way,how much will you charge and does it include the accommodations?Looking forward to your early reply.概要写作：（参考范文）Possible version:Yours, Li HuaAs the figures show,2015has taken the place of2014to be the warmest year since1880.(要点1)It’s clear that most of its months were the warmest on record,which was in line with an extraordinarily warm winter.(要点2)Actually,it’s mainly the result of a major El Nino,and increased human-made emission into the atmosphere also contributed to it.(要点3)As a consequence of the same elements,2016has been expected to be even hotter than2015.(要点4)听力原文Text1M:I’d like a single to New York,please.W:That will be fifty-six pounds at5:10.Text2M:Have you been to the new swimming pool since it opened?W:Are you kidding?Tomorrow is the deadline for my project.Text3W：I really look forward to the football game next Tuesday.M：It’sa pity that I can’t go with you.I can’t ask for leave that day.My boss wants me to accompany him to the factory.Text4W:Look at the clouds.I’m afraid it’ll rain soon.M:You could be right,but it’s not what the weather report said.W:What did the weather report say?M:It said it would be fine all day today.Text5W:Manhattan Square,please.M:All right,Madam.When are we supposed to be there?W:I’d like to go around the city,if you don’t mind.Text6M:Good evening,Madam.Can I help you?W:Yes.It’s about my son Tony.He went to school this morning,but hasn’t been back yet and it’s6:00 p.m.now.M:Just a moment,please.May I have his name?W:Tony Smith,112Broadway.M:Thank you.Now Mrs.Smith,what is the matter exactly?W:Well,Tony left home at7o’clock this morning,but just now his teacher called me and asked why Tony didn’t go to school.M:Do you think it’s possible that he went to a friend’s home?W:I don’t think so.I called his friends and our neighbors,but none of them have seen Tony today.M:I see.Now let’s move to some details.How old is he?W:10years old.And he is1.5meters tall.M:What’s he wearing?W:A blue shirt,and white shoes,carrying a green school bag.M:We’ll do our best to find him,Mrs.Smith.Let me show you out of the station.But wait!Look at the boy standing at the gate and isn’t he...?Text7M:Excuse me.May I have some information about the party you’re advertising outside?W:Well,the party will provide students with a chance to get to know each other.It’s also a chance for the seniors to share their experience with new students to help them graduate successfully.M:Who is organizing the event?W:Actually,it is the StudentUnion.M:I hope we don’t have to sit through a lot of boring speeches.Will we have any time for some fun at this party?W:Most of the time will be split between eating,dancing and singing,so don’t worry.However,some words of wisdom from former students should not be missed.And besides that,there’s another reason to go:the pie there will be delicious.I know since I’m the one who will be making it.You will love it if you taste it.It’s even better than those bought from the store.M:OK,thank you very much.Text8:M:I just can’t believe it!How much did we pay the newspaper to advertise this job?W:$90,Mr.Smith.M:Ninety dollars?It wasn’t worth ninety cents.Every single person who’s come here today has been an absolutely hopeless loser.W:Well,uh...M:Are there any applicants left now,or is that the last?W:There’s just one gentleman waiting,Mr.Smith.M:Ah well,he can’t be any worse than the rest of them.What’s his name?W:John Berry.M:OK,show Mr.Berry in then,Judy.W:OK.Please come in,Mr.Berry.Mr.Smith will see you now.Text9M:Have you ever operated a cash register?W:No.M:Well,it’s pretty simple.First,you have to make sure you have a roll of paper in there so you can print receipts.W:OK.Where can I find extra rolls of paper if I need any?M:In the back office.Make sure the cash drawer is ready.You need a drawer full of bills and coins soyou can give customers change.And if a customer pays with large bills,like fifties or hundreds,put them under the drawer.W:Alright.So,when a customer comes in,I just add up the cost of their groceries?M:Yes.Then press“Total”.The register will tell you how much the customer owes.If they pay in cash, enter the amount they give you.The screen will tell you how much change to give.Text10Good morning,everyone and welcome to Hershey!I’m going to tell you something about the history of the Hershey company before we find out how chocolate is really made.Let’s start by going back over a hundred years in time.Did you know that chocolate was something that only rich people could afford to buy?So how come we all eat it today?We have Milton S.Hershey,the founder of Hershey to thank for that. He had a dream.And his dream was to make good chocolate that didn’t cost a lot of money.One of Mr. Hershey’s first businesses was the Lancaster Carmel Company.But when he saw some German chocolate machines in Chicago in1893,he decided he wanted to make chocolate himself.In1900,Mr.Hershey sold the Lancaster Carmel Company for$1million.He used the money from this sale to build what is now the world’s largest chocolate factory.The completion of the Hershey chocolate factory in1905meant the high production of chocolate could begin.Mr.hershey’s chocolate business developed quickly and so did the area around it.A bank,a department store,a school and even a zoo were built and in1906,the village of Derry Church was renamed Hershey after its founder,Milton S.Hershey.。

浙江省名校新高考研究联盟（Z20）2019届高三第一次联考数学试题一：选择题。

1.已知集合，，则A. B.C. D. 或【答案】C【解析】【分析】利用一元二次不等式的解法化简集合,再由交集的定义求解即可.【详解】集合，，．故选C．【点睛】研究集合问题，一定要抓住元素，看元素应满足的属性.研究两集合的关系时，关键是将两集合的关系转化为元素间的关系，本题实质求满足属于集合且属于集合的元素的集合.2.设复数满足为虚数单位，则A. B. i C. D. 1【答案】B【解析】【分析】把已知等式变形，利用复数的除法运算法则：分子、分母同乘以分母的共轭复数，化简复数，从而可得结果. 【详解】由，得．故选B．【点睛】本题考查了复数代数形式的乘除运算，是基础题．复数是高考中的必考知识，主要考查复数的概念及复数的运算．要注意对实部、虚部的理解，掌握纯虚数、共轭复数、复数的模这些重要概念，复数的运算主要考查除法运算，通过分母实数化转化为复数的乘法，运算时特别要注意多项式相乘后的化简，防止简单问题出错，造成不必要的失分.3.设函数，则的值为A. B. C. D. 2【答案】C【分析】由分段函数，先求=ln2，然后根据判断范围再由分段函数另一段求出值【详解】，=ln2，ln2，即=【点睛】本题主要考察分段函数求函数值，这类题目，需要判断自变量所在范围，然后带入相应的解析式解答即可4.已知是空间中两条不同的直线，，是两个不同的平面，则下列命题正确的是A. 若，，，，则B. 若，，则C. 若，，则D. 若，，，则【答案】D【解析】【分析】利用与相交或平行判断；根据与相交、平行或判断；根据或判断；由面面垂直的判定定理得．【详解】由，是空间中两条不同的直线，，是两个不同的平面，得：若，，，，则与相交或平行，故错误；若，，则与相交、平行或，故错误；若，，则或，故错误；若，，，则由面面垂直的判定定理得，故正确．故选D．【点睛】本题考查命题真假的判断，考查空间中线线、线面、面面间的位置关系等基础知识，是中档题．空间直线、平面平行或垂直等位置关系命题的真假判断，除了利用定理、公理、推理判断外，还常采用画图（尤其是画长方体）、现实实物判断法（如墙角、桌面等）、排除筛选法等；另外，若原命题不太容易判断真假，可以考虑它的逆否命题，判断它的逆否命题真假，原命题与逆否命题等价.5.已知实数满足约束条件，则的最大值为A. 1B. 4C. 2D.【答案】B【分析】由约束条件作出可行域，化目标函数为直线方程的斜截式，数形结合得到最优解，联立方程组求得最优解的坐标，把最优解的坐标代入目标函数得结论.【详解】作出实数满足约束条件对应的平面区域如图阴影部分由得，平移直线，由图象可知当直线经过点时，直线的截距最大，此时最大由解得．代入目标函数得．即目标函数的最大值为4．故选B．【点睛】本题主要考查线性规划中，利用可行域求目标函数的最值，属于简单题.求目标函数最值的一般步骤是“一画、二移、三求”：（1）作出可行域（一定要注意是实线还是虚线）；（2）找到目标函数对应的最优解对应点（在可行域内平移变形后的目标函数，最先通过或最后通过的顶点就是最优解）；（3）将最优解坐标代入目标函数求出最值.6.已知双曲线：，则“”是“双曲线的焦点在轴上”的A. 充分不必要条件B. 必要不充分条件C. 充要条件D. 既不充分也不必要条件【答案】A【解析】【分析】根据充分条件和必要条件的定义，结合总表示焦点在轴上判断即可．【详解】双曲线的焦点在轴上或，或，或推不出，“”是“双曲线的焦点在轴上”的充分不必要条件．故选A．【点睛】判断充分条件与必要条件应注意：首先弄清条件和结论分别是什么，然后直接依据定义、定理、性质尝试.对于带有否定性的命题或比较难判断的命题，除借助集合思想化抽象为直观外，还可利用原命题和逆否命题、逆命题和否命题的等价性，转化为判断它的等价命题；对于范围问题也可以转化为包含关系来处理.7.函数的图象可能是A.B.C.D.【答案】A【解析】【分析】利用排除法,由是奇函数排除；排除；排除；从而可得结果.【详解】因为，可得是奇函数排除；当时，，点在轴的上方，排除；当时，，排除；故选A．【点睛】本题通过对多个图象的选择考查函数的图象与性质，属于中档题.这类题型也是近年高考常见的命题方向，该题型的特点是综合性较强、考查知识点较多，但是并不是无路可循.解答这类题型可以从多方面入手，根据函数的定义域、值域、单调性、奇偶性、特殊点以及时函数图象的变化趋势，利用排除法，将不合题意的选项一一排除.8.已知，是椭圆与的左、右焦点，过左焦点的直线与椭圆交于，两点，且满足，，则该椭圆的离心率是A. B. C. D.【答案】B【解析】【分析】由，，利用椭圆的定义，求得，，，可得，，由二倍角公式列方程可得结果.【详解】由题意可得：，，可得，，，，，，,可得，可得．故选B．【点睛】本题考查椭圆的简单性质的应用以及椭圆的离心，属于难题．离心率的求解在圆锥曲线的考查中是一个重点也是难点，一般求离心率有以下几种情况：①直接求出，从而求出;②构造的齐次式，求出;③采用离心率的定义以及圆锥曲线的定义来求解．9.已知实数，满足，，则的最小值是A. 10B. 9C.D.【答案】B【解析】【分析】利用基本不等式求得，则，展开后再利用基本不等式可求得的最小值．【详解】，，，，当且仅当时，取等号．则，当且仅当时，且，时，的最小值为9，故选B．【点睛】本题主要考查利用基本不等式求最值，属于难题.利用基本不等式求最值时，一定要正确理解和掌握“一正，二定，三相等”的内涵：一正是，首先要判断参数是否为正；二定是，其次要看和或积是否为定值（和定积最大，积定和最小）；三相等是，最后一定要验证等号能否成立（主要注意两点，一是相等时参数是否在定义域内，二是多次用或时等号能否同时成立）.10.已知三棱锥的所有棱长为是底面内部一个动点包括边界，且到三个侧面，，的距离，，成单调递增的等差数列，记与，，所成的角分别为，，，则下列正确的是A. B. C. D.【答案】D【解析】【分析】利用公式将问题转化为：比较与，，夹角的大小，然后判断到，，的距离，在中确定所在区域，利用数形结合可以解决．【详解】依题意知正四面体的顶点在底面的射影是正三角形的中心，则，，其中，表示直线、的夹角，，，其中，表示直线、的夹角，，，其中，表示直线的夹角，由于是公共的，因此题意即比较与，，夹角的大小，设到，，的距离为，，则，其中是正四面体相邻两个面所成角，所以，，成单调递增的等差数列，然后在中解决问题由于，结合角平分线性质可知在如图阴影区域不包括边界从图中可以看出，、所成角小于所成角，所以，故选D．【点睛】本题考查了异面直线及其所成角，以及公式的应用，考查了转化思想与数形结合思想的应用，属于难题．若直线与其在平面内的射影所成的角为，平面内任意直线与、成的角为，则.二：填空题。

浙江省名校新高考研究联盟（Z20）2019届高三第一次联考数学试题一、选择题（本大题共10小题，共40.0分）1. 已知集合{|1}A x x =<，2{|320}B x x x =++≤，则(A B ⋂= )A. ∅B. {|1}x x <C. {|21}x x -≤≤-D. {|2x x <-或1}x l -<<【答案】C 【解析】解：集合{|1}A x x =<， 2{|320}{|21}B x x x x x =++≤=-≤≤-，{|21}A B x x ∴⋂=-≤≤-．故选：C ．先求出集合A ，B ，由此能求出A B ⋂．本题考查交集的求法，考查交集定义、不等式性质等基础知识，考查运算求解能力，是基础题．2. 设复数z 满足()212(z i i i ⋅+=-+为虚数单位)，则(z = )A. i -B. iC. 1-D. 1【答案】B 【解析】解：由()212z i i ⋅+=-+， 得()()()()1221252225i i i i z i i i i -+--+====++-． 故选：B ．把已知等式变形，再利用复数代数形式的乘除运算化简得答案．本题考查了复数代数形式的乘除运算，是基础题．3. 设函数()ln ,1,1x x x f x e x ≤--⎧⎪=>-⎨⎪⎩，则()()2f f -的值为( )A. 1eB. 2eC. 12D. 2【答案】C 【解析】解：函数ln ,1(),1x x x f x e x -⎧≤-⎪=⎨>-⎪⎩， ()2ln 2ln2f ∴-=-=，()()()ln 212ln22f f f e --===． 故选：C ．推导出()2ln 2ln2f -=-=，从而()()()ln22ln2f f f e --==，由此能求出结果． 本题考查函数值的求法，考查函数性质等基础知识，考查运算求解能力，是基础题．4. 已知m ，n 是空间中两条不同的直线，α，β是两个不同的平面，则下列命题正确的是( )A. 若m α⊂，n α⊂，//m β，//n β，则//αβB. 若αβ⊥，//n α，则n β⊥C. 若//αβ，//m α，则//m βD. 若m α⊥，n β⊂，//m n ，则αβ⊥【答案】D【解析】解：由m ，n 是空间中两条不同的直线，α，β是两个不同的平面，得： 在A 中，若m α⊂，n α⊂，//m β，//n β，则α与β相交或平行，故A 错误； 在B 中，若αβ⊥，//n α，则n 与β相交、平行或n β⊂，故B 错误； 在C 中，若//αβ，//m α，则//m β或m β⊂，故C 错误；在D 中，若m α⊥，n β⊂，//m n ，则由面面垂直的判定定理得αβ⊥，故D 正确． 故选：D ．在A 中，α与β相交或平行；在B 中，n 与β相交、平行或n β⊂；在C 中，//m β或m β⊂；在D 中，由面面垂直的判定定理得αβ⊥．本题考查命题真假的判断，考查空间中线线、线面、面面间的位置关系等基础知识，是。

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2019年浙江高三一模英语试卷（浙南名校联考）-学生用卷一、阅读理解（共10小题，每小题2.5分，共25分）1、【来源】 2019年浙江高三一模（浙南名校联考）第21~23题7.5分An advertisement in the newspaper stating that the local SPCA was promoting a campaign called "Home for Christmas", appealing to local residents（居民）to give their numerous older cats a chance at a forever home, made my husband and I visit the shelter. As we were chatting with the worker, an orange cat reached for my husband's shoulder. We took this as a sign, so Cici came to be a part of our lives.It didn't take her long to fit in with our life. As the weeks went by we all adapted to our routine. It soon became evident that Cici was an extremely TAL#NBSP laid-back cat and there wasn't much that would upset her, aside from her humans serving up a late dinner. About this time, I heard about an organization called You Are Not Alone that was looking for dogs to visit the local seniors. Although cats had never been considered as appropriate visitors, Cici was reluctantly scheduled for an interview and surprisingly she was declared fit for the program.A few days later, Cici and I arrived at the Nanaimo Seniors Village and prepared to meet a resident who wanted a visit from a cat. Loan Samuels had recently moved to the Village after giving up her home and her two pets. Understandable, she was missing them terribly and I was hoping that our visits would help with the healing of her loss. I still remember Joan's face when she saw Cici for the first time. It was truly love at first sight. Every Tuesday morning after that, Cici and I had the pleasure of visiting loan. Sometimes we played cards or just chatted while the cat made herself cozy on Joan's bed. All our times together were so special.I must admit being part of the pet visitation program was truly a privilege. Playing a part in bringing two sweet souls together and watching them bond was truly a gift.(1) The writer got Cici from.A. An organization called You Are Not AloneB. A center called Home for ChristmasC. An animal shelter called SPCAD. The Nanaimo Seniors Village(2) What does the underlined word "laid-back" in the second paragraph most probably mean?A. Easy-going.B. Sensitive.C. Smart.D. Wild.(3) Which of the following would be the best title for this passage?A. A Programme to Care for CatsB. The Cat that Made a DifferenceC. The Visit that Changed My LifeD. A Home for Old Cats and Seniors2、【来源】 2019年浙江高三一模（B篇）第24~26题7.5分2019~2020学年浙江杭州西湖区杭州学军中学西溪校区高二上学期期中（B篇）第24~26题7.5分(每题2.5分)Literature reflects life. So in aging Japan there are a large number of hit books by aged authors. "Age 90: what's so great about it? " is a humorous essay on the difficulties of the elderly, by Aiko Sato, who is 95. It sold one million copies in 2017, making it Japan's bestselling book that year.In 2018 the Akutagawa literary prize went to Chisako Wakatake, 63 at the time, for her first novel "Live by Myself“. The book talks about how to live in old age." Going to Die Soon", also by Ms Uchidate, feature 78-year-old Hana, a former alcohol-shop owner trying to make the most of her remaining years. The novel has been called a book for shukatsu, or preparing for death, making readers think more deeply about what it means to age.Japan's population has the world's highest percentage of over-65s, People are living longer, so many have at least 20-30 years of retirement, for much of which they are energetic.And although the Japanese have been spending less on books, that is least true for the over-60s. Lawson, a convenience-store chain, recently decided to stock books with the older generation in mind.But the wrinkle writers' books are attracting younger readers, too. Some are preparing for their own old age or want to understand the increasing number of old people they see around them. Others find relevance in the themes explored, such as loneliness, a problem that stretches well beyond the silver-haired.The most notable feature of the new trend is that the vast majority of authors, and main characters, are women. Especially popular, are the ara-hun （"around-hundred" years-old） writers like Ms Sato, whose book, readers say, helps them be more positive. Their popularity also reflects support for strong women who are passionate about their work, a phenomenon that is all too rare in Japan today.(1) "Going to Die Soon" is a book.A. about how to stay positive in old ageB. which won the Akutagawa literary prize in 2018C. which helps people live the rest of the life to the fullestD. written by a 78-year-old Hans, an alcohol-shop owner(2) The books by aged authors are popular with younger readers because.A. young readers now have more elder relativesB. young readers face similar problems as the elderlyC. young readers are worried about becoming oldD. young readers are as lonely as the elderly(3) What can we beam from the passage?A. Aged writers might be more successful than younger writers.B. An aging society brings more chances for women writers.C. People have to work after their retirement in an aging society.D. Ambitious career women might not be popular with the Japanese.3、【来源】 2019年浙江高三一模（浙南名校联考）第27~30题10分The AlphaGo program's victory is an example of how smart computers have become. But can artificial intelligence（AI）machines act ethically（合乎道德也）, meaning can they be honest and fair?One example of AI is driverless cars. They are already on California roads, so it is not too soon to ask whether we can program a machine to act ethically. As driverless cars improve, they will save lives. They will make fewer mistakes than human drivers do. Sometimes, however, they will face a choice between lives. Should the cars be programmed to make a sudden turn to avoid hitting a child? What if the only risk is damage to the car itself, not to the passengers?Perhaps there will be lessons to learn from driverless cars, but they are not super-intelligent beings. Teaching ethics to a machine even more intelligent than we are will be the bigger challenge.About the same time as AlphaGo's victory, Microsoft's "chatbot" took a bad turn. The software, named Taylor, was designed to answer messages from people aged 18-24. Taylor was supposed to be able to learn from the messages she received. She was designed to slowly improve her ability to handle conversations, but some people were teaching Taylor racist ideas. When she started saying nice things about Hitler, Microsoft turned her off and deleted her ugliest messages.AlphaGo's victory and Taylor's defeat happened at about the same time. This should be a warning to us. It is one thing to use AI within a game with clear rules and clear goals. It is something very different to use AI in the real world.Eric Schmidt is one of the bosses of Google, which owns AlphaGo. He said people will be the winner, whatever the outcome. Advances in AI will make human beings smarter, more able and "just better human beings."(1) What's the second paragraph mainly about?A. Whether AI machines are capable to predict possible risks.B. What AI machines will do to save human lives.C. Whether AI machines can make ethical decisions.D. What AI machines will do to avoid damages to themselves.(2) What is said to be the bigger challenge facing humans in the AI age?A. How to prevent AI machines doing harm to humans.B. How to avoid being over-dependent on AI machines.C. How to ensure that super-intelligent AI machines act ethically.D. How to make super-intelligent AI machines share human feelings.(3) What do we learn about Microsoft's "chatbot" Taylor?A. She could not tell good from bad.B. She could turn herself off when necessary.C. She was not made to handle new situations.D. She was good at performing routine tasks.(4) What is Eric Schmidt's attitude towards artificial intellgence?A. Negative.B. Unconcerned.C. Positive.D. Doubtful.二、七选五（共5小题，每小题2分，共10分）4、【来源】 2019年浙江高三一模（浙南名校联考）第31~35题10分(每题2分)Why People Who Sleep Late Are Smarter Geniuses have a few things in common. Going against the grain is one of those things. Following a routine schedule is another. Above all, they sleep late. Studies have shown that people who stay up late are actually smarter and more creative.1.They drink coffee.Being a coffee drinker has some impacts on success. Coffee drinkers often take part in extra physical activity, leading to numerous health benefits.2. Reaction time, memory and general cognitive abilities thus improved by drinking coffee.3A BBC report stated that late risers are in better moods. This has something to do with the fact that early risers use more cortisol（皮质醇）, which results in more muscle aches, headaches and cold symptoms.They make better use of golden time.4. Such time-of-day variations in function are not unusual. Organisms are adapted to the continual change in light and dark during a 24-hour period to reproduce faster. So when morning people are going back home to take a rest, night owls are making use of the best time of the day to equip themselves.They remain alert for longer hours.In the study, brain activity was measured for the early birds and nightowls.5. It is explained that night time minds have a longer （生理节奏的）clock and hence can give more attention for a longer time.A. They are in better moodsB. And the latter one scored higherC. Here are some reasons which make senseD. Sleeping late does too much harm to our bodyE. This helps the brain and body both function at a higher levelF. According to an Australian study, humans learn better in the evening.G. Coffee makes morning people up at night, while all-nighters are unaffected .三、完形填空（共20小题，每小题1.5分，共30分）5、【来源】 2019年浙江高三一模（浙南名校联考）第36~55题30分One of my neighbors used to have a beautiful tree in her front yard. Her dad planted it forher1it was nothing more than a little branch and several years later it started to2towards the sky at an amazing speed. Soon it blessed her with cooling shade in the3and glorious, golden leaves in the fall.When the two-day snowstorm struck our town, heavy snow4the tree's branches that were still full of leaves. The weight split that lovely tree downthe5. It was so sad seeing half of it6on the ground after the storm. When I talked to my neighbor later, she said that the damage had been too much and that the7tree would have to be cutdown.8she had saved a few seedlings from it that she hopedto9in the future.Still, it was a shock to drive by her house the other day and see10but a stump（树桩）in her front yard. I missed that tree. I missed its branches, itsleaves11in the afternoon sun. I missed seeing its limbs reach towards the heavens. I thought that the stump would be a sad12of its loss for a long time to come. My wonderful neighbor,13, had another plan. When I drove by her home today I saw a tiny bird feeder sitting on that stump and a colorful songbird havingits14. It was such an affirmation（肯定）of15. It was such a joy to see. I could feel my heart16.Life by its very nature is a17bag. It hands us both beauty and tragedy, love and loss,18and pain. What we do with it, however, is up to us. We can let it19us in two, or we can use even is hardest times to makeour20stronger and our hearts more loving. We can spend it complaining or we can use it to help others.A. unlessB. whenC. untilD. becauseA. circulateB. windC. shootD. standA. springB. summerC. fallD. winterA. fell onB. take apartC. get throughD. broke downA. topB. middleC. bottomD. rootA. dancingB. spreadingC. tremblingD. lyingA. abnormalB. bareC. hopelessD. entireA. ConsequentlyB. EventuallyC. ThankfullyD. AccidentallyA. replaceB. renewC. replantD. removeA. everythingB. somethingC. anythingD. nothingA. fallingB. shakingC. floatingD. shiningA. storyB. expectationC. reminderD. viewA. thoughB. thereforeC. stillD. otherwiseA. restB. babyC. dateD. dinnerA. lifeB. timeC. effortD. natureA. beatB. smileC. hurtD. sinkA. mixedB. changeableC. separatedD. stableA. sorrowB. wealthC. pleasureD. wonderA. splitB. pullC. benefitD. attachA. connectionsB. familiesC. bodiesD. souls四、语法填空（共10小题，每小题1.5分，共15分）6、【来源】 2019年浙江高三一模（浙南名校联考）第56~65题15分2019~2020学年浙江杭州西湖区杭州学军中学西溪校区高二上学期期中第56~65题15分(每题1.5分)2020~2021学年3月广东广州越秀区广州市第二中学高二下学期月考第86~95题15分(每题1.5分) 2019~2020学年广东广州越秀区广州市第二中学高二下学期期中第86~95题10分People love the view of old buildings in Paris, especially Notre Dame Cathedral（巴黎圣母院）.1（fortunate）, on April 16, a fire destroyed the spire （尖顶） of the 850-year-old cathedral, and two-thirds of the roof. Flames could beseen2（rise） through the top of the monument. The citizens in Paris gathered around the cathedral, praying3the hundreds of firefighters who fought the flames.4was no wonder that people felt sad. The church itselfafter all, is a historical and artistic treasure.5（build） in 1163, it is one of the world's most famous tourist sits, attracting about 12 million6（visit）every year.Notre Dame Cathedral is home to many7（religion） artworks, paintings and sculpture. Despite its long history and many treasure, thecathedral8（need）the help of a writer to become truly famous. Victor Hugo's 1831 novel, The Hunchback of Notre Dame, presented the buildingto9wide audience of readers. He wrote the book to remind people of it, with the hope10they would protect the beautiful old buildings of Paris.五、书面表达（共1小题，共15分）7、【来源】 2019年浙江高三一模（浙南名校联考）第66题15分2019~2020学年12月江苏南京秦淮区南京市第一中学高二上学期月考第71题15分假定你是李华，上周你在海外购物网站 Amazon 上购买了一个书包，但是收到后发现存在质量问题，请你给该网站负责人 Smith 先生写一封信，内容包括：1. 反映质量问题；2. 你的诉求。

绝密★考试结束前（高三8月返校联考）浙江省名校新高考联盟（Z20联盟）2019年第一次联考化学试题命题：平湖中学审题：桐乡告急中学元济高级中学可能用到的相对原子质量：H 1，C 12，N 14，O 16，Na 23，Mg 24，S 32，Cl 35.5，K 39，Ca 40，Fe 56，Cu 64选择题部分一、选择题（本大题共25小题，每小题2分，共50分。

每小题列出的四个选项中只有一个是符合题目要求的，不选、多选、错选不得分）1、下列属于酸的是A．HClO B. CaCO3 C. CH3OCH3 D. NaOH2、下图中可用于分离油和水混合物的主要仪器是A. B. C. D.3、下列关于电解质的说法正确的是A．二氧化硫的水溶液能导电，故二氧化硫是电解质B．铜是电和热的良导体，故铜是电解质C．H2SO4是强酸，CH3COOH是弱酸，故H2SO4溶液的导电性强于CH3COOH溶液导电性D．水是弱电解质，硫酸钡是强电解质4、下列属于物理变化的是A. 煤的气化B. 天燃气的燃烧C. 烃的裂解D. 石油的分馏5、下列物质的水溶液显碱性的是A．Na2CO3 B. NHCl4 C.NaHSO4 D. FeCl36、下列说法不正确的是A．甲醛的水溶液常用于种子杀菌消毒和标本的防腐，也可用于浸泡食物B．碘是一种药用元素，含碘食品的生产也需要碘元素C．海水中提取镁、工业炼铁的过程中都用到了碳酸钙D．2CaSO4·H2O变成CaSO4·2H2O的反应在医疗上可用于制作石膏绷带7、下列表示正确的是A．二氧化硅的结构式：O=Si=O B. 葡萄糖的摩尔质量：180C．绿矾的化学式：FeSO4·7H2O D. 乙烯的比例模型：8、下列变化过程中，加入还原剂才能实现的是A．H2O→O2 B. SO32-→SO2 C. I-→I2 D. CuO→Cu9、下列说法正确是A. 用广发pH试纸测亚硫酸溶液的pH，结果试纸先变红后褪色B．工业制备镁不选择氧化镁的主要原因是氧化镁的熔点高，电解成本高C．工业制备硫酸中，三氧化硫变成硫酸的反应需要从塔顶喷水D．生物炼铜法是利用细菌将硫酸铜转化为硫化铜析出而得以富集10、下列说法不正确的是A．石油的主要成分是碳氢化合物，其中还溶有气态和固态的碳氢化合物B．可用氢气除去乙烷中混有的少量乙烯C．可通过燃烧时现象的不同来区分乙烯和乙炔D．煤隔绝空气加强热能得到煤炉气、煤焦油、粗氨水、焦炭等产品11、下列说法正确的是A．2,2-二甲基丙烷和2-甲基异丁烷是同分异构体B．16O2和18O2是同位素C．葡萄糖和麦芽糖是同系物D．氕、氘、氚是同种核素12、下列说法正确的是A．棉花、麻、羊毛、蚕丝的主要化学成分都是蛋白质B．淀粉、纤维素、氨基酸、蛋白质在一定条件下都能水解C．皂化反应完全后，向反应液中加入热的饱和食盐水，容器底部有固体析出D．可用水鉴别乙醇、苯和四氯化碳13、X、Y、Z和W代表原子序数依次增大的四种短周期元素，X原子核内没有中子，在周期表中，Z与Y、W均相邻；Y、Z和W三种元素的原子最外层电子数之和为17。