2016上海市奉贤区2016届高三第二学期(二模)化学试题(含答案)

2016年上海市奉贤区高考数学二模试卷（理科）一、填空题（共14小题，每小题5分，满分70分）1．（5分）若i（bi+1）是纯虚数，i是虚数单位，则实数b=．2．（5分）函数f（x）=的定义域是．3．（5分）在△ABC中，||=2，||=3，•＜0，且△ABC的面积为，则∠BAC=．4．（5分）双曲线4x2﹣y2=1的一条渐近线与直线tx+y+1=0垂直，则t=．5．（5分）已知抛物线y2=4x上一点M（x0，2），则点M到抛物线焦点的距离为．6．（5分）无穷等比数列首项为1，公比为q（q＞0）的等边数列前n项和为S n，则S n=2，则q=．7．（5分）在一个水平放置的底面半径为cm的圆柱形量杯中装有适量的水，现放入一个半径为Rcm的实心铁球，球完全浸没于水中且无水溢出，若水面高度恰好上升Rcm，则R=cm．8．（5分）从4名男生和3名女生中选出4人参加某个座谈会，若这4人中必须既有男生又有女生，则不同的选法种数共有．（用数字作答）9．（5分）在平面直角坐标系xOy中，将点A（2，1）绕原点O逆时针旋转到点B，若直线OB的倾斜角为α，则cosα的值为．10．（5分）已知函数f（x）=2x﹣a•2﹣x的反函数是f﹣1（x），f﹣1（x）在定义域上是奇函数，则正实数a=．11．（5分）把极坐标方程ρ=sinθ+cosθ化成直角坐标标准方程是．12．（5分）在（x++1）6展开式中的常数项是（用数值作答）13．（5分）在棱长为1的正方体ABCD﹣A1B1C1D1中，若点P是棱上一点，则满足|PA|+|PC1|=2的点P的个数为．14．（5分）若数列{a n}前n项和S n满足S n﹣1+S n=2n2+1（n≥2，n∈N+），且满足a1=x，{a n}单调递增，则x的取值范围是．二、选择题（共4小题，每小题5分，满分20分）15．（5分）平面α的斜线与平面α所成的角是35°，则与平面α内所有不过斜足的直线所成的角的范围是（）A．（0°，35°]B．（0°，90°]C．[35°，90°）D．[35°，90°] 16．（5分）已知log2x，log2y，2成等差数列，则M（x，y）的轨迹的图象为（）A．B．C．D．17．（5分）设z1，z2∈C，z12﹣4z1z2+4z22=0，|z2|=2，则以|z1|为直径的圆面积为（）A．πB．4πC．8πD．16π18．（5分）方程9x+|3x+b|=5（b∈R）有两个负实数解，则b的取值范囤为（）A．（3，5）B．（﹣5.25，﹣5）C．[﹣5.25，﹣5）D．前三个都不正确三、解答题（共5小题，满分60分）19．（12分）平面外ABC的一点P，AP、AB、AC两两互相垂直，过AC的中点D 做ED⊥面ABC，且ED=1，PA=2，AC=2，连接BP，BE，多面体B﹣PADE的体积是；（1）画出面PBE与面ABC的交线，说明理由；（2）求面PBE与面ABC所成的锐二面角的大小．20．（12分）已知椭圆C：=1（a＞b＞0）的长轴长是短轴长的两倍，焦距为2．（1）求椭圆C的标准方程；（2）不过原点O的直线l与椭圆C交于两点M，N，且直线OM，MN，ON的斜率依次成等比数列，问：直线l是否定向的，请说明理由．21．（12分）如图所示，A，B是两个垃圾中转站，B在A的正东方向16千米处，AB的南面为居民生活区，为了妥善处理生活垃圾，政府决定在AB的背面建一个垃圾发电厂P，垃圾发电厂P的选址拟满足以下两个要求（A，B，P可看成三个点）：①垃圾发电厂到两个中转站的距离与它们每天集中的生活垃圾量成反比，比例系数相同；②垃圾发电厂应尽量远离居民区（这里参考的指标是点P到直线AB的距离要尽可能大），现估测得A，B两个中转站每天集中的生活垃圾量分别约为30吨和50吨，设|PA|=5x＞0．（1）求cos∠PAB（用x的表达式表示）（2）问垃圾发电厂该如何选址才能同时满足上述要求？22．（12分）（1）已知0＜x1＜x2，求证：；（2）已知f（x）=lg（x+1）﹣log3x，求证：f（x）在定义域内是单调递减函数；（3）在（2）的条件下，求集合M={n|f（n2﹣214n﹣1998）≥0，n∈Z}的子集个数．23．（12分）数列{a n}，{b n}满足，a1＞0，b1＞0；（1）求证：{a n•b n}是常数列；（2）若{a n}是递减数列，求a1与b1的关系；（3）设a1=4，b1=1，当n≥2时，求a n的取值范围．2016年上海市奉贤区高考数学二模试卷（理科）参考答案与试题解析一、填空题（共14小题，每小题5分，满分70分）1．（5分）若i（bi+1）是纯虚数，i是虚数单位，则实数b=0．【考点】A1：虚数单位i、复数．【专题】11：计算题；35：转化思想；4A：数学模型法；5N：数系的扩充和复数．【分析】由i（bi+1）=﹣b+i，又i（bi+1）是纯虚数，即可得到实部等于0，则b 可求．【解答】解：i（bi+1）=﹣b+i，又i（bi+1）是纯虚数，则﹣b=0，即b=0．故答案为：0．【点评】本题考查了复数代数形式的乘除运算，考查了复数的基本概念，是基础题．2．（5分）函数f（x）=的定义域是[0，+∞）．【考点】33：函数的定义域及其求法．【专题】11：计算题．【分析】由题意可得2x﹣1≥0，解不等式可得函数的定义域．【解答】解：由题意可得2x﹣1≥0，解不等式可得x≥0所以函数的定义域是[0，+∞）故答案为：[0，+∞）【点评】本题考查了求函数的定义域的最基本的类型：偶次根式型：被开方数大于（等于）0，还考查了指数不等式的解法．属于基础试题．3．（5分）在△ABC中，||=2，||=3，•＜0，且△ABC的面积为，则∠BAC=150°．【考点】9O：平面向量数量积的性质及其运算．【专题】5A：平面向量及应用．【分析】由题意可得∠BAC 为钝角，再由×2×3×sin∠BAC=，解得sin∠BAC=，从而得到∠BAC的值．【解答】解：∵在△ABC中，||=2，||=3，且△ABC的面积为，∴=，即，解得sin∠BAC=，又•＜0，∴，∴∠BAC=150°．故答案为：150°．【点评】本题主要考查两个向量的数量积的定义及三角形的面积公式，考查已知三角函数值求角的大小，是基础题．4．（5分）双曲线4x2﹣y2=1的一条渐近线与直线tx+y+1=0垂直，则t=±．【考点】KC：双曲线的性质．【专题】34：方程思想；48：分析法；5B：直线与圆；5D：圆锥曲线的定义、性质与方程．【分析】求得双曲线的渐近线方程，直线tx+y+1=0的斜率为﹣t，运用两直线垂直的条件：斜率之积为﹣1，计算即可得到所求值．【解答】解：双曲线4x2﹣y2=1即为﹣y2=1，可得渐近线为y=±2x，直线tx+y+1=0的斜率为﹣t，而渐近线的斜率为±2，由两直线垂直的条件：斜率之积为﹣1，可得﹣t=±，即有t=±．故答案为：±．【点评】本题考查双曲线的渐近线方程的运用，考查两直线垂直的条件：斜率之积为﹣1，考查运算能力，属于基础题．5．（5分）已知抛物线y2=4x上一点M（x0，2），则点M到抛物线焦点的距离为4．【考点】K8：抛物线的性质．【专题】34：方程思想；35：转化思想；5D：圆锥曲线的定义、性质与方程．【分析】把点M（x0，2）代入抛物线方程，解得x0．利用抛物线的定义可得：点M到抛物线焦点的距离=x0+1．【解答】解：把点M（x0，2）代入抛物线方程可得：=4x0，解得x0=3．∴点M到抛物线焦点的距离=x0+1=4．故答案为：4．【点评】本题考查了抛物线的定义标准方程及其性质，考查了推理能力与计算能力，属于中档题．6．（5分）无穷等比数列首项为1，公比为q（q＞0）的等边数列前n项和为S n，则S n=2，则q=．【考点】88：等比数列的通项公式．【专题】34：方程思想；49：综合法；54：等差数列与等比数列．【分析】由无穷递缩等比数列的各项和可得=2，解方程可得．【解答】解：∵无穷等比数列首项为1，公比为q（q＞0）的等边数列前n项和为S n，且S n=2，∴=2，解得q=，故答案为：．【点评】本题考查等比数列的通项公式和无穷递缩等比数列的各项和，属基础题．7．（5分）在一个水平放置的底面半径为cm的圆柱形量杯中装有适量的水，现放入一个半径为Rcm的实心铁球，球完全浸没于水中且无水溢出，若水面高度恰好上升Rcm，则R=cm．【考点】LF：棱柱、棱锥、棱台的体积；LG：球的体积和表面积．【专题】11：计算题．【分析】求出球的体积等于水面高度恰好上升Rcm的体积，即可求出R的值．【解答】解：在一个水平放置的底面半径为cm的圆柱形量杯中装有适量的水，现放入一个半径为Rcm的实心铁球，球完全浸没于水中且无水溢出，若水面高度恰好上升Rcm，所以，，所以R=（cm）；故答案为：．【点评】本题是基础题，考查球的体积，圆柱的体积的求法，考查计算能力．8．（5分）从4名男生和3名女生中选出4人参加某个座谈会，若这4人中必须既有男生又有女生，则不同的选法种数共有34．（用数字作答）【考点】D5：组合及组合数公式；D9：排列、组合及简单计数问题．【专题】11：计算题．【分析】根据题意，选用排除法；分3步，①计算从7人中，任取4人参加某个座谈会的选法，②计算选出的全部为男生或女生的情况数目，③由事件间的关系，计算可得答案．【解答】解：分3步来计算，①从7人中，任取4人参加某个座谈会，分析可得，这是组合问题，共C74=35种情况；②选出的4人都为男生时，有1种情况，因女生只有3人，故不会都是女生，③根据排除法，可得符合题意的选法共35﹣1=34种；故答案为34．【点评】本题考查组合数公式的运用，解本题采用排除法较为简单．9．（5分）在平面直角坐标系xOy中，将点A（2，1）绕原点O逆时针旋转到点B，若直线OB的倾斜角为α，则cosα的值为．【考点】I2：直线的倾斜角．【专题】31：数形结合；35：转化思想；5B：直线与圆．【分析】设直线OA的倾斜角为θ，则tanθ=，tanα==，cosα=．【解答】解：设直线OA的倾斜角为θ，则tanθ=，则tanα====3，∴cosα===．故答案为：．【点评】本题考查了直线的倾斜角与斜率的关系、三角函数求值，考查了推理能力与计算能力，属于中档题．10．（5分）已知函数f（x）=2x﹣a•2﹣x的反函数是f﹣1（x），f﹣1（x）在定义域上是奇函数，则正实数a=1．【考点】4R：反函数．【专题】31：数形结合；34：方程思想；35：转化思想；51：函数的性质及应用．【分析】f﹣1（x）在定义域上是奇函数，可得：原函数f（x）在定义域上也是奇函数，利用f（0）=0即可得出．【解答】解：∵f﹣1（x）在定义域上是奇函数，∴原函数f（x）在定义域上也是奇函数，∴f（0）=1﹣a=0，解得a=1，∴f（x）=，经过验证函数f（x）是奇函数．故答案为：1．【点评】本题考查了反函数的性质，考查了推理能力与计算能力，属于中档题．11．（5分）把极坐标方程ρ=sinθ+cosθ化成直角坐标标准方程是（x﹣）2+（y ﹣）2=．【考点】Q4：简单曲线的极坐标方程；Q8：点的极坐标和直角坐标的互化．【专题】11：计算题；29：规律型；35：转化思想；5S：坐标系和参数方程．【分析】先在极坐标方程ρ=sinθ+cosθ的两边同乘以ρ，再利用直角坐标与极坐标间的关系，即利用ρcosθ=x，ρsinθ=y，ρ2=x2+y2，进行代换即得．【解答】解：∵ρ=sinθ+cosθ，∴ρ2=ρsinθ+ρcosθ，∴x2+y2=y+x，即x2+y2﹣x﹣y=0．即（x﹣）2+（y﹣）2=．故答案为：（x﹣）2+（y﹣）2=．【点评】本题考查点的极坐标和直角坐标的互化，能在极坐标系中用极坐标刻画点的位置，体会在极坐标系和平面直角坐标系中刻画点的位置的区别，能进行极坐标和直角坐标的互化．12．（5分）在（x++1）6展开式中的常数项是581（用数值作答）【考点】DA：二项式定理．【专题】32：分类讨论；35：转化思想；49：综合法；5P：二项式定理．=，（r=0，1，…，6），令的展开式的通项公式【分析】T r+1T′k+1==2k x r﹣2k，令r﹣2k=0，对k，r分类讨论即可得出．=，（r=0，1，…，6），【解答】解：T r+1==2k x r﹣2k，令的展开式的通项公式T′k+1令r﹣2k=0，k=0，r=0时，可得：T1=1．k=1，r=2时，可得：T3=，T′2=，∴=60．k=2，r=4时，可得：T5=，T′3==24，∴×24=360．k=3，r=6时，可得：T7=，T′4==160，∴×160=160．∴（x++1）6展开式中的常数项是1+60+360+160=581．故答案为：581．【点评】本题考查了二项式定理的应用，考查了分类讨论方法、推理能力与计算能力，属于中档题．13．（5分）在棱长为1的正方体ABCD﹣A1B1C1D1中，若点P是棱上一点，则满足|PA|+|PC1|=2的点P的个数为6．【考点】L2：棱柱的结构特征．【专题】15：综合题；5F：空间位置关系与距离．【分析】由题意可得点P是以2c=为焦距，以a=1为长半轴，为短半轴的椭圆与正方体与棱的交点，可求．【解答】解：∵正方体的棱长为1∴AC1=，∵|PA|+|PC1|=2，∴点P是以2c=为焦距，以a=1为长半轴，以为短半轴的椭圆，∵P在正方体的棱上，∴P应是椭圆与正方体与棱的交点，结合正方体的性质可知，满足条件的点应该在棱B1C1，C1D1，CC1，AA1，AB，AD 上各有一点满足条件．故答案为：6．【点评】本题以正方体为载体，主要考查了椭圆定义的灵活应用，属于综合性试题．14．（5分）若数列{a n}前n项和S n满足S n﹣1+S n=2n2+1（n≥2，n∈N+），且满足a1=x，{a n}单调递增，则x的取值范围是（2，3）．【考点】8H：数列递推式．【专题】15：综合题；35：转化思想；48：分析法；54：等差数列与等比数列．【分析】根据条件求出与a n的有关的关系式，利用条件，{a n}单调递增，建立条件，即可得到结论．【解答】解：由条件S n+S n=2n2+1（n≥2）得S n+S n+1=2（n+1）2+1，﹣1+a n=4n+2，两式相减得a n+1+a n+1=4n+6，两式再相减得a n+2﹣a n=4，得{a n+2}是公差d=4的等差数列，故a n+2由n=2得a1+a2+a1=9，a2=9﹣2x，从而a2n=4n+5﹣2x；n=3得a1+a2+a3+a1+a2=19，a3=1+2x，从而a2n+1=4n﹣3+2x，由条件得，解得2＜x＜3，故x的取值范围为（2，3），故答案为：（2，3）．【点评】本题主要考查参数的取值范围的求解，根据条件求出与a n的有关的关系式是解决本题的关键，有一定的难度．二、选择题（共4小题，每小题5分，满分20分）15．（5分）平面α的斜线与平面α所成的角是35°，则与平面α内所有不过斜足的直线所成的角的范围是（）A．（0°，35°]B．（0°，90°]C．[35°，90°）D．[35°，90°]【考点】MI：直线与平面所成的角．【专题】38：对应思想；44：数形结合法；5G：空间角．【分析】做出斜线与射影所确定的平面，则当α内的直线与射影平行时．夹角最小为35°，当直线与射影垂直时，夹角最大为90°．【解答】解设平面α的斜线的斜足为B，过斜线上A点做平面α的垂线，垂足为C，则∠ABC=35°，∴当α内的直线与BC平行时，直线与斜线所成的角为35°，当α内的直线与BC垂直时，则此直线与平面ABC垂直，∴直线与斜线所成的角为90°，故选：D．【点评】本题考查了线面角的定义，异面直线所成的角的计算，属于中档题．16．（5分）已知log2x，log2y，2成等差数列，则M（x，y）的轨迹的图象为（）A．B．C．D．【考点】3A：函数的图象与图象的变换．【专题】51：函数的性质及应用．【分析】根据等差中项，得到2log2y=2+log2x，继而得到y2=4x，x＞0，y＞0，问题得以解决．【解答】解：∵log2x，log2y，2成等差数列，∴2log2y=2+log2x，∴y2=4x，x＞0，y＞0，∴M（x，y）的轨迹的图象为焦点为（1，0）的抛物线的一部分，x＞0，y＞0，故选：A．【点评】本题考查了等差中项和对数的运算性质，以及抛物线的问题，属于基础题．17．（5分）设z1，z2∈C，z12﹣4z1z2+4z22=0，|z2|=2，则以|z1|为直径的圆面积为（）A．πB．4πC．8πD．16π【考点】A4：复数的代数表示法及其几何意义．【专题】35：转化思想；49：综合法；5N：数系的扩充和复数．【分析】由已知可得（z1﹣2z2）2=0，因此z1=2z2．再利用|z2|=2，即可得出|z1|．【解答】解：∵﹣4z1z2+4=0，∴（z1﹣2z2）2=0，∴z1=2z2．∴|z1|=2|z2|=4，∴以|z1|为直径的圆的面积=π×（）2=4π．故选：B．【点评】熟练掌握复数的运算和模的意义是解题的关键．18．（5分）方程9x+|3x+b|=5（b∈R）有两个负实数解，则b的取值范囤为（）A．（3，5）B．（﹣5.25，﹣5）C．[﹣5.25，﹣5）D．前三个都不正确【考点】53：函数的零点与方程根的关系．【专题】11：计算题；32：分类讨论；35：转化思想；36：整体思想；51：函数的性质及应用．【分析】化简9x+|3x+b|=5可得3x+b=5﹣9x或3x+b=﹣5+9x，从而讨论以确定方程的根的个数，从而解得．【解答】解：∵9x+|3x+b|=5，∴|3x+b|=5﹣9x，∴3x+b=5﹣9x或3x+b=﹣5+9x，①若3x+b=5﹣9x，则b=5﹣3x﹣9x，其在（﹣∞，0）上单调递减，故当b≤3时，无解，当3＜b＜5时，有一个解，当b≥5时，无解；②若3x+b=﹣5+9x，则b=﹣5﹣3x+9x=（3x﹣）2﹣，∵x∈（﹣∞，0）时，0＜3x＜1，∴当﹣＜b＜﹣5时，有两个不同解；当b=﹣时，有一个解；综上所述，b的取值范围为（﹣5.25，﹣5），故选：B．【点评】本题考查了绝对值方程的解法与应用，属于中档题．三、解答题（共5小题，满分60分）19．（12分）平面外ABC的一点P，AP、AB、AC两两互相垂直，过AC的中点D 做ED⊥面ABC，且ED=1，PA=2，AC=2，连接BP，BE，多面体B﹣PADE的体积是；（1）画出面PBE与面ABC的交线，说明理由；（2）求面PBE与面ABC所成的锐二面角的大小．【考点】MJ：二面角的平面角及求法．【专题】11：计算题；35：转化思想；41：向量法；5G：空间角．【分析】（1）延长PE交AC于F，可证F与C重合，故直线BC即为面PBE与面ABC的交线；（2）以A为原点，AB为x轴，AC为y轴，AP为z轴，建立空间直角坐标系，利用向量法能求出面PBE与面ABC所成的锐二面角的大小．【解答】解：（1）延长PE交AC于F，直线BC即为面PBE与面ABC的交线；理由如下：∵AP、AB、AC两两互相垂直，∴PA⊥平面ABC，∵DE⊥平面ABC，∴DE∥PA，∴=，∴F与C重合．∵C∈PE，C∈AC，PE⊂平面PBE，AC⊂平面ABC，∴C是平面PBE和平面ABC的公共点，又B是平面PBE和平面ABC的公共点，∴BC是面PBE与面ABC的交线．（2）∵AP、AB、AC两两互相垂直，=S梯形ADEP•AB=（1+2）×1×AB=，解得AB=．∴AB⊥平面PAC，∴V B﹣PADE以A为原点，AB为x轴，AC为y轴，AP为z轴，建立空间直角坐标系，B（，0，0），P（0，0，2），E（0，1，1），=（，0，2），=（0，1，﹣1），设二面角PBE的法向量=（x，y，z），则，取y=1，得=（﹣，1，1），平面ABC的法向量=（0，0，1），∴cos＜＞===，∴面PBE与面ABC所成的锐二面角的大小为arccos．【点评】本题考查了平面的性质，二面角的计算，属于中档题，解题时要认真审题，注意向量法的合理运用．20．（12分）已知椭圆C：=1（a＞b＞0）的长轴长是短轴长的两倍，焦距为2．（1）求椭圆C的标准方程；（2）不过原点O的直线l与椭圆C交于两点M，N，且直线OM，MN，ON的斜率依次成等比数列，问：直线l是否定向的，请说明理由．【考点】K3：椭圆的标准方程；KH：直线与圆锥曲线的综合．【专题】11：计算题；35：转化思想；49：综合法；5D：圆锥曲线的定义、性质与方程．【分析】（1）由椭圆的长轴长是短轴长的两倍，焦距为2，列出方程组能求出椭圆C的标准方程．（2）由题意设直线l的方程为y=kx+m，（km≠0），联立，得（1+4k2）x2+4kmx+4（m2﹣1）=0，由此利用根的判别式、韦达定理、等比数列、椭圆性质，结合已知条件能求出直线l不定向．【解答】解：（1）∵椭圆C：=1（a＞b＞0）的长轴长是短轴长的两倍，焦距为2，∴，解得a=2，b=1，∴椭圆C的标准方程为．（2）由题意设直线l的方程为y=kx+m，（km≠0），联立，得（1+4k2）x2+4kmx+4（m2﹣1）=0，△=16（4k2﹣m2+1）＞0，设M（x1，y1），N（x2，y2），则，，∴y1y2=（kx1+m）（kx2+m）=，∵直线OM，MN，ON的斜率依次成等比数列，∴=k2，∴﹣+m2=0，∵m≠0，∴k2=，方向向量=（±2，1）．∴直线l不定向．【点评】本题考查椭圆方程的求法，考查直线是否定向的判断与求法，是中档题，解题时要认真审题，注意根的判别式、韦达定理、等比数列、椭圆性质的合理运用．21．（12分）如图所示，A，B是两个垃圾中转站，B在A的正东方向16千米处，AB的南面为居民生活区，为了妥善处理生活垃圾，政府决定在AB的背面建一个垃圾发电厂P，垃圾发电厂P的选址拟满足以下两个要求（A，B，P可看成三个点）：①垃圾发电厂到两个中转站的距离与它们每天集中的生活垃圾量成反比，比例系数相同；②垃圾发电厂应尽量远离居民区（这里参考的指标是点P到直线AB的距离要尽可能大），现估测得A，B两个中转站每天集中的生活垃圾量分别约为30吨和50吨，设|PA|=5x＞0．（1）求cos∠PAB（用x的表达式表示）（2）问垃圾发电厂该如何选址才能同时满足上述要求？【考点】HR：余弦定理．【专题】33：函数思想；48：分析法；58：解三角形．【分析】（1）由条件可设PA=5x，PB=3x，运用余弦定理，即可得到cos∠PAB；（2）由同角的平方关系可得sin∠PAB，求得点P到直线AB的距离h=PAsin∠PAB，化简整理配方，由二次函数的最值的求法，即可得到所求最大值及PA，PB的值．【解答】解：（1）由条件①，得，∵PA=5x，∴PB=3x，则，可得；（2）由同角的平方关系可得，所以点P到直线AB的距离h=PAsin∠PAB，=，∵cos∠PAB≤1，∴，∴2≤x≤8，所以当x2=34，即时，h取得最大值15千米．即选址应满足千米，千米．【点评】本题考查解三角形的数学模型的解法，注意运用余弦定理和同角的平方关系和二次函数的最值的求法，考查化简整理的运算能力，属于中档题．22．（12分）（1）已知0＜x1＜x2，求证：；（2）已知f（x）=lg（x+1）﹣log3x，求证：f（x）在定义域内是单调递减函数；（3）在（2）的条件下，求集合M={n|f（n2﹣214n﹣1998）≥0，n∈Z}的子集个数．【考点】16：子集与真子集；4N：对数函数的图象与性质．【专题】38：对应思想；49：综合法；51：函数的性质及应用．【分析】（1）使用分析法证明；（2）设0＜x1＜x2，利用（1）的结论和对数函数的性质化简f（x1）﹣f（x2）判断其符号，得出结论；（3）由（2）的结论及f（9）=0列出不等式组，解出n即可得出M中元素的个数．【解答】（1）证明：∵x2+1＞0，x2＞0，欲证：，只需证：x2（x1+1）＞x1（x2+1），即证：x1x2+x2＞x1x2+x1，只需证：x2＞x1，显然x2＞x1成立，∴．（2）解：f（x）的定义域为（0，+∞）．设0＜x1＜x2，则f（x1）﹣f（x2）=lg（x1+1）﹣lg（x2+1）+log3x2﹣log3x1=lg+log3=lg﹣log9．∵0＜x1＜x2，∴0＜＜＜1，∴lg＞log9＞log9，∴f（x1）﹣f（x2）=lg﹣log9＞log9﹣log9=0．∴f（x1）＞f（x2），∴f（x）在定义域（0，+∞）上是减函数．（3）解：由（2）知f（x）是定义在（0，+∞）上的减函数，且f（9）=0，∵f（n2﹣214n﹣1998）≥0，∴0＜n2﹣214n﹣1998≤9．∴13447＜（n﹣107）2≤13456．∵115＜＜116，=116，n∈Z，∴n﹣107=116或n﹣107=﹣116．∴集合M有两个元素．∴集合M有4个子集．【点评】本题考查了不等式的证明，对数函数的性质，函数单调性的应用，属于中档题．23．（12分）数列{a n}，{b n}满足，a1＞0，b1＞0；（1）求证：{a n•b n}是常数列；（2）若{a n}是递减数列，求a1与b1的关系；（3）设a1=4，b1=1，当n≥2时，求a n的取值范围．【考点】8H：数列递推式．【专题】15：综合题；35：转化思想；49：综合法；54：等差数列与等比数列．【分析】（1）由题意可知a n•b n=a n﹣1•b n﹣1=…=a1•b1，故问题得以证明；（2）根据{a n}是递减数列，得到（a1﹣b1）2＞0，a n＞b n，得到a1＞b1恒成立，（3）先判断a n＞2，再根据a n+1﹣a n=，得到a n+1﹣a n＜0，{a n}是递减数+1列，即可得到a n﹣a2＜0，求出a n的取值范围．【解答】解：（1）∵，=a n+b n，=，∴2a n+1∴b n=，+1b n+1=a n•b n，∴a n+1∴a n•b n=a n﹣1•b n﹣1=…=a1•b1，∴{a n•b n}是常数列；（2）{a n}是递减数列，a n+1﹣a n＜0，∵a2﹣a1=（a1+b1）﹣a1=（b1﹣a1）＜0∴a1＞b1，∵a3﹣a2=（b2﹣a2）＜0，∴a2＞b2，∵（a1+b1）＞，∴（a1﹣b1）2＞0，猜想a n﹣a n=（b n﹣a n）＜0，+1∴a n＞b n，∴a1＞b1恒成立，﹣a k+1=（b k+1﹣a k+1）==＜0，∵a k+2∴a1＞b1时，{a n}是递减数列．=（a n+），a1=4，（3）整理得a n+1∴a2=，∴a1＞0⇒a2＞0⇒a3＞0⇒…⇒a n＞0，﹣2=（a n+）﹣2=＞0，当n≥2时，a n+1∴a n＞2，+1﹣a n=（b n﹣a n）==，∴a n+1∵a n＞2，﹣a n＜0，∴a n+1∴{a n}是递减数列，∴a n﹣a2＜0，∴a n∈（2，]【点评】本题考查了递推数列的，常数列，数列的函数特征，以及a n的取值范围，培养了学生的运算能力，转化能力，属于难题．。

2015学年度第一学期奉贤区高三调研化学试卷（考试时间120分钟满分150）2016.01相对原子质量：H—1,O—16 ,Na—23,S—32 ,Cu—64 ,Sn—119 ,I—127 ,Fe—56,N—14 ,Br-80一、选择题（本题共10分，每小题2分，只有一个正确选项。

）1、油脂是重要的工业原料。

关于“油脂”的叙述错误的是A．不能用植物油萃取溴水中的溴B．皂化反应是高分子生成小分子的过程C．和H2加成后能提高其熔点及稳定性D．水解可得到丙三醇2、下列有关氯元素及其化合物的表示正确的是A．质子数为17、中子数为20的氯原子：ClB．四氯化碳分子的比例模型：C．氯分子的电子式：D．氯乙烯分子的结构简式：H3C﹣CH2Cl3、下列有关物质结构的叙述正确的是A．在离子化合物中不可能存在非极性共价键B．由电子定向移动而导电的物质一定是金属晶体C．有键能很大的共价键存在的物质熔沸点一定很高D．只含有共价键的物质不一定是共价化合物4、下列说法正确的是A．分子式为C2H6O的有机化合物性质相同B．相同条件下，等质量的碳按a、b两种途径完全转化，途径a比途径b放出更多热能途径a：C CO+H2CO2+H2O 途径b：C CO2C．食物中可加入适量的食品添加剂，如香肠中可以加少量的亚硝酸钠以保持肉质新鲜D．生石灰、铁粉、硅胶是食品包装中常用的干燥剂5、对于复分解反应：X＋Y Z＋W，下列叙述正确的是A．若Z是强酸，则X和Y必有一种是强酸B．若X是强酸，Y是盐，反应后可能有强酸或弱酸生成C．若Y是强碱，X是盐，则Z或W必有一种是弱碱D．若W是弱碱，Z是盐，则X和Y必有一种是强碱二、选择题（本题共36分，每小题3分，只有一个正确选项）6、“轨道”2Px与3Py上电子一定相同的方面是A、能量B、呈纺锤形C、自旋方向D、在空间的伸展方向7、部分共价键的键长和键能的数据如表，则以下推理肯定错误的是共价键C﹣C C=C C≡C键长（nm）0.154 0.134 0.120键能（kJ/mol）347 612 838A．0.154 nm＞苯中碳碳键键长＞0.134nm B．C=O键键能＞C-O键键能C．乙烯的沸点高于乙烷D．烯烃比炔烃更易与溴加成8、H2SO3水溶液中存在电离平衡H2SO3H+ + HSO3—和HSO3—H+ + SO32—，若向H2SO3溶液中A．通入氯气，溶液中氢离子浓度增大B．通入过量H2S，反应后溶液pH减小C．加入氢氧化钠溶液，平衡向右移动，pH变小D．加入氯化钡溶液，平衡向右移动，会产生亚硫酸钡沉淀9、常温下，下列各组离子一定能在指定溶液中大量共存的是A．使酚酞变红色的溶液中：Na+、Al3+、SO42﹣、Cl﹣B．=1×10﹣13mol•L﹣1的溶液中：NH4+、Ca2+、Cl﹣、NO3﹣C．与Al反应能放出H2的溶液中：Fe2+、K+、NO3﹣、SO42﹣D．水电离的c（H+）=1×10﹣13mol•L﹣1的溶液中：K+、Na+、AlO2﹣、CO32﹣10、下列指定反应的离子方程式正确的是A．氯气溶于水：Cl2+H2O→2H++Cl﹣+ClO﹣B． Na2CO3溶液中CO32﹣的水解：CO32﹣+H2O→HCO3﹣+OH﹣C．酸性溶液中KClO3与KCl反应制得Cl2：ClO3﹣+Cl﹣+6H+→Cl2+3H2OD．NaHCO3溶液中加足量Ba（OH）2溶液：HCO3﹣+Ba2++OH﹣→BaCO3↓+H2O 11、在化学能与电能的转化过程中，下列叙述正确的是A．电解饱和食盐水时，阳极得到Cl2和NaOH溶液B．教材所示的铜－锌原电池在工作时，Zn2＋向铜片附近迁移C．用电解法提取氯化铜废液中的铜，用铜片连接电源的正极，另一电极用铂片D．原电池与电解池连接后，电子从原电池负极流向电解池阳极12、设N A为阿伏伽德罗常数的数值，下列说法正确的是A．23g Na 与足量H2O反应完全后可生成N A个H2分子B．1 molCu和足量热浓硫酸反应可生成N A个SO3分子C．标准状况下，22．4L N2和H2混合气中含N A个原子D．3mol单质Fe完全转变为Fe3O4，失去8N A个电子13、下列装置应用于实验室制氯气并回收氯化锰的实验，其中一个装置能达到某一实验目的是A．用装置甲制取氯气 B.用装置乙除去氯气中混有的少量氯化氢C．用装置丙分离二氧化锰和氯化锰溶液 D.用装置丁蒸干氯化锰溶液制MnCl2•4H2O14、25℃时，在10mL浓度均为0.1mol/LNaOH和NH3·H2O混合溶液中，滴加0.1mol/L的盐酸，下列有关溶液中粒子浓度关系正确的是：A．未加盐酸时：c(OH－)＞c(Na＋)= c(NH3·H2O)B．加入10mL盐酸时：c(NH4＋) ＋c(H＋) ＝c(OH－)C．加入盐酸至溶液pH=7时：c(Cl－) = c(Na＋)D．加入20mL盐酸时：c(Cl－) ＝c(NH4＋) ＋c(Na＋)15、下列设计的实验方案能达到实验目的的是A．制备Al（OH）3悬浊液：向1mol•L﹣1AlCl3溶液中加过量的6mol•L﹣1NaOH溶液B．提纯含有少量乙酸的乙酸乙酯：向含有少量乙酸的乙酸乙酯中加入过量15%的Na2CO3溶液，振荡后静置分液，并除去有机相的水C．检验溶液中是否含有Fe2+：取少量待检验溶液，向其中加入少量新制氯水，再滴加KSCN溶液，观察实验现象D．探究催化剂对H2O2分解速率的影响：在相同条件下，向一支试管中加入2mL5%H2O2和1mLH2O，向另一支试管中加入2mL5%H2O 2和1mLFeCl3溶液，观察并比较实验现象16、某同学将光亮的镁条放入盛有NH4Cl溶液的试管中，有大量气泡产生。

上海市奉贤区2014届高三下学期4月调研测试化学试卷20140418考生注意：1.本试卷满分150分，考试时间120分钟。

2.本考试设试卷和答题纸两部分，试卷包括试题和答题要求；所有答案必须凃或写在答题纸上；做在试卷上一律不得分。

3.答题前，考生务必将答题纸上用钢笔或圆珠笔在答题纸正面清楚地填写姓名、准考证号，并将核后的条形码贴在指定位置上，在答题纸反面清楚地填写姓名。

4.答题纸与试卷在试题编号上是一一对应的，答题时应特别注意，不能错位。

相对原子质量：H-1 C-12 O-8 Na-23 S-32 Ca-40 Fe-56 Ni-59 Cu-64 Br-80 Ba-137 N—14一、选择题（本题共10分，每小题2分，每题只有一个正确选项）1、2013年6月11日，神舟十号由长征二号F改进型运载火箭成功发射，其后完成了与天宫一号的对接任务，实现了中国航天史上首次太空授课。

下列操作在太空舱可以实现的是A．过滤B．分液C．氧化铁粉和铝粉混合D．酸碱中和滴定2、下列有关化学用语描述正确的是A．钠离子的轨道排布式：B．硫离子的结构示意图：C．NH4Cl的电子式：D．对氯甲苯的结构简式：3、科学家根据自然界存在的N2制取N3，其后又陆续制取出N5、含N5+的化合物及N60。

N5+极不稳定，需保存在-80℃的干冰中；N5+由于其极强的爆炸性，又称为“盐粒炸弹”；N60与C60结构相似，并在受热或机械撞击后，其中积蓄的巨大能量会在瞬间释放出。

分析上述材料，下列说法中不正确的是A.+5N常温下会剧烈爆炸，体积急剧膨胀，放出大量的热B.N60的发现开辟了能世界的新天地，将可能成为很好的火箭燃料C.N2、N3、N5、N5+、N60互为同素异形体D.含N5+的化合物中既有离子键又有共价键4、A、B、C、D、E为原子序数相邻且依次递增的同一短周期元素，下列说法正确的是(m、n均为正整数)A．若HnEOm为强酸，则D是位于VA族以后的非金属元素B．若C的最低化合价为－3，则E的气态氢化物的分子式为H2EC．A、B的最高价氧化物水化物均为碱，则碱性A(OH)n强于B(OH)n＋1D．若B为金属，则C一定为金属元素5、在测定液态BrF3导电性时发现，20℃时导电性很强，其他实验证实存在一系列有明显离子化合物倾向的盐类，如KBrF4，(BrF2)2SnF6等，由此推测液态BrF3电离时的阳、阴离子是A．BrF2+，BrF4- B．BrF2+，F- C．Br3+，F- D．BrF2+，BrF32-二、选择题（本大题共36分，每小题3分，每题只有一个正确选项）6、有一种军用烟幕弹中装有ZnO 、Al 粉和C2Cl6，其发烟过程中的化学反应如下：① 3ZnO + 2Al → Al2O3 + 3Zn ② 3Zn + C2Cl6 → 3ZnCl2 + 2C下列有关叙述不正确的是A. 反应①是铝热反应B. 反应②是置换反应C. C2Cl6属于卤代烃D. 烟幕是小液滴分散在空气中形成的7、设NA 代表阿伏加德罗常数（NA ）的数值，下列说法正确的是A ．乙烯和环丙烷（C3H6 ）组成的28g 混合气体中含有3NA 个氢原子B ．1 mol 碳酸钾晶体中含阴离子数为NA 个C ．标准状况下，22.4L 氯气与足量氢氧化钠溶液反应转移的电子数为2NAD ．将0.1mol 氯化铁溶于1L 水中，所得溶液中含有0.1NA Fe3+8、已知：① 2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ ；② H2(g) + S(g) → H2S(g) + 20.1 kJ 。

奉贤区2018学年第二学期高三年级质量调研考试化学试卷相对原子质量：O-16 S-32一、选择题(共40分，每小题2分，每小题只有一个正确答案)1.我国导航卫星的“心脏”使用的是铷原子钟，下列关于铷的说法正确的是A. 85Rb和87Rb互称为同素异形体B. 86Rb和87Rb具有相同的中子数C. 原子核外电子数是37D. 的质子数是87【答案】C【解析】【分析】85Rb和87Rb质量数分别为85、87，质子数均为37，利用质子数=电子数，质子数+中子数=质量数及同位素的概念来解答。

【详解】A. 85Rb和87Rb的质子数相同个，中子数不同，互称为同位素，A项错误；B. 86Rb和87Rb具有相同的质子数，不同的中子数，B项错误；C. 原子核外电子数=质子数=37，C项正确；D. 的质子数是37，质量数为87，D项错误；答案选C。

【点睛】元素符号的含义要牢记在心，其左上角为质量数（A）、左下角为质子数（Z），质子数（Z）=核电荷数=原子序数，质量数（A）=质子数（Z）+中子数（N）。

这是化学用语最基本知识点，需要牢记在心。

2.下列说法错误的是A. 发展太阳能发电、利用CO2制造全降解塑料都能有效减少环境污染B. 从2019年7月1日起我国全面实施垃圾分类，提高废品回收率，减少对环境影响C. 绿色化学又称“环境无害化学”、“环境友好化学”、“清洁化学”D. 石油裂化、裂解为化学变化，而煤的气化和石油的分馏均为物理变化【答案】D【解析】【详解】A. 发展太阳能发电是新能源发电，可减少环境污染，利用CO2制造全降解塑料，可降低CO2的排放，减少温室效应，有效减少环境污染，A项正确；B. 垃圾分类，提高废品回收率，减少对环境影响，B项正确；C. 绿色化学是研究利用一套原理在化学产品的设计、开发和加工生产过程中减少或消除使用或产生对人类健康和环境有害物质的科学，又称“环境无害化学”、“环境友好化学”、“清洁化学”，C项正确；D. 石油的裂化、裂解和煤的气化均为化学变化，只有石油的分馏属于物理变化，D项错误；答案选D。

奉贤区 2018 学年第二学期高三年级质量调研考试化学试卷相对原子质量：O-16 S-32一、选择题 ( 共 40 分，每题 2 分，每题只有一个正确答案)1.我国导航卫星的“心脏”使用的是铷原子钟，以下对于铷的说法正确的选项是A. 85Rb 和87Rb 互称为同素异形体C.原子核外电子数是37B.86Rb和 87Rb拥有同样的中子数D.的质子数是87【答案】 C【分析】【剖析】85Rb 和87Rb 质量数分别为 85、87，质子数均为 37，利用质子数 =电子数，质子数+中子数 =质量数及同位素的观点来解答。

【详解】 A.85Rb和87Rb的质子数同样个，中子数不一样，互称为同位素， A 项错误；B. 86Rb 和87Rb拥有同样的质子数，不一样的中子数， B 项错误；C.原子核外电子数 =质子数 =37， C项正确；D.的质子数是 37，质量数为 87， D 项错误；答案选 C。

【点睛】元素符号的含义要切记在心，其左上角为质量数（A）、左下角为质子数（Z），质子数（ Z） =核电荷数 =原子序数，质量数（A）=质子数（ Z） +中子数（ N）。

这是化学用语最基本知识点，需要切记在心。

2.以下说法错误的选项是A. 发展太阳能发电、利用CO2制造全降解塑料都能有效减少环境污染B.从 2019 年 7 月 1 日起我国全面实行垃圾分类，提升废品回收率，减少对环境影响C.绿色化学又称“环境无害化学”、“环境友善化学”、“洁净化学”D.石油裂化、裂解为化学变化，而煤的气化和石油的分馏均为物理变化【答案】 D【分析】【详解】 A.发展太阳能发电是新能源发电，可减少环境污染，利用CO2制造全降解塑料，可降低CO2的排放，减少温室效应，有效减少环境污染， A 项正确；B. 垃圾分类，提升废品回收率，减少对环境影响， B 项正确；C.绿色化学是研究利用一套原理在化学产品的设计、开发和加工生产过程中减少或除去使用或产生对人类健康和环境有害物质的科学，又称“环境无害化学”、“环境友善化学”、“洁净化学”， C 项正确；D. 石油的裂化、裂解和煤的气化均为化学变化，只有石油的分馏属于物理变化，D项错误；答案选 D。

2015学年第二学期奉贤区高三英语调研测试（2016.04）（完卷时间120分钟，满分150分）第I卷(共103分)I. Listening ComprehensionSection ADirections:In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. 7:00 a.m.. B. 7:30 a.m.. C. 8:00 a.m.. D. 8:30 a.m..2. A. Parent and child. B. Husband and wife.C. Teacher and student.D. Shop assistant and customer.3. A. See a doctor about her strained shoulder.B. Use a ladder to help her reach the tea.C. Replace the cupboard with a new one.D. Place the tea on a lower shelf next time.4. A. At Mary Johnson’s. B. In an exhibition hall.C. At a painter’s studio.D. Outside an art gallery.5. A. He helped Doris build up the furniture.B. Doris helped him arrange the furniture.C. Doris fixed up some of the bookshelves.D. He was good at assembling bookshelves.6. A. Blue. B. Red. C. Black. D. Green.7. A. He doesn’t get on with the others. B. He has been taken for a fool.C. He doesn’t feel at ease in the firm.D. He has found a better position.8. A. They’d better not go riding.B. It is not good riding in the rain.C. They can go riding half an hour later.D. Riding a bike is a great idea.9. A. The man has to get rid of the used furniture.B. The man’s apartment is ready for rent.C. The furniture is covered with lots of dust.D. The furniture the man bought is inexpensive.10.A. The man tells the woman how to get to a cinema.B. The woman lost her way in the street.C. The woman wants to know how to get to Joe’s house.D. The man tells the woman how to get to a nearest snack bar.Section BDirections: In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. She was a 19-month-old British girl.B. The hospitals in Qatar were full at that time.C. She was the daughter of a doctor in London.D. The Qatar doctors were unsure how to cure her.12. A. A doctor in Qatar. B. A nurse in London.C. Dr. Brown.D. Agatha Christie.13. A. Substance used in making glass. B. Drug found in a detective story.C. Medicine often used in hospital.D. A deadly poison easily got in Qatar.Questions 14 through 16 are based on the following passage.14. A. Ignore small details while reading.B. Read at least several chapters at one sitting.C. Develop a habit of reading critically.D. Get key information by reading just once or twice.15. A. Choose one’s own system of marking.B. Underline the key words and phrases.C. Make as few marks as possible.D. Highlight details in a red color.16. A. By reading the textbooks carefully again.B. By reviewing only the marked parts.C. By focusing on the notes in the margins.D. By comparing notes with their classmates.Section CDirections: In Section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet.Blanks 17 through 20 are based on the following conversation.Complete the form. Write ONE WORD for each answer.Blanks 21 through 24 are based on the following conversation.II. Grammar and vocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.AYour next Disney obsession has arrived! Zootopia（疯狂动物城）hit theaters on March 4, and it’s fun at the movies for the entire family. If you’re in the mood to smile, Zootopia is the movie for you. The new Disney movie __25_____ (release) on March 4, starring Jason Bateman, Ginnifer Goodwin, Idris Elba and more. Critics are obsessed __26______ Zootopia-- Check out __27______ the critics are saying about Zootopia now!---Zootopia is the best Disney animated film that I have seen in a long time. This is a movie that works for everyone. __28_____ doesn't matter whether you're 4 years old or 40 years old.---I love the characters. They do such a great job of building up Judy Hopps（兔子朱迪）from the moment she comes on screen and I think I will remember her as one of my favorite Disney characters.---Zootopia is smart, funny, and is mature in its storytelling and in its message. It’s very nice to see a movie directed at kids that has respect and treats kids as people who __29____ understand complex themes, not mindless creatures that need simple cliched（刻板的）messages.---I predict that Zootopia will be one of the __30_____ (well-known) animated films of the following years. The film does what every animated film should do; it tells a solid story with a lot of emotional depth, and a message that can hopefully teach kids important lessons. The film is also funny, smart, beautifully animated, __31_____ (fill) with great characters, and features great voice work. If you have kids, and they __32_____ (not see) this film yet, it’s safe to say they’d enjoy it, but they may possibly take an important life lesson away from seeing it too.BLeonardo DiCaprio（莱昂纳多·迪卡普里奥）was born in Hollywood. His parents met while ___33_____ (attend) college and then moved to Los Angeles. He was named Leonardo because his pregnant mother was looking at a Leonardo da Vinci painting in a museum __34_____ DiCaprio first kicked.He dropped out of high school following his third year, eventually __35______ (earn) his general equivalency diploma (GED).DiCaprio made his big screen breakthrough in 1992, __36_____ he was chosen by Robert DeNiro out of 400 young actors to play the lead role in This Boy's Life.In 1997, DiCaprio starred in James Cameron's Titanic(1997) as twenty-year-old Jack Dawson（泰坦尼克号男主人公）. The film became the highest-grossing film to date, and his face appeared on the covers of at least fourteen magazines.DiCaprio was also a dedicated environmentalist. In November 2010, DiCaprio donated $1 million to the Wildlife Conservation Society at Russia's tiger summit. DiCaprio's persistence in reaching the event after encountering two plane delays caused Prime Minister Vladimir Putin __37_____ (describe) him as a "real man".At the 2016 Oscar ceremony, DiCaprio’s __38_____ (win) the award for Best Actor impressed all the audience present. He expressed his appreciation and worry for the environment when he said: Climate change is real, and it is happening right now. It is the most urgent threat __39______ (face) our entire species, and we need to work collectively together and stop procrastinating（拖延）. We need to support leaders around the world who do not speak for the big polluters, but who speak for all of humanity, for the indigenous people of the world, for the billions and billions of underprivileged people out there who would be most affected by this, f or our children’s children, and for those people out there __40_____ voices have been drowned out by the politics of greed. Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.（Since 1952, the Queen's Christmas message has been televised in some form. The following is the one given by Britain's Queen Elizabeth II on December 25th, 2015. ）At this time of year, few sights arouse more feelings of __41____ and goodwill than the twinkling lights of a Christmas tree.The popularity of a tree at Christmas is __42______ due to my great-great grandparents, Queen Victoria and Prince Albert. After this __43_____ picture was published, many families wanted a Christmas tree of their own, and the custom soon spread.In 1949, I spent Christmas in Malta as a newly-married naval wife. We have returned to that island over the years, including last month for a meeting of Commonwealth（英联邦）leaders; and this year I met another group of leaders: The Queen’s Young Leaders, an inspirational group, each of them a __44_____ of hope in their own Commonwealth communities.Actually, ___45____ round the tree gives us a chance to think about the year ahead. It also allows us to ___46____ on the year that has passed, as we think of those who are far away or no longer with us. Many people say the first Christmas after losing a(an) __47______ one is particularly hard. But it’s also a time to remember all that we have to be thankful for. We should be thankful for the people who bring love and happiness into our own lives, and look for ways of __48_____ that love to others, whenever and wherever we can.One __49______ for thankfulness this summer was marking seventy years since the end of the Second World War. On VJ Day, we __50_____ the remaining veterans(老兵) of that terrible conflict in the Far East, as well as remembered the thousands who never returned.…I wish you a very happy Christmas.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Two Newcastle scientists are setting themselves to open our eyes to the medical truth by claiming that natural sunlight may help prevent skin cancer.Dr. Ron Laura, professor of health education at Newcastle University, and senior chemist Mr. John Ashton said their research points to a complete __51______ of the accepted scientific theory. They said that sunscreen creams may help cause skin cancer, the artificial indoor light could be __52_____ and that a range of drugs in common use could also ___53____ melanoma--a type of cancer that appears as a dark spot on the skin.The research is likely to be unwelcome in some traditional medical research circles. It is based on a new __54_____ that our bodies are protected from skin cancer by the regulation of a group of complex vitamins (Vitamin D) and immune process.The sunscreens, artificial light and drugs could all unfavorably affect the production of these vitamins and increase the skin’s __55_____ to the sun. But Dr. Laura said natural sunlight passing through the eyes helped __56_____ the production of cancer protection Vitamin D.He said recent statistics from the United States indicated that people who worked indoors all day in artificial light were more __57_____ melanomas than those who worked outdoors. Indoor workers should try to have at least one hour of __58_____ to direct sunlight every day, ___59_____ in the early morning and late afternoon when ultraviolet intensively was lower, Dr. Laura said.Sunscreens, long __60_____ as essential for beach lovers, could also __61______ the production of Vitamin D. Laura and Ashton said sunscreens give people a __62_____ sense of security in thinking they are __63______ from the sun’s rays.Dr. Laura said more statistics ___64_____ their claim had come to light since the first article was published. He believes his research findings are too important to be __65______ to the scientific world.51. A. contribution B. reversal C. combination D. recognition52. A. beneficial B. comfortable C. harmful D. favorable53. A. promote B. reduce C. remove D. eliminate54. A. assumption B. law C. concept D. theory55. A. sensitivity B. resistance C. adaptation D. response56. A. monitor B. measure C. slow D. stimulate57. A. subject to B. unrelated to C. free of D. dependent on58. A. exercise B. reveal C. exposure D. experience59. A. occasionally B. preferably C. enjoyably D. extremely60. A. received B. popular C. accepted D. identified61. A .balance B. adjust C. prevent D. enhance62. A. false B. strong C. true D. sharp63. A. separated B. protected C. guarded D. prohibited64. A. presenting B. doubting C. backing D. providing65. A. limited B. emphasized C. acknowledged D. explainedSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.AAfter researching the history of the Vietnam War, I called my mum and asked her if she knew anyone who went to Vietnam that I could interview. She thought for a while and suddenly remembered that she has a cousin who is a veteran of the Vietnam War.She says, “I know him since I was little. He was one of the greatest athletes of the school. You can not imagine how good and fast he was. Well, he was ranked number 18 in the country for being an outstanding track and field runner. But now, those things are only memories.”I called him and introduced myself. He was in an agreeable mood at the beginning, but as soon as I began to question him, his attitude changed. “What happened in Vietnam?” I asked. There was a moment of silence on the other line, and then he said he was willing to tell me about Vietnam.He said that he was drafted when he was 20 years old and that the two years later he spent there a part of his life he would rather forget. He said, “The problem is that you will never imagine how much suffering a nd pain I saw in that place.” There is not one book or article that can really describe the human disaster that took place there. There is nothing worse in this world than killing a man who you know has a family. It is very sad, but it is the truth, and it turns more complex when you realize you were part of that truth.When I returned to Puerto Rico, it was a total disaster: young kids without fathers, wives without husbands. Most of those who made it back have no legs, like me, or no arms. I was praised because of my bravery, but for me, that was and is pure nonsense, because that war decided my future, decided the future of my family. I, now, am just a veteran who has nothing. The thing that bothers me the most is that the people who decided to fight will probably never know that it is likely to kill a man, or feel pain and suffering from hunger and the absence of love. In war, every minute you are fearing because the only thing you have in your mind is that if you don’t kill first you are going to get kil led.66.Why did the author’s mother say “T hose things are only memories”?A. Because she would rather keep what happened in the past as a secret.B. Because her cousin is no longer a vigorous young man.C. Because though her cousin was a good runner, he lost his leg during the war.D. Because time passed quickly and she can’t remember much of the past days.67.Which of the following statements is TRUE according to the 4th paragraph?A. Books and articles all presented a false picture of the war.B. It takes mental strength to survive the war.C. The sufferings during the war greatly damaged the memory of soldiers,D. The author’s uncle felt very painful when he realized the truth of the war.68.Why did the author’s uncle find the praises “pure nonsense”?A. Because praises came too late.B. Because no praise could make up for his loss.C. Because he didn’t kill anyone during the war.D. Because too many praises seemed worthless.69.What can we infer from the last paragraph?A. Those who decide to fight should take part in the war by themselves.B. The veterans could stand any hardship in their life after the war.C. Those who made the war should be severely blamed.D. Firing first is the best policy on the battlefield.BLearn To Speak French With Rocket French!Who Wants to Learn to Speak French Fluently in the Shortest Possible Time? If You Answered “I Do!” Then Please Read on to Try My FREE 6-Day French Course.Why do you want to learn to speak French?★Do you live in a French-speaking country and want to communicate better?★Are you traveling to a French-speaking country?★Are you a home-schooled student or a parent who wants your children to learn more quickly and easily?★Are you a student who wants to get an A in French?★Have you learned French before and want a fun refresher course?Or, perhaps you just have an interest in learning the language of love!I’ve designed Rocket French Premium to be the easiest to follow, fastest system for learning how to speak French available. Rocket French Premium is an interactive course that makes you want to study. Also, it’s practical. You are going to be able to speak at a restaurant, at an airport, and with new friends!It’s a great experience to be able to speak with others in a dif ferent language. You will be able to enter into a different culture, a different world! Being bilingual is a very special ability, and it’s a gift that we want to give to you.So are you ready to get to know the secret of learning a new language? You’re lo oking right at it.Try our free six-day course. If you don’t, you’ll be missing a valuable opportunity to see just how much Rocket French Premium can improve your language level. Thousands of people worldwide have used our unique multimedia course to fast-track their French learning, while having piles of fun in the process. Will you be next?Your e-mail address is required for you to receive the FREE course. You can unsubscribe any time and your e-mail address will never be given to any third party.70. Who are target learners of Rocket French Premium?A. Students of French language.B. Language experts doing research into French.C. Teachers who are eager to improve their students’ French.D. Parents who want their children to learn French more quickly and easily.71. Rocket French Premium describes itself as ______.A. free and funnyB. practical and interactiveC. slow but efficientD. suitable for everyone72. The underlined word” fast-track” probably m eans _______ .A. speed upB. pick upC. influenceD. change73. According to the text, which of the following statements is TRUE?A. A complete Rocket French course lasts for 6 days.B. Thousands of people worldwide have benefited from Rocket French Premium.C. People will enroll in Rocket French Premium for different reasons, but everyone will begiven a gift eventually.D. Rocket French Premium mainly aims to introduce second-language learners to Frenchculture.CWhat we know of prenatal development makes all this attempt made by a mother to mold the character of her unborn child by studying poetry, art, or mathematics during pregnancy seem totally impossible. How could such extremely complex influences pass from the mother to the child? There is no connection between their nervous systems. Even the blood vessels of mother and child do not join directly. An emotional shock to the mother will affect her child, because it changes the activity of her glands (腺体) and so the chemistry of her blood. Any chemical change in the mother's blood will affect the child for better or worse. But we can not see how a looking for mathematics or poetic genius can be dissolved in blood and produce a similar liking or genius in the child.In our discussion of instincts we saw that there was reason to believe that whatever we inherit must be of some very simple sort rather than any complicated or very definite kind of behavior. It is certain that no one inherits a knowledge of mathematics. It may be, however, that children inherit more or less of a rather general ability that we may call intelligence. If very intelligent children become deeply interested in mathematics, they will probably make a success of that study.As for musical ability, it may be that what is inherited is an especially sensitive ear, a peculiar structure of the hands or the vocal organs connections between nerves and muscles that make it comparatively easy to learn the movements a musician must execute, and particularly vigorous emotions. If these factors are all organized around music, the child may become a musician. The same factors, in other circumstance might be organized about some other center of interest. The rich emotional equipment might find expression in poetry. The capable fingers might develop skill in surgery. It is not the knowledge of music that is inherited, then nor even the love of it, but a certain bodily structure that makes it comparatively easy to acquire musical knowledge and skill. Whether that ability shall be directed toward music or some other undertaking may be decided entirely by forces in the environment in which a child grows up.74.Which of the following statements is not true?A. Some mothers try to influence their unborn children by studying art and other subjectsduring their pregnancy.B. It is totally impossible for us to learn anything about prenatal development.C. The blood vessels of mother and child do not join directly.D. There are no connection between mother's nervous systems and her unborn child's.75.A mother will affect her unborn baby on the condition that ________.A. she is emotionally shockedB. she has a good knowledge of inheritanceC. she takes part in all kind of activitiesD. she sticks to studying76.According to the passage, a child may inherit _______.A. everything from his motherB. a knowledge of mathematicsC. a rather general ability that we call intelligenceD. her mother’s musical ability77.Which of the following is the best title for the passage?A. Role of InheritanceB. An Unborn ChildC. Function of instinctsD. Inherited TalentsSection CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.Whether it is “women and children first” or “every man for himself” in a shipwreck may depend on how long it takes the ship to sink, researchers said recently．When the Lusitania was torpedoed (用鱼雷袭击) by a German ship in 1915, it sank in 18 minutes and the majority of the survivors were young men and women who responded immediately to their powerful survival instincts．But when the Titanic struck an iceberg in 1912, it took three hours to go down, allowing time for more civilized behavior to take control--and the majority of the survivors were women, children and people with young children．Economist Benno Torgler of the Queensland University of Technology in Australia and his colleagues studied the two sinkings in order to explore the economic theory that people generally behave in a “rational” and selfish manner.The two tragedies provided a “natural experiment” for testing the idea, because the passengers on the two ships were quite similar in terms of gender and wealth．The major difference was how long it took the ships to sink.They suggested that when people have little time to react, instincts may rule.When more time is available, social influences play a bigger role.B ut psychologists noted that many factors other than following social norms (社会规范)could come into play in a disaster, including an evolutionary urge to save the species, attachments that are formed between individuals during the event and the leadership of authority figures．The extent of altruism（利他主义）and how it occurs “is a very controversial issue,”said Anthony R.Mawson, a professor of preventive medicine at the University of Mississippi Medical Center.He thinks the dominant response was attachment behavior．Psychologist Daniel Kruger of the University of Michigan, US thinks that the answer lies less in social norms and more in our evolutionary heritage.Human beings have a deep instinct to preserve our kind, he said, and that means “people are more likely to save those who have higher reproductive value, namely the young and women in child-bearing years”．Kruger also stressed the importance of leadership during a disaster, noting that the Titanic’s captain appeared to have greater control than the Lusitania’s．(Note: Answer the questions or complete the statements in NO MORE THAN 12 WORDS)78.According to Benno Torgler, what led to the different results between the two shipwrecks?_________________________________________________________________________79.Besides social norms and leadership, what other factors play a part in disaster behavior?________________________________________________________________________80.According to Daniel Kruger, Why do the young and women of child-bearing age take thepriority to survive?___________________________________________________________________________81.What does the passage mainly tell us?___________________________________________________________________________第II卷（共47分）I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1、这家超市的特色是24小时服务。

上海市奉贤区2017届年第二学期高中等级考质量抽测物理试卷考生注意：试卷满分100分，考试时间60分钟。

本考试分设试卷和答题纸，试卷包括三部分，第一部分为单项选择题，第二部分为填空题，第三部分为综合题。

作答必须涂或写在答题纸上相应的位置，在试卷上作答无效。

一、单项选择题（共40分，1至8题每小题3分，9至12题每小题4分。

每小题只有一个正确选项）1．力的单位N用基本单位可表示为（）（A）Pa·m2（B）A·m·T （C）J/m （D）kg·m/s22．下列电磁波光子能量最强的是（）（A）紫外线（B）X射线（C）可见光（D）红外线3．原子核23490Th表示（）（A）核外有90个电子（B）核内有234个质子（C）核内有144个中子（D）核内有90个核子4．如图所示，两个半径不同的轮子摩擦传动不打滑，则具有相同角速度的两点是（）（A）A和B （B）C和B（C）A和D （D）A和C5．放射性元素发生β衰变时（）（A）释放出电子，但核内不存在电子（B）释放出电子，说明核内存在电子（C）释放出氦原子核，但核内不存在氦核（D）释放出氦原子核，说明核内存在氦核6．汽车以额定功率上坡时，为增大牵引力，应使汽车的速度（）（A）减小（B）增大（C）先增大后减小（D）先减小后增大7．如图为某一弹簧振子做简谐振动的图像，在t1到t2时间内下列物理量变小的是（）（A）位移（B）速度（C）回复力（D）振幅8．已知地球的质量为M，引力常量为G，设飞船绕地球做匀速圆周运动的轨道半径为r，则飞船在圆轨道上运行的速率为（）（A）GMr（B）rGM（C）GMr（D）MGr9．如图所示，a、b是等量异种点电荷连线的中垂线上的两点，则某检验电荷在a、b处（）（A）电场力大小相等方向相同（B）电场力大小不等方向相反（C）电势能相同（D）电势能不同10．一列沿x轴正方向传播的横波，t＝0时刻的波形如图中实线所示，t＝0.2s时刻的波形如图中虚线所示，则此时质点P（）（A）运动方向向右（B）运动方向向上（C）周期一定为0.8s （D）周期可能为0.16s11．如图所示，条形磁铁用细线悬挂在O点。

2013学年奉贤区化学二模考试答案一、选择题（本题共10分，每小题2分，每小题只有一个正确选项）题号 1 2 3 4 5答案 C B C C A二、选择题（本题共36分，每小题3分，每小题只有一个正确选项）题号 6 7 8 9 10 11 12 13 14 15 16 17 答案 D B D A C D B C D A B B三、选择题（本题共20分，每小题4分，每小题有一个或两个正确选项。

只有一个正确选项的，多选不给分；有两个正确选项的，选对一个给2分，选错一个该小题不给分）题号18 19 20 21 22答案CD AB B B BD四、（12分）23.第三周期第IIA族；5种；4种（每空1分）；SiO2 + 2NaOH →Na2SiO3 + H2O（1分）24.c（1分）25．HZ +OH-→ H2O + Z-（2分）； 1×10-5（1分）；HY＞HZ＞HX（1分）26．①图1（1分）②BD，BE（各1分，全对才给分共2分）五、（12分）27.SO2、NO2（1分，全对给分）28. （2分）；②a 、d（2分）；29.①放热②K A=K B>K C（每空1分）30.d（1分）31.d （2分）32._2_Fe3+ +_8H2O+_3_Cl2→_2_FeO42-+_6_NO3-+ 6Cl- +16H+六、（12分）33.Fe(OH)3和Al(OH)3（1分）；AC（2分）34. 蒸发皿、玻璃棒（坩埚钳）（1分）35.防止温度过高H2O2发生分解（1分）36.①排尽装置中的二氧化碳（1分）②防止空气中的CO2与Ba(OH)2溶液反应（1分）③90%（2分），偏大④是（1分）；B、C装置中，为使分液漏斗中的液体顺利流出，用橡皮管将分液漏斗与锥形瓶、分液漏斗与集气瓶连接成连通装置。

（即恒压漏斗）（答案合理即可，2分）七、（12分）37.冷凝、回流（全对1分）38.abc （2分，2对1错1分，2错1对不给分） 39.Br 2（1分）40.①苯（或四氯化碳等有机溶剂）、硝酸银或石蕊试液（每空1分，答出一种即可） ②不正确；溴乙烷中也含乙基（2分，判断1分，原因1分） 41.c （1分）42.液体不分层(均匀溶液)（1分） 43.吸收乙醇（HBr ）；溴水（各1分） 八、（10分）44.氯原子、羟基（1分，全对给分） 45.氢氧化钠溶液，加热（1分）（2分）46. ①bc （2分） ②；（2分）③、、（2分）九、（12分）47、C 1分 48、氧化反应 1分49、H 3C －－NH 2＋(CH 3CO)2OH 3C －－NH －C －CH 3O＋CH 3COOH 2分氨基易被氧化，在氧化反应之前需先保护氨基 1分 （或其它合理答案也给分） 50、Br 2 / FeBr 3 或Br 2 / Fe 1分51、CHONHOHOHNHCHOOCHONH 22分52、H 2C ＝CH 2----------→H 2O催化剂，△CH 3CH 2OH -------→O2Cu ，△CH 3CHO－CHO NaOH－CH －CH 2－CHOOH---→△－CH ＝CH －CHO 4分十、（14分）53. 33.2, 1800 （2+2）分 54. 10.64L 2分55. 8 62.5 （2+2）分 56 1.8 4分。

虹口区2016年化学学科高考练习题2016 4(满分150分，考试时间120分钟)相对原子质量：H-1 C-12 N-14 O-16 Na-23 Al-27 S-32一、选择题（本题共10分，每小题2分，每题只有一个正确选项）1．核反应堆中存在三种具有放射性的微粒X 23894、Y 24094、Z 23892,下列说法正确的是（ ） A ．X 23894与Y 24094互为同素异形体 B ．X 23894与Z 23892互为同素异形体 C ．Y 24094与Z 23892具有相同中子数 D ．X 23894与Z 23892具有相同化学性质2．能证明BF 3为平面三角形而不是三角锥形分子的理由是（ ）A ．BF 2Cl 只有一种结构B ．三根B-F 键间键角都为120ºC ．BFCl 2只有一种结构D ．三根B-F 键键长都为130pm3．炼铁、炼钢过程中，先被氧化后被还原的元素是（ ）A ．炼铁过程中的铁元素B ． 炼铁过程中的氧元素C ．炼铁过程中的碳元素D ． 炼钢过程中的铁元素4．下列试剂不会因为空气中的二氧化碳和水蒸气而变质的是（ ）A ．Na 2CO 3B ．Na 2O 2C ．CaOD ．Ca(ClO)25. 化学反应的实质是（ ）A ．能量的转移B ．旧化学键断裂和新化学键生成C ．电子转移D ．原子种类与原子数目保持不变二、选择题（本题共36分，每小题3分，每题只有一个正确选项）6．硅与某非金属元素X 的化合物具有高熔点高硬度的性能，X 一定不可能是（ ）A ．ⅣA 族元素B ．ⅤA 族元素C ．ⅥA 族元素D ．ⅦA 族元素7．侯氏制碱法中，对母液中析出NH 4Cl 无帮助的操作是（ ）A ．通入CO 2B ．通入NH 3C ．冷却母液D ．加入食盐8．将一小块钠投入足量水中充分反应，在此过程中没有发生的是（ ）A ．破坏了金属键B ．破坏了共价键C ．破坏了离子键D ．形成了共价键9．下列颜色变化与氧化还原反应无关的是（ ）A ．长期放置的苯酚晶体变红B ．硝酸银晶体光照后变黑C ．氢氧化亚铁变灰绿再变红褐D ．二氧化碳气体冷却后变淡10．在测定硫酸铜晶体结晶水含量的实验中，会导致测定结果偏低的是（ ）A ．加热后固体发黑B ．坩埚沾有受热不分解的杂质C ．加热时有少量晶体溅出D ．晶体中混有受热不分解的杂质11．一种香豆素的衍生物结构如图所示，关于该有机物说法正确的是（ ）A ．该有机物分子式为 C 10H 10O 4B ．1mol 该有机物与 H 2 发生加成时最多消耗H 2 5 molC ．1mol 该有机物与足量溴水反应时最多消耗Br 2 3 molD ．1mol 该有机物与 NaOH 溶液反应是最多消耗NaOH 3 mol12．在铁质品上镀上一定厚度的锌层，以下电镀方案正确的是（ ）A ．锌作阳极，铁制品作阴极，溶液中含Zn 2+B ．锌作阳极，铁制品作阴极，溶液中含Fe 3+C ．锌作阴极，铁制品作阳极，溶液中含Zn 2+D ．锌作阴极，铁制品作阳极，溶液中含Fe 3+13．向等物质的量浓度的Ba(OH)2 与BaCl 2 的混合溶液中加入NaHCO 3 溶液，下列离子方程 式与事实相符的是（ ）A ．HCO 3-+OH - →CO 32-+H 2OB ．Ba 2++OH -+HCO 32-→BaCO 3↓+H 2OC ．Ba 2++2OH -+2HCO 32-→BaCO 3↓+CO 32-+H 2OD ．2Ba 2++3OH -+3HCO 32-→2BaCO 3↓+CO 32-+3H 2O14．向含有5×10-3mol HIO 3 与少量淀粉的溶液中通入H 2S ，溶液变蓝且有S 析出，继续通入H 2S ，溶液的蓝色褪去，则在整个过程中（ ）A ．共得到0.96g 硫B ．通入H 2S 的体积为336 mlC ．碘元素先被还原后被氧化D ．转移电子总数为3.0×10-2 N A15．饱和二氧化硫水溶液中存在下列平衡体系：SO 2+H 2O H ++HSO 3- HSO 3-H ++SO 3-若向此溶液中（ ）A ．加水，SO 3-浓度增大B ．通入少量Cl 2气体，溶液pH 增大C ．加少量CaSO 3 粉末，HSO 3- 浓度基本不变D ．通入少量HCl 气体，溶液中HSO 3- 浓度减小16．a 、b 、c 四种短周期元素在周期表中分布如右图所示，下列说法正确的是A ．若四种元素均为主族元素，则d 元素的原子半径最大B ．若b 最外层电子占据三条轨道，则a 的单质可用于冶炼金属C ．若a 为非金属元素，则c 的气态氢化物的水溶液可能呈碱性D ．若a 最外层有两个未成对电子，则d 的单质常温下不可能为气体17．某溶液中只可能含有----+++232332SO CO OH Br Al Mg K 、、、、、、中的一种或几种。

2016年上海市奉贤区高考三模试卷化学一、选择题1.化学与材料、生活和环境密切相关。

下列有关说法中错误的是( )A.食品袋中常放有硅胶和铁粉，都能起到干燥的作用B.大力实施矿物燃料脱硫脱硝技术，能减少硫、氮氧化物的排放C.明矾净水时发生了化学及物理变化，能起到净水作用，而没有杀菌、消毒的作用D.某新型航天服材料主要成分是由碳化硅、陶瓷和碳纤维复合而成，它是一种新型无机非金属材料解析：A、在食品袋中放入盛有硅胶和铁粉的透气小袋，硅胶(具有吸湿性)能吸收水分，但铁是较活泼的金属，具有还原性，能防止食品被氧化，故A错误；B、实施矿物燃料脱硫脱硝技术可以减少硫、氮氧化物排放，故B正确；C、明矾净水时，铝离子发生水解反应，生成氢氧化铝具有净水作用，发生了化学及物理变化，氢氧化铝不具有氧化性，不能杀菌消毒，故C正确；D、碳纤维的微观结构类似人造石墨，是乱层石墨结构和氮化硅属于无机物，属于无机非金属材料，故D正确。

答案：A2.4G网络让手机费起来了。

手机芯片的核心是硅板，其成分是( )A.SiO2B.SiC.H2Si03D.Na2SiO3解析：硅导电性介于导体与绝缘体之间，是良好的半导体材料，可用制造手机芯片，二氧化硅、硅酸、硅酸钠固体为绝缘体，不导电。

答案：B3.金属晶体的下列性质中，不能用金属晶体结构加以解释的是( )A.易导电B.易导热C.有延展性D.易锈蚀解析：A.组成金属晶体的微粒为金属阳离子和自由电子，在外加电场作用下电子可发生定向移动，故能导电，能用金属晶体结构加以解释，故A正确；B.金属晶体的导热是由于晶体内部，自由电子与金属阳离子的碰撞，能用金属晶体结构加以解释，故B正确；C.金属发生形变时，自由电子仍然可以在金属子离子之间流动，使金属不会断裂，能用金属晶体结构加以解释，故C正确；D.金属的化学性质比较活泼，容易被空气中的氧气所氧化，故金属易腐蚀不能用金属晶体结构加以解释，故D错误。

答案：D4.下列关于二氧化硅晶体的描述错误的是( )A.分子式为SiO2B.熔化时共价键断裂C.属于酸性氧化物D.1mol SiO2中含有4 mol Si-O键解析：A.二氧化硅晶体是由硅原子和氧原子构成的原子晶体，不含分子，因此没有分子式，故A错误；B.二氧化硅是原子晶体，熔化时破坏共价键Si-O键，故B正确；C.二氧化硅能与强碱溶液反应生成盐和水，属于典型的酸性氧化物，故C正确；D.在二氧化硅晶体中，每个硅原子形成四条Si-O键，1mol SiO2中含有4 mol Si-O键，故D正确。