四川省成都望子成龙学校2012-2013学年高二上学期期中模拟(4科6份)

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四川省成都望子成龙学校2012-2013学年高二上学期期中模拟(4科6份)

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2012—2013学年高二上学期期中化学考试题
(考试时间:90分钟 总分:100分)
选择题(每小题2分,每小题只有一个正确答案,共50分)
能够确定核外电子空间运动状态的量子数的组合是( )
n l B. n l m C. n l D. n l m
碳原子的最外电子层的各能级中,电子的排布方式通常是( )
A. B.
C. D.
3. 某原子的4d亚层中有1个电子、其第5电子层中的电子个数为( )
A. 0 B. 2 C. 3 D. 8
4. 的离子价电子排布式是( )
B. C. D.
下列物质性质的变化规律,与共价键的稳定性有关的是( )
的熔点、沸点逐渐升高。
的热稳定性依次降低。
碱金属单质的熔点随原子序数递增依次降低。
的熔点依次降低。
有人认为在元素周期表中,位于IA族的氢元素,也可以放在VⅡA族,下列物质能支持这种观点的是( )
HF B. C. D.
在人体所需的10多种微量元素中,有一种称为“生命之原”的R元素,对延长人的寿命起着重要的作用。已知R元素的原子有4个电子层,其最高价氧化物分子式为,则R元素的名称为( )
硫 B. 砷 C. 硒 D. 硅
有下列电子排布式的原子中,半径最大的是( )
B.
C. D.
按第一电离能递增的顺序排列的是( )
Li Na K B. Na Al S C. P Si Al D. Cl Br I
元素的电负性随着原子序数的增加而递增的是( )
Na K Rb B. N P As C. O S Se D. Na P Cl
下列分子含有的电子数目与HF分子相同,并且分子中只有两个极性共价键的是( )
B. C. D.
参考下列表中化学键的键能数据,下列分子中受热时最稳定的是( )
化学键
H—H
H—F
H—Cl
H—I
键能(KJ/mol)
436
565
368
297
氢气 B. 氟化氢 C. 氯化氢 D. 碘化氢
中心原子采取SP杂化方式的分子是( )
B. C. D.
在分子中,键采用的成键轨道是( )
B. C. D.
下列分子中,所有原子不可能共处在同一平面上的是( )
B. C. D.
已知1—18号元素的离子具有相同的电子层结构,下列关系正确的是( )
质子数: B. 氢化物的稳定性:
C. 离子的还原性: D. 原子半径
下列关于共价键说法正确的是( )
共价键只能在不同原子之间形成。
由共价键形成的分子可以是单质分子,也可以是化合物分子。
金属元素在化学反应中只能失去电子,因而不能形成共价键。
稀有气体分子只存在共价键。
下列关于化学键的说法正确的是( )
化学键存在于相邻分子之间。
化学键是分子内两个相邻原子之间的相互作用。
化学键是相邻两个或多个原子之间的强烈的相互作用。
离子键是相邻阴阳离子间的经典吸引。
下列叙述的距离属于键长的是( )
氨分子中两个氢原子间的距离。
氯分子中两个氯原子间的距离。
金刚石晶体中任意两个相邻的碳原子核间的距离。
氯化钠晶体中相邻的氯离子和钠离子核间的距离。
下列说法正确的是( )
氢键只存在于同一分子之间。
金属阳离子只能与阴离子构成晶体。
粒子间以分子间作用力结合的晶体,其熔点不会很高。
氮化硅是一种新型的耐高温耐磨材料,它属于分子晶体。
下列有关晶胞的叙述正确的是( )
晶胞是晶体中最小的结构重复单元。
不同的晶体中,晶胞的大小、形状都相同。
晶胞中的任何一个粒子都属于该晶胞。
已知晶胞的组成不可以推知晶体的组成。
已知晶体具有比金刚石还大的硬度,且构成该晶体的粒子间只以单键结合。下列关于晶体的说法错误的是( )
该晶体属于原子晶体,其化学键比金刚石更牢固。
该晶体中每个碳原子连接4个氮原子,每个氮原子连接3个碳原子。
该晶体中碳原子和氮原子的最外层都满足8电子结构。
该晶体与金刚石相似,都是原子间以非极性键形成的空间网状结构。
下列关于乙醇分子的说法正确的是( )
分子中共含8个极性共价键。
分子中不含非极性共价键。
分子中只含6键。
分子中含有一个键。
某离子晶体晶胞结构如右图所示,X位于立方体的顶点,Y位于立方体的中心,晶体中距离最近的两个X与一个Y形成的夹角A. 90° B. 60° C. 120° D. 109°28′
25. 下列有关化学用语表示正确的是( )
A. 乙酸的结构简式:
B. 的结构示意图:
C. 中子数为20的氯原子:
D. 的电子式:
化学试题参考答案
选择题(本题包括25个小题,每小题2分,共50分)
题号
1
2
3
4
5
6
7
8
9
10
11
12
13
答案
B
C
B
C
B
C
C
A
B
D
C
B
B
题号
14
15
16
17
18
19
20
21
22
23
24
25
答案
C
C
B
B
C
C
C
A
D
C
D
B
非选择题(共50分)
26. N 、 、 、 、三角锥 、 极 、 一 、 空 、 配位 、 、等同。
、 、
、非极 、分子 、 、原子 、高
(1)、
(2)
(3)1∶1


(1)氮 、硫 、氟
(2) 3 、 VIA
(3) 、
(4) > 、
(1)
(2) 6
(3) cm
望子成龙学校
高二化学选修Ⅲ模块测试
时间:90分钟 满分:100分
姓名: 得分:
第Ⅰ卷 选择题(共40分)
选择题(本题共20个小题,每小题2分,共40分,每题只有一个正确选项)
1.道尔顿的原子学说主要有下列三个论点:①原子是不能再分的微粒;②同种元素的原子的各种性质和质量都相同;③原子是微小的实心球体。从现代原子学说的观点看,你认为不正确的是
A.只有① B.只有② C.只有③ D.②①③
2.下列能级中轨道数为3的是
A.1s B.2p C.3d D.4f
3.下列各基态原子或离子的电子排布式错误的是
A.Al:1s22s22p63s23p1 B.O2-:1s22s22p6 C.Na+:1s22s22p6 D.Si:1s22s22p2
4.气态电中性基太原子的原子核外电子排布发生如下变化,吸收能量最多的是
A.1s22s22p63s23p2→1s22s22p63s23p1 B.1s22s22p63s23p3→1s22s22p63s23p2
C.1s22s22p63s23p4→1s22s22p63s23p3 D.1s22s22p63s23p63d104s24p2→1s22s22p63s23p63d104s24p1
5.已知X、Y元素同周期,且电负性X>Y,下列说法错误的是
A.X与Y形成化合物时,X可以显负价,Y显正价
B.第一电离能可能Y小于X
C.最高价含氧酸的酸性:X对应的酸的酸性弱于Y对应的酸的酸性
D.气态氢化物的稳定性:HmY小于HnX
6.已知X、Y均为主族元素,I为电离能,单位是kJ/mol。根据下表所列数据判断错误的是
元素
I1
I2
I3
I4
X
500
4600
6900
9500
Y
580
1800
2700
11600
A.元素X的常见化合价是+1价
B.元素Y是ⅢA族的元素
C.元素X与氯形成化合物时,化学式可能是XCl
D.若元素Y处于第3周期,它可与冷水剧烈反应
7.不能说明X的电负性比Y的电负性大的是
A.与H2化合时,X单质比Y单质容易
B.X的最高价氧化物的水化物的酸性比Y的最高价氧化物的水化物的酸性强
C.X原子的最外层电子数比Y原子的最外层电子数多
D.X单质可以把Y从其氢化物中置换出来
8.下列图示中横坐标是表示元素的电负性数值,纵坐标表示同一主族的五种元素的序数的是
9.对σ键和π键的认识不正确的是
A.σ键和π键不属于共价键,是另一种化学键
B.s-sσ键与s-pσ键的电子云均为轴对称图形
C.分子中含有共价键,则至少含有一个σ键
D.含有π键的化合物与只含σ键的化合物的化学性质不同
10.氨气分子空间构型是三角锥形,而甲烷是正四面体形,这是因为
A.两种分子的中心原子杂化轨道类型不同,NH3是sp2杂化,而CH4是sp3杂化
B.分子NH3中N原子形成三个杂化轨道,CH4分子中C原子形成4个杂化轨道
C.NH3分子中有一对未成键的孤电子对,它对成键电子的排斥作用较强
D.氨气分子是极性分子而甲烷是非极性分子
11.已知Zn2+的4s轨道和4p轨道可以形成sp3型杂化轨道,那么[ZnCl4]2-的空间构型为
A.直线形 B.平面正方形 C.正四面体形 D.正八面体型
12.下列现象与氢键有关的是
①NH3的熔、沸点比P、As元素氢化物的高;②乙醇可以和水以任意比互溶;③冰的密度比液态水的密度小;④尿素的熔、沸点比醋酸的高;⑤邻羟基苯甲酸的熔、沸点比对羟基苯甲酸的低;⑥水分子高温下也很稳定
A.①②③④⑤⑥ B.①②③④⑤ C.①②③④ D.①②③
13.下列有关数据的比较,不正确的是
A.元素的价电子数总等于所在族的族序数
B.NaOH晶体中阳离子和阴离子数目相等
C.CsCl晶体中每个Cs+周围紧邻的Cl-和每个Cl-周围紧邻的Cs+个数相等
D.[Co(NH3)6]3+中的N原子数与配位键数相等
14.碘单质在水溶液中溶解度很小,但在CCl4中溶解度很大,这是因为
A.CCl4与I2分子量相差很小,而H2O与I2分子量相差较大
B.CCl4与I2都是直线型分子,而H2O不是直线型分子
C.CCl4和I2都不含氢元素,而H2O中含有氢元素
D.CCl4和I2都是非极性分子,而H2O是极性分子
15.下列说法不正确的是
A.手性异构体性质完全相同
B.手性异构体分子组成相同
C.互为手性异构体的分子互为镜像关系
D.利用手性催化剂合成主要(或只)得到一种手性分子
16.区别晶体与非晶体的最科学的方法是
A.观察自范性 B.观察各向异性 C.测定固定熔点 D.进行X-射线衍射实验
17.下表列出了有关晶体的说明,有错误的是
选项
A
B
C
D
晶体名称
碘化钾


二氧化硅
组成晶体微粒名称
阴、阳离子
阴、阳离子
原子
原子
晶体内存在的结合力
离子键
金属键
范德华力
共价键
18.金属能导电的原因是
A. 金属晶体中金属阳离子与自由电子间的相互作用较弱
B.金属晶体中的自由电子在外加电场作用可发生定向移动
C.金属晶体中的金属阳离子在外加电场作用下可发生定向移动
D.金属晶体在外加电场作用可失去电子
19.某离子化合物的晶胞如右图所示,黑球代表阳离子(A)、白球代表阴离子(B)。下列说法错误的是
A.阳离子配位数为8 B.配位数与阴阳离子电荷比有关
C.配位数与阴阳离子半径比有关 D.阴阳离子数比为1∶4
20.石墨晶体为层状结构,每一层均为碳原子与周围其他3个碳原子相结合而成平面片层,同层相邻碳原子间距142pm、相邻片层间距为335 pm。右图是其晶体结构片层俯视图。下列说法不正确的是
A.碳以sp2杂化 B.每个碳形成3个σ键
C.石墨碳原子数与σ键数之比为2∶3 D.片层之间碳成共价键
第Ⅱ卷 选择题(共40分)
21.(6分)
(1)多电子原子核外电子能量是不同的,其中能层序数(n)越小 ,该能层能量 ,在同一能层中,各能级的能量按ns、np、nd、nf的次序 。
(2)原子的电子排布遵循构造原理能使整个原子的能量处于最低状态,处于最低能量的原子叫做 原子,当这些原子 能量后电子跃迁到较高能级后就变成激发态。
A
B
C
D
(3)如图是周期表中短周期主族元素的一部分,其中电负性最大的元素是 。第一电离能最小的元素是 。
22.(9分)
(1)基态氮原子的价电子排布式是 。氮气分子中含有一个 键,二个 键。
(2)白磷分子的空间构型为 ,若将1分子白磷中的所有P-P键打开并各插入一个氧原子,共可结合 个氧原子,若每个P原子上的孤对电子分别再以配位键连接一个氧原子,就可以得到磷的另一种氧化物 (填分子式)。
(3)金刚石的晶体结构中每个碳原子与周围的4个碳原子形成四个碳碳单键,这5个碳原子形成的是 结构,金刚石晶体中C原子数与C-C键数之比为 ,晶体中最小的环上的碳原子数为 。
23.(9分)
(1)将NH3和PH3两分子中的共价键的各种键参数的数值大小相比较的结果是,键长:NH3
PH3、键能:NH3 PH3(填“大于”、“小于”或“等于”)。已知P的电负性小于H,则H为 价。
(2)肼(N2H4)分子可视为NH3分子中的一个氢原子被-NH2(氨基)取代形成的另一种氮的氢化物。PH3分子的空间构型是;N2H4分子中氮原子轨道的杂化类型是 。
(3)二氧化硫(Cl-S-S-Cl)是一种琥珀色液体,是合成硫化染料的重要原料。
①写出它的电子式 。
②指出它分子内的键型 (填“极性”或“非极性”)
③若4个原子不共面,则该分子 (填“有”或“无”)极性。
④指出硫元素的化合价为 。
24.(15分)下表是元素周期表的一部分。表中所列的字母分别代表一种化学元素。
A
b
c
d
e
F
g
h
i
j
k
l
m
n
o
试回答下列问题:
(1)第三周期8种元素单质熔点高低顺序如右图,其中序号“8”代表
(填元素符号);
(2)由j原子跟c原子以1∶1相互交替结合而形成的晶体,晶型与晶体j相同。两者相比熔点更高的是 (填化学式),试从结构角度加以解释:

(3)元素c的一种氧化物与元素d的一种氧化物互为等电子体,元素c的氧化物分子式是 ,该分子是由 键构成的 分子(填“极性”或“非极性”);元素d的氧化物的分子式是 。
(4)i单质晶体中原子的堆积方式如图甲所示(面心立体最密堆积),其晶胞特征如图乙所示。则晶胞中i原子的配位数为 ,一个晶胞中i原子的数目为 ;晶胞中存在两种空隙,分别是 、 。
25.(8分)
晶胞是晶体中最小的重复单位,数目巨大的晶胞无隙并置构成晶体。NaCl晶体是一个正六面体(如图一)。我们把阴、阳离子看成不等径的圆球,并彼此相切(已知a为常数)。
请计算下列问题:
(1)每个晶胞平均分摊 个Na+,
个Cl-。
(2)NaCl晶体中阴阳离子的最短距离为
(用a表示)。
(3)NaCl晶体为“巨分子”,在高温下(≥1413℃时)晶体转变成气态团簇分子。现有1molNaCl晶体,加强热使其气化,测得气体体积为11.2L(已折算为标准状况)。则此时氯化钠气体的分子式为 。
26.(13分)某化学学习兴趣小组,为了研究晶体的结构和性质,查阅了一水合硫酸四氨合铜(Ⅱ)([Cu(NH3)4]SO4)·H2O相关资料:
一水合硫酸四氨合铜(Ⅱ)([Cu(NH3)4]SO4)·H2O为蓝色正交晶体,常温下,它易溶于水,受热分解产生氨气和硫酸铜,易与空气中的二氧化碳、水反应生成铜的碱式盐,使晶体变成绿色粉末。
它的制备原理是利用硫酸铜溶液和氨水反应生成,其晶体的析出一般不宜用蒸发浓缩等常规方法。析出晶体方法:向硫酸铜溶液中加入浓氨水后,再加入浓乙醇溶液使晶体析出。
请回答下列问题:
(1)请写出制备硫酸四氨合铜(Ⅱ)的反应方程式:

(2)析出晶体不用蒸发浓缩方法的原因是 ;其中乙醇的作用是 。
(3)硫酸四氨合铜属于配合物,晶体中Cu2+与NH3之间的键型为 ,该化学键能够形成的原因是 。
(4)请你设计实验方案探究确定SO2- 4为配合物外界。

参考答案:
选择题(每个2分,共40分)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
D
B
D
B
C
D
C
B
A
C
C
B
A
D
A
D
B
B
D
D
非选择题(共60分)
21.(共6分,每空1分)(1)越低;递增;(2)基态;吸收;(3)B;C或D
22.(共9分,每空1分)(1)1s22s22p3;ο;π;(2)正四面体;6;P4O10;(3)正四面体;1∶2;6
23.(共9分,每空1分)(1)小于;大于;-1价;(2)三角锥型;sp3;(3)①
②S-S键是非极性共价键,S-Cl是极性共价键;③有;④+1
24.(15分)(1)Si(2分);(2)SiC(1分);因SiC晶体与Si晶体都是原子晶体,由于C的原子半径小,SiC中C-Si键键长比晶体Si中Si-Si键长短,键能大,因而熔沸点高(2分);(3)CO2(1分);极性(1分);非极性(1分);N2O(1分);(4)12(2分);4(2分);正四面体空隙(1分);正八面体空隙(1分)
25.(8分)(1)4(2分);4(2分);(2)a/2(2分);(3)Na2Cl2(2分);
26.(13分)(1)CuSO4+4NH3+H2O=[Cu(NH3)4]SO4·H2O(2分);
(2)[Cu(NH3)4]SO4·H2O受热分解(2分);硫酸四氨合铜在乙醇中的溶解度远小于在水中的溶解度或减小水的极性(2分);
(3)配位键(2分);NH3能提供孤对电子,Cu2+能接受孤对电子(或有空轨道)(2分);
(4)取硫酸四氨合铜晶体溶于水配成溶液,向溶液中加入适量的Ba(NO3)2溶液,溶液中出现白色沉淀,说明溶液中有SO2- 4存在于外界(3分)
望子成龙学校高二上数学半期摸拟题
一.选择题:(60分)
1.在空间,异面直线a,b所成的角为α,且=
A. B. C.或 D.
2.在一个几何体的三视图中,正视图和俯视图如图所示,则相应的侧视图可以是( )
3.已知三条不重合的直线,两个不重合的平面,有下列命题:
①若∥,,则∥; ②若,,且∥,则∥
③若,,,∥,则∥
④若,=,,,则, 其中正确命题的个数为 ( )
A.1个 B.2个 C.3个 D.4个
4.如右图是某几何体的三视图,则该几何体的体积为( )
A、16  B、24  C、34  D、48
5.在空间中.l、m、n是三条不同的直线,α、β、γ是三个不同的平面,则下列结论错误的是: A.若α∥β,α∥γ,则β∥γ B.若l∥α,l∥β,α∩β=m,则l∥m
C.α⊥β,α⊥γ,β∩γ=l,则l⊥α
D.若α∩β=m,β∩γ=l,γ∩α=n, l⊥m,l⊥n,则m⊥n
6.空间几何体的三视图如图所示,则该几何体的表面积为
(A)6+2   (B)8+2
(C)8+2   (B)6+2
7.已知平面内不共线的四点O,A,B,C满足 则( )
A.1:3 B.2:1 C.1:2 D.3:1
8.已知底面是正三角形,顶点在底面的射影是底面
三角形的中心的三棱锥的主视图、俯
视图如图所示,其中为棱
CB的中点,则该三棱锥的左视图的面积为( )
A.9 B6.
C. D.
9.在三棱锥中,若O是底面ABC内部一点,满足,则( ) A. B. 5 C. 2 D.
10.某几何体的正视图和侧视图均为如图1所示,则在图2的四个图中可以作为该几何体的俯视图的是
A.(1),(3) B.(1),(3),(4) C.(1),(2),(3) D.(1),(2),(3),(4)
11.在三棱锥P-ABC中,PA=PB=PC=,侧棱PA与底面ABC所成的角为60°,则该三棱锥外接球的体积为
(A)2  (B)   (C)4  (D)
12.已知矩形ABCD的面积为8,当矩形周长最小时,沿对角线AC把△ACD折起,则三棱锥D—ABC的外接球的表面积等于 A.4π B.8π C.16π D.24π
二.填空题:(16分)
13.一个几何体的三视图如下图所示.刚该几何体的体积为_____

14.在中,,,为垂足,则,该结论称为射影定理。 如图,在三棱锥中,平面,平面,为垂足,在三棱锥内,类比射影定理,探究这三者之间满足的关系是 .
15.用若干个体积为1的正方体搭成一个几何体,其正(主)视图、侧(左)视图都是如图所示的图形,则这个几何体的最大体积是___
16..如图,将菱形ABCD沿对角线BD折起,使得C点至,E点在线段上,若二面角A—BD-E与二面角的大小分别为30°和45°,则=._______
三.解答题:(74分)
17.(本小题满分12分)如图,四棱锥P-ABCD中,面ABCD.底面ABCD为直角梯形,。E为PD的中点。
(1)求证:平面;
(2)求异面直线AB与PC所成的角的正切值。
18.(本小题满分12分)如图,在三棱柱ABC-A1B1C1中,侧棱AA1⊥底面ABC. AB=AC=l,∠BAC=120,异面直线B1C与A1C1所成的角为60°。
(I)求三棱柱ABC-A1B1C1的体积:(II)求二面角B1-AC-B的余弦值.
19. (本题满分12分) 如图,已知矩形的边与正方形所在平面垂直,,,是线段的中点。
(1)求证:平面;
(2)求二面角的大小。
20.(本小题12分)如图,在四棱锥中,平面,
四边形是菱形,,,E是PB上任意一点 .
(I)求证: AC⊥DE;
(II)已知二面角的余弦值为,若为的中点,求与平面所成角的正弦值 .
21.(本小题满分12分)如图所示, 四棱锥P-ABCD的底面是边长为1的正方形,
PA︿CD,PA = 1, PD=,E为PD上一点,PE = 2ED.(Ⅰ)求证:PA ︿平面ABCD;
(Ⅱ)求二面角D-AC-E的余弦值; (Ⅲ)在侧棱PC上是否存在一点F,使得BF // 平面AEC?若存在,指出F点的位置,并证明;若不存在,说明理由.
22.(本小题共14分)如图边长为4的正方形所在平面与正所在平面互相垂直,分别为的中点.
(1)求证:平面;
(2)求:二面角的余弦值;
(3)试问:在线段上是否存在一点使得平面平面若存在,试指出点N的位置,并证明你的结论;若不存在,请说明理由.
望子成龙学校高二上数学半期摸拟题答案
一,选择题:(60分) ADBAD CBBCA DC
二.填空题:(16分)
13. 答:32. 14. 答:
15. 答:11.∵体积最大时,底层有9个小正方体,左上面有2个小正方体。16. 答:
三.解答题:(74分)
17. 解:1)取AD的中点F.连结EF,CF.
因E为PD的中点,。所以EF//PA,CF//AB
所以面EFC//面PAB所以CE//面PAB ……….6分
2)由已知可得ABCF为平行四边形
所以AB//CF, 为所求的角,可证CF面PAD在直角三角形PCF中
tan……12分
18. 解:(Ⅰ)如图,以A为原点,AC为y轴,AA1为z轴,建立空间直角坐标系.
设AA1=a(a>0),依题意得B1(,-,a),A(0,0,0),C(0,1,0).
=(-,,-a),==(0,1,0),
由异面直线B1C与A1C1所成的角为60(,知
|cos(,(|=
==,解得a=. …4分
所以三棱柱ABC-A1B1C1的体积
V=AB·ACsin120(·AA1=×1×1××=. …6分
(Ⅱ)由(Ⅰ)知,=(-,,-).设n=(x,y,z)为面ACB1的法向量,则n·=0,n·=0,则取z=1,得x=-2,于是n=(-2,0,1).…9分
又m=(0,0,1)为面ACB的一个法向量,所以cos(m,n(==.
因此二面角B1-AC-B的余弦值为. …12分
19. 解:(1)建立如图所示的空间直角坐标系,则
… 2分
设平面的一个法向量为,则
取,得平面的一个法向量为,…6分
,所以,又因为直线不在平面内,所以平面。…6分
(2)由(1)知平面的一个法向量为,而平面的一个法向量为,所以向量与向量的夹角,从图中可以看出二面角为锐二面角,所以所求二面角的大小是。 …………… 12分
20. (1)证明:∵ 平面,平面ABCD, ∴ , 又∵是菱形 ∴ , ∴平面 ∵平面 ∴ …6分
(2)分别以方向为轴建立空间直角坐标系,设,则
由(1)知:平面的法向量为,令平面PAB的法向量为,
则根据得∴
因为二面角A-PB-D的余弦值为,则,即
…9分,∴
设EC与平面PAB所成的角为,∵,则
………………12分
21. 解:(Ⅰ) PA = PD = 1 ,PD = 2 , PA2 + AD2 = PD2, 即:PA ( AD ---2分
又PA ( CD , AD , CD 相交于点D, PA ( 平面ABCD-------4分
(Ⅱ)过E作EG//PA 交AD于G,从而EG ( 平面ABCD,且AG = 2GD ,
EG = PA = , ------5分连接BD交AC于O, 过G作GH//OD ,交AC于H,
连接EH.GH ( AC , EH ( AC , ( EHG为二面角D—AC―E的平面角. -----6分
tan(EHG = = .二面角D—AC―E的平面角的余弦值为-------8分
(Ⅲ)以AB , AD , PA为x轴、y轴、z轴建立空间直角坐标系.则A(0 ,0, 0),B(1,0,0) ,C(1,1,0),P(0,0,1),E(0 , ,), = (1,1,0), = (0 , , ) ---9分 设平面AEC的法向量= (x, y,z) , 则 ,
即:, 令y = 1 , 则 = (- 1,1, - 2 )-------10分
假设侧棱PC上存在一点F, 且= ,(0 ( ( 1), 使得:BF//平面AEC, 则( = 0.又因为:= + = (0 ,1,0)+ (-,-,)= (-,1-,),( =+ 1- - 2 = 0 , = ,所以存在PC的中点F, 使得BF//平面AEC. ---12分
22. (1)证明:连接交于点,连接由正方形知为的中点,为的中点,,平面平面平面
(2)二面角的余弦值为
(3)解,存在点当为中点时,平面四边形是正方形,为的中点,由(1)知,平面

成都望子成龙学校2012—2013学年度上学期半期考试试题
高二数学参考答案
1
2
3
4
5
6
7
8
9
10
11
12
C
D
B
B
B
C
A
C
C
D
D
A
(13) (14)文科 或 理科
(15) (16)④
17.解:(I)由 又, ∴。 …… 6分
(II),……8分 ∴……………10分
∵。………………………12分
18.证明(I):连结BD,令BD∩AC=O 1分
∵ABCD是正方形O是DB的中点,又E是DD1的中点,
∴EO∥BD1 3分
又EO平面ACE,BD1平面ACE,
∴BD1∥平面ACE 6分
(II)在正方形ABCD中,AB=2,AC=2,∴AO=
在直角△ADE中,AD=2,DE=1,∴AE= 8分
在Rt△EAO中,EO=== 10分
∴ 12分
19.解:(I)设的公差为, 的公比为,则依题意有且
由,解得,.∴, …5分
. ………………………………………………6分
(II). …7分,①………8分
,②… …… ………………9分
由②-①得…… …………10分
===. …12分
20.解(Ⅰ)因为平面,所以. 因为是正方形,
所以,从而平面.所以两两垂直,以为原点,分别为轴建立空间直角坐标系如图所示.因为与平面所成角为,即,所以.由可知,.则,,,,,所以,,
设平面的法向量为,则,即,令,则.
因为平面,所以为平面的法向量,,
所以.因为二面角为锐角,所以二面角的余弦值为. ………………8分
(Ⅱ)解:点是线段上一个动点,设.则,
因为平面,所以, 即,解得.
此时,点坐标为,符合题意. ………………12分
21.解:由,得 2分 此不等式与同解 3分 若,则 6分 若,则 8分 若,则或 11分 综上,时,原不等式的解集是;
时,原不等式的解集是;
时,原不等式的解集是 12分
22.解:(I)建立如图所示的空间直角坐标系,则各点坐标分别为:
,,,,
设(0≤x≤2), …………………2分
∵∴由PQ⊥QD得

∵ ……………4分
∴在所给数据中,可取和两个值. ……7分
(II) 由(Ⅰ)知,此时或,即满足条件的点Q有两个,…8分
根据题意,其坐标为和,……9分
∵PA⊥平面ABCD,∴PA⊥AQ1,PA⊥AQ2,
∴∠Q1AQ2就是二面角Q1-PA-Q2的平面角.……………………11分
由=,
得∠Q1AQ2=30(,∴二面角Q1-PA-Q2的大小为30(.………………………14分
成都望子成龙学校2012—2013学年度上学期半期考试试题
高二数学
注意事项:
1.答题前,请先将自己的姓名、考场、考号在卷首的相应位置填写清楚;
2.选择题答案涂在答题卡上,非选择题用蓝色、黑色钢笔或圆珠笔直接写在试卷上
第Ⅰ卷(选择题 共分)
一、选择题(本大题共小题,每小题分,共分,在每小题给出的四个选项中,只有一项是符合要求的).
1.在中,角所对的边分别是,且,则
A. B. C. D.
2.若点,,当取最小值时,的值为
A. B. C. D.
3.在空间,下列命题正确的是
对边相等的四边形一定是平面图形 B.有一组对边平行的四边形一定是平面图形
C.四边相等的四边形一定是平面图形 D.有一组对角相等的四边形一定是平面图形
4.已知点及圆,则过点且在圆上截得的弦为最长的弦所在的直线方程是
A. B. C. D.
5.函数的图像的大致形状是.
A. B. C. D.
6.若成等比数列,其公比为,则的值为
A. B. C. D.
7.设角是中的最小角,且,则实数的取值范围是
A. B. C. D.
8.已知在正方体中,分别是的中点,则异面直线与所成的角等于
A. B. C. D.
9.若长方体过一个顶点的三条棱的长度分别为且它的个顶点都在同一球面上,则这个球的表面积为
A. B. C. D.
10.如图,为正方体,下面结论错误的是
A.平面 B.
C.平面 D.异面直线与所成的角为
11.如图,已知三棱柱的各条棱长都相等,且
底面,是侧棱的中点,则异面直线
和所成的角的大小是
A. B. C. D.
12.已知各项均不为零的数列,定义向量,,.
下列命题中为真命题的是
A.若任意总有成立,则数列是等差数列
B.若任意总有成立,则数列是等比数列
C.若任意总有成立,则数列是等差数列
D.若任意总有成立,则数列是等比数列
第II卷(选择题 共分)
二、填空题:(本大题共小题,每小题分,共分.把正确答案填在题中横线上)
13.若某几何体的三视图(单位:)如图所示,则此几何体的体积是
14.(文科)已知直线与平行,则的值为 .
14.(理科)已知三棱锥中,,点在上,且为中点,则
(结果用表示)
15.等差数列的前项和为,且,.记,如果存在正整数,使得对一切正整数,都成立.则的最小值是 .
16.已知向量,对任意,恒有.现给出下列四个结论:
①; ②; ③; ④.
则正确的结论序号为_____________.(写出你认为所有正确的结论序号)
三、解答题:(本大题共小题,共分,解答应写出必要的文字说明,证明过程或演算步骤)
17.(本小题满分分)
在中,分别为角所对的三边,
(I)求角;
(II)若,求的值.
18.(本小题满分分)
如图,正方体的棱长为,为棱的中点.
(I)判断和过三点的平面的位置关系,并证明你的结论;
(II)求的面积.
19.(本小题满分分)
设数列是等差数列,数列是各项都为正数的等比数列,且,,是与的等差中项.
(I)求数列,的通项公式;
(II)求数列的前项和.
20.(本小题满分分)
如图,是边长为的正方形,平面,,,与平面所成角为.
(Ⅰ)求二面角的余弦值;
(Ⅱ)设是线段上的一个动点,问当的值为多少时,可使得平面,并证明你的结论.
21.(本大题满分分) 解关于的不等式
22.(本小题满分分)
如图所示,矩形的边平面,,现有数据:
①;②;③;建立适当的空间直角坐标系,
(I)当边上存在点,使时,可能取所给数据中的哪些值?请说明理由;
(II)在满足(I)的条件下,若取所给数据的最小值时,这样的点有几个?若沿方向依次记为,试求二面角的大小.
望子成龙学校
高中二年级第一学期期中测试
物理模拟题
(全卷分Ⅰ、Ⅱ卷。满分110分,100分钟)
第Ⅰ卷(选择题 48分)
一、选择题(共48分。本题共12个小题,在每小题给出的四个选项中,有的只有一个选项正确,有的有多个选项正确,全部选择对得4分,选对但不全得2分,有选错得0分)
1、一根放在水平面内的光滑玻璃管绝缘性能很好,内部有两个完全相同的弹性金属小球A和B,带电量分别是9Q和-Q。两球从图1位置静止释放,那么两球再次经过图中的原静止位置时,A球的瞬时加速度为释放时的
A、 B、 C、1倍 D、
2、如图2所示,一电子沿等量异种电荷的中垂线由A→O→B匀速飞过,电子重力不计,则电子除受电场力外,所受另一个力的大小和方向变化情况是
A、先变大后变小,方向水平向左
B、先变大后变小,方向水平向右
C、先变小后变大,方向水平向左
D、先变小后变大,方向水平向右
3、关于导体的电阻,下列表述正确的是
A、跟导体两端的电压成正比
B、跟导体中的电流强度成反比
C、决定于导体的材料、长度和横截面积
D、决定于导体中的电流强度和电压
4、两个质量相同的小球用不可伸长的细线连结,置于场强为E的匀强电场中。小球1和小球2均带正电,电量分别为q1和q2(q1>q2)。将细线拉直并使之与电场方向平行,如图3所示。若将两小球同时从静止状态开始释放,则释放后细线中的张力T为(不计重力及两小球间的库仑力)
A、 B、
C、 D、
5、如图4所示的电路中,E为电源电动势,r为电源的内阻,R1和R3均为定值电阻,R2为滑动变阻器。当R2的滑动触点在a端时合上开关S,此时三个电表A1、A2、V的示数分别为I1、I2和U。现将R2的滑动触点向b端移动,则三个电表示数的变化情况是
A、I1增大,I2不变,U增大
B、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
6、竖直放置的一对平行金属板的左极板上用绝缘线悬挂了一个带正电的小球,将平行金属板按图5所示的电路图连接,绝缘线与左极板的夹角为,当滑动变阻器R的滑动片在a位置时,电流表的读数为I1,夹角为;当滑片在b位置时,电流表的读数为I2,夹角为,则
A、< I1< I2 B、> I1 >I2
C、= I1= I2 D、< I1= I2
7、如图6所示,实线为不知方向的三条电场线,从电场中M点以相同速度飞出a、b两个带电粒子,运动轨迹如图中虚线所示,则
A、a一定带正电,b一定带负电
B、a速度将减小,b速度将增大
C、a加速度将减小,b加速度将增大
D、a、b加速度都将增大
8、图7中虚线所示是静电场中的等势面1、2、3、4,相邻的等势面之间的电势差相等,其中等势面3的电势为0。一带正电的点电荷在静电力的作用下运动,经过a、b点时的动能分别为26eV和5eV。当这一点电荷运动到某一位置,其电势能变为-8eV时,它的动能应为
A、8eV B、13eV
C、20eV D、34eV
9、有三个电阻的阻值及额定功率分别为R1=10、P1=10W,R2=20、P2=80W,R3=5、P3=20W,它们组成的电路如图8甲、乙、丙所示,关于各图的说法正确的是
A、图甲两端允许加的最大电压为60V
B、图乙干路允许流过的最大电流为3.5A
C、图丙两端允许加的最大电压为17.5V
D、以上说法均不正确
10、如图9所示,平行直线表示电场线,但未标明方向,带电为+10-2C的微粒在电场中只受电场力作用,由A点移到B点,动能损失0.1J。若A点电势为-10V,则
A、B点的电势为10V
B、电场线方向从右向左
C、微粒运动轨迹可能是轨迹1
D、微粒运动轨迹可能是轨迹2
11、在纸面内有一正,a、b、c分别是AB、BC和AC各边的中点,有一匀强电场平行于纸面,A、B、C、三点的电势分别是6V、0V、-6V,则有
A、 B、 C、 D、
12、在场强大小为E的匀强电场中,一质量为m、带电量为+q的物体以某一初速度沿电场反方向做匀减速直线运动,其加速度大小为0.8,物体运动s距离时速度为零,则
A、物体克服电场力做功qEs B、物体的电势能减少了0.8qEs
C、物体的电势能增加了qEs D、物体的动能减少了0.8qEs
第Ⅱ卷(非选择题 共62分)
二、填空题(共4小题,每空2分,共16分)
13、如图10所示,AB两端接直流稳压电源,UAB=100V,R0=40,滑动变阻器总电阻R=20,当滑动片处于变阻器中点时,C、D两端的电压UCD为 V,通过电阻R0的电流为 A。
14、A、B是一条电场线上的两点(图11甲),若在A点释放一初速度为零的电子,电子仅受电场力作用,并沿电场线从A运动到B,其速度随时间变化的规律如图11乙所示。设A、B两点的电场强度分别为EA、EB,电势分别为UA、UB,则可以判定EA EB,UA UB。(填“>”、“=”或“<”)
15、一匀强电场,场强方向是水平的,如图12所示。一个质量为m的带正电荷的小球,从O点出发,初速度的大小为,在电场力和重力的作用下,恰好能沿与场强的反方向成角的直线运动。小球将做 运动,小球运动到最高点时其电势能与在O点时的电势能之差是 。
16、图13为一电路板的示意图,a、b、c、d为接线柱,a、d与E=20V的电源连接,ab间、bc间、cd间分别连接一个电阻。现发现电路中没有电流,为了检查电路故障,用一电压表分别测得b、d两点间和a、c两点间的电压均为20V。由此可以判定ab间是 ,bc 。
三、实验题(共2题,17题4分,18题10分,共14分)
17、有一圆台状匀质合金棒如图14所示,某同学猜测其电阻的大小与该合金棒的电阻率、长度L和两底面直径d、D有关。通过实验测得合金电阻为R=6.72。根据电阻定律计算电阻率为、长为L和两底面直径分别为d、D的圆柱状合金棒的电阻分别为Rd=13.3,RD=3.38。他发现:在误差允许的范围内,电阻R满足R2=RdRD。由此推断该圆台状合金棒的电阻R= (用、L、d、D表示)。
18、为了测量一微安表头A的内阻,某同学设计了如图15所示的电路。图中A0是标准电压表,R0和RN分别是滑动变阻器和电阻箱,S和S1分别是单刀双掷和单刀开关,E是电池。完成下列实验步骤中的填空:
(1)将S拨向接点1,接通S1,调节 ,使待测表头指针偏到适当位置,记下此时 的读数I;
(2)将S拨向接点2,调节 ,使 ,记下此时RN的读数;
(3)多次重复上述过程,计算RN读数的 ,此即为待测微安表头内阻的测量值。
四、(共32分)解答时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。
19、(6分)如图16所示,在倾角为θ的很大的光滑绝缘斜面上,有一质量为m的小球,带电量为+q,初速度为v0,且方向与斜面底边AB平行。欲使小球在斜面上沿初速度方向做匀速直线运动,需加一个匀强电场,场强的最小值及方向如何?
20、(8分)如图17所示,电源内阻为r=0.4,,R1=R2=R3=4。当开关S闭合时,电流表和电压表的示数分别是1.5A和2V。求:
(1)电源电动势多大?
(2)开关S断开时,干路中的电流为多少?
21、(8分)如图18所示的电路中,两平行金属板A、B水平放置,两板的间距d=40cm。电源电动势E=24V,内阻r=1,电阻R=15。闭合开关S,待电路稳定后,将一带正电的小球从B板的小孔以初速度竖直向上射入板间。若小球带电量为C,质量为2×10-2kg,不考虑空气阻力。那么,滑动变阻器接入电路的阻值为多大时,小球恰能达到A板。(g=10m/s2)
22、(10分)如图19所示,在y>0的空间中,存在沿y的匀强电场,场强大小为E;在y<0的空间中,存在沿y轴负方向的匀强电场,场强大小也为E。一电子(电量为e,质量为m)在y轴上的P点以沿x轴正方向的初速度v0开始运动,不计电子重力。求:
(1)电子第一次经过x轴的坐标值;
(2)电子在y轴方向上运动的周期;
(3)电子运动的轨迹与x轴的各交点中,任意两个相邻交点的距离。
望子成龙学校
高中一年级第一学期期中测试
物理模拟题答题卷
第一卷 选择题(共48 分)
一、选择题:(本题包括12小题,每小题4分,共48分)
题号
1
2
3
4
5
6
答案
题号
7
8
9
10
11
12
答案
第二卷 非选择题(共62分)
二、填空题(16分)
13、 , ;
14、 , ;
15、 ;
16、 , ;
三、实验题(共14分)
17、 ;
18、(1) , ;
(2) , ;
(3) 。
四、计算题(共30分)解答时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。
19、(6分)
22、(8分)
21、(8分)
22、(8分)
望子成龙学校
高中一年级第一学期期中测试物理模拟题
参考答案
一、(每小题4分,共48分)
题号
1
2
3
4
5
6
答案
A
B
C
A
B
D
题号
7
8
9
10
11
12
答案
C
C
BC
BC
AD
ACD
二、填空题(每空2分,共16分)
13、80,2; 14、等于,小于;
15、匀减速,; 16、通路,断路。
三、实验题(17题4分,18题10分,共14分)
17、
18、(1)R0,标准电流表(或A0);(2)RN,标准电流表(或A0)的读数仍为I;(3)平均值。(每空2分)
四、计算题(共32分)解答时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。
19、(6分)
解:带电小球在斜面上受三个力(重力mg、支持力N、电场力qE)的作用而平衡,所以F合=0
因重力大小方向不变,支持力方向不变,由力的三角形可知,
当qE⊥N时电场力最小,
有qE=mgsinθ E= mgsinθ/q
20、(8分)
解:(1)S闭合时,R1和R2并联再与R4并联,最后与R3并联。
对R2有I2=U2/R2=2/4=0.5A
因为R1=R2,所以I1=I2=0.5A,
电流表的示数是R1和R3中的电流之和有I1+I3=1.5A,解得I3=1A
路端电压U=U3=I3R3=1×4=4V
总电流I=I1+I2+I3=2A 电动势E=U+Ir=4+2×0.4=4.8A
(2)S断开,R1和R3串联后与R4并联,最后与R2串联,且由S闭合时的条件打嗝知,R4=2,则
总外电阻为R={(R1+R3)R4/(R1+R3+R4)}+R2=5.6
总电流I’=E/(R+r)=0.8A
21、(8分)
解:小球进入板间后,受重力和电场力作用,且到A板时速度为0。设两板间电压为UAB,由动能定理得:

滑动变阻器两端的电压
U滑=UAB=8V
设通过滑动变阻器的电流为I,由欧姆定律得

滑动变阻器接入电路的电阻
22、(10分)
解:(1)由

(2)电子在y轴方向上的运动具有对称性,得T=
(3)由电子运动的周期性和对称性可得,运动轨迹与x轴的任意两个相邻交点间的距离为
望子成龙学校2012-1013学年度上期
高二英语期中模拟试卷
考试时间:120分钟 试题满分:150分
第一卷(三部分,共115分)
第一部分: 听力( 共两节,20小题;每小题1.5分,满分30分)(略)
第二部分:英语知识运用(共两节,满分45分)
第一节:单项填空(共15小题;每小题1分,满分15分)
1. The boss wanted an assistant with ______ knowledge of French and ______work experience.
A. the; a B. 不填; the C. a; 不填 D. the; 不填
2. She also learned a kind of alphabet for the blind, in which different finger positions ______ different letters of the alphabet.
A. stand for B. stand by C. stand up D. stand out
3. I don’t like such books ______ you read.
A. which B. when C. as D. where
4. ---- I need some fresh air, so I’m going out for a walk.
---- ______.
A. Have fun
B. With pleasure
C. Thanks a lot
D. It’s none of my business
5. The doctor always tried his best to cure the disease of his patients so that he was ______ by all the people here.
A. attracted B.repected C. directed D. affected
6. It ______ to me whether Garcia goes or not. Anyhow, I’ll go.
A. makes no difference
B. makes a difference
C. takes little difference
D. take some difference
7. It is known to us all that ______ you exercise regularly, you won’t keep it.
A. unless B. whenever C. although D. if
8. She felt very sad when Li Ming was talking with other girls without taking notice ______her.
A. in B. of C. on D. at
9. All passengers are ______ to show their tickets.
A. acquired B. demanded C. questioned D. required
10. ---- You were brave enough to raise questions at the meeting.
---- Well, now I regret ______ that.
A. to do B. to be doing C. to have done D. having done
11. ______ his effort, it is more successful than we have expected.
A. In case of B. Thanks to C. As to D. In favour of
12. ________ at what had happened, Tom didn’t know what to do.
A. Surprising B. To surprise C. Surprised D. Surprise
13. There children loved playing tricks ______ their teacher.
A. at B. with C. on D. about
14. We have to wait until the discussion between them ______.
A. brings to end B. comes to an end C. puts an end D. makes an end
15. Allow children the space to express their opinions ______ they disagree with us.
A. until B. even if C. unless D. as though
第二节: 完形填空(共20小题;每小题1.5分,满分30分)
Some people cannot learn in ordinary schools. Physical or 16 illness prevents a child from learning. Today new 17 are being used in special schools to help the disabled learn. A school is being 18 in New Jersey, USA. It is called Bancroft. Here the disabled will be trained to 19 themselves and to get along in the outside world. Bancroft is not surrounded by walls of any kind. Its director insists that it be 20 so that students may gradually develop 21 relations with the rest of the world. Bancroft students will 22 in apartments, cooking their own meals, and learning to perform other 23 . As they become able, they will buy their own furniture, 24 for it out of their own money. They will pay for their food, too. They will learn to expect 25 bills for the calls they make every month. As a step toward the goal of becoming 26 , each disabled person will decide what kind of work he wants to be 27 to do. While some of the training will be carried on within Bancroft itself, most of the students will receive 28 training in nearby towns. They will be trained by town people. After the training has been 29 completed, the student will work 30 an assistant and will begin to earn money. After that he will leave Bancroft, 31 the school will continue to give him help if he 32 it. How long will it take a student to 33 his training under this new system? The director says, “For some a year will be 34 . For others it might take ten years.” For all, however, this method offers new 35 . Many will learn to be useful and independent, supporting themselves in the world.
16. A. spirit B. mental C. thought D. body
17. A. plans B. decisions C. tools D. methods
18. A. turned up B. set up C. searched for D. longed for
19. A. enjoy B. teach C. push D. support
20. A. free B. open C. quiet D. different
21. A. special B. familiar C. normal D. close
22. A. live B. study C. hide D. cook
23. A. operations B. tasks C. plays D. acts
24. A. yelling B. waiting C. paying D. working
25. A. telephone B. education C. housing D. food
26. A. brave B. clever C. learned D. independent
27. A. asked B. sent C. trained D. made
28. A. teacher B. job C. body D. mind
29. A. successfully B. gradually C. quickly D. hardly
30. A. with B. for C. like D. as
31. A. and B. but C. so D. or
32. A. needs B. forgets C. gets D. offers
33. A. receive B. get C. complete D. stop
34. A. short B. enough C. good D. long
35. A. ideas B. abilities C. time D. work
第三部分: 阅读理解 (共20小题;每小题2分,满分40分)
第一节 (共15小题;每小题2分,满分30分)
A
I remember a day when I was a little kid. I was making a sandwich in the kitchen when I noticed the date on one of the wine bottles.
“ Dad, dad!” I cried. “This wine is too old to drink.”
“ Son, hold on,” he said.
“ No, you can’t drink this tonight! This bottle of wine was made 10 years ago.”
“ Wait, let me tell you…”
“ Would you like me to throw it away fro you?” I asked.
“ Son, wait a second,” he said. “Son, some wines get better over time. The longer you wait to drink it, the better it will be. Although this may seem strange, it is true.”
When I was young, I didn’t have any understanding of what this meant, but now, this would have been very helpful to remember as I went through my teenage years.
In our society, we forget this simple rule: The longer you want for certain things, the better they will be. But we want the best job as quickly as possible; we want to graduate from college in as few years as possible; we even speed through our homework just to chat with friends. When we do this, we lose something of great importance.
We all want to get to the next step so quickly that perhaps we don’t get ready enough to get there. This has a negative effect on our society. When trying to go to the right college, we will do anything to get in and when we rush through our class-work, we may not study enough for the test, and end up failing. We need to be ready for whatever comes, ready for the unexpected. Wine gets better over time, so do the things in our paths of life.
36. Why did the author tell his father not to drink the bottle of wine?
A. Because it smelt terrible.
B. Because it was mixed with something else.
C. Because his father drank too much that night.
D. Because it had been kept for years.
37. What the author’s father said __________.
A. made the author puzzled when he was a little child.
B. was too simple a rule to be meaningful to the author.
C. threw the author into deep thought then.
D. was an excuse to drink the wine.
38. In the author’s opinion, if we do things too quickly, we will __________.
A. do it better B. save much time
C. graduate from college more quickly D. miss something useful to us in life
39. Which of the following best reflects the main idea of the passage?
A. Well begun is half done. B. More haste, less speed.
C. Failure is the mother of success. D. Nothing is impossible to a willing heart.
B
If you are an e-book fan and want to download Harry Potter and the Deathly Hallows, forget it.
Author J.K. Rowling has said no to the first six Potter stories being released as e- books and has no plans to change that policy for the seventh and final work, Neil Blair, a lawyer with Rowling ‘s literary agency, told the Associated Press on Sunday.
Rowling has cited two reasons over the years: concern about online piracy(盗版) (which has never been a major problem for the Potter books), and the desire for readers to experience the books on paper. E-books, promoted as the future of publishing during the do-com fashion of the late 1990s, remain a tiny portion of the multibillion dollar industry.
The author herself writes in longhand, a preference that led to a rather amusing delay in Potter VII last April.
“Why is it so difficult to buy paper in the middle of town?” the author, a resident of Edinburgh, Scotland, sighed in a diary entry posted at the time on her website.
“ What is a writer who likes to write longhand supposed to do when she gets on well and then realizes, to her horror, that she has covered every bit of blank paper in her bag? Forty-five minutes it took me, this morning, to find somewhere that would sell me some normal, lined paper. And there’s a university here!” she wrote.
Rowling announced last week that Harry Potter and the Deathly Hallows would come out July 21. The six previous books have sold more than 325 million copies.
40. According to the passage, those who like reading on the net __________.
A. will be happy with what Rowling said about her books.
B. will have very good chances to read Rowling’s seventh story.
C. won’t have the chance to download Rowling’s first six Potter stories again.
D. still get nowhere to download Rowling’s works on line.
41. What does the underlined sentence “there’s a university here!” in the 6th paragraph mean? It means _______.
A. Rowling thinks paper should be easily bought around such a place.
B. it surprises Rowling that there should be a university nearby.
C. Rowling didn’t believe a university was around the place though there is one
D. it makes Rowling happy that there must be her keen readers in the university.
42. According to Rowling _________.
A. some people still enjoyed reading books printed on paper
B. she was considering publishing her books in e-book from in the future
C. there were more books printed on paper rather than published in e-book form
D. e-books had no good future in the publishing industry
43. What might be the author’s opinion on J. K. Rowling?
A. She is a very famous writer with many popular works.
B. She is a writer with somewhat traditional thought.
C. She likes her books published later than expected.
D. She is a popular writer all over the world.
C
We have developed a new range of first aid courses designed to meet the needs of home or leisure activities. It will help you in the everyday life.
Emergency life support
Our emergency life support course takes just three hours to complete and will give you the essential skills to save a life. It includes heart attacks, strokes, stomachache and serious bleeding.
Basic first aid
This course is about everyday first aid: cuts and scrapes, minor burns and scalds.
Family first aid
This course is for families to learn first aid together. The course covers the same content as Basic first aid and Emergency life support. The course takes six hours.
Activity first aid
This course takes 16 hours and includes all the content of Emergency life support, Basic first aid and Family first aid course. It also covers conditions caused by the extremes of temperature, low blood sugar and casualty management.
Total first aid
It is our most comprehensive first aid course taking 28 hours to complete. This course is for members of the public who want an in-depth knowledge of first aid techniques. This includes all of the content of the Activity first aid course and training in using some medical instruments.
Emergency aid for traffic
This course is specifically designed for accidents and covers emergency life support skills and head injuries.
Sports first aid
This course will give you all the necessary first aid skills to deal with a whole range of sporting injuries.
44. These first aid training courses are mainly designed for __________.
A. housewives B. members of the public
C. doctors and nurses D. school students
45. The underlined word “scrapes” probably means “_________”.
A. injuries caused by a serious accident B. sudden attack on the brain
C. injuries caused by something rough D. injuries caused by animals
46. How long does it take to complete the Basic first aid course?
A. Three hours B. Four hours C. Five hours D. Six hours.
D
Many songs, poems, and books have been written on love’s strong effect on people. The state of being in love has even been compared to being sick or mad. A study by Professor Semir Zeki of University College London has found that love does affect people’s brains by making them feel great.
Zeki studied young men and women who had recently fallen in love. He found that, when they were looking at photos of their loved ones, there was heightened activity in four areas of their brains. These areas deal with emotions, and one of them, in particular, is known to respond to drugs that cause feelings of great joy and excitement.
Interestingly, the study also found a lack of activity in two other areas of the brain when the volunteers looked at their lovers’ photographs. One of these areas is linked to feelings of sadness, while the other is often active in people suffering from depression. It seems love really can be uplifting.
The state of being in love, according to some scientists, may actually be good for your health. Although scientists know that being in love can make a person feel great, the exact influence of love on a person’s health is harder to determine. However, scientists say that people do need love in order to live healthy lives.
According to Dr. Thomas Lewis, people need to be in relationships because that is how we are designed. He says the brain can only maintain the overall stability of a person’s immune(免疫) system, bodily rhythms, and heart if it receives input from outside the body in the form of emotional connections with others.
Professor Antonio Damasio has a similar view. He says that love enriches a person’s imagination and creativity, and makes a person’s body work better. He also believes that love can even improve the body’s ability to fight against disease. Damasio’s wise words of advice are “ Choose love and you will live longer.”
47. The main idea of this passage is __________.
A. love can really cause people active B. love is linked with emotion
C. love is what we need D. love has effect on songs and books
48. What does the underlined word “uplifting” mean? It means __________.
A. inspiring B. available C. cold D. mad
49. From the passage we know__________.
A. People write songs, poems and books to say love is the sweetest thing
B. nobody but Professor Semir Zeki has found love does affect people’s brains
C. being in love sometimes may make someone act as if he/ she were mad
D. love causes heightened activity in all areas of a person’s brains
50. Which of the following statements is not true?
A. Some scientists have the idea that love does good to people.
B. It is possible for scientists to decide how much influence love can give people.
C. People are born to be in need of getting on touch with each other.
D. A person’s body health depends on partly on emotional connections with others.
第二节 (共5小题;每小题2分,满分10分)
根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。选项中多余选项。
Tips for better test taking
When you take a test, you are showing your ability to understand course material or perform certain tasks. The following suggestions may help you succeed in having your abilities properly evaluated and efforts rewarded!
● Analyze how you did on a similar test in the past.
Review your previous tests and sample tests provided by you teacher. 51
Arrive early for tests.
List what you need beforehand to avoid panic. Good preparation helps you succeed in the task at hand.
●Stay relaxed and confident.
Keep a good attitude and remind yourself that you are going to do your best. 52 . Don’t talk to other
students right before: stress can spread easily from one person to another.
●Read directions carefully and avoid careless errors.
If there is time, quickly look through the whole test for an overview. Scan for keywords. If permitted, write down any notes that come to mind.
●53 .
Easy questions first to build confidence . Then those with most point value.
●Review! If you have time .
54 . Check if you have answered all the questions, not making any errors or mis-marking any answers. Change answers to questions if you made a mistake, or misread the question!
●Decide on and adopt study strategies that work best for you.
55 . Check out your academic support centre or a trusted teacher for advice.
A. Be comfortable but careful.
B. Answer questions in a strategic order.
C. Successful test taking avoids carelessness.
D. Each test you take prepares you for the next one!
E. Resist the urge to leave when you complete the test.
F. If you find yourself panicking, take a few deep breaths.
G. Review where you have succeeded and where you have been challenged after the test.
第二卷(满分35分)
第四部分:写作(共三节,满分35分)
第一节 短文改错(共10小题;每小题1分,满分10分)
假定英语课上老师要求同桌之间交换修改作文,请你修改你同学写的以下作文。文中共有10处语言错误,每句中最多有两处。错误涉及一个单词的增加、删除或修改。
增加:在缺词处加一个漏字符号(),并在其下面写出该加的词。
删除:把多余的词用斜线()划掉。
修改:在错的词下划一横线,并在该词下面写出修改后的词。
注意:1. 每处错误及其修改均仅限一词;
2.只允许修改10处,多者(从第11处起)不计分。
Dear George,
It's been a week after we left your family and we are now back home. Thank you very much
for showing them around your city and providing us for the wonderful meals. After we said
goodbye to you, we went to Washington D.C., where we stayed for three days. My brother
was so much fond of the museums there that he begged my parents to staying another couple
of day. However, my father had to return to work on Monday, and we fly back last Saturday
afternoon. It was really a nice experience. If you'd like to make trip to our city some day, I
will be better than happy to be your guide.
Yours,
Mike
第二节 书面表达(满分25分)
你在网络上读到一篇关于“九零后”的英语文章,你打算以“Post-90’s Generation”为题,用第一人称给《二十一世纪英文报》写一篇文章,内容包括:
“九零后”的缺点
“九零后”的优点
你自己的感受
1.依赖性强;
2.自私;
3.贪图过舒适的生活等
1.渴望了解周围的世界;
2.思想独立;
3.有很多新观念等
注意:1.必须包括表格中的所有内容;2.为了使文章通顺完整,可以适当增加内容;3.词数:120左右。
___________________________________________________________________________________________
___________________________________________________________________________________________
___________________________________________________________________________________________
___________________________________________________________________________________________
___________________________________________________________________________________________
___________________________________________________________________________________________
___________________________________________________________________________________________
___________________________________________________________________________________________
___________________________________________________________________________________________
参考答案:
单选:1-5 CACAB 6-10 AABDD 11-15 BCCBB
完型填空:16~20 BDBDB 21~25 CABCA 26~30 DCBAD 31~35 BACBA
阅读理解:36-39 DADB 40 --43 DAAB 44—46 BCA 47-50 CACB
51-55.DFBEG
短文改错
Dear George,
It's been a week after we left your family and we are now back home. Thank you very much
since
for showing them around your city and providing us for the wonderful meals. After we said
us with
goodbye to you, we went to Washington D.C., where we stayed for three days. My brother
was so much fond of the museums there that he begged my parents to staying another couple
去掉 stay
of day. However, my father had to return to work on Monday, and we fly back last Saturday
days so flew
afternoon. It was really a nice experience. If you'd like to make∧trip to our city some day, I
a
will be better than happy to be your guide.
more
Yours,
Mike
书面表达
Post-90’s Generation
We,post-90’s generation,are living in a society full of fierce competition,and developing our special manners and values,which has raised people’s concern.
It’s true that the post-90’s generation have some weaknesses. Some of us,for example,rely on our parents too much and tend to live comfortable lives. Some are even selfish and care little for others.
At the same time,the post-90’s generation desire to learn more about the world around us,and we are independent in thinking. We always keep up with modern science and technology and have many new ideas.

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