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浙江工业大学物理化学历年期末考试题与答案试

浙江工业大学物理化学历年期末考试题与答案试

浙江工业大学物理化学历年期末考试题与答案试2006年2月物理化学-下试卷课程_物理化学(A)或(B)_________姓名___________________________班级___________________________授课教师___________________________1、在298K时无限稀的水溶液中,下列离子的摩尔电导最大的是:(A) H (B) Cl(C) NH (D) NO2、用对消法测定原电池电动势,是为了使待测电池的外电路(A) 电流稳定(B) 电流接近零(C) 电势差无限大(D) 电势差接近零3、已知298K 时H 的=350 10 S m mol ,Ac 的λ =40 10 S m mol 而实验测得某醋酸(HAc) 溶液的摩尔电导率Λ (HAc)=5.85 10 S m mol ,则此溶液中HAc 的电离度为:(A) 0.015 ; (B) 0.017 ; (C) 0.019 ; (D) 0.0214、下列示意图描述了原电池和电解池中电极的极化规律,其中表示原电池阳极的是:(A) 1 (B) 2 (C) 3 (D) 45、醋酸的极限摩尔电导率数值是根据下列何种方法得到的:(A) D-H 极限公式 (B) Kohlrausch 经验公式外推值(C) 离子独立运动定律 (D) 实验直接测得6、298 K、下,将直径1×10 m 的毛细管插入水中,需多大压强才能防止水面上升?已知此时水的表面张力为72×10 N·m 。

(A) 288 kPa (B) 144 kPa (C) 576 kPa (D) 316 kPa7、BET公式的主要用途之一是:(A) 获得高压下的吸附机理(B) 获得吸附等量线(C) 获得吸附等压线(D) 测定固体的比表面8、下列物系中为非胶体的是:(A) 灭火泡沫(B) 珍珠(C) 雾(D) 空气9、在一定温度下,将一个大的水滴(设为球形)分散成许多小的水滴,以下性质保持不变的是:(A) 表面能; (B) 表面张力;(C) 饱和蒸气压; (D) 液面下的附加压力Δp10、在稀的砷酸溶液通入H S 制备硫化砷溶胶(As S ),该溶胶的稳定剂是H S,则其胶团结构式是:(A) [(As S ) ·nH ,(n-x)HS ] ·xHS(B) [(As S ) ·nHS ,(n-x)H ] ·xH(C) [(As S ) ·nH ,(n-x)HS ] ·xHS(D) [(As S ) ·nHS ,(n-x)H ] ·xH11、对表面活性剂溶液,要较好发挥其作用,溶液中表面活性剂单体浓度是:(A) 大于CMC (B) 小于CMC(C) 接近、稍大于CMC (D) 等于零12、在pH <7 的Al(OH) 溶胶中,使用下列电解质使其聚沉:①MgCl、②Na 2SO 4、③K Fe(CN) 。

数分试卷

数分试卷

浙江工业大学数学分析(二)期末试卷(A)09-06 班级 学号 姓名 成绩一、填空题(21%)1、封闭曲线θ3cos =r ⎪⎭⎫ ⎝⎛≤≤-66πθπ所围的面积是 。

2、反常积分⎰∞++02312cos dx x x 是条件收敛还是绝对收敛?答: 。

3、级数ln 112n n ∞=∑是收敛还是发散?答: 。

4、幂级数1(1)2nn x n ∞=-∑的收敛域为 。

5、设)(x f 是以π2为周期的周期函数,在[)ππ,-上22)(x x f -=π,则其Fourier级数的和函数)(x S 在π27处的值72S π⎛⎫= ⎪⎝⎭ 。

6、设()22y x f z -=,其中f 可导,则=∂∂+∂∂yz x x z y 。

7、函数 xyz z xy u -+=32在点)1,1,1(处的梯度为________________;在点)1,1,1(处沿方向}2,1,0{=l 的方向导数为________________。

二、选择题(16%)1、若),(y x f z =于点()00,y x 处可微,则下列结论错误的是 ( )(A )),(y x f 于点()00,y x 处连续;(B) ),(),,(y x f y x f y x 于点()00,y x 处连续;(C ) ),(),,(y x f y x f y x 于点()00,y x 处存在;(D) 曲面),(y x f z =在()),(,,0000y x f y x 处有切平面。

2、二重极限与累次极限之间的关系正确的是 ( )(A)若二重极限存在,则两个累次极限均存在且相等;(B)若二重极限存在,且其中一个累次极限存在,则另一累次极限存在;(C)若累次极限均存在但不相等,则重极限必不存在;(D)若二重极限不存在,则两个累次极限中至少有一个不存在.3、以下命题中正确的是: ( )(A) 若1()f x dx +∞⎰收敛,则必有:lim ()0x f x →+∞=; (B) 若级数∑∞=1n n u 收敛,且1lim =∞→n n n u v ,则级数∑∞=1n n v 收敛。

浙工大物理化学期末卷1

浙工大物理化学期末卷1

物理化学(A)试卷1姓名:学号:班级:一、选择题(每题2分,20题,共40分)1、下列物系中的状态性质中不是强度性质的是:(A) ;(B) ;(C) ;(D) 。

2、下列说法正确的是:(A) 物体的温度越高,其热量越多;(B) 物体的温度越高,其热力学能越大;(C) 体系的焓等于恒压热;(D) 体系的焓就是体系所含的热量。

3、对于可逆变化有:Δ =∫ / ,下述各说法中,哪一个正确:(A) 只有可逆变化才有熵变 (B) 可逆变化没有热温商(C) 可逆变化熵变与热温商之和相等 (D) 可逆变化熵变为零4、一个热力学体系经过一个循环过程,其状态函数变化等于多少?(A) ΔA及ΔG等于零,ΔS、ΔU、ΔH不等于零(B) ΔA、ΔG、ΔU 等于零,ΔS、ΔH不等于零(C) 只是ΔS不为零(D) ΔA、ΔG、ΔU、ΔS、ΔH 均为零5、有关热力学可逆过程的下列叙述中正确的是:(A) 它是体系一定回到初态的过程;(B) 体系的状态变化无限缓慢的过程就是可逆过程;(C) A、B 状态间绝热可逆过程作最大功;(D)可逆过程中体系的状态变化可以在状态图上用曲线表示6、下列过程ΔH≠0的是:(A) 理想气体恒温膨胀过程;(B) 恒压、恒熵和不做非体积功的可逆过程;(C) 真实气体的节流膨胀过程;(D) 任一个绝热且不做非体积功的过程。

7、下列偏导数属于偏摩尔量的是:(A) ( S/ n ) , (B) ( G/ n ) ,(C) ( U/ n ) , (D) ( H/ n ) ,8、液态理想混合物的热力学定义可用体系中任一组分 B的化学势来表示,即:(A) = (T,P)+RT (P/P )(B) = (T,P)+RT (C /C )(C) = (T,P)+RT x(D) = (T,P)+RT (m /m )9、对于H O(g)+CO(g)=CO (g)+H (g) 反应,下列哪些因素改变不影响其平衡:(A)升高温度(B)通入氢气(C)降低温度同时增加压力(D)通入惰性气体10、对于同一化学反应,若反应方程式中计量系数写法不同,则其标准平衡常数K 和标准摩尔吉布斯自由能变 G :(A) K 相同, G 不同; (B) K 不同, G 相同;(C) K 和 G 都不同; (D) K 和 G 都相同11、对于孤立体系中发生的实际过程,下列关系中不正确的是:(A) =0 ;(B) =0 ;(C) Δ =0 ; (D) Δ =0 。

浙工大线性代数2020-2021第二学期试卷A

浙工大线性代数2020-2021第二学期试卷A

浙 江 工 业 大 学 线 性 代 数 期 末 试 卷( 2020 ~ 2021 第 二 学 期 )任课教师 学院班级: 选课班中编号:学号: 姓名: 得分:一.填空题(每空3分, 共30分)1. 已知222⨯=−A ,则13−−=A .2. 若对任意的3维列向量123=⎛⎫⎪⎪ ⎪⎝⎭x x x x ,121232−⎛⎫= ⎪+−⎝⎭x x Ax x x x 则=A .3. 120⎛⎫ ⎪= ⎪ ⎪⎝⎭A ,R()A *= ,A *= . 4. 将2阶矩阵A 的第一列乘以3,再将第二列的 -2倍加到第一列得矩阵B ,则满足B = AP 的矩阵P 为 .5. 如果向量组1110,,100−⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭k k αβγ线性无关,则参数k 满足 .6. 向量空间{}(,,)|23==x y z x y z 的维数是 ,一组基为 .7. 实向量空间2R 中的向量51⎛⎫= ⎪⎝⎭β在基1213,24αα⎛⎫⎛⎫== ⎪ ⎪⎝⎭⎝⎭下的坐标为 .8. 若112−⎛⎫= ⎪⎝⎭P AP ,其中(),αβ=P ,令()+,=αββQ ,则1−=Q AQ .二. 单项选择题(每小题2分,共10分)1. 设n 阶矩阵, , A B C 满足关系式ABC =E ,则以下一定正确的是 ( ). (A) =ACB E (B) =CBA E (C) =BAC E (D) =BCA E2. 设A 为可逆方阵,*A 是A 的伴随矩阵,则-1(2)*=A ( ).(A)-112A A (B) 12A A (C) -12A A (D) 2A A 3. 向量组12,,...,(2)≥S s ααα线性无关的充分必要条件是( ). (A) 12,,...,S ααα中任意向量非零 (B) 12,,...,S ααα中任意两个向量线性无关 (C) 12,,...,S ααα中任意s-1个向量线性无关(D) 12,,...,S ααα中任意向量都不能由其余向量线性表示 4. 若0⨯=n n A ,但0*≠A ,则0=AX 的解空间维数为( ). (A) n (B) 1 (C) n-1 (D) 0 5. 设3阶矩阵A 的特征值为1,-1,2,则下列矩阵中( )是可逆矩阵. (A) 22E -A (B) 2-E A (C) 2E +A (D) +E A三. 计算题(每题10分,共50分)1. 已知11111200=10301004⎛⎫⎪⎪⎪⎪⎝⎭A,求A.2. 已知111011,001A−−⎛⎫⎪= ⎪⎪⎝⎭211,012B⎛⎫= ⎪⎝⎭矩阵X满足2,−=XA B X求X.3. 已知向量组12341-1123,,-1,25316⎛⎫⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪ ⎪==== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭λαααα的秩为2.(1) 求参数λ;(2) 求该向量组的一个极大无关组,并用该极大无关组表示其余向量.4. 已知线性方程组1232123123 + +4+24=⎧⎪−+−=⎨⎪−=−⎩x x kx x kx x k x x x 问:(1)当参数k 满足什么条件时,方程组有唯一解?无解?有无穷多解? (2)有无穷多解时,求方程组的通解.5. 设111111111⎛⎫ ⎪= ⎪ ⎪⎝⎭A ,(1)求A 的特征值和特征向量;(2)求正交矩阵Q 及对角矩阵Λ,使得1=−Q AQ Λ.四、证明题(共10分)1. (6分)设,A B 均为n 阶矩阵,其中A 为可逆矩阵,证明:AB 与BA 相似.2. (4分)设A 为n 阶方阵,证明:存在A 的如下分解:=A BP ,其中2=B B ,P 为可逆矩阵.。

浙江工业大学2011-2012_2_期末试卷A及答案

浙江工业大学2011-2012_2_期末试卷A及答案
(4)控制电路中缺少应串联于其中的热继电器的常闭触点FR,否则将无法实现过载保护。(3分)
A. 50Hz B. 48Hz C. 2Hz
19.熔断器对三相异步电动机起(A)的作用。
A.短路保护B.欠压保护C.过载保护
20.关于TN-S供电系统,下列说法中错误的是(A)。
A. TN-S适用于三相三线制供电系统B. TN-S系统中工作零线N和专用保护线PE严格分开C. TN-S系统中用电设备的金属外壳与零线连接
12.在感性的正弦交流电路中,下列方法可提高电路的功率因数(C)
A.在电路上并联电感B.在电路上串联电感C.在电路上并联电容
13.在R、L串联的正弦交流电路中,R =40,XL= 30,电路的无功功率Q= 480 var,则视在功率S为( B )。
A. 866 V·AB. 800 V·AC. 600 V·A
(2分)
2.电路如图2-2所示,在开关S闭合前电路已处于稳态,求开关闭合后的电压 。
图2-2
解:由换路定则:
(3分)
(3分)
(2分)
根据三要素法, 时:
(2分)
3.有一线圈的电阻 、感抗 与容抗 的电容串联后接至正弦交流电源,通过的电流为0.2A,求总电压U、线圈端电压 、有功功率和视在功率。
解:令 A,
16.在电源电压不变的情况下,空心线圈中引入铁心后,则交流励磁电流(C)。
A.变大B.不变C.变小
17.三相异步电动机是由于( C )而产生电磁转矩的。
A.定子磁场与定子电流的相互作用
B.转子磁场与转子电流的相互作用
C.旋转磁场与转子电流的相互作用
18.某三相异步电动机在额定运行时的转速为1440r/min,电源频率为50Hz,此时转子电流的频率为(C)。

2006-2007学年浙江工业大学第二学期期末试卷

2006-2007学年浙江工业大学第二学期期末试卷

浙江工业大学2006/2007学年第二学期有机化学C 期末考试试卷班级 姓名 学号 任课教师一、 命名或写结构式。

(共10分,每小题1分)O HO 2NN O 2N O 21.C H OO H2.O C H 34. 蒽醌N H OO5.2, 2-二甲基丙酸乙酯7.(E ) - 4 -甲基-2- 戊烯二酸9.苯甲醛-2,4-二硝基苯腙10.3. 3- 甲基四氢呋喃氢氧化苄基三甲铵8.6.NNO H二、完成下列反应 (20分,每空1分)°OCHCH 3H 3C H 3CCH 3ONa CH 3O H2.+HI+1.O CH 3C H 3C H O+32-2 ) H +1) P C l 32) N H 35.6.C O O C 2H 5C O O C 2H 5C 2H 5O N a C 2H 5O HC H 37.NH O OC H 3C H 2BrEtO Na H 2O O H 323.BrC H 2C H 2C CH 3O22H +M g/Ether, C H3CHO34.C HOO C H 3浓 碱+2三、理化性质比较(12分,每小题2分)1、将下列化合物按其与Lucas 试剂作用由快到慢的顺序排列( )。

a. 2-丁醇b. 2-甲基-2-丁醇c. 2-甲基-1-丙醇2、下列化合物酸性由强到弱的排列次序( )。

a. 苯酚b. 对硝基苯酚c. 对氯苯酚d. 对甲苯酚3、下列化合物与NaHSO 3加成反应活性由大到小的次序为( )。

4、下列化合物能发生碘仿反应的是( )。

5、下列化合物碱性由强到弱的排列次序( )。

8.N O 2Fe+ H C lN aN O 2CH 9.OHCNLiAlH 43++O HC H 3N (C H 3)310. a. (CH 3)2CHCHOb. CH 3CH 2OHc. C 6H 5CHOd. CH 3COCH 2CH 3N H 2a. b.N H 2c.O C H 3N H 2d.N O 2N O 2CN H2OCHOa. F 2CHCHO b. ClCH 2CHO c. CH 3CHOd.6、下列化合物中具有芳香性( )四、基本概念题(8分)1、为什么乙二醇及其甲醚的沸点随分子量的增加而降低 ?(2分)2、一般羰基化合物的烯醇式异构体都不稳定,为什么乙酰乙酸乙酯的烯醇式异构体却具有较大的稳定性 ?(3分)3、糠醛是不含α氢原子的醛,其化学性质与苯甲醛或甲醛相似,试完成以下反应。

浙江工业大学《无机化学》(上册)期末考试模拟试卷及答案(原稿)[1]

浙江工业大学《无机化学》(上册)期末考试模拟试卷及答案(原稿)[1]

《无机化学》(上册)期末考试模拟试卷试卷(一)一、填空题(每空1分,共30分)1.测定溶液酸碱性的方法很多,定性测定可用_______________.定量测定可用____________.2.原子轨道组成分子轨道的原则是: 对称性原则、________________原则、_________________原则。

3。

已知下列元素在周期表中的位置,写出它们的外层电子构型和元素符号.(1)第四周期第ⅣB族________________________;(2) 第五周期第ⅥA族________________________。

4.空气中的SO2与水作用并经氧化,生成___________ 使大理石转化为_________, ________,因而使雕象遭到破坏.5.分子分散系是一种单相体系,粗分散系是一种_________,胶体分散系是一种_________。

6.已知反应: 2NO+Cl2=2NOCl,当C NO增大一倍,其反应速度为原来的4倍,而C(Cl2)增大一倍,其反应速度为原来的2倍,其反应速度方程为______________,该反应为___________级反应。

7。

含醋酸,醋酸钠各0.1mol·L—1的溶液90mL,加入10mL0.01mol·L-1的NaOH,加NaOH前溶液的pH值是________,加入NaOH后溶液的pH是_______. 已知KHAc=1。

8×10-58.从水的E-pH图可知:(b)线代表_________________________________________处于电化学平衡时电极电势和pH值的关系线,(a)(a)线代表________________________________________处于电化学平衡时电极电势和pH值的关系线。

9.稀有气体的第一个化合物—————-———,,是由——-———-—于————————年制备的。

浙江工业大学大学物理期末考试试卷(含答案)

浙江工业大学大学物理期末考试试卷(含答案)

浙江工业大学大学物理期末考试试卷(含答案)一、大学物理期末选择题复习1.一个质点在做圆周运动时,则有()(A) 切向加速度一定改变,法向加速度也改变(B) 切向加速度可能不变,法向加速度一定改变(C) 切向加速度可能不变,法向加速度不变(D) 切向加速度一定改变,法向加速度不变答案B2.对质点组有以下几种说法:(1) 质点组总动量的改变与内力无关;(2) 质点组总动能的改变与内力无关;(3) 质点组机械能的改变与保守内力无关.下列对上述说法判断正确的是()(A) 只有(1)是正确的(B) (1) (2)是正确的(C) (1) (3)是正确的 (D) (2) (3)是正确的答案C3.下列说法正确的是()(A) 闭合回路上各点磁感强度都为零时回路内一定没有电流穿过(B) 闭合回路上各点磁感强度都为零时回路内穿过电流的代数和必定为零(C) 磁感强度沿闭合回路的积分为零时回路上各点的磁感强度必定为零(D) 磁感强度沿闭合回路的积分不为零时回路上任意一点的磁感强度都不可能为零答案B4.人造地球卫星,绕地球作椭圆轨道运动,地球在椭圆的一个焦点上,则卫星的()(A) 动量不守恒,动能守恒(B) 动量守恒,动能不守恒(C) 对地心的角动量守恒,动能不守恒1、(D) 对地心的角动量不守恒,动能守恒答案C5.两根长度相同的细导线分别多层密绕在半径为R和r的两个长直圆筒上形成两个螺线管,两个螺线管的长度相同,R=2r,螺线管通过的电流相同为I,螺线管中的磁感强度大小B R 、B r 满足( )(A ) r R B B 2=(B ) r R B B =(C ) r R B B =2(D )r R B B 4=答案C6.一个半径为r 的半球面如图放在均匀磁场中,通过半球面的磁通量为()(A )B r 2π2 (B ) B r 2π(C )αB r cos π22 (D ) αB r cos π2答案D7. 质点作曲线运动,r 表示位置矢量,v 表示速度,a 表示加速度,s 表示路程,t a 表示切向加速度。

浙江工业大学《自控》期末试题及解答2011-12考研复习资料

浙江工业大学《自控》期末试题及解答2011-12考研复习资料

浙江工业大学2011/2012(1)期终考试《自动控制原理》试题课程 自动控制原理 姓名 班级 学号第一部分 简答题(共4题,每题4分,共16分。

答题必须简明扼要。

)1从控制的观点分析飞机在气流中能保持预定航向行驶的原因。

飞行控制系统是采用反馈的闭环控制系统,对于外部的扰动有一定的抑制能力。

气流相当于对于系统的扰动。

在扰动作用下,闭环反馈系统能够保持原来的性能。

2列举控制理论中的几种数学模型。

(4种以上)(1)传递函数,(2)微分方程,(3)结构框图,(4)频率特性,(5)状态模型。

3简述滞后串联校正改善系统性能的原因。

(1)降低截止频率,有助于增加相位裕度;(2)降低高频增益,有助于抑制高频扰动和噪声;(3)在不太影响暂态性能情况下,可以提高增益减低稳态误差。

4如何测量得到一个含积分环节(例如,1(1)s Ts +)的频率特性?构成一个稳定的闭环控制系统。

在闭环系统中分别测量环节的输入和输出处信号,而获得频率特性。

第二部分 填空题(共20分。

)5图示闭环系统稳定时参数K 的取值范围为 42>K 。

(5分)第5题图6零初始条件下,某系统在单位脉冲()t δ作用下的响应为25()510tt k t ee --=+,该系统的输入输出间的传递函数为21545710s s s +++ ,该系统输入()u t 与输出()y t 间的微分方程为 ()7()10()15()45()y t y t y t u t u t ++=+。

(每空2分) 7已知单位负反馈系统的开环传递函数为(1)()()(21)(31)K s G s H s s s s +=++,0,K >在单位阶跃1()t 、单位斜坡t 、单位加速度2t 作用下系统的稳态误差分别为 0 , 1/K , 无穷大 。

(每空2分) 8某离散系统及初始状态如下:则该离散系统在初始状态作用下的系统输出响应y(k)为:。

(5分)第三部分 计算分析和证明题(共4题,共64分。

0809第二学期复习卷2

0809第二学期复习卷2

0809第⼆学期复习卷2浙江⼯业⼤学2008 / 2009学年第⼆学期期终复习卷2课程姓名班级[⼀]单选择题(1) 在⾮等压过程中加热某体系, 使其温度从T 1升⾄T 2, 吸热Q , 则此过程的焓增量ΔH 为: (a) ΔH =Q (b) ΔH =0(c) ΔH =△U +△(pV ) (d) ΔH 等于别的值(2) 凝固热在数值上与下列哪⼀种热相等: (a) 升华热 (b) 溶解热 (c) 汽化热 (d) 熔化热(3) 下列四种表述(1) 等温等压下的可逆相变过程中,体系的熵变ΔS =ΔH 相变/T 相变 (2) 隔离体系经历⼀⾃发过程总有 d S > 0(3) ⾃发过程的⽅向就是使隔离体系混乱度增加的⽅向 (4) 在绝热可逆过程中,体系的熵变为零两者都不正确者为:(a) (1),(2) (b) (3),(4) (c) (2),(3) (d) (1),(4)(4) 吉布斯⾃由能的含义应该是(a) 是体系能对外做⾮体积功的能量(b) 是在可逆条件下体系能对外做⾮体积功的能量 (c) 是恒温恒压可逆条件下体系能对外做⾮体积功的能量 (d) 按定义理解 G =H - TS 。

(5) 2 mol 液态苯在其正常沸点(353.2 K)和101.325 kPa 下蒸发为苯蒸⽓,该过程的ΔvapA等于(a) 23.48 kJ (b) 5.87 kJ(c) 2.94 kJ (d) 1.47 kJ(6) (1) 处于标准态的 CO 2(g) 和 O 2 (g),其标准燃烧焓值为零(2) 因为m ln r p G RT K ?=-$,⽽p K是由平衡时的组成表⽰的,所以m r G ?$表⽰平衡时产物的吉布斯⾃由能与反应物的吉布斯⾃由能之差 (3) ⽔在25℃, p 下蒸发,求算熵变的公式为()m m mS H G T ?=?-?$$$(4) 在恒温,恒压下可逆电池反应,求算熵变的公式为 r m r m S H T ?=?上述说法正确的是:(a) 1 , 2 (b) 2 , 3 (c) 1 , 3 (d) 3 , 4(7) ⼆组分理想溶液总的蒸⽓压:(a) 与溶液的组成⽆关; (b) 介于⼆纯组分蒸⽓压之间;(c) ⼤于任⼀组分的蒸⽓压; (d) ⼩于任⼀组分的蒸⽓压(8) 2O 在⽔中的亨利常数(a )因升温⽽增⼤(b )因加压⽽增⼤(c )因降温⽽增⼤(d )因溶⼊2H ⽽增⼤(9) 理想⽓体反应 M +N =3P ,在某温度下,KP =100 kPa ,现测得在某时刻时,各物质的分压均是100 kPa ,这时系统的化学亲和势 A = -ΣB B µν应该:(a )⼤于零(b )⼩于零(c )等于零(d )⽆法确定上述关系(10) 设可逆反应M??→←??P,由等温式求出某⼀指定状态时Δr G m <0,可表明在该指定状态:(a )M能转化成P(b )M转化成P的同时,P转化成M(c )事实上M和P达平衡(d ) 1mol M 转化成 1mol P 后,⾃由能的变化值(11) 在298K 时,⽓相反应 H 2+I 2=2HI 的Δr G m $=-16778 J ·mol -1,则反应的平衡常数K 为:(a ) 2.0×106 (b ) 5.91×106(c ) 873 (d ) 18.9(12) 对于理想⽓体反应:N 2(g)+3H 2(g) ==2NH 3(g),下列结果正确的是:(a )K c =K (RT /p) (b )K = K p (p)2(c )K p = K c (RT )2 (d )K x =K(p/p)-1(13) 对于⼀个化学反应,测得不同温度时的平衡常数值,可以确定该反应的:(a ) m r G ?,r m H ? (b )m r G ?,r m S ?(c ) r m H ? ,r m S ? (d ) m r G ?(14) 对于反应C(s)+H 2O(g) = CO(g)+H 2(g),298K 时,r m H ?=131.31kJ ·mol -1,下⾯哪种情况对正向反应有利:(a )增⼤总压(b )通⼊H 2(c )提⾼温度(d )降低温度(15) 当反应 C 2H 6(g)===CH 2CH 2(g)+H 2(g) 达到平衡后,向其中加⼊不参与反应的惰性⽓体,则:(a )等压时平衡向左移动(b )等容时平衡向右移动(c )等压时平衡向右移动(d )等容时平衡向左移动(16) 当反应系统中加⼊不参与反应的惰性⽓体时,下列叙述中不正确的是:(a )使平衡常数改变,平衡发⽣移动(b )总压⼀定时,惰性⽓体的加⼊和减⼩反应系统总压的效果相同(c )对于∑νB >0 的反应,增加惰性⽓体,反应正向移动(d )对于∑νB <0 的反应,增加惰性⽓体,反应逆向移动(17) 已知红⾠砂α-H g S 与⿊⾠砂β-H g S 的转化反应α-H g S====β-H g S ,其Δr G m $=980-1.456 T (J),那么常温下判断正确的是:(a )⿊⾠砂稳定(b )红⾠砂稳定(c )两种⾠砂可互相转化(d )⽆法判定(18) (1) 处于标准态的 CO 2(g) 和 O 2 (g),其标准燃烧焓值为零(2) 因为Δr G m $= -RT ln K $,⽽K $是由平衡时的组成表⽰的,所以Δr G m $表⽰平衡时产物的吉布斯⾃由能与反应物的吉布斯⾃由能之差(3) ⽔在25℃, p ?下蒸发,求算熵变的公式为ΔS m $=(ΔH m $-ΔG m $)/ T(4) 在恒温,恒压下可逆电池反应,求算熵变的公式为Δr S m = Δr H m / T上述说法正确的是:(a ) 1 , 2 (b ) 2 , 3 (c ) 1 , 3 (d ) 3 , 4(19) 巳知反应 CH 3OH(g)=HCHO(g)+H 2(g); r m H ?=92.0 KJ ·mol -1HCHO(g)= CO(g)+H 2(g); r m H ?=8.4 KJ ·mol -1若要提⾼平衡混合物中 HCHO 的含量, 应采取:(a )升⾼温度(b )降低温度(c )增加压⼒(d )减⼩压⼒(20) 对于完美晶体, 在 00ln S k =Ω中, 应当是_______(a) 01S = (b) 00Ω= (c) 01Ω= (d) 01Ω>(21)室温下,10p 的理想⽓体绝热节流膨胀⾄ 5p 的过程有(1) 0W > (2) 12T T > (3) 0Q = (4) 0S ?> 其正确的答案应是(a) (3),(4) (b) (2),(3) (c) (1),(3) (d) (1),(2) (22)纯物质单相系统的下列各量中不可能⼤于零的是(a )(/)p H S ?? (b )(/)p G T ?? (c )(/)V U S ?? (d )(/)S H p ??[⼆]计算题有⼀⽓体反抗外压202.65kPa 使其体积从10L 膨胀到20L ,从环境吸收了1255J 的热量,求此⽓体内能的变化?[三]计算题0.1kg C 6H 6(l)在Op ,沸点353.35K 下蒸发,已知66(C H )gl m H ?=30.80 kJ mol -1。

浙工大09-10(II)线性代数期末卷-参考答案

浙工大09-10(II)线性代数期末卷-参考答案
此时,方程组有无穷多解。
2
对应的最简方程为 ⎨
பைடு நூலகம்
x1 = 1 ⎩ x2 − x3 = −1 ⎧
⎛ x1 ⎞ ⎛ 1 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 通解为 ⎜ x2 ⎟ = ⎜ − 1⎟ + t ⎜ 1 ⎟, t ∈ R . ⎜ x ⎟ ⎜ 0 ⎟ ⎜1⎟ ⎝ 3⎠ ⎝ ⎠ ⎝ ⎠ ⎛1 0 1⎞ ⎟ ⎜ (1)求 A 的特征值和对应的特征向量; 4. (14 分)设 A = ⎜ 0 1 1 ⎟ , ⎜1 1 2⎟ ⎠ ⎝
四、证明题(略)
4
T T T T
(1)求 p 的值使该向量组线性无关; (2)求 p 的值使该向量组线性相关,并在此时求出它的秩和一个极大无关组,同时将其余 向量用该极大无关组线性表示出来。
1
解: (1)令 A = (α1
α 2 α3
⎛1 −1 3 ⎜ ⎜0 − 2 −1 α4 ) → ⎜ 0 0 −7 ⎜ ⎜0 0 0 ⎝
浙 江 工 业 大 学 线 性 代 数 期 末 试 卷 参 考 答 案
( 2009 ~ 2010 第 二 学 期 )
一、选择题(每题 2 分,共 10 分) : 1. B; 2. D; 3. C; 4. A; 5. D
二、填空题(每空 3 分,共 30 分) 2. B(B − E )
−1
1.
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2
λ
−1
− 0.8 − 1 1 ⎞ ⎛ 2 ⎜ ⎟ ~ 1 2⎟ , A = ( A, β ) → ⎜ − 0.8 − 1 ⎜ 0 0 0 9⎟ ⎝ ⎠
此时,方程组无解。 (3)当 λ = 1 时,增广矩阵为
⎛2 1 −1 1 ⎞ ⎛1 0 0 1 ⎞ ⎜ ⎟ ⎜ ⎟ ~ 2 ⎟ → ⎜ 0 1 − 1 − 1⎟ , A = ( A, β ) = ⎜ 1 − 1 1 ⎜ 4 5 − 5 − 1⎟ ⎜ 0 0 0 0 ⎟ ⎝ ⎠ ⎝ ⎠

浙江工业大学物理化学2009- 2010(2)学年复习题2

浙江工业大学物理化学2009- 2010(2)学年复习题2

浙江工业大学2009 / 2010(2)学年期终复习卷2一、选择题答案1、将CuSO 4水溶液置于绝热箱中,插入两个铜电极,以蓄电池为电源进行电解,可以看作封闭体系的是(a) 绝热箱中所有物质 (b) 两个铜电极 ;(c) 蓄电池和铜电极 (d) CuSO 4水溶液 。

2、当热力学第一定律写成d U = δQ – p d V 时,它适用于(a).理想气体的可逆过程 (b). 封闭体系的任一过程(c). 封闭体系只做体积功过程 (d). 封闭体系的定压过程3、1mol 理想气体经历可逆绝热过程,功的计算式有下列几种,其中哪一个是错误的(a ) C v (T 1- T 2) (b )C p (T 2-T 1)(c ) (P 1V 1- P 2V 2)/(r-1) (d )R(T 1-T 2)/(r-1)4、将某理想气体从温度T 1加热到T 2。

若此变化为非恒压过程,则其焓变ΔH 应为何值?(a) ΔH =0 (b) ΔH =C p (T 2-T 1)(c) ΔH 不存在 (d) ΔH 等于其它值5、下述说法哪一个正确?(a) 热是体系中微观粒子平均平动能的量度(b) 温度是体系所储存热量的量度(c) 温度是体系中微观粒子平均能量的量度(d) 温度是体系中微观粒子平均平动能的量度6、在一个绝热的刚性容器中发生一个气相反应,使系统的温度从1T 升高到2T ,压力从1p 增大到2p ,则(a) Q >0, W >0, U ∆>0 (b) Q >0, W ﹦0, U ∆>0(c) Q ﹦0, W ﹦0, U ∆>0 (d) Q ﹦0, W ﹦0, U ∆﹦07、关于基尔霍夫定律适用的条件,确切地说是(a ) 等容条件下的化学反应过程(b ) 等压条件下的化学反应过程(c ) 等压或等容且不做非体积功的化学反应过程(d ) 纯物质在不同温度下的可逆相变过程和等压反应过程8、封闭体系中,若某过程的R A W ∆=,应满足的条件是(a )等温、可逆过程 (b )等容、可逆过程(c )等温等压、可逆过程 (d )等温等容、可逆过程9、可逆机的效率为η,冷冻机的冷冻系数为β,则η和β的数值满足(a) η<1,β<1 (b) η≤l ,β≤1(c) η<1,β>1 (d) η<1,β可以小于、等于、大于110、一卡诺热机在两个不同温度之间的热源之间运转, 当工作物质为气体时, 热机效率为42%, 若改用液体工作物质, 则其效率应当(a ) 减少 (b ) 增加(c ) 不变 (d ) 无法判断11、熵变∆S 是:(1) 不可逆过程热温商之和(2) 可逆过程热温商之和(3) 与过程无关的状态函数的改变值(4) 与过程有关的状态函数的改变值以上正确的是(a ) 1,2 (b ) 2,3(c ) 2 (d ) 412、理想气体从状态 I 经自由膨胀到状态 II ,可用哪个热力学判据来判断该过程的自发性?(a) ∆H (b) ∆G(c) ∆S (d) ∆U13、根据熵的统计意义可以判断下列过程中何者的熵值增大?(a) 水蒸气冷却成水 (b) 石灰石分解生成石灰 (c) 乙烯聚合成聚乙烯 (d) 理想气体绝热可逆膨胀14、1mol 的单原子理想气体被装在带有活塞的气缸中,温度是 300K ,压力为 1013250Pa 。

浙江工业大学高等数学期末03-04(二)卷A标准答案

浙江工业大学高等数学期末03-04(二)卷A标准答案

a 2 - x 2 - y 2 a 2 - x 2 - y 2 ⎰ν xy⎝⎰ ν ⎪ 0 ⎭ ∞ 03-04(二)期终试卷答案一.1、 z x - z ;2、2 3 ;3、 ν 3 d 0 4 ν 4 d ϕ 0 0 f (ψ 2 )ψ 2 sin ϕd ψ 4、2; 5、 (-1)n -1 n二.A ;B ;C ;B ;D∂z三.1、 ∂x= 2 f u + yf v ,∂z = ∂y f u + xf v ,dz = (2 f u + yf v )dx + ( f u + xf v )dya x2a 2ax - x 2 2、 ⎰0 dx ⎰- xf ( x , y )dy + ⎰a dx ⎰- 2ax - x 2 f ( x , y )dy a a + a 2 - y 20 a + a 2 - y 2 ⎰0 dy ⎰y f (x , y )dx + ⎰- a dy ⎰- y f (x , y )dx3、∑ : z = D : x 2 + y 2 ≤ a 2 a ⎛ a ⎫ ds = dxdy = dxdy 或 = dxdy ⎪ ⎪ ⎝ ⎭⎰⎰ zds = ⎰⎰ z ⋅ a dxdy∑ D xy z= νa 3sin ν四.1、 lim = lim 1 ⋅n +1 n →∞ n →∞ ν sin ν n = lim 1 ⋅ n = 1 < 1n →∞ ν n +1 ν 所以绝对收敛。

2、 ∑bn sin( n ν ) 是x 在(- ν ,ν )上的付氏级数 n =1因为x 是奇函数, 所以a n = 0b =2 νx sin( nx )dx n ν ⎰0= - 2 ⎛ x cos nx ν - 1 sin (nx ) ⎫ n ν 0 n ⎪ 2(-1)n -1 n 1 + + x 2 y 2 z 2 z 2 u n +1 u n ⎰ 1 z =2 ⎪ ∂z ⎩ ⎩⎫ 23、 lim = lim n + 2 ⋅ 2n n ! = 0 所以收敛区间(- ∞,∞)n →∞ n →∞ 2n +1 (n +1)! n +1 ∞ 1 ⎛ x ⎫n ∞ 1 ⎛ x ⎫ns ( x ) = ∑ (n -1) ⎪ + ∑ ⎪ n =1 x ∞ !⎝ 2 ⎭ 1⎛ x ⎫nn =1 n !⎝ 2 ⎭ x = ∑ ⎪ + e 2 -1 2 n = 0 n !⎝ 2 ⎭=⎛ x + ⎝ x 1⎪e 2 -1 ⎭⎧ ∂z = 14 - 8y - 4x = 0 ∂x 五、(1) ⎨ ⎪ = 32 - 8x - 20 y = 0 ⎩∂y⎧x = 1.5 驻点⎨ y = 1∂2 z ∂2 z ∂2 z2 因为∂x 2 = -4 , ∂y ∂x = -8 , ∂y 2 = -20 所以 AC - B = 16 > 0. A < 0所以当 x = 1.5, y = 1 时, z 取唯一极大值,为最大值。

浙工大期末物理化学卷2

浙工大期末物理化学卷2

浙⼯⼤期末物理化学卷2浙⼯⼤期末物理化学卷2浙江⼯业⼤学物理化学期末考试06/07第⼆学期(卷A)姓名:学号:班级:任课教师:总分:题号⼀⼆1 ⼆2 ⼆3 ⼆4 ⼆5 ⼆6 总计分数⼀、选择题(每题2分,20题,共40分)注意:题号后标有“A”的题⽬只要求修物理化学A的同学解答;题号后标有“B”的题⽬只要求修物理化学B的同学解答。

请将答案填在表格中:1 2 3 4 5 6 7 8 9 10 11 12 14 14 15 16 17 18 19 201、下列过程ΔH≠0的是:(A) 理想⽓体恒温膨胀过程;(B) 恒压、恒熵和不做⾮体积功的可逆过程;(C) 真实⽓体的节流膨胀过程;(D)任⼀个绝热且不做⾮体积功的过程。

2、下列说法正确的是:(A) 物体的温度越⾼,其热量越多;(B) 物体的温度越⾼,其热⼒学能越⼤;(C) 体系的焓等于恒压热;(D)体系的焓就是体系所含的热量。

3、封闭体系在不可逆循环中,热温商之和Σ(Q/T):(A) ⼤于零 (B) 等于零(C) ⼩于零 (D) 不可能⼩于零4、对于孤⽴体系中发⽣的实际过程,下式中不正确的是:(A) W = 0 (B) Q = 0 (C) ΔS > 0 (D) ΔH = 05、恒压下纯物质体系,当温度升⾼时其吉布斯⾃由能:(A) 上升 (B) 下降(C) 不变 (D) ⽆法确定6、体系的状态改变了,其内能值(A)必定改变(B)必定不变(C)不⼀定改变(D)状态与内能⽆关7、体系进⾏⼀次循环过程(A)W=0 (B)|Q|=|W|(C)Q=0 (D)U=08、理想⽓体绝热向真空膨胀,则(A) ΔS= 0,W= 0 (B) ΔH = 0,ΔU = 0(C) ΔG = 0,ΔH = 0 (D) ΔU = 0,ΔG = 09、下述说法哪⼀个错误?(A) 封闭体系的状态与其状态图上的点⼀⼀对应(B) 封闭体系的状态即是其平衡态(C) 封闭体系的任⼀变化与其状态图上的实线⼀⼀对应(D) 封闭体系的任⼀可逆变化途径都可在其状态图上表⽰为实线10、下述说法中哪⼀个错误?(A) 体系放出的热量⼀定等于环境吸收的热量(B) 体系温度的降低值⼀定等于环境温度的升⾼值(C) 热⼒学平衡时体系的温度与环境的温度相等(D) 若体系1与体系2分别与环境达成热平衡,则此两体系的温度相同11、A,B,C 三种物质组成的溶液,物质 C 的偏摩尔量为 (A) ()A BC ,,,T p n n n µ?? (B) ()A BC ,,,T p n n G n ?? (C) ()A BA ,,,T p n n A n ?? (D) CA n n p TB n H ,,,)/(??12、下列各式中哪个是化学势?(a)CB ,,T S n H n ??? ???? (b)CB ,,T p n A n ??(c)CB ,,T V n G n ??(d) CB ,,S V n U n ??? ????13、饱和理想溶液中溶质的化学势µ与纯溶质的化学势µ*的关系式为(A )µµ=* (B )µµ>* (C )µµ<* (D )不能确定14、在⼀定温度压⼒下,对于只有体积功的任⼀化学反应,能⽤于判断其反应⽅向的是:(A )rmG ?(B )K (C ) mr G ? (D ) mr H ?15、在302 K 时,标准压⼒下, A →B 过程 ?r H = -102kJ, ?r S = -330 J/K ,可判断该过程:(A )⾃发(B )⾮⾃发(C )平衡(D )⽆法判断16、在等温等压下,当反应的 rmG ?= 5 kJ ·mol -1时,该反应能否进⾏?(A )能正向⾃发进⾏(B )能逆向⾃发进⾏(C )不能判断(D )不能进⾏17、在α,β两相中均含有A 和B 两种物质,当达到平衡时,下列种哪情况是正确的:A B A A (a) (b) αααβµµµµ== A B A B(c) (d) αβββµµµµ==18A 、A 、B ⼆组分组成理想溶液,在⼀定温度下,若A Bp p **>,则(A )A A x y > (B )A A x y = (C )AA x y < (D )⽆法判断19A 、封闭体系经过⼀循环过程后,则(A )体系的熵增加(B ) U= 0(C ) Q = 0 (D )体系的T 、P 都不变20A 、在某压⼒下, ⼀个化学反应的Δr H m$=0, Δr S m$>0 ,表⽰该反应:(A )平衡常数 K > 1 (B )平衡常数K=1(C)平衡常数K< 1 (D)处于化学平衡状态18B、下列体系属于单相的是:(A) 极细的斜⽅硫和单斜硫混合物(B) 漂⽩粉(C) ⼤⼩不⼀的⼀堆尿素碎粒(D) 墨汁19B、关于相的说法正确的是:(A) 不同的相间存在界⾯(B) 同⼀相间不存在界⾯(C) 同⼀物质形成的固体⼀定是单相(D) 两种物质形成的固体混合物⼀定是两相体系20B、盐⽔三组分体系相图中, 最多能⼏种固体盐共存在 :(A) 有⼆种固态盐(B)有三种固态盐(C) 有⼀种固态盐(D)多于四种固态盐⼆、计算题(每题10分,6题,共60分)注意:题号后标有“A”的题⽬只要求修物理化学A的同学解答;题号后标有“B”的题⽬只要求修物理化学B的同学解答。

浙江工业大学2012-2013_2_期末试卷A及答案(加密)

浙江工业大学2012-2013_2_期末试卷A及答案(加密)

浙江工业大学期终考试命题稿2012 /2013 学年第 2 学期命题注意事项:一、命题稿请用A4纸电脑打印,或用教务处印刷的命题纸,并用黑墨水书写,保持字迹清晰,页码完整。

二、两份试题必须同等要求,卷面上不要注明A、B字样,由教务处抽定A、B卷。

三、命题稿必须经学院审核,并在考试前两周交教务处。

浙江工业大学2012 /2013 学年第 2 学期试卷课程电工技术基础A 姓名___________________________班级_______________________________________________________________U︒U︒143.1U︒53.1U︒53.136.9(D) 36V图1-6直流电路中,当电路发生换路时,下列说法错误的是:( A 电感电压可以突变 (C) 电容电压不能突变所示直流电路中,电路原已稳定,开关S 在t =0瞬间闭合,则 (C) 0 V (D)-ω=-U I R j C 1=+UI R时的电路得:()3010660C u IR V -==⨯= ()060V -= (2分) (2分) ]3//)10R C ms = (2分) t2I ,I ,以及电路总的有功功率。

U 为参考相量,即2200oU V =∠122001160103o o U I A R jX ∠===∠-+ (22200110o U I A R ∠==∠ (1211330o I I I A =+=∠ 由于负载对称,故:3cos L L P U I ϕ=34.5100.683322017.3⨯=⨯⨯ (2分)22022317.33L L U I ===Ω (2cos 220.68315R Z ϕ==⨯=Ω (ZZZm(1分)0.89=0.9m(1(2分)m(2中的错误,并改正(可用文字说明或直接标在图中)图2-6位置接反,应互换,否则无法在断电情况下更换熔断器。

应接在主触点KM的上方(即主触点KM分)KM应并联在启动按钮SB2的两端,控制电路中缺少应串联于其中的热继电器的常闭触点FR,。

工程图学_浙江工业大学2中国大学mooc课后章节答案期末考试题库2023年

工程图学_浙江工业大学2中国大学mooc课后章节答案期末考试题库2023年

工程图学_浙江工业大学2中国大学mooc课后章节答案期末考试题库2023年1.对于一个点的水平投影和侧面投影,其相等的坐标是()答案:Y坐标2.已知点A (20, 25, 30),则与V面对称的点B应是()答案:B(20, -25, 30)3.已知点E(30, 20, 40)和点F(30, -20, 40),关于点E和点F的相对位置应是()答案:对V面对称4.已知点A(30, 20, 45)及点B(30, 30, 35),则A、B两点的相对位置是()答案:B在A之前、在A之下5.图中的直线AB应是()答案:侧垂线6.已知直线BF两端点的坐标B(45, 60, 30),F(45, 5, 30),则此直线应该是()答案:正垂线7.图中表示的是直线AB和CD互相交叉,关于重影点的标注,下面答案中哪一个是正确的()答案:1、2、3、4 四点均标注正确8.关于图中两直线的相对位置,正确的说法是()答案:两直线相交9.过直线AB最多能作哪几种平面?()答案:正平面、正垂面、一般位置平面;10.这里四个图中,直线AB与平面P的交点,求得正确的是()答案:(c)11.关于两平面(AB//CD)、(EF//GH)的相对位置是()答案:斜交12.关于△ABC与△EFG的相对位置,正确的答案是()答案:相交13.已知立体的两个投影,关于它的侧面投影,画得正确的是()答案:14.已知立体的两个投影,哪一个侧面投影画得正确?答案:15.已知立体的两个投影,哪一个侧面投影画得正确?答案:16.已知立体的两个投影,哪一个侧面投影画得正确?答案:17.已知立体的两个投影,哪一个侧面投影画得正确?答案:18.关于球体得侧面投影,画得正确得是答案:19.关于三个侧面投影,其相贯线画得正确是答案:20.已知立体的两个投影,哪一个侧面投影画得正确?答案:21.已知立体的两个投影,正确的侧面投影是答案:22.已知立体的两个投影,正确的侧面投影是答案:23.关于穿孔棱柱体的侧面投影,正确的是答案:24.已知两立体相贯,关于它的侧面投影,哪一个图画的正确答案:25.关于相贯体的侧面投影哪一个图画的正确答案:26.以下属于正投影基本性质的是()答案:E.积聚性A.从属性C.显实性D.类似性27.挑出图中标注错误的尺寸答案:(3)(4)(7)28.判断图中红色尺寸标注的正误答案: 错误。

浙江工业大学 2020-2021高等数学(上)期末考试题及答案

浙江工业大学 2020-2021高等数学(上)期末考试题及答案

浙江工业大学高等数学(上)期末考试一、单项选择题 (本大题有4小题, 每小题4分, 共16分) 1.设f (x ) cos x (x sin x ),则在x 0处有( ).(A)f (0) 2 (B)f (0) 1(C)f (0) 0 (D)f (x )不可导.2.设 (x ) 1 x1 x , (x ) 3 33x ,则当x 1时( ).(A) (x )与 (x )是同阶无穷小,但不是等价无穷小; (B) (x )与 (x )是等价无穷小;(C) (x )是比 (x )高阶的无穷小; (D) (x )是比 (x )高阶的无穷小.x 3.若F (x )(2t x )f (t )dt,其中f (x )在区间上( 1,1)二阶可导且f (x ) 0,则( ).(A)函数F (x )必在x 0处取得极大值;(B)函数F (x )必在x 0处取得极小值;(C)函数F (x )在x 0处没有极值,但点(0,F (0))为曲线y F (x )的拐点;(D)函数F (x )在x 0处没有极值,点(0,F (0))也不是曲线y F (x )的拐点。

4.设f (x )是连续函数,且f (x ) x 2 1f (t )dt , 则f (x ) (x 2x 2(A)2 (B)2 2(C)x 1 (D)x 2.二、填空题(本大题有4小题,每小题4分,共16分)25.lim (sin xx 01 3x ).6.已知cos x x 是f (x )的一个原函数,则 f (x ) cos xxd x.lim (cos 2cos 227.n nn ncos2n 1n ) .12x 2arcsin x 18.-11 x2dx2.三、解答题(本大题有5小题,每小题8分,共40分)9.设函数y y (x )由方程e x ysin(xy ) 1确定,求y (x )以及y (0). )1 x 7求 d x .7x (1 x )10.x1 xe , x 0设f (x ) 求f (x )dx .322x x ,0 x 1 11.1012.设函数f (x )连续,g (x )并讨论g (x )在x 0处的连续性.g (x )f (xt )dtf (x )lim A x 0x ,且,A 为常数. 求1y (1)xy 2y x ln x 9的解. 13.求微分方程满足四、解答题(本大题10分)14.已知上半平面内一曲线y y (x )(x 0),过点(0,1),且曲线上任一点M (x 0,y 0)处切线斜率数值上等于此曲线与x 轴、y 轴、直线x x 0所围成面积的2倍与该点纵坐标之和,求此曲线方程. 五、解答题(本大题10分)15.过坐标原点作曲线y ln x 的切线,该切线与曲线y ln x 及x 轴围成平面图形D.(1)求D 的面积A;(2) 求D 绕直线x = e 旋转一周所得旋转体的体积V .六、证明题(本大题有2小题,每小题4分,共8分)16.设函数f (x )在 0,1 上连续且单调递减,证明对任意的q [0,1],q 1f (x )d x q f (x )dx.17.设函数f (x )在 0, 上连续,且0xf (x )d x 0,0f (x )cos x dx 0.证明:在 0, 内至少存在两个不同的点 1, 2,使f ( 1) f ( 2) 0.(提F (x )示:设f (x )dx)参考答案一、单项选择题(本大题有4小题, 每小题4分, 共16分) 1、D 2、A 3、C 4、C二、填空题(本大题有4小题,每小题4分,共16分)1cos x 2 () c 6e 35. . 6.2x .7. 2. 8.三、解答题(本大题有5小题,每小题8分,共40分)9.解:方程两边求导x ye (1 y ) cos(xy )(xy y ) 0e x y y cos(xy )y (x ) x ye x cos(xy )x 0,y 0,y (0) 177x 6dx du 10.解:u x 1(1 u )112原式 du ( )du7u (1 u )7u u 11(ln |u | 2ln |u 1|) c 712 ln |x 7| ln |1 x 7| C 77.11.解: 30 31f (x )dx xe dx30 x12x x 2dxxd ( e )x11 (x 1)2dx0 2xxxee 3 2(令x 1 sin )cos d 412.解:由f (0) 0,知g (0) 0。

浙江工业大学之江学院 微积分 期末试题2 期终试卷

浙江工业大学之江学院 微积分 期末试题2 期终试卷

之江学院04级(上)普通班微积分A 期末试卷(A 卷)班级 学号 姓名1、若)(x F 是)(x f 的一个原函数,C 为常数,则下列函数也是)(x f 的原函数的是() A 、)(Cx F B 、)(C xF C 、)(x CF D 、)(x F C +2、下列函数中在区间[-1, 1]上满足罗尔定理条件的是( )A、21()f x x = B 、()f x x = C 、3()f x x = D 、2()f x x =3、设)(x f 为可导函数,且2)0(,0)0(='=f f ,则200)(lim x dtt f x x ⎰→之值为( )A、0 B、1 C、2 D、不存在 4、下列微分方程中是线性微分方程的是( )A、x y y =' B、x y y y =-'+''sinC、x y y 2sin ='-'' D、y y =''5、下列极限结果错误的是( )A、01sin lim 0=→x x x B 、11sin lim =∞→x x xC 、1arctan lim 0=→x xx D 、1sin lim =∞→x xx二、填空题(每小题3分,共21分):1、0lim ()x x f x →存在是()f x 在0x 处连续的 条件()f x 在0x 处可导是在该点连续的 条件()f x 在0x 处可导是在该点可微的 条件2、设x e y sin = ,则=dy3、曲线3y x =在点(1, 1)处的切线方程为4、微分方程20y y y '''++=的通解是y =5、设()f x 是连续函数,则⎰=dx x f dx d )( ;⎰=x a dt t f dxd )( ; ⎰=b adx x f dx d )( 6、若,)(2c x dx x f +=⎰则⎰=dx e f e x x )(7、曲线2332x y =上从点)0,0(到点)32,1(一段的弧长=三、计算题(每小题6分,共42分):1、 求极限30sin lim x x x x -→2、 求方程e xy e y =+所确定的隐函数的导数10==y x dx dy3、 求方程⎩⎨⎧-=+=tt y t x arctan )1ln(2所确定的参数式函数的导数22,dx y d dx dy4、 计算定积分⎰+31)1(x x dx5、 计算定积分⎰-+22)cos sin (423ππdx x x x6、 广义积分⎰∞+e dx xx 2ln 1是否收敛?若收敛,求其值7、 求微分方程2x xy dx dy =-的通解四、求函数162923++-=x x x y 的单调区间和极值。

浙江工业大学2019-2020(2)通用学术英语期末考试(回忆版)

浙江工业大学2019-2020(2)通用学术英语期末考试(回忆版)

浙江工业大学2019/2020第2学期通用学术英语期末考试(回忆版)Part I Listening(37%)选择题:(2013.12CET6卷3)13.A)Bad weather.B)Human error.C)Breakdown in the engine.D)Failure of the communications system.14.A)Two thousand feet.B)Twelve thousand feet.C)Twenty thousand feet.D)Twenty-two thousand feet.15.A)Accurate communication is of utmost importance.B)Pilots should be able to speak several foreign languages.C)Air controllers should keep a close watch on the weather.D)Cooperation between pilots and air controllers is essential.听力原文:M:Did you hear about the air crash that occurred in South America recently?It was quite a tragic accident!No,I didn’t see anything in the news about it.What happened?W:A foreign airliner was attempting to land at night in a mountainous area of Argentina and flew into a hill!That sounds really terrible!Did anyone survive?M:No,everyone aboard,including the crew,was killed instantly.W:What were the circumstances?Was there bad weather,a fire,or engine failure?M:Apparently,there were some low clouds in the area,but mostly it was just miscommunication between the pilots and the air traffic controllers.W:Weren't they both speaking in English,the official international aviation language?M:Yes,they were.But the transmission from poor quality radios was slightly distorted and the accents of the Spanish speaking controllers were so strong that the pilots misunderstood a vital instruction.W:How could a misunderstanding like that caused such a serious accident?M:The pilots were told to descend to two-two thousand feet.The instruction actually meant22000feet,but they thought they heard descend to2000feet.That’s a huge difference,and it should have been confirmed,but it was not.Unfortunately,the terrain of the mountains in Norweija extends of2000feet.W:So the pilots did descend to the wrong altitude then,thinking they were following the air controllers' instructions.M:Sadlyenough.Yes-they did.It was a really bad mistake.Many people died as a result of the simple misunderstanding.W:Wow,that’s a powerful lesson on how important it can be to accurately communicate to each other.13.What was the cause of the tragedy?14.How high are the mountains in Norweija?15.What lesson could be drawn from the accident?(2013.06CET卷3)22.A)He has been seeing doctors and counsellors.B)He has found a new way to train his voice.C)He was caught abusing drugs.D)He might give up concert tours.23.A)Singers may become addicted to it.B)It helps singers warm themselves up.C)Singers use it to stay away from colds.D)It can do harm to singers'vocal chords.24.A)They are eager to become famous.B)Many lack professional training.C)Few will become successful.D)They live a glamorous life.25.A)Harm to singers done by smoky atmospheres.B)Side effects of some common drugs.C)Voice problems among pop singers.D)Hardships experienced by many young singers.听力原文:W:Rock stars now face a new hazard---voice abuse.After last week's announcement that Phil Collins might give up touring because live concerts are ruining his voice,doctors are counseling stars about the dos and don'ts of voice care.Here in the studio today,we have Mr.Paul Phillips,an expert from the High Field Hospital.Paul, what advice would you give to singers facing voice problems?M:If pop singers have got voice problems,they really need to be more selective about where they work.They shouldn't work in smoky atmospheres.They also need to think about resting their voice after a show.Something else they need to be careful about is medicines.Aspirin,for example,singers should avoid aspirin.It thins the blood.And if a singer coughs,this can result in the bruising of the vocal cords.W:And is it true that some singers use drugs before concerts to boost their voices when they have voice problems? M:Yes,this does happen on occasion.They are easily-available on the continent and they are useful if a singer has problems with his vocal cords and has to sing that night.But if they are taken regularly,they cause a thinning of the voice muscle.Most pop singers suffer from three things:lack of training,overuse and abuse of the voice, especially when they are young.They have difficult lives.When they go on tour,they do avast number of concerts,singing in smoky places.W:So,what would you advise the singers to do?M:Warm you voice up before a show and warm it down after.22.What does last week's announcement say about rock star,Phil Collins?23.What does Paul Philips say about aspirin?24.What does Paul Philips say about young pop singers?25.What are the speakers mainly talking about?判断题:(题目略)(U8,Listening3)Today,Chinese Americans make up less than1%of the U.S.population,but roughly a third of all ethnic restaurants in the U.S.are"Chinese,"and every supermarket carries a line of"Chinese"food.It started with the gold rush of1849.As thousands of Chinese streamed into California in search of gold,hundreds of thousands of Chinese migrated to the United States.By1882,when Congress curtailed Chinese immigration,there were morethan300,000Chinese nationals living on the West Coast.Most came from Guangdong Province.So most Chinese restaurants served Cantonese-style food.In Cantonese cuisine,very little goes to waste:nearly every part of an animal that can be eaten is used in one dish or another.Chinese cooks,however,quickly learned to modify their dishes to make them more palatable to a wider American audience.The result,a Chinese-American cuisine that looked and tasted Chinese,but was actually invented in the U.S.and was unknown in China.One good example of Chinese-American dish is Chop Suey.No one knows for sure when it was invented,or how it got its name.It likely started in1850when a bunch of hungry miners busted their way into a Chinese restaurant late at night and demanded to be fed.The chef just stirred all the table scraps and leftovers he could find into a big mess and served it.The miners loved it.When they asked what it was,the chef replied"Chop Suey",which means"garbage bits"in Cantonese.The dish remained virtually unheard of in China until after World War II.Today,it's advertised as American cuisine.Well,what other contributions have immigrants made to our cuisine?(U6,Listening3)In recent years,the death rate among American centenarians—people who have lived to age100or older—has decreased,dropping14percent for women and20percent for men from2008to2014.The leading causes of death in this age group are also changing.In2000,the top five causes of death for centenarians were heart disease, stroke,flu,cancer and Alzheimer's disease.But by2014,the death rate from Alzheimer's disease for this age group had more than doubled—increasing from3.8percent to8.5percent—making the progressive brain disease the second leading cause of death for centenarians.One reason for the rise in deaths from Alzheimer's disease in this group may be that developing this condition remains possible even after people beat the odds of dying from other diseases such as cancer.People physically fit enough to survive over100years ultimately give in to diseases such as Alzheimer's which affects the mind and cognitive function.In other words,it appears that their minds give out before their bodies do.On the other hand,the death rate from flu dropped from7.4percent in2000to4.1percent in 2014.That pushed flu from the third leading cause of death to the fifth.Overall,the total number of centenarians is going up.In2014,there were72,197centenarians,compared to50,281in2000.But because this population is getting larger,the number of deaths in this group is also increasing—18,434centenarians died in2000,whereas 25,914died in2014.填空:Passage One:Passage Two:Part II Reading(20%)段落匹配:(2016.12CET6卷3)Countries Rush for Upper Hand in AntarcticaA)On a glacier-filled island with fjords(峡湾)and elephant seals,Russia has built Antarctica’s first Orthodox church on a bill overlooking its research base.Less than an hour away by snowmobile.Chinese laborers have updated the Great Wall Station,a vital part of China’s plan to operate five basses on Antarctica,complete with an indoor badminton court and sleeping quarters for150people.Not to be outdone,India’s futuristic new Bharathi base,built on stills(桩子)using134interlocking shipping containers,resembles a spaceship.Turkey and Iran have announced plans to build bases,too.B)More than a century has passed since explorers raced to plant their flags at the bottom of the world,and for decades to come this continent is supposed to be protected as a scientific preserve,shielded from intrusions like military activities and mining.But an array of countries are rushing to assert greater influence here,with an eye not just towards the day those protective treaties expire,but also for the strategic and commercial that already exist. C)The newer players are stepping into what they view as a treasure house of resources.Some of the ventures focus on the Antarctic resources that are already up for grabs,like abundant sea life.South Korea,which operates state-of–the-art bases here,is increasing its fishing of krill(磷虾),found in abundance in the Southern Ocean,while Russia recently frustrated efforts to create one of the world’s largest ocean sanctuaries here.D)Some scientists are examining the potential for harvesting icebergs form Antarctica,which is estimated to have the biggest reserves of fresh water on the planet.Nations are also pressing ahead with space research and satelliteprojects to expand their global navigation abilities.E)Building on a Soviet-era foothold,Russia is expanding its monitoring stations for Glonass,its version of the Global Positioning System(GPS).At least three Russian stations are already operating in Antarctica,part of its effort to challenge the dominance of the American GPS,and new stations are planned for sites like the Russian base, in the shadow of the Orthodox Church of the Holy Trinity.F)Elsewhere in Antarctica,Russian researchers boast of their recent discovery of a freshwater reserve the size of Lake Ontario after drilling through miles of solid ice.“You can see that we’re here to stay,”said Vladimir Cheberdak,57,chief of the Bellingshausen Station,as he sipped tea under a portrait of Fabian Gottlieb von Bellingshausen,a high-ranking officer in the Imperial Russian Navy who explored the Antarctic coast in1820. G)Antarctica’s mineral,oil and gas wealth are a longer-term prize.The treaty banning mining here,shielding coveted(令人垂诞的)reserves of iron ore,coal and chromium,comes up for review in2048.Researchers recently found kimberlite(金伯利岩)deposits hinting at the existence of diamonds.And while assessments vary widely, geologists estimate that Antarctica holds at least36billion barrels of oil and natural gas.H)Beyond the Antarctic treaties,huge obstacles persist to tapping these resources,like drifting icebergs that could jeopardize offshore platforms.Then there is Antarctic’s remoteness,with some mineral deposits found in windswept locations on a continent that is larger the Europe and where winter temperatures hover around minus55 degrees Celsius.I)But advances in technology might make Antarctica a lot more accessible three decades from now.And even before then,scholars warn,the demand for resources in an energy-hungry world could raise pressure to renegotiate Antarctica’s treaties,possibly allowing more commercial endeavours here well before the prohibitions against them expire.The research stations on King George lsland offer a glimpse into the long game on this ice-blanketed continent as nations assert themselves,eroding the sway long held by countries like the United States,Britain. Australia and New Zealand.J)Being stationed in Antarctica involves adapting to life on the planet’s driest,windiest and coldest continent,yet each nation manages to make itself at home.Bearded Russian priests offer regular services at the Orthodox church for the16or so Russian speakers who spend the winter at the base,largely polar scientists in fields like glaciology and meteorology.Their number climbs to about40in the warmer summer months.China has arguably the fastest growing operations in Antarctica.It opened its fourth station last year and is pressing ahead with plans to build a fifth.It is building its second ice-breaking ship and setting up research drilling operations on an ice dome13,422 feet above sea level that is one the planet’s coldest places.Chinese officials say the expansion in Antarctica prioritises scientific research.But they also acknowledge that concerns about“resource security”influence their moves.K)China’s newly renovated Great Wall Station on King George lsland makes the Russian and Chilean bases here seem outdated.”We do weather monitoring here and other research.”Ning Xu,53,the chief of the Chinese base, said over tea during a fierce blizzard(暴风雪)in late November.The large base he leads resembles a snowed-in college campus on holiday break,with the capacity to sleep more than10times the13people who were staying on through the Antarctic winter.Yong Yu,a Chinese microbiologist,showed off the spacious building,with emptydesks under an illustrated timeline detailing the rapid growth of China’s Antarctic operations since the1980s“We now feel equipped to grow,”he said.L)As some countries expand operations in Antarctica,the United States maintains three year-round stations on the continent with more than1,000people during the southern hemisphere’s summer,including those at the Amundsen Scott station,built in1956at an elevation of9,301feet on a plateau at the South Pole.But US researchers quietly complain about budget restraints and having far fewer icebreakers the Russia,limiting the reach of the United States in Antarctica.M)Scholars warn that Antarctica’s political drift could blur the distinction between military and civilian activities long before the continent’s treaties come up for renegotiation,especially in parts of Antarctica that are ideal for intercepting(拦截)signals from satellites or retasking satellite systems,potentially enhancing global electronic intelligence operations.N)Some countries have had a hard time here,Brazil opened a research station in1984,but it was largely destroyed by a fire that killed two members of the navy in2012,the same year that a diesel-laden Brazilian barge sank near the base.As if that were not enough.a Brazilian C-130Hercules military transport plane has remained stranded near the runway of Chile’s air base here since it crash-landed in2014.O)However,Brazil’s stretch of misfortune has created opportunities for China,with a Chinese company winning the$100million contract in2015to rebuild the Brazilian station.P)Amid all the changes,Antarctica maintains its allure.South Korea opened its second Antarctic research base in 2014,describing it as a way to test robots developed by Korean researchers for use in extreme conditions.With Russia’s help,Belarus is preparing to build this first Antarctic base.Colombia said this year that it planned to join other South American nations with bases in Antarctica.Q)“The old days of the Antarctic being dominated by the interests and wishes of white men from European. Australasian and North American states are over.”Said Klaus Dodds,a politics scholar at the University of London who specialises in Antarctica.“The reality is that Antarctica is geopolitically contested.”36.According to Chinese officials,their activities in Antarctica lay greater emphasis on scientific research.37.Efforts to create one of the world’s largest ocean sanctuaries failed because of Russia’s obstruction.38.With several monitoring stations operating in Antarctica,Russia is trying hard to counter America’s dominance in the field of worldwide navigational facilities.39.According to geologists’estimates.Antarctica has enormous reserves of oil and natural gas.40.It is estimated that Antarctica boasts of the richest reserves of fresh water on earth.41.The demand for energy resources may compel renegotiation of Antarctica’s treaties before their expiration.42.Many countries are racing against each other to increase their business and strategic influence on Antarctica.43.Antarctica’s harsh natural conditions constitute huge obstacles to the exploitation of its resources.44.With competition from many countries,Antarctica is no longer dominated by the traditional white nations.45.American scientists complain about lack of sufficient money and equipment for their expansion in Antarctica.36.[J]37.[C]38.[E]39.[G]40.[D]41.[I]42.[B]43.[H]44.[Q]45.[L]仔细阅读:(2019.12考研英语一)Grade inflation--the gradual increase in average GPAs(grade-point averages)over the past few decades—is often considered a product of a consumer era in higher education,in which students are treated like customers to be pleased.But another,related force—a policy often buried deep in course catalogs called“grade forgiveness”—is helping raise GPAs.Grade forgiveness allows students to retake a course in which they received a low grade,and the most recent grade or the highest grade is the only one that counts in calculating a student’s overall GPA.The use of this little-known practice has accelerated in recent years,as colleges continue to do their utmost to keep students in school(and paying tuition)and improve their graduation rates.When this practice first started decades ago,it was usually limited to freshmen,to give them a second chance to take a class in their first year if they struggled in their transition to college-level courses.But now most colleges save for many selective campuses, allow all undergraduates,and even graduate students,to get their low grades forgiven.College officials tend to emphasize that the goal of grade forgiveness is less about the grade itself and more about encouraging students to retake courses critical to their degree program and graduation without incurring a big penalty.“Untimely,”said Jack Miner,Ohio State University’s registrar,“we see students achieve more success because they retake a course and do better in subsequent contents or master the content that allows them to graduate on time.”That said,there is a way in which grade forgiveness satisfies colleges’own needs as well.For public institutions,state funds are sometimes tied partly to their success on metrics such as graduation rates and student retention—so better grades can,by boosting figures like those,mean more money.And anything that raises GPAs will likely make students—who,at the end of the day,are paying the bill—feel they’ve gotten a better value for their tuition dollars,which is another big concern for colleges.Indeed,grade forgiveness is just another way that universities are responding to consumers’expectations for higher education.Since students and parents expect a college degree to lead a job,it is in the best interest of a school to turn out graduates who are as qualified as possible—or at least appear to be.On this,students’and colleges’incentives seem to be aligned.26.What is commonly regarded as the cause of grade inflation?A.The change of course catalogs.B.Students’indifference to GPAS.C.Colleges’neglect of GPAS.D.The influence of consumer culture.27.What was the original purpose of grade forgiveness?A.To help freshmen adapt to college learning.B.To maintain colleges’graduation rates.C.To prepare graduates for a challenging future.D.To increase universities’income from tuition.28.According to Paragraph5,grade forgiveness enable colleges to_________.A.obtain more financial supportB.boost their student enrollmentsC.improve their teaching qualityD.meet local governments’needs29.What does the phrase“to be aligned”(Line5,Para.6)most probably mean?A.To counterbalance each other.B.To complement each other.C.To be identical with each other.D.To be contradictory to each other.30.The author examines the practice of grade forgiveness by________.A.assessing its feasibilityB.analyzing the causes behind itparing different views on itD.listing its long-run effectsPart III Vocabulary(15%)英英匹配:(英文释义可能有部分与试卷上不一致)asymmetric—not equal;having two sides or parts that are not the same in size or shapecompliance—the action or fact of being in accordance with rules or standardsencompass—to include a wide range of ideas,subjects,etc.monopoly—exclusive control or possession of somethingvocation—a job that you do because you feel it is your propose in life and for which you have special skills scenario—a situation that could possibly happenmaximum—as large,fast,etc.as is possible,or the most that is possible or allowedunderestimate—to think or guess that sth.is smaller,cheaper,easier,etc.than it really is coordination—the process of organizing people or things in order to make them work together effectively reside—to exist in or belong to sb.or sth.选词填空:(句子顺序可能不一样,大致意思以及关键词应该没错)1.He paused,not for an answer,but to let his words sink in.2.Previous studies have suggested that preschool education programmes and better nourishment can raise IQ scores.e of green building materials for construction and sustainable development of the environment is particularly important.4.The film uses the footage to coral reefs and show fragile environment under threat to global warming.5.Patients tested negative for COVID-19were transferred to a…….6.The committee passed on her application and unanimously approved of it.7.Like most Japanese companies with a rigid hierarchy,workers and managers have strictly defined duties.8.The entrepreneur's speech is enlightening for college students to start their own business9.Much of the book's appeal derives from the personality of its central character.10.Knowing that he is coping with stress in his own way is extremely helpful but does not always help heralleviate the pain.Part IV Translation(18%)1.The signaling theory of gift giving is consistent with another observation:People care most about the custom when the strength of affection is most in question.送礼的信号传递理论与另一种观察结果一致:当感情的深浅受到怀疑时,人们非常在乎习俗。

物理化学 - 复习题2答案

物理化学 - 复习题2答案

= 0.871
14、解: 解:设计下列三步变化过程。
冰(-10℃) ¾D¾H1,1¾DS1¾® 冰(0℃) ¾D¾H2,2¾DS¾2 ® 水(0℃) ¾D¾H3,3¾DS¾3 ® 水(-10℃)
DH = DH1 + DH2 + DH3 = [37.6 ´10 + 6020 - 75.3´10] J = 5643 J
15、c,体系经历的变化为绝热不可逆变化,所以 ΔS 体> 0;环境与体系间没有热交换,压力
亦无变化,体积的变化可忽略,所以环境的状态未变, 即 ΔS 环= 0 (环境的体积变化可忽略
是基于下述认识,即一般情况下,总可以认为环境相对于体系是无穷大的)。
16、c,17、a ,18、b,19、d,
20、c 因为 p=RT/(Vm-b)
= 182.4 J·K-1·mol-1
[2]
DcGmy =
D
c
H
y m
-
TDcSmy =
-2.862
MJ·mol-1
[2]
DcGmy = Wr' 即为恒温恒压下所能作的最大功。
[2]
( ) ( ) 8、解:(1)
D
r
H
y m
=
DC
H
y m
金刚石
-
DC
H
y m
石墨
= 1.88 kJ·mol-1
[2]
( ) ( ) Dr Smy = Smy 金刚石 - Smy 石墨 = -3.263 J·K-1mol-1
4、解:
{ } Q = DH = -9916 + (122.6 -126.8)´ éë(-5) - 5ùû J= - 9874J
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取代乙酰乙酸乙酯的成酮水解
1) I 2 / NaO H 2) H +
CH 3 I +
CH 2 COOH
碘仿反应
CHO 2) 2 COOH 浓 NaOH
C H 2O H COOH
+
COOH
无α-H 的醛Cannizzaro反应
COOH
期末复习
O 3) CH 3 CCHO
O
+ 2 CH 3 OH
HCl
N
N H
3
2
1
期末复习
五、鉴别题 1. 用简便的化学方法鉴别以下化合物: (A) 乙醛
乙 醛 2-己 酮 苯酚 苯甲醇 FeCl 3 紫红色 浑浊 lucas
(B) 2-己酮
(C) 苯酚
(D)苯甲醇
Ag Tollens
期末复习
2. 用简便的化学方法鉴别以下化 3 CH 2 CO O
HO O CCH 2 CO O H
pK a1 = 2.9
+ HO O CCH 2 CO O + H pK a2 = 5.7 + O O CCH 2 CO O + H
期末复习
下列化合物加热时放出CO2的有:( 1, 2, 3, )
吡咯,吡吡啶和六氢吡啶碱性比较?为什么?
N H
期末复习
一、命名或写结构 3-苯基丙烯醛 4, 4’-偶氮苯二甲酸
CH=CHCHO
肉桂酸?
HOOC
N=N
CO O H
CH 3
3-甲基四氢呋喃
O
CH 3
+
C 12 H 25 Cl
氯化三甲基十二烷基铵
H 3C
N CH 3
期末复习
4-戊酮醛
O
(R) -2-氯丙酸
二氢吡咯
O
CH 3 CCH 2 CH 2 CHO
CH 3 CO O H + CH 3 CH 2 O H 2 CH 3 CO O C 2 H 5
1) CH 3 CH 2 O Na 2) CH 3 CO O H
期末复习
CH 3 CH 3 Br
CO O H
H N O 3 , H 2 SO 4
1) C uC N , K C N 2) H 3 O +
CH3
(B) 乙酰氯
气体
(C) 乙酰胺
AgNO 3
期末复习
六、有机合成
ClCH 2 CH 2 CHO
H 2C HCl H 2C OH OH O C lC H 2 C H 2 C H O
CH 3 CHCH 2 CH 2 CHO OH
HCl
M g, Ether O M gClCH 2 CH 2 CH O
CH 3 CHO H 3O + O CH 3 CHCH 2 CH 2 CH OH O
Br
H 3C C CH 3 C CH 3 CH 3
H 3C
C CH 3
H C CH 3
CH 3
B
H2C C CH 3 H C CH 3 CH 3
A
C
+
O
O
CH 3 CCH 2 COC 2H 5
乙酰乙酸乙酯的合成
三、理化性质比较 1、将下列化合物按pKa值大小排列次序:(
OH OH Cl
c>b>a

注意pKa和酸性,以及
COOH COOH COOH
Pkb和碱性的关系,看 清题目
a
b
c
期末复习
2、将下列化合物按沸点高低排列成序:(
(A) CH 3 CH 2 CH 2 COOH (B) CH 3 CH 2 CH 2 CH 2 OH
期末复习
某化合物的分子式为C11H14O2,红外光谱表明在1720cm-1有强吸 收峰。核磁共谱振数据如下: δ=1.0(二重峰,6H),δ=2.1(多重峰,1H),δ=4.1(二重峰,2H), δ=7.8(多重峰,5H)。确定该化合物的构造式。
O C
CH 3 CH CH 3
δ=1.0(二重峰,6H),
NH 2
甘油
6) NO 2 NO 2
,
O 2N
NO 2
O 2N N
H 2 SO 4
多硝基苯的选择性还原
取代喹啉的合成
期末复习
7) H 3 C CHO + (CH 3 CO) 2 O CH 3 COOK
H3C
CH=CHCOOH
Perkin reaction
1) LiAlH 4 2) H 2 O
H 3C CH=CHCH 2 OH
期末复习
N HOOC
CH 3 Cl, AlCl 3
N
NH 2
NH 2
+ N 2 Cl HO OC
CH 3
KM nO 4
NaNO 2 , HCl
HNO 3 , H 2 SO 4 HOOC Fe + HCl NO 2 HOOC NH 2
CO O H
期末复习
CH 3 4个 碳 以 下 的 原 料 O CH 2 CH 2 C CH 3
OCH 3 OCH 3
CH 3 CCH
生成缩醛,保护醛基
C 2 H 5 MgBr H2O
+
C 2H 5
H 酮和格氏试剂反应生成叔醇,同时还原醛基 O H O O
C
ether
CH 3
C
CHO
C
NK
4)
(CH 3 ) 2 CHCH 2 Br +
C O
NCH 2 CH(CH 3 ) 2 C O
Gabriel 合成法合成伯胺, 注意条件!
FeBr 3 , Br 2 CH 3 Br
CH 3 Fe + HCl Br NaNO 2 , HCl
CH 3 Br
NO2
NO 2
NH 2
N 2 Cl
+
期末复习
七、结构推导
1.分子式为C5H10O的化合物可还原为正戊烷,可与苯肼作用生成
苯腙,但没有碘仿和银镜反应。试写出这个化合物的构造式。
O CH 3 CH 2 CCH 2 CH 3
(A) CH 3 CH 2 CH 2 NH 2 (B) CH 3 CH 2 CONH 2 H 2C H 2C CO NCH 2 CH 3 CO
(C)
CH 3 CH 2 NCH 2 NH 2 H
(D)
期末复习
7.下列化合物能发生碘仿反应的是( a. 2-甲基丁醛 b. 异丙醇 b, c c. ) d. 丙醇 2-戊酮
期末复习
四、基本概念题 指出下列化合物哪些能发生碘仿反应,哪些可发生银镜反应, 哪些可发生自身羟醛缩合反应。
碘仿反应: 2, 3
银镜反应: 1, 2, 4
自身羟醛缩合反应: 2
期末复习
试解释丙二酸pKa1比丙酸pKa小,而pKa2却比丙酸pKa大的事实。
+ + H
CH 3 CH 2 CO O H
O
CH 2
CH
δ=7.8(多重峰,5H)
CH3 CH CH3
1720cm-1
δ=4.1(二重峰,2H) O
C O CH 2
期末复习
化合物A(C6H13Br),与硝酸银的醇溶液迅速反应得到淡黄色的
溴化银沉淀;A在氢氧化钠的乙醇溶液中加热主要得到产物 B(C6H12)和其微量的异构体C;将B用酸性高锰酸钾溶液氧化后 只生成唯一一种酮,而C经氧化后可生成一分子酮、一分子CO2 和H2O。试推断A,B,C的构造。
Cl 2 , hv
CH 3 COCH 2 COOC 2 H 5
1) 稀 碱 , 2) H + 3)
CH 2 Cl
CH 3 COCHCOOC 2 H 5 CH 2
C 2 H 5 ONa
CH 3 CH 2 O H
KM nO 4 H
+
CH 3 CO O H H+ CH 3 CO O C 2 H 5 CH 3 CO CH 2 CO O C 2 H 5
(A) C 6 H 5 SO 3 H (B) C 6 H 5 COOH (C) C 6 H 5 OH (D) C 2 H 5 OH
期末复习
5. 将下列负离子按碱性大小排列成序:(
CH 3 OH CN
b>a>c

CO O
CO O
CO O
a
b
c
6. 将下列化合物按碱性大小排列成序:( c > a > b > d ).
8.下列羰基化合物对NaHSO3加成反应速率由快到慢的 顺序为( c > d > b > a ) a. 苯乙酮 b. 苯甲醛 c. 2-氯乙醛 d. 乙醛
9. 比较下列酯类水解的活性大小
A) O 2 N COOCH 3
a>d>b>c
B) COOCH 3
C) CH 3 O
COOCH 3
D)
Cl
COOCH 3
DMF
O H C N CH3 CH3
NH O
水杨酸
OH
邻苯二甲酰亚胺
COOH
期末复习
二、完成反应
1) Br CH 3 COCH 2 COOC 2 H 5 CH 3 COCHCOOC 2 H 5 C 2 H 5 ONa
1) 稀 NaoH 2) H + , H 2 O
取代乙酰乙酸乙酯的合成 O
C H 3C C H 2
α,β-不饱和酸的选择性还原
NaNO 2 HCl, 0 C
o
8)
NO 2
Fe + HCl
NH 2
+
N 2 Cl
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