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新科粤版九年级化学下册《九章现代生活与化学 9.3 化学能的利用》教案_2

9.3 化学能的利用◆教材分析：生活在现代社会，学生对“电”有着丰富而又强烈的感性认识。

当学生们了解了化学反应中能量转化的原因，并感受了探究化学能与热能的相互转化过程之后，会对化学能与电能之间的转化问题产生浓厚的兴趣。

正是基于学生的这种心理特征，教材开始出示几幅图片，把学生带进了“化学能与热能之间相互转化”的研究之中。

从思维角度看，“将化学能直接转化为电能”的思想，是对“化学能→热能→机械能→电能”思维方式的反思和突破。

教材呈现的思路是首先介绍化学电池的功能；接着介绍了氢氧电池的组成、工作原理、应用和优点。

◆教学目标：1.知识与技能（1）了解常见的能源。

（2）知道电池中的能量转化。

（3）能够了解社会的发展与能源利用的关系。

2.过程与方法利用网络信息、图片，拓展学生的视野；并通过学生自学，培养学生的表达能力、合作意识和自学能力。

通过对能源现状和使用的了解，培养学生节约能源、合理使用化学物质的观念。

3.情感、态度与价值观（1）了解学习化学的价值，认识化学能在当今社会的重要应用。

（2）树立学生节约能源观点，认识到能源的短缺，并合理使用能源。

◆教学重点：知道常用的能源；化学能的使用。

◆教学难点：化学能转化为电能。

◆教学准备：多媒体、电脑。

◆教学过程一、导入新课化学反应都有新物质产生，同时还伴随着能量变化。

人们利用化学反应，有时主要是为了制取所需要的物质，有时却主要是为了利用化学反应所释放出的能量。

例如，人们利用氢氧焰来焊接金属，主要就是利用氢气和氧气化合时所放出的能量。

在当今社会中，人类所需能量的绝大部分是由化学反应产生的，特别是由煤、石油和天然气等化石燃料或它们的制品的燃烧所产生的。

由此可见，研究化学反应及其能量变化对于我们是多么重要！今天我们一起学习相关知识。

二、推进新课活动1：了解常见的能源［导入］从人类祖先发现野火起，人类便叩开了能源之门，开始利用化学反应所释放的能量。

如今人类所需能量的绝大部分是由化学反应产生的，如燃料的燃烧。

初中化学教学论文多媒体技术在九年级化学课堂教学改革中的应用

多媒体技术在九年级化学课堂教学改革中的应用摘要：化学课堂教学是一个富有逻辑性的教学过程，教师与学生在不断的交流中获得进步。

多媒体技术是当代教育发展有效的促进工具，代表着教育变革的方向。

化学教学的发展逐渐与多媒体技术联合在一起，形成了高效的多媒体化学教学模式。

由于化学教学形式的不同，多媒体技术在化学教学中的应用有得有失。

在新课程的改革中，多媒体条件下的化学教学具有广泛性、实效性。

关键词：化学教学多媒体技术演示实验探究性实验二十一世纪是信息化的时代，人们在虚幻的空间中寻找自己，多媒体信息技术已经成为人类生活的一部分。

多媒体信息技术的发展正改变着人们的生活方式。

教育是一种社会责任，教育促进了人类文明发展。

教育的方方面面都会受到社会发展的影响。

当教育和信息技术联系起来，教育的新模式出现了，借助多媒体技术，教育正向着更高更远的空间发展。

化学是社会发展的催化剂，快速地推动着社会的发展。

中学化学是中学阶段锻炼学生思维模式的有效课程，化学思维决定着中学生的认知水平。

化学课堂传递着化学知识，化学课堂教学经过长期的发展，逐渐形成了专业性的教学模式。

化学课堂教学最有效模式是以学生自主学习为主，教师演示引导为辅，课后巩固为补充。

相继经历了传统的教学方式；改革性的以演示实验为主题的教学尝试；应用多媒体的开放性的以实验探究为主题的教学改革，最后发展成为借助多媒体技术，灵活实用的实验探究型教学模式。

化学教学的每一次变革，都是化学教学认知水平的一次提升，都是化学教学认识的升华。

化学教学过程中，博学多才的教师与追求真知的学生密切配合，化学的教学不断进步，学生自主学习的能力不断提升。

一、传统的化学教学方式。

传统的化学教学方式比较单调，教师在课堂上拥有绝对的权威，学生没有任何的发言权。

教师在讲台上慷慨激昂地讲述，学生在座位上小心翼翼地听着，手中的笔不停地写着。

我们通常称为“填鸭式”教学。

教师依据自己的教学思路，一步步将知识塞给学生，通过大量的习题加以强化。

数学：人教版九年级上-22(衡中内参)2021年版.3-实际问题与一元二次方程(教案)

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课题：22.3实际问题与一元二次方程一、教学目标1.会利用一元二次方程解决简单的图形问题.2.培养分析问题解决问题的能力，发展应用意识.二、教学重点和难点1.重点：利用一元二次方程解决简单的图形问题.2.难点：根据图形问题列方程.三、教学过程（一）创设情境，导入新课师：前面我们学习了有关一元二次方程的知识，我们学习了什么是一元二次方程，学习了什么是一元二次方程的根，学习了如何解一元二次方程.现在，老师要同学们想这样一个问题：为什么要学习这些知识？学习这些知识的目的是什么？（稍停后再叫学生）生：……（多让几名同学发表看法）师：和一元一次方程一样，一元二次方程也是解决实际问题的工具.学习一元二次方程不是为了什么，而是为了解决实际问题.从这节课开始，我们来学习如何利用一元二次方程解决实际问题（板书课题：22.3实际问题与一元二次方程）.师：下面我们来看一个例子.（二）尝试指导，讲授新课（师出示下面的例题）例扎西家有一个长方形院子，它的长比宽多3米，面积为54平方米，院子的长和宽各是多少米？师：大家把这个题目默读几遍.（生默读）师：题目要求院子的长和宽，我们设院子的长为x米，则院子的宽为多少米？生：(x-3)米（师板书：解：设院子的长为x米，则院子的宽为(x-3)米）.师：读了题目，又设好了未知数，你能按题目的意思画一个图吗？大家试一试.（生画图，师巡视）师：我们一起来画图.扎西家有一个长方形的院子（边讲边画一个长方形），现在设这个院子的长为x 米（边讲边标：x 米），则宽为(x-3)米（边讲边标：(x-3)米），院子的面积为54平方米（边讲边标：面积54平方米，画好的图如下所示）. （x-3）米x 米面积54平方米师：根据这个图，大家列一列方程.（生列方程，师巡视）师：（板书：根据题意列方程，得）列出的方程是什么？生：x(x-3)=54.（多让几名同学回答，然后师板书：x(x-3)=54）师：（指方程）列出的方程是一个一元二次方程，大家把它整理成一般形式.（生整理方程）师：整理后的方程是什么？生：x 2-3x-54=0（师板书：整理，得x 2-3x-54=0）.师：（指x 2-3x-54=0）大家用公式法解这个方程.（生解方程，师巡视）师：方程的两个根x 1等于什么？x 2等于什么？生：x 1=9，x 2=-6（师板书：解方程，得x 1=9，x 2=-6，如有必要师可在黑板的其它地方板演解方程过程）师：（指准x(x-3)=54）这里的x 表示什么？（稍停）表示院子的长，院子的长不能是负数，（指准x 1=9，x 2=-6）所以x 2=-6不符合题目的意思，要舍去（板书：（不合题意，舍去））.所以院子的长为9米（板书：答：院子的长为9米）.师：院子的宽为多少米？生：宽为6米.（师板书：宽为6米）师：这道题目做完了，做了这道题目，谁来归纳一下怎么利用一元二次方程解决实际问题？（让生思考一会儿后再叫学生）生：……（让几名同学回答）师：（指准例题）利用一元二次方程解决实际问题，第一步要读题，反复地读题，有的时候还可以画一画图，通过读题画图弄清题目的意思；第二步设未知数；第三步根据题目的意思列出一元二次方程；第四步解一元二次方程，一元二次方程的根有两个，要根据题意来取舍解出的根，-6这个根不符合题目意思，要舍去；第五步答.师：利用一元二次方程解决实际问题就这么五步，实际上与利用一元一次方程解决实际问题的步骤是一样的.师：下面就请同学们自己来做两个练习.（三）试探练习，回授调节1.完成下面的解题过程：一个直角三角形的两条直角边相差5cm，面积是7cm2,求两条直角边的长.解：设一条直角边的长为 cm，则另一条直角边的长为 cm.根据题意列方程，得 .整理，得 .解方程，得 x1= ，x2= （不合题意，舍去）.答：一条直角边的长为 cm，则另一条直角边的长为 cm.2.一个菱形两条对角线长的和是10cm，面积是12cm2，(1)求菱形的两条对角线长；(2)求菱形的周长.（提示：菱形的面积=两条对角线积的一半）（四）归纳小结，布置作业师：（指例题）本节课我们学习了一个例题，大家再看一看这个例题，回顾一下利用一元二次方程解决问题有哪几个步骤.（作业：P48习题1(1)(2)2.3.）四、板书设计（略）课题：22.3实际问题与一元二次方程（第2课时）一、教学目标1.会利用一元二次方程解决传播问题.2.培养分析问题解决问题的能力，发展应用意识.二、教学重点和难点1.重点：利用一元二次方程解决传播问题.2.难点：根据传播问题列方程.三、教学过程（一）基本训练，巩固旧知1.填空：(1)有一人得了流感，他把流感传染给了10个人，共有人得流感；第一轮传染后，所有得流感的人每人又把流感传染给了10个人，经过两轮传染后，共有人得流感.(2)有一人得了流感，他把流感传染给了x个人，共有人得流感；第一轮传染后，所有得流感的人每人又把流感传染给了x个人，经过两轮传染后，共有人得流感.（(1)题答案为11，121，(2)题答案为1+x，1+x+x(x+1)，先让生自己做，然后师进行讲解）（二）创设情境，导入新课师：和一元一次方程一样，利用一元二次方程可以解决实际问题，上节课我们做了一个例题，本节课我们再来看一个例题.（三）尝试指导，讲授新课（师出示下面的例题）例有一人得了流感，经过两轮传染后，共有121人得了流感，每轮传染中平均每一个人传染了几个人？师：大家把这个题目好好默读几遍.（生默读）师：谁能不看黑板说出题目的意思？生：……（让几名同学说）师：这个题目怎么设？生：设每轮传染中平均一个人传染了x个人.（师板书：解：设每轮传染中平均一个人传染了x个人）师：（在黑板的其它地方板书：第一轮后）设平均一个人传染了x个人，那么第一轮后，共有多少人得了流感？生：1+x.（多让几名同学回答，然后师板书：1+x）师：（在黑板的其它地方板书：第二轮后）那么第二轮后，共有多少人得了流感？（让生思考一会儿再叫学生）生：1+x+x(1+x).（多让几名同学回答，然后师板书：1+x+x(1+x)）师：下面大家根据题目的意思列一列方程.（生列方程，师巡视）师：（板书：根据题意列方程，得）列出的方程是什么？生：1+x+x(1+x)=121（生答师板书：1+x+x(1+x)=121）.师：（指方程）这是一个一元二次方程，怎么解这个方程？大家试着解一解.（生解方程）师：解出来的结果是什么？生：x1=10，x2=-12（生答师板书：x1=10，x2=-12）.师：（指方程）解这个方程是有讲究的，很多同学用公式法解，发现数字比较大，解起来比较麻烦.实际上我们可以用直接开平方法来解.怎么用直接平方法来解？（稍停）师：（指准1+x+x(1+x)=121）1+x+x(1+x)有公因式1+x，我们把1+x提取出来，得到(1+x)(1+x)（边讲边在其它地方板书：(1+x)(1+x)），可见方程可以化成(1+x)2=121（边讲边在其它地方板书：(1+x)2=121），用直接开平方法解这个方程，容易求出x1=10，x2=-12.师：方程中的x表示每个人传染的人数，所以x2=-12不符合题目的意思，要舍去（板书：（不合题意，舍去））.师：最后还要答.（板书：答：每轮传染中平均每个人传染了10个人）师：下面请大家自己来做一个练习.（三）试探练习，回授调节2.完成下面的解题过程：有一个人知道某个消息，经过两轮传播后共有49人知道这个消息，每轮传播中平均一个人传播了几个人？解：设每轮传播中平均一个人传播了x个人.根据题意列方程，得 .提公因式，得( )2= .解方程，得 x1= ，x2= （不合题意，舍去）.答：每轮传播中平均一个人传播了个人.3.一个人知道某个消息，设每轮传播中一个人传播了x个人，填空：(1)经过一轮传播后，共有人知道这个消息；(2)经过两轮传播后，共有人知道这个消息；(3)经过三轮传播后，共有人知道这个消息；(4)请猜想，经过十轮传播后，共有人知道这个消息.（五）归纳小结，布置作业师：本节课我们学习了利用一元二次方程解决传播问题.俗话说：一传十，十传百.这一传十，十传百是怎么么传的？（指准方程）用方程来表示就是(1+x)2=121.如果传了三轮，就成了(1+x)3；如果传了十轮，就成了(1+x)10.（作业：P48习题1(3)(4)4，4题中91改为81）四、板书设计（略）课题：22.3实际问题与一元二次方程（第3课时）一、教学目标1.会利用一元二次方程解决增长问题.2.培养分析问题解决问题的能力，发展应用意识.二、教学重点和难点1.重点：利用一元二次方程解决增长问题.2.难点：根据增长问题列方程.三、教学过程（一）基本训练，巩固旧知1.填空：(1)扎西家2006年收入是2万元，以后每年增长10％，则扎西家2007年的收入是万元，2008年的收入是万元；(2)扎西家2006年收入是2万元，以后每年的增长率为x，则扎西家2007年的收入是万元，2008年的收入是万元.（(1)题答案为2.2，2.42，(2)题答案为2(1+x)，2(x+1)2，先让生自己做，然后师进行讲解，并写出过程）（二）创设情境，导入新课师：上节课我们学习了利用一元二次方程解决传播问题.什么是传播问题？就是像“一传十，十传百”这样的问题.与传播问题类似的还有一种问题，叫什么问题？叫增长问题.师：下面我们就来看一个增长问题.（三）尝试指导，讲授新课（师出示下面的例题）例扎西家2006年收入是2万元，2008年的收入是2.6万元，求扎西家收入的年平均增长率.师：大家把这个题目好好看几遍.（生默读）师：谁能不看黑板说出题目的意思？生：……（让几名同学说）师：这个题目怎么设？生：设扎西家收入的年平均增长率为x.（师板书：解：设扎西家收入的年平均增长率为x）师：（指准板书）扎西家2006年收入是2万元（板书：2006年 2万元），年平均增长率为x，那么，2007年扎西家的收入是多少万元？（板书：2007年）生：2(1+x).（生答师板书：2(1+x)万元）师：（指准板书）2007年收入是2(1+x)万元，年平均增长率x，那么，2008年扎西家的收入是多少万元？（板书：2008年）生：2(1+x)2.（生答师板书：2(1+x)2万元）师：知道了扎西家2008年的收入可以表示成2(1+x)2，下面大家根据题目的意思列一列方程.（生列方程，师巡视）师：（板书：根据题意列方程，得）列出的方程是什么？生：2(1+x)2＝2.6（生答师书：2(1+x)2＝2.6）.师：接下来解方程（板书：解方程，得）用什么方法解这个方程比较简单？（稍停）用直接开平方法.（以下师在其它地方板书解方程过程）师：得到x1≈0.14，x2≈-2.14（生答师板书：x1≈0.14，x2≈-2.14）.师：扎西家的收入是增加的，所以增长率应该是正数，x2≈-2.14不符合题目的意思，要舍去（板书：（不合题意，舍去））.师：扎西家收入的年平均增长率约为0.14，也就是14％（板书：答：扎西家收入的年平均增长率约为14％）.师：下面请大家自己来做一个练习.（三）试探练习，回授调节2.完成下面的解题过程：某公司今年利润预计是300万元，后年利润要达到450万元，该公司利润的年平均增长率是多少？解：设该公司利润的年平均增长率是x.根据题意列方程，得 .解方程，得 x1≈，x2≈（不合题意，舍去）.答：该公司利润的年平均增长率是 %.3.某公司今年利润预计是300万元，设该公司利润的年平均增长率是x，填空：(1)明年该公司年利润要达到万元；(2)后年该公司年利润要达到万元；(3)第三年该公司年利润要达到万元；(4)第十年该公司年利润要达到万元.（五）归纳小结，布置作业师：本节课我们学习了利用一元二次方程解决增长问题，增长问题在现在生活中很常见，它与传播问题类似，希望大家掌握解决这两个问题的方法.（作业：P48习题1(5)(6)7）教学反思1 、要主动学习、虚心请教，不得偷懒。

Ehrlich-Schwoebel barrier controlled slope selection in epitaxial growth

a r X i v :c o n d -m a t /9905134v 1 [c o n d -m a t .s t a t -m e c h ] 11 M a y 1999Ehrlich–Schwoebel barrier controlled slope selection in epitaxial growth S.Schinzer,S.K ¨o hler,G.Reents Universit¨a t W¨u rzburg,Institut f¨u r Theoretische Physik,Am Hubland D-97074W¨u rzburg,Germany We examine the step dynamics in a 1+1dimensional model of epitaxial growth based on the BCF–theory.The model takes an-alytically into account the diﬀusion of adatoms,an incorporation mechanism and an Ehrlich–Schwoebel barrier at step edges.We ﬁnd that the formation of mounds with a stable slope is closely re-lated to the presence of an incorporation mechanism.We conﬁrm this ﬁnding using a Solid–On–Solid model in 2+1dimensions.In the case of an inﬁnite step edge barrier we are able to calculate the saturation proﬁle analytically.Without incorporation but with inclusion of desorption and detachment we ﬁnd a critical ﬂux for instable growth but no slope selection.In particular,we show that the temperature dependence of the selected slope is solely determined by the Ehrlich–Schwoebel barrier which opens a new possibility in order to measure this fundamental barrier in experiments.Pacs:81.10.AjTheory and models of crystal growth;physics of crystal growth,crystal morphology and orientation 1Introduction Molecular beam epitaxy (MBE)has attracted much interest from both,theo-retical and experimental physicists.On the one hand it allows the fabricationof high quality crystals with arbitrary composition and modulated structures with atomically controlled thickness [1].On the other hand it represents a model of nonequilibrium physics which still lacks a general theory [2].In par-ticular,the appearance and the dynamics of three dimensional (3D)struc-tures (pyramids or mounds)in crystal growth are not well understood in terms of the underlying microscopic processes.A long time ago Burton,Cabrera and Frank introduced the BCF–theory of crystal growth[3].Within this theoretical approach the crystal surface is described by steps of single monolayer height.The evolution of the surface is calculated by solving the diﬀusion equation on each terrace.Within this framework the growth of spirals and the stepﬂow has been investigated.Elk-inani and Villain investigated such a model including the nucleation probabil-ity of new islands[4].They found that the resulting structures are unstable. Towers appear which keep their lateral extension and grow in height only. They called this eﬀect the Zeno–eﬀect.The same observation has been made with a“minimal model”of MBE where fast diﬀusion together with a high Ehrlich–Schwoebel barrier has been implemented[5].Even though the Zeno–eﬀect has been observed recently on Pt(111)[6],quite typically a coarsening process with appearance of slope selection emerges which has been reported for such diverse systems as Fe(001)[7,8],Cu(001) [9,10],GaAs(001)[11,12],and HgTe(001)[13].In addition,slope selection seems to be the generic case of solid–on–solid computer simulations[14,12, 15].In terms of continuum equations the selection of a stable slope has been related to the compensation of uphill and downhill currents[16,17].An uphill current can be generated by an Ehrlich–Schwoebel barrier[18,19].The barrier hinders adatoms to jump down a step edge.Hence,more particles attach to the upper step edge which leads to a growth–instability[20]and 3D–growth.Another process,which constitutes a downhill current,has been recognized using molecular dynamics simulations[21,22].Such diverse mechanisms as downward funneling,transient diﬀusion or a knockout process at step edges lead to the incorporation of arriving particles at the lower side of the step edge.In addition,it has been suggested that such a process is responsible for reentrant layer–by–layer growth[23]Recently we have proposed a simpliﬁed model of epitaxial growth quite sim-ilar to the“minimal model”of Krug[24].In particular we found that an incorporation mechanism is crucial to achieve slope selection.However,one simplifying assumption of the model is an inﬁnite Ehrlich–Schwoebel barrier. In this article we will present in more detail the argument leading to slope selection and we will generalize our results using a continuous step dynamics model analogous to[4].In sec.2we will introduce our extension of the BCF theory and will discuss the relation to existing results(sec.2and3).Typical mound morphologies and the growth dynamics are compared in section4.Figure1:Density of diﬀusing adatoms on a terrace of sizeℓ.The origin of the x-axis is chosen to be in the middle of the terrace.Afterwards we will investigate the emergence of slope selection within the framework of this model(sec.5).We will show that the selected slope has a temperature–dependence which is solely determined by the Ehrlich–Schwoebel barrier.Hence,the determination of selected terrace widths in experiments would give direct insight into microscopic properties such as the Ehrlich–Schwoebel barrier.We conﬁrm the predicted importance of the incorporation mechanism using a kinetic Monte–Carlo simulation of a Solid–On–Solid model in sec.6.Another eﬀective downward current could be due to detachment from steps and subsequent desorption.We will show in sec.7 that slope selection cannot be achieved by these two processes alone.In section8we will calculate the saturation proﬁle in the limiting case of an inﬁnite Ehrlich–Schwoebel barrier.2BCF theoryThe model is based on the Burton–Cabrera–Frank model in1+1dimensions. Within this framework the crystal surface is speciﬁed by the position and direction(upward or downward)of steps.Figure1shows the crystal surface from the point of view of the BCF–theory.It is a coarse grained view–the detailed positions of atoms are not important.However,the terraces of the height of one atomic monolayer(ML)can still be distinguished.The most fundamental assumption is that at each time t the adatom concentration ρis a function of the step positions only.In other words,the diﬀusion of adatoms is considerably faster than the step velocity.Thus,the diﬀusionequation becomes∂ρa(1)where D is the diﬀusion constant and F/a is the ﬂux density with a denoting the lattice constant.Hence,1/F is the time necessary in order to deposit one monolayer.Up to now,this equation was solved with special boundary conditions at x =−ℓ/2and +ℓ/2in the literature.Clearly,the boundary conditions are chosen depending on whether the terrace is a vicinal,a top,or a bottom terrace.In the following we will discuss the typical case of a vicinal terrace.The extension to top and bottom terraces is straightforward.To include an incorporation mechanism it is necessary to extend the theory.We assume that there exists an incorporation radius such that all particles arriving close to a downward step within this radius immediately jump down the step edge.Hence,one has to split the density of diﬀusing particles into two regions.The ﬁrst region close to the upper edge where eq.(1)holds,and thesecondonegiven by the incorporation radius close to the downward step where no particles arrive (F =0).To describe the motion of steps the ﬂux of incorporated particles must be taken into account separately.In the following we will discuss in detail the situation ℓ>R inc as sketched in ﬁg.1.For smaller terraces only one region exist and the calculations are much easier.Since our analytical calculations will show that ℓ>R inc is the generic case we concentrate on this situation.The general one-dimensional solution of eq.(1)is a parabola characterized by three parameters.In addition to the two diﬀusion equations,four boundary conditions are necessary to determine the two distributions ρ1and ρ2.ρ1(−ℓ/2)=0(2)ρ1(ℓ/2−R inc )=ρ2(ℓ/2−R inc )(3)ρ′1(ℓ/2−R inc )=ρ′2(ℓ/2−R inc )(4)−Dρ′2(ℓ/2)=D ℓ1=1k B T(6)Figure2:Arrangement of steps around the bottom terrace of widthℓbetween two mounds.where a stands for the lattice constant.The resulting density distribution has the form as indicated inﬁg.1:a parabola in the upper region and linear close to the downward step.The detailed expressions ofρ1andρ2are not of much interest since the evolution of the crystal is determined by the currents at the edges.In the following we will call u(ℓ)the upward current,i.e.u(ℓ)=−Dρ′1(−ℓ/2)F=−(ℓ−R inc)2.(8)2a(ℓ+ℓ1)Note,that these results are very similar to the corresponding equations(2.2) and(2.3)of reference[4]where no incorporation was considered.Setting R inc=0we regain their results.The absence of a dependence on D reﬂects the ansatz of a quasi–stationary distribution.All arriving particles are compensated for by the loss of particles at the borders and hence the currents are proportional to F.The density itself is proportional to the ratio F/D which again is intuitively clear.3Closure of bottom terracesIn the following,we will reinvestigate the discussion of[4]concerning the closure of a bottom terrace(c.f.ﬁg.2).In the limiting case of an inﬁniteEhrlich–Schwoebel barrier the dynamics of the steps become very simple. We denote by x(t)the position of the right step which of course will depend on the time t.The origin is chosen to be in the middle of the bottom terrace.Due to the inﬁnite Ehrlich–Schwoebel barrier the movement of the right and the left step are symmetric.The evolution is then described by ˙x(t)=−F x(t)−F R inc.Theﬁrst term corresponds to the particles which do fall on the bottom terrace and diﬀuse to the right.The second term is the contribution of particles which are incorporated from the step above(which is valid as long as the bottom terrace is more than a distance R inc away from a top terrace).As a result x(t)evolves asx(t)=(x0+R inc)exp(−F t)−R inc.(9) As long as R inc>0there exists a closure timet c=1R inc.(10)Without an incorporation mechanism(R inc=0)the bottom terrace will never be closed.This is the reason why Elkinani and Villain called their model the Zeno–model to remind the greek philosopher and his paradox.Even though the situation is changed if the discrete structure of the terraces is considered1they showed that this trend still holds which gives rise to the formation of deep cracks.Likewise they found that evenﬁnite values of the Ehrlich–Schwoebel barrier do not change this growth scenario which has been investigated in more detail in[25].Once mounds are built up they remain forever with aﬁxed lateral size.Our discussion of this limiting case shows that the inclusion of an incorporation mechanism changes the growth in a fundamental manner.4Growth dynamicsTo set up the basic ideas of the behaviour during crystal growth we show two typical surface proﬁles according to the numerical integration of the step system.Inﬁg.3we compare the resulting structure of the Zeno model [4]without an incorporation mechanism and with the inclusion of such a mechanism.a)b)Figure 3:In (a)we show the typical height proﬁle of the Zeno model.With inclusion of an incorporation mechanism in (b)we obtain structures with slope selection and prevent the formation of deep cracks.In both cases we have deposited (from bottom to top)20,50,100,and 200ML where the height proﬁles have been shifted by +25,0,-45,and -140ML respectively.The regions are subsections of 200a out of a surface of size 485a .The simulations were carried out on on a system of 485a width with param-eters corresponding to the model used in sec.6.D =1012exp −0.9eVs ≈5664a 2550K k B a ≈8.2aR inc =1aF =1ML s −1As in [4]the Ehrlich–Schwoebel barrier has been suppressed for bottom ter-races of one lattice constant width.Without an additional incorporation mechanism the appearance of trenches is unavoidable in accordance to [25].The incorporation mechanism gives rise to a well deﬁned slope which does not change with time.Another fundamental diﬀerence is the coarsening be-haviour.Without an incorporation mechanism the trenches are stable and the number of mounds remains constant.The additional incorporation mech-anism leads to a coarsening behaviour.In lattice models as well as for continuum equations the coarsening is driven by ﬂuctuations [26,24]and in 1+1dimensions the corresponding exponent is 1/3.This is in accordance to Ostwald-ripening which has been predicted from the similarities of the relevant continuum equations [17].However,since we treat the step evolution in a deterministic manner we do not obtain a scaling behaviour.The only way ﬂuctuations come into play during the simulation is when new islands are nucleated.As a consequence the evolutionof e.g.the width of the height distribution w is characterized by jumps(data not shown).A jump in w appears each time when two mounds merge.These ﬁndings are a direct conﬁrmation of the relevance of theﬂuctuations for the coarsening behaviour.5Slope selectionThe inclusion of an incorporation mechanism leads to slope selection which is apparent fromﬁg.3b.Siegert and Plischke[16]required a cancelation of upward and downward currents in the continuum equations.Again,in the case of an inﬁnite Ehrlich–Schwoebel barrier the calculations are straight-forward.In this case the downward current on a vicinal terrace of sizeℓis solely due to the incorporation mechanism,i.e.proportional to F R inc.All the remaining diﬀusing adatoms will contribute to the upward current and hence the current will be F(ℓ−R inc).As a consequence the slope selection will be achieved with a mean terrace width of sizeℓ∗=2R inc(11) in accordance to theﬁndings in[24].It remains to calculate the terrace widths for arbitrary parameters.Since we know the currents(equations(7),(8)and the incorporation mechanism)we obtain the overall slope(resp.terrace width)dependent currentJ(ℓ)=u(ℓ)+d(ℓ)+F R incF=ℓ1R inc2=2R inc+The stable slope does not depend on the diﬀusion constant.However,itshould be clear from the derivation,that in order to achieve slope selection the typical diﬀusion length should be much larger thanℓ∗.Otherwise the vicinal terraces would not proceed via stepﬂow.Rather new nucleationevents on the terraces would lead to a rugged surface structure.6Solid–On–Solid modelIn order to verify the predicted importance of the incorporation mechanism we use computer simulations of the Solid–On–Solid(SOS)model on a simple cubic lattice.All processes on the surface are(Arrhenius-)activated processes which are described by one common prefactorν0and an activation energy which is parameterized as follows:E B is the barrier for surface diﬀusion,at step edges an Ehrlich–Schwoebel barrier E S is added.However,this barrier is not added for a particle which sits on top of a single particle or a row[27,23]. Each next neighbour contributes E N to the activation energy.Within this framework the diﬀusion constant becomes D=ν0·exp(−E B/k B T). Here,we concentrate on a particular set of parameters even though other parameter sets were used as well.We chooseν0=1012s−1,E B=0.9eV,E N=0.25eV,and E S=0.1eV.This model was already investigated in[28]and reproduces some kinetic features of CdTe(001).The deposition of particles occurs with a rate F.The two simulations shown inﬁg.4are carried out on a300×300lattice at560K and started on a singular(ﬂat)surface. Inﬁg.4the resulting surfaces with and without the inclusion of the incor-poration mechanism are shown.Without an incorporation mechanism no slope selection occurs.Clearly,without incorporation the conﬁguration of the towers remains unchanged whereas the inclusion leads to coarsening. The number of mounds diminishes with time.Hence,without an incorpora-tion mechanism no coarsening can be identiﬁed.We want to mention that it seems that at higher temperatures the attachment/detachment kinetics of atoms at step edges yields a coarsening eﬀect(data not shown).However, still no slope selection has been observed.Atﬁrst glance ourﬁndings contradict previous results obtained with a very similar model.ˇSmilauer and Vvedensky obtained a formation of mounds with slope selection irrespective of the inclusion or exclusion of an incorporation mechanism[14].However,they implemented the Ehrlich–Schwoebel barrier in a diﬀerent way.Rather than to hinder the jump over a step edge they im-pede the jump towards a step edge.Their motivation for this implementationFigure4:We compare the morphology of the surfaces without(left)and with(right)an incorporation mechanism.Note that the two grey–scales are diﬀerent.The heights in the left picture range from1208to1326whereas the right surface only spans a height diﬀerence of12from minimum to max-imum.The contour lines are drawn for the same surface at an earlier stage where only300ML have been deposited.Without incorporation the mounds (towers)are nearly unchanged despite the deposition of1000ML.was to allow the adatoms to leave a small line of particles of width one which has been tested as a cause for reentrant layer-by-layer growth[27,23].In our simulations the same goal is achieved by suppressing the Ehrlich–Schwoebel barrier in such a situation.However,in their simulations particles arriving directly at a step edge have a probability of1/4to jump down the edge,1/4 to jump away from the edge and1/2to jump along the step edge.Eﬀectively this leads to an incorporation radius of length1/2.Other simulations of SOS-models used bcc(001)[29,15]in order to study the growth of typical metals.In these simulations the SOS–restriction is implemented in such a way that an adatom must be supported by the four underlying atoms.Hence,the downward funneling process is directly im-plemented.Again,as a result slope selection is achieved,which has already been discussed in great detail in[30].7Detachment and desorptionOne might assume that other mechanisms could lead to a zero in the slope dependent current.In the following we will carry out an analogous calcula-tion with an adatom–detachment rate from steps and inclusion of desorption [31].One might assume that both processes generate an eﬀective downward current which can compensate for the Ehrlich–Schwoebel eﬀect.To inves-tigate whether they are suﬃcient to obtain slope selection(and to simplify notation)we exclude the incorporation mechanism.Thus,the distinction of the two regions on a terrace is not necessary.The desorption of diﬀusing adatoms is easily incorporated including a term −ρ(x)/τin the diﬀusion equation(1)[31].In order to include detachment from steps we have to replace boundary condition(2)by−Dρ′(−ℓ/2)=γ−Dℓ1ρ(ℓ/2)−γaτ−DFDτ(∆−1)+D(∆−1)(16)where∆=e2ℓ/√If we assume a positive E S in the general case we obtain a critical currentF C=aγd t (t)=−F(x1(t)+R inc)d x2d t (t)=−F(x i(t)−x i−1(t))(19) ...Measuring the time in units of1/F(i.e.setting F=1in the above equations) the time to grow one monolayer is equal to one.In addition,to simplify notation we will measure all lengths in units of R inc.i x i(0)−x i−1(0)e−1e2−e−1e3−2e2+1 1.9949346e−1e5−4e4+93e2+1 2.00007Table1:Analytical expressions of the saturation proﬁle step positions and the numerical values of the terrace widths.The solution for the lowest terrace has been given in eq.(9).Thesolutionfor all steps isx n(t)=ni=1t n−i(n−i)!e(23)which can be solved as described in the appendix using the generating func-tion.As a result one obtains the initial positions of the steps on an inﬁnite symmetric step proﬁle.Every time the bottom terrace is closed the steps (with new indices)are located at these positions.In table1we show the analytical expressions for the step positions derived from the generating function.In addition,the numerical values of the ter-race widths are shown.With growing index the terrace widths are rapidly approaching two.Even though they oscillate around this value it can be shown thatlimi→∞(x i(0)−x i−1(0))=2.(24)Note that we measure the lengths in units of R inc.Hence we predict a slope selection with slope1/(2R inc)as derived from the simple argument of section 5.The derivation shows that the selected slope is controlled by the closure of the bottom terrace only.Another length scale is the nucleation length.It is deﬁned by the typical length of a top terrace at which nucleation of an island occurs[33].This length scale is responsible for the rounding of the towers in ﬁg.3b and causes a perturbation of the steady–state saturation proﬁle.9ConclusionWe have investigated the eﬀect of an incorporation mechanism on the mor-phology of growing surfaces.The inclusion of an incorporation mechanism in a1+1dimensional BCF–theory as well as in SOS computer simulation in 2+1dimensions is necessary in order to obtain slope selection and a coarsen-ing process.We were able to derive analytically the temperature dependence of the selected slope.We found that the Ehrlich–Schwoebel barrier alone controls the temperature dependence.In the limit of an inﬁnite step edge barrier we derived the steady state saturation proﬁle.In this case the result-ing mound morphology is controlled by the closure of the bottom terrace.A Generating functionTo simplify notation we introduce the shifted step positions b j=x j(0)+1. We will try to extract the generating functionf(z)=∞j=0b j z j(25)for the shifted step positions.Clearly,the b j are only of physical meaning if j>0and b0can be chosen arbitrarily.Starting from equation(23)b n−1=ni=1b i(n−i)!(27)⇒e z∞m=1b m z m=∞n=2ni=1b i z nChoosing b0=0and using b1=e we arrive ate z f(z)=∞n=0ni=0b i z ne z−1−z.(31)The lowest coeﬃcientsb j=1∂z jz=0(32)derived from the generating function are shown in table(8).In addition,the generating function can be used to formally prove equation(24).***This work has been supported by the Deutsche Forschungsgemeinschaft DFG through SFB410.References[1]M.A.Hermann and H.Sitter.Molecular beam epitaxy.Springer(Berlin),1996.2.,rev.and updated ed.[2]A.-L.Barab´a si and H.E.Stanley.Fractal concepts in surface growth.Cambridge University Press,1995.[3]W.K.Burton,N.Cabrera,and F.C.Frank.Philosophical Transactionsof the Royal Society of London,243:299–358,1951.[4]I.Elkinani and J.Villain.Journal de Physique I France,4:949–973,1994.[5]J.Krug.Journal of Statistical Physics,87(3/4):505–518,1997.[6]M.Kalﬀ,P.ˇSmilauer,sa,and Th.Michely.Surface ScienceLetter,in press,1999.[7]K.Th¨u rmer,R.Koch,M.Weber,and K.H.Rieder.Physical ReviewLetters,75(9):1767,1995.[8]J.A.Stroscio,D.T.Pierce,M.D.Stiles,A.Zangwill,and L.M.Sander.Physical Review Letters,75(23):4246–4249,1995.[9]H.-J.Ernst,F.Fabre,R.Folkerts,and pujoulade.Physical ReviewLetters,72(1):112–115,1994.[10]J.-K.Zuo and J.F.Wendelken.Physical Review Letters,78(14):2791–2794,1997.[11]M.D.Johnson,C.Orme,A.W.Hunt,D.Graﬀ,J.Sudijono,L.M.Sander,and B.G.Orr.Physical Review Letters,72(1):116–119,1994.[12]C.Orme,M.D.Johnson,K.-T.Leung,B.G.Orr,P.ˇSmilauer,and D.D.Vvedensky.Journal of Crystal Growth,150:128–135,1995.[13]S.Oehling,M.Ehinger,T.Gerhard, C.R.Becker,ndwehr,M.Schneider,D.Eich,H.Neureiter,R.Fink,M.Sokolowski,and E.Um-bach.Applied Physics Letters,73(22):3205–3207,1998.[14]P.ˇSmilauer and D.D.Vvedensky.Physical Review B,52(19):14263–14272,1995.[15]K.Th¨u rmer,R.Koch,P.Schilbe,and K.H.Rieder.Surface Science,395:12–22,1998.[16]M.Siegert and M.Plischke.Physical Review Letters,73(11):1517–1520,1994.[17]M.Siegert and M.Plischke.Physical Review E,53(1):307–318,1996.[18]G.Ehrlich and F.G.Hudda.Journal of Chemical Physics,44(3):1039–1049,1966.[19]R.L.Schwoebel and E.J.Shipsey.Journal of Applied Physics,37(10):3682–3686,1966.[20]J.Villain.Journal de Physique I,1:19–42,1991.[21]J.W.Evans,D.E.Sanders,P.A.Thiel,and A.E.DePristo.PhysicalReview B,41(8):5410–5413,1990.[22]Y.Yue,Y.K.Ho,and Z.Y.Pan.Physical Review B,57(11):6685–6688,1998.[23]P.ˇSmilauer,M.R.Wilby,and D.D.Vvedensky.Physical Review B,47(7):4119–4122,1993.[24]M.Biehl,W.Kinzel,and S.Schinzer.Europhysics Letters,41(4):443–448,1998.[25]P.Politi.Journal de Physique I,7:797–806,1997.[26]Lei-Han Tang,P.ˇSmilauer,and D.D.Vvedensky.The European PhysicalJournal B,2:409–412,1998.[27]B.Poelsema,R.Kunkel,N.Nagel,A.F.Becker,G.Rosenfeld,L.K.Verheij,and sa.Applied Physics A,53:369–376,1991.[28]S.Schinzer and W.Kinzel.Surface Science,401(1):96–104,1998.[29]J.G.Amar and F.Family.Physical Review B,54(20):14742–14753,1996.[30]J.G.Amar and F.Family.Physical Review B,54(19):14071–14076,1996.[31]J.Villain and A.Pimpinelli.Physique de la croissance cristalline.´Editions Eyrolles,1995.[32]J.Krug and M.Schimschak.Journal de Physique I France,5:1065–1086,1995.[33]J.Villain,A.Pimpinelli,L.Tang,and D.Wolf.Journal de Physique I,2(11):2107–2121,1992.。

肃州区七中九年级化学下册第九章现代生活与化学 9.1 有机物的常识教学课件新版粤教版

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9.从50 g 20%的硝酸钾溶液中取出25 g溶液后,剩余硝酸钾溶液中溶质的质量分数是( B ) A.40% B.20% C.15% D.10% 10.以下物质溶于水后液体温度没有明显变化的是( C ) A.NaOH B.浓硫酸 C.NaCl D.NH4NO3 11.将10 g某物质溶于90 g水中形成溶液,所得的溶液中溶质质量分数是( D ) A.一定等于10% B.一定大于10% C.一定小于10% D.三种情况均有可能
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科粤版九年级下册9.1有机物的常识优秀教学案例

2.鼓励学生互相评价、互相学习，培养学生的批判性思维和自我改进的能力。
3.教师对学生的学习情况进行评价，关注学生的学习进步和问题，为学生提供及时的指导和帮助。
四、教学内容与过程
（一）导入新课
1.利用生活实例，如食品、药品、日常用品等，引入有机物的概念，激发学生的学习兴趣。
2.向学生提出问题，如“有机物和无机物有什么区别？”、“有机物在生活中的应用有哪些？”等，引导学生思考和讨论。
二、教学目标和基本性质，掌握常见有机物的结构和特点。
2.理解有机化学反应的机理，能够分析并解释一些简单的有机化学反应。
3.能够运用有机化学的基本知识，解决一些实际问题，如有机物的鉴别、合成和应用等。
（二）过程与方法
1.通过观察、实验、讨论等方式，培养学生的观察能力、实验能力和团队合作精神。
（三）学生小组讨论
1.让学生分成小组，每组选择一个有机物，如乙醇、醋酸等，进行观察和讨论。
2.引导学生从有机物的结构、性质、反应等方面进行探究，提出问题并寻找答案。
3.鼓励学生分享自己的观点和发现，培养学生的团队合作精神和沟通能力。
（四）总结归纳
1.引导学生对所学知识进行总结归纳，如有机物的分类、性质、反应等。
2.引导学生运用科学的方法，进行观察、实验、推理、验证等，培养学生的科学思维和探究能力。
3.培养学生自主学习的能力，鼓励学生提出问题、寻找答案、表达交流，提高学生的学习效果。
（三）情感态度与价值观
1.培养学生对有机化学的兴趣和好奇心，激发学生对科学的热爱和追求。
2.培养学生的科学素养，使学生认识到有机化学在生活中的重要性和应用价值。
3.设计具有启发性的问题，引导学生主动思考、探究，激发学生的思维能力和创造力。

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(2)火炬内部结构 ①燃烧系统采用多孔燃气管输出燃气。与单孔燃气管相比，等量的燃气从多孔燃气管输出产生的火焰更旺，原因是__增__大__了__燃__气__与__空__气__的___
接触面积，燃烧更充分；关闭燃气开关可熄灭火炬，其灭火原理是清除或隔离可燃物。
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(3)“深地一号·顺北油气田基地”成功勘探出全球最深的油气田，获取丰富的石油和天然气资源。其中天然气主要成分燃烧的化学方程式为
CH4++2O2
CO2+2H2O
。
(4)5G基站的建立是深蓝网络空间领域数字基础的有力支撑。5G铁塔表面的冷喷锌涂层达到防锈的目的，其防锈原理是 ___覆__盖__保__护__层__，______
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7.(2023·潍坊)中华传统文化典籍中包含很多化学知识。下列解释错. 误. 的是 (B ) A.《华阳国志》:“取井火煮之，一斛水得五斗盐”——加热蒸发可得食盐 B.《天工开物》:“凡石灰，经火焚炼为用”——煅烧石灰石可生成熟石灰 C.《梦溪笔谈》:“熬胆矾铁釜，久之亦化为铜”——铁的金属活动性强于铜 D.《周易》:“泽中有火，上火下泽”——池中有机物发酵产生可燃性气体
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知识归纳
中考名题点拨中考精品题训练
中P考AR名T题0点2 拨
让学习变的简单
学考精练·化学·九年级全一册·配粤科版
【例题1】(情景题)科技助推中国梦。我国正在深海、深空、深地、深蓝等领域积极抢占科技制高点。

安徽省长丰县实验高级中学人教版高中数学五教案：3.3.2简单线性规划问题

3。

3.2 简单线性规划问题3。

3。

2 简单线性规划问题(3课时）一、知识与技能1。

掌握线性规划的意义以及约束条件、目标函数、可行解、可行域、最优解等基本概念;2.运用线性规划问题的图解法，并能应用它解决一些简单的实际问题.二、过程与方法1。

培养学生观察、联想以及作图的能力，渗透集合、化归、数形结合的数学思想，提高学生“建模”和解决实际问题的能力;2。

结合教学内容,培养学生学习数学的兴趣和“用数学”的意识,激励学生创新.三、情感态度与价值观1.通过本节教学着重培养学生掌握“数形结合”的数学思想，尽管侧重于用“数"研究“形”,但同时也用“形"去研究“数”，培养学生观察、联想、猜测、归纳等数学能力；2。

结合教学内容,培养学生学习数学的兴趣和“用数学”的意识，激励学生勇于创新.教学重点重点是二元一次不等式(组）表示平面的区域.教学难点难点是把实际问题转化为线性规划问题,并给出解答。

解决难点的关键是根据实际问题中的已知条件,找出约束条件和目标函数,利用图解法求得最优解.为突出重点,本节教学应指导学生紧紧抓住化归、数形结合的数学思想方法将实际问题数学化、代数问题几何化。

第1课时导入新课师前面我们学习了二元一次不等式A x+B y+C ＞0在平面直角坐标系中的平面区域的确定方法，请同学们回忆一下。

（生回答）推进新课 [合作探究］师在现实生产、生活中，经常会遇到资源利用、人力调配、生产安排等问题。

例如，某工厂用A 、B 两种配件生产甲、乙两种产品，每生产一件甲产品使用4个A 产品耗时1小时，每生产一件乙产品使用4个B 产品耗时2小时，该厂每天最多可从配件厂获得16个A 配件和12个B 配件，按每天工作8小时计算,该厂所有可能的日生产安排是什么? 设甲、乙两种产品分别生产x 、y 件，应如何列式？生由已知条件可得二元一次不等式组：⎪⎪⎪⎩⎪⎪⎪⎨⎧≥≥≤≤≤+.0,0,124,164,82y x y x y x师如何将上述不等式组表示成平面上的区域？生（板演）师对照课本98页图3.39，图中阴影部分中的整点（坐标为整数的点）就代表所有可能的日生产安排，即当点P （x,y ）在上述平面区域中时,所安排的生产任务x 、y 才有意义。

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（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）个性化辅导教案授课时间：2015--7---15日备课时间：2014---7---14年级: 五升六年级课时：2小时课题：unit3及其现在进行时态学生姓名:李皓滨教研老师：谢老师教学目标与重难点1.重要单词及其词组2.知识点辨析3.语法：现在进行时态讲解公开试讲时间及教务主任签名所有授课内容必须在本中心公开试讲由教务主任签字核实之后方可授课，欢迎各位家长监督我们的教学工作！教务主任签字：【课前开心一刻】读短文，判断对错。

正确的写“T”，错误的写“F”。

My name is Peter. I have a happy family. My father is a doctor. He is tall and strong. Hisbirthday is in March. My mother is a teacher. She is kind. Her birthday is in November. I am a student. My birthday is in May. I love my family very much.( )1. There are three people in Peter's family.( )2. Peter is a student.( )3. Peter's mother is a nurse.( )4. Peter's birthday is in May.( )5. His mother's birthday is in March.【上节课知识点回顾】回顾一般现在时【授课内容】重点句子1. We have a few fun things in spring. After the sports meet, we have an Easterparty. 在春天我们有一些趣事。

（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）2. –When is the party? 派对在什么时候？– It’s in April. 在四月份。

3. Autumn is my favourite season.秋天是我最喜欢的季节。

4. We’ll go to the Great Wall. 我们将去长城。

5. It’s usually in September or October. I’ll eat mooncakes with my family.通常在九月或十月。

我将和我家人一起吃月饼。

5. We will roll Easter eggs. We will look for eggs. We will eat chocolate eggs. You will meet the Easter Bunny!我们将滚复活节彩蛋。

我们将找蛋。

我们将吃巧克力蛋。

你将见到复活节兔子！6. Will you come to the party? Please send me an email at robin@ by March 23rd.你回来参加派对吗？请在3月23日前发送电子邮件至robin@。

7. That’s very kind of you. 你真是太好了。

8. – What will you do for your mum on Mother’s Day? 母亲节那天你将为你妈吗做些什么？– I’ll cook for my mother. 我将为我妈妈做饭。

语音辅音字母组合ch, sh的发音：[ t。

课外补充：绝对大数一般发[ t，如ch icken lun ch tea chch 个别发[ k ]，如s ch ool个别发[ ，如ma ch ine （机器）重点知识及语法1、询问活动或节日在几月份：– When is…? – It’s in+月份.2、表示将要做某事：We’ll/We will + 动词原形3、介词in, on, at的用法：in+大致时间（年月，季节，早中晚等）on + 具体某一天（年月日，星期等）at + 具体时刻（……点钟）重点作文1、活动邀请函，如Birthday party思路导引（1）交代几个Wh-信息点：When（时间）, Why（原因）, Where（地点）, What （活动）。

（2）结尾:：邀请对方参加活动Will you come to the party? 如需回复要注明RSVP，并给出回复的联系方式Please send me an email at…2、范文：（1）课本P29 Read and write（2）Birthday partyWHEN: May 17th,7 p.m.WHY: Chen Jie’s birthdayWHERE: Chen Jie’s homeWHAT: We will play many games. We will sing and dance. We will eat a lot of food.（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）RSVP: Will you come to the party? Please send me an email at Chenjie@126. com by May 30th.练习：选择填空。

（）1. My birthday is ___________ May.A. atB. inC. onD. of（）2. ___________ is the first month of a year.A. FebruaryB. DecemberC. OctoberD. January（）3. ___________ is Children's Day?A. WhenB. WhatC.WhichD. Why( ) 4. ——What's the date?——__________________.A. In JulyB. April 4thC. MarchD. 9th（）5. ——Is your birthday in June?——__________________.A. Yes, it is.B. Yes, it isn't.C. No, it isn't.D. No, it is.（） 6. ——How old are you?——I'm ______________.A. firstB. twelveC. secondD. fifth（）7. ——When is your birthday?——It's in ____________.A. thirdB. JanuaryC. 20thD. 8:30（）8.When _______ your birthday?A. amB. isC. areD. be（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（）9.My birthday is in July. Aunt Mary's birthday is in July,_________.A. eitherB. toC. tooD. two（）10.——When is National Day?——____________________.A. It's September 10th.B. It's March 12th.C. It's December 25th.D. It's October 1st.阅读理解。

1.阅读短文，选择正确的答案。

Hello, my name is Li Ming , yesterday I made a birthday chart for my family.My birthday is June 9th. My mother’s birthday is June 14th. M y grandpa’s birthday is June 23th.My gandpa is in America, my grandpa lives with my uncleBill and aunt Mary in America, so I usually make an e-card to him for his birthday.There are 2 birthdays in November, my father and my cousin Alice. Alice likes dolls, so I often send her some beautiful dolls .My grandma’s birthday is in April , it is very warm and nice in April, thenwhen her birthday is coming ,we often go hiking for her birthday. There is none in December.This is my chart ,I love my family.( )1.How many people are there in Li Ming’s family?A. 6B. 8C. 9( )2.How many birthdays are there in June?A. 2B. 4C. 3（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）（☆☆）教学过程（☆☆）( )3 .How many birthdays are there in December?A. 0B. 2C. 3( )4.What will Li Ming do for his grandpa’s birthday?A. send an e -mailB. send an e-cardC. send a birthday cake( )5.When is Li Ming's birthday?A.June 14thB. June 9thC. June 23th【重难点内容】现在进行时态精讲精练1. 用法：①现在正在进行或发生的动作例：I am reading an English book now. 我现在正在读一本英语书。