解析化学中简答题简化方法.doc

解析化学中简答题简化方法简答题是近年化学高考中常呈现的题型。它首要考察学生对所学常识了解的精确性，思想的完好性，推理的紧密性和表述的条理性。近几年化学高考题中简答题的分值占到左右，在总分值中已占有必定的份量。简答题看起来好像不难，但要精确答复确不易，学生多感到有力无处使，形成失分较多。学生在简答题中常见过错是：基础常识不结实，对有关概念、基本理论了解不透彻，不能答复出常识要害；思想混乱，缺少紧密的逻辑思想才能；表达不标准，不能用精确的化学用语答复疑问。怎么才干精确、完好、简练、谨慎地答复此类题呢？我以为，除应加强基础常识教育外，还应培育学生仔细审题、捉住答题的要害和要害、运用精确化学用语表述疑问的才能。此外，还要加强此类题解法的辅导。下面就以近年高题为例，剖析这类题的答复办法。The simple answer is often appears in recent years the chemical in the college entrance examination questions. It mainly inspects the student to understand the accuracy of knowledge, ideological integrity, logic reasoning tightness and expression. In recent years, the chemical examination questions of short answer scores accounted for about 10%, has occupied a certain amount in the total score. The simple answer seems not difficult, but it is not easy to accurately answer, students feel have no place to make a strong, the formation of lost more. The student is a common fault in the short answer: basic common sense is not strong, the related concepts, basic theories are not well understood, not answer common sense; the ideological confusion, lack of close logic thinking ability; the expression is not standard, not accurate answer questions in chemistry. How to accurate, complete, concise, careful to answer such questions? I think, should strengthen basic education, but also should cultivate students to carefully examines the topic, catch the answer key and key, the use of accurate chemical expression question to. In addition, but also to strengthen the solving methods of counseling. Following on recent high problem as an example, analysis of this kind of question answer way.例时，纯水的值小于，为什么？Cases of 1.80 , pure water pH value is less than 7, why?答案：水的电离是一个吸热反响。室温时，纯水中摩升，因此。但温度升高到时，水的电离度增大，和均大于摩升，故。Answer: water ionization H2O HOH is an endothermic reaction. At room temperature, pure water ( H + ) = ( OH ) = 10 - 7 friction / L, pH = 1g ( H + ) = 7. But the temperature to 80, the degree of ionization of water increases, ( H + ) and ( H ) are more than 10 - 7 friction / L, pH = LG ( H + 7 ) 10 7 , ( OH - ), ( H + ) ( OH - ) 10-14 and fault, is attributed to the knowledge base.例当化学反响（气）（气）（气）处于平衡状况时，向其间参加一种含量较多的氯气，平衡发作移动，在树立新平衡曾经，中所含的百分含量比原平衡状况时能否会添加？请阐明理由。Example 2 when the chemical reaction of PCl5 ( gas ) PCl3 ( gas ) Cl2 ( gas ) in the balance, to participate in a 37Cl between the chlorine content of more, balanced attack move, in the establishment of a new balance once, percentage of 37Cl containing PCl3 than the original balance situation will add? Please explain the reason.答案：参加含量较多的氯气后，平衡向左移动，使的分化反响也在进行，所以，中含的百分含量也会增大。Answer: chlorine in the 37Cl content of more, balance move left, the differentiation of PCl5 responses are, therefore, percentage of PCl3 containing 37Cl will increase.剖析：本题是用同位素示踪法考察学生关于可逆反响中的化学平衡是动态平衡这一基本概念。“动态平衡是化学平衡的三个基本特征之一，喷灌设备是中学教育重复强调的要点。标题没有直接问，而是问的改变状况；不是问树立平衡后而是问树立平衡前；不只要答复能否会添加，并且需求阐明理由。这样，把基础常识作了两次变换，答题难度加大。因此，在教育中应加强学生思想灵活性、变通性的练习。Analysis: This is by isotopic tracer chemical equilibrium method was used to investigate the students about reversible reaction is one of the basic concepts of dynamic balance. Dynamic balance is one of the three basic characteristics of chemical equilibrium, are the main points of middle school education repeatedly emphasize. The title is not directly ask PCl5, ask PCl3 to change the situation; dont ask to establish balance but ask to establish equilibrium; dont answer will add, and explain the reason. So, the basic common sense made two transformation, answer difficulty. Therefore, in the education should strengthen the students ideological flexibility, flexibility exercises.例甲、乙两瓶氨水的浓度分别为摩升和摩升，则甲、乙两瓶氧水中之比（填大于、等于或小于），阐明理由。3 A, B concentration with two bottles of ammonia were 1 friction / L and 0.1 M / L, a, B two bottles of oxygen in water ( OH ) ratio ( greater than, less than or equal to fill ) 10, explain the reason.答案：在同一温度下，关于同种弱电解质，浓度越小，电离度越大。甲瓶氨水的浓度是乙瓶氨水浓度的倍，故甲瓶氨水的电离度比乙瓶氨水的电离度小，所以，甲、乙两瓶氨水中之比应小于。Answer: at the same temperature, the same weak electrolyte, concentration is small, the greater the degree of ionization. A bottle of ammonia concentration is 10 times the B bottle of ammonia concentration, so a bottle of ammonia ionization degree than B bottle of ammonia ionization degree is small, so the, a, B two ammonia water bottle ( OH ) ratio should be less than 10.剖析：本题首要考察电解质浓度对电离度的影响。考生常常把浓度对电离度的影响和对电离平衡常数的影响相混杂，形成错解。有些考生虽对“同一弱电解质，浓度越小，电离度越大”熔铝炉这个大前提明白，但要使用这一大前提剖析详细疑问时，却显得思想混乱、表达的逻辑联系不清。其实“答案”中用到的推理办法是咱们思想中常见到的方式逻辑推理办法“三段论”。除此而外，还有因果、先总后分或先分后总等思想办法在近年的高考简答题中均有表现。 因此，教师在教育中应加强学生逻辑思想、推理才能的练习。Analysis: effects of this primary investigation electrolyte concentration on the degree of ionization. Candidates often concentration mixed effects on ionization degree and influence on the ionization equilibrium constant, form error. Some candidates have to with a weak electrolyte, concentration is small, the degree of ionization is the premise to understand , but to use the detailed questions, but is ideological confusion, expression of the logic connection is not clear. In fact, reasoning method used in the answer is often seen in our thinking way of logical reasoning way - syllogism . In addition, there are causal, general the first or after total way of thinking in the college entrance examination in recent years displayed in short answer questions. Therefore, teachers should strengthen students logical thinking, reasoning ability in the education practice.例在时，若个体积的某强酸溶液与体积的某强碱溶液混和后溶液呈中性，则混和之前该强酸与强碱的值之间应满意的联系是。Cases of 4 at 25 , if a solution of a strong acid solution 10 volume with the 1 volume of the mixed solution was neutral, mixing before between the acid and alkali and pH value should be satisfied with the contact is.答案：酸碱Answer: pH acid + pH base = 15剖析：本题首要考察学生对溶液酸碱性和值之间联系等常识的知道。时，体积的某强酸溶液与体积的某强碱溶液混和后溶液呈中性，阐明反响中强酸的离子和强碱中离子物质的量持平。令强酸中离子物质的量为摩，体积为升，则强酸中摩升，酸，强碱中摩升，强碱中摩升，碱，因此，酸碱。Analysis: know of solution acidity and pH value links sense this first inspects the student. At 25 , a solution of a strong acid solution 10 volume with the 1 volume of the mixed solution was neutral, clarify the OH - ion H + ions and alkali substances reaction acid in the amount of flat. The strong H + ion in the amount of substance for 0.1 Mo, 1 volume of 1 litres, the acid ( H + ) = 0.1 friction / L = 1, pH acid, alkali ( OH ) = 1 friction / liter, alkali ( H + ) = 10 - 14 friction / L, pH base = 14, therefore, pH acid + pH base = 15.解此题的要害是先要把通常联系转化成详细数值，再把由详细数值推出的特别联系推及到通常。因为答题中涉及到由“通常特别通常”这两个推理进