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1、高中化学计算题 (一)（High school chemistry problem (1)）The second method USES the law of the conservation, so the computational complexity is much less, also do not need chemical equation is listed first, balancing, thereby significantly shorten about the time, not I dont know according to which one is more
2、equations to panic, as a result of nitric acid. Then look at the topic: example 2 in a 6 litres of airtight container, add 3 X (gas) and 2 litres Y (gas), under certain conditions the following reaction: 4 X (gas) + 3 (gas) Y (gas) 2 q + nR (gas) after reaching balance, container, constant temperatu
3、re of mixed gas pressure increase 5% than the original, X concentration decreases to a third, is the value of n in the reaction equation A. 3 B. 4 C. 5 D. 6 Solution one: grasp the concentration of X decreases by 1/3, the coefficient ratio of the combination of chemical equation is equal to volume r
4、atio, can list the beginning state, variable and final state of each substance separately: 4 x 3 y 2 q nR The initial state is 3L, 2L, 0, 0 Variable - a third * 3 l = 1 l - 3/4 + 2/4 = 3/4 l * 1 * 1 l l = 1/2 l = n + n / 4 * 1 l / 4 l The final state is 3 minus 1 is equal to 2L 2 minus 3/4 is equal
5、to 5 over 4L 0 plus 1/2 is equal to 1 over 2L 0 plus n over 4 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 1 页，共 22 页 - - - - - - - - - is equal to n over 4 liters Relation by above knowable, balance (final) after mixing the volume of a gas of 5/4 (2 + + 1/2 + n /
6、4) L (15 + n) / 4 L, according to the question than the original mixed gas pressure increased by 5% (15 + n) / 4 to 5 = 5 * 5%, n = 6 are obtained. Method 2: choose difference method, according to the question of mixed gas pressure than the original increased by 5%, according to the question than th
7、e original mixed gas pressure increased by 5%, the volume of the mixed gas increases the (2 + 3) * 5% = 0.25 L, according to the equation, 4 X + 3 y can only generate 2 q + nR, namely every 4 X reaction volume, total volume change quantity for the (2 + n) - (4 + 3) = n - 5, existing = 1/3 * 3 L L X
8、reaction, namely the volume change of 1 L * (n - 5) / 4 = 0.25 L, so as to calculate n = 6. Method 3: seize the mixed gas pressure than the original 5% increase, it is concluded that at the beginning of the reaction by X + Y, balance must shift to the right first, after generating Q and R, the press
9、ure increase, show positive reaction is certainly increase the reaction volume, the sum of the reaction equation coefficient of X and Y will be less than the coefficient of the sum of Q and R, so 4 + 3 5, in the four options that accords with a requirement only D n = 6, for the persons to be the ans
10、wer. Subject test is about the content of the chemical equilibrium. A solution is to follow the law of chemical equilibrium, as according to the specification, although can certainly work out the correct answer, but did not hold choice, dont ask, dont 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - -
11、 - - - - 名师精心整理 - - - - - - - 第 2 页，共 22 页 - - - - - - - - - process, as long as the results, the characteristics of calculated as a topic to do, the ordinary student also should use at least 5 minutes to complete, spend more time. Method using difference method, with the volume of n variables (delt
12、a), to establish the equation for 豱 RanXia worked out the value, but still failed to make full use of the choice characteristics of multiple choice, available to about 1 minutes. Solution three understanding on the relationship between the balance of mobile and volume change, not half a minute can b
13、e concluded that the only correct answer. Thus, in the process of calculation according to the characteristics of the subject chooses different problem solving methods, often could help in the process of reducing operation time and the chance of error, to achieve fast and accurate the effect of prob
14、lem solving, and use more problem solving method usually has the following kinds: Quotient method: this method is used to solve the molecular weight of organic compounds (especially hydrocarbons). Because of alkanes general formula CnH2n + 2, and the molecular weight of 14 n + 2, the corresponding p
15、araffin base general formula CnH2n + 1, and the molecular weight of 14 n + 1, olefins and naphthenes general formula CnH2n, molecular weight of 14 n, the corresponding alkyl the general formula CnH2n - 1, the molecular weight of 14 n - 1, alkynes and alkadiene general formula CnH2n - 2, molecular we
16、ight of 14 n 2, corresponding hydroxyl general formula CnH2n - 3, the molecular weight of 14 n - 3, so can be known molecular weight minus the oxygen-containing functional groups of the organic matter type 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 3 页，共 22 页 - -
17、 - - - - - - - quantity, difference divided by 14 (hydrocarbon directly in addition to 14), is one of the largest business for the number of carbon atoms (i.e., the value of n), the remainder into the general formula, and molecular weight is of its class. example 3 the total reaction of a single cha
18、in of monool 14 grams to a metal sodium is to generate 0.2 grams of hydrogen. The isomer of this alcohol is the number of isomers A. 6 B. 7 c. 8 d. 9 As one yuan glycol containing only one - OH, per mol, alcohol can turn out 1/2 molh2, generated by the 0.2 grams of H2 infer 14 grams of alcohol shoul
19、d be 0.2 mol, so the molecular weight of 72 grams/mo, the molecular weight of 72, after deducting amount of hydroxy type 17, remaining 55, 14, divided by the biggest dealer of 3, 13, is unreasonable, should take business for 4, to 1, generation into the general formula, and molecular weight should b
20、e 4 carbon olefin base or naphthenic base, combined with the linear, so as to deduce the number of isomers for six. 2. This approach is most suited to qualitatively average method is used to solve the components of the mixture, the possibility for the mixture of ingredients, without considering the
21、content of each component. According to the mixture of various physical quantities (such as density, volume, molecular weight, concentration of amount of substance, such as mass fraction), the definition of the type or subject to conditions, mixture can calculate the average of certain physical quan
22、tities, but the average must be between composition mixture between the 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 4 页，共 22 页 - - - - - - - - - various components of the same physical quantity value, in other words, the mixture of the two components of the physic
23、al quantity must be a larger than average, a smaller than average, to meet the requirements, which can determine the mixture of may. example 4 a mixture of two metallic mono mixtures of 13g, which were added to the full amount of dilute sulfuric acid, emits 11.2 L of gas in the standard condition. T
24、he two metals may be A.Z n and Fe B.A l and Zn C.A l and Mg d. m g and Cu Mixture as A kind of metal, because is adequate dilute sulphuric acid, 13 grams of all metal reaction generated 11.2 L (0.5 mole) are all hydrogen gas, that is, the metal must 26 grams per release 1 mole hydrogen, if all is +
25、2 valence metal, its average atomic weight is 26, comprise A mixture of + 2 valence metal, its atomic weight A greater than 26, A smaller than 26. The generation of selected items, in replacement of the hydrogen reaction, + 2 price with zinc, atomic weight is 65, Fe atomic weight is 56, Mg atomic we
26、ight of 24, but for Al, because in the reaction + 3 valence, to replace the 1 mol hydrogen, as long as 18 grams of Al is enough, can be seen as A + 2 prices its atomic weight is 27 / (3/2) = 18, if have the Na + 1 price to participate in the same reaction, it will be considered A + 2 price when its
27、atomic weight is 23 * 2 = 46, for Cu, because it cant replace the H2, so as the atomic weight of infinity, and got an A in both metal atomic weight are more than 26, C in two kinds of metal atomic weight are less than 26, so A, does not meet the requirements, C B Al the atomic weight of smaller 名师资料
28、总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 5 页，共 22 页 - - - - - - - - - than 26, zinc is greater than 26, D Mg in atomic weight is smaller than 26, Cu atomic weight is greater than 26, B, D for the persons to be the answer. 3. Limit method. Contrary limit method and
29、average method, this method is also suitable for the qualitative or quantitative solution of mixture composition. According to the mixture of various physical quantities (such as density, volume, molar mass, the amount of substance concentration, mass fraction, etc.) the definition of the type or su
30、bject to conditions, the mixture as only one of the component A, namely the quality score or gas volume fraction is 100% (maximum), another component B or corresponding to the mass fraction of gas volume fraction is 0% (minimum), can the value of one of the components of A physical quantity N1, usin
31、g the same method can be calculated only contain B does not contain A mixture of N2 physical quantities of the same values, and the mixture of the physical quantities N is average, must be between the mixture of each component of A, B, between the same quantity of numerical namely N1 N ping N2 can m
32、eet the requirements, which can determine the mixture of may. In 5 four students at the same time analysis of a mixture of KCl and KBr, they all match into aqueous samples from 2.00 grams, add enough HNO3 again after adding suitable amount of AgNO3 solution, after being precipitated completely filte
33、r to get the quality of the precipitation of silver halide dry as shown in the following four options, of which data is reasonable C. a. b. 3.06 g 3.36 g 3.66 g D. 3.96 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 6 页，共 22 页 - - - - - - - - - Subject such as accord
34、ing to the method, usually mixture containing KCl and KBr, can have an infinite variety of ways, the calculated data also has many possibilities, to verify that the data is reasonable, four options must be plug in to see if there is a solution, also is equivalent to four questions to do the math, ve
35、ry much. It takes time to use limit method, all set to 2.00 g for KCl, according to KCl - AgCl, every 74.5 grams of KCl can generate 143.5 grams of AgCl, precipitation is available for (2.00/74.5) * 143.5 = 3.852 grams, to the maximum, the same can be obtained when the mixture all for KBr, every 119
36、 grams of KBr available precipitation 188 grams, so due to precipitation of (2.00/119) * 188 = 3.160 grams, as minimum, is in between numerical accords with a requirement, so can only choose B and C. 4. Estimation. Chemical topic choice involved in the calculation, in particular, to examine is a che
37、mical knowledge, rather than the operational skills, so the calculation of the amount should be smaller, generally do not need to gauge the exact value, can be combined with the conditions in the topic of the operation of the numerical estimates that meet the requirements can be selected. The solubi
38、lity of a certain salt at different temperatures is known. If the salt solution with a mass fraction of 22% is gradually cooled by 500C, the temperature range of the crystal is started Temperature (0C) 0, 10, 20, 30, 40 Solubility (gram / 100g water) 11.5 15.1 19.4 24.4 37.6 名师资料总结 - - -精品资料欢迎下载 - -
39、 - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 7 页，共 22 页 - - - - - - - - - A. 0-100 - C b. 10-200 - C c. 20-300 - C d. 30-400 - C Subject test solution crystallization and solubility of the solute and the relationship between the saturation solution. Solution precipitation crystals, means
40、that the concentration of the solution is beyond the current temperature of the saturated solution concentration, based on the definition of solubility, solubility/(100 grams of water solubility +) * 100% = mass fraction of saturated solution, if the solubility under various temperature numerical su
41、bstitution, compare the saturated solution mass fraction and the size of 22%, can be obtained as a result, but the computation is too big, not accord with the characteristics of the multiple choice questions. Known from the table, the salt solubility increases with temperature rise, Can, in turn, wi
42、ll be 22% of the solution as a temperature of saturated solution, as long as the temperature is lower than the temperature, precipitation will crystal. In solubility/(100 grams of water solubility +) * 100% = 22%, available: solubility * 78 = 100 * 22, namely the solubility = 2200/78, division of tr
43、ouble, using estimates, should be between 25 and 30, the solubility in the 30-400 - c, so choose D. Method of 5. Poor. To be involved in the amount of material in the process of reaction, the concentration, particle number, volume and quality of arithmetic amount change a specific response, using th
44、e numerical dispersion change contribute to fast and accurately and to establish the quantitative 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 8 页，共 22 页 - - - - - - - - - relationship, so as to eliminate interference, rapid problem solving, and even some questions
45、 cannot be resolved due to lack of conditions are solved. example 7 the sum of HA, H + and A - in the 1 liter concentration of the weak acid HA solution is nC A.n * 100% b. (n / 2) * 100% c. D.n (n - 1) * 100% % According to the concept of degree of ionization, needs only the ionized HA of amount of
46、 substance, then the value and the amount of HA (l * C mo/liter = 1 C mo) by its percentage is the degree of ionization HA. Request ionized HA of amount of substance, but according to the HA H+ A -, due to the original weak acid for 1 l * C mo/liter = C mo, degree of ionization of X, the ionization
47、HA of amount of substance for XC , namely the ionization of H + and A - also CXmol, respectively, in the solution of ionization for (C - CX) mol, HA, HA, so H + and A - the sum of amount of substance of (CX) C - CX + + CX moab, namely (CX) C + = the nC, which can be concluded that 1 + X = n, so the
48、value of X is n - 1, take A percentage so choose c. issues involved in the number of particles more easily confused, using the differential method is helpful to quick problem solving: according to the ionization of HA type, each HA ionization generated after A H + and A -, namely particle number inc
49、reases A, particle number C mo has now nC moab, increased (n - 1) * C mo, immediately known have (n - 1) * C mo HA ionization, the degree of ionization of the C/C (n - 1) = n - 1, choose the C items more quickly. 6. Substitution method. All options are a particular material 名师资料总结 - - -精品资料欢迎下载 - -
50、- - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 9 页，共 22 页 - - - - - - - - - one by one, into the original problem to calculate the results correctly, this originally is the most helpless when solution choice methods, but as long as the appropriately combining subject to conditions, the scope