高中化学计算题

1、高中化学运算题 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 equ

2、ations 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 temperature of mixed gas

3、 pressure increase 5% than the original, X concentration decreases to a third, A. 3 B. 4 C. 5 D. 6 is the value of n in the reaction equation Solution one: grasp the concentration of X decreases by 1/3, the coefficient ratio of the combination of chemical equation is equal to volume ratio, can list

4、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 to 5 over 4L 0

5、plus 1/2 is equal to 1 over 2L 0 plus n over 4 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 / 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、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 the original mixed gas pressure increased by 5%, the volume of the mixed gas increases the 2 + 3 * 5% = 0.25 L, according to the equat

7、ion, 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 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

8、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 pressure increase, show positive reaction is certainly increase the reaction volume, the sum of the reaction equation coefficient of X and Y will

9、 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 answer. Subject test is about the content of the chemical equilibrium. A solution is to follow the law of chemical equilibrium, as according to

10、 the specification, although can certainly work out the correct answer, but did not hold choice, dont ask, dont 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 diff

11、erence method, with the volume of n variables delta, 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

12、mobile and volume change, not half a minute can be 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

13、, to achieve fast and accurate the effect of problem 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 mo

14、lecular weight of 14 n + 2, the corresponding paraffin 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 al

15、kadiene general formula CnH2n - 2, molecular weight 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 quantity, difference divided by 14 hydrocarbon

16、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 chain of monool 14 grams to a metal sodium is to generate 0.2 grams o

17、f 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 should be 0.2 mol, so the molecular weight of 72 grams/mo, the molecula

18、r 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 be 4 carbon olefin base or naphthenic base, combined with the linea

19、r, 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 content of each component. According to the mixture of various phy

20、sical 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 quantities, but the average must be between composition mixture between

21、the various components of the same physical quantity value, in other words, the mixture of the two components of the physical 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 mix

22、tures of 13g, which were added to the full amount of dilute sulfuric acid, emits 11.2 L of gas in the standard condition. The 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 g

23、enerated 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 + 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

24、, in replacement of the hydrogen reaction, + 2 price with zinc, atomic weight is 65, Fe atomic weight is 56, Mg atomic weight 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 2

25、7 / 3/2 = 18, if have the Na + 1 price to participate in the same reaction, it will be considered A + 2 price when its 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

26、of metal atomic weight are less than 26, so A, does not meet the requirements, C B Al the atomic weight of smaller 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 m

27、ethod and 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 ty

28、pe or subject 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,

29、using 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 N1 N ping N2 can meet the requirem

30、ents, the mixture of may. of numerical namely which can determine 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 f

31、ilter 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 Subject such as according to the method, usually mixture containing KCl and KBr, can have an infinite variety of ways, the calc

32、ulated 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, very much. It takes time to use limit method, all set to 2.00 g for KCl, according to KCl - AgCl, every 74.

33、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 grams of KBr available precipitation 188 grams, so due to precipitation of 2.00/119 * 188 = 3.160 grams, a

34、s 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 chemical knowledge, rather than the operational skills, so the calculation of the amount should be smaller, gene

35、rally 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 solubility of a certain salt at different temperatures is known. If the salt solution with a mass fraction of 22% i

36、s 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 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 relatio

37、nship between the saturation solution. Solution precipitation crystals, means 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

38、saturated solution, if the solubility under various temperature numerical substitution, 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

39、table, the salt solubility increases with temperature rise, Can, in turn, will 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: s

40、olubility * 78 = 100 * 22, namely the solubility = 2200/78, division of trouble, 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, vo

41、lume and quality of arithmetic amount change a specific response, using the numerical dispersion change contribute to fast and accurately and to establish the quantitative relationship, so as to eliminate interference, rapid problem solving, and even some questions cannot be resolved due to lack of

42、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 substance, then the value and the amou

43、nt 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 HA of amount of substance for XC , namely

44、 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 value of X is n - 1, take A percentage so choos

45、e 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 increases A, particle number C mo has now nC moab,

46、 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 one by one, into the original problem to calculate the results correctly, this

47、 originally is the most helpless when solution choice methods, but as long as the appropriately combining subject to conditions, the scope of the narrow to plug in can also use the method of substitution problem quickly. 8 some alkane 11 grams is completely burnt, and oxygen 28L is required for the

48、standard condition. The molecular formula of this alkane is A.C 5 h12 biggest 4 h10 Arthur c. 3 h8, dc 2 h6 Alkanes is because it is over, composition CnH2n + 2, the molecular weight of 14 n + 2, namely every 14 n + 2 grams hydrocarbon combustion generated n mo and n + 1 the H2O and CO2 would consum

49、ption to n + n + 1 / 2, or 3 n / 2 + 1/2 the O2, existing 11 grams of alkanes, oxygen is 28/22.4 = 5/4 mo, its ratio is another shall subscribe, four options in the value of n in 14 n + 2 : 3 n / 2 + 1/2, does not need solving equations can quickly learn that n = 3 for the persons to be the answer.

50、7. Relation method. For a multistep reactions, according to the relationship of various mainly chemical equation, conservation, etc., listed corresponding relation, quickly in the requirements of the material and the number of cases given to establish a quantitative relationship between the amount o

51、f material, which involves the intermediate process from a large number of operations, not only saving the operation time, also to avoid the computation error of the calculation results, the influence of is one of the most frequently used method. 9 a certain amount of iron powder and 9 grams of sulp

52、hur powder are mixed and heated. After the reaction, excess hydrochloric acid will be added, and the generated gas will be completely burnt. The amount of iron powder will be collected for 9 grams of water 14 g b. a. 42 g c. d. 56 g 28 g Because in the title indicates the amount of iron powder, iron

53、 powder, therefore, may be excessive, also may be less than, after reaction with sulfur powder, add excessive hydrochloric acid gas generated when there are many possible: or only H2S iron all into FeS2, or there are both H2S and H2 iron besides generating FeS2 have left, so only by sulfur powder an

54、d generate the quality of the water quality, not easy to establish equation. According to the quantitative relationship of each steps listed equation 1 the Fe - FeS iron of conservation - H2S sulfur of conservation - H2O hydrogen conservation, 2 the Fe - H2 chemical equation - H2O hydrogen conservat

55、ion, thus learned that regardless of iron which involved in the reaction, each one iron eventually generates a H2O, so fast it is concluded that the quantity of the material of iron is the amount of substance of water, basic has nothing to do with sulfur, so should be as 9/18 = 0.5 mo iron, namely,

56、28 grams. 8. Comparison. An organic molecular formula is known, according to the requirement of the topic related to calculate the amount of isomers, for example, the structure of reactants or products, reaction equation coefficient ratio and so on, often should use structure comparison method, the

57、key is to understanding of the structure characteristics of organic matter thoroughly, the relevant functional location, nature of the master, generation into the corresponding conditions are determined. CH3 The molecular formula is C12H12 hydrocarbon, the structure is, if there are 9 types of dibro

58、mide on the naphthalene ring CH3 In isomers, the number of isomers of tetrabromodes in naphthalene rings A. 9 kinds of B. 10 kinds of c. 11 species d. 12 species Ontology is pray with naphthalene ring four bromine content in the number of isomers, does not need to consider functional group heterogen

59、eous and heterogeneous carbon chain, but for the position of functional groups heterogeneous, such as according to the usual practice, the four bromine atoms individually into the hydrogen of naphthalene ring position, then count out the number of isomers, but because of the quantity, structure is v

60、ery difficult, very easy wrong number, number. Seize title - dibromo generation conditions given nine, analysis for organic Mao solid di. It is not hard to see, naphthalene rings only six hydrogen atoms can be replaced by bromine, that is to say, every place of four hydrogen atoms, is sure the rest