The paper is divided into two parts: multiple choice and non-multiple choice questions. The exam lasts 30 minutes out of 56.
Possible relative atomic masses: h—1 li—7 c—12 n—14 o—16 k—39 cl—35.5 mn—55 fe—56 zn—65
Multiple choice questions
1. Multiple Choice Questions (This question has a total of 7 sub-questions, each sub-question is 6 points. Of the four options given for each sub-question, only one of them meets the requirements of the question. )
7． Chemistry is closely related to science, technology, society and the environment. The error in the following relevant statements is ( )
a． Marine water resources are abundant, and bromine, potassium, magnesium, caustic soda and other substances can be obtained from seawater
b． Baking soda is a leavening agent for making pastries such as steamed buns and bread, and is also a treatment for excessive stomach acid
c． Benzene is an important chemical raw material that can be obtained from coal distillation
d． So2 and no2 in the atmosphere are acidic oxides, which are closely related to the formation of acid rain
Analysis From seawater, bromine, potassium, magnesium and caustic soda and other substances can be obtained, item a is correct; baking soda is nahco3, can be used as a leavening agent for pastries, can also treat excessive gastric acid, b is correct; benzene can be obtained from coal distillation, c is correct; so2, no2 is closely related to the formation of acid rain, but no2 is not an acid oxide, item d is wrong.
Answer d
8． Let na be the value of the Avogadro constant, and the following description is correct ( )
a． The neutron number contained in 2.0 g heq \o\al(18,2)o and 2.0 g d2o are na
b． Hydrocarbons with the formula cnh2n of 14 g contain a carbon-carbon double bond number of na/n
c． The number of hydrogen atoms in an aqueous solution of 100 g mass 17% is na
d． At room temperature, the molecular number of so2 generated by putting 56 g of iron flakes into sufficient amounts of concentrated sulfuric acid is 1.5na
Parse a. 2.0 g heq \o\al(18,2)o contains neutron number eq \f(2.0 g, 20 g·mol-1) (18-8)na=na and 2.0 g d2o contains the neutron number eq \f(2.0 g, 20 g·mol-1)[(2-1)×2+(16-8)]na=na, so a is correct; b.14 g The amount of olefins with the formula cnh2n is eq \f(14 g,14n g·mol-1) =eq \f(1,n)mol, containing carbon-carbon double bonds of na/n, but the hydrocarbon with the molecular formula cnh2n is not necessarily an olefin, if it is a naphthene, there is no carbon-carbon double bond, so b is wrong; c. water and hydrogen peroxide contain hydrogen atoms, 100 g mass fraction of 17% h2o2 aqueous solution in the number of hydrogen atoms than na, so c error; d. at room temperature, iron passivation in concentrated sulfuric acid, will be 56 g Iron tablets are put into a sufficient amount of concentrated sulfuric acid to generate so2 molecules less than 1.5na, so d is wrong; therefore, a is chosen.
Answer a
9． W, x, y, and z are known to be short-period elements, and the atomic number increases sequentially. w and z are of the same primary family, x, y, z are of the same period, where only x is a metallic element. The following statement must be correct ( )
a． Atomic radius: x>y> z>w
b． The acidicity of the oxygen-containing acid of w is stronger than that of the oxygen-containing acid of z
c． The stability of the gaseous hydride of w is less than the stability of the gaseous hydride of y
d． If the difference in atomic number between w and x is 5, the chemical formula for forming the compound is x3w2
Analysis According to the information provided, it can be inferred that w is in the previous cycle of x, y, z, so the atomic radius: x>y>z>w, a term is correct; w, z is the same main family, non-metallicity: w>z, but the acidity of the oxygen-containing acid of w is not necessarily stronger than the acidity of the oxygen-containing acid of z, b term is wrong; w, z is the same main family, non-metallicity: w>z, y, z is the same period, non-metallicity: y<z, so non-metallicity: w>y, The stability of the gaseous hydride of w is greater than the stability of the gaseous hydride of y, and the term c is wrong; if the atomic number difference between w and x is 5. And only x is a metal element among the four elements, then w is n when x is mg, w is o when x is al, and the chemical formula of the compound formed by the two is mg3n2 or al2o3, and the d term is wrong.
10． (2018 · Hebei Five Schools Joint Examination) The structure of 5 carbon atoms on the main chain of carbon and hydrogen in the molecule with a mass ratio of 21:4 has a common structure of 5 carbon atoms ( )
a． 3 types of b. 4 kinds
c． 5 types of d. 6 types
Analysis The mass ratio of carbon to hydrogen is 21:4, that is, the ratio of the amount of matter is eq \f(21,12):eq \f(4,1), simplified to 7:16, the molecular formula is c7h16, belongs to the alkanes, and there are 5 carbon atoms on the backbone:
(The other methyl group can move in the (1)(2)(3) position, there are 3 types),
、
, there are 5 types, so option c is correct.
Answer C
11． (2018 · Shandong Jining high school end of the third period) high iron (such as na2feo4) has been widely used in water treatment, with iron-based materials as the anode, in a high concentration of strong alkali solution using electrolysis can be prepared by way of high-lying iron, the device as shown in the figure. The following statement is incorrect is ( )
a． a is the anode, and the electrode reaction formula is fe-6e-+8oh-===feoeq \o\al(2-,4)+4h2o
b． In order to prevent the diffusion of the methenic acid is reduced, the ion exchange membrane is a cation exchange membrane
c． The cations in the solution during electrolysis move towards the a pole
d． The cause of a small amount of gas on the iron electrode may be 4oh--4e-===o2↑+2h2o
Analysis Of the iron-based material is an anode, high-iron can be prepared by electrolysis in a high concentration of strong alkali solution, so iron is an anode, and the electrode reaction formula is fe-6e-+8oh-===feoeq \o\al(2-,4)+4h2o, so a is correct; the cation exchange membrane can prevent feoeq \o\al(2-,4) from entering the cathode region, so b is correct; the cations in the solution move to the cathode during the electrolysis process, so the cations move to the b pole, so c is wrong Oxidation occurs on the iron electrode, so the gas generated may be oxygen, and the electrode reaction formula is 4oh--4e-===o2↑+2h2o, so d is correct.
12． At 298 k, there is a relationship between the concentrations of hxoh and hcoo- in the mixed solution of formic acid (hcooh) and sodium format: c(hcoo-)+ c(hcooh) = 0.100 mol·l-1, and the relationship between the concentration of carbon-containing particles and ph is shown in the figure. The following statement is correct ( )
a． At 298 k, add distilled water to dilute the p-spot solution, and n(h+)·n(oh-) remains unchanged in the solution
b.0.1 mol·l-1 hcoona溶液中有c(hcoo-)+c(oh-)=c(h+)+c(hcooh)
c． At 298 k, the ionization constant of hcooh is ka = 1.0 × 10-3.75
d． After mixing 0.1 mol·l-1 hcoona solution and 0.1 mol·l-1 hcooh solution in equal volumes, the pH of the solution is 3.75 (the change in the volume of the solution after mixing is negligible)
Analysis According to the p point, ph = 3.75, c (hcoo-) = c (hcooh), the ionization constant of hcooh can be obtained ka = eq \ f(c(hcoo-)·c (h+), c(hcooh)) = c(h+) = 1.0×10-3.75, and the c term is correct. At 298 k, the distilled water was added to dilute the p-spot solution, and the solution c(h+)·c(oh-)=kw remained unchanged, but because the volume of the solution increased, n(h+)·n(oh-) increased, a term was wrong; 0.1 mol·l-1 hcoona solution, according to charge conservation, c(hcoo-)+c(oh-)=c(h+)+c(na+), because hydrolysis was weak, hcoona solution c(na+)) >c(hcooh), Therefore, c(hcoo-)+c(oh-) >c(h+)+c(hcooh), item b is wrong; 0.1 mol·l-1 hcoona solution and 0.1 mol·l-1 hcooh solution are mixed in volumes, and the degree of ionization of hcooh and the degree of hydrolysis of hcooh are different, so c(hcoo-) ≠c(hcooh), so the ph≠ of the solution is 3.75, and the d term is wrong.
13． Marine animal sea squirts contain a variety of secondary metabolites with novel structure, which is one of the important sources of marine anti-tumor active substances. The process for extracting natural products with antitumor activity from sea squirts is as follows:
If the process is simulated in a laboratory, the following statement about the steps in the process is incorrect ( )
Analysis Step (1) Separation to obtain filtrate and insoluble substances, obviously separating solid-liquid mixtures, its operation is filtration, requiring funnels, beakers and other instruments, a correct; step (2) Separation to obtain an organic layer solution and an aqueous layer solution, obviously separating the incompatible liquid mixture, its operation is to dispense liquids, need to dispense funnels, beakers and other instruments, b items are correct; step (3) is to obtain solids from an aqueous solution, its operation is evaporation, need to evaporate and other instruments, and the crucible is an instrument for burning solids, and item c is wrong Step (4) is to obtain toluene from the solution of the organic layer, which is obviously a liquid mixture with different boiling points, and its operation is distillation, which requires distillation of flasks, alcohol lamps, thermometers, condensing tubes and other instruments, and d is correct.
Non-multiple-choice questions
26． (14 minutes) (2018· Beijing No. 4 Middle School Phase III) A chemical team simulated the preparation of high-purity manganese carbonate in the laboratory (the main component mno2, compounds with impurities of iron and copper, etc.), and the process is as follows (some operations and conditions are slightly):
(1) Slowly into the flask (see above) into the excess mixture of gas for "manganese immersion" operation, the main reaction principle is:
so2+h2o===h2so3 mno2+h2so3===mnso4+h2o (after iron leaching, excess so2 will reduce fe3+ to fe2+)
(2) Add a certain amount of pure mno2 powder to the flask after the end of the "manganese immersion".
(3) Then adjust the ph to about 3.5 with na2co3 solution and filter.
(4) Adjust the filtrate ph to 6.5 ~ 7.2, add nh4hco3, there is a light red precipitation generated, filtering, washing, drying, to obtain high purity manganese carbonate.
(1) "Manganese immersion" reaction often has the formation of by-product mns2o6, the influence of temperature on the "manganese immersion" reaction is as follows, in order to reduce the generation of mns2o6, the appropriate temperature of "impregnated manganese" is __________
(2) Consult Table 1, (3) The main component of precipitation when the pH is 3.5 is ________________ (2) The main role of adding a certain amount of pure mno2 powder is ___
The ion equation for the corresponding reaction is ____
___________________________________________________________。
Table 1: Ph for the generation of the corresponding hydroxide
(3) (3) The filtrate obtained in (3) contains CU2+, which can be added to remove CU2+ by adding excess insoluble electrolyte MNS, and filtered to obtain pure MNSO4. Explain the role of adding mns with the principle of balanced movement ___
______________________________________________________。
(4) (4) The ion equation that reacts after the addition of nh4hco3 is ________
_____________________________________________________________。
Analysis (1) From the figure to get more than 150 °C, the by-product mns2o6 will not be generated, and the leaching rate of manganese is close to 100%, so choose 150 °C (or more than 150 °C).
(2) According to the data in the table, when ph = 3.5, the result is mainly iron hydroxide precipitation, so the purpose of adding mno2 in the previous step is to oxidize ferrous ions (the title says: after iron leaching, excess so2 will reduce fe3+ to fe2+). The reaction is: mno2+2fe2++4h+===mn2++2fe3++2h2o, considering that the topic believes that the ferrous ions are obtained because of excessive sulfur dioxide, so the added manganese dioxide can also oxidize the excess sulfur dioxide, the reaction is: mno2 +so2===mn2++soeq \o\al(2-,4).
(3) The sedimentation dissolution equilibrium of mns into the water is: mns(s)
mn2+ (aq)+s2-(aq), the copper ions in the solution will bind sulfur ions to form a CUS precipitate, so that the reaction balance is constantly moving to the right, the MNS precipitate is converted into a CUS precipitate, and then the purpose of removing CU2+ can be achieved by filtration. Of course, the premise of the above conversion is that the solubility of cus is smaller.
(4) Ammonium bicarbonate is added to form a manganese carbonate precipitate, then mn2+ binds the carbonate ionized by bicarbonate to make the ionization equilibrium: hco
h++coeq \o\al(2-,3) moves forward, and the ionized h+ is combined with hcoeq \o\al(-,3) to obtain co2, the equation is: mn2++2hcoeq \o\al(-,3)==mnco3↓+co2↑+h2o.
Answer (1) 150 °C (or 150 °C or more)
(2) fe(oh)3 (2) Oxidize fe2+ to fe3+, remove excess so2 oxidation mno2+2fe2++4h+===mn2++2fe3++2h2o, mno2+so2===mn2++soeq \o\al(2-,4)
(3)mns(s)
mn2+(aq)+s2-(aq)-cu2+(aq)+s2-(so)
Cus(s) is more difficult to dissolve than mns, promotes mns to dissolve continuously, shifts to the right in equilibrium, and removes CU2+
(4)mn2++2hcoeq \o\al(-,3)===mnco3↓+co2↑+h2o