If you are preparing for the national competition, the topic of the preliminary competition is hardly a problem.
The preliminaries of all provinces in China are the most difficult, close to the national preliminaries. Other provinces prefer the college entrance examination, and people who have studied the competition will not be a problem.
I am the first in Sichuan, but for that roll of noodles, I would be the third in Sichuan.
Respondent: Couror 888-Manager Level 4 6-20 14:52
Why did you post last year's national competition? This is a provincial competition and can't be put on the map.
/Soft/chem/GS shx/200805/ 1053 . html
The problems in Fujian are much more difficult than those in Shaanxi.
In 2008, Fujian senior high school chemistry competition examination questions.
(May 8, 2008, 8: 30 to 1 1:30 * *)3 hours.
Total score of title 1234567890
score
1.( 10) Fill in the blanks.
1. In the textbook, aluminate surface is described as a strong acid, while silica gel is a very weak acid. The reason why Al _ 3+ enhances the lattice acidity of SiO2 _ 2 _ 2.
2. It is known that boron nitride has two forms, the common one is a slippery gray substance, and the second one is synthesized under high pressure, which is one of the hardest substances known. Below 300℃, both of them remain solid, and these two configurations of boron nitride are similar to and respectively.
3. It is known that yellow solid is a simple compound of main group elements, which is insoluble in hot water, but soluble in hot dilute hydrochloric acid to form an orange-red solution. When the solution is cooled, a white crystalline precipitate is precipitated. When the solution is heated, the white precipitate is dissolved again. This compound is.
4. Acidify a solution to obtain S and H2SO3. What sulfur-containing compounds may be contained in the original solution?
.
(10) The preparation of celestite by green chemical method effectively overcomes the shortcomings of traditional production methods, such as low conversion rate, serious pollution of three wastes, high energy consumption and complicated production. The chemical purity of the prepared strontium carbonate product reaches 99. And high-quality ammonium sulfate can be prepared, thus greatly reducing the production cost. Celestite, produced in a certain place in eastern Hubei, China, has an average mass fraction of 80% and contains a small amount of impurities such as Ba, Ca, Mg, SiO _ 2, Fe2O3 and FeO. Celestite can react with ammonium bicarbonate solution under certain conditions to produce crude strontium carbonate products containing some impurities. In the above reaction process, adding appropriate amount of NH4Cl as catalyst can obviously speed up the reaction. Dissolve the precipitate in (NH4)2SO4 solution with HCl to prepare strontium salt solution containing soluble SrCl2 and some impurities, and then purify it to remove impurities such as Ba2+ and Ca2+ to prepare high-purity SrCl2 solution. At this time, pure ammonium bicarbonate solution is added, and pure strontium carbonate precipitate is generated after reaction. Precipitated strontium carbonate is dehydrated, washed, dried and crushed to obtain high-purity strontium carbonate products.
(It is known that ksp (srso4) = 3. 2× 10-7，ksp (srco3) = 1。 1× 10- 10.)
1. Try to explain the principle of green reaction simply and write the equation of main reaction.
2. What is the reason for adding appropriate amount of NH4Cl as catalyst?
3.( 10) Carbon compounds are very common. Answer the following questions:
1. Thick limestone deposits are often found in shallow sea areas under warm climate conditions, but rarely seen in deep sea areas. Why?
2. The formation of large ∏ bonds in CO2, and draw the structural formula of CO2.
3. It is known that the Ka 1 of H2CO3 found in the manual is 4.45× 10-7, which is the apparent dissociation constant of H2CO3. In fact, only 0. 166% of CO2 in the solution is converted into H2CO3, and the real first-step dissociation constant of H2CO3 is calculated.
Four. (10)
1. A middle school student takes pure Na2SO3? 7H2O 50.00 g, heated above 600℃ to constant weight, analysis and calculation show that the sample quality after constant weight is equivalent to the calculated value of anhydrous sodium sulfite, and the composition of each element is also in line with the calculated value. However, when it was dissolved in water, it was found that the alkalinity of the solution was much higher than expected. What do you think is the reason? (Please use chemical equation);
2. How to verify your ideas with simple experiments? (Please use chemical equation);
3. Sulfite and sulfate are commonly used in papermaking and pulping. Sulfite pulp is white and slightly yellowed. Why?
4. Pulp should be bleached with bleaching powder solution, and its temperature should be controlled at 35-38℃ and not more than 40℃. Why?
5. If the pulp whiteness has been reached, but there is still chlorine Shang Gao residue, bleaching must be terminated. What compounds are usually added? And write the chemical reaction.
5.( 10) Try to write the Lewis structure of COCl2; The geometric configuration of COCl2 _ 2 _ 2 is predicted by the theory of (VSEPR) valence electron pair repulsion. Explain the bonding process of molecules by (VB) valence bond method, discuss the properties of bonds in molecules, and predict and explain the sizes of ∠ClCO and ∠ClCCl.
6.( 10) In order to solve the energy crisis, it was proposed to use CaCO3 to make C2H2 as fuel. The specific reaction is:
( 1)
(2)
(3)
1. How much C(s) is needed to prepare 1 mol C2H2(g), and how much heat can be released by burning these carbons?
2. If 1 mol C2H2 can give off 1298 kJ of heat after complete combustion, is it economical to use C2H2(g) as fuel?
3. Is the entropy change of the above three reactions greater than zero or less than zero? Why do reactions (1) and (2) need high temperature, while reaction (3) only needs normal temperature?
It is known that the δ FH (298k)/kj mol- 1 of related substances is:
Calcium chloride (s):-60, carbon dioxide (g):-393, H2O(l):-285, C2H2(g): 227,
Cao (s):-635, calcium carbonate (s):- 1207, carbon monoxide (g):-11
Seven. (10) An organic weak acid HA 1.000g was dissolved in appropriate amount of water, and the end point was determined by instrumental method. When titrating to the stoichiometric point with 0. 1 100 mol/L NaOH, it consumes 24.60ml. When adding NaOH solution 165438,
8.(5 points) The system in which unsaturated bonds (double bonds or triple bonds) and single bonds are alternately arranged in organic molecules is called the p-p*** yoke system, and the wavelength of light absorbed by this * * * yoke system is in the ultraviolet region, which can be used for qualitative or quantitative analysis. Compound A is a drug that acts on the central nervous system. When A is heated in the presence of sulfuric acid, two compounds, B and C, are obtained. They are isomers, both of which have absorption in the ultraviolet region, and their molecular formulas are both C7H7Cl.
1. Please draw the structural formulas of B and C and point out which isomers B and C belong to;
2. Compound A is named by systematic nomenclature;
3. Is there any optical isomer (enantiomer) in Compound A?
9.( 12) Benzene is an important organic chemical raw material, from which various organic compounds can be prepared. The reactions for preparing some organic compounds from benzene are listed below:
1. Write the structural formula of the main organic product compound A-J in the above reaction;
2. Among the above reactions, what reaction types (substitution reaction, addition reaction, elimination reaction and oxidation reaction) do C→D, C→E, C→F, E→G, E→H and H→I belong to respectively?
3. In the process of preparing compound A by the above method in the laboratory, hydrogen chloride gas continuously escapes. What should be done to avoid air pollution? Please write down the names of the experimental devices and reagents needed to treat hydrogen chloride.
X.( 13) the synthetic method of the intermediate compound 5 of the antihypertensive drug Viprostol is as follows:
1. Please write down the reagents needed for spaces A-C in the above synthetic route;
2. The compound 1 can be synthesized from ethylene and propylene by the following steps. Please write down the structural formula of compound 6-8 and the reagents needed for D-G space. ..
3. The compound 1 can also be transformed from 1- pentyne by the following two methods: (1) 1- pentyne reacts with reagent J and then is treated with reagent K to obtain the compound1; (2) 1- pentyne reacts with reagent l to obtain compound 9, and then it is treated with dilute acid solution to obtain compound 1. Please write down the reagents represented by J, K and L, and name compound 9 by systematic nomenclature.
Scratch paper
Answers and grading standards of the preliminary test questions in the chemistry competition of senior high schools in Fujian Province in 2008
1. Fill in the blanks (10 score)
1.( 1)Al3+ replaces Si4+ with extra protons to achieve charge balance (2 points).
2. Graphite and diamonds. (2 points)
3. Lead chromate. (2 points)
4.(4 points) There are several possibilities:
(2 points) (1) contains+2h+= H2SO4 3+s.
(1) (2) contains S2- and +2S2-+6H+ = 3S+3H2O.
n()/n(S2-)& gt； 1/2
(1) (3) contains S2- and n ()/n (S2-) >; 1/2
Two. (10)
1. According to the principle of solubility product (2 points), the reaction can occur, and the reaction equation is
Srso4+2nh4CO3 = srco3 ↓+(NH4) 2SO4+CO2 =+H2O, (2 points)
Srco3+2hcl = srcl2+H2O+CO2 = (2 points)
Srcl2+2nh4CO3 = srco3 ↓+2nh4cl+H2O+CO2 = (2 points)
2. The reaction equation under the action of catalyst is: (2 points)
SrCO3 + 2NH4Cl = SrCl2 + (NH4 ) 2CO3
src L2+2 NH 4h co 3 = src O3↓+2 NH4Cl+CO2 ↑+ H2O
Three. (10)
1. The formation of limestone is the result of CaCO3 deposition, and a certain amount of CO2 is dissolved in seawater, so there is the following balance among CaCO3, CO2 and H2O:
Calcium carbonate+carbon dioxide +H2O calcium bicarbonate (AQ) (1)
The solubility of CO2 in seawater decreases with the increase of temperature and increases with the increase of pressure. In shallow sea area, the bottom pressure of seawater is low and the water temperature is high, so the CO2 concentration is low. According to the principle of equilibrium motion, the above equilibrium moves to the direction of producing CaCO3, so there is more CaCO3 precipitation in shallow water. (2 points)
The situation in the deep sea is just the opposite, so there is little CaCO3 deposited at the bottom of the deep sea. ( 1)
2. Two ∏34( 1 min)
( 1)
3. Assume that the real first-step dissociation constant of carbonic acid is Ka 1'
Ka1= [h+] [HCO3-]/[H2CO3+CO2] = 4.45×10-7 (2 points).
According to the meaning of the question: [CO2]/[H2CO3] = 60 1.
The real ka1'= [h+] [hco3-]/[h2co3] = 602× 4.45×10-7.
= 2.68× 10-4 (2 points)
Four. (10)
1.4Na2SO4 = 3Na2SO4+Na2S, because Na2S is more alkaline. (2 points)
2. Dissolve the heated product in water, and check whether the solution contains SO4 2-and S2-
Detection of SO42- ion: SO42-+Ba2+= baso4 ↓ insoluble in hydrochloric acid. ( 1)
Detection of S2 ion: (1)
Methods 1: Adding precipitant: S2-+Pb2+= PBS (black) Other characteristic precipitants can also be scored.
Method 2: Lead acetate test paper turns black.
Method 3: Add hydrochloric acid S2-+2h+= H2S ↓ (you can smell the special smell of H2S).
3. Because sulfite pulp uses the mixed solution of sulfurous acid and acidic sulfite as cooking agent, SO2 will be produced in the cooking process, which will combine with organic pigments to form colorless compounds. (2 points)
4.3ca(OCL)2 = ca(clo 3)2+2 calcl 2( 1)
The disproportionation rate of ClO- is related to temperature. The higher the temperature, the faster the disproportionation. When the temperature is high, the bleaching solution decomposes and the chlorine consumption increases, so the temperature cannot be high. ( 1)
5. Add Na2S2O3 (1 min).
2ca (OCL) 2+Na2S2O3+H2O = 2calcium chloride +Na2SO4+H2SO4 (1)
Verb (abbreviation of verb) (10 points)
1、
Logarithm of bonding electrons: 4 (1 point)
Unbonded electron logarithm: 8 (1 point)
2. Triangle (1 point)
3.VB method
(1) central atom c hybridizes with SP2 (1)
(2) Forming σ bond (1 min)
(3) forming π bond (1 min)
c 2s 22p 2(SP2) 1(SP2) 1(SP2) 1(2Pz) 1
Cl 3S23Py23Pz23Px 1
Cl 3S23Py23Pz2 3Px 1
O 2S22Py2 2Px 1 2Pz 1
4. Adhesion: 3 σ (1 min)
1 π( 1 point)
5. due to electron repulsion, c = o > c-cl (1)
6. forecast ∠ clco > ∠ clccl, that is ∠ clco >12oo, and ∠ clccl <12oo (1point).
Six (10)
When printing this topic, the following known conditions will be ignored:
It is known that the δ FH (298k)/kj mol- 1 of related substances is:
Calcium chloride (s):-60, carbon dioxide (g):-393, H2O(l):-285, C2H2(g): 227,
Cao (s):-635, calcium carbonate (s):- 1207, carbon monoxide (g):-11
During the examination, the areas that failed to notify each examination room urgently were not scored, and the total score of the examination paper was 90 points. This change does not affect the number of cities entering the summer camp rematch.
1. Preparation of 1 mol of C2H2(g) requires 3 mol of C(s)(36 g) (1 min).
Reaction 3C (g) +3O2 (g) →3CO2 (g) (1 min)
δ RH (298k，(4))=- 1 179 kj( 1)
2. Although burning 1 mol C2H2(g) releases more heat than burning 3 mol C(s). But the reaction
( 1) δ RH (298k，( 1)) = 179KJ ( 1)。
(2) δ RH (298k，(2)) = 464kJ ( 1)
(3) δ RH (298k，(3)) =-63kJ ( 1)
That is to say, the production of 1 mol acetylene still needs (179+464)kJ of heat, with a heat loss of 63kJ, which cannot be fully utilized. Therefore, it is not cost-effective if it is simply used as fuel. ( 1)
3. The entropy changes of the three reactions are all greater than zero. ( 1)
Reaction (1) and (2) are endothermic reactions, so high temperature is beneficial to the reaction. ( 1)
Reaction (3) is exothermic, so normal temperature is beneficial to the reaction. ( 1)
Seven. (10)
The total amount of HA is: 0. 1 100? 24.60=2.706 (million moles)
When NaOH solution 1 1.00ml is added dropwise, it is known that the pH is 4.80, that is, it is in the solution.
The amount of A- is 0. 1 100? 11.00 =1.210 (millions of moles) (2 points)
So the amount of HA is: 0. 1 100? 24.60-0. 1 100? 11.00 =1.496 (millions of moles) (2 points)
4.80=pKa+ (4 points)
\ pKa=4.80+0.09=4.89 (2 points)
Eight. (5 points)
4. 3 points for this little question
B and C belong to Z, E isomer (or A: cis-trans isomer) (1 min).
The score of this short question is 1, and no one who is different from this answer will be graded. (E)-3- ethyl-1- chloro-1- pentene -4- alkyne -3- ol
6. The score of this small question is 1. Yes or Answer: Optical isomers (enantiomers) exist.
Nine. (12)
1. This small question has 8 points, including A ~ F 1, G ~ J 0.5(2 points).
2. 3 points for this small question, and 0.5 point for each reaction.
C→D oxidation, C→E substitution, C→F substitution, E→G substitution, E→H elimination, H→I addition.
3. This little problem is 1.
Treat hydrogen chloride with gas absorption device (0.5 points). (Just answer "gas absorption device" and score);
Required reagents: sodium hydroxide (or potassium hydroxide, molecular formula: NaOH and KOH) (0.5 points).
X.( 13 points)
4. 3 points for this small question, 1 points A to C.
5. There are 7 points in this small question, in which the structural formula and space D-G of compounds 6-8 have 1 minute respectively.
6. 3 points for this little question