Junior High School Chemistry Competition Questions. Urgent!! A total of three questions, just for i

pwd=1234 Extraction code: 1234 Introduction: High-quality materials for junior high school chemistry**, suitable for teachers at all stages, daily tutoring for students, sprint for the high school entrance examination, and skill improvement learning.

The landlord is also competing today.。。。 The first one is actually a math problem. Just set two unknowns!

1.An equation based on the valency comes out.

The five-valent ratio is 1:4 to the six-valent

2 should be the ratio according to the equation n2 + 3 h2 = 2 nh3 mass ratio 14:3

3. The same mass Fe and Zn react with acid to consume different amounts of acid, and it may be that according to the total mass and the mass of the acid, the mass of Zn and Fe is introduced, and then the thickness of the acid is introduced.

Let's make a reference.

1. Na, NaOH, Na2CO3, NaHCO3, etc. (Na2O also works, but NaH can't).

2. C, because AgNO3 and BaCl2 can be soluble in dilute hydrochloric acid and do not react with dilute hydrochloric acid, while K2CO3 can react with dilute hydrochloric acid to produce a large number of bubbles and obtain a clear solution.

Therefore, the first package of powder can be dissolved in dilute hydrochloric acid to obtain a clear solution, and no other phenomena are seen, indicating that it does not contain K2CO3; The second package of white powder is also soluble in hydrochloric acid, which produces a large number of bubbles, and a clear solution is obtained after the reaction, indicating that it must contain K2CO3. Only option C does.

3. A, because the ratio of the number of Na+ to SO42- in each Na2SO4 is 2:1, the ratio of the total number of Na+ to SO42- is 2:1.

The relative mass of H2O is 18, and the relative mass of Na2SO4 is 142, and according to the mass fraction of Na2SO4, it can be concluded that the content of water molecules in the solution is 75 times that of Na2SO4 particles, so the ratio of the number of the three particles is 2:1

75, choose A.

This is a simple question, one is that there is not enough CO2, and only CaCO3 is generated

co2 --caco3

x x = There is also an excess of CO2, which generates part of Ca(HCO3)2 and CaCO3

So first CO2 reacts with all Ca(OH)2, and then excess CO2 reacts with CaCO3.

co2---ca（oh）2 --caco3

x y x= y=1g now there is a precipitate for, then dissolved so.

co2---caco3 --2ca(hco3)2

z z= so total consumption.

1) because new impurities NO3 2- will be introduced;

2) Remove a small amount of the solution after step 1 into a clean test tube, drop a few drops of H2SO4 dilute solution, if there is a white precipitate generated, that is, BaCl2 has been excessive;

3) removal of Ca2+ and excess Ba2+;

4) Beakers, glass rods, evaporation dishes, funnels, iron frames, alcohol lamps, rubber head droppers, filter paper;

5) Filtration operation: the filter paper should be close to the funnel wall; Place the glass rod against one side of the three-layer filter paper; The filtrate is slowly drained with a glass rod; The bottom of the funnel is against the wall of the beaker.

Evaporation operation: the evaporation solution should not be more than 2 3 of the volume of the evaporation dish, stir with a glass rod during evaporation to prevent boiling; When a crystalline film appears, immediately stop heating and evaporate dry with residual temperature.

1) What is the reason why BACL2 in step cannot be changed to BA(NO3)2?

A: No, you cannot. Because it will bring in NO3- (impurities).

2) After the steps, how can I tell if there is an excess of BACL2?

Answer: Let it stand, add BaCl2 solution dropwise in the supernatant, if there is no turbidity, it means that BaCl2 has been excessive, otherwise you need to continue to add dropwise.

3) What is the purpose of joining NA2CO3?

Answer: Remove Ca2+, Ba2+ ions in solution.

4) What are the main instruments that need to be used in the above operations of coarse salt purification?

Answer: Medicine spoons, beakers, glass rods, funnels, filter paper, iron racks, iron rings, evaporation dishes, alcohol lamps.

5) The experimental protocol needs to be improved, please write out the specific operation steps.

The steps are almost complete, and I don't think there is any need to add any more steps. What is lacking should be how to judge whether the added impurity removal reagent is sufficient and the problem of precautions in filtration and evaporation.

1) Because this is a misunderstanding of purification, excess ions cannot be added, resulting in unnecessary separation processes.

2) Extract a small amount of the solution after step 1, and then add a small amount of the original crude salt solution, if there is a sedimentation generation, it means that there is an excess, if not, the opposite.

3) The calcium ions in the filtered solution are settled to achieve the purpose of removing Ca ions.

4) Beakers, test tubes, droppers, glass rods, filters, alcohol lamps.

5) Do not purify and filter, otherwise the refined salt will not be pure.

1) Because nitrate impurity ions will be introduced.

2) No more white precipitates.

3) Remove calcium ions and excess barium ions.

4) It is mainly a three-step operation of dissolution, filtration and evaporation. The main instruments are beakers, glass rods, funnels, filter papers, iron frames, evaporation dishes, alcohol lamps, etc.

5) When adding three excess reagents, it is best to filter them gradually.

1. Nitrate cannot be removed.

2. Add sulfuric acid.

3. Remove barium ions and calcium ions.

4. Beakers, funnels, glass rods, iron frames, crucibles, alcohol lamps. 5，

1.Bring in new impurities.

2.No more precipitation generation.

3.Precipitate calcium and barium ions.

4.Droppers, filter beakers, glass rods, funnels, filter paper, iron racks, iron rings, evaporation dishes, alcohol lamps.

5.Heating process.

1.Nitrate ions will be introduced; 2.Take the upper layer overnight, continue to add BACL2 dropwise, and see if there is a precipitation; 3. Remove calcium and barium ions in solution; 4.

Glass rods, funnels, beakers, iron stands (with iron rings), crucibles 5Step 6, appropriate amount of dilute hydrochloric acid, and compare with pH test paper until neutral.

The answer is c

If the ratio of the number of cations is Cu2+:K+=3:4, then we can set the number of Cu2+ as 3 and the number of K+ as 4, then the sum of the number of charges carried by the cations is (+2)*3+(+1)*4=+10

The solution is electrically neutral, so the total number of charges carried by the anion is also 10, so that the number of SO2- is x, and the number of CI- is y

then (-2)*x+(-1)*y=-10

Solution: 2x+y=10

Substituting the four answers of ABCD, only C is true.

So the answer is c

Let's look at the possible combinations of CuCl2 CuSo4 Kcl K2SO4, then if they are all Cu combinations, SO4:Cl=

3：6=1:2。In the same way, if K is combined with SO4 and CL, the ratio is 1:1, and these two values are extreme hypothetical values, so the answer is between 1, and the answer is D, I don't know if it's right, but can it help

DB, potassium hydroxide reacts with dilute hydrochloric acid, no gas is generated, and there will be no bubbles in the B test tube, so A is wrong; a. Sodium carbonate can react with dilute hydrochloric acid to form gas, which makes the B test tube bubble, but the generated carbon dioxide will press hydrochloric acid into the B test tube, and the sodium chloride generated by the reaction of hydrochloric acid and sodium hydroxide is a substance that is easily soluble in water, and will not make the B test tube turbid, so a is wrong; d. Zinc can react with dilute hydrochloric acid to generate hydrogen, so that the phenomenon of bubbles appears in the B test tube, the generated hydrogen gas increases the pressure in A, and the dilute hydrochloric acid is pressed into the B test tube, and the silver chloride generated by the reaction of hydrochloric acid and silver nitrate is insoluble in water and acid precipitation, making the solution turbid, so D is correct; c. Calcium carbonate reacts with dilute hydrochloric acid to generate calcium chloride, water, carbon dioxide, and reacts, causing bubbles to appear in the B test tube, and the generated carbon dioxide increases the pressure in A, pressing dilute hydrochloric acid into the B test tube, and dilute hydrochloric acid reacts with calcium hydroxide to generate calcium chloride, which cannot make the solution turbid, so C is wrong.

It should be c, there is carbon dioxide gas on the left, and carbon dioxide will make the clarified lime water on the right turbid.

Because the solution in B becomes turbid, carbon dioxide gas is produced for a, but the sodium hydroxide solution cannot be turbid b will not produce gas when dilute sulfuric acid is added.

dAlthough gas is generated, the hydrogen produced does not make silver nitrate turbid.

No, continue to ask.

Choose C sulfuric acid meets calcium carbonate to produce carbon dioxide, so it bubbles. Carbon dioxide meets water to produce carbonic acid, and carbonic acid and calcium hydroxide produce calcium carbonate, so it precipitates.

Answer: c can be used by the method of elimination, the method of verification.

Classmates, this is not a chemistry question, this is an authentic high school physics question Jiangsu Elective 3-5!

The charge number of the element = the ordinal number of the atom.

The sum of masses and charges before and after the reaction does not change.

The decay releases a HE nucleus with a mass of 4 and a nuclear charge of 2, that is, the original nucleus has a mass of 4 less and a charge of 2

Y occurs 4 times in a row - decays into an isotope of element DB No. 105 with mass 268, so the mass of y is 268 + 4 * 4 = 284 and the number of nuclear charges is 105 + 2 * 4 = 113

Finding x by y, x mass 284 + 4 = 288 nuclear charge number 113 + 2 = 115 What is the target in the question, "X is obtained by hitting the target with high-energy 48ca"? What elements?

There must be one of the elements in this question, you can look up the periodic table to get the mass number and the number of nuclear charges.

04 semi-final questions.

In recent years, the rematch has basically not been evaluated.

If you haven't taught yourself this part, it's useless to explain it. No matter which inorganic textbook has this content in the last few chapters, it is very simple, that is, the equation can balance the mass number and the proton number, because the problem itself is too single and too simple, so the chemistry competition basically gives up this body type.