Middle School Chemistry A super easy question

Molecular makeup. Because it is bonded with covalent bonds, it is a covalent compound.

HCI gas is composed of molecules, because it is an acid, so it can electrolyze hydrogen ions and chloride ions in water, (in fact, electrolysis is a chemical reaction, because it is too simple, so there is no such equation in junior high school chemistry).What hydrochloric acid is made of depends on what state it is, if it is a gas, it is a molecular composition; If it is an aqueous solution, it is ionic composition.

,,, according to the amount of zinc, referring to the above reaction order, when n(zn) n(feso4), the reaction is 1 3.then a is cu, which does not fit the topic.

When N(FeSO4) N(Zn) N(FeSO4)+N(CuSO4), the reaction is , then A is Fe, Cu, and the second group is compliant.

N(Zn) N(FeSo4)+N(CuSO4), the reaction is , then a is Zn, Fe, Cu, and the fourth group is compliant.

Since the metallicity of Zn is stronger than that of Fe, when Zn exists, Fe in FeSo4 solution must be replaced, and then Cu in CuSo4 solution will be replaced, and the first group is wrong.

Similar to the above, the metallicity of Zn and Fe is stronger than that of Cu, so it is inevitable that the replaced Cu will be wrong.

That is, the metallicity of one metal is more metallic than that of another, and if the latter exists, the former must exist).

For reference, it should be right.

The first one is unreasonable: Zn is added to the solution, and according to the table of metal activity sequence, a displacement reaction will occur, and the remaining solid in the solution cannot be only Zn.

The second is reasonable. If the amount of Zn is just enough to replace Fe and Cu in the solution;

The third is unreasonable. According to the order of metal activity, if Zn and Fe are present, both can be displaced.

The fourth is reasonable. If zn is excessive, it will do.

a.Unreasonable, there will be at least the presence of cu in the composition.

b.Reasonable, Zn first replaced Cu, and then replaced Fe, at this time the Zn powder was not excessive.

c.Unreasonable, there will be at least the presence of cu in the composition.

d.Reasonable, at this time zn is sufficient.

Choose A. Zinc is the most reactive metal here. Replace the copper first; Replace the iron again (there is no question).

1.Add some zinc powder to the mixed solution of ferrous sulfate and copper sulfate, and Zn reacts with CuSO4 first, because Cu is the least active among the three. Then substance a has Cu2

Put A into a small amount of dilute sulfuric acid, there is gas to produce Cu does not react with H2SO4, so it is possible to replace Cu in the replacement of Fe, then A has Cu and Fe It is also possible to have excess, then A substance may have Zn, Fe, Cu!!

1. Aluminum is added to dilute hydrochloric acid to produce bubbles, and the bubble rate is faster than that of iron.

2. Cu, Al, first insert Fe to generate FeSo4, copper insertion has no phenomenon, Fe is greater than Cu, Al insertion replacement, solution fading, etc. prove Al Fe, in summary, Al Fe Cu (PS; Inserting Al first does not determine the activity of Cu versus Fe because FeSO4 is reacted).

3. Cu, Fe, Al, Cu insertion has no phenomenon, the weakest, Fe insertion generates FeSO4, which can be replaced with AL, indicating Al Fe

2.Iron, aluminum, copper.

Iron is added first, and the solution turns light green with bubbles.

Then put aluminum, the solution becomes colorless, it can be proved that aluminum is more active than iron and then put in copper, there is no obvious phenomenon, you can know that the order of activity of copper is after iron and aluminum3Iron, copper, aluminum.

Iron is added first, and the solution turns light green with bubbles.

Then copper is added, and there is no obvious phenomenon, it can be seen that the order of activity of copper is after iron and then aluminum is added, and the solution becomes colorless, which proves that aluminum is more active than iron.

The second question is Fe, Al, Cu, copper does not react at all, the activity is the worst, after putting in iron, after rectifying iron, putting in Al, Al will first replace the iron and then react with hydrochloric acid.

The third question is cu, fe, al

1. Aluminum has a large number of bubbles coming out, less iron, and copper is basically non-reactive.

2. Cu, Al3, Cu Fe Al.

(1) Aluminum reflects more violently than iron.

2) Copper aluminum.

3) Copper, iron, aluminum.

(1) More bubbles are AL, less FE

The topic is too complicated, simply put, because the test tube wall is hot, there is sodium hydroxide or calcium oxide or mixed, and the two conjectures are not valid.

If there are bubbles in the filter residue, it means that it is calcium carbonate, and the calcium oxide has deteriorated.

There can be bubbles in the filtrate with hydrochloric acid, indicating that the sodium hydroxide has deteriorated.

The key is in the last two steps, it is more difficult to consider, now it needs to be assumed, assuming that there is sodium carbonate and calcium hydroxide in the liquid, then the sixth step is not established, because after adding hydrochloric acid, it becomes calcium chloride and sodium chloride, if the sixth step is still true, then there is a premise at this time, there must be sodium carbonate in it, if there is sodium carbonate, the fifth step should not be clear, but turbid, because calcium chloride and sodium chloride and sodium carbonate are impossible to coexist, will generate precipitation. Therefore, the assumption is not true, there is no calcium hydroxide in it, that is, there is no possibility of containing calcium oxide.

You can also think of it this way, if there are calcium ions in the filtrate that should precipitate as early as the fifth step, how can there be a sixth step? If the carbonate is left without precipitation, it means that there are no calcium ions in the fifth step.

The sodium hydroxide is partially deteriorated, and I didn't find any positive information in the title, so I can only push it back, because the calcium oxide is all deteriorated, so the initial heat is calcium hydroxide.

What is Lu Slag? By the way, learn it? Thank you. I understand it by filter residue.

After writing it, I found out that there was a problem with this question, which was really dizzy! This ** experiment is seriously flawed, and it needs to be improved, if you have to give C an answer, it is "calcium hydroxide has deteriorated" (if the slag is understood by the filter slag).

First of all, the key to CAO lies in Luli.

First push back: drop dilute hydrochloric acid (bubbles). Description: The liquid contains CO32 - drops into the clarified lime water (white precipitate). It was precipitated after adding clarified lime water, indicating that Ca2+ did not exist

Therefore, it means that Ca2+ is all in precipitation.

Then push forward according to the topic:

If there is no deterioration of the cao, then the cao and the water will produce calcium hydroxide that is partially soluble in water, which is impossible according to the inversion just now, so the cao is completely spoiled (and it is all caco3).

As for naoh

First of all, the outer wall of the test tube is hot, indicating the presence of NAOH

Then drop dilute hydrochloric acid (bubbles) to indicate that the liquid contains CO32-, and the absence of Ca2+ indicates the presence of NaCO3

No, I think. Your answer is wrong.

In my opinion: 1 Add enough water The heat on the outer wall of the test tube indicates the presence of CA because quicklime reacts with water and releases a lot of heat while NaOH and water do not.

2 Naoh is definitely completely spoiled. If there is partial deterioration, there will also be a step phenomenon. And there are some problems with the title.

Step 5: If you add too much dilute hydrochloric acid, then step 6: Add clarified lime water according to the meaning of the title, and it will not precipitate unless you pass the generated gas into the clarified lime water. Even if it is partially deteriorated as you say, then add dilute hydrochloric acid, and the hydrochloric acid must first react with NaOH so that no bubbles are formed. As for why it reacts first with NAOH.

You can think of it the other way around, if you add sodium carbonate, the CO2 will be absorbed by NaOH and will be neutralized with NaOH first. According to my knowledge of three years of junior high school.

First of all, the first chemical formula: CaO+H2O=Ca(OH)2A: does not hold, because if it is all deteriorated, the outer wall of the test tube will not heat up. However, in the test, the outer wall of the test tube was heated, so it was proved that II was not established.

b：caco3+2hcl＝cacl2+h2o+co2；

na2co3，ca(oh)2；Not established (if there is no deterioration, operation 5 will not produce gas, since gas is generated, it proves that there is deterioration).

C: Partial deterioration.

Shame I didn't see it was ca(oh)2,caco3,naco3...

So the guy above is right.

Some of them have deteriorated, and some of them have heat, and some of them have alkali metal oxides that emit heat when they meet water, and there is turbidity that makes the clarified lime water muddy, indicating that there are carbonates, and they have deteriorated.

cao+h2o=ca(oh)2

a.Not true.

Na2CO3 does not hold.

c.The slag has air bubbles, or the filtrate produces air bubbles, indicating partial deterioration.