A chemistry question is not very clear. A chemistry question, I don t know

Limestone and dilute hydrochloric acid react to form CO2, carbon dioxide dissolves in water to form weak acid, so litmus reagent is reddened when it encounters acid, A is wrong, B as long as the copper wire is lifted to the top, limestone and dilute hydrochloric acid are separated, the reaction stops immediately, C quicklime can react with moist CO2 to generate CaCO3, D This can not prove that its density is greater than air.

A turns red. b Correct, move up and down to control the occurrence and stop of the reaction.

c Calcium hydroxide produced by the reaction of quicklime with water can react with carbon dioxide.

d The only carbon dioxide that is produced is from above.

B Analysis: A: Carbon dioxide encounters litmus solution, which can react with water to form carbonic acid, making purple litmus red.

B: Put the coarse copper wire down, the limestone reacts with dilute hydrochloric acid to generate carbon dioxide, pull the coarse copper wire upward, and the separation reaction between limestone and dilute hydrochloric acid stops.

C: Carbon dioxide reacts with water to produce carbonic acid, which is an acidic gas that can react with limestone, so carbon dioxide cannot be dried with quicklime.

D: This experiment does not verify the density of carbon dioxide.

A cotton will turn red, CO2 dissolves in water and is acidic first.

b Correct, move up and down to control whether the limestone is in contact with dilute hydrochloric acid to control the occurrence and stop of the reaction.

c No, quicklime reacts with CO2.

d It only proves that CO2 is less dense than water, so it will come out;

So choose B

Question 1: The condition under which iron rusts.

In the textbook experiment, it takes a long time for iron to rust, so an experimental team designed the following improvement experiment.

1) Check the air tightness of the device: connect the device, close the piston of the separating funnel, connect the C catheter to a beaker filled with water, and slightly heat the A device, indicating that the phenomenon of good air tightness is: there are bubbles escaping from the C catheter, stop heating, and the C catheter will form a water column.

Because the volume of the gas expands when heated, more than the gas will escape, and the heating will stop, the gas will rot and shrink, and the water will enter the conduit to form a water column.

2) After the inversion of the ballast should begin, the reaction phenomenon in device A is: there is a large number of bubbles generated. 2H2O2 = 2H2O + O2 (MnO2 catalysis). MNO2 is the catalyst.

3) After 4 minutes, the wire is still bright, the surface of the wire is gray at B, and the wire is still bright at D. This experiment shows that iron rust is mainly related to H2O and O2.

The comparison of experimental phenomena at B and D shows that an important factor determining the speed of iron rusting is the concentration of O2. Because the higher the concentration of reactants, the faster the reaction rate.

Little classmate: 1Wooden strips with sparks can only burn in pure oxygen, which is a common method to test pure oxygen, so 1 is wrong.

2. Bright blue-purple flame, light blue ...... in the airThat's right.

3. According to the different boiling points of oxygen and nitrogen, separate oxygen ......That's right.

4. It should be right!

False, a wooden strip with sparks can only test whether a bottle of gas is oxygen, and cannot be re-ignited in the air.

Right. Yes, in industry, oxygen is often produced by separating liquid air method.

No, the lab used potassium chlorate kclo3 instead of potassium perchlorate kclo4

1.Sticks with sparks rekindle in oxygen.

2.Sulfur is burned in pure oxygen, producing a bright blue-purple flame. 3.Right.

1.This experiment can check for pure oxygen; 2.The phenomenon is not right; 4.No catalyst required; The addition of mnO2 as a catalyst is for potassium chlorate.

Below I revise my answer again. [The original answer is attached below].

Analysis: The given conditions of the question are constant temperature and constant pressure, we assume that the gas in container B is divided into two parts: 2molc occupies 4 volumes, and 1molhe occupies 2 volumes [it will never change], so that when the B container is not considered to HE, A and B can achieve exactly the same equilibrium state, but is this like my original answer "The decrease of A and the increase caused by gas C in B are exactly equal"?

We can try to assume an equilibrium state, assuming that equilibrium is reached when the formation of the armor is generated, at this time, the gas in the armor will decrease, and the volume will decrease the volume; And B should also be in equilibrium c is, so that the reacted c is, the gas will increase, and the volume will increase the volume. Therefore, my original answer "the decrease of A and the increase caused by gas C in B are exactly the same" is wrong! In this way, it can be stated that:

1. The x=1 in the second half of my original answer is correct; 2. The answer to the previous question should be "yes", when the piston stops moving, both sides reach balance.

P.S. This question is a bit ambiguous: are the partitions and pistons moving freely? [The previous question seems to be like this]; Or is the bulkhead moving freely and the piston under control? [The latter question seems to be so]. The following are all free to move in accordance with:

No. A and B can achieve exactly the same equilibrium state, the volume of A will decrease, and the volume of B will increase. The partition is no longer centered.

However, the decrease in A is exactly the same as the increase caused by gas C in B. So, the piston does not move. Since there are no other known conditions to determine the equilibrium state, it is only known that the separator is between 0 and 2 to the left of 0 (excluding 0 and 2).

Let's determine the position of k in the form of the answer: [First, determine that k will move to the left, and position x is the volume of A's decrease - that is, 2 times the amount of matter that the gas has decreased, that is, the volume of B's increase.] At this time, the gas of system A decreases by mol, [At this time, the gas of system A is totaled (; The volume is 6-x] and the resulting c is xmol; Because the two equilibrium are equivalent, system B should have xmolc left, then the reacted c is (2-x)mol, and the gas of system B is increased by (2-x) 2 mol, [at this time, the gas of system B is total[3+(2-x) 2]mol= mol; The volume is 6+x], and (6-x) can be obtained according to the equal pressure on both sides

6+x)=(:( can be solved x=1. In fact, if the gas of system A is reduced by mol=the gas of system B, the gas of system B is increased by (2-x) 2 mol, and x=1

I was dizzy, it was a competition question, and our teacher said that the answer was a bit problematic, which was very nonsense.

Of course, the empty delicate is the two substances that generate the other two substances, one of which is the vertical of the precipitation.

1) Between x and y is the neutralization reaction of acid and base.

The grinding between Ba(OH)2+H2SO4=BASO4 +2H2O2)X and Y is the reaction between salt and salt, and there is sodium nitrate in the product.

NaCl + Agno3 = AgCl + Nano33) x and y is the reaction between acidic oxidation and alkali.

SO2 + BA (slow deficiency OH)2 = BASO3 + H2O

Gas B must be CO2, so gas A is SO2 or H2S, and it is concluded that there must be S2- or SO32- ions or both in the original solution.

It can also be concluded that there must be no ag+

1.Light yellow precipitate: S solution must contain S2- and SO32-, and there must be no Ag+ and Al3+ ions Ba2+ ions.

2.Excess hydrogen bromide and excess ammonia bicarbonate produce CO2, so gas B must be CO2White precipitate Al(OH)3.

In the first step, ALO2-+4H+=Al3++2H2O in the second step, AL3++3HCO3-=AL(OH)3+3CO2

So it must contain alo2-

3.Gas C must be NH3. The white precipitate may be BaCO3 or Baso4 or both. Therefore, it is not certain whether it contains SO42-

In summary: it must contain: Na+ S2- SO32- ALO2-

Must not be included: ag+ ba2+ al3+

May contain: CO32- SO42-

The yellowish precipitate is sulfur, and under acidic conditions, sulfur ions and sulfites undergo a centering reaction and the gas is carbon dioxide.

There are sulfur ions, so silver ions, barium ions, aluminum ions, metaaluminate are not present, gas b is carbon dioxide, the first white precipitate may be sodium bicarbonate gas, c is ammonia, and the white precipitate may be barium carbonate, or it may be a mixture of barium carbonate and barium sulfate, a good topic.

Have a problem with your wordless book in heaven?

It's because you misunderstood, and it was right to choose D.

Combustion produces water vapor, carbon monoxide and carbon dioxide, indicating that the organic matter is not completely burned, and the number of carbon and hydrogen atoms can be directly considered. As for a, when it is fully combusted, only water vapor and carbon dioxide are generated, the relationship between oxygen consumption (1:5).