Chemistry Puzzle Bonus points for detailed process!!!!

1.Analysis: The reaction of sufficient nitric acid does not produce gas, indicating that nitric acid is only acidic.

In middle school, the insoluble nitric acid with SO4 2- generation is BaSO4, and the 50 ml of barium sulfate in the question is.

The reddish-brown precipitate is apparently Fe(OH)3After calcination, it contains 100ml of Ba, Fe respectively because of ferric oxide) to calculate the quality of oxygen in the sample: oxygen content.

And there is oxygen in the sample, so the ratio of the number of atoms in the sample is ba:Fe:O=:2:4, and the sample: Bao-Fe2O3 can be obtained

Quite simply, the final reddish-brown precipitate is Fe(Oh)3, and the gram solid is Fe2O3You can find the quality of the iron, too much to type.

The white precipitate formed by condition 1 is BaSO4, and the number of moles of barium can be calculated from the gram solid.

From condition 2, it is known that the reddish-brown precipitate is Fe(OH)3, and the solid gram after combustion is Fe2O3, from which the number of moles of iron can be calculated.

After the molar number of the two metals is calculated, write the molecular formula according to their proportions.

Assuming that both metals are all grams of a, the amount of hydrogen produced is respectively due to the sufficient amount of sulfuric acid.

Mg:24 2=a x gives x=a 12;In the same way, aluminum: 54 6 = a y to get y = a 9, compare x and y, y is greater than x, so aluminum produces more hydrogen.

mg+h2so4=mgso4+h2

24 2m m1=12m

2al+3h2so4=al2(so4)3+3h22*27 6

m m2=9m

So it is magnesium that emits more H2.

Write the chemical reaction equation well.

mg+h2so4=mgso4+h2

2Al+3H2SO4=Al2(SO4)3+3H2 It can be seen that one magnesium atom produces one hydrogen molecule, and two aluminum atoms produce three hydrogen molecules.

If both metals are one gram, there are 1 24 magnesium atoms and 1 27 aluminum atoms.

1 24 mg to generate 1 24 H2

1 27 AL generates 1 27 3 2 1 18 h2 so the mass ratio is 18 to 24, which is 3 to 4. So the mass of aluminum reaction to produce hydrogen gas is large.

1.There is a weight A hanging under the spring scale, and the beaker contains solution B

1) If A is a block of iron and B is dilute sulfuric acid, then put A into B, and after a while, the reading of the spring scale will become smaller (larger, smaller, unchanged).

2) If A is an iron block and B is a copper sulfate solution, then put A into B, and after a while, the spring scale reading will become larger.

2.There is a spiral bright iron wire at the bottom of the large test tube, the test tube is poured into the water, and after being placed for a week, it is observed that there is a red substance on the surface of the wire, and the water surface in the test tube will rise, the reason for the above phenomenon is that the iron rusts and consumes oxygen, and the gas in the test tube decreases and the pressure becomes smaller

3.Displacement reaction is one of the basic types of chemical reactions.

1) The displacement reaction between the metal and the salt solution, generally the more active metal can replace the less active metal from its salt solution, such as copper and silver nitrate solution, the chemical equation is Cu+2AGnO3=2AG+Cu(NO3)2

2) The non-metallic element also has a similar replacement reaction law between the metal and the salt solution, that is, the more active sub-metals can replace the less active non-metals from their salt solution, such as the following reactions may occur in the solution:

cl2+2nabr=2nacl+br2;i2+na2s=2nai+s↓；br2+2kl=2kbr=i2, from which it can be judged:

The order of S, Cl2, I2, and Br2 activity from strongest to weakest is Cl2> Br2>I2>S

1 unchanged. Constant (due to the conservation of mass).

2 reddish-brown rust.

Less Iron reacts in moist air to form rust (iron reacts with water, and oxygen in the air forms rust).

cu+2agno3==2ag+cu（no3）2cl2、br2、i2、s

1. Becoming smaller (dissolving) and becoming larger (displacement).

2. Rust Reduction Iron is easy to oxidize with oxygen under wet conditions to form Fe2O33 Cu+2AGno3=2AG+ Cu(NO3)2Cl>br I> S br>i So Cl>br>i>S

1. (1) Smaller (2) Larger.

2. Rust rises The rust of the wire consumes oxygen, the gas volume decreases, the pressure decreases, and the water surface rises.

3 （1） cu+2agno3=cu(no3)2+2ag2) cl2、br2、i2、s

1. (1) Smaller: The reaction quality of iron and sulfuric acid is reduced.

2) Becoming larger, the mass of copper is increased by replacing iron.

2. Rust Drop Iron reacts with oxygen and water in the air to form rust.

3(1)Cu 2agNO3=Cu(NO3)2 2AG2)Cl2 br2 I2 S The oxidation of the oxidant is greater than that of the oxidation product.

1 (1) becomes smaller; (2) become larger;

2 rust; Rise; The rust of the iron wire needs to consume the oxygen in the air and the water in the test tube, so a negative pressure (less than the outside atmospheric pressure) is generated in the test tube, so that the water level in the test tube rises.

3（1）cu+2agno3==2ag+cu（no3）22）cl2 br2 i2 s

1. Smaller Bigger 2. Red solid rises Iron rust consumes oxygen, and the pressure in the test tube becomes smaller 3. Cu+2agNO3=2AG+Cu(NO3)2

cl2 br2 i2 s

1) Get smaller. (2) become larger;

2.Rust, falling. Oxygen-absorbing corrosion.

3(1)cu+2agno3=cu(no3)2 +2ag2)cl2>br2>i2>s

1 The mixture of insoluble solids and water, such as ——,, can be separated by filtration2 Soluble solids and insoluble mixtures, such as ——,, can be separated by the dissolution method3 The mixture of soluble solids and intolerant solids, such as ——,, can be separated by the dissolution method4 If the solubility of the two soluble solids varies with temperature, if the solubility of them varies with temperature, if ——, can be separated by the recrystallization method.

5 Two immiscible liquids, such as ——,, can be separated by liquid separation6 Two liquids that dissolve each other but have different boiling points, such as distillation7 The mixture of colloids and solutions, such as ——,, can be separated by dialysis.

There is 1 H in the bicarbonate, and when it reacts with a strong base, it mainly depends on the relationship between the ratio of H and OH.

When NaHCO3 reacts with Ca(OH)2, because Ca(OH)2 can provide 2 OH-, when NaHCO3 is insufficient (or when Ca(OH)2 is sufficient), the H provided can only neutralize one OH-, so the amount of the substance is 1:1; When NaHCO3 is sufficient (or Ca(OH)2 is insufficient), the H provided neutralizes both Oh- and so the amount of the substance is 2:1

Ca(HCO3)2 reacts with NaOH, and NAOH is insufficient, and can only neutralize one HCO3-, 1:1 relationship.

ca(hco3)2 + naoh == caco3↓ +nahco3 + h2o

NAOH overdose, neutralizes 2 HCO3-, 1:2 relationship.

ca(hco3)2 + 2naoh == caco3↓ +na2co3 + 2h2o

hco3- +oh- ==co32- +h2o

When Ca(OH)2 is sufficient, sodium bicarbonate is small, and only 1 bicarbonate can only consume 1 OH-, so there is still one OH- left to form NaOHThe resulting carbonate is combined with calcium ions to produce calcium carbonate.

nahco3+ca(oh)2 === naoh + caco3↓ +h2o。

When Ca(OH)2 is insufficient, sodium bicarbonate is excessive, and there is enough HCO3- to react with OH-. 2OH- consumes 2HCO3- to produce two carbonates and two H2O, one carbonate combines with calcium ions to form calcium carbonate, and the other becomes sodium carbonate.

2nahco3+ca(oh)2 === na2co3 + caco3↓+ 2h2o。

NAOH overdose.

ca(hco3)2+2naoh==caco3+na2co3+2h2o

Ca(HCO3)2 (adequate).

ca(hco3)2+naoh==nahco3+caco3+h2o

The reason is that the first step in the reaction between Ca(OH)2 and NaHCO3 is the reaction of HCO3 root and OH root to form CO3 root and H2O, which is the second equation you wrote; And if there is ca(oh)2 left after this step of the reaction is completed, ca(oh)2 will continue to react with na2CO3, and the combination of these two reactions is the second equation you wrote.

As for the second question, NAOH is a strong base, and the reaction between a strong base and a weakly basic compound is not limited by the quantity, and the resulting substance is the same.

For example, when Ca(OH)2 is sufficient, NaHCO3 + Ca(OH)2 === NaOH + CaCO3 + H2O

This is because HCO3+OH==H2O+CO3 is at 1; 1 time! When Ca(OH)2 is sufficient, it means that there is an excess of OH2, and HCO3 is reacted to completion, that is, only excess OH is left, and the generated CO3 is combined with Ca to form CaCO3 precipitation!

Then the latter situation can also be obtained from the above method:

Ca(HCO3)2+2NaOH (sufficient) ===CaCO3 + 2H2O+Na2CO3

Ca(HCO3)2 (sufficient) + NaOH===CaCO3 + 2H2O+NaHCO3

1l correct solution, don't mislead you. It should be 27 kinds.

21 species. First of all, it is not difficult to calculate that the unsaturation is 14 + 1-20 2 = 5The benzene ring already accounts for 4, and there is still one left, which can only be a double bond or ring. In addition, there are 8 C's that are useless.

Let's start with the case of double bonds. For chain-saturated substituents, the end must be methyl. If the substituent still has a branched chain, the number of methyl groups will increase.

Whereas, a double bond can only "remove" one methyl group, therefore, there can only be one substituent, and the substitution itself has no branched chains. This substituent is unique: 1-octene removes one H on No. 8C.

There are 3 types of inter-neighbor pairs.

Consider looping again. Let's start with the monosubstituent. As with the way of thinking just now, for each ring, a maximum of two methyl groups are "removed". Therefore, the substituent group has at most one branched chain before ringing, and it is a straight branched chain.

When the substituents are unbranched, there are 6 possibilities. (8 - self - adjacent c).

When the branched chain is methyl, it is repeated with the 6 types just now (think about why).

When the branch chain is ethyl, there are 4 types.

When the branch chain is n-propyl, there are 2 types.

Let's look at the long substituents. As I just said, the substituent terminal must be a methyl group, and a ring can go to a maximum of 2 methyl groups. In this way, there are at most 2 substituents, and neither substituent is branched.

In fact, this situation is equivalent to two C's on the benzene ring connected by 8 C's. There are 6 types.

There are 21 species in total.

Your question is wrong, the number of h of phenolic compounds is even.

If there is no methyl group, there is no branched chain, and phenolic compounds must have a phenolic hydroxyl group, if you do not consider the cyclic shape, there is a c=c double bond, and there are only three kinds of inter-neighbor pairs, if you want to consider the cyclic shape, there is no double bond, and there can be no branched chain, and only three cases of inter-neighbor pairs can be formed. There should be 6 in total.