Urgent: Help to get started with English and chemistry in the third year of high school.

Like you, you have to endure hardships now, and you have to do more questions, and you may have to do 2 or even 3 test papers a day when others do it. Chemistry is relatively simple, just do more questions, English is more troublesome, but more reading and listening is also effective. The college entrance examination is not about who is smarter, but who is more proficient, as long as you do more questions, there are so many routines in the college entrance examination, all the questions are familiar, practice makes perfect, you will definitely be able to do well in the college entrance examination.

As for listening, I don't think there's anything to listen to in English, it's not easy to learn grammar, it's better to do more reading, listen more and recite a few model essays like lz, as long as you have a sense of language, you can get a high score for a single completion form (provided that you memorize the words well, if you don't memorize the words well, it is difficult to take the English completion test with a single choice.) Chemistry is a preview before class, I feel that chemistry is not very difficult, at least read the book to understand, the problem can not be done can be seen to solve the problem, through the solution to learn the solution to the problem, class listening is also necessary, do not understand can be copied down after class and slowly study.

I remember that when I was in my third year of high school, I did at least 4 papers a day, one for English, one for mathematics, one for science, one for Chinese, and all kinds of review materials. English listening should be listened to every day, and only by listening every day can you maintain a better state. In short, it's more questions, and that's my experience.

I also wish LZ a good review and a good score in the college entrance examination.

Read the book, then do the questions, do the questions carefully, and then read the book to consolidate when you are done! There is no other way but to teach yourself!

The reason is simple: One of the keys to ion separation and purification is not to introduce impurities. And BC has impurity cations to produce a denier balance. So new impurity ions are introduced.

In D, iron ions were precipitated by the double hydrolysis of MGCo3 and ferric ions. Since the solubility of MGCo3 is less than that of magnesium hydroxide, magnesium ions will not settle to form magnesium hydroxide. So it remains in solution.

Finally, an appropriate amount of hydrochloric acid is added to remove possible excess MgCO3

I am also in my third year of high school, and I always heard about the tactics of the sea of questions before my third year of high school, but I still think that the most important thing is that we must concentrate 100% attention in class, because we have loopholes in our previous learning, and the third year of high school is to make up for the loopholes. We need to systematize our exercises and knowledge, and only after the system will know that it is a loophole, so as not to do the questions blindly. Regarding the specific knowledge points, different regions may have different emphasis, I don't know which region you are a candidate, and I can't say.

I hope we can work together and be worthy of ourselves.

The amount of the substance is that the amount of the substance added to HCl is, and all the dissolution needs to be known, and the amount of the remaining HCL substance is AL2O3+6HCl=2ALCl3+3H2O

Start adding NaOH, NaOH first to neutralize the remaining HCl in the first step of the reaction, continue to add dropping, and the Al3+ generated by the first step reaction to generate a precipitate, continue to add dropwise, NaOH reacts with the generated Al(OH)3, so 1, neutralize the remaining needs, then.

When 150 ml of NaOH solution is added, precipitation begins.

The amount of precipitation is the largest, and the precipitation is Al(Oh)3, that is, when all Al3+ is converted into Al(Oh)3, it is the maximum, so it is needed, and the previous reaction is needed, that is.

When 600 ml of NaOH solution is added, the precipitated lead volume reaches its maximum.

al3+3oh-=al（oh）3↓

The precipitate happens to be completely dissolved, i.e., Al(OH)3 is completely converted to ALO2

al（oh）3+

oh-alo2

According to the above ion equation, 2H2O can be consumed, and the last two steps of the reaction are required, namely.

When 750 ml of NaOH solution is added, the resulting pellet happens to dissolve again.

al3+3alo2

6h2o=4al（oh）3↓

According to the above ionic equation, it can be obtained that all aluminum elements can be converted into Al(OH)3, and the amount base of AlCl3 is required.

The addition of AlCl3 just converts all the aluminum elements into aluminum hydroxide.

From the chemical equation Al2O3+6HCl=2AlCl3+3H2O, gram Al2O3 is .

600ml2mol Qingzhi bent l hydrochloric acid is Mo, Al2O3 is completely reacted and the hydrochloric acid is left, so the reaction of NaOH is Mo, plus the reaction with the remaining hydrochloric acid before NaOH, for. So when the reputation is stuffy.

When 150 mL of NaOH solution, a deep and simple lake began to appear; When milliliters of NaOH solution are added, the precipitate reaches its maximum.

Question 3, Al(OH)3+NaOH=Naalo2+2H2O, from the above question, we can know that Al(OH)3 is, so when the precipitate is just dissolved, the NaOH is co-reacted, so it is 750ml

The fourth question is AlCl3+3Naalo2+6H2O=4Al(OH)3+3NaCl

From the above question, it can be seen that naalo2 is, so alcl3 is needed to be grams.

1) c10h10，8

The structural computer in the back is not easy to draw.

Benzene-ch=ch-ch=ch2

1. Miswriting of oxides suspected to absorb sulfur.

so2 + 2nh3 + 2h2o = 2nh+ +so32-，so2 + nh3 + h2o = nh4+ +hso3-。

Second, it is suspected to be chalk powder.

caco3 + 2h+ +so42- =caso4 + co2↑ +h2o，caco3 + 2h+ =ca2+ +co2↑ +h2o。

3. The reaction is: 2CO + O2 = 2CO2, CH4 + 2O2 = CO2 + 2H2O (the conditions are all ignition).

1. Let CO be XML and CH4 be YML, according to the topic and reaction equation, X + Y = A = B, X 2 + 2Y = B, you can solve: Y (x+y) = 1 3.

2. The volume of gas after combustion is 4 a ml suspected to be: the volume of gas after combustion has decreased by 4 a ml.

It can be seen that the volume of CH4 combustion gas does not change, and the volume decreases because of the reaction of CO.

x + y = a, x 2 = a 4, we get: x = , y = , and oxygen consumption is x 2 + 2y = b, i.e.: = b, 5a = 4b.

Since oxygen must be sufficient, the actual value should be: 4b 5a.

Fourth, the data is incorrect. 3 Take 120ml of solution A and react with 440ml of solution B, then the precipitate should be produced: 3 Take 120ml of solution A and react with 400ml of solution B, then the precipitate will be generated.

The amounts of Al(OH)3 are: 78g mol = , 78g mol = .

From the addition of 400ml of solution B from 3 to 440ml of solution B produced less precipitation than the addition of 440ml of solution B to 2, it can be seen that A is ALCL3 solution and B is NaOH solution.

Take 440ml of solution A from 1 and react with 120ml of solution B to produce a precipitate and 2 take 120ml of solution A and react with 440ml of solution B to produce a precipitate, it can be seen that: 1 NaOH is insufficient, not completely precipitated, 2 NaOH is excessive, and the precipitate is partially dissolved.

According to 1, c(NaOH) = 3 = .

According to 2, it can be seen that Al(OH)3 + OH- =ALO2- +2H2O, the remaining needs are dissolved, and the total NaOH is dissolved from the beginning of the precipitation to the precipitation: += , and then by Al3+ +4OH- =AlO2- +2H2O, we know that C(AlCl3) = 4 = .

5. Ag - E- = Ag+, 1Molag is oxidized and loses 1mol electrons, NO3- +4H+ +3E- =No +2H2O, and the reduced NO3- is 1Mole- 3E- =.

Do you still need to ask this kind of question? Think more about yourself, otherwise, you won't have room to think independently, how to face the exam papers in the future.

With Na: alcohol, phenol, carboxylic acid (all generate H2).

with NaOH: phenols, carboxylic acids (to generate H2O); esters (alcohol + sodium carboxylate); Halogenated hydrocarbons (heated in aqueous solution to generate alcohol + Nax, and heated in alcohol solution to form unsaturated hydrocarbons + H2O + Nax).

with NaHCO3:phenol (bicarbonate generated); Carboxylic acids (to generate CO2) and Na2CO3: carboxylic acids (to generate CO2).

...The organic compounds that can react with sodium are: alcohol (hydrogen on substituted hydroxyl group).Carboxylic acid (needless to say)...Phenol.

The organic substances that can react with sodium hydroxide are: carboxylic acid, phenol can react with sodium bicarbonate, and the organic substances that can react with sodium bicarbonate are: carboxylic acid (note that phenol cannot react with sodium bicarbonate).

Organic compounds that can react with sodium carbonate are: phenol (phenol reacts with sodium carbonate to produce only sodium bicarbonate.

Acidity table: formic acid, acetic acid, carbonic acid, carbon, phenol, bicarbonate.

If you don't understand, you can continue to ask--