Seeking experience, the National Junior High School Physics and the National Junior High School Chem

The junior high school applied physics knowledge contest is not too difficult, if your goal is just to win a prize, it is easy to get the award, you only need to have a solid foundation in physics, and then look at a special tutorial book (called the junior high school applied physics knowledge contest or something) before the exam in the multiple-choice questions, note that it is a multiple-choice question, because the application of the multiple-choice question is very strong, what to test what bicycle tires, paving, pipe repair, etc., if you don't look at the previous questions, you will be confused on the spot, so you must look at the multiple-choice questions. Secondly, the amount of calculation is very large (be careful), as for the review focus on work and energy, must be examined, especially the question of electrical power and mechanical energy, often give you a nameplate of electrical appliances and then ask you to ask for something, focus on review.

I'm a senior one, and I'm engaged in a physics competition, which is a simple competition, that is, you must be careful and patient. Good luck.

That book teaches junior high school applied physics knowledge competitions.

The 3rd National Mathematics, Physics and Chemistry Proficiency Competition for Middle School Students (Guangdong Province) Preliminary Round Winning Score Query Grade 7 Mathematics: First Prize: 86 points (inclusive) and above.

Second prize: 78 points (inclusive) - 85 points.

Third prize: 70 points (inclusive) - 77 points.

Grade 8 Mathematics: First Prize: 85 points (inclusive) and above.

Second Prize: 78 points (inclusive) - 84 points.

Third prize: 70 points (inclusive) - 77 points.

Grade 9 Mathematics: First Prize: 83 points (inclusive) and above.

Second prize: 73 points (inclusive) - 82 points.

Third Prize: 68 points (inclusive) - 72 points.

Senior 1 Mathematics: First Prize: 107 points (inclusive) and above.

Second prize: 96 points (inclusive) - 106 points.

Third Prize: 85 points (inclusive) - 95 points.

Senior 2 Mathematics: First Prize: 92 points (inclusive) and above.

Second Prize: 79 points--- 91 points.

Third prize: 70 points--- 78 points.

Grade 8 Physics: First Prize: 97 points or above.

Second Prize: 92 points (inclusive) - 96 points.

Third Prize: 87 points (inclusive) - 91 points.

Grade 9 Physics: First Prize: 87 points (inclusive) and above.

Second Prize: 82 points (inclusive) - 86 points.

Third prize: 75 points (inclusive) - 81 points.

Senior 1 Physics: First Prize: 105 points (inclusive) and above.

Second prize: 92 points (inclusive) - 104 points.

Third Prize: 81 points (inclusive) - 91 points.

Senior 2 Physics: First Prize: 96 points (inclusive) and above.

Second Prize: 88 points (inclusive) - 95 points.

Third Prize: 79 points (inclusive) - 87 points.

Grade 9 Chemistry: First Prize: 95 points (inclusive) and above.

Second prize: 88 points (inclusive) - 94 points.

Third Prize: 81 points (inclusive) - 87 points.

Senior 1 Chemistry: First Prize: 96 points (inclusive) and above.

Second prize: 90 points (inclusive) - 95 points.

Third prize: 85 points (inclusive) - 89 points (points).

Senior 2 Chemistry: First Prize: 106 points (inclusive) and above.

Second prize: 99 points (inclusive) - 105 points.

Third prize: 94 points (inclusive) - 98 points.

Chemistry competitions in junior high school are divided into Olympiads, and there are some regular competitions.

You have a guarantee for this competition, because I was the last one...

1.Find a good competition teacher, an experienced teacher will tell you how to do a competition 2Buy books on your own.

Bibliography Purchase Order:

Inorganic Chemistry, Third Edition, Jilin University, (up and down), General Inorganic Chemistry, Peking University, Yan Xuanshen.

Fundamentals of Organic Chemistry" Xing Qiyi, Pei Weiwei, Pei Jian (and one forgot, nicknamed Xing Daben) "Fundamentals of Structural Chemistry" (I forgot whether it was called this) Duan Lianyun Whether the content in these books should be read should be considered, it is best to ask the teacher, sometimes don't read what Schrödinger equation is, don't test and can't understand...

The above number of supporting exercises is best to buy, and there is also an exercise (also considered a reference book) that is also needed, that is, "Analytical Chemistry" (the version is casual, as an exercise).

In addition, it is best to finish the college entrance examination knowledge before the winter vacation, and then enter the competition.

This is called "cultivating feelings".

Hope you have a bright future!

Yes, it is normal to add points to the Olympiad college entrance examination.

Yes, there is a guarantee. However, it is recommended that students master the knowledge of the college entrance examination first. Also, in the college entrance examination, don't solve the college entrance examination questions according to the idea of competition.

Chemistry competition. I didn't understand it when I was in junior high school. And it's basically impossible to go to the competition and win a prize.

1.I'm wrong. How can the mass of the metal be greater than the mass of the metal salt.

2.The meaning of the title is that the first time you throw Cu into it, you will just completely react FeCl3 or partially react FeCl3, and then add Fe and Cu2+ will be replaced, and B must be correct, A, I'm not sure, after all, if you replace it, the copper will adhere to the surface of the iron, preventing the reaction from continuing in junior high school. I guess it's fine.

If there is Fe3+, then the precipitated metal will continue to react with it, so if it is not true, D, the residual solid is Fe, that is, all the cations in the solution (except H+ Fe2+) have been reacted, and Cu should be replaced, and the residual solid will definitely have Cu Contradictory answer.

ab bar. 3.Did you learn mol in junior high school?

First calculate the mass fraction of Na2CO3 in ten crystalline water Na2CO3 is about 37% If the original solution is thicker than it, then 10 water Na2CO3 will not be completely dissolved in it, so the mass fraction must be smaller than 37% After selecting in CD, you can think like this If m=n then it must be as much as the precipitated solids But if there are ten crystalline waters, isn't it equivalent to adding Na2CO3 at the same time and pouring water into it to dilute Then if m = n The former must have fewer crystals so m>n choose C

4.Let the mass fraction of formaldehyde be x total mass of solution 100g100g*10%=100gx*(2 (2+12+16))+100g*(1-x)*(2 (2+16)).

Solution x=25%.

1.……The mass of the metal decreases = = after the acid ions are obtained. Emotional acid ions are the best medicine

2.One situation is that the iron of BG is not enough to replace all the copper ions, that is to say, all the iron becomes copper, and there are still copper ions in the solution at this time, then there are A>C; When BG has too much iron, all the copper is replaced and there is a surplus, then C>A, so it is not valid; Exactly when the reaction a=c, in summary, choose a

The mass fraction of Na2CO3 in China is 37%, which means that to form a solution, the mass fraction of water must be greater than 63%, so A% may be 30% ......Then you can think like this [pure crookedness], add Na2CO3*10H2O to precipitate Na2CO3*10H2O, that is to say, the amount of precipitated Na2CO3*10H2O is m mol, and after the addition of Na2CO3 in N mol, it wants to seize water as its crystallization water, and after these waters are less, there are some solute crystallization that were originally dissolved in these waters, so it does twice the result with half the effort, and the same amount of crystals is precipitated, and the effect of adding Na2CO3 is obviously better, so m>n, choose c.

4.Let the mass of formaldehyde be x, the total mass is m, and (10%m-x 15) is the mass of the hydrogen atom contained in the water molecule, so the mass of water is, (10%m-x 15) 1 9, and the mass of water is the total mass minus the mass of formaldehyde, so.

10%m-x 15) 1 9=m-x, solution x=1 4m, choose c

1. Maybe the number is wrong, but let's talk about the idea:

mg+h2so4=mgso4+h2↑

zn+h2so4=znso4+h2↑

fe+h2so4=feso4+h2↑

So the total mass of the metal + the total mass of sulfuric acid = the total mass of the metal sulfate + the mass of hydrogen.

The total mass of the metal and the total mass of the metal sulfate are now known, so the subtraction of the two is the mass of hydrogen minus the total mass of sulfuric acid.

According to the equation, the mass ratio of sulfuric acid and hydrogen can be obtained as 49:1

Now that we know the mass difference and the mass ratio between the two, we can solve it by a system of binary equations.

2. Select b2fecl3+cu=2fecl2+cucl2

cucl2+fe=fecl2+cu

In fact, there is a reaction that occurs: 2FeCl3+4Fe=3FeCl2, but it is not relevant to this problem).

Let's start with a: if the residual solid is all copper, then c=a, which does not match the title.

C: The remaining solid, whether copper or iron, or a mixture of the two, can react with the solution of the trivalent iron salt, so C is wrong.

d: CuCl2+Fe=FeCl2+Cu, this reaction must occur, so the solid must contain copper, so D is wrong.

From the above analysis, it can be seen that there are three possibilities for solids: iron, copper, iron and copper. We have already excluded and in our discussion and are therefore correct, i.e. b is right.

3. Well, I'll think about it again, but what the second floor said seems to make a lot ...... sense

4. Select C to set the quality of HCO as AG and the quality of H20 as BG

The mass fraction of hydrogen in HCO is 1 15, and the mass fraction of hydrogen in water is 1 9

So the equation can be listed: (a 15 + b 9) (a + b) = 10%.

The solution is b=3a

So the solute mass fraction of HCO = A (A + B) 100% = A 4A 100% = 25%.

Hope the above answers are helpful to you.

The frictional force is equal. The right hand is subjected to the sliding friction force exerted by the wooden stick to the right, because the force is mutual, the wooden stick will be subjected to the same large sliding friction force to the left, that is, the thrust to the left, acting on the left hand, the left hand remains stationary, and the balance force is acted, and the static friction force on the left hand is equal to the thrust, and the thrust is equal to the friction force on the right hand.

On the left side. The right hand is displaced, the left hand is not moving, so to the left.

There are two types of friction, stationary friction (generally its maximum value, i.e., maximum resting friction) and sliding friction.

And one thing is very clear, once from stationary to sliding, the sliding friction must be less than the maximum resting friction, so the first empty must be the left hand friction is large and the right hand is small.

The second empty didn't see the picture, and didn't dare to talk nonsense, hehe.

Less than the stick moves from rest to the right relative to the ground, so the friction of the left hand is greater than that of the right hand.

The second one is left.