Chemistry Multiple Choice One or Two Options, How High School Chemistry Multiple Choice

1 Option D No matter what the oxide of sodium is, the final product after reaction with hydrochloric acid is NaCl.

The amount of hydrochloric acid is , so according to the conservation of chlorine, the amount of NaCl species produced is also , let the oxide of this sodium be Na2x, so the original Na2x is, in Na2X, the mass of Na is 23*, so the mass of x can be calculated as, and then divided by the amount of the substance of x to get the molar mass of x is 24g mol.

Therefore, the molar mass of oxygen bound to Na in the oxide should be greater than 24 g mol and less than 24 g mol of two or more substances. Based on this, it can be concluded that the two options of Cd are true, but since sodium cannot form sodium superoxide NaO2 in dry air, C is wrong, so D is selected.

2. Select AB First, add enough water to dissolve all the samples, and it can be concluded that the sample does not contain BA(NO3)2.

The 9 g precipitate generated by the reanalysis should all be CaCO3, which is, assuming the sample is pure K2CO3, the CaCO3 would be generated.

If only Na2CO3 is mixed in the sample, the amount of carbonate in the mixture solution of the same mass will be increased, so the amount of CaCO3 produced will be higher, but the final mass of CaCO3 will be higher, so it can be judged that Kno3 must be present, so that the carbonate in the mixture solution of the same mass will be reduced. Therefore, it can be concluded that:

There will definitely be Kno3 in the sample, there may be Na2CO3, and there will definitely be no Ba(No3)2.

d, with extreme assumptions, it is necessary to neutralize 20 mmol of HCl anyway.

Be sure to have Na2O2.

a, b, let's say pure K2CO3 has 100 mmol of CO3, but the actual amount is only 90 mmol (known by the amount of precipitation).

Let's just sauce purple.

The structure of multiple-choice questions consists of two parts: the question stem and the options, which are carefully set and often have a certain pertinence and strong confusion. After the analysis of the college entrance examination papers in recent years, the multiple-choice questions have the following characteristics:

1) The knowledge capacity of the examination is larger, the coverage is wide, and there is a good degree of discrimination, which is conducive to selection.

2) A small number of "basic questions" increase the base of candidates' scores, which is conducive to stabilizing students' examination psychology and helping students to play a normal level.

3) Retain certain types of common questions and adjust the total difficulty.

4) The question type remains relatively stable, which is conducive to the preparation of candidates.

High School Chemistry Multiple Choice Test Content.

1.Chemical terminology, classification of substances, colloids, energy, chemistry and society, life, high technology, etc.;

2.Avogadro's constant;

3.The correspondence between the ion equation and the chemical facts, the coexistence of ions;

4.Judgment of oxidation and reducibility, electrochemistry;

5.Atomic structure, periodic law, chemical bonds;

6.chemical reaction rate and chemical equilibrium;

7.Determination of weak electrolytes, ionization equilibrium, hydrolysis equilibrium, dissolution equilibrium, and the relationship between particle concentration in electrolyte solution;

8.The basic operation of chemical experiments (instrument use, substance separation and inspection), and the safety of chemical experiments;

9.structure and properties of organic matter;

10.Small calculations.

1. List special cases and fast row options.

For some conceptual judgments, propositional judgments, correct and false questions, if you can't make a direct judgment from the front, you can list counter-examples, special cases, and quickly judge the right or wrong options.

2. Clever use of assumptions to static braking.

When answering questions about the movement of the four equilibrium (chemical equilibrium, ionization equilibrium, precipitation-dissolution equilibrium), the influence of multiple factors (such as temperature, hydrolytic equilibrium, concentration or pressure) sometimes occurs, and for such questions, the hypothetical method can be used to determine the effect of the change of the first factor on the result, and then judge the effect of the change of the second factor on the result, and then arrive at the correct answer.

3. Grasp the constant and distinguish ions.

In the electrolyte solution, some students cannot distinguish between ionized ions and ionized ions of electrolyte, and they are very confused about solving such problems. If the concentration of H+ and OH- in water ionization is constant, and the ions are identified, it is easy to break through this kind of problem. That is, all the ionization equilibrium and hydrolysis equilibrium equations in the solution, including the water ionization equilibrium equation, are written first, and then the ions** are analyzed to find the correlation between ion concentrations.

1. Many high school students have a bad foundation in chemistry, so they must first find the root cause. Whether you don't remember the common formulas and concepts, or you don't have a good grasp of the basic problem-solving methods. These are the core contents of high school chemistry, and if these things are not clear, it will cause difficulties in high school chemistry learning.

In addition, if you have any doubts about chemistry learning, you should consult your teacher immediately to make sure that you know what to expect.

2. If high school students want to improve their chemistry scores, they must deal with the relationship between reading books and doing problems. Many students just read books and don't do questions; There are also students who blindly do questions and do not read books, so they are not able to learn chemistry. High school students are required to do some matching practice questions after reading the book.

Only by doing questions can you test whether you really understand the textbook.

3. High school chemistry is much more difficult than junior high school, and the degree of abstraction is also relatively high. These contents have also been considered to be the key knowledge that causes students to differentiate and learn chemistry. In addition, high school students should also have the habit of sorting out their mistakes and finding out the reasons for their mistakes, so as to improve their chemistry scores.

4. If high school students want to improve their chemistry scores, they should pay attention, listen carefully to lectures, and study actively. When the teacher starts to teach a new lesson, high school students should pay attention to how the teacher analyzes the problem, especially when the teacher is demonstrating the experiment, and the students should also carefully watch how the teacher operates.

5. If you want to learn high school chemistry well, no matter how good the teacher says, high school students must implement their understanding and mastery of knowledge in their homework. Homework is an important way to test students' understanding of textbook knowledge, so completing homework carefully is also an important part of learning chemistry well. According to statistics, most of the students who do not achieve satisfactory results in chemistry cannot complete their homework according to the teacher's requirements.

Commonly used indicators are mostly weak acids or bases, such as litmus; Phenolphthalein and methyl orange are more complex organic acids. The molecules and ions of the indicator have different colors, and the acid or base solution can affect the ionization balance of the indicator, so the indicator will show different colors in the acid or base solution. The following aspects should be considered when selecting indicators for neutralization titration:

1) The narrower the discoloration range of the indicator, the better, and the indicator can change color with a slight change in pH. Due to the wide range of discoloration and the difficulty of observing the color change at the equivalence point, litmus solution is not used in neutralization titration. (2) The change of the color of the solution from light to dark is easy to observe, while it is not easy to observe from dark to light.

Therefore, an indicator should be chosen that changes the color of the solution from light to dark at the end of the titration. When strong acid and strong base are neutralized, although phenolphthalein and methyl orange can be used, when the alkali is titrated with acid, methyl orange is added to the alkali, and when the equivalence point is reached, the color of the solution changes from yellow to red, which is easy to observe, so methyl orange is selected. When the acid is titrated with alkali, phenolphthalein is added to the acid, and when the equivalence point is reached, the color of the solution changes from colorless to red, which is easy to observe, so phenolphthalein is selected (3) When the neutralization of strong acid and weak alkali, strong base and weak acid reaches the titration endpoint, the former solution is acidic, and the latter solution is alkaline, and the alkaline color change indicator (phenolphthalein) should be selected for the latter, and the acid color change indicator (methyl orange) should be selected for the former.

4) In order to prevent the discoloration of the indicator from abnormality and error, the dosage of the indicator should not be too much during neutralization and titration, the temperature should not be too high, and the concentration of strong acid or strong alkali should not be too high.

If the pointer points to that side, it means that side is heavy.

A because the reaction rate of zinc is fast and more hydrogen is released, so when the two sides have not completely reacted, the hydrogen released on the side of zinc powder is fast and light! It should be pointed to the side of the iron powder.

The relative atomic mass of BZN is 63, Fe is 56, and ZN consumes less sulfur jujube acid at the same mass! It may occur that Zn is completely reacting, but because Fe consumes a lot of sulfuric acid, the amount of sulfuric acid is insufficient, and Fe is left!

When the amount of sulfuric acid is very small, there is also a surplus of zn! At this time, the sulfuric acid is completely reacted, and the amount of hydrogen released is the same because the starting state is the same, so it points to the center.

With reference to the explanation of item b, it can be concluded that there is a surplus in FE when Zn is left in the remaining rock!

If you have any questions, please ask. Hope.

There are several scenarios:

1. Zn and Fe are not excessive, that is, the sulfuric acid in the beaker is not reactive, and the quality of zinc powder decreases faster than Fe, and the balance will first be biased to the side of iron powder, but because the relative atomic silver content of Fe is less than Zn, the amount of Fe is greater than Zn, and more hydrogen is replaced, and finally the balance is biased to the side of zinc powder. None of the above options can be observed in this case.

2. Zn is not too excessive Fe, that is, there is a surplus of iron powder, and the zinc powder just replaces the hydrogen in the sulfuric acid, which can be observed in B and C.

3. Both Zn and Sibozhou Fe are excessive, that is, Fe and Zn are surplus, and all the hydrogen in the sulfuric acid is replaced, and the balance will eventually be balanced, and C can be observed.

a, d are impossible to observe.

First of all, you should understand that there are the same number of hydrogen ions in the solution, so the amount of hydrogen that is reduced in the end is also the same. b should be based on the complete reaction of zinc, because zinc and iron have the same mass, but the mass of zinc of the same amount of matter is greater than that of iron bonds, so the amount of zinc (the muffled ridge is the number of zinc atoms) is small, and when the zinc is completely infiltrated, there is iron left. Got it, dear

Five selections, six selections, and bs need to be explained in detail, which can be asked.

Five's detailed explanation: C2H4 can be ethylene and C4H8 can be Cyclobutane, so it is wrong.

Detailed explanation of six: Explanation: If the symmetry of the alkane structure is the highest, the highest symmetry of the alkane of 6 Cs can only be: CH3-CH(CH3)-CH(CH3)-CH3

If the alkane's symmetry is as low as possible, 6 C's are in the main chain, left and right, 3 C's symmetry, and there are 3 kinds of monochlorogenated compounds in C C C C

If there is 1 CH3 as a branch hall and on position 2c: CH3-CH(CH3)-CH2-CH2-CH3, there are 5 monochlorinated substances

If 1 CH3 is moved to position 3C, there are 4 monochlorinated substances: CH3-CH(CH3)-CH2-CH2-CH3

Categories: Education Science >> Entrance Examination >> College Entrance Examination.

Problem description: Under standard conditions, 100ml of NaOH solution will be passed into the reaction to complete the reaction, then the concentration relationship of the microparticles or commas in the solution is correct ( ).

a．c（na+）＞c（hco3-）＞c（h2co3）＞c（co32-）

b．c（na+）＞c（hco3-）＞c（h+）＞c（oh-）

c．c（na+ ）c（hco3-）+c（co3 2-）+c（h2co3）

d．c（oh-）+c（co3 2-）=c（h+）+c（h2co3）

Please give a reason. Analysis:

The selection and sum just completely generated NaHCO3, which was weakly alkaline. Therefore A is true and B is false.

Judging from the original shirt book sales, na is also for it.

C atom in CO2 = C atom in (HCO3- +CO2- +H2CO3). The concentration of a substance is nothing more than dividing it by a volume at the same time. Of course, the results will not change, they will still be equal.

So C(Na+)C(HCO3-)+C(CO2-)+C(H2CO3) therefore. c The answer is incorrect.

Final NaHCO3 solution.

HCO3- ionized H+ and CO32-; H2CO3 and OH- are electrolyzed

So c(h+)=c(CO3 2-); c（h2co3）=c（oh-)

Thus C(OH-)+C(CO2-)=C(H+)+C(H2CO3).

d The answer is correct.

The main thing in this type of question is to analyze hydrolysis and ionization. As well as grasping something that is certainly equal and unchanging. Basically, this kind of question is very easy.