Chemistry freshman judgment questions. Yes, come in, please 3Q

1, D 2, did you miss the question? BC is right 3, A 4, C 5, D 6, D 7, B 8, and the answer to question C is wrong, except for the C answer, D》B》A

According to A to D, MR decreases by 2, A to E, MR decreases by 4, A has at least two - chain liter OH, and then according to A's MR try to write the chemical formula of A, we can get that A is CH2(OH)CH(OH)CH2OH, D is oxidized by one hydroxyl group, E is oxidized by two hydroxyl groups, B is esterified with 3 parts of acetic acid, and C is nitrification reaction.

To do this kind of organic inference problem, you should pay attention to first inferring the functional groups contained in the organic matter, and boldly guess the possible structural formula, but you still have to pay attention to the details and don't lose points because of low-level errors.

A is ethylene tetraglycol, D is aldehyde, triethanol E is dialdehyde, diethanol, and B is ethylene tetraacetate.

According to the difference in molecular weight between A and the virtual jujube sources B and C, it can be seen that A has 3-OH except for the increase in the differential esterification reaction

The general binding molecular weight of A:CNH2N+2O3 was used to obtain n=3

This question is easy to solve.

1. Let the following masses be: air - m2, ammonia - m3 by the formula "mass = quantity of matter molar mass" and Avogadro's law m3 m2 = 29 17, after changing from air to ammonia, the mass lost is: (29-17) v 22400 = 3v 5600, so the answer is a- 3v 5600

na×v/22400=vna/5600

3. The answer to the first empty is approximate, regardless of the change in the mass of the water after absorbing the gas, the process is as follows:

Quality of water: e v; Ammonia mass: 17 V 22400;Mass fraction: 17 V 22400 (E V) = 17 22400E

Concentration of the substance: V 22400 (V 1000) = 1

1. A + B (black powder) has Cl2 production.

The black powder should be mno2+4HCl (concentrated) ==mnCl2+Cl2+2H2O

A is HCL concentrated.

b is mno2

2, D (potassium salt solution) +1 in Cl2 - E solution + F (dark red solution).

Pushed back by 3: there is KBR+CL2====KCl+BR2 (dissolved in water is orange-red).

So d is KBR and E is KCL

3. Put g (potassium salt solution) +f - d (add agno3 to produce precipitation, the precipitate is insoluble in hno3) + x (add starch solution to turn blue).

It means that X is an I2 elemental, and D must be Br (because F is not Cl2 nor I2 elemental, it can only be Br2, and AGBR is insoluble in Hno3).

ki+ br2===kbr+i2

G is ki and x is i2

4. Add the E solid obtained by evaporation and crystallization of the E solution in 2 to the concentrated H2SO4 to obtain A

KCl+H2SO4 (concentrated) ==K2SO4+2HCl (refractory volatile acids to prepare volatile acids).

To sum up: J is such a happy new year.

A is hydrochloric acid (HCl), B is MNO2, C is Cl2, D is KBR, E is KCL, F is BR2, G is Ki, and X is I2 (iodine).

This question is not troublesome without a diagram, as long as you are familiar with the properties of each substance, it is easy to judge.

1) Because both NO and NO2 are derived from the reduction of nitric acid to obtain electrons, and in the redox reaction, the gain and loss of electrons are conserved, in this problem, only copper is increased by the valence of electrons, because copper becomes copper ions, which increases the 2 valence, and because copper is, it can be assumed that it has gained electrons, so nitric acid also loses electrons, and what does nitric acid become when it loses electrons? There are only no and no2, so let no be and no2 be ymol, and the equation x+y = sum can be listed.

x+3y= solution to get x=, y=, and finally no2= no=.

2) List the equations to solve, according to the equations.

Cu + 4 Hno3 (concentrated) = Cu (No3) 2 + 2 No2 + 2H2O3Cu + 8Hno3 (dilute) = 3 Cu (No3) 2 + 2 No + 4H2o According to the first equation, because No2 is then nitric acid is.

According to the second equation, nitric acid is because no is.

Since the volume is 140ml, C = per liter.

Landlord, please look at my hard typing, my answer!!

1) First of all, Cu must be oxidized to 2-valent Cu ions, and the valency increases by 2 and loses 2 electrons. CU totally. So cu loses electrons in total, and the electrons are n elements, n in Hno3 is +5 valence, no is +2 valence, no2 is +4 valence, total no2 + no=, let no2 = xmol, then no=( According to the previous electrons, we get:

x+( so x=, so no2= no=

2) If the first question is solved, the chemical equation column is solved. It is advisable to do it yourself.

The equation gives you:

The equation for the reaction of metallic copper with concentrated nitric acid and dilute nitric acid is as follows:

Cu + 4 Hno3 (Thick) = Cu (No3) 2 + 2 No2 + 2H2O3Cu + 8Hno3 (Diluted) = 3 Cu (No3) 2 + 2No + 4H2O Pure Handmade, I hope you can understand, and I hope so.

1) Let the volumes of NO and NO2 in the reaction be x, y (according to the conservation of electrons of gain and loss) x+y=, respectively

x/ ×1=(

The solution is x= l y= l

2) According to the fact that there is only one solute nano3 in the solution after the reaction, the amount and concentration of the substance of the original nitric acid is c

Conservation of nitrogen) then 10-3 V a + the concentration of orthonitric acid is: c= (10-3 va +3) From the initial and final state of the reaction, copper loses electrons in the reaction, hydrogen peroxide gains electrons in the reaction, and 30% of the mass of hydrogen peroxide is required to be m (according to the conservation of electrons gained).

Solution: m = g

Cu stands for conservation of electrons: Cu loses electrons.

No gets 3mol electrons and No2 gets 1mol electrons.

So a total of electrons are obtained, and there are 3n(no)+n(no2)= and n(no)+n(no2)= according to the conservation of elements.

The solution is n(no)=,n(no2)=

v(no)=,v(no2)=

Question 1. Pick B

The condition of the question is that a large number of ions can coexist in an acidic colorless transparent solution.

Then H+ is naturally present in the solution

S2- in A can form H2S with H+, H2S is a weak electrolyte and does not ionize easily, so S2- is not present in large quantities in this solution.

Not in group B solution.

Question 2. The sub-outer shell of the electron shell refers to the outermost layer that is reversed, and n represents the number of electron layers.

Question 3. A: An ionic bond is a chemical bond, which is formed by electrostatic action, it is not a force, so it is wrong to say that it is an electrostatic gravitational force.

B: It's a decrease in energy

Sodium chloride is an ionic crystal, so many people think that sodium chloride is reflected in ions in chemical reactions, and when writing chemical equations in middle school, it is indeed written in ions, but in real reactions, chlorine's electron-gaining ability is quite strong, and it can snatch the outermost electron of sodium over, so the reaction between sodium and chlorine can be regarded as atoms in the reaction

The first question is B, because S2- in A will react with H+ in an acidic solution to form hydrogen sulfide.

The second question n represents the number of electronic layers, and the secondary outer layer refers to the second layer from the outside to the inside.

Thirdly, ionic bonding is not an electrostatic attraction.

When sodium reacts with chlorine, it is not an ionic reaction, and there will be ions there.

That is true.

a.Phenol addition H2 may only be saturation of the benzene ring, so there is no doubt that A is cyclohexanol BAfter knowing that there is a hydroxyl group in the functional group of A, and then seeing that the condition is concentrated Hno3, generally speaking, it is a elimination reaction, you may wish to guess that it is cyclohexene first, and then go back and think about whether there are other possibilities if there is a contradiction.

But since the next condition is br2, it is likely that the olefin will be used for the addition reaction.

c.Seeing that the following becomes a diolefin, it must be obtained by removing the substituents from the dibrominated substances, so c is 1,2-dibromocyclohexane.

D-diolefin addition BR2 can be added to one or two parts. But if you look at the next step, you have to add H2, which means that you have to leave an unsaturated key, so br2 must only add one. Diolefins add a BR2-1,4 addition, so get that answer.

e.Nature is to saturate the carbon-carbon double bond. Get the result.

In the end, I went through it and found that it all made sense, indicating that the question was answered correctly.

That's it.

H2 Addition Concentrated H2SO4 eliminates H2O is generally also a bonus, but pay attention to the application of the Marhalanobis rule.

It is the negative group that is added to the side with less hydrogen.

If you listen carefully, you can understand this.

Everybody is from here.

Why are the current questions like this, people are speechless.