Questions about high school chemistry Questions about high school chemistry

It doesn't seem to be a contradiction. The structure of the electron layer is the same, which means that it is the same three-layer and the same four-layer ......For example, Na and Cl, the radius of Cl is much smaller.

Are you wrong with the same cycle and the same main clan?

That chemistry problem is the number of homo-electron layers, and since it is an ion, then the number of electrons. Then the higher the positive valence state, the more electrons are lost, then the number of nuclear charges it has is larger, and the radius is smaller. For the negative valence, the larger the absolute value of the negative valence, the more electrons are obtained, then the smaller the original number of nuclear charges, the smaller the binding on the electrons, and the larger the radius.

Personal opinion. So it's +(n+1)<+n<-n<(n-1)-.

1) For ions with the same electron shell structure, the larger the number of nuclear charges, the smaller the radius.

2) Elements of the same main group, whether metallic or non-metallic, whether atomic radius or ionic radius, increase from top to bottom.

These two sentences do not contradict each other. Because the electron shell structure of atoms or ions of the same main group elements is different. The structure of the electron shell is the same, and the number of electrons on each layer is the same, for example, the structure of Na+, Mg2+, Al3+ has two electron shells, and the number of electrons is respectively

and a chemistry question

The four ions of x, y, z, and r have the same electron shell structure, and their valence states are n+, (n+1)+, n, (n-1)-., respectivelyThen the order of these four ionic radii from large to small is.

According to x, y, z, r are ions with the same structure of the electron shell, and according to the number of charges they carry, it can be judged that their nuclear charge number relationship is: zr>x>y.

Elemental bromine reacts with NaOH to form a NABR solution, which is separated from upper and lower bromobenzene (bromobenzene in the lower layer). Bromobenzene is an organic substance, and the Nabr and Nabro solutions are inorganic substances with chemical structures that are not similar to each other and incompatible with each other.

SO2 is introduced into BA(OH)2, and SO2 reacts with water to form H2SO3, and then reacts with barium hydroxide to form BASO3 + H2O. The concentration of hydrogen ions increases during the reaction from an alkaline solution to a neutral solution.

2NaOH + BR2 = NaBR + NABR + H2O (NaBR, Nabro are soluble in water, while bromobenzene is insoluble in water, so stratified and NaOH solution does not react with bromobenzene).

The stronger the ability of non-metals to gain electrons, it means that the ability to attract electrons is stronger, so that when non-metal atoms get electrons, the chemical bonds formed are stronger, and the energy of chemical bonds is relatively low, and more energy will be released (the lower the energy of the compound, the more stable it is).

The stronger the electron loss ability of a metal element, the weaker the ability to attract electrons, and the easier it is to lose electrons, so less energy is required. Seek adoption.

The process of synthesis is a process of releasing energy, and the process of decomposition is a process of absorbing energy. The electrons are absorbed and the electrons are separated.

Because the ionization of electrons is carried out in the first stage (the loss of the first is called first-stage ionization, and the loss of the second is called second-stage ionization), generally the energy consumed by the first-stage ionization is lower than that of the second-level ionization, because the electroneutral atom is positively charged after losing one electron, and the attraction ability to electrons is enhanced, so it is not easy to lose the second. The metallic elements involved in high school are generally outer electrons with very few electrons, no more than 3 electrons. Therefore, it is easy to lose electrons to reach the noble gas-type full-shell structure of the previous element.

Non-metals, on the other hand, tend to get electrons to achieve a full-shell structure because they have too many electrons and are not easy to lose all of them.

The more energy released by the electrons of the non-metallic elements, the lower their own energy will become, that is, the tighter the binding force with the electrons, and the electrons are not easy to lose because they have to absorb so much energy released by them, and the greater the energy, the more difficult it is to obtain, so it is more difficult to lose again. That is, non-metals have strong electron ability.

For metals, volatile electrons, electron loss absorbs energy, the less it is absorbed, the easier it is for environmental conditions to meet this small energy demand, so it is easy to lose, so it is said that the electronic ability is strong, and if the energy absorbed is a lot, the environmental requirements are more harsh, it is not easy to meet the energy demand, it is not easy to lose electrons, so it is said that the electronic ability is poor.

1.(Select) The following statements about alkanes are false: c

In the alkane molecule, all the chemical bonds are single bond --- pairs, C-C single bond and C-H single bond.

b All alkanes have substitution reactions with Cl2 (chlorine) under light conditions--- photohalogen substitution is a characteristic reaction of alkanes.

The general formula of the C alkane molecule is CNH2N+2, which is not necessarily the --- of alkanes, but the alkanes.

With the increase of the number of carbon atoms, the melting and boiling point of alkanes gradually increases--- pairs, and the number of c atoms increases, the molecular weight increases, and the melting and boiling point increases.

2.Should concentrated HCl be heated for the reaction with mnO2 (chlorine production)? --To heat.

A correct alkane is a saturated hydrocarbon, and all bonds are saturated single bonds.

b Correct substitution reaction is a characteristic reaction of alkanes.

C False The general formula of the alkane molecule is CNH2N+2, and it must be an alkane that conforms to this general formula.

d error. The boiling point cannot be said together.

For the boiling point, the more branched chains, the greater the molecular distance due to steric hindrance, the weak intermolecular force, the low boiling point, and the small density;

However, for the same number of branches, the more symmetrical the molecules, the tighter the molecular binding, the stronger the intermolecular force, the higher the boiling point and the higher the density. For the melting point, the more symmetrical the branched chain, the tighter the molecular bonding, the stronger the intermolecular force, the higher the boiling point, and the higher the density. Want.

1b only reacts with the gas alkane light.

2 Heat.

1 Option C I believe that no one can give a counterexample Isoalkanes are also alkanes.

2 Heating is required.

1 Choose c, because the carbon atom is already in the tetravalent saturation structure at this time, and it is impossible to form other functional groups.

2 Heating is required, which is not a question of how fast or slow the reaction is, but of the reaction mechanism.

1 should be choose B, I did it by the elimination method, the melting point of n-alkanes, the same series C1-C3 is not so regular, but C4 above is increased with the increase of the number of carbon atoms. To be heated.

Add excess acid to the Naalo2 solution and the ionic equation is.

AlO2- 4H+ Al3+ 2H2O add a small amount of acid to the Naalo2 solution, the ionic equation is.

AlO2- H+ H2O Al(OH)3 state nucleus.

CO2 leakage and merger: CO2 + 2alo2- +3H2O ==2Al(OH)3 (precipitation) +CO3 2-

CO2 excess trace: CO2 + AlO2- +2H2O ==Al(OH)3 (precipitation) +HCO3

First, it's alo-

Second, alo- 4H+ Al3+ 2H2O [excess acid].

Let's start with the problem of ion coexistence, if it's an alkaline solution, it means that it contains OH- then H+, HCO3- and so on don't exist, in fact, as long as it's reactive with OH, it doesn't exist, and the rest is the same as in a neutral solution. The same goes for acidic solutions, which means that the solution contains H+, so as long as it reacts with H+, it does not exist.

In fact, you can just consider that when CO2 is introduced into NaOH, the reaction starts with 2NaOH + CO2 ===Na2CO3, and when the NaOH reaction is gone, CO2 is introduced again, and then Na2CO3 + CO2 + H20 = 2NaHCO3 occurs

The Ca(Oh)2+CO2 you said only occurs Ca(Oh)2+CO2 =CaCO3+H2O and the reaction concentration has nothing to do with it.

Ions that can undergo ionic reactions cannot coexist in large quantities in solution.

Acidic conditions can not coexist: 1) carbonate ions, sulfite ions, and weak acid acid acids, these are the reaction of long-lived gases. 2) Hydroxide ions, fluoride ions, acetate ions, hypochlorous acid and ions are long-lasting and difficult to ionize substances.

3) Silicate ions.

After hydrolysis, no precipitate is formed, and gases, water ions can coexist in large quantities. For example, H+ and OH- together form water, so they cannot coexist. N-divalent barium ions and sulfate ions will form barium sulfate precipitates, which cannot coexist.

A small amount of carbon dioxide in water will form carbonate ions, and an excess of carbon dioxide will form bicarbonate ions, and the two ions will combine with calcium ions at different concentrations to form calcium carbonate.

1. No, Na2O2 is a peroxide, and Ko2 and the like are called superoxides, and they are not basic oxides.

2,A only shows oxidation. b only shows acidity, if it is ferrous oxide, it is in line with the topic. c is only acidic. d is in line with the topic.

3. Because ammonia is a weak base, it cannot dissolve aluminum hydroxide, cannot generate metaaluminate ions, and naturally cannot be clarified.

1。No, it's peroxide, and sodium oxide is a basic oxide.

Dilute nitric acid produces NO, and concentrated nitric acid produces NO23. Not.

1. Sodium peroxide belongs to peroxide and is not classified as alkaline oxide.

2, d, the reaction produces copper nitrate, water and nitrogen oxides, which exhibit oxidation and acidity.

3. No, you can't.

1. I don't remember much in high school, it shouldn't be because sodium peroxide is an alkaline substance, but it is not an oxide.

2. When reacting with copper.

3. Although aluminum hydroxide is an amphoteric oxide, the alkalinity of ammonia is not strong enough, so it cannot be clarified.

a, o b, na c, al d, si e, s f, cl1) naf (write the electronic formula yourself, na is an alkali metal, and one electron is lost.

f is a halogen, to get an electron, both become 2 electron shells, forming ionic compounds) 2) Al2O3 +2Oh- =2alo2-+H2OA, E is the same main group, the number of electrons outside the nucleus of the atom of element E is twice the number of protons in the nucleus of element A, let the number of protons of A be X, then the number of protons of E is X+8 or X+18, if 2X=X+8, then X=8 is the oxygen element, then E is S. (If 2x=x+18, then x=18, which is not in place).

The number of outermost electrons of an atom of element d is half of the number of electrons in the subouter shell. , we can know that d is si, then b and c are na, al (the sum of the outermost electrons of b and c is equal to the outermost electrons of d d).

f is cl

I want to vomit when I did this kind of question in high school, and tell you what elements each ABCDEF represents should be enough for you to answer the question, right?

A: Oxygen B: Sodium C: Aluminum D: Silicon E: Sulfur F: Chlorine.

The key to solving the problem lies in the last condition given, "the outermost electron number of the atom of element d is half of the number of electrons in the secondary outer shell", which can determine that d is li or si, and then according to the "sum of the outermost electrons of the atoms of b and c is equal to the outermost electron number of the atom of d d", the possibility of li is excluded, and d is determined to be si. Knowing that d is si, the following is just fine. In addition, you can guess the answer based on the questions given.

For example, in question 3, if you do more questions, you can see that d and e are likely to be non-metallic elements, and because the atomic number of abcdef increases sequentially, you can immediately lock the range to the non-metallic elements in the third row of the periodic table.

1) Naf2)Al2O3 +2OH- =2alo2-+H2OA, E is the same main group, the number of electrons outside the nucleus of the atom of element E is twice the number of protons in the nucleus of element A, let the number of protons of A be X, then the number of protons of E is X+8 or X+18, if 2X=X+8, then X=8 is the oxygen element, then E is S. (If 2x=x+18, then x=18, which is not in place).

The number of outermost electrons of an atom of element d is half of the number of electrons in the subouter shell. , we can know that d is si, then b and c are na, al (the sum of the outermost electrons of b and c is equal to the outermost electrons of d d).

f is cl

A is oxygen. b is sodium.

c is aluminum. d is silicon.

e is sulfur. f is chlorine.

The question should be fine.