A question in senior one chemistry compulsory 2 , please write the specific solution process, thank

3 types. Fill padding ch3

Fill fill I1 ch2cl-c-ch2-ch3 fill fill i fill fill ch3

Fill padding ch3

Fill fill I fill fill 2CH3-C-CCLH-CH3 fill fill fill I fill fill CH3

Fill with CH3

Fill I3 CH3-C-CH2-CH2Cl fill I fill CH3

There are 3 equivalent hydrogen in 2,2-dimethylbutane. So there are 3 types of monochlorides.

CH3-C(CH3)2-CH2-CH3 The hydrogen on the first carbon and the hydrogen in parentheses are equivalent hydrogen and are structurally symmetrical. The hydrogen on the first bomb after the parentheses is a category. The last bullet is also of a class. So there are a total of 3 monochlorine substitutes.

They were: ch2cl-c(ch3)2-ch2-ch3ch3-c(ch3)2-chcl-ch3

ch3-c(ch3)2-ch2-ch2cl

2,2-dimethylbutane CH3-C(CH3)2-CH2-CH3 and chlorine gas are replaced by monochlorine under light conditions, and three products are formed.

The basis for judging the type of monochlorinated alkane is to find the type of hydrogen atom in the alkane, 2,2-dimethylbutane CH3-C(CH3)2-CH2-CH3 has three hydrogen atoms, so monochlorosubstitution is three.

ch2cl-c(ch3)2-ch2-ch3ch3-c(ch3)2-chcl-ch3

ch3-c(ch3)2-ch2-ch2cl

The amount of sulfuric acid substances is:

Note: One mol of any substance contains particles of this substance na) contained as Avogadro's constant = sulfuric acid molecules, i.e. sulfuric acid molecules.

Since each sulfuric acid molecule contains 2 h atoms, it contains h atoms, i.e., h atoms.

Since each sulfuric acid molecule contains 4 O atoms, it contains O atoms;

NH3 contains the number of hydrogen atoms as:

Because the relative molecular mass of NH3 is 14+3=17

So the amount of matter is.

Each NH3 contains 3 hydrogen atoms, and the H atom is, i.e., H atoms.

If not, you won't be able to contact me.

Hope it works.

Armor. C2H6 B. N2H4 C. H2O2 butin. HCL Answer AIt should be located in the third cycle, group VIIA.

b. ac.The first half of the sentence is true, they both contain polar and non-polar bonds, but H2O2 is the polar molecule DCorrect, such as CH4O methanol, is also an 18-electron compound.

Upstairs, you guys do the questions carefully, don't mistake people's children, Ding can still be HF, HF has a total of 10 electrons, and the others have 18 electrons.

The least dense is H2, so A is the H element.

The hardest is diamond, so B is element C.

The outermost electrons of d are three times that of the subouter shell, so it is an element O and because B, C, D, and E can all form a molecule with an equal number of electrons with A, so C is an element of N and E is an element of F.

It can be known that A is ethane, B is hydrazine, C is hydrogen peroxide, and D is hydrogen fluoride A is fluorine, so it is correct.

b atomic radius b is greater than c than d is greater than a

cHydrogen peroxide is a polar molecule.

d cannot form compounds equal to ethane electron number.

By the question: ABCD are: h c n o f

a. Correct (fluorine).

b. Error (decreasing in the same period).

c. Error (hydrogen peroxide is a polar molecule).

d. (A is ethane CH).

1. First use dilute nitric acid to test carbonate ions, then use barium nitrate to test sulfate, and finally use silver nitrate to test chloride ions.

2、al3+、no3-、cl-、

Sulfur ions, carbonate and water undergo double hydrolysis to produce carbon dioxide and hydroxide aluminum.

3. Students, you must master the chemical equation by yourself: sodium and water to produce sodium hydroxide and hydrogen, aluminum and sodium hydroxide to produce sodium metaaluminate and hydrogen, and iron and water vapor to react to produce iron hydroxide and hydrogen.

4. Both.

5、c6、mno4-7、c

AL is put into a colorless solution to produce hydrogen, which may be an acid or a base.

2AL+2NAOH+3H2O=2NAALO2+3H2 (gas)2AL+6HCl=2ALCl3+3H2 (gas)A Hydrogen ions, calcium ions, sodium ions, nitrate ions Nitric acid is used as a strong oxidizing agent, and aluminum is oxidized to form a dense oxide film, and the reaction is stopped, and no hydrogen is generated.

b Sodium, magnesium, chloride, aluminum, no acid and no alkali.

c Copper ions, magnesium ions, chloride ions, hydroxide ions Combination ions have precipitation formation, which is not good.

d Barium ions, chloride ions, hydroxide ions, nitrate ions There is alkali barium hydroxide. Pick D

Well, I forgot about D for a long time

Select D is colorless, so it does not contain copper ions, excludes nitric acid in C A, reacts with aluminum, does not produce hydrogen, and B does not react.

A is not the outermost layer, the nth layer can hold up to 2n*n electrons, and the m layer is the third layer, so it can hold up to 18 electrons, which is correct.

b Except for the first shell, the outermost shell of the other layers is to hold up to 8 electrons, this statement is correct c The outer electron energy is lower than the inner shell electron, the m layer is the third layer, the L layer is the second layer, the second layer is lower than the third layer, and the L layer is lower than the M layer, this statement is wrong.

Both D and Mg contain electrons in the M layer, the electronic configuration of Na is 281, and the electronic configuration of Mg is 282, and the M layer, which is the third layer, contains different e-books, which is correct.

The final answer is C

According to: the outermost layer does not exceed 8e, and the subouter layer does not exceed 18e, so a pair.

The further outward, that is, the farther away from the nucleus, the higher the energy, so C is wrong.

Different atoms have different electron numbers, so the m-layer is different.

Because the addition of KOH to the original solution just dissolves the precipitate completely, indicating that OH- is completely reacted!

Reaction formula: Al3+ +4OH-===ALO2- +2H2O We know that Al3+ reacts completely with OH - at 4:1 and dissolves! So if the added KOH is, the original solution then Al3+ is.

Then in the original solution, according to the conservation of charge, the total charge concentration of the cations and cations is equal, there is 2·c(so4 2-)=c(k+)+3c(al3+)- the coefficient in front is the number of charges carried by each ion.

Substituting value: 2·

The original solution C(K+) = is obtained

Because the same volume of koh is added, there is c(k+)=c(k+) total =(

Since the volume of the solution has been doubled, the concentration is halved).

Hope it helps!

If you can understand, give it an adoption! Hee-hee.

The resulting precipitate happens to be completely dissolved, indicating that Al3+ is completely converted into AlO2- assuming that the original solution volume is 1L, and the reaction is 2L, the amount of K+ species in the solution after the reaction is X, and the amount of AlO2- is Y

Conservation of solution charge before reaction:

SO4 2- The amount of substance, the new KOH is.

c(k+)+3c(al3+)=2c(so4 2-)(1)

Conservation of solution charge after reaction:

c(k+)=2c(so4 2-)+c(alo2-)x= (2)

Find the system of equations to get:

x=,y=so the concentration of k+ c(k+)=

According to the reaction of 10FEO-2kmNO4, the amount of potassium permanganate substance is, the consumption of NFeO=, and the mass is grams,