200 additional points in high school chemistry!!!!!!!!!!!!!!!!!

There are 2 types of monochlorine substitution and 5 types of dichloride substitution.

The molecule is centrally symmetrical, and monochloride substitution is only two types of substitution for 2 nodes and 3 nodes for carbon atoms.

Dichloride substitution should be seen in the six-membered ring, and one substituent is placed on 3 nodes, which are three substitutions between neighbors.

The substituent group is placed on the 2 node, and the second substituent group has only one intermediate position, and the adjacent and pairs are the same as the upper one. Then there is the second substituent group that is placed on the other six-membered rings, and there is only one. So there are two substitutions of substituents on 2-node carbon atoms.

There are 5 types in total.

You can think of it as 3 six-membered rings on the side and a six-membered ring on the bottom, and the 3 six-membered rings on the side are equivalent and symmetrical, so just look at a six-membered ring, there are 4 chlorine substitutes in a side six-membered ring, and the six-membered ring on the bottom side happens to be not considered, because all of them are already included in the side six-membered ring.

As for dichloride substitutes, you can assume that 2 chlorines are in the same six-membered ring (note that some carbon atoms can only replace one chlorine atom, and some can replace 2), and there are 8 kinds in total. Then consider that two chlorine atoms are in different six-membered rings, which is when you find that if the chlorine atoms are not in the lower top corner of the six-membered ring, they are actually in the same six-membered ring, so they are not in the same six-membered ring, as long as they are considered in the lower top angle, there are 2 kinds.

There are 10 types in total.

Set socks to sell x reacted a

2x + 2y ==z + 2w

a a a/2 a

2-a 7-a a/2 a

2-A + 7-A + A 2 + A = A= When opened, a and b are connected, and the pressure is equal to the outside atmospheric pressure, which is of course equivalent to isothermal isobaria.

3.In this way, it is equivalent to the following two equilibrations in two isothermal containers, and the X conversion rate is the same when the two finally reach equilibrium.

2x + 2y ==z + orange 2w

2x + 2y ==z + 2w

Using the above method, we can tease to find the total amount of each substance in equilibrium after opening, compared with the equilibrium state in the original B, which is better than 2+7 instead of all, because A is not equivalent to B and K before K is opened, and the pressure is different.

4 methyl and ethyl groups are respectively numbered, then the chlorine atom can replace the hydrogen atom on the bit;

4 methyl and ethyl groups are respectively numbered, then the chlorine atom can replace the hydrogen atom on the bit;

2 methyl and ethyl groups are respectively numbered, then the chlorine atom can replace the hydrogen atom on the bit;

directly offered a reward of 200 points, and the gods of affirmation rushed to answer.

Your answer is incorrect, because the volume of the three clusters of atoms in themselves cannot be adjacent positions.

In this problem, first determine that the unsaturation is 3, that is, there are 3 double bonds or a double bond and a triple bond.

Then think about how to build the carbon chain, determine that there are 6 carbons left after the central carbon atom can be allocated, in order to maintain chirality, the number of carbons in the carbon chain linked by the four bonds of the central carbon atom can only be 0,1,2,3 (because after hydrogenation, it is still chiral, and hydrogenation will be fully saturated, so for chirality, only the number of carbons in different positions can be different, this is only applicable to this question, if the number of carbons is more, it should be considered, isomerism of the same carbon number hydrocarbon group), that is, the carbon chain in the example.

Then assign the unsaturated bonds, and find that there are no three double bonds (the same carbon cannot be connected with two double bonds), so it is a double bond and a triple bond, and then put it on the top, there are four ways to put it (two kinds of three bonds on the double carbon chain, that is, the two given by the title, and two kinds of three bonds on the three carbon chain), and the fifth? At this time, it is necessary to consider that the carbon chain of the three carbons is isomeric, that is, there is an isopropyl group, so there is a fifth type.

2.This is a saturated hydrocarbon alcohol, four carbons have a total of two carbon chains, the basic linear carbon chain, and one carbon hangs three carbons, and the two carbon chains have four hydroxyl hangs, namely butanol, isobutanol, tert-butanol, and 2-methylpropanol. And then is this an ether-forming reaction or an elimination reaction?

If all are counted, the elimination reaction produces a total of 3 olefins, namely: 1-butene, 2-butene, and 2-methylpropylene. In the case of ether formation reaction, any two alcohols can be ether-formed, and the permutations and combinations of 4+3+2+1=10 kinds, there are a total of 13 kinds of products.

3.Hydrogen bromide is soluble in water, and hydrogen bromide is a polar molecule and should be soluble in water. I can't remember if it's soluble in organic solvents, and I forgot about it for many years after graduation

Amino groups are easily oxidized, so they should be protected in nitro form and reduced back in the last step.

For the transformation of the organic diagram, the entire number of the benzene ring and the position of the substituents are written down according to the number. In the future, practice makes perfect.

The rotation of the bond involves the structure of the matter, the angle of the benzene bond is 120, and the bond between the two benzenes is also 120. Since the configuration of benzene is planar, you can get the four carbon collinearity that cross it, that is, no matter how you turn these four carbons, they will always be coplanar. After determining the coplanar carbon on the benzene ring, even the carbon of the modulo substituents on it is also coplanar.

11 carbons.

Figure 1 first verifies the water in the product, and if the product passes through the solution first, it is not possible to determine whether the water has formed or not. Big in and small out I-H

In addition, the balance depends on the reaction of other substances generated with the solution. Magenta can combine with sulfur dioxide to form an unstable colorless substance, and magenta fades; Acidic potassium permanganate oxidizes sulfur dioxide gas to remove sulfur dioxide; Neither of these reagents reacts with carbon dioxide; Both sulphur dioxide and carbon dioxide react with lime water to form a white precipitate.

Therefore, the sulfur dioxide should be verified first, magenta is introduced, the magenta fades, then potassium permanganate is used to remove the remaining sulfur dioxide, and finally the remaining gas is passed into lime water. Long in short out, c-b-g-f-d-e.

The first one: choose wrong, ag=lg([h+] [oh-])=lg1=0, b wrong, ag 0, c wrong, ag 0

The second one: choose the right one, generate strong acid and weak alkali salt, the solution is acidic, B pair, because the two happen to be completely reacted, C pair, the ammonia concentration of pH 11 is, and the degree of ionization is 1%.

The third sheet: the mass of the selected liter of water = 1000 grams, the amount of the substance = 1000 18 = attached:

AG is actually equal to the POH-pH=(14-pH)-pH=2(7-pH)NH3 ionization equilibrium constant.

c(oh-)=10 -3, c(nh4+)= original c(nh3)= so the nitrogen ion is about c(nh4+) c(nh3) = water has 18g, volume.

c(h+)=10 -7mol l, n(h+)=10 -7*1mol water is ionized, and how many water molecules are ionized into one hydrogen, and there is na about up and down in the formula.

1mol of water is ionized, 10ml is 10 18* for one.

<> as shown in the figure above: the structure has two axes of symmetry, A and B, with a total of 8 H atoms.

Rotate 180 degrees around the A axis: h6 h7 coincides, h2 h3 coincides, h8 h5 coincides, h4 h1 coincides.

Rotate 180 degrees around the b axis: h6 h3 coincides, h7 h2 coincides, h8 h1 coincides, h4 h5 coincides.

That is, the description is: h1 h4 h5 h8 as a group can be coincident by rotation; h2, h3, h6, h7 are Li Yuntong, and the other group can be coincident by rotation.

By rotation, the structure is the same as before rotation, which means that the chemical positions of the H atoms that can coincide are quietly the same, which means that they themselves are the same. So h1 h4 h5 h8 is one group, h2 h3 h6 h7 is another group, and there are two different hydrogen genogens.

The benzene ring is a regular hexagonal shape with no long or short sides, and you can draw it.

You haven't given me any additional points for the physics questions in front of you, so you can't be a man without keeping your promises.

Because there is also bicarbonate involved in the reaction (complete), the ions are conserved. The remaining hydroxide group does not react, but forms sodium hydroxide with sodium ions.

A small amount of sodium bicarbonate reacts with calcium hydroxide to form water, sodium hydroxide, and calcium carbonate.

Sodium hydroxide can be ionized, and aluminum hydroxide is a precipitate, so it is not about in the ion equation.