How to break the third option of the 12 questions in the second year of high school?

The answer to this question is d.

Options A and B are clearly wrong and are not explained.

Option C: Press C, the volume fraction of C should be reduced. Here's why:

1) It turns out that the volume fraction of c is:

2) After removing c and d, the balance shifts to the right, i.e. c and d are generated again, and the key is to analyze how many c's are generated. Suppose that the new generation becomes a mol of c, then the volume fraction of c is (

Because the amount of ABCD is8 is equilibrium, so they are4 is also in equilibrium, then the operation of option c can be seen as adding and in equilibrium.

Because the conversion rate of the original B is 40%, and the conversion rate of the sum of more additions must be less than 40% (because there is already a product in the system), the amount of newly generated C A is <, so the conversion rate of C (< (that is, the conversion rate of C becomes smaller.

Option D: Suppose that after adding 1 mol of B, the reaction consumes x mol of B, then 3 x mol of A is consumed at the same time, and the conversion rate of A is (3x+ x+, and the conversion rate of B is (, so A is 2 times that of B.

The collected gas hydrogen is: You may wish to assume that there is magnesium in the original alloy and aluminum in the cover nmol. There are two formulas, m+3n2= 24m+27n= solution, m=, n=.

So the mass with magnesium is grams, and the mass of aluminum is grams.

2.The amount of substances used to remove hydrochloric acid is 2m+3n=, so the concentration of raw hydrochloric acid is 3In order to obtain pure magnesium hydroxide, it is necessary to convert all aluminum into metaaluminate, and sodium hydroxide of 2m+4n= is required, and at least sodium hydroxide v=

Answer: The mass of magnesium and aluminum in the alloy is grams, grams, and the concentration of hydrochloric acid is 4mol l, and at least 100ml of 5mol l of NaOH is added to obtain pure magnesium hydroxide precipitation.

The remaining OH-: l*2=, c(oh-)=, c(h-)=1*10-14, pH=13-LG4=

Step 1: First add with hydrogen to turn the benzene ring into a structure of cyclohexane Step 2: Eliminate halogen atoms under the condition of heating NaOH alcohol solution to form carbon-carbon double bondsStep 3: Add with halogen elements to form dihalogenated compounds.

Step 4: Hydrolysis under the condition of heating of NaOH aqueous solution, that is, the product of this kind of univariate substitution becomes binary substitution, which generally needs to go through: first eliminate the ethylene, then add to obtain dihalogenated compounds, and then hydrolyze to obtain the process of binary substitution.

The amount of koh is:

2KOH + CO2 = K2CO3 + H2O, the amount of CO2 produced by butane combustion is, which contains C also.

Butane C4H10, butane containing C is equal to 1 16 mol, which is 1 16 mol of butane burned and released QKJ

That 1mol butane put 16qkj,

Choosing A, it can be obtained from the question, 1mol of butane is burned to produce 4mol of carbon dioxide, and 2mol of koh is required to absorb 1mol of carbon dioxide

After using it, it can be reversed to absorb 1 4 mol of carbon dioxide, that is, the amount of butane is 1 16 mol, and the heat q. is releasedSo 1mol of butane can release 16q.

Select A, process: KOH reacts with CO2 to generate positive salt, K2CO3, there is a dosage of KOH to obtain the generated CO2, 1mol of butane (C4H10) has 4mol of carbon (C), then butane is divided by 4 is equal to, that is, butane combustion releases QKJ heat, according to the proportional relationship, the combustion heat is 16Q

The atoms of benzene are all in one plane, the carbon-carbon double and triple bonds are not rotatable, the ethylene atom is in one plane, the methyl group can be rotated, and up to three atoms are coplanar.

According to the title, in addition to benzene, there can be 3 atomic groups - NH2, -COOH, -CH31, methyl group and carboxyl group together, and there can be 2-amino2-phenylacetic acid; o-aminophenylacetic acid, p-aminophenylacetic acid, m-aminophenylacetic acid.

2. Amino and methyl groups are combined, o-aminomethylbenzoic acid, p-aminomethylbenzoic acid, m-aminomethylbenzoic acid.

3. There are ten kinds of amino groups, methyl groups and carboxyl groups on the benzene ring at the same time.

There are 17 kinds in total, choose D

The description of the title is indeed easy to misunderstand. I guess the meaning of the title is that I want you to write about isomers containing benzene rings, amino groups, and carboxyl groups. I think there are 17 types.

Ammonium bisulfate is a strong acid base, it is obvious that the aqueous solution is acidic, because the solvent water will ionize itself into hydrogen ions and hydroxide ions, so the concentration of hydrogen ions is greater than that of sulfate, and ammonium is a weak alkali in the water will also be partially hydrolyzed into ammonia and hydrogen ions, so it is less than, the hydroxide in the strong acid solution is naturally very little, so the smallest, of course, can be roughly calculated, because the concentration of hydroxide in the water at room temperature multiplied by the concentration of hydrogen ions = ten to the minus fourteenth.

He is a cube, and the two chlorine atoms of the dichloride are respectively located.

1. Both ends of the edge.

2. Face both ends of the diagonal.

3. Both ends of the body diagonal.

Body diagonal, face diagonal, , proximal, a total of three.