Chemistry Contest Questions Please analyze 10 in detail

The larger the pka value, the smaller the ionization equilibrium constant ka, (pka=-logka, you can push out ka=10 to the power of -pka, that is, 1 10 pka power) and the initial concentration is constant, that is, the smaller the concentration of this form of substance. The ionization constant of the carboxyl group in the round is between that of the substituent phosphoric acid mentioned above.

Between the first and second ionization constants, it indicates that this substance is relatively acidic, the third form is under neutral conditions, and the fourth form is under alkaline conditions. The reason why the melting point is as high as 200 should be because of the formation of peptide bonds Co-NH-related, I estimate that the B structure diagram should be the fourth form (alkaline), and the C structure diagram should be the second form (weakly acidic), in addition, for D answer, Ka is the acid ionization constant = [A-]*H+] [Ah], for weak acids, [A-] is a strong base (Lewis acid-base theory), so only convert into a strong salt, by finding its hydrolysis constant (equivalent to the basic ionization constant kb) kh, and then according to kw =ka*kb came to ask ka, I thought it should be like this, I hope it will help LZ, hehe··

Now the questions for high school students are really difficult, and I am a junior in the chemistry department, and I am still very broken when I look at this kind of question.

I read the question and chose answer A, but I didn't expect it to be more than one choice.

The amount of reading in this question is very large, pay attention to the concentration of hydrochloric acid used There are 2 kinds, and, so at the beginning it is actually roughly neutralized, notice, at this time, you have to add a few drops of koh to light red, and then use hydrochloric acid for titration, the phenolphthalein at the beginning is only a rough indicator, and the real endpoint is in v2, which is also described in the title, and this is the titration endpoint, so the data should use v2 instead of v1

The blank experiment is like doing the above experiment without esters, in order to eliminate the interference of other impurities in the water, so the same end point is V4 instead of V2

Finally, the v4 of the blank experiment should be deducted from v2 to ensure that the hydrochloric acid consumed at this time is used for titration.

Remember that there is also a supplementary problem to this question, that is, the ester will contain some acid after being oxidized, so the acid in the ester should be deducted through the corresponding blank experiment.

Saponification is a reaction in which esters are decomposed into organic salts and alcohols under the action of potassium hydroxide (or sodium), and one hydroxide group corresponds to one organic salt. The organic acid is alkaline, and the addition of phenolphthalein is an indicator, and the solution turns red. The addition of hydrochloric acid is to replace weak acid with strong acid, and turn organic salts into organic acids.

Generally, organic acids are almost all weak acids, so a slight excess of hydrochloric acid is its complete transformation, which is the solution becoming colorless. In the addition of sodium hydroxide is to neutralize the excess acid, because when alkaline, phenolphthalein becomes light red (pH> indicating that potassium hydroxide is also slightly excessive, and organic acid is converted into sodium organic acid, the following titration with hydrochloric acid is the real focus, titration to colorless is to neutralize the excess alkali, plus bromophenol blue is an acidic indicator, the discoloration range is pH green, pH is greater than blue, when the solution turns green, that is, pH <4,6. At this time, organic salts are completely converted into organic acids.

v3 and v4 were blank comparison trials.

Choose a, the unit of molar mass is g mol.

a.The chemical formula of melamine is C3H6N6 and the molar mass is 12*3+1*6+14*6=126g mol

The original number ratio of n h is 3:6:6=1:2:2

n The mass ratio of the two elements is 12*3:14*6=36:84=3:7dThe mass fraction of nitrogen is 84 126*100%=

First 2ash3=2as+3h2 is decomposed by heat.

Therefore, the decomposition reaction gas increases, and the increase amount is half of the decomposed ash3. In addition, the amount of gaseous substances in a container is proportional to the pressure.

Now the gas volume increases by 33-25=8kpa, so the ash3 participating in the reaction is 8*2=16kpa

That is, the percentage is 16 25 * 100% = 64% to choose d

Choose d2ash3(g) = 2as(s) + 3h2(g) to start.

End Total. By the gas equation of state pv=nrt

When t,v is constant, p is proportional to n.

Therefore there is a solution x=

I think D should be chosen

First of all, ash3 is decomposed into two elemental elements, and he has only two elements, so it can only be as and h2ash3(g)==2as+3h2(g).

Original n1 = n1 n2 = p1 p2

After the reaction is n2=

Added according to the reaction equation that there are participating reactions.

So this is the topic of the chemistry competition? It should only be a question for the general exam, but the question for the real chemistry competition is much more difficult than this.

In the first question, the phenolphthalein discoloration point is 8, and the pH of the sodium bicarbonate solution is exactly 8, and there is a mutation point.

option a, if there is only sodium carbonate, then titration to sodium bicarbonate with phenolphthalein as an indicator, and then titration to sodium chloride with methyl orange as an indicator, are all 1 to 1 reactions, and the volume consumed twice according to the equation should be equal, which is inconsistent with the title.

Option B, V1 only converts sodium carbonate into sodium bicarbonate, V2 is to convert the reaction of sodium bicarbonate and the original sodium bicarbonate into sodium chloride, obviously the latter consumes a large volume, which does not match the question.

Option C, V1 consumes the total amount of neutralized sodium hydroxide and sodium carbonate, and V2 only converts sodium bicarbonate after sodium carbonate reaction into sodium bicarbonate, so the former obviously consumes a large volume, correct.

Option d, which reacts with sodium hydroxide, has only one mutation point, which does not match the question.

Question 2: Being able to directly titrate indicates that there is a mutation point.

A and B, one is a strong base and a weak salt, and the other is a strong acid and weak alkali salt, which cannot be directly titrated, such as titration of sodium acetate with hydrochloric acid, which will form a buffer solution of sodium acetate and will not have an endpoint mutation.

c hcn, k a (hcn) = , which is weakly alkaline and cannot be titrated directly.

d is a weak acid, and the endpoint is titrated with sodium hydroxide to generate sodium formate, and the endpoint can be indicated with phenolphthalein.

B reacts with NaHCO3 solution to release CO2, then B has a carboxyl group.

After the reaction of C with NaOH , this is due to the formation of hydrochloric acid due to excess HCl and hydrolysate.

After neutralizing the acid, HNO2 reacts to a yellow oil, which reacts with benzenesulfonyl chloride to form an alkali-insoluble precipitate, which indicates that it is a secondary amine after neutralization.

That is, there is an H on N, and the aromatic substance is obtained by reacting with CH3CL, then there is a benzene ring, and C is only 7 C's in total, then C is [C6H5-NH2-CH3]Cl and NaOH is C6H5NHCH3 after the reaction

The unsaturation of b is 5, and there is a carboxyl group, then there are 4 unsaturations left, which may be a benzene ring.

Then b can be cl-c6h5-cooh with inter-adjacent pairs in three cases.

A is the amide formed by B and C6H5NHCH3. There are three kinds of B, and there are three kinds of A.

Select D, MXOY, the solid mass is reduced by about 50% after the reaction is complete, and the solids are M after the reaction, so the reduced mass is the mass of the oxygen element in the original MXOY, so the mass of the oxygen element in the original MXOY is about 50%, that is, the relative molecular mass of M in MXOY The relative molecular mass of O.

Bringing D,M2O7 in, the relative molecular mass of M is 110, and the relative molecular mass of O is 112, which is the most similar.

After the D reaction, only M remains, and the mass of the solid decreases by 50%, indicating that the mass fraction of M in the oxide is close to 50 percent, and the D option is the closest. abcd

d option is the closest.

After the reaction, only m remains, and the mass of the solid decreases by 50% 55x=16y x y=2 7