When Na3 Co NO2 6 reacts with potassium salt, what is the effect if the solution is more alkaline?

In this case, if the alkalinity is too strong, the reaction will be complete, and some alkaline substances will remain.

Because if the alkalinity of the solution is relatively strong during the reaction of the fish pseudo-salt in his case, then the big impact is that it may cause a comprehensive reaction of acid and alkali, so it will affect its test data.

Halogenated hydrocarbons react with NaOH to form alcohols or alkenes.

Phenol reacts with NaOH to form sodium phenol.

Ethanol reacts with sodium metal to form sodium ethanol.

Carboxylic acids can react with NaOH and sodium carbonate (hydrocarbonate) to form carboxylate and water (and carbon dioxide can also be formed by reacting with carbonate (hydrocarbon)).

The ester is hydrolyzed under NaOH catalysis to produce the corresponding alcohols and carboxylates.

R-OH and R-O-R are chemically stable and do not react with alkalis.

Hey three, CO26 reacts weakly with false salts, and is easy to be alkaline, so he will switch to acidity.

Acidic acid: sulfuric acid, hydrochloric acid, nitric acid, acetic acid (acetic acid), carbonate and other salts: ammonium chloride, silver nitrate and other alkaline bases

Ammonia, sodium hydroxide, calcium hydroxide (lime water), potassium hydroxide and other salts: sodium carbonate and other things are very basic things that need to be summarized by yourself, and no one can replace them.

What about when you react with fake salt? If it is stronger, it may affect one of his deteriorations.

The key is to understand that the increased weight comes from **, cao+h2o=ca(oh)2, nahco3+naoh=na2co3+h2o, ca(oh)2+na2co3=caco3+2naoh, the first reaction to gain weight is a molecule of water, the second reaction is to lose weight of a piece of water, and the third reaction has no weight gain or decrease. In the end, the weight gain of the substance can only be more than the weight gain of the first reaction compared to the weight loss of the second reaction. Grams of water are moles, indicating that CAO is more moles than NaHCO3, note that the NaCO3 that reacts with Ca(OH)2 is composed of two parts, that is, the number of moles of the original NaCO3 plus the number of moles generated by the second reaction, so it can be known that the moles of the first reaction are more than the second reaction is exactly the molar number of the original NaCO3.

i.e. molar NaCO3 is present in the original mixture.

The answer is b.

2naclo2=naclo3+naclo

3naclo=naclo3+2nacl

Given that Naclo cannot exist in a hot base in the stem, a, c are wrong.

CL in Naclo2 is +3 valence. A disproportionation reaction occurs in a hot base, and the Cl in the product is higher than +3 valence. One is below +3.

d in +5 and +7 respectively, contradictory.

There are two reactions that occur:

naclo2=naclo3+naclo

NaClo = NaClO3 + NaCl (untrimmed) so the final product is NaCl as well as NaClO3

So choose B.

3naclo2＝nacl+2naclo3

The products are NaCl and Naclo3, where Cl is +1 valence, and the given NaCl cannot exist in a hot alkali solution, so B is chosen

B, the valence of Cl rises and falls.

Na2CO3 is alkaline, and there is a more complicated problem here, I don't know if you have learned it.

First of all, according to the principle of hydrolysis, the weaker the more hydrolyzed, but here you can't directly compare HCO3 and H2CO3, because the acid corresponding to Na2CO3 is HCO3-, and the acidic order is H2CO3>HCO>HCO3-, HCO3- is the weakest, so it corresponds to the salt Na2CO3 is the most alkaline, there are Na2CO3>Naclo>NaHCO3, this problem is a bit difficult for those who have just come into contact, if you don't understand, remember it, It's better to ask the teacher directly.

Sodium carbonate is weaker than hypochlorous acid because carbonic acid is less acidic.

Sodium bicarbonate is completely ionized in an aqueous solution to give sodium ions and bicarbonate ions.

HCO3- can be ionized and hydrolyzed.

The ionization process is: HCO3-=H++CO32- (reversible) hydrolysis reaction. HCO3-) H2O=H2CO3+(OH-) (reversible) hydrolysis intensity is greater than ionization intensity, and OH- makes the solution alkaline. So the solution is alkaline.

Na2SO3 is a strong electrolyte, all ionized, Na2SO3 = 2Na+ +SO32-

Na2SO3 can only be hydrolyzed and can only be alkaline.

SO32- +H2O HSO3- +OH-HSO3- +H2O H2SO3 + OH-So, sodium sulfite is alkaline.

This question should be analyzed as follows:

Add HA solution to Na2CO3 solution or release gas in liters", launch acidic: HA>H2CO3;

The "NH4A solution is neutral", using "strong acid and strong alkali salt, weak acid and weak alkali salt are neutral, strong acid and weak alkali salt are acidic, and strong alkali and weak salt are alkaline", combined with the above analysis, (NH4)2CO3 is alkaline.

Take phenolphthalein. As an indicator, only NA2CO3 can be changed to NAHCo3, with methyl orange.

to change it to naclSo, sodium hydroxide.

The acid consumed under both indicators is the same, and Na2CO3 is twice as acid consumed with methyl orange as an indicator with phenolphthalein. NaHCO3 reacts only with methyl orange as an indicator. (Phenolphthalein discoloration does not allow NAHCO3 to continue the reaction, methyl orange can).

The composition is only naoh, v1 = v2Only Na2CO3, V2= can also be NAOH: NAHC3=1:1