Chemistry topics in the first year of high school. Urgent. Chemistry questions in the first year of

2naoh+so2=na2so3+h2o naoh+so2=nahso3

The composition is sodium bisulfite = y, sodium sulfite = x

2x+y=x+y=

x=y=

Solution: Naoh:

so2＋2naoh＝na2so3＋h2o

SO2 as well.

Na2SO3 SO2 H2O 2NAHso3NA2SO3 and:

Therefore, the components of the solute in the obtained solution are NaHSO3 and Na2SO3, and the amounts of substances are respectively.

The quantity ratio of their substances is 5:4, and the reaction equation is written: 5NaOH + 4SO2 = = Na2SO3 + 3NaHSO3 + H2O, then sodium sulfite is, sodium bisulfite is.

In fact, you can set sodium sulfite xmol and sodium bisulfite ymol, which can be solved according to the conservation of sodium and sulfur.

2naoh+so2=+h2o naoh+so2=nahso3

Let Na2SO3 be XMOL and NaHSO3 be ymol, and use the conservation of sodium atoms and sulfur atoms to set up a binary equation 2x+y=, x+y=, and solve the equation to obtain the answer.

Analysis: Before adding water, it was MGO and MG3N2. And the later one is MGO. The difference between the difference between the two is the difference between mg2n3 and its corresponding mgo.

mg3n2→3mgo △m

100 20 i.e. m(mg3n2) = =

So the quality score.

Mg3N2 - 3MGO - difference.

100 120 20 Weight gain before and after the response.

So n(mg3n2)=

m（mg3n2）= 100 =

So the quality score.

We are in dire need of wealth, thank you).

The best way to understand is here!!

Let mgo be xmol and mg3n2 be y mol and start with the total mass of 40x+100y=

Later, Mg3N2 is converted to 3MGO Total mass 40x+3*40y= It can be seen that Mg3N2 is m(mg3N2)= 100 =

So the quality score.

Actually, the difference above is pretty good, but mine understands it better.

Sodium Aluminum Hydroxide Reaction:

2al2naoh

2h2o==2naalo2

3H2 Aluminum and Hydrochloric Acid Reaction:

2al6hcl==2alcl3

The 3H2 reaction produces hydrogen, which consumes more than 2NaOH

6hcl=1:3

This is very simple, as long as the chemical equation is written, the trim can be obtained, it should be HCl:NaOH=3:1

1mol Mg3N2 is converted into 3MGO, the mass is increased by 20g, and the weight is increased before and after the reaction, then the Mg3N2 in the original mixture is, that is, so the mass fraction of magnesium nitride in the mixture obtained by combustion is, i.e.

122-116=6g

44-32=12g

116-114=2g

2/32-4=28g

The relative molecular mass of this gas is (28g).

100 moles.

Amount of concentration = mass fraction =

The amount and concentration of a substance is not related to the volume size, but only to the original concentration and the number of chloride ions in the compound.

So the ratio of the quantity to concentration of the substance is (1:2:3).

2co + o2 == 2co2

So the ratio of the amount of CO to O2 reacted = 2:1 Now there are 2molCO to 2molo2, indicating that there is an excess of O2. Only 1mol O2 and 2mol CO2 react completely to form 2mol CO2

So after the reaction, the gas mixture is measured to be CO2 and O2 containing CO2 molecule number = 2Na

Contains the number of O2 molecules = 1Na

Number of oxygen atoms = 2*2 + 2*1 = 6na

CO2 and O2 coexist, 2CO+1O2=2CO2, so there is 1mol02 and the atom is 2Na+4Na=6NaEach contains 1NaO2 molecule and 2NaCO2 molecule.

2co + o2 === 2co2

The amount of matter 2 1 2

Therefore, 2mol CO and 1mol O2 completely react to form 2mol CO2, and after the reaction, there is 1mol of oxygen remaining and 2mol of carbon dioxide generated, and the number of molecules of 1mol of oxygen is 1 Avgadro constant (that is, multiplied by 10 to the 23rd power), and the number of molecules of 2mol of carbon dioxide is 2 Afgadro constants. There are a total of 6 mol of oxygen atoms, which is 6 Avgadro constant atoms.

2co+o2=2co2

From this, it can be seen that O2 has a surplus, and 1 mole of O2 remains, and 2 moles of CO2 can be generated to find the amount of matter *Na for the molecular number.

O2 nA, CO2 NA.

The oxygen atom is 2*Na + 2*2Na = 6Na

Carbon dioxide 2mol. Oxygen 1mol.

3* .The oxygen atom is 6mol i.e. 6*

There must be none: SO42-, Ca2+, H+, Fe3+, possibly: Cl-;

Detailed answer: The solution is colorless and clear, and there is no Fe3+, because the Fe3+ solution is light purple;

It can be seen that the solution does not contain SO4 2-, contains CO3 2-, Ca2+, H+ and CO3 2-, and cannot coexist, so there is none; Since the added BaCl2 solution contains Cl, it is impossible to determine whether Cl- exists in the original solution, and the solution contains K+ because the solution is neutral

There must be: CO3 2-, K+, there must be none: SO42-, Ca2+, H+, Fe3+, there may be: Cl-

Since it is a solution that must have cations, this is questionable, the cations should be K+ or H+, (Fe3+, there is color removal), but the title has stated that the ions can exist in large quantities, so H+ excludes (CO3 2-, must exist, due to the dropwise addition of excessive BaCl2 solution, there is a white precipitate generated, and the precipitate can be completely dissolved in dilute nitric acid, so it is CO3 2-,, and at the same time excludes insoluble SO42-,)

Agno3 solution was added to the filtrate, and a white precipitate was formed, which was insoluble in dilute nitric acid. The precipitation is:

AGCL, since the added BaCl2 solution contains Cl-, it is not possible to determine Cl- in the original solution

There must be no Fe3+ in the colorless clear solution, add excess BaCl2 solution, there is a white precipitate that can be completely dissolved in dilute nitric acid, Baso4 is insoluble in dilute nitric acid, BaCO3 is soluble in dilute nitric acid, Ca2+, H+ will react with CO32-. There must be: co32 - there must be none:

so42- ,ca2+, h+

Add Agno3 solution to the filtrate, there is insoluble white precipitate in dilute nitric acid for AgCl, but it may be ** in the BaCl2 solution may have: Cl-, cation only K+, there must be: K+, there must be: CO32-, K+

There must be none: Fe3+, SO42-, Ca2+, H+ and possibly: Cl-

There must be: Fe3+, H+, Ca2+, SO4 2- There must be: K+, CO3 2- from the meaning and conditions of the title. (Potassium ions because there must be cations in the solution).

There may be: cl-

Ferrite ions appear red, and are the first to be excluded. The resulting white precipitate is completely dissolved in dilute nitric acid and can only be BaCO3, indicating that there must be CO3-, there must be no SO4 2-, and there can be no CA +2 and H+ if there is CO3-

cl- may have been present inherently, or it may have been added with baci2.

Got it?

Because it is a clear solution, the colored ions Fe 3+ are excluded first, because there is a white precipitate after adding barium chloride, so there must be one of SO4 2-, CO3 2-, because the precipitate is dissolved in dilute nitric acid, so there is no SO4 2 -,,BaSO4 does not react with dilute nitric acid).

Since there is CO32-, there must be no Ca2+, CaCO3 is the precipitate).

Since there is a white precipitate after the addition of AGNO3 and it is insoluble in nitric acid, it is not certain whether there is Cl- introduced in the previous step.

Since H+ reacts with CO3 2- to form carbonic acid, and carbonic acid easily decomposes into carbon dioxide and water, there must be no hydrogen ions.

Since the solution should be neutral and other cations have been eliminated, there must be potassium ions.

Answer: It can be judged that there must be CO3 2-, K+ in this solution, there must be none: CO3 2-, SO4 2-, Ca2+, H+, Fe3+ There may be: Cl-,

Colorless means no Fe 3+

It shows that there must be no SO4, there must be CO3 2-, and if the question is colorless and clear, then there is no Ca2+, H+, and Cl- is introduced, so the original solution Cl- is dispensable.

In addition, positive and negative ions must appear in pairs, so there must be k+

To sum up, there must be K+, CO3 2-, there must be no Ca2+, H+, Fe 3+, SO4

There may be: cl-

16GA reacts exactly exactly with 20GB to generate WGC and, W=16+

When 15GB reacts with 8GA, 8GA can only consume excessively, which is calculated as A.

16ga and 20GB react exactly to generate completely.

8GA and 10GB react exactly to the exact size.

Molar mass of c =

Magnesium carbonate is added to a sufficient amount of hydrochloric acid, which is equivalent to:

MgCO3+2HCl==MgCl2+H2O+CO2 into a solution dissolved with calcium hydroxide (, Ca(OH)2+CO2==CaCO3+H2O

Calcium hydroxide is insufficient, and then dissolved.

CaCO3 + H2O + CO2 = = Ca(HCO3)2 generated, remaining.

Magnesium carbonate is.

Carbon dioxide is released.

Calcium hydroxide is.

Excess carbon dioxide dissolves half of the precipitate.

Ca(OH)2 + CO2 ==== CaCO3(s) +H20CaCO3 + H20 + CO2 *****= Ca(HCO3)2, so the remaining precipitate grams.

Simply calculated, the amount of magnesium carbonate is the amount of calcium hydroxide, and it can be seen that the excess carbon dioxide formed will be partially converted into calcium bicarbonate, specifically the carbon dioxide to form calcium carbonate, and then the remaining calcium carbonate will be consumed. Remaining calcium carbonate. So the final precipitation mass is 5g

Magnesium carbonate is added to a sufficient amount of hydrochloric acid to produce the gas CO2

Excess CO2 in a solution dissolved in calcium hydroxide.

Therefore, the mass of precipitated calcium carbonate is 10g

MgCO3+2HCl==MGCl2+CO2+H2OCA(OH)2+CO2===CaCO3+H2OWhat is not enough Ca(OH)2, so CO2 does not react completely, and finally 10gCaCO3 is generated

mgco4 + 2hcl *****mgcl2 + co2 + h2o

CO2 + Ca(OH)2 *****CaCO3 + H2O calculates the amount of magnesium carbonate according to the mass of magnesium carbonate, and it can be seen from the equation relationship that 1mol of magnesium carbonate produces 1mol of carbon dioxide, and 1mol of carbon dioxide can generate 1 mole of precipitation.

Calculate the amount of precipitated substance, multiply it by the molar mass (molecular weight), and get the precipitated mass.

Let's do the math yourself.