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According to the equation for the magnesium oxide reaction produced by magnesium and oxygen under the condition of ignition;

Conjectures and hypotheses] analysis based on the mass relationship between the reaction of magnesium and oxygen, the reaction with carbon dioxide and the reaction with nitrogen;

Experiment**] According to the principle of determining the oxygen content in the air: it only reacts with oxygen and does not produce new gas analysis options;

Experimental conclusions] According to the mass relationship between magnesium reaction with oxygen and nitrogen reaction.

Evaluation and Reflection] Since magnesium strips can be burned in nitrogen and carbon dioxide, it can be seen that oxygen is not necessary for combustion, and a new understanding of the properties of nitrogen and carbon dioxide that cannot support combustion is carried out

Analysis] Magnesium oxide is produced by magnesium and oxygen under the condition of ignition, and the equation of the reaction is: 2mg+O22mGO;

Conjectures and assumptions] According to the chemical inverse equation, grams of magnesium react with oxygen to produce grams of mgo, and react with carbon dioxide to produce grams of magnesium oxide and grams of carbon. The reaction of gram magnesium with pure nitrogen produces Mg3N2 as grams, and the content of nitrogen in the air is large, so the solid mass is less than that before the reaction, so the solid mass after the reaction is smaller.

Experiment**] The principle of determining the oxygen content in the air: it only reacts with oxygen and does not produce new gas analysis options; Both charcoal and sulfur powder react with oxygen to produce new gases, so red phosphorus is chosen;

Experimental conclusions】 According to the chemical inverse equation, it can be seen that the reaction of gram magnesium with oxygen produces MG3N2 as gram, and the content of nitrogen in the air is large, so the reason for the reduction of solid mass compared with before the reaction is due to the mixture of magnesium oxide and magnesium nitride.

Evaluation and Reflection] Since magnesium strips can be burned in nitrogen and carbon dioxide, it can be seen that combustion does not necessarily require oxygen to participate or nitrogen and carbon dioxide can also support combustion under certain conditions

Therefore, the answer is: 2mg+O22mgo;

Conjectures and assumptions] ; According to the inverse chemical equation, grams of magnesium react with oxygen to produce grams of mgo, and react with carbon dioxide to produce grams of magnesium oxide and grams of carbon. However, the reaction of gram magnesium with pure nitrogen produces Mg3N2 as gram, and the content of nitrogen in the air is large, so the solid mass is less than that before the reaction.

Experiment**] b; Red phosphorus reacts only with oxygen and does not produce new gases;

Experimental conclusions: magnesium oxide and magnesium nitride;

Evaluation and Reflection] Combustion does not necessarily require oxygen (or nitrogen to support combustion, etc.).

2mg+O2=2mgo (reaction conditions are written as "ignition") After the reaction of magnesium and nitrogen, the solid mass is less than that of magnesium and carbon dioxide, and the solid mass is greater than that of red phosphorus. Red phosphorus and oxygen are burned to produce solid phosphorus pentoxide, and the nitrogen produced is relatively pure; The combustion of charcoal and sulfur powder will produce gas oxides, and the nitrogen produced will be mixed with impurities.

MGO with Mg3N2

Combustion does not necessarily require the involvement of oxygen.

1) Not true; Metaaluminate alo2-, mno4-permanganate.

2) The structure of N60 is similar to that of diamond, each N is surrounded by 4 N atoms, and 28 14=2mol N atoms are in 28gn60.

Using the amortization method, each n has 4 bonds, and each bond belongs to 2 n common.

Therefore, each n corresponds to 2 n-n bonds separately.

So n(n-n bond) = 2*2 = 4mol

There are a total of 4na n-n bonds.

3) Fe3+ does not coexist with Fe in solution.

Therefore, all the original Fe is converted into Fe2+ and Fe3+

m（fe）=

Conservation by atoms, there is n(Fe2+) = n(Fe3+) = H+ completely consumed because there is Fe2+ left in the solution.

According to the conservation of charge, there is a residual no3- after the reaction.

Whereas, Fe dissolves and loses electrons.

NO3- becomes NO, the valency decreases by 3, and the electrons are obtained, so some NO is conserved according to N atoms, and the original dilute nitric acid N(HNO3) = at the same temperature and pressure, the volume ratio is equal to the ratio of the amount of matter.

The molecular weight of CO and N2 is the same, so no matter how much CO is, it has no effect on the result.

So the average molecular weight of CO2 and H2 = 28

Let CO2 have x and h2 have y

then 44x+2y x+y=28

The solution yields 16x=26y

So x:y=13:8

So in the gas mixture, the volume ratio of CO2 to H2 is 13:8, while CO is uncertain.

Not true, e.g. mno4-

ALO2- Contains metal ions but is not a cation.

2.In the N60 molecule, each N atom forms 3 nitrogen-nitrogen single bonds, and each nitrogen-nitrogen single bond belongs to 2 N atoms common, so on average, each n atom has a nitrogen-nitrogen single bond; 28gn60, is a 2molN atom, so a total of 3mol nitrogen and nitrogen single bonds.

3.The amount of Fe3+ and Fe2+ substances in the solution is exactly equal, indicating that the nitric acid reaction is complete. And you can write the following general equation:

6Fe20Hno3 (dilute) = 3Fe (NO3) 2 + 3Fe (NO3) 3 + 5NO + 10H2O

6mol20mol

n(hno3)

n(hno3)=[20×(

The molar mass of the molar mass of N2 is equal, so the ratio of Co is fine, but the ratio of Co2 to H2 must be 1 to 1

Let the amount of methane substance x and the amount of carbon monoxide substance y

In the standard case, vm=. According to Bi Ma jujube n=v vm,n=m mx+y)·

x·16g object bridge mol + y·28g hand split mol = x=, y=

Hence methane volume.

Carbon monoxide quality.

Methane molecular weight 16, carbon monoxide 28. The total volume of the gas, the mass of the gas, the mass of grams, so the molecular weight of the draw is 19, and the alkane of Jiachaling accounts for 75% of the mixed gas, and the carbon monoxide is 25% (g).

NaOH relative molecular mass = 40

H2O relative molecular mass = 18

According to the conservation of mass before and after the reaction, there is: a + 40 = b + 18 so b - a = 22 choose c

naoh 40 h2o 18

The relative molecular weight difference of b a is 44-18 = 22

c Using the limit method If all N02 is N02, 46g is 1mol, and N02 contains three atoms (one nitrogen atom and two oxygen atoms) which is 3Na

If all of them are N2O4, 46g is 3Na, and N2O4 contains six atoms (two nitrogen atoms and four oxygen atoms).

Therefore, no matter what ratio the two are mixed, 46g of the gas mixture will always contain an atomic number of 3 Na

There is also the chemical equation 2NO2===N2O4 which can be converted into each other.

D Fe powder and sufficient amount of water vapor reaction at high temperature should be 3Fe + 4H2O(G) === High temperature ===Fe3O4 + 4H2 According to the chemical equation calculation, it can be seen that the number of molecules generated should be D, so D is wrong.

Write out the electrode reaction:

The positive electrode Cu2++2E-==Cu

Negative zn-2e-==zn2+

Draw a diagram of the installation.

zn graphite |

--cuso4 solution |

The positive electrode Cu2++2E-==Cu

Negative zn-2e-==zn2+

The device diagram requires copper and zinc to be connected by wires and in the same electrolyte.

The positive electrode Cu2++2E-==Cu

Negative zn-2e-==zn2+

Draw a diagram of the installation.

Let the valency of r be x with a relative atomic mass y

The relative molecular mass of acetate is m, i.e., y+59x=mThe relative molecular weight of nitrate is n, i.e., y+62x=nx=(n-m)3

The relative molecular mass of acetate is 59 and the relative molecular mass of nitrate is 62, giving the equation:

m-59x=n-62x

In the solution, x=(n-m) 3 chooses d

Let the valency of r be +x and the relative atomic mass m

Then the acetate chemical formula of r is r(ac)x

The nitrate chemical formula is R(NO3)X

then m+59x=m

m+62x=n

Solution (n-m) 3

There seems to be no answer.

2naoh+cuso4=cu(oh)2↓+na2so480---98 --14240*10%--y---xx=7,y=

The mass fraction of the solute in the resulting solution = 7 (40+.)

2naoh+cuso4=cu(oh)2↓+na2so480 98 142

4. The mass fraction of the solute in the resulting solution.

cabcbcc

If you're satisfied, remember!