A chemistry problem, help with 5

According to the conservation of mass, the amount of hydrogen produced should be equal, so find the iron that is reacted off, then the iron remains.

Zinc is, release hydrogen, reduce the mass gram, so dilute sulfuric acid should also release hydrogen requires iron, that is, so the iron remains, it is easier to solve from the perspective of mass conservation.

16 lessons CH4 is fully burned in the air to produce 44gCO2 and 36GH2O, 2CO2+2Na2O2=2Na2CO3+O2, 2H2O+2Na2O2=4NaOH+O2

Sodium peroxide gains weight by 28g + 4g = 32g

You use the principle of conservation. Both CO2 and H2O produced by CH4 combustion can react with Na2O2, so you write the chemical formula, and calculate it directly according to the conservation of C and H, I won't calculate it for you, you practice it yourself, and you will do it later.

Both CO2 and H2O produced by the combustion of CH4 can react with Na2O2, so in the end, the mass of sodium peroxide increased is the same as that of CH4, so it is 16 grams.

According to the conservation of c and h, it is calculated directly.

na2o2na2co3

Increase by 28c:h=1:4

2na2o2

4naoh increases by 4

This is equivalent to 12 c, an increase of 32 copies.

Therefore, 16gch4 increases the mass by 32g

Because the zinc reacts completely, you can find the mass of the hydrogen that is generated, and the mass of the hydrogen produced by the two reactions is equal, so you can use the mass of the hydrogen that you just found to find the mass of the iron that has been reacted, and you can get the mass of the remaining iron by subtracting the amount of the iron that has been reacted, and you should be right, and two 98g are useless.

(1) Dilute sulfuric acid and iron filings to boil.

2) A 2H2O+2E-==H2 +2Oh- or 2H++2E-==H2

Isolate the air to prevent ferrous hydroxide from being oxidized.

3) The test tube is filled with hydrogen, which is not easy for outside air to enter.

Close the water stop prematurely to neutralize the sodium hydroxide solution by pressing the sulfuric acid from the left tube into the right tube.

Analysis: (1) When preparing FeSo4 solution, dilute sulfuric acid and iron filings need to be added to inhibit the hydrolysis of Fe2+ and prevent Fe2+ from being oxidized to Fe3+ by O2 in the air. Boiling distilled water removes the O2 dissolved in it.

2) During electrolysis, the anode reaction is: Fe==Fe2++2E-, to prepare pure Fe(Oh)2, the electrolyte solution is required to provide OH-, and the electrolyte solution does not contain ions reacting with Fe(OH)2, so only A is correct according to the above requirements. The cathodic reaction formula is:

2h++2e-==h2 or 2h2o+2e-==h2 +2oh-. The surface of the electrolyte solution is covered with a layer of benzene to isolate the air and prevent Fe(OH)2 from being oxidized.

3) Open the water stop clamp, FE reacts with H2SO4 to generate H2 to fill the whole device, close the water stop clamp after a period of reaction, the air pressure in the left test tube increases, and the Fe2+ generated by the reaction enters the right test tube along the catheter and reacts with NaOH to generate white precipitate Fe(OH)2. Since the device is filled with H2, it is not easy for outside air to enter, so the precipitated white color can be maintained for a long time. If the water stop clamp is closed prematurely, the sulfuric acid in the left tube is pressed into the right tube, and the NaOH is neutralized, and the Fe(OH)2 solution is not obtained.

(1) 1 iron block.

2. Boiling (2) Can you give a picture?

From the chemical formula of ferrous sulfate (FeSO4) and iron sulfate (Fe2(SO4)3), it can be seen that the ratio of the number of S to O is 1:4, then their mass relationship is S:O=32:

16 4 = 1:2, where the mass fraction of oxygen is 42%, so the mass fraction of sulfur is 42% 2 21%, so the mass fraction of iron is 1-42%-21% 37%.

The ratio of the number of sulfur atoms to oxygen atoms is 1:4, so the ratio of the mass fraction of sulfur to the mass fraction of oxygen is 1:2, so the mass fraction of sulfur is 21%.

Therefore, the mass fraction of iron is 1-42%-21%=37%.

Molecular formula: Ferric sulfate Fe2(SO4)3

Ferrous sulfate FeSO4

Let the amount of ferric sulphate in the mixture be a, and the amount of ferrous sulphate b

Then the mass of the oxygen element in the ferric sulphate is.

12*16/(112+96*3)*a

The mass of iron is (56*2) (112+96*3)*a, the mass of oxygen in ferrous sulfate is 4*16 (56+96)*b, and the mass of iron is .

56/(56+96)*b

The mass fraction of oxygen in the mixture is:

12*16/(112+96*3)*a+4*16/(112+96)*b]/(a+b)=42%

Simplification a = 2b

The mass fraction of the element iron is:

112/(112+96*3)*a+56/(56+96)*b]/(a+b)*100=(=

This is a question that we can't ask for directly, but only indirectly.

Because in the mixture, element O exists in the form of sulfate, that is, the ratio of elements S to O is a certain number, and the number of S and O is 1:4

So the mass ratio is 1:2

So the mass fraction of s is 21%.

So the mass fraction of iron is 1-42%-21%=37%.

I think I did it when I was in junior high school. It seems to be used in proportions to find the constant proportional relationship in the mixture. Of course, this question is S and O. Breakthrough.

Analyzing the chemical formulas of ferrous sulfate (FeSO4) and ferric sulfate (Fe2(SO4)3), it is not difficult to see that the ratio of S to O is 1:4, and the mass relationship is as follows: M(S):

m(o)=32:16 4=1:2, where the mass fraction of oxygen is 42%, so the mass fraction of sulfur is 42% 2 21%, so the mass fraction of iron is 1-42%-21% 37%.

caco3+2hcl=h2o+cacl2

mg(oh)2+2hcl=mgcl2+2h2o

Because it is an aluminum kettle, and hydrochloric acid is to react with aluminum, if there is too much hydrochloric acid, if you are not careful, the kettle will be pierced.

The equation for removing limescale with hydrochloric acid.

CaCO3 + 2HCl = CaCl2 + H2O + CO2 mg(OH)2+2HCl=MGCL2+2H2O can not add too much hydrochloric acid because:

2al+6hcl=2alcl3+3h2↑

CaCO3+2HCl=CaCl2+H2O+CO2 Gas symbol MG(OH)2+2HCl=MGCL2+2H2O Because of the addition of hydrochloric acid and scale, it will also react with aluminum.

2al+6HCl=2alcl3+3h2 gas symbol.

The number of particles contained in 1mol of any particle is the Avogadro constant.

The total volume of the drops is VD*d At this time, the weight of the stupid solution of the grease is [VD*d*(m v)]mg From the ratio of the surface area, the number of lipid molecules can be obtained, and the number of oil molecules is sa The mol number of greases is mol

So n= s a).

Hope it helps

This is inorganic chemistry in college After a rough look, the question is relatively simple, but I haven't touched it for a year, and I forgot it.

The inner capacity is almost ==

Feel free to help you see something

Analysis: The d orbital can hold a total of 10 electrons to be semi-filled, and there must be 5 electrons. Take b

For example, the outermost arrangement of Fe is 3D64S2, so Fe3+ is 3D5, and according to Hunt's rule, it is half full.

10, a analysis: the reaction quotient of 1 is to reach the equilibrium, an equilibrium is δg = 0 (under constant temperature and pressure), don't ask me why, wait until you learn physical chemistry to know.

11 Analysis: What is DG? I guess it's the Gibbs function. If the galvanic cell works spontaneously, it should be that the Gibbs function change should be less than zero, not increased.

13 Correct.

Analysis: After a cursory look, I think it's very correct.

14, wrong.

Analysis: Take 4s and 3d as examples, they are a group of energy levels, but not equivalent orbitals, 4s < 3d

Inorganic learning is the basis of other chemistry

Oxidation is the reduction of the valency of S The occurrence of the nature of acidity The combination of H and OH- so 1 choose A

Electron transfer is a valence change, so you just need to mark the elements of the valence change.

There are no requirements for electronic transfer, you have to pay attention to the book, and you will do more questions.

One. Two.

Three. Four. Although the aqueous solution of 1 and 2 also conducts electricity, they are pure because the substances formed by the reaction of Sakura Shen dissolved in water and water are used to conduct Songho electricity, so 1 and 2 cannot be counted as electrolytes.