Senior Chemistry, ask for help, Senior Chemistry, ask God for an answer, thank you

The relative atomic mass of helium is 4, so the relative molecular mass of the gas mixture is 4*8=32 (the density is 8 times the density of helium under the same conditions).

co2 44 4

co 28 12

The formula is universal, and there are not many chemical calculation questions themselves, so it doesn't matter whether it is important or not, and the test is at most a multiple-choice question, when we took the exam, we didn't take it, but it was also a test point.

1) Anhydrous copper sulphate.

2) Beakers, measuring cylinders, glass rods.

Take a small amount of sample, add enough dilute H2SO4, if the solution is dissolved by colorless blue and the sample is partially dissolved, it means that the red substance contains Cu2O, otherwise, there is none.

2nh3 + 6cuo = = 3cu2o + n2 + 3h2o3) copper and cuprous oxide copper , cuprous oxide.

Analysis: (1) Because anhydrous copper sulfate is a white powder, it turns blue when exposed to water.

2) Because copper does not react with dilute sulfuric acid, when cuprous oxide can react with dilute sulfuric acid, cuprous oxide itself undergoes redox reaction, and the reaction produces copper sulfate solution that is blue, 3) 2NH3 + 6CuO= = 3Cu2O + N2+3H2O

x2nh3+ 3cuo=△= 3cu + n2+3h2oy yx+y=8g/80g/mol=

The solution gives x=y=so the amount of matter of Cu2O is.

cu2o+h2so4=cuso4+cu+h2oz

z=yields solid copper mass=

I hope mine will help you in your studies!

Hope it solves your problem.

Hydrosulfuric acid is a 2-element weak acid, partially ionized, then there are 2 major ionizations and 2 minor ionizations in the hydrosulfuric acid solution.

Primary ionization 1: Hydrosulfuric acid ionization is hydrogen ion and sulfur ion, the degree of ionization of this is the largest, the main ionization 2: Ionization of hydrogen ion and sulfur ion, this degree of ionization is secondary to ionization 1:

Hydrolysis of sulfur ions, ionization of water, maximum degree of hydrolysis Secondary ionization 2: Hydrolysis of sulfur ions, ionization of water, the degree of hydrolysis is secondaryIf pH is considered, then in the presence of primary ionization, the secondary ionization (hydrolysis) is negligible (Lechatre's principle); If ion concentration is considered, secondary ionization (hydrolysis) is generally a consideration.

a, with the addition of sodium hydroxide, the hydrogen ions in the solution are consumed, and the concentration of the product decreases, prompting the two major ionization equilibrium to move forward (to the right), resulting in an increase in the concentration of thibion ions and sulfur ions. Then for secondary ionization, it is the concentration of reactants that increases, so the equilibrium continues to move to the right, so that the degree of ionization of water increases. A correct.

Chemistry is a multiple-choice question, so you don't need to look at it later.

Option A is correct. Item A, according to the structural formula in the figure, it can be judged that the chemical formula in A is correct. In item b, hydrocarbons are compounds composed of carbon and hydrogen, and the organic matter contains oxygen, which does not belong to hydrocarbons, so it does not belong to aromatic hydrocarbons.

C, only the carbon atom in the molecule that is directly attached to the benzene ring is in the same plane as the benzene ring, so C is wrong. D, this compound undergoes a elimination reaction, and only one aromatic olefin can be obtained, so D is wrong.

<> give it to you, just after the exam, I will do it for you.

The third year of high school chemistry asks for a big answer, and you can find it by checking it on the Internet.

There may be NaOH, NaCl, AlCl3, and Naalo2 in solution B, which is neutral, that is, there is no AlCl3 (acidic) and Naalo2 (basic), so there is only NaCl

That is, na:cl=1:1

That is, Na2O2:HCl=1:2, that is, Al2O3 does not matter.

The solution is neutral, so there is no ALCL3 (hydrolyzed acidic) and Naalo2 (hydrolyzed alkaline).

Na2O2 and HCl react completely, Al2O3 does not react, it is a precipitate!

Therefore, satisfying NA2O2:HCl:Al2O3=1:2:n is the OK regression option that satisfies the above relation, and it is B!

So the answer is b

Hello! The oxidation of the three ions is from strong to weak: Cl->Br->Fe2+ can be seen from the title, Br- has a surplus and participates in the reaction, so all the chlorine participates in the reaction and Fe2+ becomes Fe3+, where chlorine is mol, so the electron is mol multiplied by 2;

Let the amount of Febr2 participating in the reaction be x, divalent ferrous (x mol) loses electrons first, then bromide ions (2x times 1 3) lose electrons, and finally the mol is conserved by the conservation of gains and losses by 2 = x + 2x times 1 3x = moles of solution.

And then the concentration of the amount of the substance can be found.

The ferrous ions have all turned into iron ions, what else do you not understand???

Because chlorine all participates in the reaction, and chlorine is moles, the electron is moles multiplied by 2;

Let the amount of Febr2 participating in the reaction be x, divalent ferrous (x mol) loses electrons first, then bromide ions (2x times 1 3) lose electrons, and finally the mol is conserved by the conservation of gains and losses by 2 = x + 2x times 1 3x = moles of solution. And then the concentration of the amount of the substance can be found.

The conservation of electrons is equal to the total number of electrons in the whole state.

Mole multiplied by 2 is the electron obtained by chlorine gas.

x+2x times 1 3 x is the electron lost when the ferrous ion is first oxidized to the ferric ion.

When ferrous ions are oxidized, chlorine reacts with bromine ions, then the total number of electrons lost by x bromide ions and ferric ions is equal to the total number of electrons obtained by bromide ions.

The crux of this topic is really the question of the sequence of redox reactions.

2-valent iron is more reducible than bromine ions, then 2-valent iron is oxidized first.

I'm sure LZ will calculate the rest.

All ions in the ion are acidic coexisting ions, CO2 is acidic when introduced into water, so it does not change the properties of the solution, and ions can coexist in large quantities. b is in line with the title The two ions of Na+ and C6H5O- can be regarded as the free state of sodium phenol. Because carbonic acid is stronger than phenol, a reaction of strong acid to weak acid occurs when CO2 is introduced.

The solubility of phenol in water at room temperature is very small, and the premise of the topic is that a large number of ions coexist, indicating that the ion content of Na+ and C6H5O- in the solution is very high, and phenol insoluble substances will appear after CO2 is introduced. So d doesn't fit the topic.

a. As can be seen from the figure, the device has no external power supply, and forms a closed loop by itself to undergo redox reactions to form galvanic batteries, which can provide electrical energy.

So a is correct.

B, A electrode: p-chlorophenol + 2E + H ion = phenol + Cl ion B electrode: CH3COO- 7E + 4H2O = 2HCO3-+9H ion A to get electrons is the anode potential is high, B electron loss is the negative electrode potential is low.

So b is correct.

c. Write the electrode equation according to the trim principle (see above). So C is correct.

D. It can be seen from the reaction of the two electrodes of AB that pole A consumes H+, and pole B produces H+, so it is automatic from the generation end to the consumption end.

Or a electrode undergoes a reduction reaction, which is equivalent to the cathode in the cell, where the cations move towards the cathode.

So H+ moves from the B pole to the A pole, that is, the negative pole moves to the positive pole.

So d is wrong.

The positive charge inside the battery should move from the negative pole to the positive pole, right? So D may be wrong, if C's reaction is correct, then the hydrogen ions inside the battery, that is, the positive charge flows to the A pole, and the electrons flow to the A pole through the above loop, then the A pole should be the positive pole of the power supply, and the first three options should be right, this piece of knowledge is a bit vague, just for reference.

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