Solve a high school chemistry problem about the quantity of a substance and find a detailed solution

The answer is abNote: The requirement of the question is that the number of chloride ions is equal, that is, the amount of chloride ions is equal.

The amount of chloride ion in 50 ml of 1 mol l of aici3 in the question is 1 mol l 50 ml 3=

A 150 ml of 1 mol L of Naci chloride ion substance amount is 1 mol L 150 ml =

B 75 ml of 2mol l of NH4CI chloride ion amount of 2mol l 75 ml =

The amount of chloride ion of C 150ml of 3mol L of KCl is 3mol L 150ml =

The amount of chloride ion in D 75 ml of 2mol l of Caci2 is 2 mol l of 75 ml 2=

nacl mgcl2 alcl3

The mass ratio of sodium, magnesium and aluminum is 23:16:9

Rule. The ratio of the amount of sodium chloride, magnesium chloride, aluminum chloride substances is:

2.The ratio of the amount of sodium chloride, magnesium chloride, and aluminum chloride = 3:2:1 so. The ratio of the amount of chloride ion substances in the three = 3:2*2:1*3=3:4:3, so 1mol chloride ion has:

3 (3 + 4 + 3) = NaCl, 4 (3 + 4 + 3) = Mg Chloride, 3 (3 + 4 + 3) = Aluminium Chloride.

I don't understand hi

1、n(al3+)=a/27 mol

c(al3+)=1000a/27v

C(SO42+)=(3 2)*(1000A, 27V)=500A, 9V4V(V4)=16, indicating that the dilution is 16 times, so after dilution, C(SO42+) = (500A, 9V) 16=125A, 36V, mol L

C should be selected for the original question, but A is omitted from the C answer

2、（1）、n(cuso4·5h2o)=

n(cu2+)+n(so42+)=2*

2), v (solution) = m d = (

c(cuSO4)=solution)=

3)v=ml+150ml=390ml

Medium c(H2SO4) = 1000*

Therefore, when used up, n(H2SO4)= can be provided

To set up a 500ml dissolving volume flask, n(H2SO4)=6* is also required. 31molv(h2so4)=

390+128=518ml >500ml indicates that the 500ml volumetric flask is not correct.

Therefore, a 1000ml volumetric flask is used to prepare the solution, but the amount of concentrated sulfuric acid required is still 128ml.

Hope it helps. Ha ha.

so4(mol) (a/27)*3/2

1000*so4(mol)/16v

Contains 8 grams of GuSO4 i.e. SO4 shared.

20v 1000 1mol l*1000=nml The number of moles of sulfuric acid is unchanged to find the number of moles, and then calculate the volume.

mNO2 4HCl (concentrated) == mnCl2 Cl2 2H2OX mol

x = 2mol

The formation of chlorine gas is mol

The calculated participation in the reaction is 2 mol

The amount of HCl that is oxidized is *2 = 1 mol, hopefully, it will help you.

Good luck welcome to consult my hi Please continue to support our team.

MnO2 4HCl (concentrated) ===MnCl2 Cl2 2H2O

So x==2mol, the HCl participating in the reaction is 2mol, of which only the chlorine gas is oxidized, and the oxidized is 1mol.

49GH2SO4:

1) The number of molecules is.

2) o The amount of atomic matter is 2mol

3) The mass of the H element is 1g

4) The number of s atoms is.

5) The amount of electron is 25mol

6) The number of protons is 25na=

H2SO4, multiplied by Avogadro's constant to get the number of molecules, 1mol H atom, S atom, 2mol O atom.

1. The number of molecules contained in a gas is about, what is the molar mass of this gas?

The amount of substance of the gas: n = n na = ( The molar mass of the gas is: m = m n = = 64g mol

2. It is a small number in the lower right corner of NA) contains sodium ions, what is the molar mass of Na2R; What is the relative atomic mass of r; What is the amount of substance contained?

The molar mass of the substance: n= = Na2r is: m= m n = = 62g The relative atomic mass of mol r is:

23*2+m(r) =62 m(r)=62 - 46 =16 The amount of substances contained is: n= m m = ) = what is the relative atomic mass of nitrogen and the number of molecules contained in a certain elemental rx is the same and the ratio of the number of molecules is 3:2; What is the value of x?

n(n)=( = n(r)=( mr=16 ( = 3/2 x=3

An equation can be drawn based on the amount of the substance that does not change before and after mixing.

Let the volume of NaOH solution be VThe volume of the solution of 50mol L of NaOH is V2

then the solution is v1:v2=

Therefore, the volume ratio of the front and back solutions is 3:1

v1：v2=3：1

The "sodium hydroxide solution" in your question should be.

Let the volumes be V1 and V2 respectively, and the amount of sodium hydroxide before and after remains unchanged.

v1 x 3v2 = v1, that is, v1:v2=3:1

Your title must have been written incorrectly, not a sodium hydroxide solution, but. Otherwise, it can't be.

If yes, it can't be.

But if it is, it's simple.

Let the volume of NaOH solution be XL, and the volume of NaOH solution is yl, and the solution is X:Y=3:1

1. The amount of CuCl2 is the amount of chloride ion and the amount of water is nAccording to the number of particles proportional to the amount of matter, there is 1 100 = n = 20 molTherefore, the quality of water is 20mol*18g mol=3602, and D. is selected

The number of hydrogen atoms is equal, and the amount of matter that is both hydrogen atoms is equal. Let the quantities of hydrogen and ammonia be a and b respectively, and according to the same conditions, the volume ratio of the gas is equal to the quantity ratio of the substance, and there is a*2=b*3

Get a b = 3 2

3. The volume of hydrogen produced by the two metal elements is equal when the ratio of the amount of the substance is 3:2, indicating that the valency of the two metal elements A and B is +2 and +3 respectively, and the amount of the substances of the two metals A and B is 3x and 2x respectively, according to the law of hydrogen produced by the reaction of metal and acid, there is 6x= to obtain x=

If the molar masses of A and B are 8A and 9A respectively, then 8A* is solved to obtain A=3 and the molar masses of A and B are 24g mol and 27g mol respectively

a The number of electrons transferred is 2Na. Wrong. From the change of Cl valency, it can be seen that each Cl atom gets 1 electron.

b False. NH4+ is an ion containing 11 protons and 10 electrons.

c should be na. Wrong. The gas contains 1 mol of molecules under the noisy condition of the standard hole.

d Right. Because the molar mass of CO2 and N2O is 44g mol, the gas molecule contained in the 44g mixture is 1mol, which is 3 atomic molecules.

2al+3cl2==2alcl3 transfer 6 electric beats and subs The attack should be 6na

C. Because Cl2 and HCl are not anti-Changla, the number of molecules should be 3, and NAD should be "under standard conditions".

To sum up, B should be chosen

ps: I'm also in my first year of high school, so I don't know if it's right (》3").

1. The question requires the gas to be fully dissolved, so there is a limit to the m and n given in the question, if you want to be fully dissolved, you can only become an acid root, and the conservation of n element is the answer. In many cases, chemistry should not be drilled, but should be based on facts and topics.

2. NO2 will self-synthesize into nitrogen tetroxide: 2NO2=N2O4, which is not 1mol.

3. Under the translation of that formula: 2 times potassium sulfate + sodium hydroxide + (barium hydroxide - potassium sulfate - ammonium carbonate) x2 = hydrochloric acid, that is: potassium ion + sodium ion + 2 times of non-precipitated barium ion = chloride ion, charge conserved.

In the final solution, the cations are potassium ions, sodium ions, and barium ions, and the anions are chloride ions, and the positive charge of the cation is equal to the negative charge of the anion, which is the principle of this equation. (Barium ions come from barium carbonate dissolved with acid, so use the total subtraction to become barium sulfate.) ）

In the end, the better ones have to ask for money... So just ask

lz You wrote the second chemical formula of the first question incorrectly, and the left and right sides can't be 2 parts NAOH

The second chemical equation of the first problem is problematic, there is n before the reaction, but n is not present in the reaction product. That's a problem.

Question 2: No2 will automatically synthesize N2O4 in a dynamic equilibrium, Question 3: