The chemical solution calculation helped me solve the solution

1. Solution: The mass of H2SO4 in the configured solution is: 500ml*20%*, the volume of concentrated sulfuric acid is x, and the solute mass of the solution before and after dilution remains unchanged, is: x*98%* x=

The volume of water required is.

2. The mass of H2SO4 in the configured solution is: 500g*20%=100g, the volume of concentrated sulfuric acid is x, and the solute mass of the solution before and after dilution remains unchanged, including: x*98%*

Get x=

This is the quality of this sulfuric acid.

This is the quality of sulfuric acid that is required.

Since the mass of sulfuric acid should be the same, then the concentrated sulfuric acid also contains 114g of sulfuric acid.

Then the mass of this concentrated sulfuric acid is approximately equal to 114.

The body is only about equal to the answer.

1.The calculation is based on the constant mass of the solute before and after dilution.

v= The volume of water required is about: 2v=

The quality of the solute before and after dilution is not repentant.

Assuming that a mass fraction of 98% x grams of sulfuric acid is required, there is.

490g*20%=98%x

x=100g

The amount of repentant limbs that need to be treated before opening = 490g-100g = 390g

First: Take a 500ml volumetric flask. Weigh 100 g of NaCl with a scale. Put it in a volumetric flask, add water, and dissolve it into a volumetric flask with a scale of 500 ml.

Group 2 simple species: add grams of NaCl to 250 grams of sodium chloride solution with 25% mass fraction of dissolved sedan ears, and then add 250 grams of water. Principle: 250*25%=

g nacl gram. 250 + 250 = 500 ml.

The third; Add 70 g of NaCl to 200 g of sodium chloride solution with 15% solute mass fraction, and then add 300 g of water. Principle: 200*15%=30g.

NaCl 30 + 70 = 100 grams.

200 + 300 = 500 ml.

at a certain temperature.

When the solution is saturated.

Dissolved solute M1

The solubility of g dissolved m2g is equal to 100 m1 m2

gThis is the solubility of the solid-liquid.

The gas is at a certain temperature pressure.

6 by volume

Ferrous sulfate FeSO4 Fe mass fraction w (fe) = 56 152 = 7 19 ferrous sulfate m (feso4) = 560kg (7 19) = 1520kg fe + h2so4 = feso4 + h2560kg x * 30% x =

1 Hno3 contains 1 H+, so the amount of H+ substance is 1 HCL contains 1 H+, so the amount of H+ substance is the total amount of H+ is.

1 Hno3 contains 1 NO3-, so the amount of NO3- is the amount of Cu substance.

3cu+8h+ +2no3-=3cu2+ +2no↑+4h2o3 8 2

As can be seen from the above equation, the Cu requires H+ with No3-, the fact is that H+ only, insufficient, No3- is, sufficient, and the reaction calculation should be based on insufficient substances.

3cu+8h+ +2no3-=3cu2+ +2no↑+4h2o8 3n

The amount of a substance with n=cu2+ is.

In mixed acids, the amount of substances of H+ = mol, and the amount of substances of NO3- = mol

According to the ionic equation of the reaction 3 Cu+ 8 H+ +2 No3- = 3 Cu2+ +2 No +4 H2O

It can be seen that when copper is mol, mol H+ and mol no3- are required to dissolve completely.

So H+ is insufficient, calculated according to H+.

3 cu+ 8 h+ +2 no3- = 3 cu2+ +2 no↑ +4 h2o

Mol x is solved to get x = mol

That is, the amount of Cu2+ in the solution after a sufficient reaction is mol

The ion equation must be written here to solve it correctly, because when HNo3 is consumed, the H ions produced by HCl ionization can continue to react with the acidic NO3- to form nitric acid, and this problem exists with the ion equation.

The amount of substance of the H ion is.

3Cu + 8H (ion) + 2NO3 - == 2NO + 3Cu (+2 valence) + 4H2O

It is known that the excess of Cu is calculated by H ions.

According to the density of the water.

1ml=1g

So a. The quality score is:

Because what is made is a solution.

So the mass fraction remains the same.

Then move on. Then the mass fraction of c is (

This is followed by d's { (

So the mass fraction of d is above.

It's too small to be counted.

Set configurable XG

1/x=｛［（

(1) Dissolve the drug in water and dilute it into a 4ml solution with water (2) take solution A, dilute it to 1ml with water, and get the solution diluted to 10% of the original) (3) Take solution B, dilute it to 1ml with water, and get solution C: same as above) (4) Take solution C, dilute it to 1ml with water, and get solution D: dilute to 20% of the original (1) The mass fraction of the solute in the final solution D?

2) What is the volume of 1g of the configurable D solution? 1g/

100---x

The solubility of the substance at x=60t, s=60g

No water. Copper sulphate hunger.

Into. cuso4

5H2O then the mass becomes: very well calculated), i.e. the combined water, which is the original.

Saturated solution. Due to the loss of rot residue mu to water, it becomes supersaturated, and there is a precipitation of sulfuric acid beam trillion copper, and the precipitation is xg, then x is compared to (x + solution x =

So the answer is.

But me. There is a question.

Could it be that the original saturated solution is not precipitated by CuSO4·5H2O, but according to the answer, it is precipitated by CuSO4