High One Chemistry. Regarding dilution, the amount of concentration of the substance.

The answer is 2g,,1 10 times.

Question 2 Take the CuSO4 obtained by pyrolysis and prepare 100ml of CuSO4 solution 1 n=250*,m=mn=

Diluted to 250ml is equal to 10 times, so yes, the amount of substance is 1 10 times the original.

2. n(cuso4)=,m=n*m=

Question 1. 2g x 40g mol 40g mol is the test amount of NaOH.

x (25ml/250ml)

25ml/250ml

Question 2. x x 250g/mol

mnaoh=

Because the concentration of the substance in the original solution is, take 25ml, and dilute it into 250ml, and the concentration of the substance is one-tenth.

Fold. cuso4·5h20

1.According to the conservation of elements, it can be obtained: 10mol of oxfy contains n(o)=10xmol, n(f)=10ymol, and because n(o)=15x2=30mol after the reaction, n(f)=12x2=24mol (here I look at the volume as the quantity of matter, because the volume is proportional to the amount of matter, so it can be calculated in this way).

10x=30...=>x=3，10y=24...=>y=then the chemical formula of a is .

Landlord, is there no problem with this data? Isn't it 10ml of f2? When I did it before, the answer was o3f2. If you make a mistake, just change the data above, the method is correct)

Analysis] Let the concentrations of these three substances be x, y, z, respectively, and the volume of the solution is 1l, then:

n(bacl2)=n(baso4)=>n(so42-)=10^-3mol

n[fe2(so4)3]=x·1=x...=>n(so42-)'=3x=10^-3...=>x=1/3000

n(znso4)=10^-3mol=y

n(k2so4)=zmol=10^-3mol

So x:y:z=(1 3000):(10 -3):(10 -3)=1:3:3

The amount of the substance of HCl = mol l * 10 1000 l = mol

The amount of substance of the diluted solution concentration = mol l = mol l

The principle is that during dilution, the amount of solute substances does not change.

The amount of the substance concentration of the diluted solution.

The amount of grasping solute material is constant. 10×10-3l×

Find the amount of solute first, change the unit to 10ml = liter, liter. Finally c=

Under standard conditions, the amount of 336L HCl gaseous substance is 15mol, the density of hydrochloric acid is, the mass of 1L of water is 1 kg, the mass of HCl is grams, and the total mass is grams, so the total volume of the solution is 1323ml, and the amount concentration of hydrochloric acid is 11mol L.

The number of moles of 1,336L HCl is 336 mol

2. Then calculate the volume of the solution after HCl dissolution is (15* ml

3. Then calculate the concentration of the substance: c=n v=15 mol l

Wrong solution: c = 336l

Wrong reason: Treat the volume of the solution as 1L.

Correct analysis: The volume of the solution is calculated based on the mass and density of the solution.

C hydrochloric acid = 336 L mol-1 (336 L mol-1

First the l is converted into mol, and the mass is also turned into mol

Then before c dilution times v before dilution = c dilution after v dilution.

n(hcl)=336l/

hcl%=15mol/1l=15mol/l

The volume of dissolved HCl is not counted. )

Apply formulas.

c (concentrated solution) v (concentrated solution) = c (dilute solution) v (dilute solution) 100ml= and substitute the data into the formula.

c×2lc= mol/l

It is found by the amount of the substance before and after the reaction. If there is 18 moles before the reaction, then the concentration of the amount of substance after screening is.

Grasp the mass of the substance before and after dilution, unchanged C1 V1=C2 V2 to obtain C2=

A volume and 1 volume can be understood in this way.

A volume can be considered as a L HCl, and 1 volume can be considered as 1 L H2O, so the amount of HCl substance = 1 mol

Solution mass = + 1000 =

Solution volume = ml = l

So the amount concentration of the substance of hydrochloric acid = (1 mol l

n(hcl)=a/ mol

m (solution) = + 1000 = grams.

v(solution) = (ml=(l.)

c(hcl)=(a/ )/[( =1000ab/(22400+ mol/l

A volume and 1 volume can be understood as liters.

That is, AlHCL is dissolved in 1 L of water.

There is hydrochloric acid A hydrochloric acid volume v'=m p=(v ml.)

So the hydrochloric acid concentration is (a'

One part containing a BMOL barium chloride precipitate completely indicates that there are BMOL sulfate ions Two parts have a total of 2BMOL sulfate ions, one part has a demol of sodium hydroxide indicates a 2mol of hydroxide A 2mol of magnesium ions A total of two parts of a total of a total of a magnesium ion can bind AMOL sulfate ions The remaining 2b-amol sulfate ions are completely bound to potassium ions then contain 4b-2amol of potassium ions, so the concentration is 4b-2a v mol l

I don't know if it matches your answer, it should be accurate, I hope it helps you.

A solution containing a mol NaOH is added to completely precipitate magnesium ions into magnesium hydroxide, then Mg2+ has a 2 mol, and magnesium sulfate is a 2 mol.

The other part is added to the solution containing B mol BaCl2, which happens to completely precipitate the sulfate ions into barium sulfate, so the sulfate ions have B mol.

So potassium sulfate has (b-a2) mol, and potassium ions have 2(b-a2) mol = (2b-a) mol.

The concentration is (2b-a) v mol l.

Pay attention to the first sentence of the question, "Divide the mixed solution into two equal parts", do not ignore it.

(b-2a) 2 V mol L, the charge is conserved, and the positive charge is equal to the negative charge.

hcl~~agcl

The HCl reaction produces an AGCL precipitate.

To make the mass of the precipitate equal to the mass of the original hydrochloric acid, the mass of the hydrochloric acid solution containing the HCl solute =

So the volume of hydrochloric acid v = (ml) = (l) n (HCl) = = 1mol

So the amount concentration of the substance of the original hydrochloric acid = n v = 1 ( = choose c

Let the HCl participating in the reaction be 1mol, that is, the mass is, and observe the equation:

agNO3+HCl = AGCL+HNO3 The amount of the substance of the original hydrochloric acid can be obtained due to the equal mass of the precipitated mass and the mass of the original hydrochloric acid, and the concentration = 1 (

There is no data in the calculation questions, and reasonable values can be taken arbitrarily. Let's assume that the solute in the original hydrochloric acid is 1mol [of course, you can also set it to amol], then we just need to find a way to find the volume of these hydrochloric acids, we can find the concentration, because there is "the mass of the precipitate is known to be equal to the mass of the original hydrochloric acid", so we first find a way to find the quality of these hydrochloric acids, then the mass of hydrochloric acid containing solute HCL of 1mol should be equal to the mass of AGCL generated by this 1mol HCL, that is, the mass of 1mol of HCL, that is. In this way, the volume of these hydrochloric acids is Noted, the unit is ml, and to become l, it takes 1000, i.e. l, so the amount of the substance of the original hydrochloric acid concentration = 1 (

==Well, you're getting the ratio wrong.

C(K+)=C(Cl )=1 2C NA+ =C SO42) Let's look at it this way.

a=a=a=a

a=a=a：a：2a：a=1：1：2：1

The ratio of the quantity and concentration of the four substances should be 1:1:2:1, and the ratio of the amount of the substance should also be 1:1:2:1

I don't explain.

c is 2nacl na2so4.K2SO4 is clear with a 2 before sodium chloride.