About the minimum solubility of gases in the fountain experiment

The result is about 1:17, but I don't know if it's right or not......

At the beginning, the atmospheric pressure in the bottle is p0

When the gas in the bottle dissolves in water, the pressure decreases, and the atmospheric pressure causes the water to enter the bottle from the catheter. Therefore, it is a critical condition that the water can just enter the bottle after the gas is dissolved. At this time, the sum of the air pressure in the bottle and the water pressure of the catheter is exactly the atmospheric pressure. Namely.

p1 + gh = p0

By the formula pv=nrt, the water can be disregarded compared to the total volume of 250ml, so v is considered constant, and r is constant, and t is likewise considered constant, then the change in p is proportional to n, i.e.

p1 n1 = p0 n0, variant, n1 n0 = p1 p0

The part of n1 that is less than n0 is dissolved in water. The volume of this part (set to va) should be at p0.

va=（n0-n1）rt/p0

In the initial state, p0v=n0rt, i.e., rt p0=v n0

va=（n0-n1）*v/n0=v*（1-n1/n0）=v*（1-p1/p0）=v*[1-(p0-ρgh)/p0]

where, v=250ml, p0=103400pa, =1000kg m3, g=10m s2, h=35cm=

Substituting the value va=, the solubility is va, which is about 1:17

When we reviewed the experiment a while ago, 1:40 could be generated.

But unlike HCL1, the phenomenon of hundreds is obvious.

1:1 carbon dioxide is counted.

40 should be the bottom line.

If high school.

Actually, theories are fine.

A NH3 is very soluble in water and can be used for fountain experiments, so A is correct;

B HCl is very soluble in water and can be used for fountain experiments, so B is correct;

C SO2 can react with NaOH solution and be absorbed, thus forming a pressure difference, which can be used for fountain experiments, so C is correct;

D no is insoluble in water, nor does it react with alkali or acid, because the jujube residue is not a salt oxide, can not be used as a fountain stool to repent the experiment, so d is wrong

Therefore, the former finch d

NH3, HCl, SO2, NO2 and water can form fountains. These gases are characterized by being (very) soluble in water. When dissolved in water, the difference in air pressure is obvious, resulting in the formation of fountains.

Fountains can also be formed by acidic gases and NaOH solutions, such as CO2 and Naoh, SO2 and NaOH, etc. Because these gases are more soluble in solution.

The result is about 1:17, but I don't know if it's right or not......

At the beginning, the atmospheric pressure in the bottle is p0

When the gas in the bottle dissolves in water, the pressure decreases, and the atmospheric pressure causes the water to enter the bottle from the conduit. Therefore, after the gas is dissolved, the water can just enter the bottle, which is the critical condition for slag annihilation. At this time, the sum of the air pressure in the bottle and the water pressure of the catheter is exactly the atmospheric pressure. Namely. p1gh

P0 is determined by the formula pv=nrt, compared to the total volume of 250ml, the water can be disregarded, so v is considered constant, and r is constant, and t is likewise regarded as constant, then the change in p is proportional to n, i.e.

p1 n1p0 n0, variant, n1 n0

The part of p1 p0n1 that is less than n0 is dissolved in water. The volume of this part (set to va) should be at p0.

va=(n0-n1)rt/p0

In the initial state, p0v=n0rt, i.e., rt p0=v n0

va=(n0-n1)*v/n0=v*(1-n1/n0)=v*(1-p1/p0)=v*[1-(p0-ρgh)/p0]

where, v=250ml, p0=103400pa, =1000kg m3, g=10m s2, h=35cm=

The solubility is va, which is about 1:17

Solution (1) Let the amount of NH3 be X, then the volume of NH3 dissolved in water = the volume of NH3 and the volume of gas = XMOL*

NH3 dissolved in water to obtain the amount of the solution solute concentration = n v = xmol mol l =

2) NO and O2 (volume ratio 4:3).

Let the amount of Nuo in the mixed gas be 4x, and the amount of nitric acid generated is Y, then the amount of O2 is 3x, and the volume of the resulting solution = (4x+3x)mol*

4no+3o2+2h2o=4hno3

4 3 44x 3x yy=4x

The amount of substances in the resulting solution solute concentration = n v = 4xmol mol l = (3) NO2 and excess O2

Let the amount of NO2 in the gas mixture be 4x

4no2+o2+2h2o=4hno3

4 1 44x x 4x volume of the resulting solution = the sum of the volume of NO2 and the volume of reaction O2 = (4x+x)mol * the amount of substance of the resulting solution solute concentration = n v = 4xmol 112xl =

Solution (1) Let the amount of NH3 be X, then the volume of NH3 dissolved in water = the volume of NH3 and the volume of gas = XMOL*

NH3 dissolved in water to obtain the amount of the solution solute concentration = n v = xmol mol l =

2) NO and O2 (volume ratio 4:3).

Let the amount of Nuo in the mixed gas be 4x, and the amount of nitric acid generated is Y, then the amount of O2 is 3x, and the volume of the resulting solution = (4x+3x)mol*

4no+3o2+2h2o=4hno343

44x3xy

y = 4x the amount of substances in the solute of the solution obtained = n v = 4xmol mol l =

3) No2 and excess O2

Let the amount of NO2 in the gas mixture be 4x

4no2+o2+2h2o=4hno341

44xx4x

The volume of the resulting solution = the sum of the volume of NO2 and the volume of the reaction O2 = (4x+x)mol * the amount of substances of the solute of the resulting solution, the concentration = n v = 4xmol 112xl =

Choose A fountain formation to have enough pressure difference, so the gas should be very soluble in the liquid (or reaction), and the solubility of CO2 in A is too small to form a fountain, B and C The solubility of gas is very large, yes, D will react, so it can also.

In the case of standard conditions, c=1 (mol l).