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Novay_Z [31]
3 years ago
14

If molecule dissociates in water what does it do

Chemistry
1 answer:
natita [175]3 years ago
4 0
It either forms a base or an acid...I think that's the answer you're looking for
You might be interested in
How can we make use of acids or bases to remove metals from soils?
gtnhenbr [62]
Adding acid and and catching the solution that drains through.
3 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
3 years ago
Ground water has some minerals dissolved in it. If you heat this water, collect the vapor in another container, and then cool it
liq [111]
The answer would be the last one- it separates dissolved substances.

Have a great rest of your day!
6 0
3 years ago
Steam at 200 °C is cooled to 105°c. What happens to the volume? Mass? Average kinetic energyy of the molecules and the particle
taurus [48]
The mass stays the same because if you have the same amount of steam then it can't change. The volume will get slightly smaller because the average kinetic energy of the molecules is less, so they move around less, so they take up less space. The particles are moving less fast. 
8 0
3 years ago
Be sure to answer all parts.
sergey [27]

The  moles of O2 are needed to react completely with 1.00 mol of C₂H₂ is 5 mol , 0.23  moles of C₂H₂ are needed to form 0.46 mol of CO₂ .

<h3>What is a Balanced Reaction ?</h3>

A balanced reaction is a reaction in which the number of atoms present in the reactants is equal to the number of atoms in the products.

It is given in the question that

the combustion of Acetylene is given by

2 H―C≡C―H + 5 O₂ → 4 CO₂ + 2 H₂O

Given moles of Acetylene = 1 mol

Moles of oxygen needed = 5 mol

as the mole fraction of Acetylene to Oxygen is 1 :5

The mole of Acetylene needed to form 0.46 mol of CO₂ is ?

mole fraction of CO₂ to Acetylene is 4:2

Therefore 0.23 moles of CO₂ is required.

To know more about Balanced Reaction

brainly.com/question/14280002

#SPJ1

5 0
2 years ago
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