1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nikitich [7]
3 years ago
9

24. According to the author, the term avant-garde

Chemistry
1 answer:
REY [17]3 years ago
5 0
F.a particular art movement founded intentionally in the early 1900s
You might be interested in
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
2 years ago
To study earths interior, geologist often rely on indirect methods, such as evidence from fossils.
igomit [66]

Answer: false

Explanation:

It is false that to study Earth's interior, geologists often rely on indirect methods, such as evidence from fossils. They rely on seismic wave.

6 0
3 years ago
Read 2 more answers
NEED HELP ASAP PLEASE
frutty [35]
C. NO2 is covalent. Hope this helps!
6 0
3 years ago
Read 2 more answers
How many particles are in 13.5 grams of Beryllium ?
Marina86 [1]

Explanation:

I don't know this answer what Hydrogen Helium lithium

4 0
2 years ago
Read 2 more answers
A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m
ioda
In this kind of exercises, you should  use the "ideal gas" rules: PV = nRT
P should be in Pascal: 
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = \frac{PV}{nT} = constant
\frac{P1 V1}{n T} = \frac{P2 V2}{n T}
==> P1 * V1 = P2 * V2
V2 = \frac{P1 V1}{P2} = \frac{445 0.016}{1225}
V2 = 0.00581 m3 = 5.81 L


7 0
3 years ago
Other questions:
  • Please help fast!!! What does the weight of an object depend on?
    7·2 answers
  • I need help can you help
    12·1 answer
  • The mineral rhodochrosite [manganese(II) carbonate, MnCO3] is a commercially important source of manganese. Write a half-reactio
    7·1 answer
  • What is the molarity of a solution made with 400 grams of copper II chloride in 3.5 liters of water?
    6·1 answer
  •  why don't we ever see solid or liquid oxygen on earth?
    14·1 answer
  • What best describes the reactants and products in a reaction at equilibrium?
    15·1 answer
  • Difference between pure and applied research
    6·2 answers
  • PLEASE I NEED HELP!!! I WILL MARK YOU AS BRAINLEST!!!!
    12·1 answer
  • Justify the statement nuclear energy is harmful as well as useful
    8·1 answer
  • imagine an atmosphere exists that contains every element in the universe in its atomic, rather than molecular form. which elemen
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!