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KengaRu [80]
3 years ago
8

Which of the following atoms would have the smallest atomic radius?

Chemistry
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

Lithuim is the smallest

Explanation:

i hope this helps!!

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A gas has a volume of 52.1 L at 1.55 atm. What is the new volume at a pressure of 2.00 atm?
Deffense [45]

Answer: 40.38 L

Explanation:

Formula: P1V1 = P2V2

(1.55)(52.1) = (2.00)(x)

80.755 = 2x

40.38 = x

6 0
3 years ago
State whether the following statements are true or false. If false, explain why. (a) A reaction stops when equilibrium is reache
Elden [556K]

Answer:

(a) False;

(b) False;

(c) False;

(d) True.

Explanation:

(a) When equilibrium is reached, the forward reaction rate becomes equal to the reverse reaction rate, that's why the molarity of each species remains constant, but reactions don't stop.

(b) According to the principle of Le Chatelier, an increase in molarity of either reactants or products would lead to a disturbance of equilibrium. This disturbance would lead to the shift of equilibrium towards the side which would minimize such a disturbance.

(c) Equilibrium constant is only temperature-dependent, it's independent of molarity, pressure, volume etc. of any species present in the reaction.

(d) The greater the initial molarity of reactants, the more products can be formed, e. g., since the ratio of products to reactants should be kept constant, the larger the amount of reactants, the greater the amount of products formed to keep a constant ratio.

7 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
Which description best characterizes the motion of particles in a solid
bixtya [17]
<span>The movement of particles within a solid is extremely slow when compared to that of a gas. It is also significantly slower in movement than that of the movement found within the particles of liquid. The more movement present and the faster the movement of the particles the more space will be present between each particle. This causes the material to spread out as they become less densely packed within a solid material.</span>
8 0
3 years ago
Read 2 more answers
Study the figures below which illustrate the steps in the following chemical reaction:
Tanya [424]

Answer:

Hello - this is Mrs. Gussman, your chemistry teacher.  I wrote this exam question and posting it online is a violation of the academic integrity policy.  Remove this post immediately.

Explanation:

7 0
2 years ago
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