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aleksley [76]
3 years ago
15

Which statement describes clean air as a natural source

Chemistry
1 answer:
Vaselesa [24]3 years ago
6 0

Answer:

as I guess it is last statement

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For a reaction at equilibrium, which change can increase the rates of the forward and reverse reactions?a decrease in the concen
Monica [59]
Question requires a change resulting in an increase in both forward and reverse reactions. Now lets discuss options one by one and see there impact on rate of reactions.

1) <span>A decrease in the concentration of the reactants:
                                                                                       
When concentration of reactant is decreased it will shift the equilibrium in Backward direction, so resulting in increasing the backward reaction and decreasing the forward direction. Hence, this option is incorrect.

2) </span><span>A decrease in the surface area of the products:
                                                                               Greater the surface Area greater is the chances of collision and greater will be the rate of reaction. As the surface area of products is decreased it will not favor the backward reaction. Hence again this statement is incorrect according to given statement.

3) </span><span>An increase in the temperature of the system:
                                                                             An increase in temperature will shift the reaction in endothermic side. Hence, if the reaction is endothermic, an increase in temperature will increase the rate of forward direction or if the reaction is exothermic it will increase the rate of reverse direction. Hence, this option is correct according to given statement.

4) </span><span>An increase in the activation energy of the forward reaction:
                                                                                                   An increase in Activation energy will decrease the rate of reaction, either it is forward or reverse. So this is incorrect.

Result:
          Hence, the correct answer is,"</span>An increase in the temperature of the system".
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
2 years ago
Which two compounds are classified as bases by the Brønsted-Lowry definition, but not by the Arrhenius definition, and why?
konstantin123 [22]

Answer: Ammonia (NH3) and sodium carbonate (Na2CO3), because they accept hydrogen ions but lack hydroxide ions.

Explanation:

i took the test and got it correct :) hope this helps

6 0
3 years ago
HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
jekas [21]

Answer:

Abiotic is non living thing, while biotic is a living thing

hope this info helps

4 0
2 years ago
Why does beryllium lose electrons when forming ionic bonds,while sulfur gains electrons?
Nadya [2.5K]
Beryllium is a metal so it needs to loose electrons to have a full outer shell while sulfur is a non-metal so it needs to gain electrons to fill its outer shell.
8 0
3 years ago
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