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
zaharov [31]
2 years ago
6

Which is the answer?

Chemistry
2 answers:
Digiron [165]2 years ago
8 0
I believe your answer is B, but im not completely sure
professor190 [17]2 years ago
3 0
The answer is B.Steel
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
In what sense are photons quantized?
lara [203]

Answer:

D. Each photon has a specific amount of energy​

Explanation:

5 0
2 years ago
7
Umnica [9.8K]
Answer: Volume im pretty sure
3 0
2 years ago
Read 2 more answers
Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2
lapo4ka [179]

Answer: Concentration of CO = 0.328 M

Concentration of Cl_2 = 0.328 M

Concentration of COCl_2 = 0.532 M

Explanation:

Moles of  CO and Cl_2 = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of CO and Cl_2  =\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M

The given balanced equilibrium reaction is,

                            CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial conc.          0.860M     0.860M           0

At eqm. conc.    (0.860-x) M  (0.860-x) M     (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

Now put all the given values in this expression, we get :

4.95=\frac{x}{(0.860-x)^2}

By solving the term 'x', we get :

x =  0.532 M

Thus, the concentrations of CO,Cl_2\text{ and }COCl_2 at equilibrium are :

Concentration of CO = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of Cl_2 =  (0.860-x) M  =  (0.860-0.532) M = 0.328 M

Concentration of COCl_2 = x M = 0.532 M

3 0
3 years ago
How many significant digits are in the measurement 64,000 mg<br> A.2 <br> B.3<br> C.4<br> D.5
Y_Kistochka [10]
5 because the 0's all matter.
8 0
3 years ago
Other questions:
  • I ONLY NEED 1 EXAMPLE for each one PLZ HELP ME I WILL MAKE BRAINLIST HELP ME FAST PLZ!!!! Using a skillet on the stove to cook s
    5·2 answers
  • The lock-and-key mechanism refers to
    11·2 answers
  • Within each group of four atoms or ions presented below. select the species that are isoelectronic with each other:Note: you are
    5·1 answer
  • On a graph, which type of line shows a direct proportion? A. upwardly curved line B. downwardly curved line C. straight line D.j
    7·2 answers
  • Woodpeckers find insects to eat by pecking holes in trees with their beaks. One day, a woodpecker finds a particular tree that o
    14·1 answer
  • How man grams of cl2 are consumed to produce 12.0 g of KCl
    7·1 answer
  • For the Balanced Reaction: 3 Mg + Fe2O3 → 3 MgO + 2 Fe
    5·1 answer
  • 2 examples of when scientists used space probe technology to find out more about comets.
    11·1 answer
  • IF UR GOOD AT CHEM PLZ ANSWER MY PREVIOUS QUESTIONS ON MY PROFILE, I NEED HELP!!
    13·1 answer
  • How many moles are in 16 grams of nitrogen molecule
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!