A 0.0300 mol sample of NO2(g) is placed in a rigid 1.00 L reaction vessel and allowed to reach equilibrium at a certain temperature. What is the value of Kc at this temperature if 0.00500 mol of N2O4(g) is present at equilibrium

12.5

Explanation:

Step 1: Write the balanced equation

2 NO₂(g) ⇄ N₂O₄(g)

Since the reaction takes place in a 1.00 L vessel, the initial concentration of NO₂ is 0.0300 M and the concentration at equilibrium of N₂O₄ is 0.00500 M.

Step 2: Make an ICE chart

        2 NO₂(g) ⇄ N₂O₄(g)

I          0.0300        0

C           -2x            +x

E     0.0300-2x        x

Step 3: Find the value of x

Since [N₂O₄]eq = 0.00500 M, x = 0.00500

Step 4: Calculate the concentrations at equilibrium

[NO₂] = 0.0300-2(0.00500) = 0.0200 M

[N₂O₄] = 0.00500 M

Step 5: Calculate the concentration equilibrium constant (Kc)

Kc = [N₂O₄]/[NO₂]²

Kc = 0.00500/0.0200²

Kc = 12.5

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

: )

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.