At a certain temperature, 0.660 mol SO3 is placed in a 2.00 L
container.

2SO3(g)↽−−⇀2SO2(g)+O2(g)
2
SO
3
(
g
)
↽
−
−
⇀
2
SO
2
(
g
)
+
O
2
(
g
)

At equilibrium, 0.150 mol O2
0.150
mol
O
2
is present. Calculate c.
K
c
.

0,15

Explanation:

Kc is the constant of equilibruim of concentration.

[ ] that symbol means concentration.

Kc = Products/Reagents

Kc = [SO2]² . [O2]/[SO3]²

The questions says that in equilibrium we have:

[0,15mol] of O2

Now, looking to the balancing of the equation, we see that SO3 and O2 are in a disposition 2:1. Which means that for every molecule of O2 formed, we need to use 2 molecules of SO3.

Then, we used 0,30 mol of SO3 for that convertion. That also means that we have 0,30mol of SO2, because he is in a disposition of 1:1.

[SO2] = 0,30mol

[SO3] = 0,30mol

Kc = [0,30]². [0,15]/[0,30]²

Kc = 0,15

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

increased stability in the completion of fluorine's outer shell.