 Okay, so let's attempt this one. It says the equilibrium constant Kc for the reaction hydrogen gas plus carbon dioxide gas goes to water gas plus carbon monoxide gas is 4.2 at 1650 degrees Celsius. Initially a 0.80 mole of hydrogen and 0.80 mole sample of carbon dioxide were injected into a 5 liter flex. Calculate the concentration of each of these species in equilibrium. Okay, so let's go about doing this. So they're giving us the Kc, so that means the concentration in molarity. Okay, so if we look here, we've got number of moles, number of moles and volume. So those are molarity, but from those numbers we can get the molarity. Okay, so how do we do that? Well, let's take these n's and replace them with n's. So what's the molarity? It's going to be the number of moles divided by the classical. Okay, so when we do that, we're going to get 0.16 molar hydrogen and 0.16 molar carbon dioxide. Is everybody okay with doing that first one? So let's now, so we've got the value for Kc, but we don't have the Kc expression. Okay, and remember we get that Kc expression from the balanced reaction equation, which we have a balanced reaction equation up here. Let's go ahead and write out what the expression for Kc would be. Well remember it's going to be the product raised to their coefficients. So concentration of water times the concentration of carbon oxide divided by the reactants raised to their coefficients. So this one is the fairly straightforward Kc. So I'm actually going to need this middle piece, this room in the middle for writing an ice table, as you guys have predicted already, I see. Would you mind if I raised this portion? So would you okay with that? Write our ice table, initial change equilibrium. Okay, so it said we started out with only hydrogen gas and only carbon dioxide. Yes, so when we put those in, we're going to put 0.16 there, 0.16 there. So that means we didn't start in with any water and we didn't start with any carbon dioxide. The change, of course, well, since we only started with reactants and no products and this is in equilibrium, it's going to be pushed that way. So that means minus x there, minus x there, plus x there, plus x there. So what are we going to get in equilibrium? 0.16 minus x. So this one's nice in setting up to be perfect squares, so that's really nice. Okay, so let's go ahead and plug in our equilibrium because this is Kc, right, so it's at equilibrium. So let's plug in our equilibrium values from our ice table. So we've got 1.16, or 0.16 minus x, 0.16 minus x, x and x. So let's put them in the proper positions. So water, x. Carbon monoxide, x. Hydrogen gas, 0.16 minus x. Carbon dioxide, 0.16 minus x. So this, of course, also equals x squared divided by 0.16 minus x squared, like that. So we can take those squares out and make the whole thing squared. So let's just plug in 4.3 because that's the value of 4.2, sorry, 4.2 because that's the value for Kc. So we're going to have 4.2 equals x divided by 0.16 x and we're going to square the whole thing. Okay, so how do we get the square out of there? We take the square root of both sides. Okay, so the one side is going to be x divided by 0.16 minus x and the other side is going to be x divided by 0.16 minus x. So the other side, square root of 4.2, which is 2.05. It's actually going to be two significant figures, so we'll clear that up in a little bit, okay? But let's just keep the three significant figures right. So now let's try to isolate x. So how do we do that? Multiply both sides by 0.16 minus x. So times 2.05 equals x, like that. So let's multiply that by 2.05 times 1.6. So that's going to be 0.3. So what we're going to do is we'll reduce this now finally to two significant figures. So 0.33 minus 2.05x equals x. So move this over to the other side. So this is going to be 3.05x equals 0.33, figuring out the value of x. Now x is going to be 0.33 divided by 3.05, and that should give us 266. So I get 0.11 molar for x. So these here are the concentrations at initial, or these initial concentrations. So what are the equilibrium concentrations? Let's figure that out now. So that's going to give us all we need to figure out concentrations. Hydrogen gas concentration at equilibrium and equal 0.16 molar minus 0.11 molar. So that's going to give us all we need to figure out the concentration. So that equals 0.05 molar. Hopefully that makes sense, right? The concentration of CO2 and equilibrium is going to be the same, 0.16 molar, 0.05 molar, and then the concentration of H2O at equilibrium, well that equals x, right? So that's going to be 0.11 molar. And the concentration of carbon monoxide at equilibrium is also x, so it's also 0.16 molar. So that was a nice table one, pretty involved, but not too terribly complicated, okay? Thank you.