 Okay, so let's do this one. It says, consider the following reaction. The cool thing is it's already balanced for you, everything. It asks, is this a redox or non-redox reaction? It's real easy to figure it out because elements that are un-combined with anything else, like aluminum here or fluorine here, fluorine is un-combined with another type of element, those are always zero oxidation state. Okay, anything that's un-combined. Here we've got something that's combined, right? So its oxidation state is going to be different, okay? Collagens, group 7s are always negative, not always, fluorine is always negative 1, aluminum is always plus 3, okay? The oxidation number is the same thing as the charge effect, okay? Sometimes you'll see like the bigger halogens change their oxidation number, like chlorine, iodine, things like that, but fluorine is too small. So that has an oxidation number of negative 1? Negative 1, yeah. Oh, because there's three of them, right? Oh, okay. Right, so what you effectively are doing is bringing that up into, you know, what their formal charges would be, you know? So you say aluminum is, while fluorine is always going to be minus 1 and aluminum is always going to be plus 3, right? So you're kind of doing what we were doing with the net ionic and total ionic equation if you want to think about it that way. But not necessarily, you know? So that's how you figure out what the oxidation numbers of these things are, okay? So, aluminum fluoride has different oxidation numbers than aluminum and fluorine, so, yes, the oxidation numbers change, so it's a redox reaction, that's how you know. So if the oxidation numbers change, it's a redox reaction. Name the reaction type, it's a decomposition, right? Because you go from one thing to two things and then you can do the, yeah, that's, you can do the moles, the grams to moles thing. Yes, but I was confused. Should we do it? Yeah. Okay, so that's, you always get your conversion factors from your reaction equation, the coefficient. Oh, we're still using the same. Uh-huh. Yeah, so let's just do the thing. Okay. Guys, if 64 grams of aluminum fluoride react, how many moles of fluorine will be equal? So aluminum fluoride, 64 grams, so the number of moles of aluminum fluoride is going to be 64.0 grams, and then you're going to multiply that by the inverse of the molar mass of aluminum fluoride, and thank goodness it's all added up for me. So 83.98 grams per one mole. Cancel, cancel. That gives you the number of moles, right? And in this case, to three, six, six is going to be 0.762 moles of aluminum fluoride. And then, uh, ask how many moles of fluorine will be formed? Well, the number of moles of fluorine, you use that number of moles of aluminum fluoride, and you know there's a two to three ratio, and that's that conversion factor that you get from the reaction equation. So it's going to be 0.762 moles A L F3. Multiply that by two moles A L F3 on the bottom, three moles F2, cancel, cancel. 0.762 times three divided by two equals 1.14 moles F. Now that I know that we were using the same. Yeah, that makes sense today. So all of those things, I have like a lot of those where it's like keep doing the same thing, you know?