So we're given several thermochemical reactions, and the, we have the reaction and the delta H for them, and we're trying to calculate the standard heat of formation of this strontium carbonate. Now, what does that mean? The heat of formation of a reaction is the energy changed to produce 1 mol, or the enthalpy change to produce 1 mol of this substance from the elements in their most stable form. So Sr is a solid that's a metal, it's a solid, that's its most stable form. Carbon's most stable, stable form is graphite, and oxygen's most stable form is O2. And we need to balance it for 1 mol, so we have to keep a one on this side, because that's the definition of heat of formation. So we are going to put, let's see a, three halves here to give me 3 oxygens. So we have 1 carbon, 1 strontium already. So what we're trying to do is, determine the delta H of this reaction. The delta H of this reaction would then be, by definition, the heat of formation. So let's try to construct this reaction from these puzzle pieces. We need strontium on the left-hand side of the equation, 1 mol of it. We look up here and the only equation that has strontium is this first one, but it's 2 mols on the right-hand side. So we have to reverse the reaction and divide by 2. So we will reverse the reaction and divide by 2, okay. If we do that, we can determine the delta H of this reaction. By changing the sign, because we reversed the reaction, and dividing it by 2, because we cut it in half. All right so that gets our strontium in place. Now we're ready to get a carbon in place. So here's the graphite equation, this third equation here, it's got 1 mol of graphite on the left-hand side, that's what we needed, so we use it exactly as we see it. Carbon graphite plus O2 gas producing carbon dioxide gas. Since we used it just as we saw it, take the delta H just as it comes off the table there 394 kilojoules. So then I got the graphite in place. The O2, so far we've got one and a half mols of O2, okay? So it looks like it's good. Now let's make sure we get the SrCO3 in place, okay? So we look up here and SrCO3 come from the second equation. It has 1 mole on the right-hand side, so we're going to use it just as we see it. SrO solid plus CO2 gas producing CrCO3. Since it's just Sr, sorry CO3 and that's not a gas. It's a solid, okay. So the delta H of this reaction, I read it right off of here is a negative 200, boy, I'm having a hard time writing, 234 kilojoules. Now, let's verify that all of these do add up. To give me the reaction I'm looking for. I'm trying to obtain this reaction. So we've got the strontium. We've got one and half mols of oxygen. We have Sr0, cancelling with Sr0. We've got carbon dioxide, cancelling with carbon dioxide. And that leaves me with exactly what we see up here, one Strontium, one carbon, and a total of one and a half oxygens, three half oxygens. And the SrCO3. So yes, that does construct this reaction. So these have to add up, to give me the delta H of the reaction. When you add all of these up you will get a negative 1220 kilojoules. So, that is the delta H of this reaction, and by definition that would be the calculating standard heat of formation of the strontium carbonate.