So we have here an unusual experiment of this type. Uh, when we look in the data table for the different concentrations and rates, we see that the Hypo chloride ion, the OOCL minus, is held constant and experiments one and two and then again in three and four. It's a 40.90 in three and four, so we can pick either pair. Experiments wanted to or experiments three and four toe. Look at the effect of changing the iodide. But then, when we look at the iodide concentration, there are no experiments where it's helped. Constance. We'll see what we need to do in this situation. Uh, what we need to do first is what we can do. Which is, uh, I picked Teoh. Compare experiments three and four. As I say, we could have done one or two or three and four, so I've set it up. The rate ratio and three and four is equal to the iodide concentration ratio and three and four, where the Hypo chloride is 0.90 molar in both three and four, and so have plugged in the rates for experiments three and four and the ah iodide concentrations in three and four. And basically those are both ratios of eight. And so that tells me that the, uh, reaction is first order in iodide. Once I know that, then I can pick any two of the other experiments. Even though I can't hold the iodide constant, I can take account of its effect because I know the order with respect to it. So now when I write the ratio, I picked experiments three and two. Um, I include the iodide concentration ratio in those two experiments, as well as thehyperfix chloride concentration ratio. And so, uh, plugging in the rates in experiments two and three, I find that the ratio of the rates is four and that both of the reactant is doubled in concentration in those two experiments. And so that gives me a known for the iodide to the first powers to and dividing both sides by two. Then I have That, too, is to to the Z Z being the letter I chose for the experiment on the hypochlorite. And so the reaction is first order with respect to both reactive. Now we need the rate constant here. And so what we do is, we write the great law. Uh, and so for K divide both sides by the iodide. Yeah, and the hyper correct. And so picking. Experiment 17.9 times 10 to the minus two. And that is Moeller per second. Our units canceled everywhere else. Units were the same in all of those ratios. But in calculating K, we want to take account of the units, uh, and the concentrations and experiment one or 0.12 Moeller and she report 18 Mueller. And so I can calculate the value of K, and we get 3.66 Eso won power of Moeller and two powers of Muller per muller per second. And so that then is theory rate constant. And for part c, the rate it's going to be I'm going to take a plug in given concentrations. And so, uh, both art. 0.15 Moeller very much like trial number one. We're going to get a rate. It's about the same as we had in trial number one. So notice Moeller and Moeller Moeller squared. Uh, the units on K need to combine with those on the concentration. So is to give a rate in units of polarity per second. And so, without rounding off as I did to write this down, I'm just going to go ahead and multiply 8.2 four times 10 to the negative, too. Muller per second. That is how you get an initial rate with given concentrations. You've gotta have the rate law with the dependence on concentrations, work up the value of the rate constant so that you can plug that into the rape law and then plug in any concentrations and you can calculate the right.