So this is the sort of question that if you don't really know the tricks, it's going to feel impossible. But it's really not too challenging, if you do know. And so the first big thing you have to realize is that if the plot of something of one over a versus time where a is the concentration of reactive is linear, that tells us that this is second order with respect to a and since the reaction just a becoming be that also immediately tells us what the rate lot should be. The rate it's always first eagle toe, a rate constant. But since it's the only thing we have is a and we know it's to the second order who take a we square it. This is our rate lock. And so now we're already really along that well, since we also know this is a second order reaction, we can just copy the integrated rate law one over the concentration of a she looks a k T plus one over the concentration of a not where que is the rate constant. A. Not is the amount we started with, and this constellation of a is the amount at Time T. And so that's our integrated rate long and finally to figure out the rate constant K. We don't actually to do any work because we're told with the slope of the line is and another fact that comes associate with that's in this thing. You should really study the plots and when they're linear, meaning what? But anyways, the slope of this, if it's linear, is equal. Okay, that's the case for a second reaction. So we're told with the slope is and so we can just read off what Kate should be. It's 3.6 times 10 to the negative. Second leader squared her moles. It should not be leaders where it's just the leaders leaders promote for a second. It's now we have K. We want to figure out the House life for this reaction when he'd use the half life formula for a second order reaction, which is one over K times. The initial concentration we have all these values. The initial concentration were given as 2.8 times attention any of third. I have k right here and I can just plug this in and pump out a number, and it's equal to 99 to 0 0.6 seconds part of it. And so from here, if we want to figure out how long it would take to get to a certain concentration, we use the integrated rate lock. So remember, let's go right here. What's bugging? We're looking to get to the concentration of seven times 10 to the negative fourth. And we, as in all these cases, were starting at 2.8 times tense. Negative third. Okay, we know that. And so be careful with how you plug this in, but you can find the time pretty easily. It is equal to 29761 seconds. That's our final answer.