(a) The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20 g>mL. The density of mercury is 13.6 g>mL. What do you predict for the height of a barometer column based on 1-iodododecane, when the atmospheric pressure is 749 torr?

Video transcript

everyone in this example, we need to calculate the height and meters of a column of benzene and it's told that we're exerting a pressure similar to that of a column of water. Were also given the density of our benzene as well as the density of water. So what we want to recall is that because we're dealing with the column of benzene as well as a column of water, we're dealing with the pressure head formula and we want to recall that our formula for the pressure of the head of a liquid is going to be equal to our pressure head times or rather equal to our density, multiplied by gravity, multiplied by the height of our liquid. And so according to this question, our pressure exerted by both bending and our column of water should be equal. And so we would say that our pressure head of our benzene is going to equal the pressure head of our column of what And so plugging in our unit. What we should have is for the left hand side of our equation, we're given our density of our benzene column of liquid, Which is given as .877 g per middle leader. We want to multiply this by our value for gravity, which we should recall is equal to nine . m/s. And we're solving for our height in meters of density. So we'll just plug in H Benzene. So this is going to be set equal to the right hand side of our formula. Where we're plugging in our given density of water as 1.00 g per milliliter. And then this is then multiplied by our gravity constant, which we recall is 9.81 m/s and then multiplied by the height of our column of water. In this question is given as 15 centimeters. However, because they want our final answer to be in units of meters for our column of benzene, we want to convert our height for water to also meters. So we're going to convert from m and two m in our numerator by recalling that our prefix anti tells us that for one centimeter we have 10 to the negative two power. And so this allows us to cancel our units of centimeters, leaving us with meters as our final unit for the height of water. And so now we want to go ahead and multiply our left hand side out. So taking the density times gravity, we're going to get a value on our left hand side for Benzene equal to 8.60337. And multiplying those units, we would have grams times meters divided by milliliters times second. And this is still multiplied by our height of Benzene. And then on our right hand side we can simplify it by taking the product of all of our parentheses here. So for our height of our water we would have 0. m. And then multiplying it by our density value of water and gravity We would have 1.47, sorry, 1.47, And then multiplying all of our units out. We would have grams times meters, which will be meters squared because of our multiplication of the meters times meters over here. And this is then going to be divided by milliliters times six. And so now to isolate for the height of our benzene column, we would have eight points 033. Sorry about that. Three, 37 g times meters divided by milliliters times seconds divided on both sides. And so we would be able to cancel this out on the left hand side. And what we would have for the height of our benzine fluid is that it's equal to a value Of 0.171. And then for our units we want to go ahead and on the right hand side cancel out our units of graham as well as our meters squared with one of the units of meters, which leaves us behind with Just one unit of m. And we can also get rid of middle leaders and our inverse. And so that leaves us with just our unit of meters in our numerator as our final unit for our height of our column of our benzene fluid. And this will complete this example as our final answer. So I hope that everything that I reviewed is clear. If you have any questions, please leave them down below. And I will see everyone in the next practice video.