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Fhancnk DODatiGhl R01?Ihntor"Radon: The Problem No One Wants t0 Face is the title of an article appearing in Consumer Reports_ Radon is gas emitted from the gr...

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Fhancnk DODatiGhl R01?Ihntor"Radon: The Problem No One Wants t0 Face is the title of an article appearing in Consumer Reports_ Radon is gas emitted from the ground that can collect in houses and buildings At certain levels it can cause lung cancer: Radon concentrations are measured in picocuries per liter (pCi/L): radon level of 4 pCi/L is considered "acceptable_ Radon levels in house vary from week to week_ In one house, sample of 8 weeks had the following readings for radon level (i

Fhancnk DODatiGhl R01? Ihntor "Radon: The Problem No One Wants t0 Face is the title of an article appearing in Consumer Reports_ Radon is gas emitted from the ground that can collect in houses and buildings At certain levels it can cause lung cancer: Radon concentrations are measured in picocuries per liter (pCi/L): radon level of 4 pCi/L is considered "acceptable_ Radon levels in house vary from week to week_ In one house, sample of 8 weeks had the following readings for radon level (in PCi/L). 24 5.7 55 81 19 (a) Find the mean, median and mode: (Round your answers to two decimal places:) Fnaan median mode (b) Find the sample standard deviation_ coefficient of variation and range_ (Round your answers to two decimal places:) rnac (c) Based on the data would you recommend radon mitigation in this house? Explain_ Yes_ since the average and median values are both over acceptable" ranges Yes_ since the average value is over acceptable' ranges, although the median value is not. No, since the average and median values are both under acceptable" ranges yes, since the median value is over acceptable' ranges_ although the mean value is not: cuomt Aneaor



Answers

Radon, a radioactive noble gas, is an environmental problem in some areas, where it can seep out of the ground and into homes. Exposure to radon-222, an $\alpha$ emitter with a half-life of 3.823 days, can increase the risk of lung cancer. At an exposure level of $4 \mathrm{pCi}$ per liter (the level at which the EPA recommends action), the lifetime risk of death from lung cancer due to radon exposure is estimated to be 62 out of 1000 for current smokers, compared with 73 out of 10,000 for nonsmokers. If the air in a home was analyzed and found to have an activity of $4.1 \mathrm{pCi} \mathrm{L}^{-1}$, how many atoms of ${ }^{222} \mathrm{Rn}$ are there per liter of air?

For this question. They are asking us to find the volume in leaders of Radan produced by 1 g of radium per day. So to do this they tell us that 1.0 times 10 to the 15th Adams of radium abbreviated are a produces 1.373 times 10 to the fourth Adams of Radan r n per second. So for this question, we're looking for the volume of Radan and we want that in leaders. And it says that the conditions for this reaction are at standard temperature and pressure. And they gave us the volume of radium, which is equal to 1.0 g. And we also know that the molar mass well, abbreviate that our A M m is equal to 226 g per mole and we're also gonna be using avocados number. So I'm was going abbreviate that with a big letter A and that's equal to 6.0 to three times 10 to the 23rd. All right, so the first start this question, we need to figure out the Adams that we have of rape are radium. So are a Adams. So to do this, We're gonna take Abu Cadres number. I wanna times that by our amount of radium that we do have so 1.0 g I want to divide this by the molar mass. So 226 scrims per mole and this will give us the atoms of radium that we have. So when we calculate itself, it's going to be two point 67 times 10 to the 21st Adams of radium that we have. All right, so now that we have the atoms of radium, let's get our Adams of radium or right on. So we have 2.67 times, 10 to the 21st Adams of radio and this is gonna equal between 373 times 10 to the fourth Adams because that's how much are produced from the radium. I wanna times this all right. And then this is the Adams of radium because that is how many atoms of Radan are produced from that many atoms of rating. Radiant. I would divide this by one point times 10 15 Adams of raping. So our Adams of Radan is equal to 3.67 times 10 to the 10 Adams of right on per second. All right, so we do want the amount of leaders in a day, so we're gonna calculate that out one day, his equal to 24 hours And when times that by 60 minutes, cause they're 60 minutes in every hour on then. We're also now times that by 60 again because they're 60 60 seconds and every minute. So we calculate this out, it's gonna be 86,400 seconds. So we know that 1.0 g of radium produces three points a seven times 10 to the 10 Adams of Radan per second. All right, so now we want the amount of zero point 1 g of radium for every day, so that's just gonna be 3.67 times 10 to the 10 are in Adams. And when the times that by that's 86,000 400 seconds and this is gonna equal 3.17 times 10 to the 15 are in Adams per day and then at standard temperature and pressure, we know that the volume of one bowl of Adams of any substance the volume is going to be equal to 22.4 leaders. That's based on avocados hypothesis. So to get the volume of our Radan, we're gonna take 22.4 leaders. We're not times that by our number of rate on Adam's, so 3.17 times 10 to 15 are in Adams per day. I'm gonna divide that by avocados number again so that we we just have the amount of leaders. So we're going to buy that by a 6.0 to 3 times 10 to the 23rd Adams. So when we calculate this out, the volume of Radan per day is gonna equal one point 18 times 10 to the negative seven leaders. And that is the volume of Radan that's produced by 1 g of radium per day.

All right. Pretty standard problem here. The very first thing I asked you to do is find the median. And so I've gone ahead and plug them. How to do that? You click equals median. Select all your data and it'll spit it out for you. Similar with quartz tiles. One in three. I like to type in court. Tell that E x c and select the data comma and put a one. And you do the exact same thing for quartile three. Except for the three in where that one is. And so on the next screen, I'll have the five data summary just from taking the highest and lowest values as well. Please, Please, please, please ignore this. Lowest value. That is incorrect. That low value is 48 28. I think it took a quick glance from the bottom up and saw that valued, uh, neglected this one right here. But there are there is your five data summary right there, correctly done. And this I had to find the mean in the centre deviation to see if there are any outliers. Because you use the Z test to check for outliers where you do that you select a value to track the mean divide by the Syrian deviation. So I went ahead and did that for this 1st 1 And then once you do it for the 1st 1 a little box will come up in the bottom right corner of the cell and you can drag it down and we'll do it for all the other ones. And so that's what's happened here. And you can see that there is not a single Z value less than negative three. But there is exactly one Z value greater than positive three, which makes sense. You can see this 23 25. That means $2,300,025 for that home and say that for sure, is an outlier because Izzy values greater than three. And so they wanted us to find the mean and compare it to the media. And the mean is 482.1, and the median is right here at 215 point nine. And the reason they wanted us to see this discrepancy is a thing like this outlier is gonna cause you're mean to be much, much fire. But the median is actually a better measure of the middle value of these home prices, like 215.9, is closer. Two more prices of the homes than the 482,000 is. So that's why they used this value as opposed to this value.

In part a of this problem. We want to determine what the density of Radan is at a temperature of 298 Kelvin in a pressure of one atmosphere. We can determine this by manipulating the ideal gas law in such a way. Isolate ratio of the number of moles of raid on, divided by the volume on one side of the equation. The reason that we want to isolate this ratio so that we can easily convert it into the density, which is a mass per volume of raid on by multiplying that ratio by Mu Moeller, Mass of Radan were given values in the problem for the pressure and the temperature. And we know what the universal gas constant are is so we can go ahead and directly solve for that and over v reef you in the following way. The pressure is one atmosphere, and we know that our is 0.8 206 leaders times atmospheres, her mole times, Kelvin and we're at a temperature of 298. Calvin. When we perform that math, we should get final answer for the ratio of an over V to be 0.0 for 09 Moles of Raid on for a leader. We know that the units will be most per leader because the other units cancel out toe leave this leaders per mole ratio and the denominator, which has flipped two moles per liter when it's brought up to the numerator. And that corresponds to this reissue of the number of moles per unit volume. Now, as stated before, all we have to do to go from that ratio to the density of Radan is multiply by the molecular weight, the molar mass of raid on which can be found on a periodic table. So now all we have to do to find that density is multiply that 0.409 moles per liter of raid on I. It's smaller mass which, when we look at the periodic table we can see is 222.8 grams per mole. When we multiply those two values together and round off 23 sig figs, we see that the density of raid on at the conditions stated in the problem is 9.8 rams. Her leader, which again is inappropriate unit for density as we have a mass invited by a volume, Part B is more of a conceptual question. It asks us if we would expect greater rate on concentrations to be found in the top floor of the building or in the basement. If we look at the density that we calculated in part A, we noticed that it is quite large. Moller Mass of Radan is relatively large compared to other elements in the periodic table. Larger objects in in terms of Adam's ones with higher molecular weights than others are expected to be heavier. And because of that, greater force of gravity pushes down on those heavier molecules in Adams, and we would expect that heavier objects sink. And because of that, the answer to Part B is that we would expect to find greater concentrations of raid on in the basement as opposed to in the top floor, because in the top floor of a building, we would expect to find later gas articles compared to ones with greater densities and that are there for heavier than those later molecules in raid on would be an example with that, with a density over nine grams

To describe radioactive decay. There are several different units that we sometimes use. One unit is called the Curie, and it's symbolized. See, I. Another unit that we use is the Becca Row, and one Curie is equal to 3.70 times 10 to the 10 Beck Carell's B que. So if we have description of radioactive decay in terms of 4.0 p. C. I for a leader and we want to change it into becquerels per leader, we first changed the P stands for Pete Go, which for every curie there are, tend to the 12 PICO curies. And then we can change it to pectorals, using our relationship from above. So this gives us a rate of 0.1 for eight. Back corrals her leader to describe how the activity changes over time. So given a certain amount of activity and trying to reach a new activity, we use the equation. A CI equals a zero times 1/2 G over t 1/2 where a zero or a not is the initial activity. 80 is the activity after a certain time has passed, and so ci is the time that's gone by and t 1/2 is the half life. So if we know that we have on initial decay an initial activity and 1/2 life, we can determine what the activity is after a certain amount of time. So if, for example, I have an activity of 41.5 PICO curies her leader and I want to know what the activity is after 9.5 days in the half, life is 3.82 days. I can solve for the activity by multiplying 45 times, 1/2 raised to the 9.5 divided by 3.82 from this equal 7.40 He could curies her leader, which, if I want to express it and Becca rolls, I do the same conversions as I did above. So for every one Curie, they're one times 10 to the 12 he could Curie's and again. For every Curie, there are 3.70 times 10 to the 10 Becker owls, so the activity after 9.5 days is zero went to seven for Becker girls for leader. We can also use this equation to find the time it would take to reduce the activity to a certain level. So instead of solving for activity, I'll solve for T. I want to know when the activity reaches the safe level of 4.0. Peek a curious per leader, and I know that I have an initial level of 7.40 people. Curie's for leader times 1/2 a t which is what I'm solving for over the half life, which is 3.82 days solve for T. I divide both sides by 7.4 just 0.5405 It was 1/2 seeing over 3.82 Because she is in the exponents, I have to take the natural log of both sides. So the log of 0.5 for 05 equals the natural log of 1/2. 4.5 Trump's t over 3.82 dividing both sides by the natural log of 1/2. Yes, quite eat a 76 equals t over 3.82 or solving for tea. Three went 39 days


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