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5. A nationwide study of 139 stream waters found a commonly used detergent surfactant; nonylphenol, in 52% of the 139 streams. The median stream concentration of 4-...

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5. A nationwide study of 139 stream waters found a commonly used detergent surfactant; nonylphenol, in 52% of the 139 streams. The median stream concentration of 4-nonylphenol was 700 ng/L_ Controlled studies to estrogen (namely 17a-ethinyl estradiol) found feminization of fish populations exposed to 2 ng/L_ Consider the information in the following table, would you expect feminization of fish populations exposed to 700 ng/L of 4-nonylphenol? Show all calculations needed to support your answer

5. A nationwide study of 139 stream waters found a commonly used detergent surfactant; nonylphenol, in 52% of the 139 streams. The median stream concentration of 4-nonylphenol was 700 ng/L_ Controlled studies to estrogen (namely 17a-ethinyl estradiol) found feminization of fish populations exposed to 2 ng/L_ Consider the information in the following table, would you expect feminization of fish populations exposed to 700 ng/L of 4-nonylphenol? Show all calculations needed to support your answer. Chemical 17a-ethiny estradiol 4-nonylpheno _ Estrogenic activity 1.0 0.0oo1 A scientific study reported the half life of NP in soil as 1.4 days: Calculate the time needed for degradation of 99.9% of NP applied to soil based on the above half life



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The first-order rate constant for the decomposition of a certain hormone in water at $25^{\circ} \mathrm{C}$ is $3.42 \times 10^{-4}$ day $^{-1}$. (a) A $0.0200-\mathrm{M}$ solution of the hormone is stored at $25^{\circ} \mathrm{C}$ for two months. Calculate its concentration at the end of that period. (b) Calculate how long it takes for the concentration of the solution to drop from $0.0200 \mathrm{M}$ to $0.00350 \mathrm{M}$. (c) Determine the half-life of the hormone.

This question will give him the first order rate constant for the decomposition of a certain uh reacted 1st. We want to find out What the concentration of the reaction will be at the end of two months if we start off with a 0 to Mueller solution. So to do that, we know that since it's the first order reactant reaction, the relationship between uh concentration and time should be Lynn I'm a zero over a. She equal K. T. So then plugging things in. We get that then uh 0.02, Mueller over X Should equal. Okay, just 3.4, 2 times 10 to the negative, four times 60. And this is because we're assuming that there's 30 days per month. About. So then we get the X 50 010196 more at the end of that month that you want period Next, we want to determine how long it will take for the concentration of the solution to go from .02 Moeller 2.0035 solar. So again, we're using that same equation and we find that it should take Approximately 5096.4 days. Finally, we want to find out what the half life is. So, since it's the first order reaction, The half life should be equal 2.0693 over K. So it should be 2026.32 days. Yeah.

Nuclear chemistry is the sub field of chemistry that deals with radio activity, nuclear processes and transformations in the nuclei of atoms. The first thing we need to do here is convert the half life from days two seconds and then plug it into our rate constant equation. Why don't we have K. Is equal to 9.693 divided by T. To the power of a half. So that gives us 123 times 10 to the six. That's just a half life in seconds. What we should get out is 5.63 times 10 to the minus seven seconds to the minus one. And so what we do with this now is plug it into the following equation. The following equation that we need to use is and is equal to M. Multiplied by N. A. Over big M. Big M. Is the molecular mass of the atom. Little M. Is the mass of the sample. We can calculate an we get 1.88 times 10 to the 17 atoms Following on from this. This is where we now use R. K. Constant value where we have A. Is equal to kate and multiply the two together to get a 1.59 times 10 to the 11 atoms per second. So now moving on to part B, what we have is Ln a knot of A is equal to Katie. So I would like to solve for his tea as we plug in the values that we know soul for tea. When we get 6.85 times 10 to the five seconds. Then what we need to do is convert the factor for time from seconds to days, divide by 3600 Multiplied by 24. We got 7.93 days and then continuing on for the final part of this video. What we have is the conversion factor from for the time. For days, two seconds time is equal to 652 times 10 to the seven seconds. We can calculate the activity after 755 days, we have L. N. A. North over A. It's equal to Katie. So what we need to do is solve for A and that is equal to 1.21 times 10 to the minus five atoms a second. Finally, we can calculate the number of atoms present A is equal to K. And you want to solve for N. Here. And what we get is 22 atoms.

Okay. So we're given that the half life is 14 0.3 days. So let's go ahead and use that to find RK All right. It's okay. Is the Ellen of two over our half life? So that's going to be .693 Divided by 14.3. So okay. Comes out to be .0485 days To the -1. Okay. So we know that for first order Ellen of a zero over a equals Katie. Now we can use concentrations or moles or grams. We can use almost anything in there. So, we're just going to take the L. N. Now it says that we went from Where we were to 68%. So I don't need the actual mass here. I'm just going to say if it started at one and ended at .68. Okay. And then we'll plug in our okay here With 0485 and there's your time. So this would take 7.95 days two. get down to 68% of what we started with no part B. We'll do the same idea here. Okay we're going to use the same equation. Ellen of A zero over a is Katie. In this case I am going to use our amounts. So our original amount again, I can use concentrations, moles or grams. Since I've got grams, I'm gonna go ahead and stay in there 10 times 10. The -6 g is what we started with. We'd like to know what we end up with here. So that's going to be RK0485. And they said this is 755 days. So the L. N. Of 10 with zero times 10 to the negative six minus the Ln of X Equals 36.6. So negative the Ln of X was 48 0.13 So we should get here at the allen of X negative 48 13 And our X is going to come out to be 1.25 Times 10. The Negative 21 g. But we've been asked to report it in adam's. So I'm just gonna go ahead and change grams to moles here. Okay, We use the molar mass of phosphorus and then we'll just multiply buy avocados number. Okay, Atoms in one more. Then we'll get about there's about 24 atoms remaining.

Hello, Bargains. My solution to problem number 23. Okay, so, um what we talked. We told this substance has an initial value off 100. That's 2.8 grams aunt. It has 1/2 life with two points here for five minutes. And we need to know how many half guys have passed when the substance amount of this substance reaches 8.3 grams on also the total time it takes for it too decayed to 8.3 grams. So before answering any of this, we need to find an appropriate formula which we can call after tea which models the decay off this substance. So we can obtain this by recognizing what we have in the question that we can take and drive this everything. So we know that the initial value of this substance is 132.8 grams. So when time is zero, so we can say tears. There are efforts zero The amount off this substance is 132.8 grounds. Perfect. So we also know that the half life off this substance is equal to 2.45 minutes. Now, how can we write this in terms off effort tape, we can say no. That's the value or f off t the value of tea when f of tears. He called 1/2 of its original amount. So 132.8 divided by two, which is in fact 66.4. But we don't need to worry too much about that for now. Um, value of tea when this happens is 2.0 full five, so we can just write him off. 2.0 awful. Fight is equal to 132.8. Bye bye, t amazing. So now let me share. Show you how it can use these two facts here to determine a formula Rafferty, which represents the decay off this substance. So a generic formula, which represents exponential decay, is off the full pay north times. Exponential functions part of Katie. Where Kay is it the less than zero a case? So we're gonna use the first cracked, but before zero is equal to 132.8 sub in T as zero into this formula here, we'll get a New York Times eat power off K time zero which is equal to the power off. I know. And we know that is equal to 130 to point A. So we now consulting a note to this formula here, Cassandra, 2.8 times, Ese Parro. Katie, no. Let's use the fact here that f boils 2.5 is equal to 132.8, divided by two. Okay, so we're gonna sub in t as two point Syria for five into this form your head. So we have 100 owners of this formula, hip 132.8 times eat power off 2.5 Subbing in tears that times came on, Then we're going to equate it to this value 132.8 or divided by a tick. Amazing. Okay, so we're gonna try and solve this equation for K. I slept K and we'll have a value so we can have a value off Paul. Um, Formula Head. First thing we want to do is divide by 132.8 on both sides. So dividing on this sidewalk canceled 132.8 and also dividing on this side or cancel the 132.8 on the top on the numerator. So we will be left with eats the power. Oh, two point Syria, five times k is equal to one pump. Okay, Next, we're going to take the natural look of both sides so natural local feet hurt. 2.45 times K is equal to the natural log off half. Okay? And if you remember your look rules which state that the natural log off £8 big is it would be times natural. Log off a you'll see that we can bring this whole power to the front of the natural look So we can hope. Looks, we can say, Do you point? Tearful five times. Okay, oh, times the natural look e equal to the natural. Look off. Ah, half on. If you also remember that the natural off e is equal to one. We can rewrite the US two points here for five is e cool chair. Um oh. To preserve for five k is equal to the natural log or 1/2 Amazing on dhe. Um, then we can divide both sides by 2.45 We can say OK is equal to the natural log off a heart or divided by 2.45 And if you work this out on your calculator, you should get an answer off. Negative. 0.33 8924 significant figures. So we've got value for Katie. We can sub that into this formula here. Uh uh, formula for the decaying off this substance is 132.8 times E. It's the power of negative 0.3389 tape. Amazing. Okay, so you got this formula. Now we can We want to know the time it takes for this substance to decay. Um, to reach amount off zero at 8.3 grounds. So essentially, we want to find T when f t is equal to 8.3. So we can equate this formula here, which is equal to everybody. Thio 8.3. Um, we should be able to solve the tea and get our answer. Um, so let's do you now. So 132.8 times east heart. Nick too. 0.33 point nine Steve Support to 8.3. Amazing. Okay, so first, we're going to divide both sides by 132.8. Cancel on the left hand side. Oh, on. We will just get this fraction on the right hand side. Okay? And then we could take the natural look of both sides on just as I did in the previous step. I'm gonna bring this to the front of the natural look. So gonna do that? Here, pick nine. Team, um, sequel to the natural log or 8.3 to put by 100 heads to point a and then the natural or E um is equal to one. So we can rewrite this after a 10.3389 t is equal to the natural log off. Um, what was it? A 0.3 by 100 on DDE. That's 2.8. Amazing. So we're gonna divide both sides by zero point next to 0.3389 to isolate the tea on. We will get him. I said what boy? So I put a negative in the numerator on dhe a negative. Avlynn deserve a point through 389 on the two nominees that If you type this into your calculator, you will get an answer off. 8.18 Um, yes. Now, Uh, yes. Sorry. Minutes. Um, so that answers the part of the question that says, What is the total time taken for the substance to decay to 8.3 grams? So that's that part of the question. But then it also asks, how many half lives does it supports to reach a 0.3 grabs? So we know that 1/2 life is equal to two points here for five minutes. That we have to do is divide 8.818 by the number for of 1/2 lives, which is to put cheerful five on. We get an answer off. Four. So the answer is, it surpasses. It goes through 4/2 lives to get to a weight off eight, amass off 8.3 grams. Um, yeah. Thank you for watching


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