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Question

Eurorg aach noclln arelos: lomarderdomcm ricaled oporijon:ExnressEna Ncicalin { mcles bom ty canmiaFdcaled LuetelnJen2YTUR - 2V5R "2v40a 2+n8R + 2-50R - 24n3R Isimpinac_ ansiar |on axidansnnr Lima NcIciaarneeded354*& _ 82a*& Tha €mclnad Icrm 0E 1645 92*t smcimvour ~ucanswcr vroradkzals natlea51251 PtJenzants2844"14ad 27R5*4#faq 47_Findu = conjunale ol aach nrenrunbSnatUutol Aume vanablos rcprezent ncn-ncgatne walve?67 * Ys-y "Vt 57T7(36-74m)? (JYm-746)" D 'Sie

Eurorg aach noclln arelos: lomarderdomcm ricaled oporijon: ExnressEna Ncicalin { mcles bom ty canmia Fdcaled Lueteln Jen 2YTUR - 2V5R "2v40a 2+n8R + 2-50R - 24n3R Isimpinac_ ansiar |on axidansnnr Lima NcIciaarneeded 354*& _ 82a*& Tha €mclnad Icrm 0E 1645 92*t smcimvour ~ucanswcr vroradkzals natlea 51251 Pt Jenzants 2844"1 4ad 27 R5*4# faq 47_ Findu = conjunale ol aach nrenrunb Snat Uutol Aume vanablos rcprezent ncn-ncgatne walve? 67 * Ys-y "Vt 57T7 (36-74m)? (JYm-746)" D 'SieIa #Que Jrile1704 a0 e{ac aneval vuing(drak 49 044040 Tr= Coniuoaia 0i /6 (TFc vour inW0 Intho @DMjtol [Simallyeel ansahi Use ensava Hnnent eclya coniuqulc (Wyp? YoUr ATwt in Ihe bin4 +5 )



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(II) Rubidium-strontium dating. The rubidium isotope
$\frac{87}{37} \mathrm{Rb}$, a $\beta$ emitter with a half-life of $4.75 \times 10^{10}$ yr, is used to
determine the age of rocks and tossils. Rocks containing
fossils of ancient animals contain a ratio of $\frac{87}{38} \mathrm{Sr}$ to $\frac{87}{37} \mathrm{Rb}$ of
$0.0260 .$ Assuming that there was no $\frac{87}{38}$ sr present when the
rocks were formed, estimate the age of these fossils.

Using the isotopes and atomic mass. That simulation we're looking at mixed isotopes and specifically the element boron eso setting up the simulator for a right symbols of the isotopes of boron that has shown us naturally occurring in significant amounts. The two isotopes of boron would be born on 10 and born 11 for be predict the relative amounts of percentage of these boron isotopes found in nature and explained the reasoning behind your choice. If we look at the average atomic mass of boron on the periodic table, we'll find that this is equal to 10.8 g. So I would predict knowing the two isotopes airborne on 10 and born 11, I would say that this would be, um, I guess he would be 80% born 11 and 20% boron 10 because the masses slightly shifted towards the born 11 isotope for part, C had isotope to the black box to make a mixture that matches your prediction. You may drag isotopes remove the sliders, so let's go ahead and make this mixture. Here, I'm going to drag in. I ate four on 11 isotopes from the box, so that would be here so 12345678 And I'm gonna draw to bore on 10 and for see, I will see if I could take a picture of this year for us. So this would be our mixture in See where we've got eight boron 11 and to bore on. 10 isotopes now for D were asked to reveal the correct amounts, uh, reveal the percent composition and average atomic mass. How well does your mixture compare? So let's go back and will reveal. And, um so I can see that by revealing this. Let's take a picture here. When I revealed percent composition an average atomic math way, I see that my prediction is, um my prediction is correct. 20% for the isotope or on 10 8% for the isotope born 11 Thea Bridge. Atomic mass from a periodic table was 10.8. So, uh, I'm going to write here that my prediction in parts be waas um extremely close. Uh huh. You know, or I'm gonna say it was correct because the ratio was 80 20. And lastly, for part e, select nature's mix of isotopes and compare it to your prediction. Zil will select the nature's mix of isotopes, and we'll compare it to our prediction. So for part E, when we select the nature's mix and on our screen, it looks here and way use this. So for nature's mix, um, mhm, you can't see. We can see it here, um, em for nature's mix. We can see that the, um percent composition we have 19.9% of boron 10 and 80.1% boron 11 and the average atomic mass is 10.811 to the average atomic mass is 10.811 00 am you and we'll compare this to our prediction. So my prediction was extremely close to the nature's mix given in the simulation mhm.

This question asked us to rank these dying's in terms of reactivity in a deal. Older reaction. So the first thing we need to know about a deal's older reaction is that the S sis confirmation is the only way that it can react. So if a compound cannot be and the SS confirmation if it is locked into s Trans, it is not capable of doing a deal's holder. So the only one here we have that is locked into the S trans position. Is this one right here? So, just for contacts, Estes looks like this and s Trans looks like this on and we need the esus or me this for a deal's older reaction. So the only one here that we have this locked into S Trans is this one. So I was going to be our at least reactive at number four on then. The only one that is locked into the S. Trans is this ring over here into this is going to be our number one, our most reactive. Then we have these two in the middle. Until these come both interchange between the Estes and S Trans. Um and so the only real difference between them is when they're in the excess position as they're both drawn here. Is Derek hindrance. So, this one, it does have methyl groups on and put their out of the way of the side that deals all the reaction occurs on that happens over here, and all of our methyl groups are over here. So these aren't gonna have a big effect on the Syria Kendrys in this reaction. But on this one this month, a group is right in the way of where the reaction happens. That's definitely going to inter introduced since Derek and, uh, hindrance. So because of that, this one will be less reactive than this one. So in order, this is number ones are most reactive. And then this one this one and the mists are least reactive.

Anybody. So we want to find aides of the rocks were given half life of the Rubio or video? Yes. Video. Sorry. Half life as 4.75 hides on a year. Okay? And we're also given 87 Binyam and 87 great sodium. Yeah, and I'm just gonna process off so we'll look at it. And a fine time Work on a out and, uh, are year times over You time half life? Yeah. And for this, we're actually gonna Tane inside here. You out in of the number of Michael's inside. You are a believer. GM Media A. There were no near the original and the ones that we found minus out. And you And now we want to solve this part right this second that we're gonna do subs some algebra, and I want to this part of you. So you want to know? Number of rhodium president. Our listen all or video minus video. Then we have the original video. Hi. Um, plus and medium and video People's 0.0 six. The end up and it computes not here. We end up where in a r E equals the original. Over the video here. And then we have ADM and the video plus one and was a plug this into here because this part from here from these two places Ok, uh, here, we're gonna have 0.26 plus one. That's gonna eagle 1.0 six. Now, we're gonna go back up. There are equation that we wrote and in this place right here. What? We originally had video video, and you have to time half life we're gonna put And how it is, Alan, too. Time, half life. And they gave us the half life numbered up before. 0.75 finds 10 years. Yeah, And it's all of this is your populate, er you're going to get 1.7 610 through the night here. Okay? And this is your answer, and I hope that helps. I know it got kind of weird right in the but sometimes you'll see this happen, especially to simplify. Okay. Thank you, guys.

So we have the following organic molecule. We have won our to our one to dive aroma one to die final ethane and we want to draw a new map projection. We also want to show what would happen if we performed in 82 elimination on this organic compounds. Um, we just want a double check to make sure that we have the right, uh, configuration. So it's going to look at the left carbon here and a sign our priorities so quickly going here we know that our berm Oh, he's gonna be our priority one. We know that this carbons gonna be already too, because I was a connecting roaming to it. Ah, the finals gonna be party three and our hajj is gonna be proud of for all right, so that looks like it's going Ah, counter clockwise, which would appear s, however, says we want the hydrogen going away from us, and right now we haven't going towards us. We would say it's the opposite. So it's actually our configuration. If we do the same thing for the right carbon, uh, we would assign our highest party group here, which is the roomy legal that one, Uh, this carbon view to because of the connecting bro Mean the fennel would be three and hydrogen with before here it looks like it's also going counterclockwise, but since again we have the hydrogen going towards us. But we wanted to go away from us. It's actually are as well. So now that we have, ah established that it is indeed the right configuration, how we draw on human projection of this. So let's say we haven't I hear and we're looking down the organic molecule in that direction, looking down that bond so the resulting Newman projection would look something like this. So we'll go ahead and draw the ah, front carbon donated by, ah period here. And we know that the bones were 120 degrees from each other so well, look, something like this looks like the final group is pointing upwards. The drumming group, disappointing to left and behind your group, is pointing to the right. Now we want to draw the carbon in the back and it's resulting groups as well, or it's connecting groups. So you do note that with a circle and the falling lines, so it appears that the remain is on our left. Hydrogen is on the right. And this Ah, the group is going down. All right, so we want to perform a two reaction or any to elimination reaction. An important thing to know is when we're doing it to elimination. Reaction is that we want Thea reacting hydrogen and Al Kyohei Light, which is in this case roaming to be an anti pere plainer geometry from each other. So we want them to be 180 degrees from each other, in which case we see that this particular bro mean circle and this particular hydrogen circled in red are 180 degrees from each other. And this particular bro mean in this particular Hodgins, circled in red are also 180 years from each other. But since we know that there is a lot of symmetry here, we could essentially say that these are essentially the same. Ah, same groups. It would undergo the same e to reaction, elimination, reaction. So we're just gonna focus on the green green bromate and green hydrogen groups for now. So in order to a more easily visualize the ante para planner geometry, we go ahead and rotate these bonds. So would say we're focusing on the front carbon here, and we want to rotate the front cover and by 120 degrees. So, in other words, we want Well, im sorry we're looking at Thea. We're looking at the ray we're looking at this one circled on bread. So the bro mean And the hydrant circled in red. Since those are in the front, let's say we wanted to rotate 100 that 120 degrees. So this particular bro mean? What I mean by that is this particular bro me would go here to this position of the federal group. This group would go down. Disposition of hydrogen group in this hydrogen would go to this position of the Roaming group. Now, I could do the same thing for the back carbon as well. I want to rotate everything by 120 degrees so I could perform the same thing for the back. I won't draw that out, but I'm gonna go and draw the resulting human projection of what that would look like. So now for our front carbon, we now have our roaming group on the top here our fellow group on the bottom right and our hydrogen on the bottom left. And now for our backs, Carbon. The groups coming out of that since we also rotated that 120 grease which I didn't illustrated here. But if you were to do that on your own, you would see you would have something like this. We have the final group on the top left here. You would have the roaming group on the top right here, and you would have the hydrogen on the bottom here. Now we could easily see that those that bro me and a hydrogen that was circling red. They exhibit ante para planner geometry and they are good to undergo any to elimination now so we can go ahead and draw. Hurry to elimination, which is a one step reaction in which a base deep resonates this hydrogen. And when the Basie permeates the hydrogen, the electrons go get shared between the carbon here, the front carbon and the back carbon and this Roman leaves. And as a result, we know that we for Manal Keen and when we foreman all cane, we get ah, plainer geometry. So the resulting Newman projection will look something like this where this fennel and brew mean are now plainer. And this hydrogen and fennel are now plainer. And when are we? Ah, the bro mean. And the hydrogen that were circling, read or no longer in the equation are no longer in the Newman confirmation. Human projection. Rather Sorry. All right, so now how we would draw this in a bond line structure is something like this. Notice how the fennel in, bro mean are on the same side. And how the hydrogen fennel are on the same side here and now the question asks us, Teoh determined steri chemistry are this Al came, And in order to do that, we signed priority. Um, so we compare high carbon one here which having a little one here and in this carbon, it's gonna be a little too. But we're looking at carbon number one here, and we see that the final group has a higher party. So priority Group goes to the final. And I was looking at carpet number two and we see that bro Ming has priority over the final group because so that's carbon. The, uh, connecting carbon to the Al Kane is a carbon. So the brewing has higher already group here, and we see that the private groups are on the same side as each other. So we would say that is Zeus, Amun or Z, and therefore the resulting Al Kane is in the Z exhibit C stare a chemistry.


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