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%ne Inwqel od} (5 Kt DATA TABLE 1iid2 Xl0 Aal Inilln | mess coppe' elcctrdosigh Fira m7t coboci Doclrodc? Average € Jrrent (A} 0. 468 cureni JDyCbiion 180 DATA ANALYSIS (nthinae clatchorje Incovamon C) Ihe: passed 'hrolgh (he aloctrolyti cellbol each tal Tunit Ava-Je Usceou Nrlann Fromi Hoc e Rocz tromtne latroin Fean cJ cula;0 Ie nnbefolecctons In the electc /8 > lcr cuch bal ~J,op exoerimgtt Ihat Inc chargc 0l 07 elceatcn i 1,002 cnuOmt eecuda Determnz Ine aunbar cumni Alam Lecan Aft AnJde Kcess U9ey io ekcloe nroolcd One Ccnpem 07 CU eJchE Rcmomnor (nat [hia € eztchysi5 Cuculale Ine @umer Cofcen aictisnarGran copner iost alir noco Lacet The mass bosl al Inetanodr eolam coth Ine Ma ccojel Hlarna Icsl &ld Mc [ass copper icns produced (t10 trae One Macitcnt neg gible} Calculate the numte~ Aonldro $ nunblt copper atoms cocecr eacn Irial Cornpirc this valttc Advonced Chemitr with Vcrrult Tinc



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You want to find out how many atoms of the isotope 65 $\mathrm{Cu}$ are in a small sample of material. You bombard the sample with neutrons to ensure that on the order of 1$\%$ of these copper nuclei absorb a neutron. After activation, you turn off the neutron flux and then use a highly efficient detector to monitor the gamma radiation that comes out of the sample. Assume half of the 66 $\mathrm{Cu}$ nuclei emit a 1.04 -MeV gamma ray in their decay. (The other half of the activated nuclei decay directly to the ground state of $66 \mathrm{Ni} . )$ If after 10 $\mathrm{min}$ (two half-lives) you have detected $10^{4}$ MeV of photon energy at 1.04 MeV, (a) approximately how many 65 $\mathrm{Cu}$ atoms are in the sample? (b) Assume the sample contains natural copper. Refer to the isotopic abundances listed in Table 44.2 and estimate the total mass of copper in the sample.

So the studies question off us. Right on. What did what a different processes take you occur? So first off, you have some amount off. A couple of 65. We did the usual copper sample, 1% off. This is activated Bible body in grief neutrons. And this gives us compass 66. Copper 66 would be key by a two t For past ways. I want go stare Italy to the ground states off Geico and the other half actually meets a guy Marie in that case, And he got Maria said energy or for boys for me. And you detect right over to half lives. Daddy Total. Come on, energy. It's given this sure s O d c total amount come energy that was detected to have lives. Everyone work backwards to find out. What is the amount off a couple of 65 that we initially have. So, first off, you would need to find a number of karma. I've got my race, their work, you detector divide the total energy by the energy off each car. Marie, get my 0.62 instead of poetry to find out how much waas Actually, Kate, all right. How much decayed A curse in this process, Number two key will be quest to number off a couple of 66. What a play. By what whiteness? Half a pound off to Since 2/2 life has occurred, right? So to half life would be multiplying. The orginal amounts off campus e six way half twice, right. You got out much behalf by twice because to half lives. And then you take the origin of ah minus T final amount You get to you want to ask the kid in a modest kit would be twice off this value since he has two pathways, right? And we argue about that Pull off this pathways they take up half the proportion so above two Key became created 2.62 times 10 poetry times two and you will be able to get the number off a couple of 66. Then they should give us 2.56 instead of all four. Finally, we can get the number of copper. 65. Since you know, the only 1% is activated. So a total of all Kappa Psi's by 2 300 times off 2.56 full and this gives us 2.5 stop. That's all of six. No, to find. What is the mess off copper that we have in total? You will meet our We need the proportion they were given for natural copper for natural Ka Po, we have a couple of 63 taking up about 69%. A couple of 65 taking up about 30%. So we have found, actually, what is the number off atoms for copper 65. So we can actually use it to induce Deduce what is the number of couple 60 treat. So we know couple number Kopassus trees, 69.1% percent off the total couple. Let's just say offs. You. This is the total number of copper. Maybe in order it's number 65/65 s 30% to carry off the total number couple, we take the issue. No, we can find what is T both couple 60 tree. This very We can just substitute for my previous part which is 2.56 sometimes 12 6 Now we can find out what it's all total mess. Total mess of copper will be taking the mess this auto might mess for couple 60 tree. What part by it? Eaten up off Papa. 63 atoms Plus for a couple of 65. And then we convert mass in terms Auto mess units to kilograms 1.66 instead of all my I stood seven kg per you, you know, final and so is 8.77 stand to Paul minus 19 groups.

In this example, we have a consumption matric see which provide is provided here and this is a seven by seven matrix of the United States economy as it was in 1958. So what we have here next is a particular demand factor. This vector D is given to be first 74,000 then 56,000. For the third sector of the economy, we have 10,500 for the fourth sector, we have 25,000. The next sector is 17,500. Next we have 196 100,000 and the final is 5000 from the seventh sector. So the units here are in millions of dollars. And the long Tom Fif model for this particular situation is of the form, see or rather, I minus c times the production Vector X, which is are unknown, which is equal to the demand vector d. So we consult this, but what we would need to do is form a new matrix, which is going to be C minus. I augmented with the demand vector d. This takes quite a while to compute such a matrix. Even if it's just working out to this portion. Hope and it should be I minus c here. Not a C minus I, but putting this into a computer algebra system. We see that this is equal to this rather large matrix that we see here. So informing that matrix I took, I dont matrix that we see here we went to the main diagonal. It took one minus all of these entries. So, for example, in this entry, one mice 10.12 gives us the 0.9988 then the remainder entries in the and the consumption matrix C We just took the opposites. That's why you see all these negative signs and that gives us I minus c. Then here we have a augmented with the vector d. What we need to do next is ro reduce this entire seven by eight matrix, but this is not something that should be done by humans. So let's put this back into the computer algebra system and roll reduce. So, in my computer algebra system, I found that this matrix is row equivalent to this monstrosity we see below. But don't forget that we have augmented. Oh, dear. That was a wrong matrix. It's Rocca linked to this matrix we see displayed here. We have ones on the main Dagnall going this way, which tells us that our consumption matrix see when we for my my a c we did get a convertible matrix. Now this column here tells us that the production level X that's required is X equals 99,575 0.7 and these are in units of millions of dollars from the first sector of the economy. Then it's 97,000 703. 51,230 0.5 from the third sector, then 131,570 49,000 488 0.5, then 329,000 554. And lastly, 13,000 835.4 from the last sector of the economy. So this vector X is the production vector required, given the consumption matrix and the demand vector

Hello. And in this question here we want to calculate in the first part of the question we want to calculate the atomic mass in atomic units of mass for each of the nuclei that were given in our table. Okay. And to do this we've been given the number of uh we've been given the mass number and we've given being given the mass excess. Okay. So the atomic mass. Okay, I'm going to use subscript a for a ton of mass is equal to the mass number. Okay. Times the one unit. One atomic mass. Okay. Which is so this is a value you has a value of 1.67 times 10 to the power of minus 27 kg. Okay, so the atomic mass is equal to a times you plus the excess mass. Okay. And this will give the mass of the the mass of the nucleus in whatever units we we didn't push. Okay now we want to calculate this in atomic units. Okay. So what we're going to do is we're going to divide by so we're going to have We're going to divide by U two calculated in atomic units. Or we're going to factor out we're going to factor out to you. So this is equal to one plus any over you multiplied by you. And if we're able to calculate this quantity here we'll be able to calculate the mass and atomic units. Now we're also told in the question that u is equal to 100 and 34.49 mega electron volts divided by C squared. Which is helpful because we're given the mass excess in units of kilo electron volts are divided by C squared. So all we need to do for each of virus masses. Okay, I'll do two examples and then I'll just give you the answers for the others. Okay, So the mass, the atomic mass, the mass and atomic units of um C F is equal to. So we're told the mass number OK. Is equal to 252 152. And we're told that the mass access is equal to uh 70. It's equal to 76,000 uh huh. And 34 times 10 to the power of three electron volts over C squared. And we need to divide by you which is equal to 309 131.49 times 10 to the power of six electron volts over C squared. Ok. And this is multiplied by you. So we just type this into our calculator and get the mass in the public units is equal to 252.081. You okay then we'll do the example for the next one which is uh the mass of F M. Okay, So we're told that the atomic mass is equal so that the mass number sorry, is 256. We're told that the mass defect is 85,496 times 10 to the power of three electron volts divided by c squared. We divide this by you which is equal to 931.49 times 10 to the power of six electron volts divided by C squared. And this is all multiplied by you. Okay. So we type this into our calculator and this is equal to 256.197841. You so we just repeat the exact same process for the other ones. And I'll give you the answers here. So the mass of barium and atomic atomic units is 100 and 39.906045. You the mass of zen in is equal to 139.9216417. You the mass of ah P. D. is 111.9073141. You and the mass of Molybdenum is equal to 108.9278038. You? Okay. And that was just by calculating the same method as before. Now in the next question we're asked um calculate the energy released in these reactions. So the first reaction is um five 98 CF. And that decays into barium and um aluminum between neutrons. Okay, so to calculate the energy released. Okay so we're going to have energy released here. Okay. This is the energy released to the environment. E. And we're gonna calculate this energy we're gonna apply conservation of momentum. So the initial momentum that sorry, conservation of energy and conservation of momentum. So the initial energy is the rest mass energy on the left hand side and the final energy is the rest mass energy of barium plus the rest mass energy of molybdenum plus the rest mass energy of the neutrons and there's three of them. So we multiply by three plus E. Which is the energy released to the environment. And this energy could be in the form of kinetic energy of the barium, the neutrons and the militant. Um Okay so we can rearrange this equation here for E to get that E. Is equal to the so this is what he is equal to. And we multiply this all by C squared. Now we've calculated these masses earlier on in the question. Okay, so we just look up to where we calculated them and we can sub in to guess that this is equal to 0.21661 you times C squared and we can fill in for you and for C. So I've said what you is in the very beginning of the question sees the speed of light which has a value of three times 10 to the power of eight m per second. So this gives in in jewels, the energy released is 3.355 times 10 to the power of minus 11 jules. Okay and the second reaction we want to um we want to discuss is so we have this decaying into zen in and uh palladium and for neutrons. Okay, so again the energy that's going to be released. Okay, so the energy is equal to so we can apply a conservation of energy and then you can rearrange for the energy. And on doing that, you're going to get that this is equal to the mass of FM minus the mass of xenon minus the mass of palladium minus four times the mass of the neutron all multiplied by c squared. Ok, so that's just by doing the the same as we did for the other reaction we can sub in for the masses. So we've calculated all these masses earlier on in the question and we get this is equal to 0.2281 you times C squared fill in for you and for C to get 3.429 times 10 to the power of minus 11 jewels. Okay, so that is the energy released. And then the final part of the question we want to know are these, can these reactions occur spontaneously? And the answer is yes they can. And the reason for this. Okay, is that energy is released to the environment. Okay, If this energy here was negative, that would mean we'd have to supply energy into the system and that would mean that they can't a correspondent a Gnaeus lee as we need to add energy. However, these release energy to the system, so we don't need to supply anything in the beginning for this reaction to happen so they can occur spontaneously. So this is the solution to the question, and thank you for listening.

So if we use our linear aggression tool, we, um, see that this pink artist purple graph here represents, uh, a an exponential model for the number of cell phone users. And the black quadratic graph represents the number of phone lines over the period from 8 1985 to 1998. Um, and, uh, we used a graphing utility to pot their aggressions on. And we also found this appropriate viewing angle to see, um, the year in which there would be the same A number of cell phone users as phone lines. And that appears to be, UH, 15.198 So it would be 15 years, which would be the year 2000.


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