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Question 3In the following nuclear equation, identify the missing product: Ca + u HOA 46 2 Ti OB " Sc Oc #Ti OD " Ar E:none of these...

Question

Question 3In the following nuclear equation, identify the missing product: Ca + u HOA 46 2 Ti OB " Sc Oc #Ti OD " Ar E:none of these

Question 3 In the following nuclear equation, identify the missing product: Ca + u H OA 46 2 Ti OB " Sc Oc #Ti OD " Ar E:none of these



Answers

Complete each of the following nuclear equations by supplying the missing particle. a. $\quad \frac{226}{88} \mathrm{Ra} \rightarrow \frac{222}{86} \mathrm{Rn}+?$ b. $\frac{222}{86} \mathrm{Rn} \rightarrow_{84}^{218} \mathrm{Po}+?$ c. $_{1}^{2} \mathrm{H}+_{1}^{3} \mathrm{H} \rightarrow_{2}^{4} \mathrm{He}+?$

Problem. We need to find the mystery element that IHS created in east reactions. So for this 1st 1 there is just one product. So we know that you need to take the numbers on the left and add them together in order to find out iconic mass and the atomic number of the compound. So if the oven together you get 1 13 over 49. The 49th element on the periodic table is India. So the mystery element would be Indian. 1/13 thank you into the next problem. And this when we have a new turn on the left side. For me to account for that through subtraction, the least attract one from 10 plus 14 we'd get 13 and five. Quest seven plus two is seven. The seventh element is nitrogen, so the question mark is nitrogen 13.

Okay, so for part A, we have p 0 to 10 forms, 206 p. B A. Rep you piece to two or six plus and unknown. So the first thing I want to do is determined the top numbers of PB N P O. So PB have untimely number of 82 P O has an atomic number of 84 so we can look this up on the Peoria table and then what we do is we gotta balance out the number of D A timely numbers on both sides and the atomic mass on both sides. So the atomic number on the left side is 84. So far. On the right side, we have 82. So in order pounds that we need to add a two. Since the missing particle is on the right side, we need to add a to two D right side. If the missing particle was on the left side in this case and we would be subtracting from the left side so to here and when we look on the periodic table, we see that that is helium, right? And then I was like the mass. So we have to 10 on the left, two of six on the right. So we need out of four. So it's me that this is a four. So four to helium can also be written. As for two Alfa particle, cause it's an Alfa particles there. Same thing now for part B, Aren't we have unknown? Oh, no, we don't. So we have treat me, um And then on the right time, we have three to helium plus an unknown. Right. So we have one on the left side. We have two on the right side in order. Bounce it out. We need to minus one on the right side. And then the weight. We have three on the left, three on the right. So this is a zero. So zero in the negative one. That is a beta particle. And for see, we have carbon 14 forms. An unknown plus Hey, beta. Now let's look a t a timeline number. We have six on the left. We have negative one on the right. You don't bounce it out. We needed seven, right? And when were the coming period table? We see that that is nitrogen. Now let's look at the weight. We have 14 on the left. We have zero from the right from the beta particle. So we need 14 more on the right. So that is 14 right there. Now let's look at D. We have unknown plus a neutron and then on the right side we have 59 Cayenne. Okay, so I think there's a type of here because just iron is 26. That is the untimely number of iron. It's either dad or it's not iron and it's so fun. And we would have changed the mass on the top. So I'm pretty sure this is iron. So most look at the atomic numbers. We had 26 on the right side. We have zero on the left side, right? And in order to balance this out, we need to add a 26 on the last time. So we have 26 year and that is also iron. And let's look at the weight. So we have 59 on the right and so far we have one on the left, so we need to have 50 age. So this is the answer right here.

This question asked us to complete the blank spots and figure out the nuclear bombardment reactions. So in the first case, we see Calcium 44 being bombarded by a proton, and there's going to be a neutron produced in this reaction, along with a different daughter nucleus. So let's start with trying to figure out the daughter nucleus produced in this first nuclear bombardment reaction. So to do this, what we have to do is balance the mass numbers, which is the number given on the top left hand side corner on the two sides of the equation and then balance the atomic numbers on the two sides of the equation. So if we tried to do that, they see that on the left hand side of the equation, calcium has a mass off 44 the Proton has a mass off one, and so it has a total of 45 on the left hand side. But there's only one on the right hand side, so if we balance the mass there, we have to have a nucleus with the mass number 44 next to the neutron. Same thing with the atomic numbers. We have 20 and one on the left hand side, 21 and so far, only a zero on the right hand side. So to balance them, we have to have Ah, nuclear's with an atomic number 21 and this belongs to Skandia. So that's how we figure out the daughter nucleus produced in that bombardment reaction if we move wanted the second equation, which is again another nuclear bombardment reaction. We see that California is bombarded with Laura, and we know that five neutrons are produced as a result alarm with her unknown daughter nucleus. So let's try to figure out what this unknown daughter nucleus is by again balancing the mass numbers on the two sides of the reaction and by balancing the atomic numbers on the two sides. Of the reaction, 2 42 and 10 is going to be to 62 and there's only five on this side. So if we take the difference between 2 62 and five, we get to 57 and that is going to be the mass number off the unknown Got a new please. And if we also try to balance the atomic numbers on the two sides, yeah, it's 98 plus five. And the charge of the neutron is zero. So all five is are also going to have a charge total of zero. So we have 98 plus five on the side, which is equal to one or three and zero on the side. So we'll have, uh, not the nucleus with the atomic number one or three on the right hand side. And if we take a look at the periodic table, we see that this belongs to Lawrence E. Um, a lot. And if we move on to the next equation, you see that palladium on No. Six is bombarded with an Alfa particle, and as a result, we're told that silver one or nine is produced along with possibly a different type of particle. So let's start to figure out what this particle is. So on the right hand side, we only have a mass number off one or nine. And on the left hand side we have 106 plus four. So the different is actually born, and that should be the mass number off the unknown particle. And if we take a look at the atomic numbers, atomic numbers on the, uh, left hand side. We have 46 plus two, which is 48 and only 47 on the right hand side. So we'll have to have a particle with the charge one in. This corresponds to a proton. This is how we figure out the unknown component off each nuclear bombardment reaction.

For each of these reactions, we have to find the missing particles. And the main idea for all of these is that the mass number and atomic number on both sides should be equal because of the conservation of mass and conservation of charge. The for part A on the left, the mass numbers 226 on the rate, it's only 222. So to find our missing mass number 226 minus 222 gives us for no for missing particle. The master emerald before the atomic number on the left is 88 on the right, it's 86. Do 88 minus 86 gives us. So are missing. Particle is for two, and maybe you recognize that this is our Alfa particle For part B. We just follow the same ideas before on the left side, the overall mass number is nine plus one is 10 and on the right. It is only four do 10 minus four. It's a six are missing. Particle has amassed number of six on the life. Our overall atomic number is four plus one is five on the rate. It's only two. You five minus two is three. So we have 63 And you think the atomic number. We can look on the periodic table and see that element with an atomic number of three is lithium for part C. We have on the right. Now. These are completed side. We're missing something on the left. We have 17 on the right in 17 on the lift. So the top is already balanced in our missing particle as a mass number of zero on the rate we have atomic number of seven on the left is eight. Do seven minus A. It says negative one. So are missing particles zero minus one e. Because this is our beta particle.


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