Ordinary potassium contains 0.012 percent of the naturally occurring radioactive isotope 4OK, which has half-life of 1.3X 109What is the ctivity of 3.6 kg of potass...

Ordinary potassium contains 0.012 percent of the naturally occurring radioactive isotope ${ }^{40} \mathrm{~K}$, which has a half-life of $1.3 \times 10^{9} \mathrm{y} .(a)$ What is the activity of $1.0 \mathrm{~kg}$ of potassium? (b) What would have been the fraction of ${ }^{40} \mathrm{~K}$ in natural potassium $4.5 \times 10^{9} \mathrm{y}$ ago?

As we all know that activity R. is equal to lend a multiplication end. And here's the value of the half is given a 1.28 multiplication tend to depart nine years. Changing the unity. Just multiply it by this value Which is equal to 4.04 multiplication than to the power. 16 seconds. Now here I can write the value of image, 0.0 12% multiplication, zero point Toledo, multiplication, regional multiplication, 60 Kg. So on solving it further I can write the value of a med Euro 0.216 multiplication and 10 to depart minus three kg. Changing it in graham, I can write your own 30.216 Graham. Now the value of and can be detained at 6.023 multiplication, 10 to deport 23 multiplication and 0.0216 g By 40 grand. Which on simplification I get the value of energy 3.25 multiplication, 10 to depart 20. Now we all know that R. Is equal to learn to buy T half multiplication end, so I will just put the value to calculate the activity so I can write, learn to buy 4.04 multiplication and 10 to the power 16's, again Multiplication 3.25, multiplication, 10 to the 420, Which is equal to 5.6 multiplication 10 to the power three BQ as the activity.

For this question, I have rewritten now, all the information we are given. We have the Mueller masses of the three isotopes Of potassium potassium 39, potassium 40 and potassium 41. And we know that the average of these three is 39.0983 and the abundance of potassium 39 is 93.2581%. But to solve the percent abundance of these two, we can create two equations. So we take the average, we know that the average is the each individual mass multiplied by its abundance. So 39.0983, which is the average is equal to 38 .963707, which is the lightest isotope. And we know that it's abundance is 93.2581%. So we will write that um As a fraction or not, as a percentage will divide it by 100. So that gives us 0.93 2581. And then we can add The abundance of the second isotopes. So 39.963999. And we don't know that it's percentage. So we will call this X. And for the last I said hope 40.9618- five. And we don't know what's abundance. So we will call this Why? So this it's going to be X. And this is going to be why? Then from this equation, since we have two unknowns, we won't be able to solve X and Y. But we can make another equation. We know that if we add ex N. Y, it will equal 1 -0.93- 581. Since that is a big chunk of our 100%. So x plus y is equal to 0.0674 19. We can rearrange this equation to make why 0.067419 X. So now we can sub this in for why here? So now the equation when we set this in we will have only access and no wise in our equation so we can actually solve for X. So let's do that. So 39.09 83 is equal to um 38.9637 oh seven multiplied by 0.93 to 581, Which is equal to 36.33 68128. So that's this first term here is this number plus 39.963 999 x Plus 40.9618- five. And on the place of why we will put 0.067419 -1. So now we can bring this term over here and subtract it. So 39.0983 -36.3368128 is equal to 2.7614 872. That is equal to 39.963 999 X. again at this term, then we can expand this. So we will do The 40.96, 1, 8- five, multiplied by 0.067419. Not as equal to 2.7616, minus, Expand that 40.9618- five x. So now we have two terms with X in two terms without X. So we will put the ones without X on the left. So we will do the 2.76 -2.7616. This term here, 2.761487 to minus this term. Since we're bringing it over to the left, 2.761605- eight. That is equal to 39.963 999 x -40.9618- five X. And then we can simplify this. That is equal to negative 0.00011808. And that is equal to -0.997 8 to 6 X. So we can solve for X. Bye, Dividing both sides by negative 0.9978-6 X. is equal to 0.000 11834. You can plug it into our first equation, sulfur. Why y is equal to 0.67419 minus x 0.11834. Why is equal to 0.0673 so that we can express those As percentages by multiplying them by a 100 and potassium 40 has an abundance of 0.011834%,, And potassium 41 has an abundance of 6.73%,, and those are the answers.

Hey guys leads to problems 16. In this problem, we need to find out how many neutrons and how many protons the potassium isotopes contained. The mass number is given. We need to also find out if the average atomic mass of potassium is consistent with the relative abundance of the isotope. The evidences are also given here. Mhm. Mhm. No, let's do that. We know that the atomic number of photos um is 19. Mhm. And we know the mass numbers are given. Yeah, yeah. Therefore the Isotopes of Potash in our petition, 39 41 and 40. With the nuclear symbols given here. In this case, we write the mass number on the top left in the middle. We write the symbol, they element symbol. And on the bottom left corner we read that to make numbers entities are the nuclear symbol for potassium. Now we need to count the number of problems. We know that the mass number is is equal to the number of protons plus number of neutrons. Therefore the number of neutrons is a signal to mass number minus number of problems. The mass number is 39 and the number of program is is a girl to 19. Therefore 39 -19 gives us 20 For protection 41 I set up, we have 41 -19 which gives us 22. And for protection 40, they have 40 -19, which gives us 21. And these are the number of neutrals for that given isotopes. And finally, we need to count is the average atomic mass of Potosi um is consistent with the relative abundance of di. Sotto. We know that Average at week mass of potassium is 39.1 g. And according to the relative abundance, we see that. What a shame 39 Is having the relative evidence of 93.25 Since for the Shem 39 is the most abundant I sort of rash. Um From the relative abundance, we see that The every legitimate mess should be close to 39 And therefore the average atomic mass of potassium, which is 39.1 g, is consistent with the relative abundance of the isotope. These two are close. These two messes are close, which is expected. Therefore it is consistent.

Hey, guys, let's do problem 99 in this problem, the average atomic weight of what a shame is given and its three naturally occurring I start ups. Maximum mess is also given, and only the percentage abundance off partition with 41 mass number is given. Now we need to calculate the percentage dependence off the other two, the other twice eaters. So let us write down the information that is given. The first information given is the atomic weight off photo Shame and the other two informations are the the mass off each three off the Assad jobs, which are region here, and only the percentage abundance off the petition with mass number of 41 is given, which is 6.73% age. No, we need to figure out what is the percentage of abundance off audition with mass number of 39 40. So we know the total natural abundance off any element is 100% age. Therefore, we can write down the one. The first run is having the national abundance off say later we can take any Number X, and the other one then should have the natural abundance off 100 miners, 6.73 minus X. So we need to figure out what is developed X and then we can calculate what are the national abundance off? 39 K and 40 days. They just right on one thing. 100 minus 6.73 means 93.27 Okay, so these should be done. Natural abundance of these elements. No, we know that atomic weight is the weighted average off done masses of national organizer troops. What does that mean? That means that Adamik wait is the some off the It is the sum we need to take the summer off. The abundance off any isotope multiplied by its mass. Therefore, abundance off. I set up. What a shame I started with my similar 29 divided by 100 multiplied by mass off this I sort of. And when we take the summer off the similar thing for all the three and sit ups, we get that to make weight off that element. Therefore, abundance off petition with mass number 40 divided by 100 multiplied by the same it's mess and then when it trade and take the same or some abundance off petition with US number or 41 divided, 100 multiplied by its mess. Now most of the informations are given. Most of the values are given. We need to calculate eggs. Let us place the numbers that weight off for the shame is 39.98 in you. Evidence off petition with mass number 39 is X divided by 100. It's I started with Mess is 38.964 Hey, emu, then we need to take the some, then the abundance off petition with mass number off 40 es. The evidence is 93.27 minus X divided by 100. We need to want to play the whole thing with I set up smells, which is 39.964 and finally 6.7 20 very 100 multiplied by 40.962 am you Now we need to do this simple calculation. We can use the calculator and just please the numbers. How can I 0.98 a. M. U and I just went to divide this a mess by 100. Therefore, it will be 0.38964 x I am you. Then let's break down this part. If you break it down What we will get ease. 37.27 am U minus 0.39964 in you. And finally the last room. This one, we'll give you two point 757 in you. Now I will take the values reaching having eggs on the left hand side off the creation and the other one is this one. This should be our next. So I will just take these two on the left hand side which will re 0.39964 minus 0.38964 x in you and in on the right hand side. What we believed with is 37.27 I am you plus 2.757 I am you and minus this one will go to the right hand side. Therefore, there should be a minus sign before how to 9.98 AM you then what we even get is finally if you do the calculations zero wine 01 x in you each other zero point 9 to 9 he and you. Therefore, what we can say is we can say is X equal to 0.929 divided by 0.1 which is 93% age. Therefore, the natural evidence off. What a shame with 39 mass number is 93 percentage. Therefore, the other one with mass number of 40 should have the value off 93.27 Minus 93 equals you zero point 27 or we can say that it is 0.3 percentage. Therefore, our answer to this problem is the percentage abundance off rotation with 39 mass number is 93% age and the other one with 40 mass number is 0.3%.