In this very simple chemistry question, you need to determine the atomic mass of a given element. The information provided will be the mass of particular isotopes of the atom, and the fractional abundance of the atom. All you need to know is the information provided in the last post, about how to solve this type of problem. This problem will feature the atom tin, abbreviated Sn on the periodic table of elements. Now, the problem with this question is not that it is difficult in any way, because you would know that it is not, if you read the last post, which tells you exactly how to solve a problem like this one. The real problem lies in the complexity of the question, because this particular element has many more isotopes than most other elements. Therefore, the real problem with this problem (haha) is being careful to not make mistakes. In a problem like this, one mistake can make the entire answer completely wrong and waaaay off!
I will start off using the same mechanism that I did for the first chemistry problem. 1) Given information: I am given the mass of an atom of each particular isotope, as well as that isotope’s fractional abundance, for all of the isotopes of tin. So as not to be entirely redundant, I will not list all of this information individually, because it is all written out plainly in the table above. 2) What I must do: I must find the atomic mass of tin. But to do so, I need to first multiply the atomic mass of each isotope of tin by its fractional abundance. Then I must sum all of these values, and finally I will get the atomic mass of tin. 3) Solving the problem: Sn112: 111.904821 amu × 0.0097 = 1.0854767637 amu Sn114: 113.902782 amu × 0.0066 = 0.7517583612 amu Sn115: 114.903346 amu × 0.0034 = 0.3906713764 amu Sn116: 115.901744 amu × 0.1454 = 16.852113577 amu Sn117: 116.902954 amu × 0.0768 = 8.9781468672 amu Sn118: 117.901606 amu × 0.2422 = 28.5557689732 amu Sn119: 118.903309 amu × 0.0859 = 10.2137942431 amu Sn120: 119.902197 amu × 0.3258 = 39.0641357826 amu Sn122: 121.903440 amu × 0.0463 = 5.644129272 amu Sn124: 123.905275 amu × 0.0579 = 7.1741154225 amu Now that I have done the first part of what is required for me to solve this problem, all I have left to do is add all of the answers. After adding all of the values obtained from multiplication above, the final answer is 118.7101106389 amu. Normally, after any chemistry question, you would follow the rules for rounding and significant figures. In this particular example, I did not need to apply these rules because I considered all of the values to be exact numbers, previously defined values with no assumed uncertainty. A quick check of the periodic table of elements shows that my numerical answer is completely correct! This problem is actually a very good one to do on your own, with any other element of your choice, because all of the information (isotopic masses, fractional abundances, and atomic masses) are easily accessible on the Internet. Try it yourself!
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Time for our first MCAT question! Two solids are placed in front of you and you are given some information about each solid. You are given the information that Solid #1 has a volume of 56 cm^3 and a density of 0.35 g/cm^3, and that Solid #2 is a sphere with a radius of 4.32 mm and a density of 0.65 g/cm^3. You are asked to find the difference in mass between the two solids. You must give your answer in SI units and show your work. First, I will think about what I have been asked... And now, I will solve the problem... I hope you tried solving the problem yourself, first. And that's it for the first chapter of introductory chemistry! Happy studying everyone.

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