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Problem

Average atomic masses listed by IUPAC are based o…

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61

Problem 22 Easy Difficulty

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99 amu, and 8.82% abundance with 21.99 amu. Calculate the average atomic mass of this element

Answer

Average atomic mass is 20.15 amu

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Chemistry 101

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Chapter 2

Atoms, Molecules, and Ions

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Video Transcript

this question asked us to find the average atomic mass of an element given the natural abundances of its ice atopic masses. So the isotope, with a mass of 19.99 atomic mass units, has an abundance of 90.92%. That's right. Now's a decimal here. The isotopic, the mass of 20.99 atomic mass units, has an abundance of 0.26% in the institute, with the mass. 21.99 atomic mass units has an abundance of 8.82%. So to find the average atomic mass, we multiply each of the ice, a topic topic masses by their abundance. So 19.99 times 0.9092 is 18 17 in 20.99 times 0.26 is 0.5 and 21.99 times 0.0 82 is 1.94 Then, finally, to find the average, uh, time, ask you some these weighted averages we did that you can answer of 20 point 15 and units atomic mass units. I recommend that instead of doing each of these equations one by one like this and some of them that you do all three together in your calculator so you don't round in between steps of Ray to this final answer.

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Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson

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