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Chemistry
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OpenStudy (anonymous):
Use the information for the isotopes of X to calculate the average atomic mass of X, and identify the element.
Isotope Abundance Mass (amu)
6X 7.5% 6.015
7X 92.5% 7.016
I'm kinda confused :/
11 years ago
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OpenStudy (zale101):
\(Average ~atomic~ mass~ of~ an ~element\)
=
\(
(Abundance~percentage_1*Mass~of~Isotope~ 1)\)
+
\((Abundance~percentage_2*Mass~of~Isotope~ 2))\)
11 years ago
OpenStudy (anonymous):
So would I have to calculate that like 7.5 * 6.015, and the same for the other one?
11 years ago
OpenStudy (photon336):
You would add the two masses together
11 years ago
OpenStudy (zale101):
Correct. You'll add it with Isotope 7x to get the average atomic mass of the element
11 years ago
OpenStudy (photon336):
but first you need to multiply the atomic mass by the % abundance I believe
11 years ago
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OpenStudy (anonymous):
So it's 45.1125 + 648.98 ?
11 years ago
OpenStudy (anonymous):
@Photon336 would it be 694.0925?
11 years ago
OpenStudy (photon336):
@Zale101 6X and 7X are just the names of the isotopes right?
11 years ago
OpenStudy (zale101):
I believe so.
11 years ago
OpenStudy (anonymous):
Yeah they are...
11 years ago
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OpenStudy (photon336):
yeah.. so convert 7.5% to decimal form which is 7.5/100 then you multiply this by 6.015 amu
11 years ago
OpenStudy (photon336):
remember @flatouthero its' a percentage
11 years ago
OpenStudy (anonymous):
Oh, so it's 6.94
11 years ago
OpenStudy (zale101):
Correct!
11 years ago
OpenStudy (zale101):
:)
11 years ago
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OpenStudy (zale101):
What element is that?
11 years ago
OpenStudy (anonymous):
Thank you both for your help!
11 years ago
OpenStudy (photon336):
no problem
11 years ago
OpenStudy (anonymous):
Lithium, btw xD
11 years ago
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