plz answer this question before 10am
Answers
Chlorine consists of two isotopes, 75% chlorine-35 and 25% chlorine-37, so using these two mass numbers ...
... again think of the data based on 100 atoms, so 75 have a mass of 35 and 25 atoms have a mass of 37.
The average mass = [ (75 x 35) + (25 x 37) ] / 100 = 35.5
So the relative atomic mass of chlorine is 35.5 or RAM or Ar(Cl) = 35.5
Note: 35Cl and 37Cl are the most common isotopes of chlorine, but, there are tiny percentages of other chlorine isotopes which are usually ignored at GCSE/IGCSE and Advanced GCE AS/A2 A level.
BORON
So this problem has given you the percent abundances, and the final average atomic mass. Let's go ahead and set this up as an equation:
(80.2100)(B1)+(19.80100)(B2)=10.81
amu
B1 and B2 are the atomic masses of each isotope. So is there any additional information we can use? Well, we have the atomic mass of the more abundant isotope, Boron-11. Let's plug that in for B1:
(80.2100)(11.01)+(19.80100)(B2)=10.81
amu
And that's honestly the most confusing part. From here on, it's just a simple process of algebra as we solve for B2:
⇒(19.80100)(B2)=10.81−(80.2100)(11.01)
⇒(B2)=10.81−(80.2100)(11.01)19.80100=9.99
amu
HOPE IT WILL HELP YOU
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amu
B1 and B2 are the atomic masses of each isotope. So is there any additional information we can use? Well, we have the atomic mass of the more abundant isotope, Boron-11. Let's plug that in for B1:
(80.2100)(11.01)+(19.80100)(B2)=10.81
amu
And that's honestly the most confusing part. From here on, it's just a simple process of algebra as we solve for B2:
⇒(19.80100)(B2)=10.81−(80.2100)(11.01)
⇒(B2)=10.81−(80.2100)(11.01)19.80100=9.99 amu
Hope it helps...
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