Saph00x
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Can someone explain how to do this, please!!

Bromine has two isotopes, Br–79 and Br–81. The relative atomic mass of bromine is 79.9.
Calculate the percentage of Br–79 atoms in a sample of bromine.
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Pigster
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Do you know how to calculate relative atomic mass from abundances of isotopes?

Can you do algebra?

If the answer to both is yes, then you should be able to do this. If it is a yes, what have you tried so far?
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T0mas
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Call x the proportion (as a fraction or decimal) of isotope Br-79 and y the proportion (as a fraction or decimal) of isotope Br-81.

The average RAM would be given by 79x + 81y.
We are effectively told 79x + 81y = 79.9

79x + 81(1-x)=79.9

x= 0.546 (or 54.6%), y= 0.454
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Saph00x
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(Original post by T0mas)
Call x the proportion (as a fraction or decimal) of isotope Br-79 and y the proportion (as a fraction or decimal) of isotope Br-81.

The average RAM would be given by 79x + 81y.
We are effectively told 79x + 81y = 79.9

79x + 81(1-x)=79.9

x= 0.546 (or 54.6%), y= 0.454

I expanded the brackets so i got:
79x + 81 -81x = 79.9
79x - 81x = -1.1
x(79-81) = -1.1

I know this is definitely wrong but im not sure what to do
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T0mas
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(Original post by Saph00x)
I expanded the brackets so i got:
79x + 81 -81x = 79.9
79x - 81x = -1.1
x(79-81) = -1.1

I know this is definitely wrong but im not sure what to do
x & y percentage (= 100% total)
mass of 100 atoms = 100 x relative mass = 7990.8
79x + 81y = 7990.8
and y = (100 - x)
thus: 79x +81(100-x) = 7990.8
so: 79x +8100 -81x = 7990.8
gather terms: 79x-81x = 7990.8-8100 = -109.2
and: -2x = -109.2
so: x = 54.6%
and y = 45.4%
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Saph00x
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Thankk youuuu.

I just don't get this line:
"mass of 100 atoms = 100 x relative mass = 7990.8"

why do you want to use the mass of 100 atoms and then multiply it by the RAM
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