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    A naturally occurring sample of the element boron has a relative atomic mass of 10.8. In this sample, boron exists as two isotopes, 10B and 11B
    (i) Calculate the percentage abundance of 10B in this naturally occurring sample of boron.
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    anyone
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    I believe the answer is 20%
    As (10*20)+(11*80) divided by 100 equals 10.8,the relative atomic mass.
    Hope this helps.
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    (Original post by KirstyPayton23)
    I believe the answer is 20%
    As (10*20)+(11*80) divided by 100 equals 10.8,the relative atomic mass.
    Hope this helps.
    thanks i got to 10x + 11y/x+y = 10.8 where x and y are the different relative abundances. where did you get the 20 and 80 from?
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    Well as the relative atomic mass is 10.8 it suggests that the the isotope B11 is in greater abundance as it is closer to 10.8 than B10 is. Assuming this I just plugged different numbers in using trial and error until i got the answer of 10.8.
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    (Original post by KirstyPayton23)
    Well as the relative atomic mass is 10.8 it suggests that the the isotope B11 is in greater abundance as it is closer to 10.8 than B10 is. Assuming this I just plugged different numbers in using trial and error until i got the answer of 10.8.
    without trial and error do you know how you'd get a ratio from the formula i created? ^^
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    anyone else actually know
 
 
 
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