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    Ok i seriously have no clue how to do this, please explain in the simplest terms thanks

    Question: What are the relative abundances of Iridium-191 and Iridium-193 if the R.A.M is 192.2?
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    iridium 191 is x
    193 is y

    so xs percentage plus ys =100
    so x + y =100
    and rearranged y = 100 -x

    (191 X x / 100) + (193 X 100-x)/100 =192.2
    get the 100s off the bottom by multiplying by 100 and then rearrage to find x
    and this is x = 40
    so y = 60
    and they are the abundances
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    (Original post by affleming)
    iridium 191 is x
    193 is y

    so xs percentage plus ys =100
    so x + y =100
    and rearranged y = 100 -x

    (191 X x / 100) + (193 X 100-x)/100 =192.2
    get the 100s off the bottom by multiplying by 100 and then rearrage to find x
    and this is x = 40
    so y = 60
    and they are the abundances
    Ok thanks dude. Can u check if i did this one is correctly pls:

    RAM of Br=79.9 , two isotopes Br-81 and Br-79
    i got : Br-81=45% and Br-79=55%

    Is that right? thanks
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    Chemistry questions go here: http://www.thestudentroom.co.uk/forumdisplay.php?f=130
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    (Original post by Hanz_a93)
    Ok thanks dude. Can u check if i did this one is correctly pls:

    RAM of Br=79.9 , two isotopes Br-81 and Br-79
    i got : Br-81=45% and Br-79=55%

    Is that right? thanks
    yep thats right
 
 
 
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