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anonymous

  • one year ago

Please Help Fan & Medal

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Photon336

  3. anonymous
    • one year ago
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    @DarrenMadx

  4. DarrenMadx
    • one year ago
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    i don't even have a clue. i never took chemistry so i don't know. my head hurts just looking at that question

  5. anonymous
    • one year ago
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    lololololol

  6. DarrenMadx
    • one year ago
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    sorry i couldn't help you

  7. anonymous
    • one year ago
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    its okay bro thanks alot

  8. DarrenMadx
    • one year ago
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    no prob. good luck with your question

  9. Photon336
    • one year ago
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    @princestonA hey, for each isotope do the following. Step#1 write down the atomic mass for one of the isotopes. Step#2 convert the % abundance to decimal by putting the number over 100 Step#3 multiply the two numbers you'll get a number. just write this down and put a box over it to remember it. do this for each isotope and then add the numbers up to get the total average mass. Start with the first isotope they give you and tell me what you get. |dw:1444334821299:dw|

  10. Photon336
    • one year ago
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    This looks complicated but once you get it it's not that bad.

  11. anonymous
    • one year ago
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    Oh thanks alot is 28.09

  12. anonymous
    • one year ago
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    @Photon336 can you you do me one favor calculate the average atomic mass of rubidium. Rubidium has two isotopes, 85Rb and 87Rb. 85Rb has an atomic mass of 84.912 amu and occurs at an abundance of 72.17%. 87Rb has an atomic mass of 86.909 amu and occurs at an abundance of 27.83%. I wanna see the work so i know how to do it.

  13. Photon336
    • one year ago
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    Yeah sure

  14. anonymous
    • one year ago
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    huh?

  15. Photon336
    • one year ago
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    ok i'll try this in a sec

  16. Photon336
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @PrincestonA @Photon336 can you you do me one favor calculate the average atomic mass of rubidium. Rubidium has two isotopes, 85Rb and 87Rb. 85Rb has an atomic mass of 84.912 amu and occurs at an abundance of 72.17%. 87Rb has an atomic mass of 86.909 amu and occurs at an abundance of 27.83%. I wanna see the work so i know how to do it. \(\color{#0cbb34}{\text{End of Quote}}\) |dw:1444335901174:dw| FOR 85 RB \[84.912amu *\frac{ 72 }{ 100 } = 61.280 amu, 85RB\] For 87 RB \[86.909amu*(\frac{ 27.83 }{ 100 }) = 24.186amu\] now we add them up \[Rb^{87} + {Rb ^{85}}\] \[(24.186+61.280)= 85.467 amu \] @PrincestonA you can check to see if this number is right by going to the periodic table. every element is the average atomic mass for isotopes. we check our value of 85.467 by looking at the periodic table and we see that these two values match so that's how we know our answer is correct. hope this helped.

  17. Photon336
    • one year ago
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    it's probably better that you see it done step by step,

  18. anonymous
    • one year ago
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    thank you sooooooo much pho Your the best dude

  19. Photon336
    • one year ago
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    yeah no problem, once you see how it's done the problem is not that bad

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