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GamerChick333 Group Title

Okay, so I'm new to chemistry and absolutely confused. Can anyone please explain to me how to solve the following?? Mg -> ____ + 2e- ....Help would be very much appreciated!!

  • one year ago
  • one year ago

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  1. cozuwhoiluv2 Group Title
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    \[mg \rightarrow mg ^{2+} + 2 e\]

    • one year ago
  2. GamerChick333 Group Title
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    thats awesome thank you sooooo much. I still have three more to figure out, could you explain how to do it?

    • one year ago
  3. cozuwhoiluv2 Group Title
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    i'm sowry.. i don't know how to explain it...

    • one year ago
  4. GamerChick333 Group Title
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    thats okay, thanks for the help you did give me though. :)

    • one year ago
  5. febylailani Group Title
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    we must make it neutral :) do u understand?

    • one year ago
  6. GamerChick333 Group Title
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    well, that i get, im just not sure how to do so. For example: Ba -> Ba^(2+) + _______ ??? How would i get rid of the two? is that what yhou mean im supposed to do? sorry, chemistry is very confusing to me...

    • one year ago
  7. febylailani Group Title
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    ah.. how to explain it. if 2e = debt (-2) so, we must pay it by +2. just like that. but, we must write it on the right-top the element (Mg)

    • one year ago
  8. UnkleRhaukus Group Title
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    well, magnesium has 12 protons - by definition, to be a neutral atom the positive charges of the protons is canceled out by negatively charged electrons , (-)elections and (+)protons have a charge that is equal in magnitude and opposite in sign, \[ _{12}\text{Mg}^0 \longrightarrow \_\_ + 2e^- \] \[ _{12}\text{Mg}^0 -2e^- = \_\_ \] taking away electrons from magnesium will effect the overall charge \(0-(-2)=2\) the nucleus of the atom will remain the same

    • one year ago
  9. febylailani Group Title
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    yup, u must pay it by ...... electron(s). to make it neutral :)

    • one year ago
  10. GamerChick333 Group Title
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    So, would it be Ba^(2-) ?? Sorry for asking so many questions. I just want to be sure.

    • one year ago
  11. UnkleRhaukus Group Title
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    \[_{12}\text{Mg}^0 -2\times_0e^{-1} =~ _{12}\text{Mg}^{2+}\]

    • one year ago
  12. GamerChick333 Group Title
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    Thanks everybody =D

    • one year ago
  13. febylailani Group Title
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    uwc :)

    • one year ago
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