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sloppycanada
 one year ago
I can't for the life of me figure out the oxidations state for oxygen is Al(OH)4^1
2Al + 8H2O → 2^+1 + 3H2 + 2Al(OH)4^1
One oxygen atom in Al(OH)4^1
sloppycanada
 one year ago
I can't for the life of me figure out the oxidations state for oxygen is Al(OH)4^1 2Al + 8H2O → 2^+1 + 3H2 + 2Al(OH)4^1 One oxygen atom in Al(OH)4^1

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Photon336
 one year ago
Best ResponseYou've already chosen the best response.1For starters it's an ion so whenever you have an ion the oxidation states sum up to whatever number that ion is. So in that case they sum up to 1

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1Al(OH)4 Step one count up the atoms for each element Al = 1 O = 4 H = 4 Charge = 1 so They must add up to that. Then multiply #aroms by oxidation state: 4 atoms of H x 1 =+4 4 atoms of O x 2 = 8 Add them to find the remaining charge +48 = 4 At this point we know that 1 must be the sum since it's an ion so we need what number added to 4 gives us 1? X 4 = 1 x = +3 +3 = one of aluminum oxidation state.
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