Chemistry

For the following,
2HBr (aq) + Ba(OH)2 (aq) ---> 2H2O (l) + BaBr2 (aq)
write the net ionic equation including the phases

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  1. I showed you how to do the last one. You try this one. Just follow the rule.
    weak acids, weak bases, water, insoluble substances(solids; i.e., ppts), gases all written as molecules.
    All others written as ions.

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  2. 2HBr (aq) +OH^-(aq)----> 2H2O (l)+Br2^-(aq)???

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  3. 2HBr (aq) + Ba(OH)2 (aq) ---> 2H2O (l) + BaBr2 (aq)
    No, HBr is a strong acid. Ba(OH)2 is a strong base and you've handled that ok. Bromide exists as Br^- and not Br2^- so the way you do that is 2Br^-
    Why don't we do this the long way?
    You have the balanced molecular equation. Next step is to convert it to the total ionic e3quation like this.
    2H^+ + 2Br^- + 2K^+ + 2OH^- ==> H2O + Ba^2+ + 2Br^-
    Now cancel the ions common to both sides; i.e., 2H^+ does not cancel, 2Br^- does, 2K^+ does, 2OH^- does not.. Net ionic equation is
    2H^+(aq) + 2OH^-(aq) ==> 2H2O(l)

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  4. Consider the following precipitation reaction (balanced).
    precipitation reaction:
    2NH_{4}Br(aq) + Pb(C_{2}H_{3}O_{2})_{2}(aq) -> 2NH_{4}C_{2}H_{3}O_{2}(aq) + PbBr_{2}(s)
    2
    NH
    4
    Br
    (
    aq
    )
    +
    Pb
    (
    C
    2
    H
    3
    O
    2
    )
    2
    (
    aq
    )

    2
    NH
    4
    C
    2
    H
    3
    O
    2
    (
    aq
    )
    +
    PbBr
    2
    (
    s
    )

    Enter the balanced net ionic equation, including phases, for this reaction.
    net ionic equation:

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