Science

1. Remember all of our questions pertain to the reaction below:
Br-(aq) + NO3- ==> NO(g) + Br2(l)
Options: Br, N, O, -2, -1, 0, +1, +2, -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5
[Select] is oxidized, it's oxidation number changes from [Select] to [Select].
[Select] is reduced, it's oxidation number changes from [Select] to [Select].
[Select]'s oxidation state doesn't change, its oxidation number is always [Select]
2. Fill the coefficients for the four original reactants and products:
Options: 8, 7, 6, 5, 4, 3, 2, 1, 0
[Select] Br- (aq) +
[Select] NO3- (aq) ==>
[Select] NO(g) +
[Select] Br2 (l)
3. If you were balancing this reaction in acidic conditions how many of each additional species would there be and where in the reaction would they be located.
Options: 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 and options: the reactant, products, neither
Protons (H+): There would be
[Select] on the [select] side
Water (H2O): There would be
[Select} on the [select] side
Hydroxide (OH-): There would be
[Select #] on the [select] side
4. The same thing as 3 but this time if you were balancing this reaction in basic conditions.

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  1. With no spacing this is really difficult to understand. I'm not all that interested in going through this in detail and answering everything. Let me get you started, then if you have trouble tell me the trouble you are having and what you don't understand and I can help you through it. In other words, I'm willing to HELP but I'm not willing to do all of the work for you. You need to learn how to do this.
    Br-(aq) + NO3- ==> NO(g) + Br2(l)
    Definitions you need to know. Oxidation is the loss of electrons; reduction is the gain of electrons. Br^- has an oxidation number (ON) of -1 on the left and zero on the right. So it lost e which makes it oxidized. N in NO3^- on the left has an ON of +5 and it changes to +2 on the right. That's a gain of electrons so it is reduced. The ON of Oxygen goesn't change. It is -2 on the left (for EACH O atom) and -2 on the right in NO.
    The balanced equation is
    6Br-(aq) + 2NO3 + 8H^+ ==> 2NO(g) + 3Br2(l) + 4H2O
    I'll leave you here. I shall be happy to help you understand anything I've done above but you must be clear in what you don't understand. The same applies for the remaining questions.

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