Addition of nitric acid, HNO3, and ammonium molybdate, (NH4)2MoO4, to a solution containing aqueous sodium phosphate, Na3PO4, results in the formation of a yellow precipitate of ammonium called 12-phosphomolybdate, (NH4)3PMo12O40 and two aqueous nitrate solutions: ammonium nitrate, NH4NO3 and sodium nitrate, NaNO3. Water is also formed in this reaction. Note that 12-phosphomolybdate is a very big anion! If needed, heating in a boiling water bath can increase the rate of this reaction. You will use the formation of a yellow solid on addition of HNO3 and (NH4)2MoO4 to a sample to indicate that the phosphate ion, PO43-, is present.

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes