Summarize your results by preparing a table with the formulas from Group A as pa

Question

Summarize your results by preparing a table with the formulas from Group A as partings of die columns and the formulas of Group B as headings of the rows. Within the table, fill in the formula, systematic name and color of each precipitate foppied. if no precipitate was observed, write 'No Reaction'. Write the molecular, complete and net ionic equations for the following reactions: AgNO_3 with Na2CrO_4 Cr(fNO3)3 + Na3PO_4 Pb(NO3)2 -+ Nal Pb(NO3)2m +Na3PO4 Three of the seven tested reactions from part B were acid/base reactions For each acid/base reaction include the following: a. Molecular equation Complete ionic equation Net ionic equation For each reactant, indicate whether it was a strong acid, strong bas weak acid or weak base. Three of the seven tested reactions from part B were oxidation/reduction reactions For each oxidation/reduction reaction include the following: Molecular equation Below each molecular equation, state which element was oxidized and indicate the oxidation number the element had as a reactant and product. Zine was oxidized. The oxidation number changed form 0(reactant) to +2 (product).

Explanation / Answer

NEt ionic equation include only the ions that form precipitate so:

a)

Molecular

2AgNO3(aq) + Na2CrO4(aq) = Ag2CrO4(s) + 2NaNO3(aq)

Ionic

2Ag+(aq)+ 2NO3-(aq) + 2Na+(aq) + CrO4-2(aq) = Ag2CrO4(s) + 2Na+(aq) + 2NO3-(aq)

Net ionic

2Ag+(aq) + CrO4-2(aq) = Ag2CrO4(s)

b)

Molecular

Cr(NO3)3(aq) + Na3PO4(aq) = CrPO4(s) + 3NaNO3(aq)

Ionic

Cr+3(aq) + 3NO3-(aq) + 3Na+(aq) + PO4-3(aq) = CrPO4(s) + 3Na+(aq) + 3NO3-(aq)

Net ionic

Cr+3(aq) + PO4-2(aq) = CrPO4(s)

c)

molecular

Pb(NO3)2(aq) + 2NaI = PbI2(s) + 2NaNO3(aq)

ionic

Pb+2(aq) + 2NO3-(aq) + 2Na+(aq) +2I-(aq) = PbI2(s) + 2Na+(aq) + 2NO3-(aq)

net ionic

Pb+2(aq) + 2I-(aq) = PbI2(s)

d)

molecular

3Pb(NO3)2 + 2Na3PO4 = Pb3(PO4)2(s) + 6NaNO3(aq)

ionic

3Pb+(aq)+ 2NO3-(aq) + 6Na+(aq) + 2PO4-3(aq) = Pb3(PO4)2(s) + 6Na+(aq) + NO3-(aq)

net ionic

3Pb+2(aQ) + 2(PO4-3) =Pb3(PO4)2 (s)

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