[Kax0r]

13) Write a balanced net ionic equation for the reaction of Pb(NO3)2(aq) with NaI(aq).

A) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 I-(aq)  PbI2(s) + 2 Na+(aq) + 2 NO3-(aq)
B) Pb(NO3)2(aq) + 2 NaI(aq)  PbI2(s) + 2 NaNO3(aq)
C) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 I-(aq)  Pb2+(aq) + 2 I-(aq) + 2 Na+(aq) + 2 NO3-(aq)
D) Pb2+(aq) + 2 I-(aq)  PbI2(s)


15) Write a balanced net ionic equation for the reaction of NiBr2(aq) with (NH4)2S(aq).

A) Ni2+(aq) + 2 Br-(aq) + 2 NH4+(aq) + S2-(aq)  NiS(s) + 2 NH4Br(s)
B) Ni2+(aq) + S2-(aq)  NiS(s)
C) NiBr2(aq) + (NH4)2S(aq)  NiS(s) + 2 NH4Br(aq)
D) Ni2+(aq) + 2 Br-(aq) + 2 NH4+(aq) + S2-(aq)  NiS(s) + 2 NH4+(aq) + 2 Br-(aq)


Okay, I've gotten several answers, a couple say D first B second, but these threads seem to disagree:
http://answers.yahoo.com/question/index?…
http://jchemed.chem.wisc.edu/JCESoft/CCA…

Who is right? I'm so confused

Edit: I'm pretty confident in D and B but I wanted to make sure. It looks like the full ionic equation for the first one would be:
Pb2+2NO3+2Na+2I = PbI2 + 2Na + 2 NO3.
which, when removing spectator ions would be- Pb2+ 2I = PbI2.

For the second one, it would be much the same. Ni+2Br+2NH4+S=NiS+ 2 NH4 + 2Br, removing spec ions, Ni+S=NiS

[Kax0r]

yeah look this is due in like a half hour so can I please get some answers?

[Borek]

Try here: forum rules.