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AgNO3 + KCl
Na2S (aq) + Ni (NO3)2 (aq) ----> 2 NaNO3 + NiS
Yes i got that but i am looking for the "net ionic equation"
First, let's split up the first reaction into its component ions...\[Ag(aq)^++NO_3^-(aq)+K^+(aq)+Cl^-(aq)\]Then, find out what the products will be. In this case, it's a precipitation reaction...\[AgNO_3(aq)+KCl(aq) \rightarrow AgCl(s) + KNO_3(aq)\]Now, this is the overall equation. The net ionic equation involves breaking each compound into its component ions, and dropping all ions that appear in aqueous solution on both the reactants and products side. Doing this to the above equation...\[Ag^+(aq)+Cl^-(aq) \rightarrow AgCl(s)\]This is the net ionic equation for a.
oh srry i will help u in c
2) Write out the total ionic equation and cancel anything that is common on both sides Ca+2(aq) + 2Cl-(aq) + 2Na+(aq) + CO3-2(aq) --> CaCO3(s) + 2Na+(aq) + 2Cl-(aq) so the net ionic equation becomes: Ca+2(aq) + CO3-2(aq) --> CaCO3(s) this is net eqn for c
Xishem why did you get rid of the K and NO3
The net ionic equation is the total equation with all spectator ion removed. Spectator ions are ions which are present in aqueous solution on both sides of the equation. In other words, they didn't participate in the reaction (spectators). I'll show it for part a...\[Ag^+(aq)+NO_3^-(aq)+K^+(aq)+Cl^-(aq) \rightarrow\]\[Ag^+(s)+Cl^-(s)+K^+(aq)+NO_3^-(aq)\]The potassium ion and the nitrate ion exist in aqueous solution on both sides of the equation, therefore they are spectator ions and are to be removed to give...\[Ag^+(aq)+Cl^-(aq) \rightarrow AgCl(s) \]Which is the net ionic equation. Make sense?
OHHHHH yessss okay that makes a lot more sense now...thanks guys! i will try b) ill post what I got
would it be Ni-2(aq) + S-2 (aq) ---> NiS(s)
:D!!!! THANKS Xishem and AravindG! I wish i could give more than 1 medals hah