Which is the correct net ionic equation for the reaction of ammonium iodide with lead (II)...
Which is the correct net ionic equation for the reaction of ammonium iodide with lead (II) nitrate to form lead (II) iodide and ammonium nitrate.
I tried selecting
NH4I (aq) + PbNO3‒ (aq) ⇌ PbI2 (s)+ NH4NO3 (aq)
and
NH4+ (aq) + Pb2+ (aq) +2I‒ (aq) + NO3‒ (aq) ⇌ PbI2 (s)+ NH4NO3 (aq)
but I'm not sure what I'm doing wrong.
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Which is the correct net ionic equation for the reaction of
ammonium iodide with lead (II)...
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Lead(II) nitrate and ammonium iodide react to form iodide and
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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq) + 2NH,I(aq) — Pl(s) + 2NH, NO, (aq) What volume of a 0.270 M NH I solution is required to react with 487 mL of a 0.680 M Pb(NO3)2 solution? volume: How many moles of Pbly are formed from this reaction? moles: mol Pbl2
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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3), (aq) + 2 NH,I(aq) + PbI,(s) + 2 NH, NO3(aq) What volume of a 0.150 M NH4I solution is required to react with 591 mL of a 0.540 M Pb(NO3)2 solution? volume: mL How many moles of Pbly are formed from this reaction? moles: mol PbI2
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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3), (aq) + 2NH I(aq) → Pb1,(s) + 2 NH, NO, (aq) What volume of a 0.310 M NH I solution is required to react with 193 mL of a 0.680 M Pb(NO3)2 solution? volume: How many moles of Pbly are formed from this reaction? moles: mol PbI,
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