Pb(NO3)2(aq) + 2NaOH(aq) -->Pb(OH)2(s) + 2NaNO3(aq)
PbO2 -------> Pb + O2 for Lead IV Oxide
2PbO -------> 2Pb + O2 for Lead II Oxide
You cannot make diphosphorous pentoxide (P2O5) from lead(II) nitrate, Pb(NO3)2. Where would the phosphorous come from? There is no phosphorous in the reactants.
2NO + O2 → 2NO2
Pb(No3)2------>2Pbo+No2+o2
2Pb(NO3)2 ----> 2PbO + 4NO2 + O2
A typo error.
2Pb(NO3)2 ===> 2PbO + 4NO2 + O2
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Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq) Aqueous lead II nitrate reacts with aqueous sodium iodide to form solid lead II iodide precipitate and aqueous sodium nitrate.
Pbs + 2Ag(NO3) -> 2Ags + Pb(NO3)2
Yes, of course: 4 (NH4)3PO4 + 3 Pb(NO3)4 -> Pb3(PO4)4 + 12 NH4NO3 for lead (IV) nitrate and 2 (NH4)3PO4 + 3 Pb(NO3)2 -> Pb3(PO4)4 + 6 NH4NO3 for lead (II) nitrate.
Chemical formula for water is H2O and for lead nitrate is Pb(NO3). Lead nitrate simply dissolves in water and making water poisonous.
2NH4Cl (aq) + Pb(NO3)2 (aq) ----> 2NH4NO3 (aq) + PbCl2 (s)
Equation is Zn + Pb(NO3)2 --> Zn(NO3)2 + Pb
Nh3+4hj
For lead (II) nitrate: H2SO4 + Pb(NO3)2 -----> 2HNO3 + PbSO4 ...................................................(white)
Pb + 2NO3 - --> Pb(NO3)2
The balanced equation is 2 KI + Pb(NO3)2 -> 2 KNO3 + PbI2.
2NaCl+Pb(NO3)2==== 2NaNO3+PbCl2
Pb(NO3)2(aq) + 2KI)aq) = 2KNO3(aq)+ PbI2(s) PbI2 is a yellow coloured solid and will precipitate out of the solution.
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq) Aqueous lead II nitrate reacts with aqueous sodium iodide to form solid lead II iodide precipitate and aqueous sodium nitrate.
Lead(II) Nitrate is Pb(NO3)2
Die
There is no reaction between lead nitrate and sodium nitrate, because both compounds contain the same anion (nitrate). The reaction, if written, would look like this...Pb(NO3)2 + NaNO3 ==> NaNO3 + Pb(NO3)2
Pbs + 2Ag(NO3) -> 2Ags + Pb(NO3)2