would there be a reaction when lead nitrate and sodium iodide are mixed together?
971239
When the two solutions are mixed, you'll get a bright yellow precipitant.\n\nPb(NO3)2 + 2 NaI ---> PbI2 + 2 NaNO3
817375
lead acetate test is used for detection of iodide ion. \nso, let the sample salt be ferric iodide.This ferric iodide doesn't react easily with lead acetate . sample is boiled first in 50% NaOH solution as lead acetate reacts with NaOH to form NaI, which reacts easily  with lead acetate to form lead iodide(yellow ppt.)
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
1044812
Diluted potassium iodide
790197
The solute would the the silver nitrate because it is desolved in the water.  Water would be the solvent.
485711
What are you asking here?  If there is an equation for the half-reactions of a pencil dipped in cupric chloride solution?  If so, it is easy, since there is no reaction.\n\nCuCl2 ============> Cu2+  + 2 Cl-  in solution\n\nPencil lead  = graphite and clay.  No ions in solution.\n\nThus, no reaction.\n\nOTOH, if you are referring to lead ions in solution (say, from lead nitrate) then you get:\n\nCuCl2 + Pb2+ ===========> PbCl2(s) + Cu2+\nThe half-reactions are:\n\nCuCl2 ===========> Cu2+  + 2 Cl-\nPb(NO3)2 ==========> Pb2+  + 2 NO3-\nPb2+  + 2Cl- ============> PbCl2 (s)\n\nThe copper ions and nitrate ions would remain in solution.
793687
could be a chemical reaction to processed salt (iodine...etc)...try switching to sea salt.
1008732
Ksp = [Pb+2]*[Cl-]^2/[PbCl2]\n\nby definition, [PbCl2] = 1\n\nAlso, based on the reaction:\n\nPbCl2 =====> Pb+2  +  2 Cl-\n\n[Cl-] = 2 [Pb+2]\n\nSubstitute this into the equation, and you get:\n\nKsp = [Pb+2] * [Cl-]^2 = [Pb+2]*(2[Pb+2])^2 = 4[Pb+2]^3\n\nNow, since you know that 0.99 grams of PbCl2 is dissolved in your liter of solution, you can calculate the molarity from:\n\n(0.99 g)/ (278.1 grams/mole) = 3.56x10^-3 moles = [Pb+2]\n\nKsp = 4*(3.56x10^-3)^3 = 1.8 x 10^-7\n\nThe text value for the Ksp of lead (II) chloride is 1.7 x 10^-5\n=================\nOkay, now for the second part of your question:\n\nIf you have 0.5M lead nitrate, which is soluble, you will have 0.5M Pb+2\n\nSo, let's plug 'n chug (we'll use the text value for Ksp to show the process):\n\nKsp =  1.7x10^-5 = [Pb+2]*[Cl-]^2\n\nbut, with the lead nitrate in solution, [Pb+2] = 0.5 moles/liter (the amount you'd get from the dissolution of lead chloride will be so small as to be negligible), so:\n\nKsp = 1.7x10^-5 = [0.5]*[Cl-]^2\n\n[Cl-]^2 = (1.7x10^-5)/(0.5) = 3.4x10^-5\n\nso [Cl-] = sqrt(3.4x10^-5) = 5.83 x 10^-3  and since [Cl-] = 2 [Pb+2], we halve that value to get:\n\nsolubility = 5.83x10^-3/2 = 2.92x10^-3 moles per liter\n\n(2.92x10^-3 moles/liter) * 278.1 grams/mole = 0.811 grams/liter\n\nIf you use the value of the Ksp calculated in the first half of the question, then the solubility of PbCl2 in your solution is 3x10^-4 moles/liter, or 0.083 grams/liter
610699
Pb(NO3)2(aq) + H2SO4(aq/l)[depending on concentration] --> PbSO4(s) + 2HNO3(aq)\nPb2+(aq) + SO42-(aq) --> PbSO4(s)\n\nI don't see how an redox reaction could occur.
642853
The pH can vary widly depending on the type of electroplating.  The nitrates will be the species associated with the metals used and of course sodium.\nCan you be more specific on which proceses you are looking at  See \n\nhttp://es.epa.gov/techinfo/facts/metalmfg.html\n\nhttp://es.epa.gov/cooperative/topics/metfactsheets.html\n\nhttp://www.epa.gov/ttn/chief/ap42/ch12/final/c12s20.pdf
834819
Without knowing what reaction you're trying to get, or everything else present, thats a difficult question to answer.\n\nIn a sense. you're forming a silver nitrate solution, but you have to knw all the spectator ions to know what reaction will actually occur.  The silver cation in solution needs to have a anion associated with it.  \n\nThe purpose may also have something to do with the variable solubilities of the silver compound that may be formed.  \n\nAnother potential reaction involves the nitrate as an oxidizing agent.
387399
Pb2+ + Cl- ---> PbCl2\n\nWe know that the lead in Pb(NO3)2 has to have a 2+ charge because a nitrate ion has a 1- charge, so two of them is a 2- charge.  The entire compound must have no charge, so a 2+ charge is assigned to the lead. \n\nBoth compounds Pb(NO3)2 and NaCl dissociate (they are both soluble), and because nitrate and Na+ ions are always soluble, they can be ignored in a net ionic equation.
750179
NaCl2 + NaCO3 ==> NaCO3 + NaCl2 NOTE: these two reactants will not react because the reactants and the products are identical\n\nCu + O2 ==> CuO2   This is already balenced\nCuSO3 + Fe ==> FeSO3 + Cu   The charge is not given so procuts could be different\n\n2HCl + 2Zn ==> 2ZnCl + H2\nMnO2 + H2O2 ==> No reaction: similar to NaCl2+NaCo3\n\nGood luck
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
831216
Cu (s) + Pb(NO3)2 (aq) ---------> No Reaction. \nCopper will not replace lead.\n\nreaction VI in the attached link.
279857
I have no idea what you are asking, would you like to rephrase the question?\n\nNitrate is an ion....a poly atomic ion with chemical formula NO3 with a super script -1, indicating it has a negative 1 charge over the entire ion.  Nitrate is an anion, meaning it has a negative charge, it will combine with cations (+) to form ionic compounds.\nCations and anions combine so form an overall neutral compound.\n\nFor example, in the case that Nitrate combines with a Potassium Ion to make Potassium Nitrate, the formula is,\nK+ + NO3- --> KNO3\nSince Potassium has a +1 charge, only 1 Nitrate is needed to make an overall neutral compound, however, in the case of a Calcium ion combining with Nitrate,\nCa+2 + NO3- --? Ca(NO3)2\n2 Nitrates are needed to balance the +2 charge on Calcium.
947511
no there isn't but it will last about 1-2 hours. ammonium nitrate and water it the reaction.