"Golden Rain" Pb(NO3)2 + KI --> Anyone can explain this reaction?
970329
Its a double replacement reaction. It is really simple to do. All you do is add the aqueous forms of both reactants and the reaction happens instantaneously.\n\nInsoluble PbI is formed and KNO3 is also formed and precipitates out.. Any compound with K+ or NO3- in it is soluble so it remains aqueous.
998581
The answer is A\n\nCoBr2 + Pb(NO3)2 ==> PbBr2  + Co(NO3)2\nPb(C2H3O2)2 + 2KBr ===> PbBr2 + 2KC2H3O2\n\nIn both cases, the net ionic equation is:\n\nPb+2  + 2 Br-  ====> PbBr2(s)
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
712912
BaI2(aq) + MgSO4(aq)  --> BaSO4(s) + MgI2(aq)\n\nSo BaSO4 will precipitate.
829175
AB + CD ---> AD + CB\nlogic doesn't work here as Si doesn't react/bond with Ca or Phosphorus.and Ca(PO) should be Ca3(PO4)2\nand SiO should be in the form of SiO2. \nso unless you want hypothetical compounds SiO2 won't form with either of the two other reactants.\nbased on the forumlae from an SI data book, the above reaction would not balance.\ngood luck solving this one. :-)
896646
"The final experiment was to observe the reaction when one small crystal of NH4SCN was added to 0.1 M FeCl3. The color of FeCl3 was yellow, but after NH4SCN was added, the color of contents changed to red.\nIn second experiment, when the Mg was ignited, the resulting flame was hot. Therefore, the release of thermal energy through the interaction of Mg and O was shown.\nFiguring out the type of the last two reactions was difficult, because the chemical equations were not given. However, the possibility that the type of both reactions were replacement reactions was high. The reason being that the chemical reactions were observed as a color change and this means that the composition of the substance is changed.\nThis experiment was very successful, and it was a good opportunity to observe chemical reactions directly."\n\nI would change your paragraph to that.  I would also expand on the other experiments if you know more about them.
836198
OK....You need to understand how this is all working, and why.  So, the formula equation for this reaction would be something like:\n\nKOH(aq) + HNO3(aq) --> H2O(l) + KNO3(aq)\n\nSince KOH, HNO3 and KNO3 are all strong electrolytes, they are totally ionized when they're dissolved in water.  So, written in complete ionic form, that equation would be:\n\nK+(aq) + OH-(aq) + H+(aq) + NO3-(aq) --> H2O(l) + K+(aq) + NO3-(aq)\n\nTo get to the net ionic equation, you just cross out the things that are the same on both sides of the arrow.  K+(aq) and NO3-(aq) are on both sides, so the net ionic equation is:\n\nH+(aq) + OH-(aq) --> H2O(l)
750811
Your question makes no sense.\nIf I understand correctly you are making artificial sea water by making a solution of KCl, HNO3,Hg(+2), then you are adding Pb(+2) and Cd(+2) to known concentrations... measuring potential (what method?) and you want to determine the concentration of...what? You know the concentrations of all species (unless a reaction is taking place)
887064
The main thing you need to know to figure out the ionic equation is whether the product is soluble or not.  In this case, PbSO4 and Mg(NO3)2 are your products, and lead salts are insoluble.  Thus, PbSO4 and any ion that is used to make this molecule is removed from the ionic equation:\n\nPb(+2) + 2(NO3)(-1) + Mg(+2) + (SO4)(-2) -> PbSO4 + Mg(NO3)2 (aq)\n\nbecomes:\n\n2(NO3)(-1) + Mg(+2) -> Mg(NO3)2 (aq)
284634
3K4Fe(CN)6 + 4FeCl3 -> 12KCl + Fe4[Fe(CN)6]3
422969
AgNO3(aq)+2Na2S2O3(aq)--> [Ag(S2O3)2]3-(aq) + 4Na+(aq)+ NO3-(aq)
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
502725
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
165589
Why is lead Pb3? It should be Pb. Lead doesn't form Pb-Pb bonds, except possibly in organometallics (not applicable here).\n\nLead's (IV) oxidation state is known and stable. The equation ought to be:\n\nPb + O2 --> PbO2\n\n\n======\nUMiamiChemist:\n\nBut what is Pb4O2? Balancing chemical equations is one thing ... but the results should be meaningful.
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
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
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)