If we add PbNO3 liquid to water, can we observe the precipitate of Pb(OH)2?
344076
actually, the only possible solution is FLAVEN, with the hey-hey, and the hey LAY-deeeeez!!
851017
B)   PbS  use the solubility rules. all leads are insoluble except with NO3, ClO4, and SO4....there is a list of solubility rules.
875344
1) Litmus is red in acid, blue in base...so it would turn red in rainwater (which is sl acidic, more or less depending upon atm contamination), blue in ammonia.\n2) AgCl precipitates out...KCl + AgNO3--->AgCl(ppt) + K+ + NO3-\n3) Zn is more active than Fe, so it plates out at the expense of Fe...Zn(+2) + Fe--->Zn(ppt) + Fe(+2)\n4) Depends on the type of battery...for the lead/acid type it is a reversible reaction between PbO in the plates and the H2SO4 electrolyte (in overly simple terms)\n.....Discharge: 2H(+) + SO4(-2) + PbO--->PbSO4 + H2O\n.....Charge: PbSO4 + H2O--->PbO + 2H(+) + SO4(-2)
517995
67.  AgCl <---> Ag+ + Cl-  Ksp = [Ag+][Cl-] = 1.8 x 10^-10\n[Ag+] = 0.001/2 = 0.0005 M; [Cl-] = 0.0025 M; [Ag+][Cl-] = 1.25 x 10^-6, as this is greater than Ksp, a precipitate will form.\n\n68.  SrCrO4; Ksp = [Sr2+][CrO42-] = 3.5 x 10^-5; molar solubility = [Sr2+] = [CrO42-] = sqrt (3.5 x 10^-5) = 0.0059 M\n\n69.  Total volume = 3L  [Fe2+] = 0.150 * 1/3 = 0.050M; [OH-] = 0.0333 * 2 / 3 = 0.0222 M.  Ksp Fe(OH)2 = 4 x 10^-38.  (0.050) * (0.0222)^ 2 >>> 4 x 10^-38, so a precipitate will form.\n\n73.  Pb3(AsO4)2 <---> 3 Pb2+ + 2 AsO43-.  If solubility = x, then [Pb2+] = 3x, [AsO43-] = 2x.\n4.0 x 10^-36 = [3x]^3[2x]^2 = 108 x^5 ==> x = 3.26 x 10^-8
779131
1. PbO + PbS\nmoles PbO = 0.257 mol\nmass of Pb from PbO: 0.257 mol * 207.2 g/mol = 53.25g\nmoles PbS = 0.141 mol\nmass of Pb from PbS: 0.141 mol * 207.2 g/mol = 29.21g\ntotal mass of Pb: 53.25g + 29.21g = 82.46g\n\n2. BaCl2 + Na2SO4 --->  2NaCl + BaSO4\nmoles Na2SO4: 0.042 L * 0.453 mol/L = 0.019 mol\n- means you also need 0.019 moles BaCl2\nvolume BaCl2 required: \n0.019 mol/ 1.131 mol/L = 0.0168 L or 16.8 mL\n\n3. 4FeCl3 + 3O2 --->  2Fe2O3 + 6Cl2\nbased from the balanced equation above, 4 moles of FeCl3 are needed for every 3 moles of O2 for the oxidation reaction. Hence, the excess reagent is FeCl3. Only 2.66 moles FeCl3 is needed to react 2 moles of O2. There is 0.33 mole of FeCl3 remains unreacted.
791931
Precipitation:: Precipitation is the formation of a solid in a solution during a chemical reaction. When the chemical reaction occurs the solid formed is called the precipitate. This can occur when an insoluble substance, the precipitate, is formed in the solution due to a reaction or when the solution has been supersaturated by a compound. The formation of a precipitate is a sign of a chemical change. In most situations, the solid forms ("falls") out of the solute phase, and sinks to the bottom of the solution (though it will float if it is less dense than the solvent, or form a suspension).\n\nThis effect is useful in many industrial and scientific applications whereby a chemical reaction may produce a solid that can be collected from the solution by various methods (e.g. filtration, decanting, centrifuging). Precipitation from a solid solution is also a useful way to strengthen alloys.\n\nAn important stage of the precipitation process is the onset of nucleation. The creation of a hypothetical solid particle includes the formation of an interface, which requires some energy based on the relative surface energy of the solid and the solution. If this energy is not available, and no suitable nucleation surface is available, supersaturation occurs.\n\n\n\nAggregation::Particle aggregation in materials science is direct mutual attraction between particles (atoms or molecules) via van der Waals forces or chemical bonding.\n\n\n\nCoagulation::The transformation of liquid into jelly like semisolid mass.\n\n\nFlocculation::Flocculation refers to a process where a solute comes out of solution in the form of floc or "flakes." The term is also used to refer to the process by which fine particulates are caused to clump together into floc. The floc may then float to the top of the liquid, settle to the bottom of the liquid, or can be readily filtered from the liquid.
727503
The pH value would increase, by how much would depend on the volume and concentration of the sodium hydroxide used.\n\n2.0Molar would have a much greater effect than 0.1Molar in the same volumes added to your water.
861717
2 HNO3(aq) + Sr(OH)2(aq) --> Sr(NO3)2 (aq) + 2 H2O\n\n2 H3PO4(aq) + 3 Ba(OH)2(aq) --> Ba3(PO4)2 (cr)* + 6 H2O\n\n* Barium Phosphate will precipiate out of an aqueous solution in this reaction
41331
Hg2Br2, BaSO4, AgCl, PbCl2, Al(OH)3, are insoluble\nKBr is soluble.\nYou can refer to this link for details on solubility of salts.\nhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/ch18/soluble.html
113958
In any reaction, white ppt formed implies that there is a compound formed in the solution  during the reaction which is white in colour and insoluble in water. So, it remains undissolved in water at the bottom like a white powder. This is called as white precipitate (ppt)
1099415
I don't know how you are precipitating out the gold, but I don't think a colorimetric test for gold in solution is possible.  It is non-reactive and does not ionize.  Given that gold has filled upper electron orbitals, you might be able to run a UV-VIS spectrum and get some UV absorption for solvated gold versus clean aqua regia.  The UV apsorption is probably going to be in the lower UV wavlengths, and it might not be possible to differentiate between trace levels of gold and trace levels of transition metal impurities.\n\nIf you are precipitating the gold by an equilibrium process, just load up on your precipitation compound to push the equilibrium as far as possible to one side.  Another possibility is to electrolytically remove any trace gold from the aqua regia.  Sorry I don't have any better suggestions.
157642
Firstly, the purer the water, the worse it is to use as a conductor. Mainly, the dissolved conductive elements we find in tapwater increase its conductivity, but are not as effective as say the ones found in saltwater. So, unless you wanted to spend the time and money to dissolve the copper, zinc, lead, and other conductive materials in water that could more easily be ducted into wire, it won't be a good idea.
990586
You cant.\nThe key word here being "completely".\n\nNo matter how insoluble a given precipitate is, it is always soluble to some, very minor, degree.\n\nHowever, by choosing to precipitate a compound with a very low, practically zero, solubility, the ion is (for all intensive purposes) removed from solution.\nWhen dealing with such low-low solubilities, the Ksp value for the compound becomes important...the lower the Ksp, the lower the solubility.
831216
Cu (s) + Pb(NO3)2 (aq) ---------> No Reaction. \nCopper will not replace lead.\n\nreaction VI in the attached link.
135415
That's a difficult question to answer because magnesium hydroxide is practically insoluble in water with a ksp value of like 1.8 x 10^-11 so very little magnesium hydroxide is dissolved in water.  But if you were to compare the solubility of magnesium hydroxide at 25C with 80C my guess is that it would be more soluble at a higher temperature but only by an almost insignificant amount.  \n\nIn short, Mg(OH)2 is pretty much insoluble in water so my direct answer after all the reasoning would probably be: no, Mg(OH)2 does not fit the pattern due to its extremely low solubility.
715971
Gold would be deposited on the surface of the lead.\n\n3Pb + 2Au(NO3)3 ---> 3Pb(NO3)2 + 2Au
387221
B2O3(s)+Mg(s) yields MgO(s)+B(s)\nNotes: yields should be substituted with an arrow that is read as yields and in diboron trioxide the numbers are subscripts not superscripts.