what will happen if you mix lead nitrate and potassium iodide?
82368
U get precipitate of Lead Iodide. The reaction is already mentioned!!!!\n\nPb(NO3)2 (aq) + 2KI (aq) --> 2KNO3 (aq) + PbI2 (s)
771965
I can't tell you the name because i can't remember that stuff but the product will be K(NO3) PbI2\nif you're writing it in the long form it will be\n2KI + Pb(NO3)2 yeilds (arrow) 2K(NO3) + PbI2
1087032
try Vogel's "Qualitative Inorganic Analysis", I'm pretty sure it has all the answers 4 u.\n\ndont care editions, but for 7th ed, ISBN is 0582218667\n\n****\nokey then, from Vogels\n\n1) when heated, lead nitrate decomposes to white solid lead oxide (PbO), brown gas (NO2) and oxygen gas. Brown gas decolorizes when contact with water. \n\n2) ammonia odur. ammonium compounds are all volatile towards heat and decomposes to ammonia and an acid but quickly recombine if the acid, in this case, hydrochloric acid, is a strong acid.\n\n3) anhydrate copper sulphate is white, hydrate copper sulphate is blue. I have no idea which one you actually heated. If it's anhydrate, no change; if hydrate, blue ---> white.\n\n4) blue precipitate forms initally, but resolve upon excess reagent, clear blue soln results due to the formation of copper complex [blue precipitate: Cu(OH)2*CuSO4, complex: [Cu(NH3)4]^2+]\n\n5) brown precipitates froms [Cu(OH)2], no change with excess reagents.\n\n6) metal-gray stipe (magnesium stipe) gains a redish brown color due to formation of copper metal. Possible decolorization of the soln (depends on how much magnesium u actually put in)\n\n7) volatile reaction, solid metal solves,ignitable gas is formed. (if phenolphthalein is added to water initally, soln becomes pink)
799013
This one is easy as well... THERE IS NO REACTION!\n\nPB is lower on the activity series, so it cannot replace the Zn in the zinc nitrate...
950743
theoretical amount of product=10g lead acetate/ mw lead acetate x mw lead iodide\n\ntake the result of this formula as the denominator and divide 5.00 grams product/theoretical x 100%
796075
At 10°C, saturated solution of potassium chloride is more concentrated than potassium nitrate. At this temperature, potassium chloride completely dissolves while potassium nitrate would form some precipitates. At 60°C, it would be the same since both would completely dissolve at this temperature.
828284
The first thing that happens is that it melts.  If you continue heating the molten salt, you will see bubbles appearing.  These are bubbles of oxygen:\n2 KClO3(l)   --->   3 O2(g)   +   2 KCl(l)\nThe main product left behind is potassium chloride, but I believe that you also get some potassium perchlorate (tetraoxochlorate(VII)), produced by disproportionation:\n4 KClO3(l)   --->   3 KClO4(l)   +   KCl(l)\n\nIt wouldn't surprise me if the reaction also produced very small amounts of oxides of chlorine, such as chlorine dioxide, ClO2.\n\nAllan Deeds.
508878
Iodate would be reduced to iodine or even further to iodide by metabisulfite (S2O5)2-.
712812
1.   Ag + KNO3 --> AgNO3 + K\n      single replacement means that a single element breaks a compound apart and joins with part of it--but which part--the positive or negative part?  It depends on what the single element is.  If it's a metal (#1 above), it will join with the negative part of the compound.  If it's a nonmetal (# 4 below), it will join with the positive part of the compound.\n4.  Cl2 + 2 KI --> 2 KCl + I2\nI'll leave you to have fun with the rest!
975754
I'm not sure if "catalyzes" is the right word, but what you're doing by adding iodide (as KI or NaI) to the reaction mixture is adding another nucleophile to the mix.  This would then speed up the SN2 reaction of your starting material.  You would probably run the risk of iodide becoming another (competing) nucleophile.  Since you don't specify what the initial nucleophile actually IS, this may or may not be a bother to you; the gyst of your question was whether the reaction would speed up.  It probably will, because\n\nrate of SN2 rxn = k [substrate][nucleophile]\n\nAdding I- is like adding more nucleophile, unless of course  your existing nucleophile is more reactive towards the substrate than iodide is.  And as the relation above says...more nucleophile means a faster reaction.  The question is, what's your product?\nIs it the desired one?  or do you care?  (are you content with just seeing the reaction speed up....)\n\nBest I can do with what you gave me.
1187925
You have two ionic substances, it will undergo a double displacement reaction.\nPb(C2H3O2)2 + 2 NaCl --> PbCl2 + 2 NaC2H3O2\n\nNow since Lead(II) acetate, Sodium chloride, and sodium acetate are soluble, they break up in water, and the net ionic equation is\nPb2+ + 2 Cl- --> PbCl2\n\nPbCl2 or Lead (II) chloride, is the precipitate formed, and it's generally yellowish.
471546
A.  Iodine is being reduced from +5 in HIO3 to 0 in I2; oxygen is being oxidized from -1 in H2O2 to 0 in O2.\n\nB.  There is no oxidation/reduction going on; this is simple metathesis, producing the barium sulfate precipitate.
1143109
lead, poisonous gas emmissions...you are talking about damage...this stuff can kill you
954959
Potassium nitrate, or any other nitrate, is an oxidizer.  Combustion is an oxidation, so you need something oxidizable (combustable).  This would be called a reducing agent.  A typical reducing agent is any carbon molecule with lots of hydrogens.  \n\nYou need to know the reaction to get the right proportions.  Potassium nitrate mixtures can behave unpredictably, and can be dangerous.  A common one is black gunpowder.  It is a mixture of salpeter (potassium nitrate), charcoal, and sulfur.
399979
No, Potassium Nitrate will not react with water, it will only form an aqueous solution.\n\nKNO3(s) --H2O--> K+(aq) + NO3-(aq)\n\nSince Potassium ions and Nitrate ions form substances which are strong electroltes, both these ions will stay dissassociated in solution and not react with the water.
1044812
Diluted potassium iodide
173273
You should not expect such an outcome. Iodine forms a complex with starch. This is because the iodine intercalates within amylose coil. Bear in mind starch is a polymer and sucrose isn't. So you do not expect iodine to form a the dark blue complex with Sucrose.