A strip of tin (Sn) is placed in an AgNO3 solution?
339232
Yes, the reaction would spontaneously occur because Sn is higher on the activity series than Ag.\n\nSn(s) + 2AgNO3(aq) => Sn(NO3)2(aq) + 2 Ag(s)\nThe net ionic is\nSn(s) + 2Ag+(aq) => Sn+2(aq) + 2 Ag(s)\n\nI don't remember the oxidation number for Sn.  2 sounds about right.\n\nIt would turn a silver nitrate solution to a tin nitrate solution, and you'd see metallic silver precipitate out.
760563
N=aM where a is the valency (depending on the type of chemical and reaction, e.g. number of H+/OH- for acid/bases per molecule, number of electrons gained/lost in redox reactions).\n\nFor AgNO3 a=1 (it is Ag+ so for redox or precipitation reactions it gives one equivalent group per molecule)\n\nSo M= N/a = N/1 =N\n\n1 M is 169.87 g/L =16.987g/100mL =16.987%\n\nUse M1*V1=M2*V2\n16.987*V1=1*V2\nV1 is the volume in mL of the stock solution that you will use to prepare V2 mL of 1%\nYou didn't specify how much you want to prepare...\n\nAlternatively you could have converted 1% to mole/L and do the math with molarities.
953082
Okay, it's been a looooooong time since I had to use my chemistry skills, so lets see what I have:\n\nCopper is more magnetic with electrons than silver is because it's electrons (with fewer of them) are close to the nucleus (or whatever you call it) of the element.  Silver is more apt to lose its electrons with them being so far away.  \n\nCopper is more often found in a +2 state.  So I'd say that's a safe bet to go with.  I'd only assume a +1 state if your teacher said so.\n\nSo therefore:  2AgNO3 + Cu -> Cu(NO3)2 + 2 Ag would probably be your best bet.\n\nBut once again, it's been a long time since I used this so I don't know how right I might be.
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
261798
The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.
945093
The subscrips indicate that mulitples of the symbol they follow.  Like the AgNO3 the three means that there are 3 oxygens.  The two after the () means there are three whole molecule though, that can be placed infront as a coeffient.
782516
a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)
564401
Silver Carbonate  - AgCO3 (3 being the small number on teh O)\nSilver Nitrate - AgNO3\nSilver Sulphate - Ag2SO4
790197
The solute would the the silver nitrate because it is desolved in the water.  Water would be the solvent.
993350
easy!\n\nadd silver nitrate\n\nit will make silver bromide precipitate.\n\nthis is pale cream in colour.\n\nfurther confirmation is done by adding conc ammonia to it, and if it dissolves, it proves that the original solution had bromide ions.
729762
The photomicrograph of pb-sn eutectic that you were wanting is at the link below:
1082457
Anode is a silver bar and cathode is iron spoon.Electrolyte used is aqueous silver nitrate (AgNO3)\n\nAgNO3 in water ionizes into Ag+  and NO3-\nPositively charged Ag+ ions are attracted by the spoon to be plated.Ag+ reach the spoon and electrons flow from the spoon to the Ag+ ions.Thus, at Cathode:\n[Ag+]  +  [e-] =Ag\nAt the anode, electrons are removed from the Silver metal bar, oxidizing it to the Ag+ state. Thus the silver metal dissolves as Ag+ into the solution, supplying replacement silver for that which has been plated out.\nThus at anode:\nAg - [e-]= [Ag+]\nThts all.
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
419109
I don't remember the ordering, but there's probably some metal out there that won't react?  Gold, maybe?  For regular metal, that sounds a bit iffy.
778417
Find the moles of AgNO3:\n\nn = C*V, n = 0.14*25x10^(-3) mol\n\nAccording to the chem. equation:\n\n2AgNO3 + Na2S  --> 2NaNO3 + Ag2S\n\n2 moles of AgNO3 reacts fully with 1 mole of Na2S, so\n0.14*25x10^(-3) moles of AgNO3 reacts with 0.07*25x10^(-3) moles of Na2S\n\nNow you can find the volume of the Na2S solution:\n\nn = C*V  =>  V = n/C = 0.07*25x10^(-3)/0.105 = 0.05/3 L or\n\nV = 16.7 mL (approx)
129188
Because the water in Silver Nitrate acts as the hydrolysing agent.  \n\nAnd, because Halide ions react with silver nitrate to give precipitates of differing color to help distinguish between:  chlorides, bromides and iodides.
737939
Yes. It produces a bright yellow precipate:\nAgNO3 (aq) + KI (aq) → KNO3 (aq) + AgI (ppt)