If a 25.0mL H20 sample requires 27.2mL of 0.104 M AgNO3 in a titration, what is the [] of Cl- in the sample?
342141
ok well, the equation is balanced so you need equal moles of Ag and Cl to make the products. They give you the concentration and volume of Ag, so you can find the amount of moles by multiplying 0.104mol/L * 0.0272L = 0.0028288 mols of Ag. so 0.0028288 mols of Cl were there originally. and you have the volume of water, so you divide moles by volume and you get the concentration. 0.0028288 mols/0.025L = 0.113152 M. \n\nanother way to do it is the easy way, use c1v1 = c2v2. you have c1 = 0.104M, v1 = 27.2mL, v2 = 25mL, so you can find c2. you dont have to worry about units in this case so its easier. \n\nhope this helped. let me know if u need more help. laterz
760435
You are looking at a displacement reaction, in which the more reactive metal (Cu) is displacing the less reactive metal (Ag) from its salt solution (AgNO3).\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nHope this helps. :)
1030709
1 H2 +1 Cl2 ------> 2 HCl\n\n     2 - H - 2\n     2 - Cl - 2
133194
This usually happens in solution, the Chlorine is a chloride ion.\n\nAgNO3(aq) + Cl-(aq) --> AgCl(s) + NO3-(aq)
706927
a. Silver nitrate\nb. isopropanol? methanol?\nc. Potassium chloride\nd. Benzene\ne. Carbon monoxide\nf. Sodium hydroxide
307824
dissolve the sample in KOH. Add AgNO3 and weigh the ppt. to obtain the halide content.
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)
363082
2Ag+ + CO_33-2 -> Ag_2CO_3
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
724819
To determine the concentration of a reagent.
1081818
1) 2 moles of NH3 need 1 mole of H2SO4\n\nSo, 146x1.72 mmol of MH3 needs (146x1.72)/2 mmoles of H2SO4\n\nVol. of .23M H2SO4 reqd. = (146x1.72)/(2x.23) ml \n\n2) Molar mass of AgCl = 108+35.5 = 143.5 g\n\nSo, moles of AgCl = 16.9/143.5\n\n1 mol of KCl gives 1 mol of AgCl \n\nSo 16.9/143.5 moles of KCl in 584 ml is equivalent to\n\n16.9x1000/(143.5x584) M\n\n3) 1.725 g is in 51.7 g of soln.\n\nSo, % conc is 1.725x100/51.7 = 172.5/51.7\n\nI will leave it to you to do the math.
709021
All chlorides are water soluble\n\nEXCEPTIONS: AgCl, Hg2Cl2
270139
If you mean conductometric TITRATION, it won't work.  Assuming they're both at the same concentration, you'll get a flat line.  Conductance is a rough measure of the ion concetration.  The start (H+ + Cl-) has the same number as equivalence (Na+ + Cl-).  Adding more (Na+ + OH-) past equivalence will not change the concentration either.
252970
Argentometry refers to quantitative analysis using silver (remember the symbol for silver is Ag comes from the Latin word "argentum" meaning "silver") For example, silver nitrate solution could be used to determine the concentration of hydrochloric acid by precipitating the chloride as silver chloride, then drying and weighing it. Silver's high atomic mass is useful here to give high precision.
903130
Na2SO4(aq)+ 2 AgNO3(aq)-->Ag2SO4(s)+ 2 NaNO3(aq).
658363
This is a limiting reagent problem. The first job is to figure out how many moles of each reagent you have and then figure out which one will run out first because that will determine how much product you can make. It is easiest to write out the balanced equation first to see what you are dealing with:\n\n2Na + Cl2 --> 2NaCl\n\nYou have 23g x 1mol/23g the molar mass of sodium = 1 mol of sodium\n\nYou plug the standard pressure, 1 atm, the volume, 0.56 L, the R constant, 0.082, and the standard temperature, 273 K, into the ideal gas law formula PV = nRT and solve for n.\n\nYou get 0.0250 moles chlorine gas.\n\nThen figure out how many moles of chlorine are needed for the amount of sodium you have.\n\nfor 1 mol of Na you would need 1 x 1/2 = 0.5 moles of chlorine. But you only have 0.025 moles so this must be the limited reagent.\n\nNow you must figure out how many moles of NaCl form from 0.025 moles of Cl2.\n\n0.025 x 2/1 = 0.05 moles of NaCl.\n\nNow multiply this by the molecular weight for NaCl, 23 g/mol for Na and 35.5 g/mol for Cl = 58.5 g/mol total\n\n0.05 mol x 58.8 g/mol = 2.93 or about 3 grams of NaCl formed
788070
Well its easy dear.\n\n2Li (s) + Cl2 (g) ----> 2 LiCl (s)