Please what is the procedure for synthesizing ammonium persulfate (ammonium peroxydisulfate)? \nWhat are the reactants and the proportion in which theyare mixedThe reactants?\nThe necessary weblinks?
122250
add:\n\nH2S2O8(ag)+ NH3(aq) + ? -->(NH4)2S208(aq)  + ?\n\nAmmonia can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:\n\nthe net ionic equation to this reaction is:\n\n NH4+(aq) +  OH-(aq) ---> NH3(g) + H2O(l) in system 1\n \nand\n\n NH3(g)   = NH3(aq) in solution in system 2\n\nyou finish it ,if you need more help just e-mail me
310261
2.\nBy decreasing the temperature and increasing the pressure. As ussually done in Haber;s process\nAnother way is to keep on removing the ammmonia gas to make reaction go forward\nRead th Le Chatelier's principle for more detail.\n\n13.\n2 HCl + Na2CO3  -----> 2 NaCl + H2CO3\nH2Co3 can exist as water and carbon dioxide.\n\n14.\nBecause acid-base reaction is spontaneous.\n\n11.\nBecause of the formation of copper(II) carbonate which will settle down as precipitate.
484007
I'll try to explain this without going too much into chemistry.\n\nBefore adding peroxide, your mixture is fairly basic with baking soda;\n\nhydrogen peroxide is an acidic compound [though it's fairly weak], as we know, acids react with bases; it will react with the base in the mixture, in our case, baking soda. This reaction will give out gases and heat, with dish soap, you will have lots of air bubbles. Another reason why you're using H2O2 is because H2O2 decomposes on its own to give out gas, so you will have a better effect than just adding acids; this decomposition is catalysed by trace amount of metal ions [transition metals], so if you could find some copper salt [for example, CuSO4] and add them in the mixture, the reaction will be even more violent.\n\n*******\nNot all ingradients have significant purpose in your case; for example, you can remove the yeast while still having the same effects [hydrogen peroxide will kill them anyways].
378223
If the reaction goes to fast, i.e.the blue colour appears instantly, you can slow it down by adding more thiosulfate to the solution.\n\nI hope that helped, because otherwise the meaning of your question wasn't very clear to me.
379839
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)
481578
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
8720
Well first figure out how many moles of NaOH you have and then figure out how many moles of Sulfuric necessary for neutralization.  Since you are assuming that this reaction is complete and ideal as well as complete disassociation of the sulfuric (this is not really true only the K1 for sulfuric is very large)...ignoring all of this as well as activities of solutions you can go ahead and solve this.  \n\nStart here:  \nYou have .02630*.3=.00789 mol NaOH\n\nThen take your reaction equation (assuming properyly balanced) and use the ratio to determine the number of moles necessary to neutralize the NaOH so this ratio: \n1 Mol H2SO4 / 2 Mol NaOH \n\nso .00789 * 1/2=0.003945 mol H2SO4\n\nNow you now how many moles used and given this information and the volume used you can get mol/volume\n\nso 0.003945mol H2SO4/.02696L =0.146 Molar Solution\n\n(I calculated these numbers quickly...there shouldn't be a mistake....however if there is the method of solving this type of problem is always the same)
257537
I think you have a notational error: did you mean Al2(SO4)3 for aluminum sulfate?\nIf that is the case, then proceed as follows: \nAl(OH)3 + H2SO4 -> Al2(SO4)3 +H2O\n1. First, balance the Aluminum atoms:\n\n2Al(OH)3 + H2SO4 -> Al2(SO4)3+H2O (2 Al)\n2. Next, balance the Sulfur:\n2AL(OH)3 + 3H2SO4 -> Al2(SO4)3+H2O (3 S)\n3. Next, balance Oxygen:\n2AL(OH)3 + 3H2SO4 -> Al2(SO4)3+6H2O (18 O)\n4. Hydrogen is already balanced (12 H)\nGenerally speaking, the strategy is to balance the least represented element first until you move to the most represented.(here it was aluminum, followed by sulfur). Hydrogen is usually the last one you balance.
433279
CuSO4 + NH4S = CuS + NH4SO4\n\n1)Write a rough equation without balancing anything. The 2 after copper, which I presume is to be written as copper (II), indicates that copper has a charge of +2. It is a transition metal and can have a differing charge.\n\n2)Because ammonium has a charge of +1 and the sulfate has a charge of -2, there needs to be 2 ammoniums. SAme thing is for the left side. The ammonium is +1 and the sulfide is -2. There needs to be two ammoniums on the left side. \nCuSO4 + 2NH4S = CuS + 2NH4SO4\n\n3) Because there are two sulfides on the left side, there needs to be two sulfides on the right side. \n\nCuSO4 + 2NH4S = 2CuS + 2NH4SO4\n\n4) Now there are two sulfates on the right side, but not on the left. So we place a two in front of copper (II) sulfate.\n\n2CuSO4 + 2NH4S = 2CuS + 2NH4SO4\n\n5) I believe that we are done with balancing so let's check our work.\n\n6) 2 coppers are on the left and right side.\n2 sulfates are on the left and right side.\n2 ammoniums are on the left and right side. \n2 sulfides are on the left and right side.\n\n7) In most cases, not all compounds will have the same number on each side. It's just a coincidence that there are all 2 of each. If you need anymore help, email me ^_^
196659
Magnesium Sulfate\n\nMgSO4\n\nMg(OH)2 + H2SO4 -> MgSO4 + 2HOH\n\nHOH = H20\nJust in case...
548556
Fe(NH4)2(SO4)2 .6H2O is highly insoluble in non polar organic solvents like chloroform, dichloromethane, etc. Might be slightly soluble in alcohols. Better you dissolve it in minimum amount of water (make highly saturated solution) and add drop by drop alcohol  in a period of time and keep aside without shaking. Crystals will grow slowly.\nIt might be helpful.
616474
Everything is here at this link:\nhttp://genchem.chem.wisc.edu/sstutorial/Text8/Tx84/tx84.html
526510
You need to make sure that there is the same number and kind of atoms on each side of the arrow.\n\nCa(OH)2 + MgCl2 ---> Mg(OH)2 + CaCl2, both Mg and Ca stay+2\n\nBaO2(s) + H2SO4 ---> BaSO4(s) +H2O2 \n\nAg2S --->2 Ag+  +  S-2\n\nHC2H3O2 + MgCO3 (s) ---> H2O + CO2 + Mg+C2H302-\n\nI may not have written net ionic form, but I did get all the atoms on each side I am unsure of the reaction producing hydrogen peroxide, but you do have to account for all atoms.
537551
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
488217
"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.
440498
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
188775
It is prepared by sulphation of 1-dodecanol (lauryl alcohol, CH3(CH2)10CH2OH) followed by neutralisation with sodium carbonate.\n\nIt is an ionic surfactant that is used in household products such as toothpastes, shampoos, shaving foams and bubble baths for its thickening effect and its ability to create a lather.