chemical equation involved in the process of artificial rainfall process?
545618
There's no chemical action. The AgI particles act as nucleation centers around which the moisture aggregates and thus falls as rain. It is more of Surface Chemistry.
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
294143
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
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)
775000
2 MnO4-(aq ) + 5 H2C2O4(aq ) + 6 H3O+(aq ) --> 2Mn2+(aq ) + 10 CO2(aq ) + 14 H2O
340908
precipitation ,condensation & evaporation
919726
Sodium and Potassium are correct.  They are extremely reactive with water, forming Hydroxides and releasing Hydrogen gas, which is usually ignited from the exothermic heat of the reaction.\n\n2 Na + 2 H2O -> 2 NaOH + H2\n\n2 K + 2H2O -> 2 KOH + H2
750811
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)
777672
S2O3(2-)-->SO4(2-)          balance S first\nS2O3(2-)-->2SO4(2-)         balance O with water\n5H2O+S2O3(2-)-->2SO4(2-)        balance H with H+\n5H2O+S2O3(2-)-->10H++2SO4(2-)       balance charge with e-\n\n5H2O+S2O3(2-)-->10H++2SO4(2-)\n                                        +8e-\nthe 8e- is with the product side
744573
You added 0.04 liter * 0.07891 mol/liter = 3.156*10^-3 moles of Ag as AgNO3 to the dissolved unknown sample.  \n\nHowever, there was excess Ag added.  The "extra" Ag amounted to:\n\n0.01127 liter * 0.1 mol/liter = 1.127*10^-3 moles Ag\n\n\nSo the net amount of Ag that reacted with the AsO4(3-) ion was \n\n3.156*10^-3 - 1.127*10^-3 = 2.029*10^-3 moles Ag.\n\nIt takes three moles of Ag for every one mole of As to make Ag3AsO4, so the precipitate consisted of \n\n2.029*10^-3 moles/3 = 6.765*10^-4 moles Ag3AsO4, which, of course, contains 6.765*10^-4 moles of As.\n\nThe molecular mass of As2O3 is 197.8414 gm/mol, but each formula unit of this compound contains 2 As atoms, while the Ag3AsO4 precipitate only contains 1 As atom per formula unit.  If we rewrite the formula of As2O3 as AsO1.5, it would have a molecular mass of 197.8414/2 gm/mol = 98.921 gm/mol.\n\nThe precipitate contains 6.765*10^-4 moles of As, and this amount of As as an oxide would have a mass of:\n\n6.765*10^-4 moles * 98.921 gm/mol =  gm As2O3\n\nIf the initial sample had a mass of 1.223 grams, then this mass of As2O3 constituted a mass fraction of:\n\n6.692*10^-2/1.223 = 5.472*10^-2 = 5.472 wt %
1172866
this can be a problem to you if you don't know anything about simultaneous equation or if u don't know how to form simultaneous equation. another thing is u should at  least have an  idea about chemistry. look\n  \nm1v1+m2v2=MV.............1\nm1+m2=total volume after dilution=150Lt..............2\n now\n              take m1 is the initial conc of saline soln=32%=0.32\n                   ''  v1                      volume                 =?\n                 m2 is the conc of water =0\n                 v2 is the volume of water =?\n               M is the final conc of the soln = 20%=0.20\n                V is the final volume of the soln obtained after mixing=150Lt\n\nnow\n             0.32v1+0*v2=0.2*150......................1\n\n             v1+v2=150.................2\nfrom eqn 1\n                      0.32v1=30\n           so v1= 93.75Lt's\n   so the remaining amount is of water\n                v2=150-93.75=56.25lts\nso you need 56.25litres of water and 93.75litres of 32%saline to prepare the solution you need.\n\nif you have these solution just try to mix at that proportion then u will tell me.
138848
Here is one example ,\n\nAgNO3(aq) + NaCl(aq)-> AgCl(s)white precipitate + NaNO3(aq)
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
639379
The boiling point elevation constant of water is required\n\nThe formula is T = Kfm where m is molality and T is rise in temp.
812302
You might just use the formula: \nEntropy of the reaction  = \n(sum of the entropy of products) - (sum of the entropy of reactants)\nIt's as simple as that :)\nThanks.