what is the fehling reaction? What is the chemical formula for this reaction?
358679
Fehling's solution is a solution used to deferentiate between aldehyde or ketone functional groups. The substance to be tested is heated together with Fehling's solution; a red precipitate indicates the presence of an aldehydes. Ketones do not react. An example for its use is to screen for glucose in urine, thus detecting diabetes. It was developed by German chemist Hermann von Fehling.\n\nFehling's solution is always made just prior to the test. Two equal volumes of solutions are mixed:\n\n   1. 69.28 grams copper(II) sulfate pentahydrate and fill to 1 litre of solution with distilled water\n   2. 346 grams Rochelle salt (potassium sodium tartrate tetrahydrate) and 120 grams sodium hydroxide and fill to 1 litre of solution with distilled water\n\nThe cupric ion is complexed with the tartrate ion. Contact with an aldehyde group reduces it to a cuprous ion, which then precipitates as red Cu2O (copper(I) oxide).\n\nFehling's solution is nowadays often replaced by Benedict's solution.
374219
For the cell reaction: FeO(s) + H2(g) = Fe(s) + H2O(l) \n\nDG0(298 K) = DGf0 Fe(s) + DGf0 H2O(l) – DGf0 FeO(s) – DGf0 H2(g)\n\nFrom any table of Thermodynamic Properties:\nDGf0 Fe(s) = 0 KJ/mol\nDGf0 H2O(l) = -237.192 KJ/mol\nDGf0 FeO(s) = -251.4 KJ/mol\nDGf0 H2(g) = 0 KJ/mol\n\nThen: \nDG0(298 K) = 14.204 KJ/mol\n\nFurthermore: E0 = -DG0/(nF) = -14204/(2 x 96500) = -0.074 V vs NHE (normal hydrogen electrode), where n: number of electrons in each half reaction (FeO/Fe and 2H+/H2, n=2) and F is the Faraday constant (96500 C/mol).
555396
try this\n\nhttp://antoine.frostburg.edu/chem/senese/101/inorganic/faq/iron-reactions.shtml
918374
Fe2O3 -- Fe+O2\n\nthis is the balanced form\n\n2(Fe2O3)--- 4Fe+3O2\nwithout the parentheses
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
806654
F2 + HCl ---> HF + Cl2\n\nThe oxidation numbers are:\nF2 :  0\nHCl: H: +1, Cl: -1\n\nHF: H: +1, F: -1\nCl2 : 0\n\nOxidation is loss of electrons.\nReduction is gain of electrons.\nF2 ---> F(-) gains electrons so it is reduced.  If it is reduced then it is the oxidation agent.\nCl(-1) ---> Cl2 loses electrons so it is oxidized.  If it is oxidized then it is the reducing agent.\n\nTo balance write the two half reactions and balance the atoms undergoing redox:\n\nF2 ---> 2F(-)\n2Cl(-1) ---> Cl2\n\nAdd the electrons to balance the charge:\n\nF2 + 2e ---> 2F(-)\n2Cl(-1) ---> Cl2 + 2e-\n\nSum the two equations:\n\nF2 + 2Cl(-) ---> 2F(-) + Cl2\n\nAdd 2 hydrogens to both sides\n\nF2 + 2HCl ---> 2HF + Cl2\n\nand that's it.
887797
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.
946990
FeCl3 + NaCl -> FeCl2 + NaCl3\n\n\nFeCl3 + NaCl -> FeCl2 + NaCl3\n\nthese reactions are not possible \nas this reaction can take place if iron(fe) was more reactive than sodium or fe was less electropositive than na\nhence \nthese reactions are not possible
860351
The right chemical reaction is as follow:\n      6CO2+12H2O+energy=C6H12O6+6O2+6H2O\nPlz do not get confused with this reaction\n     6CO2 + 6H2O + light (energy) --> C6H12O6 + 6O2\nBecause 1st one is latest one and you can't get in some conventional books.Later one is wrong.
947778
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.
777428
well actually me and you are in the same situation. im trying to find out how to write wrod equations into a reaction. lol but neway i just learned how to balance equations its very simple.\nFe2O3+CO--Fe+CO2\n:draw a box around each set of elements:Fe2O3 would b in a box CO would be in a box and so forth.\n:then write out a diagram of how many atoms each element has:\nEx: Element/ Before/After\n      Fe          2         1\n      O            4         1\n      C            1          1\n:Then Write numbers in front of each of the boxes until the inventory for each element is the same both before and after the reaction. Now, what happens when we put a number in front of a formula? Basically, anything in that box is multiplied by that number, because we're saying that we have that many of that kind of molecule. \n:Now do this until the diagram is equal on both sides\nAND YOU HAVE YOUR BALANCED EQUATION
1049588
I think that you are trying to find the net ionic equation. \nWhich in this case would be \nAg+ (aq) + Br- arrow AgBr (solid)\nif you are trying to find a single replacement reaction?\nthen you need to put Br- instead of NO3- and cancel out the NO3 - on the product side? \nTried my Best.
834501
First let's find how many moles of each one we have\n\nmole= M*V, so for \nFeCl3 =0.234*25*10^-3= 5.85*10^-3\nNaOH =0.453*42.5*10^-3 =19.25*10^-3\n\nFeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl\n\nso\n\n1mole FeCl3 reacts with 3mole NaOH\n5.85*10^-3 mole react with x\n\nThus x =3*5.85*10^-3=17.55*10^-3 mole NaOH but we have 19.25*10^-3 mole NaOH, therefore NaOH is in excess\nTherefore the amount of Fe(OH)3 will be calculated according to the limiting reageant FeCl3.\n\nso from the reaction\n\n1 mole FeCl3 gives 1 mole Fe(OH)3\nor better yet, since the molecular weight of Fe(OH)3 is 106.85 \n1 mole FeCl3 gives 106.85 g Fe(OH)3\n5.85*10^-3 mole give x g\nx= 106.85*5.85*10^-3= 0.625 g\n\nWe calculated that 17.55*10^-3 mole NaOH reacted.\nSo we have remaining (19.25-17.55) *10^-3= 1.7*10^-3 mole NaOH\n\nThese are in a total volume of 25.0+42.5=67.5 ml= 67.5*10^-3L.\n\nSo the concentration of the excess NaOH is\nC=mole/V= 1.7*10^-3/67.5*10^-3= 0.025 M
696705
So the half reaction thing is for redox equations (which I assume is given).  You basically put the like things together and start to balance.\n\nAu+ + Cl-  >  AuCl4                  NO3-  >  NO2\nBalance everything except O and H\nAu+ + 4Cl-  >  AuCl4                NO3-  >  NO2\nBalance O with H20\nAu+ + 4Cl-  >  AuCl4                NO3-  >  NO2 + H2O\nBalance H with H+\nAu+ + 4Cl-  >  AuCl4                2H+ + NO3-  >  NO2 + H2O\nBalance charges with e-\nAu+ + 4Cl-  >  AuCl4 + 3e-         e- + 2H+ + NO3-  >  NO2 + H2O\nMake the e- equal\nAu+ + 4Cl-  >  AuCl4 + 3e-/    3(e- + 2H+ + NO3-  >  NO2 + H2O)\nAu+ + 4Cl-  >  AuCl4 + 3e-/ 3e- + 6H+ + 3NO3-  >  3NO2 + 3H2O\nCancel and add\nAu+ + 4Cl- + 6H+ + 3NO3-  >  AuCl4 + 3NO2 + 3H2O\nMake sure everything balances (charges too!)\n\nOxidizing agent is the substanced being reduced (or gaining electrons) and the reducing agent is being oxidized (or losing electron)  For Au, the charge is going from a +1 to a +4, which means its losing electrongs, so the reducing agent is Au+.  The N in NO3- is going from +5 to +4, which means its gaining electrons, so NO3- is the oxidizing agent.
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
84780
!!!!!!!!!!!!!!!! \nI think rather u wanted to ask how can u predict the formula of a chemical reaction!!\nFirstly, we have to determine the moles of reactant taking part.(It can be done easily, think A reacts with B, just mix 1 mole of A with the same amount of B.)\nThen determine what are the different products and their amounts.\nAll u have to do is some calculations..................\nRemember its not as easy as it sounds.
436191
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.