I cannot find the balanced chemical equation for this situation... I don't even know where to start! \n\n"Dissolve 10 g of CuSO4 into 100 mL of H2O in a beaker. Place a piece of Aluminum Foil into solution and observe. After 3 minutes, place 1 g of NaCl into the beaker and observe. Define the word, catalyst (I've already done that part), and relate this word to the reaction you observed. Write and balanced equation and identify the type of reaction that takes place."\n\nNow I'm assuming the salt (NaCl) is the catalyst in this situation, as it most likely speeds up the process, and that during the process a gas is released (maybe Oxygen) because when this experiment is done, bubbles are seen. What happens in the experiment is that the aluminum starts to dissolve and turn brown, then you just have brown specs in water. I have the beginning of the equation so far, I just can't find out what it yields:\n\nCuSO4 + H2O (water) + Al (aluminum) + NaCl (salt) ----> ???\n\nCould you help me with this?
213781
I suspect that the product is AlSO4.\n\nI'm thinking that the Al swaps with the Cu.  That would leave a free Cu+ which might displace the H from H20 and form either CuOH or CuO, either one of which would probably produce H2.  \n\nSo, I guess you would first get CuOH and AlOH, as intermediate reactions.\n\nI think the equation is:\n\n                          NaCl (goes over the arrow)\n6CuSO4 + 4Al + 12H2O ---------------> 2Al2(SO4)3 + 6Cu(OH)2 + 6H2
645855
1)\nSodium Sulfate (Na2SO4) will break into ions in solution,\nNa2SO4 (s) --H2O--> 2 Na+1 (aq) + SO4-1 (aq)\nFor every 1 mole of Na2SO4 dissolved, 3 moles of ions are formed.  If the concentration of Na2SO4 is 1.15 mol/Liter, then the ion concentration must be 3 times as much, 3.45 mol/Liter.\n\n2)\nTlCl3 is an ionic compound which when dissolved in water will break into ions since Chloride compounds are generally soluble,\nTlCl3 (s) --H2O--> Tl+3 (aq) + 3Cl- (aq)\n\nC2H5OH, otherwise known as ethanol, will easily dissolve (easily mincible) in water, but will not break into ions (as it is not an ionic compound).\n\nFe2(CO3)3 (although it is an ionic compound) in considered insoluble in water.  Carbonates in generally are considered insoluble unless they are bonded with a group 1 element or the Ammonium ion.\n\n3)\n.356 mol/Liter HCl means that for every 1 Liter of solution, there are .356 moles of HCl dissolved in it.\nMolarity  = moles / Liter --> moles  = Molarity * Liters\nSo if you originally had a 6.5 M solution with a volume of .0572 Liters, you originally had .3718 moles of HCl.\n\n.356 M = .3718 moles HCl / x Liters\nx Liters = .3718 moles / .1.044 Liters.\n\nSo if then old volume was 57.2 mL and the new volume should be 1044 mL, then 986.8 mL shoudl be added to solution.
529383
BaCl2 + ZnSO4 => BaSO4 + ZnCl2\n\nthe moles of BaCl2 in the solution 0.05 * 2.1 = 0.105\nthe moles of ZnSO4 in the solution 0.0456 * 1.92 = 0.087552\n\nbecause the moles of BaCl2 is greater than that of ZnSO4 so all the moles of ZnSO4 will react completely\n\nthe moles of BaSO4 produced if the percent yield is 100% : 0.087552\nthe grams of BaSO4 is produced with 100% of yield : 0.087552 * 233 = 20.399g\n\nthe percent yield is 16.775 / 20.399 *100% = 82.23%
466646
Well, for a start your equation shows copper(II) sulphide, not sulphate, and your equation is wrong even then.  That makes a bit of a difference!\n\nLet us imagine that you've got the formula right, but the name wrong.\n\nThe equation is simply Cu + S ---> CuS\n\nThen 64g will give (64+32)g.\n\nThus 64g will give 96g.\n\nSo 64 x (200/96)g of Cu will give 200g of product.
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)
183108
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
374006
Cu(s) + 2HNO3(aq) = Cu(NO3)2(aq) + H2(g)\n\nCu + 2H+ + 2NO3- = Cu+2 + 2NO3- + H2 (leaving off the s,aq,g)\n\nCu + 2H+ = Cu+2 + H2 (net ionic equation, the NO3-'s just cancel out, they don't change in the reaction).
370512
I have forgotten all the chemical formulas & whatnot... but I'm going to do this assuming that I'm right for the following:\n\n(note that if i get any of these wrong you'll have to make appropriate changes, this is just to show you HOW to do this)\nCu = Copper\nS = Sulfur\nCuS2 = Copper Sulfide\n\nSo first step is to balance the equation\n\nCu + 2S -> CuS2\n\nso then you can set up a problem like this:\n\n0.90 mol Cu * 1 mol CuS2  = # mol copper sulfide yielded\n(this is because there was 1 mole of Copper in 1 mole of CuS2.. so if CuS2 is not the correct formula change the "1 mol CuS2" accordinginly)\n\n\nThen add up the # of grams in 1 mol of copper sulfide\n\ndo that by taking the # of grams in 1 mol of copper + the number of grams in 2 mol of sulfur (again, this is just assuming that copper sulfide IS CuS2).  Once you've gotten that figure, multiply it by the number found from the earlier problem (# mol copper sulfide yielded).  This gives you the amount of grams.  To change this number to kilograms, divide it by 1000.
147226
Half reactions are:\nOxidation: Cu ----> Cu2+ + 2e    Eo = -0.34V\nReduction: NO3- + 4H+ + 3e --->  NO + 2H2O   Eo = +0.96V\n\nAdd the two voltages (do not multiply this number by the coefficients you use to cancel the number of electrons out) and you get the +0.62V for the Eo of the cell.\n\nThe Nernst equation will take this form:\nE = Eo - 0.0592/n * log {(PNO)^2 * [Cu2+]^3 / [NO3-]^2 * [H+]^8}\n\nn will equal 6 here, not 8, as that will be the number of electrons that will be cancelled out when you add the oxidation and reduction reactions together (multiplied by the correct number to cancel those electrons out).\n\nThe concentration's portion simplifies, once you plug all of the numbers in, to about 18.94, whose log is 1.278.  Multiply that by 0.0592 and divide by 6 and you get 0.01, so only 0.01 V will be lost from the standard potential for that reaction at those concentrations.
272537
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
334714
Okay, this is trivial stuff here.\n\nThe trick is to have enough + and - to add up to zero.  so, for example:\n\na) Cu2+ and NO3-;  you have two + and one -, thus to make a neutral compound you will need twice as much -, so the formula for the salt is: Cu(NO3)2\n\nYour problem in d) is a little trickier, but it is the same process:\n\nd) Mg2+ and PO4(3-): how can you get an equal number of + and - here?  use three Mg and two PO4!  Mg3(PO4)2\n\nI will leave the rest for you to do by yourself so you understand the process.
146620
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
130921
I think you should keep it simple yet smart. First get a piece of pure elemental Sodium and then let it drop on warm water. Afterwards explain how the chemical reaction released Sodium Oxide and Hydrogen Gas.
387093
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. :)
492041
Na2S (ag) + 2 AgNO3 (ag) --> 2 NaNO3 (ag) + Ag2S (cr)\n\nB. Double displacement.\n\nA. Ba(NO 3)2 and H2O\n\nB. If there is a double arrow (pointing in both directions), then it is an equilibrium reaction.\n\nC. Ba(OH)2 and HNO3\n\nD. Barium\n\nE. 6 oxygen atoms. \n\nF. Ba(OH)2
377286
Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?
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.