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Everything posted by battousai

  1. Accorind to the table of standard electrode potentials, it is easier to reduce a hydrogen ion than a transition metal ion, such as nickel or zinc. Yet, sources such as the Electrochemistry Encyclopedia for nickel, and here on page 9, for nickel, cobalt and zinc. So, why are the metal ions reduced as oppose to the hydrogen ion?
  2. I don't think it is a typo. After further search on google, I have found the book in which wikipedia sourcesseen here, and on page 125, it says, "Brandes finally succeded in decomposing lithia with a powerful battery...". Then from wikipedia again, you can see that lithia means lithium oxide. Still, the book has no mention on the process used... I've yet to research the sources given by this book, perhaps they will explain. In any case, the book also states, "Davy also obtained a small amount of lithium in the same manner", and since we know Davy fused his salts, it is likely that the book means Brandes also fused lithia to perform electrolysis. However, I am determined to find out what really happened.
  3. I originally looked on wikipedia for the history of lithium, and how it was first isolated. It says, "William Thomas Brande performed electrolysis on lithium oxide", but without any specifics, as in chemical equations, voltage potentials, apparatus, and if the salt was molten or not. Now, the reason I am interested, is because I *think, that William Thomas Brande isolated lithium by dissolving lithium oxide in a polar aprotic solvent; thus allowing electrolysis without fusing the lithium oxide. Anyway, back to the point, I cannot seem to find the specific's for Brande's process anywhere. On the internet, I only find the same sentence repeated on many websites, and I even checked my municipal library, but sadly, they do not have any books that seem to be relevant to my question, nor any books on William Thomas Brande himself. Does anyone know of a reliable source that does state the specifics of Brande's process?
  4. Ok, I think I see where this is going. I thought with a power supply, you could set both the voltage and current magnitudes. For example, on an AC adapter it says there's a DC output of 7.5V, 5.7A, but that's with respect to the resistance of the machine it's designed to power. In my electromagnet scenario, it would be : Stage 1. The power supply is set to 6V, and the electromagnet has a resistance of 1 ohm, therefore the current is 6A. Stage 2. The power supply is set to 30V, and thus the current is 30A. I'm sorry for my mistake. Thank you for helping me solve this.
  5. It's the same electromagnet. Would it not have the same resistance? If you need a number, let's just say it's 1 ohm.
  6. It's not homework; it's my own curiosity. What I mean by constant, is that in the example, the two different supplies provide the same current of 5A. I'll clarify my example. Stage 1. There is an electromagnet with a DC current powering it. The current is a constant 5A, and the voltage is a constant 6V. Stage 2. The electromagnet is disconnected from the power source. Stage 3. The electromagnet is connected to a second DC power supply. The supply provides a constant current of 5A, and a constant voltage of 30V. Now, will the electromagnet be stronger in Stage 3, as opposed to Stage 1. Will more heat be produced in Stage 3 than Stage 1? Is there a significant voltage drop in an electromagnet? Will this drop be significantly larger in Stage 3 than Stage 1?
  7. How does an increase in voltage with a constant DC current affect an electromagnet? Is the increased voltage just transformed into heat, or extra strength in the electromagnet? For example, at start there is an electromagnet with a current of 5A at 6V. The current is shut off, and a new supply of 5A at 30V is applied to the same electromagnet. How does the electromagnet react?
  8. I know an AC current can be rectified to a DC current with a diode bridge, but does the magnitude of the voltage and/or current change? If so, how. For example, if one had an AC current at 2 A and 12 V, and one converted it to DC, what would the current and voltage of the new DC current be? Btw, let's assume that the process is 100% efficient.
  9. dlzc, could you explain please? For the chem experts, would H2, be a strong enough reducting agent with BrO3- to produce, H2O and Br-?
  10. If I were to use aluminum or zinc, would the majority of the BrO3- be reduced, forming solid aluminum or zinc oxide, with the respective oxide remaining inert to the solution?
  11. In a solution with both BrO3- and Br-, does anyone know how to reduce the BrO3- to Br- without oxidising the pre-existing Br- to Br2? Also, only potassium ions can be introduced into the solution. My first thought is, to use hydrogen peroxide or potassium permanganate, but I believe this would oxidise the Br- to Br2. Does anyone have any ideas?
  12. Sodium iodide dissolves in acetone (see here, here and here [under "Uses"]) From there, I assume one could just electrolyze the solution, and obtain sodium metal and iodine. However, I assume iodine would coat the +ve electrode, and thus slow or even stop electrolysis. So, one would have to constantly clean the electrode.
  13. According to wikipedia the reaction of calcium oxide and water produces 63.7kJ/mol of CaO. This reaction is more exothermic than the dissolution of NaOH which produces 44.5 kJ/mol, seen here
  14. As stated in the first post, this circuit does not include a switch. In this case, how would one reverse the direction of the ac fan? A switch can be used, but what type of switch is this? Does it have a specific name?
  15. Yes, but the reaction is very exothermic (gives off heat). You can see this here. Also, I don't know if you are just inquiring or thinking of this reaction, but if you are, I HIGHLY suggest staying away from chlorine gas.
  16. Is there any way to reverse the direction of an ac fan?
  17. In a previous post, CaptainPanic stated that the addition of water could be causing your strange results. To prove/discount this possibility, I would repeat the original experiment but with a solution of NaOH with a lower and then higher H2O content, and compare the results. Comparing to the original titration, if the pH of the the solution drops when a smaller quantity of NaOH solution of a higher H2O concentration is added, and if the pH drops after a larger quantity of NaOH solution of the lower H2O concentration is added, then water is most likely the problem. In addition I would suggest retrying the original titration, but by using a base that does not need to be in aqueous solution. If you can get the temperature low enough, try using liquid ammonia, or, if you can, try Lithium diisopropylamide. According to wikipedia it will dissolve in non polar organic solvents. From wikipedia as well, it says that n-Butyllithium dissolves in diethyl ether, which is a polar organic solvent, so that might work. Finally, maybe one of the weak bases: Alanine, Methylamine, or Pyridine; mentioned here, might work in your solution. I'd highly suggest retrying the original titration without the use of any water, so if you feel that it's worth it, give it a try.
  18. Let's say one has a basic fan (hot wire, stator, rotor, neutral wire), and that when a current is applied from the hot wire to the neutral, the fan spins to the left. If one were to apply a current from the neutral wire to the hot wire, would the fan spin to the right?
  19. I got the reaction from http://www.chemthes.com/rxn.php?entity=14&page=4&type=&side=RHS . Upon further reading of the reaction (click "Reaction Data Page"), it says that this is carried out by conc. H2SO4, but then it also says that the H2SO4 oxidises the Br- to Br2. So, I assume if I use dilute H2SO4 and specific stoichiometric quantities according to the equation, it should work out. If not, I assume using NaHSO4 to produce HBr will work just as well as when it is carried out to procude HCl (by using NaCl, instead of NaBr). I would likw to produce CaBr2, becausw it can be dissolved in acetone, and thus hopefully electrolyzed to give Ca(s).
  20. I'm interested in making CaBr2 using NaBr. This is the current idea I've come up with: 2 NaBr(aq) + H2SO4(aq) ---> 2 HBr(aq) + Na2SO4(aq) From here, I would cool the solution in an ice bath to precipitate the Na2SO4 (wikipedia says 4.76g of Na2SO4 dissolves in 100ml), and remove the precipitate. So that would mean in a 1L solution, 47.6g would be dissolved, which is 0.3351 moles of Na2SO4 dissolved. From here, mainly HBr would remain in the solution. Next, I'd add Ca(OH)2: 2 HBr(aq) + Ca(OH)2 ---> CaBr2(aq). Does this method seem sound? Also, any other suggestions would be greatly appreciated.
  21. How did you perform this? Were the reagents dissolved in water or another solvent? I also tried this with distilled water, grocery store NaHCO3, and pharmacy citric acid, and it was a clear solution for me. I looked for some vegetable oil (don't have any at the moment), but I added canola oil, in case oil somehow interferes with the reaction, but nothing happened out there was no change.
  22. From my understanding of wikipedia, an ac induction motor induces a magnetic field in the rotor, and then it moves. So if one had a clear plastic can of aluminum shavings and placed the stator around the can, could observe a reaction in the aluminum shavings? Would they be rotating like the rotor would?
  23. Could you elaborate on your question? Do you mean solubility in water and/or why it hardly dissolves?
  24. Is there any way to magnetize aluminum or at least repel aluminum in any way using electromagnetism? Could eddy currents somehow accomplish this efficiently?
  25. Can one of these tables be made in a smaller version? Can one just make a small cardboard table let's say (so it can be bent to make minute angles) and attach a vibrator, lets say like an automatic razor? Does anyone think something like this is possible? Could a straining sieve work, like one that is used to strain water from boiled rice? Has anyone tried this before? Im also trying to think of a way to do this in a way to prevent the NaOH from converting to Na3CO3. Would performing the process in a small sealed room help this? From chem class we learned that a mole of gas at SATP is 24.8 L and carbon dioxide is less than 1% of the atmosphere right? So wouldn't a small sealed room only have a very limited amount of CO2. Would this work, or would an average room be too ventilated, making this all trivial? Any ideas?
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