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| I need help | |
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| Topic Started: May 30 2009, 08:31 PM (87 Views) | |
| Brian | May 30 2009, 08:31 PM Post #1 |
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Homosexual Ogre
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I'm trying to study for an acceptance exam, and I've come across a couple of science problems that stump the hell out of me. Well here goes. Balance These Equations: (a) NaClO = NaCl + O ----------3--------------2 (b) MnCl + Al = Mn + AlCl ----------2------------------3 (c) C H + O = H O + CO -----4-10--2-----2-------2 In case that was too confusing, I'll try again. (a) NaClO (divided by 3) = NaCl + O (divided by 2) (b) MnCl (Divided by 2) + Al = Mn + AlCl (divided by 3) (c) C (divided by 4) multiplied by H (divided by 10) + O (divided by 2) = H (divided by 2) multiplied by O + CO (divided by 2) Alright, I'm already confused just trying to explain this, and I doubt I even explained it right. Help me? Edited by Brian, May 30 2009, 08:40 PM.
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| Rebel X | May 31 2009, 01:03 AM Post #2 |
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Alright, from what I understand your trying to balance the atoms of the Chemical For the First One, "(a) NaClO = NaCl + O ----------3--------------2" means 1 atom of NaCl, 3 atoms of Oxygen equals 1 atom of NaCl and 1 atom of Oxygen. Which should look like : NaClO3 = NaCl + O2 in order to balance the atoms on both sides of the equation first you balance the Oxygens, so on the both sides you multiply by 2. This gives you NaCl2O6 = 2NaCL+ O2. The leaves you NaCl balanced and you Oxygen unbalanced, so then you multiply by 3 to balance the Oxygen. Your answer being, 2NaClO3 = 2NaCl + 3O2 The Second, (b) MnCl + Al = Mn + AlCl ----------2------------------3" Looks like: MnCl2 + Al = Mn + AlCl3 You answer being, 3MnCl2 + 2Al = 3Mn + 2AlCl3 As for the last one, it's a bit complicated. I don't think I can do it, but I can try and give you an indefinite answer later |
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| Trickster | May 31 2009, 01:23 AM Post #3 |
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Wow, i remember doing these in school, and they were somewhat easy, but it has been many years, i remember a little bit, but not a lot. |
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| Kotetsu | May 31 2009, 01:46 AM Post #4 |
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(c) C H + O = H O + CO -----4-10--2-----2-------2 -------------------------------- 2C(4)H(10) + 13O(2) = 10H(2)O + 8CO(2) The easiest way to solve this problem is by starting at a multiple of 1 at the CH compound. Work your way up to about 4. You'll notice that both 2 and 4 work when you multiply everything out by balancing it. 4 is not correct because it can be reduced. When doing these problems, starting at the compound with the largest amount of molecules is the best method. |
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| Brian | May 31 2009, 01:54 AM Post #5 |
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Alright, these seem pretty accurate from what I can see. I could barely remember them either, Josh. Thanks a lot Dan, Kotetsu. |
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| Kotetsu | May 31 2009, 02:12 AM Post #6 |
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No problem. I'm terrible at chemistry unless it's the math portions. PV=nRT / Ideal Gas laws are my favorite questions. |
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