I need you help for my chemistry homework?
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Look at this link and you can learn about converting at the end of site you can enter your number and check your answer http://www.csgnetwork.com/tempconv.html\n, but try anyway to do your homework by yourself, at the exam you will not have the yahoo answers to answer for you!
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This site should help you.\n\nhttp://sparknotes.com/chemistry\n\nGood luck.
165728
do your own homework this is not Yahoo homework hotline
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DO YOUR OWN HOMEWORK!
980626
ChemDraw from CambridgeSoft Corporation may be help you. Look at www.cambridgesoft.com
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It doesn't give me room to list them all and explain them on here so let me direct you to a website I found that does a pretty good job:\nhttp://vergil.chemistry.gatech.edu/notes/quantrev/node20.html
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2. 2 Na + 2 H2O --> H2 + 2 NaOH\n3. 2 HNO3 + Ca(OH)2 --> Ca(NO3)2 + H2O\n4. C5H12 +  8 O2 --> 6 H2O + 5 CO2
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DO YOUR OWN HOMEWORK. I'll expect you to show YOUR work when you hand it in.
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Send me an IM at jammin246 on yahoo I'll walk you through any chemistry problems you might be facing.
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(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
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www.google.com\nsearch " Chemisty Help "\n\nwow we got a smart person here.
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I agree with birchardv here...........we need the rest of the problem, and you would learn a lot more if you did the homework yourself.  But do whatever floats your boat.  Add the rest of the problem and I'll get back to you.\n\nOkay Chemist, this problem is really long; if you want a good answer with plenty of steps, I'll need more room.  Email me and I'll send you the solution.
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I have a question for you: WHAT THE HECK DOES THIS HAVE TO DO WITH CHEMISTRY?
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I am not a chemistry student but I make this research for you and maybe it helps you http://www.chem.latech.edu/~deddy/chem104/104Antacid.htm\nhttp://www.chemcool.com/regents/acidsbasessalts/aim3.htm\n\njust copy those links and you should be on your way to solve ur problem.
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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)
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1) this one is already balanced for you!!\n\n2)Count the number of Mg and O on the right side, and count the number of Mg and O on the right side.  On the left, you have 2 O and 1 Mg, and on the right you have 1 Mg and 1 O.  To make the oxygen on both sides equal, you need to double the MgO.  Now on the left, you have 2 O and 1 Mg, and on the right, you have 2Mg and 2 O.  Now you need to multiply the Mg by two so the Mg on both sides is equal.  and you're done!!!! \n2Mg + 02 --> 2MgO\n\n3) On the left, you have 1 Zn, 1H, and 1 Cl.  on the right you have 1 Zn, 2Cl and 2H.  the Zn are balanced and you just need to multply the HCl on the left to balance the whole equ.\nZn + 2HCl --->ZnCl2 + H2\n\n4) On the left, you have 1 P and 2O, and on the left you have 4 P and 10 O. If you multiply the P on the right by 4, and the O on the right by 10, you have 4P and 10 O on both sides\n4P + 5O2 ---->P4O10\n\n5)  On the left, you have 1 Fe 2 Cl, and pn the left, you have 1 Fe and 3 Cl.  The easiest way to balance this is multiply the Cl on the right by the amount of the left one and vice versa. when you do that, you have 1 Fe and 6Cl on the right, and you have 2Fe and 6Cl on the left. To finish the equation, multiply the Fe by two.\n2Fe + 3Cl2---->2FeCl3\n\nVoila!  You have balanced 5 equations!\n\ni hope this was helpful!
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Look up in Hand book of chemistry and mathematics, you school library will have it. Your professor should also have it.