Why can’t you change symbols, formulas, or subscripts in order to balance a chemical equation?
110265
The equation is meant to record the results of an experiment.  The reactants (stuff you started with) are written to the left and the products (the stuff you end up with) is written to the right.\nTo balance an equation, only the coefficients (the numbers in front of the chemicals) may be changed.\nIf you change the symbols, you change the type of atoms used.\nIf you change the formulas or subscripts, you change the chemicals (the molecules) used in the experiment.
112894
Haven't had chem in a while but maybe it would help to write out all the individual symbols.  Your second example would look like\n\nHHHPOOOO + MgOOHH  = MgMgMgPPOOOOOOOO + HHO\n\nthen you can see that the left side needs three Mg so you'd need 3 of the Mg(OH)2 to balance.  The left side also needs two P so there needs to be two H3PO4 there.  This makes a total of 12 H on the left, so to balance that you need 6 H2O on the right.  This also gives 14 O on the right to balance the 14 O you'd now have on the left.  So\n2 H3PO4 + 3 Mg(OH)2 = Mg3(PO4)2 + 6 H2O
694411
IAnd for your other request, I would request you to send a seperate mail to me through yahoo Answer so that I am able to send it part by part.These are long subjects and would take some time,The balanced equation would be                                           H3PO4+3KOH=K3PO4+3H2O
724411
Sure...\n\nfirst we want to combine like terms on each side of the equals sign.  Think of the equals like the balance point on a scale... you can do anything on one side as long as you match it (keep it balanced) on the other!\n\nso we want to combine the variables and the constants...\n\n8a - 3a + 4 = -a -4 + 2\n5a + 4 = -a - 2\n\nnow we want all the variables on one side, and constants on the other:\n\n5a -(-a) + 4 - 4 = -a-(-a) - 2 - 4\n5a + a = -2 - 4\n6a = -6\n\nnow we just divide by 6 to get a alone:\n\n6a/6 = -6/6\na = -1\n\nHope this helps! :)
1117062
if ur just solving that equation...u dont need to switch the sign.....\nwatch (im not gonna have the equal sign on the bottom takes too long). \n\nx-1<-3\n(add one to both sides now)\nx<-2\n\nthers ur answer
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
53245
you must apply oxidation numbers to the equation
40999
I am guessing you are asking about subscripts???? Subscripts is where the letter is smaller and aligned with the bottom of the other letters (such as h2O fo water but the 2 is lower). IF that is what you are talking about: Highlight the word or letters you are wanting to change, select format>fonts from the standards tool bar. In the font options put a check mark in subscript.\n\nIf this is not what you are looking for let me know I will reply again if I know the answer.
376319
Are you able to use any other numbers and operators?  If so this would be fairly easy.  Using only Three 8's and a subtraction symbol it is not possibe\n\nClosest example (8+8)/8-1=1
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
158900
Yup. They are not simple...\nChemistry...\nhttp://www.chm.davidson.edu/ChemistryApplets/equilibria/EquilibriumConstant.html\nhttp://en.wikipedia.org/wiki/Chemical_equilibrium\nThey use the equilibrium constant expression (Kc) here...\nhttp://www.cofc.edu/genchemlab/equilibrium.htm
39364
To understand balancing chemical equations better, it helps to know what it is called. It is called "stoichiometry". Just a quick web search and I found a TON of very accessible web sites on balancing chemical equations. I put a couple of them below. \n\nIt should be noted that in mathematics, there is a very easy way to solve stoichiometry equations, or do chemical balancing. It is much easier than what chemists are usually taught. To look at the mathematical version of the same problem, look for web sites and books on linear algebra. The first linear algebra web site listed below is a simple introduction, but I also included a few links to free books in linear algebra which is probably more detail than you want.\n\nHere is an easy way to start thinking about it, however. It is all about counting how many atoms of each type are on the left side of the equation, and how many are on the right side. You want to adjust the numbers in front of the molecules so that there are as many atoms of each type on the right side as there is on the left side. \n\nSuppose you are considering the burning of carbon C in oxygen O2 to make carbon monoxide, CO. \n\nThere are j C atoms and k O2 molecules that combine to make m CO molecules \n\nSo jC + k O2 ----> m CO\n\nIt is up to you to pick j, k and m so that \n\nthere are the same number of C and O atoms on both sides.\n\nIf m=2, and k=1 and j=2, then \n\n2C+O2----->2CO \n\nand you see there are 2 C atoms on each side, and 2 O atoms on each side, and you are done.\n\nOf course it gets much more complicated than this. That is addressed in the references below.
1076966
pardon me, but why is this in chemistry?
301956
The answer is B.\n\nIn your beginning reaction CO2 + C --> CO, you have 2 carbons and 2 oxygens on the reactants side, and you have 1 carbon and 1 oxygen on the products side.\n\nIn order to make the equation balanced, you have to make sure there are the same number of each element on both sides of the equation.\n\nSo to make this chemical equation balanced all you have to do is put 2 as the coefficient of the CO on the products side, making the final equation:  CO2 + C --> 2CO\nThis will give you 2 carbons and 2 oxygens on both sides.
1023121
If the formula looks correct (including the "=" at the front like the earlier suggestion) go to Tools | Option | View tab | make sure "Formulas" is not checked.\n\nIf that does not work, double-click on the formula, then click the checkmark in the formula bar.
290031
Balancing chemical equations is easy, but it's a bit of a trial and error process. Basically, you've got the the arrow in the middle, so the number of each element on one side of the arrow has to equal the number of that particluar element on the other side of the arrow. So start off by balancing any unique elements that only appear once on a particular side. Then balance your nitrogens, oxygens and carbons in that order. The last thing that should be balanced is your hydrogens. I can go into further detail if you need me to, I just don't know how much you know. \n\nFor your other question and yield is how much of something you get from a reaction. For instance, if I mix 100 moles of one substance and 100 moles of another substance, I can calculate a theoretical and experimental yield. For instance, I can say that I expect all 200 moles to react and form a single product. Since i expect it all to react, my THEORETICAL yield is 200 moles or 100%. Say I actually run the experiment and I discover that only 100 moles of my intended product formed, then my EXPERIMENTAL yield is 100 moles or 50%.
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD