when balancing a chemical equation, why can you change the coefficients but not the subscripts?
487220
Changing the subscripts will all together alterate the chemical.  It will be a completely different chemical.  The coefficients only define how much of that chemical is being used, but changing it won't change what type of chemical it is.
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.
342344
The answer is C.\n\nIn order for the equation to be balanced, you have to make sure there are the same number of each element on both sides of the equation.\n\nIn answer C, the number of each element is equal on both sides. \n\nIn answer A, the number of Sodium (Na) and Hydrogen (H) atoms are not equal on both sides of the equation.\n\nIn answer B, the number of Nitrogen (N) and Hydrogen atoms are not equal on both sides of the equation.\n\nIn answer D, the number of Sodium atoms is equal, but there is one more Sulfur (S) atom on the product (right) side of the equation.
138065
The equilibrium constant of any equation is [C]^c[D]^d/[A]^a[B]^b.\n\nthe equation is already balanced. But you have to know how many moles of each element there are in order to compute the problem
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
976256
BALANCED . . . \n\n2Sr + O2 ---> 2SrO\n\n. . . and solved . . .\n\nGet an "A".
77087
single replacement, for example: CO2+N3=CN3+O2
320739
Based on what you have given us, there are no coefficients in front of any of the compounds/elements, so the ratio would be 1:1:1.\n\nI'm guessing that you didn't provide all the information need, though.\n\nGenerally these questions give you a reaction that you must first balance to determine the mole ratios.
307574
No, it's Ag2S. The 2 is a subscript, denoting 2 Silver atoms to 1 Sulphur atom. Silver has a valency of 1+, while Sulpur has a valency of 2-, hence 2 Silver atoms are needed to balance with 1 Sulphur atom to form a compound. Besides, the equation would make sense then. 2 Silver atoms and 1 Sulphur atom on both sides of the equation, the equation is now balanced. 10 marks please, thank you!
933142
Yep, all you have to do is to make sure that the atoms shown in the left hand part of the equation appear on the left hand side.  Nothing is created or lost, you just make new compounds.  Dead easy, if only languages were so easy!!!
290716
Unless a equation is properly balanced the amount of product and reactant will not be the same, I suppose this is like a recipe where u get the proper number of ingredients(moles) and mix to together to give the you the proper number of products.
712534
used to change colligative properties of solutions
758038
omg i just had this test today too!!!!!\nu use hess's law to find the change--->write the equation, balance it \nand then use the moles (coefficients) to multiply the change is s for each element.\nhess's law says that the change in s (or h, or G, its usable for any of the thermo concepts) is equal to the sum of the entropy of the products minus the sum of the entropy of the reactants. its pretty easy after that--just plug in the values and there you go! \ngood luck! this was a hard test for me..so confusing.
83087
It depends.  Are you looking at reactions in series (where the products of the first become the reactants of the second), or reactions in parallel (where the products of each are all different from the reactants of each)?\n\nJust take all the reactants of all the equations and put them on the LHS (left-hand side) of the reaction, and all the products of all the equations and put them on the RHS of the equation.  If the reactions are in parallel, then there should be no species on both sides of the equation, so you're done.  If the reactions are in series, then you want to try to eliminate terms on both sides.  Usually, this means you have to pick a single thing to eliminate.\n\nLet's take the following theoretical reaction:\n\nA + 2B --> 2C + 2D\n\n3C + 4D --> 3F + G\n\nIn this case, you can eliminate either C or D, BUT NOT BOTH.  So you pick which one you want.  If you want to get rid of C, you look at the coefficients in front of C (2 and 3) and then identify the least common multiple (LCM) of those two numbers (6).  Now you want to multiply both sides of each equation by whatever number will make the coefficient 6 for C.  That means you mulitiply the first equation by 3 and the second by 2:\n\nA + 2B --> 2C + 2D\nx3\n____________________\n3A + 6B --> 6C + 6D\n\n3C + 4D --> 3F + G\nx2\n___________________\n6C + 8D --> 6F + 2G\n\nAdd these two resultant equations together as before:\n\n3A + 6B + 6C + 8D --> 6C + 6D + 6F + 2G\n\nBecause 6C is on both sides, we cancel it from the equation.\n\n3A + 6B + 8D --> 6D + 6F + 2G\n\nSince D is on both sides, we can subtract the lesser number from each side, but we can't eliminate it entirely from both sides:\n\n3A + 6B + 2D --> 6F + 2G\n\nAnd that's you're final answer.\n\nIf we'd wanted to eliminate D instead of C, we'd find the LCM of the coeffiecients of D (2 and 4, so the LCM is 4) and then multiply each equation by whatever is needed to make both coefficients 4.  Since the second equation already has 4D, we leave it alone and instead multiply the first equation by 2:\n\nA + 2B --> 2C + 2D\nx2\n__________________\n2A + 4B --> 4C + 4D\n\n3C + 4D --> 3F + G\n\nNow add those two and get:\n\n2A + 4B + 3C + 4D --> 4C + 4D + 3F + G\n\nCancel the 4D from both sides:\n\n2A + 4B + 3C --> 4C + 3F + G\n\nSubtract 3C from each side, and your final equation is\n\n2A + 4B --> C + 3F + G\n\nEither of these equations technically correct.  If you have only one species that is on both sides of parallel reactions, you don't have to make a decision.  In this case, if you're looking at series reactions that use a mixture of A and B (without and C or D in the initial mixture), we'd probably choose the second approach, which gives you a final equation that has only A and B on the RHS of the equation.
876161
You're right - as written it's impossible to balance.  But if you write\n\na(Ag2CO3) --> b(Ag2O) + c(CO2) + d(O2) you can balance it with the values \na=1, b=1, c=1, d=0.    Meaning that the reaction doesn't generate any oxygen.  The carbonate falls apart into the oxide and gaseous CO2.
540818
NH4Cl (note the 4 is a subscript)
970544
Here is what you do: \nDo the same operations as I did on the equations with the H values and you will be OK (Do not forget to use the same multipliers on the H as I did on the reactions).\n\n10PCl3 + 5O2 ---> 10Cl3PO\n10PCl2 + 10Cl2 --- 10PCl5\nSubstract the equations to get (keep coeff positives by moving to the other side):\n\n5O2 + 10PCl5 ---> 10Cl3PO + 10Cl2\nP4 + 5O2 ---> P4O10\nSubstract and get\n10PCl5 + P4O10 ---> P4 + 10Cl3PO + 10Cl2\nP4 + 6Cl2 ---> 4PCl3\n\nSum the two to get (P4 goes):\n\n10PCl5 + P4O10 ---> 10Cl3PO + 4Cl2 + 4PCl3\n\n4 PCl3 + 4Cl2 ---> 4PCl5\n\nSum the two equations\n\n10PCl5 + 6P4O10 ---> 10 Cl3PO.