What 3 things are conserved in a balanced chemical equation?
596362
The number of atoms/ions of an element on either side of the equation.\nThe total charge on either side of the equation.\nThe elements present are the same on either side of the equation.
814943
There is no balanced equation for that question.\n\nYou need to show your products going to reactants for the world to be able to balance the equation.\n\nGood Luck!
979796
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
900316
This is a common redox question easily balanced by the half-reaction method.   The half-reaction method is taught using a few different styles, so you'd best learn the style that your teacher or professor expects you to learn.  \n\nFirst, identify your half reactions and balance them:\n\n(Cr2O7)2- contains 2 Cr6+ ions being reduced to 2Cr3+, gaining  6 electrons:\n\n6e-  +  2Cr6+  --->  2Cr3+    (net charge now adds up to +6 on both sides)\n\nMn2+ is oxidized to Mn4+ (as MnO2) losing 2 electrons, or\n\nMn2+  ---->  MnO2  + 2e-\n\nNet electron transfer must equal zero, so we must triple the oxidizing reaction to yield 6 e- to balance the 6 e- gained, so e- gained now equals e- lost (that is, 6 e' in, 6 out)\n\nThat gives us the clue to multiply the manganese term by 3:\n\n(Cr2O7)2- + H+ + 3Mn2+ --->  2Cr3+ + 3MnO2  + H2O\n\nInspection shows the Cr, Mn, and O terms to be balanced but the left side of the equation to be down by 1 H+, so let's put a 2 by the H+....\n\n(Cr2O7)2- + 2H+ + 3Mn2+ --->  2Cr3+ + 3MnO2  + H2O\n\nWe think we're finished here, so let's make the important two final checks:\n1.  Mass Balance and 2.  Charge Balance\n\nEach side of the reaction has 2Cr, 3Mn, 7O, and 2H\n(Mass balance OK)\n\nNow add up the net charges on both sides:\n-2  +2  +6 = +6 on the left side...\n+6 + 0 + 0 = +6 on the right.  (Charge Balance OK)\n\nReaction is balanced.\n\nAgain, WARNING:  Some methods teach you to balance in ACID or in BASE.  This problem, by virtue of the H+ presence is in acid.  Balancing a redox equation in base has a couple more steps attached.  Again, consult your text or your instructor as to their preferred style.
777672
S2O3(2-)-->SO4(2-)          balance S first\nS2O3(2-)-->2SO4(2-)         balance O with water\n5H2O+S2O3(2-)-->2SO4(2-)        balance H with H+\n5H2O+S2O3(2-)-->10H++2SO4(2-)       balance charge with e-\n\n5H2O+S2O3(2-)-->10H++2SO4(2-)\n                                        +8e-\nthe 8e- is with the product side
106704
If you go to a search engine, like google, you can just type in "balancing chemical equations" and it will give you tons of info. We were doing this in chemistry a few weeks ago, and thats how I found alot of help online. \n\nGood luck!!
933519
Part A:  equilibruim to the right because more co3 added\n\nPart B:  Equilibrum to left becausemore co2 added\n\nPart C:  Equilibrium to right to replace removed h20\n\nEquilibruim dynamic because although appears stable ions constantly moving back and forth just at the same time
927756
2H2O2 → 2H20 + 02 is the balanced equation for the above decomposition
775066
You have the products, but now you have to balance the equation. The unbalanced equation is:\n\nLi2CO3 + Cr2(SO4)3 --> Li2SO4 + Cr2(CO3)3\n\nBalanced:\n\n3 Li2CO3 + Cr2(SO4)3 --> 3 Li2SO4 + Cr2(CO3)3
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.
148175
The correct balanced equation is:\n2ZnS + 3O2 --> 2ZnO + 2SO2\nThis is BALANCED\nMr chemistry teacher, you didnt balance your equation,...and i'm in the ninth grade
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%.
1029753
Within each half-reaction, the charge on each side must be balanced, as well as the number of each element.\n\nFor the overall oxidation-reduction reaction, the charge on each side must be the same, as well as the number of each element, and no free electrons should be in the equation.
749125
The unbalanced equation is:\nCH3-OH + O2 (+ heat) ---> H2O + CO2\n\nThe balanced equation would be:\n2CH3-OH + 3O2 (+ heat)  ---> 4H2O + 2CO2
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
948757
The concentrations of all the components: the reactants, the products, the catalyst.\nYou will also need the temperature.\nIf it's a gas-phase reaction, the pressure..\nIf it's a solid-liquid reaction, the surface area; or if the catalyst is a solid, the surface area of the catalyst.
539827
(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 :)