balanced equation problem?
606155
NaOH + HCl ---> NaCl + H20
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
986618
The equation is already balanced. Count the Ag's on each side = 1. Count the Na's on each side = 1. Count the Cl's on each side = 1. Count the NO3's on each side = 1. One can count the N's on each side = 1, or the O's on each side = 3.\n\nBut one could also consider the question another way:\nAg+ + NO3- + Na+ + Cl- ===> AgCl + Na+ + NO3-
729087
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!
412953
An equation that is incorrect.
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
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.
691513
Check that both sides of equation match. Here is an example:\n\n3 CH3CH2OH + 2 Cr2O7 2- + 16 H+ --> 3 CH3COOH + 4 Cr3+ + 11 H2O\n\nThere should be an equal amount of H, C, O, Cr in both sides of the equation. \nIt is sometimes difficult to know where to start balancing an equation. In such case you should try to find an element/molecule that does not appear in every molecule on the other side of the equation. Eg. when balancing this equation above, I first balance the number of Cr because it is on its own on the right side. Sometimes you just need to try and play with numbers for a while to find the right answer. Hope this helped...
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
741999
whats the equation or formula u need help with?
723776
Walk through these things slowly... First off, lets correct your first equation (?)\n\nLi + OH- + K2(2e-) <--> Li + 2e- + 2K + OH- \n\nThat doesn't make since... (is that your initial equation?)\n - Li are ballanced\n - OH- are ballanced\n - K2(2e-) goes to 2e- and 2K which is balanced\n\nWhat's the question again? I think there was a typo somewhere in your original expression. Are you going from LiOH to KOH? If so, your expression would be more like this...\n\nLiOH + K2(-2) <--> Li(+) + 2e- + 2KOH\n\nStart by balancing the OH and then the Li to get\n\n2LiOH + K2(-2) <--> 2Li(+) + 2e- + 2KOH\n\nAnd, there you go.\n - Li, K and OH are balanced
753594
It is perfectly permissible to write the products as Br2 + Br2.  It is probably easier to work this way.  After you balance the equations, just add the Br2's together to give the final reaction
196036
If you divide 2 to the second equation, you will notice that it is identical to the first one, so they are dependent to each other, and there are infinitely many solutions to this problem.\n\n^_^
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.
973095
Unless your leaving out some necessary charge, it is already balanced. On the left side if the arrow: 1 Zn, 1 S, and 1 O2. On the right side: 1 Zn, 1 S, and 1 O2. Now if there are charges or subscripts that are being left out, then it may not be balanced. The total charge on each side must be equal. Just remember that to balance, make sure that there is the same number of atoms for each element on each side of the arrow.
577084
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
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! :)