how is a balanced equation like a recipe
120806
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.
405888
I already answered your previous question :-)\n\nBTW, it is balanced because the number of atoms of each element is equla on both sides.
348952
You have to start with a balanced equation.  The subscripts will give you the correct charge and a prefix will give you the right amount of atoms on each side of the equation.  Some elements, like O, F, and I are diatomic and come in pairs.
369373
I don't have time to do them all but one will show you what you need to do.  You have to keep the first and second number in the paratheses balanced through the equation.  Work with them like a simple algebraic equation.\n\n(60,27)Co --->  (x,y)Z + (0,-1)e\n\nWhen you know y, you will have Z from the periodic table\n\n60 = x + 0\n27 = y - 1\nso y = 28, x = 60\ny = 28 is Nickel.\nSo the reaction is now: \n\n(60,27)Co --->  (60,28)Ni + (0,-1)e\n\nPiece of cake.
667148
If you do the same thing to both sides of an equation it is ok...for example:\n\n0.5x = 4\n\nok...this is pretty simple, and you can probably do it in your head, but pay attention to what you do in your head.  Likely you will multiply both sides by two:\n\n(0.5x) * 2 = 4 * 2\n\nSince 0.5 * 2 = 1, then x = 8...\n\nNow you are probably thinking "of course",  but the same is true with adding and subtracting from both sides:\n\nIf x - 10 = 60 then it is easy to do in your head....you just add 10 to both sides and get x= 70.  \n\nYour equation is just a combination of a few steps.  If x/3 +2 = 5, then subtracting 2 from both sides is an easy way to get rid of one of the numbers and simplifying the equation.
253009
Well, you have O x 2 on the left, and O x 3 on the right, so the number next to "O2" on the left has to be at least 3.\n\nThat would mean that, on the right, the number next to "PbO" has to be at least 2.\n\nAlso, since you have one "PbS" on the left, the numbers next to "PbO" and "SO2" on the right have to be the same number.\n\nIf you make both numbers on the right "2", then the number next to "PbS" on the left has to be 2.\n\nSo the complete balanced equation is:\n\n2 PbS + 3 O2 ---> 2 PbO + 2 SO2
295567
Actually, all you need is:\n\n2 H2O2 ======> 2 H2O + O2\n\nif you have water on both sides of the equation, they "cancel" out.  Just like in algebra:\n\na + b = a + c  is the same as b = c
241675
The formulas I saw by doing a web search were just mix the ingredients together.  There were no special proceses.
96681
1S = 1T + 1B\n2S +1T = 8B\n\nrewrite 1st equation as 1S-1T=1B and add 2nd equation. get\n3S=9B, so S=3B\n\n3 buttons balance 1 spool
365168
x=x is the simplest...\n\nMake up any expression, simplify it a little bit, and set them equal.
47256
So, usually the best way to do these is to just fiddle with them until you get them right, but for the sake of it, let's go with a systematic approach using algebra.  First we assign a variable in front of each of the terms on the left and right like:\n\nx Pb  +  y H2O  +  z O2  -->  a Pb(OH)2\n\nSo now we'll write down equations to make sure that we have the same numbers of both sides of each *element*.  I'll do this by writing down the variable in front of each compound in your equation (so x, y z and a) and then multiplying those by the number of times an element exists in that compound.  (So when I'm writing the equation for hydrogen, the H20 term will be 2*y, because y is the coefficient in front of H20 and H occurs twice in that compound).\n\nRemember, the number of any element on the reactants (left) side equals the number of that element on the products (right) side.\n\nFor Pb: x = a, because Pb shows up only once on each side with a subscript of 1, those two compounds will have to be present in the same molar amounts.\n\nFor Hydrogen: 2y = 2a\n\nFor Oxygen: y + 2*z = 2 * a\n\nSo now we have three equations:\n\nx = y\ny = a\ny + 2z = 2a\n\nBut four unknowns, which is fine because we'll just solve for them in terms of one of the variables and then simply choose the last variable to make all the numbers nice.  So solving the equations in terms of a:\n\nx = a\ny = a\nz = (2a - y)/2 = (2a - a)/2 = a/2\n\nSo now we have x, y and z in terms of a free parameter, a.  We'll just choose a = 2, for example, to make x, y and z all turn out as integers.  So now we've chosen:\n\nx = 2\ny = 2\nz = 1\na = 2\n\nSo the balanced equation is:\n\n2 Pb + 2 H2O + O2 --> 2 Pb(OH)2\n\nMake sense?
398021
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
513874
When you want to balance ANY chemical equation, always follow these steps:\n\n1. Balance metals\n2. Balance non-metals\n3. Balance hydrogen\n4. Balance oxygen\n\nLet's do it\n\nFirst let's balance K (metal):\n\n"2" KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + __CO2\n\ndone. now let's balance P (non metal): if you check, it is balanced already.\n\nnow C (also non metal):\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + "2" CO2\n\nnow hydrogen:\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + "2" H2O + 2 CO2\n\nfinally, oxygen (its already balanced!!)\n\nSo you end with\n\n2 KHCO3 + H3PO4 yields K2HPO4 + 2 H2O + 2 CO2
481545
Basic balancing of chemical reactions. You can always add CO2 O2 and H2O to either side. Start with the Carbon. balance the Oxygen last since it is contained in CO2 and H2O.\n\n1) 2 C6H6 + 15 O2 ==> 12 CO2 + 6 H2O\n2) C12H22011 + 12 O2 ==> 12 CO2 + 11 H20\n3) C3H6O + 4 O2 ==> 3 CO2 + 3 H2O\n4) C4H10O + 6 O2 ==> 4 CO2 + 5 H2O\n5) C4H10O + 6 O2 ==> 4CO2 + 5 H2O
91997
Balancing Chemical Equations is absolutely essential if you want to determine quantities of reactants or products. An unbalanced Chemical Equation gives only the identify of the beginning reactants and the final products using the appropriate formulas as well as the conditions of temperature, physical state, and pressure conditions under which the reaction is to operate under. However an unbalanced equation can say nothing about the quantities involved until the equation has been balanced. A balanced equation assures that the Conservation Law of matter is obeyed. The total mass of reactants must equal the total mass of products. A balanced equation is like a recipe. It tells you the proportional quantities of each substance involved. The coefficients that appear to the left of each formula can be interpreted as mole units or molecules, but for practical purposes, the coefficients will be interpreted as mole units. No coefficient is interpreted as 1 mole. The following principles should be employed when balancing a Chemical Equation by inspection:\n\n   1. Never touch subscripts when balancing equations since that will change the composition and therefore the substance itself.\n   2. Check to be sure that you have included all sources of a particular element that you are balancing on a particular side since there may be two or more compounds that contain the same element on a given side of an equation.\n   3. I would suggest that you adjust the coefficient of mono atomic elements near the end of the balancing act since any change in their coefficient will not affect the balance of other elements\n   4. When there are a group of atoms that are acting as a unit such as a polyatomic ion and they appear intact on both sides of the equation, it is best to balance them as a self contaned group. For example, if there are Phosphate groups, PO4-3, that appear on both sides, balance the phosphates as a group instead of separating the Phosphorus and Oxygens. It can be done either way, but there is less likely of a mistake if they are balanced as Phosphate groups.\n\nConsider the example of:\n\n(NH4)2CO3 ---> NH3 + CO2 + H2O\n\nThe first thing to do is choose a starting point. If we choose the carbon as the starting point, we go nowhere, since there is already one carbon on each side. Choosing oxygen is not good, because the oxygen is divided between two species on the right side. The interesting part is in the ammonium (NH4) ion. If we start with the nitrogen on the left (in the ammonium ion), we see that there are two nitrogens and therefore we must put at "2" in front of the ammonia on the right side.\n\n(NH4)2CO3 ---> 2 NH3 + CO2 + H2O\n\nSince we are already working with the ammonia, now consider the hydrogens that make up the rest of the molecule: the six hydrogens in the ammonia plus the two hydrogens in the water make eight hydrogens. Bouncing back across to the left side, we count eight hydrogens; since no other atomic species have been changed, we can count up everything and see that the reaction is now balanced:\n\n2 N, 8 H, 1 C, and 3 O on each side.\n\nNotice how the reaction itself led us through the steps of the balancing process. A more complex reaction would have involved more steps, but the process would have been the same.\n\nThis "inspection method" is the first one tried by chemists in the real world. Only when it fails to work, as in the case of many oxidation-reduction reactions, for example, are more complex procedures used.\n\nHere is another example:\n\nFe2(SO4)3 + K(SCN) ----> K3Fe(SCN)6 + K2SO4\n\nK3Fe(SCN)6 is pretty complicated. Now look at it and see what ions are in it that are not shared on the same side of the reaction. These are SCN and Fe.\n\nWe start to balance these\n\nFe2(SO4)3 + 6 K(SCN) ---> K3Fe(SCN)6 + K2SO4 Now we go left to right.\n\nThe irons are messed up, two on the Left, and one on the right. We adjust\n\nFe2(SO4)3 + 6 K(SCN) -- 2 K3Fe(SCN)6 + K2SO4 .\n\nNow t
209883
Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
32213
z = kxy/w where k is the constant of proportionality