Is there a mathematical equation that will provide me with the number of isomers I can construct for a given?
930759
There is a basic equation to calculate the number of isomers for an organic compound with chiral centers.  This equation is 2^n where n is the number of chiral centers.  Chiral centers are located at any carbon that is bonded to four different groups like a hydrogen, a methyl group, a chlorine and an ethyl group,
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
771965
I can't tell you the name because i can't remember that stuff but the product will be K(NO3) PbI2\nif you're writing it in the long form it will be\n2KI + Pb(NO3)2 yeilds (arrow) 2K(NO3) + PbI2
956142
2Rb(s)+Cl2(g)--2RbCl(s)
54224
If you have a stoichometric equation with one unknown, you can balalance the known species of both sides (based on conservation of mass) and this will tell you how much of the unknown species is on both sides.\n\nSpecies (eg specific elements) are conserved on both sides of an equation.  Hence, if you have 8 hydrogen atoms and three chlorine atoms on the left side of the equation, you know that there must be 8 hydrogen atoms and 3 chlorine atoms on the right side as well.
890894
C12H22O11+12O2-------12CO2+11H2O+E
1030709
1 H2 +1 Cl2 ------> 2 HCl\n\n     2 - H - 2\n     2 - Cl - 2
119628
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?
65886
Synthesis - a compound forms from simpler reactants
875727
Octane is an alkane with the chemical formula C8H18. It has 18 isomers.
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 :)
930815
i think there is a formula like 13.6/n^2.where n=state.find out for n=1 and n=6 states and the difference of these two energies will give u the answer u require.
810837
Just remember glucose (C6H12O6). Then using the basics of balancing any chemical equation you can figure out the rest of it.
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH
959805
HBr, -1\nHOBr, +1\nBr2, 0\nHBrO4,  +7\nBrF3  +3\n\nThere is nothing to explain.  There are rules, and you need to learn them and follow them.\n\nHere is a site with a good set of rules\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/oxidstat.html
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
967763
Equivalent to [H+]^2/([HCH3CO2] - [H+]).  You can solve for [H+] knowing Ka and [HCH3CO2].