Chemistry problem....can you solve it?
550924
I believe that at equilbrium, you would have a gradient of concentration throughout the liquid from top to bottom.  At the top, you would have primarily A and B reacting.  At the bottom, you would have primariliy AB dissassociating.  In between, you would have various degrees of A, B and AB as A and B rise up from below and AB falls from above.  The magnitude of the gradient would depend on the reaction rate.
730581
I love these kinds of problems...lol...\n\nFirst you need to find the empirical formula of the compound.  To do this, pretend that you have 100 grams of the compound.  From your elemental analysis, that means that in that 100 grams, you have 65.45 grams of C, 5.49 grams of H and 29.05 grams of O.  Convert each of those amounts into moles by dividing by each elements atomic weight.  That gives you 5.45 moles of C, 5.49 moles of H and 1.80 moles of O.  Now divide all of those by the smallest number.  (What you're trying to find is the ratio of atoms of each element in the formula.).  WHen you do that you find the empirical formula of the compound is C3H3O. From that empirical formula, the molar mass is 55.\n\nNow, go to the gas law part of the problem.  You need to somehow get to the actual molecular weight of the compound.  Using PV=nRT, you can calculate the number of moles of the compound you have.  (Remember to use the right units for each of the numbers you're given.)  I got something like 0.05 moles.\n\nNow, to get the actual molar mass, divide the mass you have in the flask by the number of moles you just calculated.  This gives a molar mass of 110 g/mol.\n\nSince your empirical formula has a mass of 55, the compound has a formula of C6H6O2.\n\nHope this makes sense...
796834
try asking this in the chemistry section next time....
784708
What is there to do to solve this?
903092
1) b)\n\n2) [H2O]^2 / 2[H2] [O2]\n  \n   [NO]^2 / [N2] [O2]\n\n3) b)\n\n4) b)
1020883
i am sure you can find a math tutor that will be happy to help you when you explain what the problem is.\n\nso what IS the problem, anyway?
727014
start with the O... they are always -2\nsince there are 2, and the compound must balance...\nPb must be +4\n\nso you get +4 -2
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...
791178
I HAVE A CHEMISTRY EXAM tomorrow. Actually a small test.lol\nwant answers go to the site below!
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
877634
Any acid, for example:\n\nHCl + H2O  --> Cl- + H3O+
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?
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!!!
826844
H\n |    \nC=O:  (You need 2 lone pairs on Oxygen)\n |   \nH\n\nIt doesnt get much better than that :)
976540
a) +1\nb) 0   (in molecule state)\nc) 0\nd) -3\ne) 0\nf) +4 of Carbon\ng) +2 of Mg\nh) +1 of Chlorine (interesting!)\ni) +6 of Sulphur
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 :)
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.