equation to maintain equilibrium?
912606
The simple answer is that the vector sum of the forces acting on an object must be zero if the object is to remain in "equilibrium" or in other words to have constant velocity.
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
694340
Let x equal the concentration of ICl at equilibrium.\n\nThen (0.076) squared /x squared = 0.11.\n\nNow solve for x.
1057769
I'm a teacher and i always give the students the "Rope example"\n\nimagine equilibrium as a rope pulled with the same force both ways! \n\nnow if you look at the your chemical reaction energy (heath) is release from the reaction -----> so if you take the heath out (so you cool the system) the reaction ------> will be more power full so equilibrium will be moved in C and D direction.\n\nIf you heath the system (you'll have a lot of heath that needs to go somehow --- and the only way it can go is by the <--------- reaction -- which is a reaction that use heath ).\n\nThe same with pressure (but you need to have different amounts of molecules -- in your reaction you have the same -- 5 in the reaction ----> and 5 in the reaction <----)\n\nif you speak about concentration use the rope example. You put more of A(increase the concentration of A) , that A need to be consumed in order to keep the equilibrium.\n\nif you take A out of system (so reduce the concentration of A) you'll encourage the <---- reaction (reaction that release more A)\n\nI don't know if you'll understand my spelling and grammar (I'm not the aching chemistry in English). \nI'm just curious what grade are you on? because  I teach this subject to 10th grade students (16 years old )\n\nIf you want me to try to rephrase this answer just let me know \njustplaincornnel@yahoo.com
701763
First, notice you have two S atoms on right hand side.\nTo get two S atoms on left, you must have two H2SO4 molecules.\nYour equation is now Cu + 2H2SO4 ---> CuSO4 + SO2 + ??\nIf you subtract right from left, you are left with 4H + 2O atoms, for 2H2O\nSo final eq is Cu + 2H2SO4 ---> CuSO4 + SO2 + 2H2O\n\nHope this helps you to approach these equation-balancing problems methodically.
410035
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.
864388
some formulas that can help :\nhttp://www.eskridgeinc.com/support/formulas.html
1047731
Assume dc/dz is constant, n(t=T)=-DA(dc/dz) T\nBut in reality dc/dz often changes.  Normally dc/dz should decrease with time.  If it is a close system, it will reach the equilibrium state.
974937
please complete the equation
444216
Write the Kc and the Kp\n\nChange PV=nRT to P/RT=(n/V) to P/RT=(concentration) note n/V is concentration.\n\n\nSubstitute P/RT for each concentration in the Kc.  You will see the Kp in the equation with extra R's and T's.  Finaih solving.
111275
Is this a question of differential equations? Just curious because that's the only time that I have ever worked with equilibrium points...if so, the equation seems to be written incorrectly. It seems confusing to have x=-3x^2-x+2 because I would just rewrite it as: 3x^2+2x+2=0. If the left side is a function it should either be y(x) or preferably, dx/dt, or some x' value. Then knowing the initial condition x(0) would make sense. With these details, I would be able to calculate the equilibrium points and classify them. You would have x=2/3 and x=-1 as your two equilibrium points, and x=2/3 would be stable and x=-1 would be unstable. Hope this helps!
691296
use that at equilibrium : n.p = ni^2\nand from charge neutrality: \n\nn + NA = p + ND
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
755565
1.) Cr(NO2)2+(NH4)2SO4\n---->CrSO4+(2)NH4NO2\n2.) (2)KNO3+H2CO3\n----->K2CO3+(2)HNO3\n3.) (3)KBr+Fe(OH)3\n----->(3)KOH+FeBr3\n4.) B2Br6+(6)HNO3\n----->(2)B(NO3)3+(6)HBr\n\nHope this helps=)\nTip: Start by balancing those more isolated atoms first, especially when you are balancing more complex equations (they may have atoms that appear in more than 1 products).\nAlso, feel free to leave numbers in fractions while you balance ur equation. You can always multiply all the numbers later to make them all whole numbers. You will find this useful later when you have to balance long equations. Itd definitely more tedious to alter every number in ur mind and writing them all out at one go. Do it progressively=)
223710
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3
720117
5.  K = [NH3][HCl] = (0.066 M)^2 = 4.36 x 10^-3\n\n9.  I'm assuming that equation should be H2(g) + CO2(g) <---> H2O(g) + CO (g)\na.  Add H2O will drive equlibrium to the left\nb.  Removing CO will drive equilibrium to the right\nc.  Adding H2 will drive equilibrium to the right\nd.  Adding something to absorb CO2 will drive equilibrium to the left\n\n6. \na.  Increasing the volume of the reaction vessel will reduce the pressure, favoring the production of more gaseous products, so the equilibrium will go to the right\nb.  Equal number of moles of gas on both sides, so increasing the volume will have no effect.\n\n1.  \na.  Decreasing the volume increases the pressure, favoring the side with fewer moles of gas.  Equilibrium would go to the left.\nb.  Equilibrium would go to the left, as there are fewer moles of gas on that side.
901005
Use the following equation and solve:\n\nP1V1/T1=P2V2/T2\n\nT2=P2V2T1/P1V1\n\nalso note the significant figures!!!!