Another equilibrium???  Show work please, thanks for the help?
924141
Keq = [C]/[A][B]\n[C] = Keq*[A]*[B]\n[C] = 0.76*[0.8]*[0.6]\n[C] = 0.36
401395
Way to go Arbitrage!  Oh to funny guy- yeah we get that E=MC2  IS a formula- but if you really want to "get it" Check out Maxwells equations. It's a great time to be a science nerd with all this info pouring in now! Figure it out & go up one on Albert!
195201
i think u typed the question wrong, can u please correct it and if u have 2 unknowns, please give 2 equations, else u cant get a definite solution
710050
You have to ad the forces and show the resulting force. The equilibrium force will have the same magnitude and opposite direction (neutralizing all 5 forces)\n\nThink of the 1 force going North, the second one going East, the 3 going S, the 4 going W and the 5  NNE.\n\n2& 4 neutralize each other, so do not worry about them\n\n1 + 3 = 20 N 270 degrees\n\n5 = 25 N 0 deg + 43.3 N 90 deg  (50 cos 60 = 25, 50 sin 60 = 43.3)\n\nso the NS component will be 43.3 - 20 = 23.3 N 90 degrees\nand the EW will be 25N 0 deg\n\nThe resulting force will then be  34.17 =sqrt(23.3^2+25^2)\nwith direction 43 degrees (arctan[23.3/25])\n\nThe magnitude of the 6th force will then be 34.17 N and will have a direction of 223 degrees (opposite the 43 degrees one 43+180)\n\nyou can only arithmetically add forces that are in the same direction (or substract forces that have opposite direction).
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
808352
The equilibrium equation is K = (pNO)^2/(pN2*pO2) = 5.0 e-5\nThe conditions for the reaction are:\ninitial: pN2 = 0.79 atm, pO2 = 0.24 atm, pNO2 = 0 atm\nchange: pN2 = -x, pO2 = -x, pNO2 = +2x\nequilibrium: pN2 = 0.79 - x, pO2 = 0.24 -x, pNO2 = 2x\n\nSo, using the equilibrium conditions in the equilibrium equation, you get:\n\nK = (2x)^2 / ((.79-x)*(.24-x))\nK = 4*x^2 / (0.1892 - 1.03x + x^2) = 5.0 E-5\nsolving for x (I'll admit, I used the solver on my calculator)\nx = 0.0015314 or -.0015443\n\nSince the pressure of NO can't be negative, x = 0.0015314 atm\n\nThe final pressures are:\nN2: 0.788 atm\nO2: 0.239 atm\nNO: 0.003 atm
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
404136
Calculate the total energy of the system before and after and make an energy balance.\n\nFor example, lets say you have an inelastic collision between two cars of equal mass m:\n\nInitial State:\nCar 1 moves at v1 to the right\nCar 2 moves at v2 to the left\n\nAssume v1 > v2\nThey hit and stick together completely.  Use conservation of momentum to find the total resultant velocity:\n\np1 + p2 = ptotal (p is the momentum of car 1 and car 2 respectively)\n\nm*v1 - m*v2 = (m+m)*vf\n\nvf is the final velocity of the clump\nthe minus sign is because I am choosing the direction "right" as positive (momentum is a vector)\n\nvf = (v1-v2)/2\n\nNow do an energy balance:\n\nEbeforecollision = 1/2*m*(v1^2) + 1/2*m*(v2)^2\nEaftercollision = 1/2*(m+m)*(vf)^2 + Efrictionlosses\n\nEbefore = Eafter (this is conservation of energy, and it is true as long as you account for all forms of energy...it wont be true if you forget about the frictional losses in Eafter)\n\nNow solve for Efrictionlosses\n\nTo get the total fractional energy loss, take the ratio Efrictionlosses/Ebefore
920693
For equilibrium of the board-and-boulder system, two conditions must be satisfied:\n\n!) Sum of vertical forces must be zero and \n\n2) The turning effects of the forces of Will and Cliff (about the center-of-mas of the system) must total up to zero.\n\nNow considering the vertical forces: \nWeight of the boulder is 71 x 9.8 N = Fwill + Fcliff\n\nFor the balancing of turning effect condition:\n(By the principle of moments) The distances Dcliff and Dwill from the CM of the two forces should be in inverse proportion  of the two forces .: \n\nDcliff = 3.15m and  \nDwill = (3.50-3.15)\nleading to: \nFcliff  = (71 x 9.8) 3.15/3.50 =. 626.22 N and\nFwill  =  (71 x 9.8) (3.50 - 3.15 ) /3.50 =.69.58 N\n\nAnswer:\nCliif applies 626.22N and Will applies 69.58 N..Got it?
288520
cmon dont do this again, give the other equation
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
799788
You are given P(t=0), Q(t=0); P'(t=0), Q'(t=0); and P''(t=0), Q''(t=0). Given that P'' and Q'' remain the same (that is, the motion of P and Q do not change), you need to show that there is a time T such that P(t=T) = Q(t=T).\n\nIt's late and I'm drunk, so you can do the actual working for yourself :D
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
396330
Another homework question.
758038
omg i just had this test today too!!!!!\nu use hess's law to find the change--->write the equation, balance it \nand then use the moles (coefficients) to multiply the change is s for each element.\nhess's law says that the change in s (or h, or G, its usable for any of the thermo concepts) is equal to the sum of the entropy of the products minus the sum of the entropy of the reactants. its pretty easy after that--just plug in the values and there you go! \ngood luck! this was a hard test for me..so confusing.
158900
Yup. They are not simple...\nChemistry...\nhttp://www.chm.davidson.edu/ChemistryApplets/equilibria/EquilibriumConstant.html\nhttp://en.wikipedia.org/wiki/Chemical_equilibrium\nThey use the equilibrium constant expression (Kc) here...\nhttp://www.cofc.edu/genchemlab/equilibrium.htm