i need to know about thermonuclear reaction rate - how can i solve this formula?
138258
try these sites: http://www.phy.ornl.gov/astrophysics/data/cf88/\n\nhttp://www.llnl.gov/science_on_lasers/07AstSpace/AS-F_Ther.html\n\nhttp://arxiv.org/abs/astro-ph/9801028\n\nI hope this helps....\n\nDon
1125729
The index-of-reaction equation (i.r.e.): dF/dtZ = kZ(l - βF)m is compared with a generalized Prout-Tompkins equation (g.P.T.e.): dF/dt = kn(l - βF)y With the boundary condition F = 0, t = 0 the equations are equivalent for z = 2 or m = 0 or β ≤ 1 and the i.r.e, is closely approximated by the g.P.T.e. for other values of the parameters by use of the following approximation ∫(1- F)-m dF ≈ AF(1-BF)5 The parameter s is a function of m. Both A and B are close to unity for 1/3 < m < 3. The i.r.e. is recommended as being more useful than the g.P.T.e. because the i.r.e. has a better theoretical basis and because it is more readily integrated and fitted to experimental data.
770398
682 J - 485 J = 196 J\n(energy going into a system is positive, energy going out is negative)
799202
for those questions, all you have to do is translate them into the equation\ncheck chemtutor.com for hellp, it explains it much clearer than i can
75087
without knowing the specifics of your problem, the formula I think you're asking for is deltaH=H(products)-H(reactants). You get the H values for your products and reactants from tables, multiply them by their coefficients, add all the reactants and all the products, the subtract reactants from products.\n\nThe only other enthalpy equation I can think of is \nH=E+PV where E=energy, P=pressure, and V=volume.
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
1032149
Q=MC (delta) T....(5000 G)(.377 J/G degree C)(200 Degree C) which should be.....i get 377 J, which isn't your answer....i just had an exam on this last week, so i thought for sure i am right....the only thing i am not sure about for certain is the specific heat of coal, i got it from a website.....need more info if i am able to help, surely they provided the specific heat of coal....also if latent heat of fusion of vaporization has a part, the answer above is going to be different....sorry dude
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
186407
One way to tell is to look up the heats of formation for the reactants and products.  Add all the reactants up and all the products up.  Subtract the reactants from the products.  If the answer is negative, the reaction is exothermic.  \n\nThe other way is to actually carry out the reaction and see if it liberates heat or not.
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
727276
Here you will find everything about the Arrhenius equation.\n\nhttp://en.wikipedia.org/wiki/Arrhenius_equation
1155911
Photoelectron energy (E) is given by\n\nE = W - hc/L\n\nWhere W is the work function of the metal and L is the excitation wavelength.
845374
Use the Nernst equation:\ndE = dE0 - 0.0592 V/n ln [C]^c [D]^d /([A]^a [B]^b)\n\nn is the number of electrons involved, 2 in this case.\nTherefore we have in all cases:\n dE = dE0 - 0.0296 V ln [Sn2+] /([Fe2+]\nFurthermore we use the standard electrode potentials to calculate  dE0 \nSn2+(aq) + 2e- -> Sn(s) -0.14 V\nFe2+(aq) + 2e- -> Fe(s) -0.41 V\nTherefore dE0 = 0.41 - 0.14 = 0.27 V\na) The concentration are 1 M/L and we have dE = 0.27 V\nb) dE = 0.27 - 0.0296 ln 0.01/0.2 = 0.27 + 0.09 = 0.36 V\nc) dE = 0.27 - 0.0296 ln 0.4/0.005 = 0.27 - 0.13 = 0.14 V\n\nOnly in the last case the reaction moves to the left: the concentration of Sn2+ is the highest.\n\nYou can take a look here for explanations and examples:\n\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
687306
Check the Michaelis-Menten equation.It's too long to write it all down for you ;)
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!