trying to figure out delta H degrees of the rxn!!?
820519
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
764542
This is an impossible figure as by your own figures ΔABC is equilateral with AB = 2AL = 10\nBC = 2BD = 10 and\nAC = 2AE = 10\nso AT = 2/3 * BT = 2/3* 5√3\n= (10√3)/3\n≠10
497024
H2S---> 2 H^+ + S^2-\nK(eq) = \n[2H^+]^2 [S^2-]\n ---------------\n     [H2S]\n\nNH4+----> NH3  + H^+\nK(eq) =\n [H^+][NH3]\n       ------------\n         [NH4+]\n\nC6H5COOH----> C6H5COO-  + H+\n\nK(eq) = \n[H^+][C6H5COO-]\n        ---------------\n         [C6H5COOH]
715037
Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
665570
I assume delta(x, y, z) = 1 is the mass density ... isn't rho usually used for that?  Anyway, you need to integrate mass times square of perpendicular distance from axis of rotation.\n\nLooks like it will be messy.\n\nMaybe outer x from 0 to 1, middle y from 0 to 1 - x^2, inner z from 0 to (12 - 4x - 3y)/2, of delta(x, y, z) (x^2 + y^2) dz dy dx.
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
789574
from the ⌠udv = uv - ⌠vdu, let u=(lnx)^2 and dv=x^4dx\nhence du=2dx/x and v=x^5/5\nsubstituing values we have:\nx^4 (lnx)^2 dx = (lnx)^2(x^5/5) - ⌠(x^5/5)(2dx/x)\n                      = (lnx)^2(x^5/5) - ⌠(2/5)x^4dx\n                      = (lnx)^2(x^5/5) - x^5/10\nthen using the limits; this yields to 1.796442507
894340
4.25 = (((2.1 rev/s) /  3.1s) * (360degrees/rev)) / 57.3 rad/degree
744648
Use Gibbs free energy equation\n\nDeltaG=DeltaH - TDeltaS\n\nYou want Delta G to be negative so if Delta H positive then you need a greater TDeltaS \nDelta S is -25 so in order for the reaction to be spontaneous the T has to be less than zero
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
324857
Found a good diagram here\nhttp://www.laynetworks.com/Block%20diagram%20of%20the%20Pentium.htm
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
469212
See if this helps.\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/atomic/hund.html
957415
Use PV = n RT, RT = constant. Therefore n = PV * constant in the three cases.\nWe have \nH2 :  P = 372 torr V = 2 liter  n_H2=  744 * constant\nN2:  P = p torr V = 1 liter  n_N2 = p * constant\nCombined: P = 303 torr V=3 liter  n_combined = 909 * constant\n\nTherefore  744 + p = 909 => p = 165 torr
856026
Use the gas law\nPV=nRT
956604
Try it out in the lewis-dot diagram\n      ..        ..\nH-O-Ca-O-H\n      ..        ..
200344
W=f.delta(x) \nx=w/f = 49821/717= 69.49 metes