Estimate the bond energy in KiloJoule/mol.?
462280
E = Fd = (2E-9 N) (200E-12 m) = x J\nx J (1 kJ / 1000 J) = xE-3 kJ\n\nI think this is already per mol...
521266
This sounds like homework, so I'll give you the ideas but not the data.\n\nThe enthalpy of the mixture at standard conditions is the sum of standard heats of formation of the components of the mix. It will be given in kJ/mol (Ar is 0 kJ/mol being an element, the other two you can look up in tables or on internet). \n\nThen you need to adjust for the sensible energy difference compared to standard conditions : you will also need to look up the specific heat of each component (usually expressed in kJ / mol ºK) and multiply the specific heat of each component by the amount and by the difference in temperature. Make sure you have everything in the same units.\n\nRemember if you are working in moles its easier because to a first approximation, one mole of any gas occupies the same volume (if you undrestand how to allow for deviations from the ideal gas laws, you wouldn't be asking this question) and I assume that the percentages are volume percent, not by weight\n\nEnjoy
132596
if I understand the question:\nto go from cal/gK to cal/gC you'd convert the temp from K to C. that's simply subtract 273. degrees C = degrees K - 273.\nif you're trying to go from j/kg to j/mol you need to know the identity of the substance you're talking about. specifically you must know the molecular weight of the substance. first convert kg to grams(multiply by 1000) then divide by the M Wt of the compound.
320537
I think for that you need the bond enthalpy for the O-H bond because that bond is being formed in H-O-H; anyways, first you see the bonds which are being broken in the reactant (in this case it is 0-0 and O-H) and add up the bond enthalpies(it should be positive). Next, see the bonds formed in the products (H-O and O=0) and add their bond enthalpies(it should be negative as energy is released when bonds are formed). Then to calculate the enthalpy change, simply add up the total enthalpies.
215160
http://www-ec.njit.edu/~skawinsk/chem243/chap11/chap11.htm\nthis may help–near the bottom
726529
2*-393.5+3*-285.8+1366.8 = -277.6 kJ/mol
407618
k e =1/2  * m *v *v\n    =1/2 *3 * (3.5*10) * (3.5*10)\n    =0.18j (ans)
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
136369
reverse reaction one(and change the sign of delta H), add the two reactions together, cancel out on both sides. then add the delta H for both reactions, should get equals 54.08kJ
701443
1 - the reaction is reversed and multiplied by 4, so the enthalpy change is 11508 kJ\n\n2 - The reaction is a combination of the other three (I'll label them A, B and C, in the order you have them listed, for simplicity)\nThe original reaction = 2A - 2B -2C (the first one is just doubled, the next two are doubles and reversed), so 2*321.8 - 2*571.7 - 2*288.6 = -2542.2 kJ\n\n3 - This one is done similarly to question 2.  A - 0.5B + 1.5C\n180.8 - 0.5*(-906.2) + 1.5*(-483.6) = -91.5 kJ\n\nHope this helps
993443
29.5kg x (0.01 m)^2 / (3600 s)^2 = 2.28 x 10^-10 joule = \n0.228 nanojoule (nJ)
870267
.4084c(80)+.4084(heat of vaporization)\nc is specific heat of water\nthat will give you J/KG C   si units
780211
By calculating. -_-\n\n1 mol is approximated to 6 x 10^23 molecules/atoms.\n\nThis can also be rephrased as:\n\n6 x 10^23 molecules make 1 mol of the compound.\nTherefore Q molecules make [1/(6x10^23) x Q] mol of the compound.
950410
Mainly H-bonding (intermolecular forces)
72400
-278.0 + 2*(-321.0) + 2*(-784.0) = -2488 kJ\n\nI am assuming that the enthalpy was actually given in kJ/mol.
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
845828
Here you go...\n\nhttp://home.planet.nl/~skok/techniques/laboratory/pka_pkb.html