what is the enthalpy of fusion of butter?
399605
32.0 mJ/mg Melting point of cow butter
434523
* The heat content of a chemical system is called the enthalpy (symbol: H)\n    * The enthalpy change (H) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.\n    * H = H(products) - H(reactants)\n    * H is specified per mole of substance as in the balanced chemical equation for the reaction\n      The units are usually given as kJ mol-1 (kJ/mol) or sometimes as kcal mol-1 (kcal/mol)\n      1 calorie (1 cal) = 4.184 joules (4.184 J)\n    * Energy changes are measured under standard laboratory conditions\n      25oC (298K) & 101.3kPa (1 atmosphere) \n\nEnthalpy changes are measured in KJ per mole, but experiments on measuring enthalpy changes always measure a temperature change. The temperature change is then converted to enthalpy change using the following formula:\n\nDH    =    m x c x DT\n\nm = mass in grams\n\nc = specific heat capacity in Joules/Kelvin/gram\n\nDT = change in temperature in degree Kelvin\n\nThe experiment  is done by simple calorimeter experiments which involve carrying out the reaction in a heat resistant vessel and measuring the change in temperature. \n\nThis is also a common practical assessment, and you should be aware of the sources of error in your measurements such as heat loss to surroundings, incorrect readings, inaccurate thermometers etc.
984026
You need to find the heats of formation of each species (will come from a standard table) and use this equation to get the heat of reaction:\n\n(delta)H = sum of heat of formation of products - sum of heat of formation of reactants\n\nEach heat of formation will be multiplied by its stoichiometric ratio, and the physical state will be important (especially that of water, as the heat of formation of steam is higher than that of liquid water).  You will not find a heat of formation for oxygen gas, because by definition there is no heat of formation for an element.
492501
From fusion of hydrogen atoms.
21495
the formula is just the heat of fusion for water (in J/kg) times the amount of water you have.\n\nThe heat of fusion in Joules per kg for water is 333548 J/kg, so you just multiply that by the mass in kg to get the joules, so you'd have 333548 J/kg x 40kg = 13341920J which, in scientific notation would be 1.33 x 10^7 J.
580457
Standard enthalpy of formation of Barium oxide crystals at 298.16K (25°C) is -548.0 kJ/mole.  For gaseous/vaporous BaO, it is -112.0 kJ/mole
358448
specific heat? that would be represented by Cp.\nHowever if you are using delta H then it would either be: delta H (fus) which is the enthalpy of fusion, used in problems concerning melting point (indicating a state change from solid to liquid)\nOr it could be delta H (vap) which is the enthalpy of vaporization, used in problems concerning a boiling point (indicating a state change from liquid to gas)\n\nHeat calculations are done using the formulas:\nq=m x (delta T) Cp <-- Used when there is a temperature change, but no state change\ng= m x (delta H fus) <-- Used for state changes from solid to liquid\nq= m x (delta h vap) <-- Used for state changes from liquid to gas
44293
Go here to find out about  "MOLAR ENTHALPY"\n\nhttp://www.carolina.com/chemistry/activities/Determining_Molar_Enthalpy_of_Solution.pdf
131859
melting point is 3550 (Degree celcious) and boiling point is 4827 (degree celcious)
274980
Since there is no temperature change in the substance, only a phase change (solid --> liquid), all you need to use is,\nQ = Lm\nwhere Q is the heat energy, L is the latent heat of fusion, and m is the mass of the substance.\nI dont happen to know the latent heat of fusion for water (I think it is like 2260 J/g, maybe), but I am sure it is in your book somewhere.\nJust plug the values into the formula and you are set.\n\nEDIT:\nThe Latent heat of fusion of water is 333 J/g.\nThe latent heat of vaporization is 2260 J/g.
570934
Based on one large egg:\n\nEgg White 17 calories\nEgg Yolk 59 calories\nFried Egg 92 calories\nHard-Boiled Egg 76 calories\nPoached Egg 76 calories\nScrambled Egg 100 calories\n\nBut don't forget theres extra (not too much) in whatever you use to fry it in (maragarine, butter, cooking oil, or things like Pam). :)
775809
I believe the units of your supposed answer are wrong.  They should be kJ/mol and not kJ.\n\nHeat = m * Cp * DeltaT = 60 * 4.186 * (13.2 - 27.2) = - 3516 J\n\nThis heat is absorbed by 28.7 g of KI.  Per mole (MW(KI) = 166 g/mol), the heat will then be:\n\n28.7 g ===> 3516 J\n166 g ===> x\n\nx = 3516 * 166 / 28.7 = 20,340 J/mol = 20.3 kJ/mol
937733
Enthalpy change of formation of 1 mole of C2H2 is the energy change when 1 mole of it is formed from its elements (2C(s) and H2(g)) in their standard states.\n\n2C(s) + H2(g) ---> C2H2(g)\n\nenthalpy of formation of C2H2= \nenergy absorbed to break H-H bond \n+ \nenergy released to form 1 C=C bond and 4 C-H bonds \n= 610 + 4(-410)\n= -1030 KJ/mol\n(If you are taking A level Chemistry, refer to your data booklet for these values under the section of bond energies)\n\nHope this helps.
509763
Try using the calculator at the webs site listed below. \nEthanol at 57.5 deg C has a heat of vaporization = 40.23 KJ/mole.\n\nThis web site has calculators for a number of other physical properties.\n\nBest wishes !
1142668
Problem \ntake D as Delta(change)\nEstimate the change in enthalpy, DH, for the following reaction:\n\nH2(g) + Cl2(g) --> 2 HCl(g) \n\nSolution \n\nTo work this problem, think of the reaction in terms of simple steps:\n\n\nFirst, the reactant molecules, H2 and Cl2, break down into their atoms\nH2(g) --> 2 H(g)\nCl2(g) --> 2 Cl(g)\n\n\nNext, these atoms combine to form HCl molecules\n2 H(g) + 2 Cl(g) --> 2 HCl(g) \n\nIn the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:\n\nDH1 = +(436 kJ + 243 kJ) = +679 kJ\n\nBond breaking requires energy, so we expect the value for DH to be positive for this step.\n\nIn the second step of the reaction, two moles are H-Cl bonds are formed. Bond breaking liberates energy, so we expect the DH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:\n\nDH2 = -2(431 kJ) = -862 kJ\n\nBy applying Hess's Law, DH = DH1 + DH2\n\nDH = +679 kJ - 862 kJ\n\nDH = -183 kJ\n\n\nAnswer\n\nDH = -183 kJ
723847
I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
610449
The latent heat of fusion -- once you cool water to 0°C, you have to remove more heat to actually crystallize it.\n\nIce at 0°C requires the addition of heat to melt into water at 0°C.  That heat comes from whatever you are cooling.\n\nThe latent heat of fusion for ice is 333.6 joules/gram