How to get kj mol
[DOC File]CHEMICAL KINETICS - prepareforchemistry
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The energy of activation is 200 kj / mol. Calculate the time required for 75% decomposition at 4500 C. 30. The ionization constant of NH4+ ion in water is 5.6 x 10 10 at 250C. The rate constant for the reaction between OH- and NH4+ to form NH3 and H2O at 250C is 3.4 x 10 10 L/ mol/ s. calculate the rate constant for proton transfer from H2O to NH3.
[DOC File]AP Chemistry Lab Manual - MOLEBUS (ALLCHEM)
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Calculate the heat of dissolving the salt in J/g and kJ/mol. Assume that the specific heat and density of the solution is the same as that of water, and that no heat is transferred to the surroundings or calorimeter. 8. In a coffee-cup calorimeter, 100 mL of 1.0M NaOH solution and 100 mL of 1.0M HCl solution are mixed.
[DOC File]Corrections to Noggle, Physical Chemistry, 3rd Edition ...
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Feb 09, 2018 · Of course, since this is a reaction that converts one mol of water, it is also acceptable to say (after the fact of calculation) that the enthalpy of this reaction is 927.0 kJ/mol, where the “/mol” is appended to mean per mole of water (or for that matter, per mol of oxygen atoms, but it is not per mol of hydrogen atoms !!!!).
[DOC File]WPHS Chemistry
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The value of ∆H =-221 kJ/mol. (You will need to write the balanced chemical equation. How many grams of magnesium will be made if 365-kJ is released? How much energy will be released if 9.45-g of magnesium reacts? Ammonium sulfate reacts with barium hydroxide endothermically. ∆H = + 127 kJ/mol.
[DOC File]Preparing for the FRQ Section of the AP Chem exam
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Calorimetry (use q=mc(T to get kJ, then divide by moles of limiting reactant consumed to get kJ/mol, which is (H) Hess’s Law. Heats of formation (products minus reactants) Bond Energy (energy required to break--positive values PLUS energy released when formed—negative values). I suggest drawing a structural formula so that you don’t ...
[DOC File]1
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Calculate the value of standard enthalpy change in kJ/mol-1, for the combustion of glucose. 2.50g*(1 mol C6H12O6/180.16g C6H12O6)=0.0139 mol C6H12O6-39.0kJ/0.0139 mol =-2.810 kJ/mol-1(1 point is earned for the correct answer) Annotated solution: The point here is that the enthalpy for the whole reaction is the same for reactants and products.
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