Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)
Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol
Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol
<em>Answer: -1367.5 kJ/mol</em>
Volume is 60 and the area is 94 have a great day
Answer:
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Explanation:
Answer:
Initially the function is symmetric with respect to the axis of the one dimensional box. In the final state it is also symmetrical, however you can envision a snapshot of the system as the light field is interacting with the wave-function wherein a node begins to develop as is shown in the middle and the wave function is evolving from the initial to final state. Now consider that the electron density during process is the square of the wave function:
Electron density during transition
As can be seen in the initial and final states the electron density is symmetrically distributed with respect to the axis of the box. However with the field on, the electron density is not symmetrically distributed and a transitory dipole moment can be present. To relate back to real molecules think of each of those orbitals as a linear combination of atomic orbitals. One important factor is the symmetry. But there may be one other factor that will be just as important as symmetry. If you treat orbital 1 as a linear combination over n orbitals and orbital 2 as a linear combinations of orbitals as well, there will be a spatial over lap between the orbital in the ground state and the orbital in the excited state. If there is no spatial overlap between the ground state and excited state orbitals there will be no transition dipole moment. However, if the electrons are in the same place spatially, a large transition dipole moment will result.
Explanation:
