Answer:
Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):
Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:
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Answer:
Kc =![\frac{[8.326x10-3]^{1} }{[1.113x10-2]^{1}[1.490x10-2]^{1} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B8.326x10-3%5D%5E%7B1%7D%20%7D%7B%5B1.113x10-2%5D%5E%7B1%7D%5B1.490x10-2%5D%5E%7B1%7D%20%20%7D)
Kc = 50.2059
Explanation:
1. Balance the equation
2. Use the Kc formula
Remember that pure substances, like H2 are not included on the Kc formula
Answer:
n = 2.208x10¹⁸ photons
Explanation:
The energy of a photon( an elementary particle) is given by the equation:
E = nxhxf
Where n is the number of photons, h is plank constant (6,62x10⁻³⁴ J.s), and f is the frequency. Knowing that the power level is 0.120mW (1.2x10⁻⁴ W), the energy in J, for a time of 78 min (4680 s)
E = 1.2x10⁻⁴x4680 = 0.5616 J
The frequency of a photon is its velocity ( c= 3x10⁸ m/s) divided by its wavelength, which is 780 nm = 780x10⁻⁹ m
f = 3x10⁸/780x10⁻⁹
f = 3.846x10¹⁴ s⁻¹
Then, the number of photons is:
0.5616 = nx6,62x10⁻³⁴x3.846x10¹⁴
n = 2.208x10¹⁸ photons.
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A wax is a type of lipid. Waxes are nonpolar, so they are not soluble in water.
