The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):
![\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol](https://tex.z-dn.net/?f=%20%5CDelta%20H%20%3D%20%5B2%2A%286%2A413%20%2B%20347%29%20%2B%207%2A498%20-%20%284%2A2%2A799%20%2B%206%2A2%2A467%29%5D%20kJ%2Fmol%20)
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Answer:
8.354 nanometers
Explanation:
To treat a diffusive process in function of time and distance we need to solve 2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:
Where Cs is the concentration in the surface of the solid
Cx is the concentration at certain deep X
Co is the initial concentration of solute in the solid
and erf is the error function
Then we solve right side,
And we need to look up the inverse error function of 0.001964 resulting in: 0.00174055
Then we solve for x:
Answer: Option (3) is the correct answer.
Explanation:
Aerobic organisms are the organisms which survive and grow in the presence of oxygen.
When oxidation of glucose occurs in the presence of oxygen then it is known as aerobic respiration.
In aerobic respiration, food releases energy to produce ATP which is necessary for cell activity. There is complete breakdown of glucose in aerobic respiration that is why more energy is released. Therefore, aerobic organisms become active.
Thus, we can conclude that characteristics very active, efficient use of energy describes aerobic organisms.
Answer:
A. is insufficient to overcome intermolecular forces
Explanation:
Just took the review
Answer:
The effects of supercritical CO2 (SC-CO2) on the microbiological, sensory (taste, odour, and colour), nutritional (vitamin C content), and physical (cloud, total acidity, pH, and °Brix) qualities of orange juice were studied. The CO2 treatment was performed in a 1 litre capacity double-walled reactor equipped with a magnetic stirring system. Freshly extracted orange juice was treated with supercritical CO2, pasteurised at 90°C, or left untreated. There were no significant differences in the sensory attributes and physical qualities between the CO2 treated juice and freshly extracted juice. The CO2 treated juice retained 88% of its vitamin C, while the pasteurised juice was notably different from the fresh juice and preserved only 57% of its vitamin C content. After 8 weeks of storage at 4°C, there was no microbial growth in the CO2 treated juice.
