Answer: Too much base was added
i guessed
Explanation:
Answer:
Gaseous nitrogen has unique chemical and physical properties that make it suitable for use in food processing. Nitrogen is inert which means it will not react with prepared food materials, which can alter their aromas or flavors. Also, gaseous nitrogen will effectively displace oxygen minimizing oxidation and the growth of microorganisms that cause foods to lose their freshness and deteriorate faster.
Explanation:
Source: https://www.generon.com/using-nitrogen-gas-in-food-packaging/
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 = \Delta H_{r} - \Delta H_{p}](https://tex.z-dn.net/?f=%20%5CDelta%20H%20%3D%20%5CDelta%20H_%7Br%7D%20-%20%5CDelta%20H_%7Bp%7D%20)
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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