How many subshells are in the n = 4 shell?
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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):
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Answer:
0.37atm
Explanation:
Given parameters:
Initial pressure = 0.25atm
Initial temperature = 0°C = 273K
Final temperature = 125°C = 125 + 273 = 398K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use a derivative of the combined gas law;
=
P and T are pressure and temperature
1 and 2 are initial and final values
=
P2 = 0.37atm