133.0873 g/mol
(NH4)3PO3 - molar mass
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:
The percent yield of this reaction is 84.8 % (option A is correct)
Explanation:
Step 1: Data given
The student isolated 15.6 grams of the product = the actual yield
She calculated the reaction should have produced 18.4 grams of product = the theoretical yield = 18.4 grams
Step 2: Calculate the percent yield
Percent yield = (actual yield / theoretical yield ) * 100 %
Percent yield = (15.6 grams / 18.4 grams ) * 100 %
Percent yield = 84.8 %
The percent yield of this reaction is 84.8 % (option A is correct)
Principal shell shows the main energy level, in that there are some more energy levels. This is how the atomic structure is defined according to the Quantum Mechanical model of atom. So exactly why do subshells exist and all i don't know