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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Option C. The object is returning to the start at a constant speed.
<h3>
Data points of the Position vs Time graph</h3>
The following data points will be used to determine the motion of the object.
<u>Position Time</u>
12 4
10 6
2 8
0 10
From the data above, the position of the object is decreasing towards zero or start point.
Thus, the object is returning to the start at a constant speed.
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Answer:
Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion.
Answer:
4.) 9, 1, and 4 5.) 4, 1, and 4
Explanation:
I am not quite sure about this because I cannot remember if the coefficient (the number before the elements) is applied to every element in the compound. If it is then your number of atoms are as follows: CORRECTION: you do not have to apply the coefficient to every element only the one that is after it. So when you back and fix the error your number of atoms will be as follows:
number 4
H: 9
P: 1
O: 4
number 5:
H: 4
S: 1
O: 4
you can calculate the number of atoms present in this compound by multiplying the coefficient and the subscripts of each atom.
hope this helped you :)
Answer:
B
Explanation:
First of all it is important to know that a half filled orbital is particularly stable. In phosphorus all the electrons occur singly in the 3p sublevel minimizing inter electronic repulsion hence it is more difficult to remove an electron from this energetically stable arrangement. In sulphur, electrons are paired in one of the 3p orbitals thereby lowering the energy of that level due to instability caused by interelectronic repulsion between two electrons in the same orbital.
