H2So4 is a strong acid and a strong electrolyte which means that when this dissociates in water, the dissociation process is complete. The first dissociation is
H2 SO4 = H + HSO4-
This can further release H+ and dissociate SO4- instead already
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
A molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
The bonding orbital, which would be more stable and encourages the bonding of the two H atoms into 
, is the orbital that is located in a less energetic state than just the electron shells of the separate atoms. The antibonding orbital, which has higher energy but is less stable, resists bonding when it is occupied.
An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as 
* orbitals.
Therefore, molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
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Hence, the correct answer will be option (b)
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if this is true or false its true hope i helped
