Answer: 48.6 kJ/mol
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_6H_6}\times \Delta H_{C_6H_6})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_6H_6%7D%5Ctimes%20%5CDelta%20H_%7BC_6H_6%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-6534.0=[(12\times -393.5)+(6\times -285.8)]-[(15\times 0)+(2\times \Delta H_{C_6H_6})]](https://tex.z-dn.net/?f=-6534.0%3D%5B%2812%5Ctimes%20-393.5%29%2B%286%5Ctimes%20-285.8%29%5D-%5B%2815%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelta%20H_%7BC_6H_6%7D%29%5D)

Therefore, the enthalpy change for benzene is 48.6 kJ/mol
Example #1 write the chemical formula for Calcium Oxide
Step 1
-Find the Atomic symbol of the metal and non-metal on the periodic table
CaO
Step 2
Find the charges for Calcium and Oxygen
which are Ca 2+ and O 2-
Step 3
Balance out the charges
They are already balanced out
Here´s the chemical formula for Calcium Oxide
CaO
Example 2 write the chemical formula for Aluminum Oxide
Step 1
-Find the Atomic symbol of the metal and non-metal on the periodic table
A
l
O
Step 2
Find the charges for Aluminum and Oxygen
Which are Al 3+ and O 2-
Step 3
Balance out the charges
You need 2 Aluminum and 3 Oxygen to balance the charges
Al 3+ O 2-
Al 3+ O 2-
= 6 + O 2-
= 6-
Step 4
If you need more then one element to balance out the charges you identify that by using subscripts
Heres your chemical formula for Aluminum Oxide

Answer:
1. 4-ethyl-1-heptene
2. 6-ethyl-2-octene
3. 1-butyne
Explanation:
The compounds are named according to IUPAC rules.
Compound 1:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain).
- The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
- The longest chain contains 7 carbon atoms, so taken the name hept.
- The double bond between C1 and C2, so take no. 1 and add the suffix ene to hept "1-heptene".
- The ethyl group is the alkyl substituent on position 4.
- So the name is 4-ethyl-1-heptene.
Compound 2:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain).
- The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
- The longest chain contains 8 carbon atoms, so taken the name oct.
- The double bond between C2 and C3, so take no. 2 and add the suffix ene to oct "2-octene".
- The ethyl group is the alkyl substituent on position 6.
- So the name is 6-ethyl-2-octene.
Compound 3:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain), there is no substituents.
- The parent chain is numbered so that the multiple bonds have the lowest numbers (Triple bond here take the lowest number).
- The longest chain contains 4 carbon atoms, so taken the name but.
- The triple bond between C1 and C2, so take no. 1 and add the suffix yne to but "1-butyne".
The statement would be considered a law.