C
SO2
5
2
X
5.44
Cross multiply to solve
2x = 27.2
X = 13.6 moles C
Answer : The atom with the same number of neutrons as 
is, 
Explanation :
- The given atom is,
Atomic mass number = 12
Atomic number = 6
Atomic number = Number of protons = Number of electrons = 6
Number of neutrons = Atomic mass number - Atomic number = 12 - 6 = 6
Now we have to determine the number of neutrons in the given options.
- In atom,
Atomic mass number = 9
Atomic number = 4
Atomic number = Number of protons = Number of electrons = 4
Number of neutrons = Atomic mass number - Atomic number = 9 - 4 = 5
- In atom,
Atomic mass number = 11
Atomic number = 5
Atomic number = Number of protons = Number of electrons = 5
Number of neutrons = Atomic mass number - Atomic number = 11 - 5 = 6
- In atom,
Atomic mass number = 13
Atomic number = 6
Atomic number = Number of protons = Number of electrons = 6
Number of neutrons = Atomic mass number - Atomic number = 13 - 6 = 7
- In atom,
Atomic mass number = 14
Atomic number = 7
Atomic number = Number of protons = Number of electrons = 7
Number of neutrons = Atomic mass number - Atomic number = 14 - 7 = 7
Therefore, the atom with the same number of neutrons as is,
Explanation:
The reaction equation will be as follows.
Using bond energies, expression for calculating the value of 
is as follows.
On reactant side, from 
number of bonds are as follows.
C-C bonds = 1
C-H bonds = 6
From 
; Cl-Cl bonds = 1
On product side, from 
number of bonds are as follows.
C-C bonds = 1
C-H bonds = 5
C-Cl bonds = 1
From HCl; H-Cl bonds = 1
Hence, using the bond energies we will calculate the enthalpy of reaction as follows.
=![[(1 \times 348 kJ/mol) + (6 \times 414 kJ/mol) + (1 \times 242 kJ/mol)] - [(1 \times 348 kJ/mol) + (5 \times 414 kJ/mol) + (1 \times 327 kJ/mol) + (1 \times 431 kJ/mol)]](https://tex.z-dn.net/?f=%5B%281%20%5Ctimes%20348%20kJ%2Fmol%29%20%2B%20%286%20%5Ctimes%20414%20kJ%2Fmol%29%20%2B%20%281%20%5Ctimes%20242%20kJ%2Fmol%29%5D%20-%20%5B%281%20%5Ctimes%20348%20kJ%2Fmol%29%20%2B%20%285%20%5Ctimes%20414%20kJ%2Fmol%29%20%2B%20%281%20%5Ctimes%20327%20kJ%2Fmol%29%20%2B%20%281%20%5Ctimes%20431%20kJ%2Fmol%29%5D)
= -102 kJ/mol
Thus, we can conclude that change in enthalpy for the given reaction is -102 kJ/mol.
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