If the formation of NaCl proceeds according to the following equation and change in enthalpy, what is the reaction and change in enthalpy for
the decomposition of 2 moles of NaCl?
*"Na" + "%2C1"_2- "Naci"., DeltaH= -411 "kj"*
OA
B.
C.
D.
2"Na" + 2"C" --2"Naci":Deltah= .411 ")"
"2"Na" + 2"C1" --2"Naci", DeltaH= -822 "J"
*2"Naci" --2"Na" + "C"; DeltaH= 822 "kj"*
2"Naci" --2"Na" + 2"C1", "DeltaH= 411 "kj"*

Explanation:

The chemical equation for the formation of NaCl is:

That equation means that 1 mole of NaCl is formed by the reaction of 1 mole of Na and 1/2 mole of Cl₂, with a release of energy of 411 kJ.

The decomposition of NaCl is the inverse of the formation reaction; thus, you swift products and reactants and inverse the sign of the change in enthalpy:

Since you want the decomposition of 2 moles you multiply the equation and the ΔH by 2:

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.