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:

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.