<span>Mass of the copper penny m = 2.6 g
Atomic mass of copper = 63.55, Atomic number = 29,
So the number of neutrons = Atomic mass - Atomic number = 63 - 29 = 34
a. Neutron mass = 34 x (2.6 / 63.55) = 1.4 grams
Copper atoms per mole = 6.040 x 10^23 atoms/mol
moles of copper = 2.6 / 63.06 = 0.04123 mol
Total atoms in the copper = 6.040 x 10^23 atoms/mol x 0.04123 mol = 0.25 x 10^23 atoms
Number of electrons in the copper = 29 per atom
Mass of the electron = 9.085 x 10^-28 g
b. Electron mass = 0.25 x 10^23 x 29 x 9.085 x 10^-28 = 65.86 x 10^-5 g</span>
Answer : The change in enthalpy of the reaction is, -310 kJ
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given main reaction is,

The intermediate balanced chemical reaction will be,
(1)

(2)

(3)

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :
(1)

(2)

(3)

The expression for enthalpy of formation of
will be,



Therefore, the change in enthalpy of the reaction is, -310 kJ
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Your answer is C. The moon orbits and rotates the same, so we NEVER see the far side of the moon from earth :)
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P=M(mass)G(Gravity)H(Height)
Gravity=9.8
M=1.5 G=9.8 H=35
so multiply all
=514.5 potential energy