Question

1. Consider the following reaction: 10 21 N e + 2 4 H ⟶ 12 24...

1.

Consider the following reaction: 10 21 N e + 2 4 H ⟶ 12 24 M g + 0 1 n.

How much energy is released in the reaction?

the masses are: Ne = 20.993849u; He=4.002603u; Mg=23.985042; and n=1.008665u

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Homework Answers

Answer #1

mass of reactants (m1) = 20.993849 + 4.002603 = 24.996452 amu

mass of products (m2) = 23.985042 + 1.008665 = 24.993707 amu

mass deffect (m) = m1 - m2 = 24.996452 - 24.993707 = 0.002745 amu

we know that,

Energy released is given as.

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