Question

50 points! The isotope shown has a mass of 14.003241 amu. Calculate how much energy is released from the binding of 2.530x10 moles of this isotope. All particles in the nucleus are shown.​

Answer 1

Answer:

E = 3.1885 E16 J

Explanation:

• E = mc²

∴ c = 3 E8 m/s

∴ m = (2.53 E1)(14.003241 amu) = 354.282 g isotope

⇒ E = (354.282 g)(3 E8)²

⇒ E = 3.1885 E19 g.m²/s²

⇒ E = 3.1885 E16 Kg.m²/s² = 3.1885 E16 J

Answer 2

Answer : The amount of energy released is, $3.189\times 10^{16}J$

Explanation :

Formula used to calculate the amount of energy released is:

$E=mc^2$

where,

E = amount of energy released

m = mass of isotope

c = speed of light = $3\times 10^8m/s$

First we have to calculate the mass of isotope.

Mass of isotope = Moles of isotope × Atomic mass of isotope

Mass of isotope = (2.530 × 10)mol × 14.003241 amu

Mass of isotope = 354.3 g = 0.3543 kg

Now we have to calculate the amount of energy released.

$E=mc^2$

$E=(0.3543kg)\times (3\times 10^8m/s)^2$

$E=3.189\times 10^{16}J$

Thus, the amount of energy released is, $3.189\times 10^{16}J$