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Bumek [7]
3 years ago
12

50 points! The isotope shown has a mass of 14.003241 amu. Calculate how much energy is released from the binding of 2.530x10 mol

es of this isotope. All particles in the nucleus are shown.​

Chemistry
2 answers:
marysya [2.9K]3 years ago
8 0

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

vodomira [7]3 years ago
4 0

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

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