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3 years ago

Tritium 3 1H decays to 3 2He by beta emission. Find the energy released in the process. Answer in units of keV.

Chemistry

1 answer:

-Dominant- [34]3 years ago

7 0

Answer:

The energy released in the decay process = 18.63 keV

Explanation:

To solve this question, we have to calculate the binding energy of each isotope and then take the difference.

The mass of Tritium = 3.016049 amu.

So,the binding energy of Tritium = 3.016049 *931.494 MeV

= 2809.43155 MeV.

The mass of Helium 3 = 3.016029 amu.

So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV

= 2809.41292 MeV.

The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV

1 MeV = 1000keV.

Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.

So, the energy released in the decay process = 18.63 keV.

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<h3><u>Answer</u>;</h3>

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The average rate of disappearance of ozone ... is found to

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This means (ignoring time)

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rate = -(1/2)*(-7.25 × 10^–3 atm)

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rate = +(1/3)∆[O2)∆t

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A 0.105 L sample of an unknown HNO 3 solution required 35.7 mL of 0.250 M Ba ( OH ) 2 for complete neutralization. What is the c

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The equation for the reaction is :

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Va = Vol. Of acid = 0.105L

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MaVa / Mb Vb = nA / nB

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Ma x 0.105 = 0.25 x 0.0357 x 2

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