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In-s [12.5K]
3 years ago
15

One of the nuclides in spent nuclear fuel is U-235, an alpha emitter with a half-life of 703 million years. How long will it tak

e for an amount of U-235 to reach 23.0% of its initial amount
Chemistry
1 answer:
VashaNatasha [74]3 years ago
6 0

Answer:

1.49 × 10⁹ years

Explanation:

Step 1: Calculate the rate constant (k) for the nuclear decay of U-235

The decay follows first-order kinetics with a half-life (t1/2) of 703 × 10⁶ years. We can calculate "k" using the following expression.

k = ln2/ t1/2 = ln2 / 703 × 10⁶ y = 9.86 × 10⁻¹⁰ y⁻¹

Step 2: Calculate the time elapsed (t) so that the final amount ([U]) is 23.0% of the initial amount ([U]₀)

For first order kinetics, we will use the following expression.

ln ([U]/[U]₀) = -k × t

ln (0.230[U]₀/[U]₀) = -9.86 × 10⁻¹⁰ y⁻¹ × t

ln 0.230 = -9.86 × 10⁻¹⁰ y⁻¹ × t

t = 1.49 × 10⁹ y

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Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             Q_{1} = mC_{1} \Delta T_{1}

Putting the given values into the above equation as follows.

           Q_{1} = mC_{1} \Delta T_{1}

                      = 27.3 g \times 1.70 J/g K \times 41

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           Q_{2} = energy required = mL_{v}

    L_{v} = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             Q_{2} = mL_{v}

                         = 27.3 \times 444

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Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            Q_{3} = mC_{2} \Delta T_{2}

Value of C_{2} = 1.06 J/g,    \Delta T_{2} = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             Q_{3} = mC_{2} \Delta T_{2}

                       = 27.3 g \times 1.06 J/g \times 34 K

                       = 983.892 J

Therefore, net heat required will be calculated as follows.

            Q = Q_{1} + Q_{2} + Q_{3}

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

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