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gregori [183]
3 years ago
5

A nuclear power plant operates at 40.0% efficiency with a continuous production of 1042 MW of usable power in 1.00 year and cons

umes 1.07×106 g of uranium-235 in this time period. What is the energy in joules released by the fission of a single uranium-235 atom?
Chemistry
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

3.00 x 10^-11 joules / atom of U-235

Explanation:

We know that the formula for Power = Work done (w)/Time (t)

We need to get the joules from power , since Joules is the SI unit of work.

From the formula P = W/t

W = Power (P) * Time (t)

The SI unit for Time is seconds, hence we change 1 year in seconds

1yr * 365 days/yr * 24hrs/day * 60mins/hr * 60 secs/min = 31536000 secs

It was stated in the question that the plant operates at an efficiency of 40%,

Thus to get the true power we divide the power provided in the question by 0.4 or 40%

= X(0.4) = 1042MW

True Power X = 1042/0.4 = 2605MW

Thus true power = 2605 * 10^6 Watts

Now we have the time in seconds and true power in Watts, we then find the work done.

From our above formula P = W/t

W = P*t = (2605* 10^6) (31536000) =

Finally, we can solve for our energy (work):

P = W / T        PT = W = (2880x10^6) (31536000) = 8.22 x 10^16 joules

We then calculate the amount of energy released by only 1 single uranium-235 atom.

= 8.22 x 10^16 joules / 1.07x10^6 g U-235 (235 g / 1 mol)(1 mol/6.0210^23 atoms)

= 3.00 x 10^-11 joules / atom of U-235

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The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis
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The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

        borneol  →  isoborneol
I         0.0972           0.0486
C         -x                     +x
E     0.0972 - x        0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832 

Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
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pOH = -log[OH⁻]

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