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NISA [10]
2 years ago
15

From the relative rates of effusion of ²³⁵UF₆ and ²³⁸UF₆ , find the number of steps needed to produce a sample of the enriched f

uel used in many nuclear reactors, which is 3.0 mole % ²³⁵U. The natural abundance of ²³⁵U is 0.72%.
Chemistry
1 answer:
Dafna11 [192]2 years ago
7 0

The number of steps required to manufacture a sample of the 3.0 mole%  ²³⁵U enriched fuel used in many nuclear reactors from the relative rates of effusion of ²³⁵UF₆ and ²³⁸UF₆. ²³⁵U occurs naturally in an abundance of 0.72% are :  mining, milling, conversion, enrichment, fuel fabrication and electricity generation.

<h3>What is Uranium abundance ? </h3>
  • The majority of the 500 commercial nuclear power reactors that are currently in operation or being built across the world need their fuel to be enriched in the U-235 isotope.
  • This enrichment is done commercially using centrifuges filled with gaseous uranium.
  • A laser-excitation-based method is being developed in Australia.
  • Uranium oxide needs to be changed into a fluoride before enrichment so that it can be treated as a gas at low temperature.
  • Uranium enrichment is a delicate technology from the perspective of non-proliferation and needs to be subject to strict international regulation. The capacity for world enrichment is vastly overbuilt.

The two isotopes of uranium that are most commonly found in nature are U-235 and U-238. The 'fission' or breaking of the U-235 atoms, which releases energy in the form of heat, is how nuclear reactors generate energy. The primary fissile isotope of uranium is U-235.

The U-235 isotope makes up 0.7% of naturally occurring uranium. The U-238 isotope, which has a small direct contribution to the fission process, makes up the majority of the remaining 99.3%. (though it does so indirectly by the formation of fissile isotopes of plutonium). A physical procedure called isotope separation is used to concentrate (or "enrich") one isotope in comparison to others. The majority of reactors are light water reactors (of the PWR and BWR kinds) and need their fuel to have uranium enriched by 0.7% to 3-5% U-235.

There is some interest in increasing the level of enrichment to around 7%, and even over 20% for particular special power reactor fuels, as high-assay LEU (HALEU).

Although uranium-235 and uranium-238 are chemically identical, they have different physical characteristics, most notably mass. The U-235 atom has an atomic mass of 235 units due to its 92 protons and 143 neutrons in its nucleus. The U-238 nucleus has 146 neutrons—three more than the U-235 nucleus—in addition to its 92 protons, giving it a mass of 238 units.

The isotopes may be separated due to the mass difference between U-235 and U-238, which also makes it possible to "enrich" or raise the proportion of U-235. This slight mass difference is used, directly or indirectly, in all current and historical enrichment procedures.

Some reactors employ naturally occurring uranium as its fuel, such as the British Magnox and Canadian Candu reactors. (By contrast, to manufacture at least 90% U-235, uranium needed for nuclear bombs would need to be enriched in facilities created just for that purpose.)

Uranium oxide from the mine is first transformed into uranium hexafluoride in a separate conversion plant because enrichment operations need the metal to be in a gaseous state at a low temperature.

To know more about Effusion please click here : brainly.com/question/22359712

#SPJ4

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Rom4ik [11]

Answer: The partial pressure of the dry oxygen is 742 torr

Explanation:

Dalton's Law of Partial Pressure states that the total pressure exerted by a mixture of gases is the sum of partial pressure of each individual gas present. Thus P(total)=P_1+P_2 .........

Given; Total pressure = 762 torr

partial pressure of water = 19.8 torr

partial pressure of dry oxygen = ? torr

Total pressure  = partial pressure of water + partial pressure of dry oxygen

762 torr = 19.8 torr = partial pressure of dry oxygen

partial pressure of dry oxygen = 742 torr

The partial pressure of the dry oxygen is 742 torr

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3 years ago
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5 0
3 years ago
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You try to measure out exactly 5.0 milliliters of water by eye into each of five test tubes. When you go back to check the volum
jeka94

Answer:

The volumes are both, accurate and precise.

Explanation:

In the measurement of a set, precision refers to how much coincidence exists in the measurements of an specific value, as the measurements are close, we can say the volumes are precise.

Accuracy means the agreement that exists between the average of one

large series of measurements and the value of  measurement

Media is 4,96 ml and I wanted to measure 5 ml. It is also close.

6 0
3 years ago
5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the
cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

q=mc\Delta T

where,

q = heat absorbed by water = ?

m = mass of water = {\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g

c = heat capacity of water = 4.186 J/g°C

\Delta T= change in temperature = 6.70^0C

q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by MgSO_4

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 5.11 g

Molar mass  = 120 g/mol

Putting values in above equation, we get:

\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol

0.042 moles of MgSO_4 releases = 2.8033 kJ

1 mole of MgSO_4 releases = \frac{2.8033 kJ}{0.042}\times 1=66.7kJ

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

3 0
3 years ago
What is the final volume (L) of a 1.00 L system at 315 K and 1.10 atm if STP conditions are established?
LenaWriter [7]

Answer:

0.95L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 1L

T1 (initial temperature) = 315K

P1 (initial pressure) = 1.10 atm

T2 (final temperature) = stp = 273K

P2 (final pressure) = stp = 1atm

V2 (final volume) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume of the system can be obtained as follow:

P1V1/T1 = P2V2/T2

1.1 x 1/315 = 1 x V2/273

Cross multiply to express in linear form.

315 x V2 = 1.1 x 273

Divide both side by 315

V2 = (1.1 x 273) /315

V2 = 0.95L

Therefore, the final volume of the system if STP conditions are established is 0.95L

5 0
3 years ago
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