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Delicious77 [7]
3 years ago
13

A process called gaseous diffusion is often used to separate isotopes of uranium-that is

Chemistry
1 answer:
algol [13]3 years ago
5 0

Answer:

<h2>uranium-235 (²³⁵U) and uranium-238 (²³⁸U)</h2>

Explanation:

The gaseous diffusion process utilizes uranium hexafluoride, UF₆, because although it is a solid at room temperature it is easily vaporized. [1] UF6 is not only convenient for its volatility, but also due to the fact that fluorine only consists of the isotope ¹⁹F, meaning the difference in molecular weights for UF6 are purely reliant on 235U and 238U.Here arises another problem however, for the masses of the two uranium isotopes are so nearly equal there is very little separation of 235UF6 and 238UF6 with one pass through a diffuser.Therefore a cascade process is needed to obtain any measurable amount of enrichment. In a cascade the feed stream at diffuser 1 is the UF6 prior to enrichment (meaning it will contain 0.711% 235U and 99.289% 238U) and marks the start of the cascade. There will be hundreds to thousands of diffusers on the upward or enriching side as well as on the downward or depleted side. The slightly enriched UF6 is sent up the cascade process to the next diffuser where it will be enriched again. The slightly depleted UF6 will be sent downward through the cascade where it will also be enriched again. In this way, the enriched uranium keeps getting enriched and sent onward, and the depleted uranium also gets enriched and sent onward. The depleted uranium always gets sent downward where it will eventually be ejected from the downward stream as depleted uranium.

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A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the
Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V

The mole fraction of nitrogen in the mixture is:

X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

Learn more: brainly.com/question/2060778

5 0
3 years ago
Why are the outer-most electrons the only ones included in the electron dot<br> diagram?
kupik [55]

Answer:

Atoms with 5 or more valence electrons gain electrons forming a negative ion, or anion. why are outermost electrons only ones included in orbital filling diagram? they are the only ones involved in chemical reactions and bonding. ... 2s orbital is farther from the nucleus meaning it has more energy.

Explanation:

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3 years ago
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