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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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Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50
OLga [1]

Answer:

\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}

Explanation:

Q1. Mass of Cu

(a) Write the equation for the half-reaction.

Cu²⁺ + 2e⁻ ⟶ Cu

The number of electrons transferred (z) is 2 mol per mole of Cu.

(b) Calculate the number of coulombs

q  = It  

\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}

(c) Mass of Cu

We can summarize Faraday's laws of electrolysis as

\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}

Note: The answer can have only two significant figures because that is all you gave for the time.

Q2. Current used

(a) Write the equation for the half-reaction.

Ag⁺ + e⁻ ⟶ Ag

The number of electrons transferred (z) is 1 mol per mole of Ag.  

(a) Calculate q

\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}

(b) Calculate the current

t = 3 h = 3 × 3600 s = 10 800 s

\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}

Note: The answer can have only one significant figure because that is all you gave for the time.

4 0
3 years ago
When one atom of carbon reacts with two atoms of oxygen, they combine to form one molecule of carbon dioxide. How many atoms of
tankabanditka [31]

The equation is: C+O2=>CO2

Since we got 10 molecules of CO2 new balanced equation would be 10C+10O2=>10CO2

from this equation we can see that we have 10 molecules of oxygen, however ,we need to find atoms. There are 2 atoms in the oxygen molecule so we need to multiply 10 by 2 which gives us 20 atoms.

The answer: there are 20 atoms of oxygen

4 0
3 years ago
How do you get hyphen notations?
laiz [17]

Answer:

In hyphen notation, the mass number is written after the name of the element. For example, in isotopic notation, the isotope of carbon that has a mass number of twelve would be represented as 12C . In hyphen notation, it would be written as carbon-12.

Explanation:

lol just took the question and looked it up online this was the first thing that i saw if its not correct im sorry

6 0
3 years ago
1. Describe the process of gene splicing.
AnnyKZ [126]

Answer:

The DNA is cut using special chemicals called restriction enzymes,.

DNA ligase are used to insert genes and other pieces of DNA into plasmids.

4 0
2 years ago
A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre
givi [52]

Answer:

\mathbf{0.02 M}

Explanation:

\text{So, from the given question:}

\text{EDTA will make complex with} V^{+3} \text{and the remaining EDTA will react with }Ga^{+3}

\text{Hence, the total concentration of} V^{+3} & Ga^{+3} \text{will be equivalent to EDTA concentration.}

V_{EDTA} = 25 \ mL

V_{V^{+3}} = 59.0 \ mL

V_{Ga^{+3}} = 13.0 \ mL

M_{EDTA} = 0.0680 \ M

M_{V^{+3}} = ???(unknown)

M_{Ga^{+3}} = 0.0400 \ M

V^{+3} + EDTA \to V[EDTA] + EDTA(Excess)  \to^{CoA} \ Ga[EDTA] _{complex}

M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})

0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18

x = \dfrac{1.18}{59}

\mathbf{x =0.02 \ M }

5 0
3 years ago
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