Are the particles in a solid motionless ?

The specific rate constant, k, for radioactive beryllium–11 is 0.049 s–1. What mass of a 0.500 mg sample of beryllium–11 remains

Sonja [21]

<u>Answer: </u>The mass of sample that remained is 0.127 mg

<u>Explanation:</u>

The integrated rate law equation for first-order kinetics:

$k=\frac{2.303}{t}\log \frac{a}{a-x}$ ......(1)

Given values:

a = initial concentration of reactant = 0.500 mg

a - x = concentration of reactant left after time 't' = ?mg

t = time period = 28 s

k = rate constant = $0.049s^{-1}$

Putting values in equation 1:

$0.049s^{-1}=\frac{2.303}{28s}\log (\frac{0.500}{(a-x)})\\\\\log (\frac{0.500}{(a-x)})=\frac{0.049\times 28}{2.303}\\\\\frac{0.500}{a-x}=10^{0.5957}\\\\frac{0.500}{a-x}=3.94\\\\a-x=\frac{0.500}{3.942}=0.127mg$

Hence, the mass of sample that remained is 0.127 mg

5 0

3 years ago

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

Dafna11 [192]

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.

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7 0

2 years ago

A compound has a percent composition of 81.71% C and 18.29% H. What is the empirical formula of this compound?

SCORPION-xisa [38]

Considering the definition of empirical formula, the empirical formula is C₃H₈.

<h3>Definition of empirical formula</h3>

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

<h3>Empirical formula in this case</h3>

In this case, in first place you know the percent composition:

C: 81.71 %
H: 18.29%

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have 81.71 grams of carbon and 18.29 grams of hydrogen H.

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

C: $\frac{81.71 grams}{12\frac{grams}{mole} }$ = 6.81 moles
H: $\frac{18.29 grams}{1\frac{grams}{mole} }$ = 18.29 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

C: $\frac{6.81 moles}{6.81 moles}$ = 1
H: $\frac{18.29 moles}{6.81 moles}$ = 2.68 ≅ $\frac{8}{3}$

To express this relationship in the form of simple integers, it is necessary to multiply by a simple number to achieve this:

C: 1×3 =3
H:≅ $\frac{8}{3}$ ×3= 8

Therefore the C: H mole ratio is 3: 8

Finally, the empirical formula is C₃H₈.

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4 0

2 years ago

According to vsepr theory, the bond angle in h2se would be predicted to be:

ella [17]

The bond angle in H2Se would be predicted should be less than 109.5 degrees. Particle shapes can be anticipated in light of Lewis dab structure utilizing the VSEPR hypothesis. VSEPR remains for Valence Shell Electron Pair Repulsion. It expresses that electron matches in the valence shell of a particle repulse each other; their sub-atomic geometry is the consequence of this shock.

4 0

3 years ago

Read 2 more answers

True or false - The electrostatic attraction of the nucleus and the outermost electron increases as you go down the group.

Mashutka [201]

Answer:

its true

Explanation:

3 0

3 years ago