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melisa1 [442]
2 years ago
7

Which of the following statements is true about atoms?

Chemistry
1 answer:
Leokris [45]2 years ago
6 0

Atoms of the same element may have different mass numbers.

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What changes are observed on heat capacity ratio and output temperatures by increasing the specific heat capacity of both the fl
BlackZzzverrR [31]

Answer:

b- The heat capacity ratio increases but output temperature don’t change

Explanation:

The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.

Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.

On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.

8 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.

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

#SPJ4

7 0
2 years ago
E the table below for the next three questions.
Sliva [168]

Answer:

Explanation:i would go wit d

6 0
3 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0°c to water at 70.0°c is __________ kj. the specific heats of ice, wa
kogti [31]
First, calculate for the amount of heat used up for increasing the temperature of ice.

      H = mcpdT
       H = (18 g)*(2.09 J/g-K)(50 K) = 1881 J

Then, solve for the heat needed to convert the phase of water.
    H = (1 mol)(6.01 kJ/mol) = 6.01 kJ = 6010 J

Then, solve for the heat needed to increase again the temperature of water.
    H = (18 g)(4.18 J/gK)(70 k)
    H = 5266.8 J

The total value is equal to 13157.8 J

Answer: 13157.8 J
8 0
3 years ago
A solution consists of 35.00 g of CuSO4 dissolved in 250.0 mL of water. The molar mass of Cu is 63.55 g/mol the molar mass of S
slamgirl [31]

The molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.

<h3>How to calculate molarity?</h3>

The molarity of a solution can be calculated using the following formula:

Molarity = no of moles/volume

According to this question, a solution consists of 35.00 g of CuSO4 dissolved in 250.0 mL of water.

no.of moles of CuSO4 = 35g ÷ 159.6g/mol

no. of moles of CuSO4 = 0.22 moles

Therefore; molarity of CuSO4 solution is calculated as follows:

M = 0.22 ÷ 0.25

M = 0.88M

Therefore, the molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.

Learn more about molarity at: brainly.com/question/12127540

6 0
2 years ago
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