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scoray [572]
3 years ago
14

A hydrated compound has analysis of 18.29% Ca, 32.37%CI, and49.34%H2O. What's its formula?

Chemistry
1 answer:
nata0808 [166]3 years ago
7 0

Answer:

CaCl₂.6H₂O

Explanation:

  • If we suppose that the compound has a mass of 100.0 g, it will contain 18.29 grams of Ca, 32.37 grams of Cl, and 49.34 grams of water.
  • We can convert the grams of each element to moles by dividing the number of grams by the atomic mass of each element.

No. of moles of Ca = (18.29 g) / (40.0 g/mol) = 0.457 mol.

No. of moles of Cl = (32.37 g) / (35.4 g/mol) = 0.914 mol.

No. of moles of H₂O = (49.34 g) / (18.0 g/mol) = 2.74 mol.

  • The no. of moles of (Ca: Cl: H₂O) is (0.457 mol: 0.914 mol: 2.74 mol).
  • To obtain the ratio of the elements in the compound, we divide over the lowest no. of moles (0.475).

So the mole ratio of (Ca: Cl: H₂O) is (1: 2: 6).

So, the formula of the compound is CaCl₂.6H₂O.

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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

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7 0
2 years ago
This is an example of a mixture that will separate______.
devlian [24]

B is the correct answer

6 0
3 years ago
_______ is the formation of alcohol from sugar.
Katarina [22]
The best and most correct answer among the choices provided by the question is the fourth choice "alcoholic fermentation"

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I hope my answer has come to your help. God bless and have a nice day ahead!
7 0
3 years ago
Read 2 more answers
Actual value= -273 Experimental value= -274
Maurinko [17]

Answer:

<h2><u><em>When measuring data, the result often varies from the true value. Error can arise due to many different reasons that are often related to human error, but can also be due to estimations and limitations of devices used in measurement. Regardless, in cases such as these, it can be valuable to calculate the percentage error. The computation of percentage error involves the use of the absolute error, which is simply the difference between the observed and the true value.</em></u></h2>

Explanation:

4 0
3 years ago
Magnesium has three naturally occurring isotopes(Mg-24, Mg-25, and Mg-26). The atomic mass and natural abundance of Mg-24 are 23
xxMikexx [17]

Answer:

The answer to your question is below

Explanation:

Data

               Atomic mass               Abundance

Mg -24 =    23.9850                         79%

Mg -25 =    24.9858                         10%  

Mg - 26=          ?                                  ?

Process

1.- Find the abundance of Mg-26

The sum of the abundance of the three isotopes equals 100%

         

 Abundance Mg-24 + Abundance-Mg-25 + Abundance-26 = 100

            79 + 10 + Abundance-26 = 100

            Abundance Mg-26 = 100 -89

           Abundance Mg-26 = 11%

2.- Find the atomic mass of Mg-26, the average atomic mass is 24.31 amu

Average atomic mass = (abundance)(atomic mass- 24) + (abundance)

                                         (atomic mass-25) + (abundance)(atomic mass-26)

    24.31 = (0.79)(23.9850)+ (0.10)(24.9858) + (Atomic mass-26)(0.11)

    24.31 = 18.94815 + 2.4986 + 0.11(atomic mass-26)

    24.31 = 21.4342 + 0.11(atomic mass- 26)

    0.11(atomic mass-26) = 24.31 - 21.4342

    0.11(atomic mass-26) = 2.8758

    Atomic mass-26 = 2.8758/0.11

 <u>   Atomic mass-26 = 26.1436 amu</u>

8 0
3 years ago
Read 2 more answers
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