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Talja [164]
3 years ago
12

BRAINLIEST!!!!!!!!!

Chemistry
1 answer:
Serga [27]3 years ago
4 0

Answer:

its 5

Explanation:

because it is

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2.56 g of hydrogen reacts completely with 20.32 g of oxygen<br> to form X g of water. X = g
Brilliant_brown [7]

Answer:

Mass of water produced is 22.86 g.

Explanation:

Given data:

Mass of hydrogen = 2.56 g

Mass of oxygen = 20.32 g

Mass of water = ?

Solution:

Chemical equation:

2H₂ + O₂   →  2H₂O

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 20.32 g/ 32 g/mol

Number of moles = 0.635 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 2.56 g/ 2 g/mol

Number of moles = 1.28 mol

Now we will compare the moles of water with oxygen and hydrogen.

                    O₂            :            H₂O

                     1              :             2

                  0.635        ;            2×0.635 =  1.27

                   H₂             :              H₂O

                    2              :              2

                 1.28            :           1.28

The number of  moles of water produced by oxygen are less thus it will be limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 1.27 × 18 g/mol

Mass = 22.86 g

 

4 0
4 years ago
Read 2 more answers
Select all statements that correctly describe hemoglobin and myoglobin structure. a. Molecular oxygen binds irreversibly to the
DIA [1.3K]

Answer:

c. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron.

e. Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron ( Fe ) (Fe) atom.

f. Hemoglobin is a heterotetramer, whereas myoglobin is a monomer. The heme prosthetic group is entirely buried within myoglobin.

Explanation:

The differences between hemoglobin and myoglobin are most important at the level of quaternary structure. Hemoglobin is a tetramer composed of two each of two types of closely related subunits, alpha and beta. Myoglobin is a monomer (so it doesn't have a quaternary structure at all). Myoglobin binds oxygen more tightly than does hemoglobin. This difference in binding energy reflects the movement of oxygen from the bloodstream to the cells, from hemoglobin to myoglobin.

Myoglobin binds oxygen

The binding of O 2 to myoglobin is a simple equilibrium reaction:

7 0
3 years ago
Which substance produces hydroxide ions in solution?
Olegator [25]

Answer:

An Arrhenius Base

Explanation:

The definition of this is a base that is a hydroxide ion donor.

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
The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.
scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg


Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0
4 years ago
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