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Xelga [282]
3 years ago
9

How many Liters are there in 51. 300 cm? (11 = 1000 cm)

Chemistry
1 answer:
Black_prince [1.1K]3 years ago
7 0

Answer:

4,572.67

Explanation:

11 divided by 1000= 90.90

44= 11x

x=4

4 x 11= 4000

44-51.300=6.300

6.300 x 90.90=572.67

4000 + 572.67 = the answer

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Calculate the volume in mL of a 0.708 M KOH solution containing 0.098 mol of solute
Olenka [21]
Molarity = mol/liter

0.708M = 0.098mol/L
Rearrange to find L:
0.098mol/0.708M = .138L

For every liter there is 1000 mL:
.138L • 1000mL =138mL KOH
7 0
4 years ago
Archer is a soil scientist and needs to examine a bacteria sample. He wants to be able to change his magnification with differen
koban [17]

Answer:

C. Dissecting microscope

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3 years ago
Neutralizing an olympic size swimming pool is conceptually very similar to performaing a massive titration experiment. Suppose a
MrRissso [65]

Answer:

6,97x10⁻³ gallons

Explanation:

pH is defined as:

pH = -log [H⁺]

Thus, you need to have, in the end:

10⁻⁷ = [H⁺]

And you have, in the first:

10^{-9,33} = [H⁺]

The volume of swimming pool is:

700'000 galllons ×\frac{3,78541 L}{1 gallon} = 2649787 L

Thus, the moles of H⁺ in the first and in the end are:

First:

10^{-9,33}mol/L × 2649787L = 1,24x10⁻³ moles

End:

10^{-7}mol/L × 2649787L = 0,265 moles

Thus, the moles of H⁺ you need to add are:

0,265 - 1,24x10⁻³ = <em>0,26376 moles</em>

These moles comes from 10M HCl, thus, the volume in gallons you need to add are:

0,26376moles*\frac{1L}{10moles}* \frac{1gallon}{3,78541L} =

<em>6,97x10⁻³ gallons</em>

<em></em>

I hope it helps!

7 0
4 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

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7 0
2 years ago
Why calcium oxide and magnesium oxide used as soil treatment ​
BartSMP [9]

Answer:

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oyster shells, and chalk) containing from 95 to 99 percent

calcium carbonate (CaCO^). Dolomitic limes are produced from

dolomitic limestone or dolomite which contains from 30 to ^+0

percent magnesium carbonate (MgCO^), the rest being calcium

carbonate.

At atmospheric pressure, calcite in limestone decomposes

at approximately 900°C to form CaO and COg. The decomposition of dolomite, CaMg is a two-stage process. At

temperatures between 650°C to 750°C dolomite decomposes to

form MgO, CO^ and CaCO^. It is necessary to raise the temperature to 900°C to decompose the CaCO^ (15, 35)» This

phenomenon is extremely important, as is shown later.

Various investigators have studied the effects of stone

size, temperature, and time of calcination of commercial

Explanation:

4 0
3 years ago
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