Which of the following is a negative effect of a volcano?

What would be the correct order of least(floats) to most(sinks) density?*

olya-2409 [2.1K]

Answer:

B

Explanation:

Water-Continental-Oceanic-Mantle

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.

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

#SPJ4

7 0

2 years ago

Can someone plz help me???? will mark as brainliest!!!!!!!!!!!

Nookie1986 [14]

1=c 2=a 3=b

i hope these answer your questions

7 0

3 years ago

Complete the reaction, which is part of the electron transport chain. The abbreviation FMN represents flavin mononucleotide. Use

Margarita [4]

Answer:

1. NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂

2. The reactant that is reduced is Q

3. The charge on iron on the right side is +2, Fe²⁺

Explanation:

NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂

The reaction above is catalysed by NADH:ubiquinone oxidoreductase (complex 1), which transfers a hydride ion from NADH to FMN, from which two electrons pass through a series of of Fe-S centers to the iron-sulfur protein N-2. Electron transfer from N-2 to Ubiquinone forms QH₂

The species in a reaction which gains hydrogen irons is reduced, Therefore, the reactant that is reduced is Q, ubiquinone to form QH₂, ubiquinol.

To determine the oxidation number of iron on the right side of the reaction below,

QH2 + 2cyt c ( Fe3+) ⟶ Q + 2cyt c(Fex) + 2H^+

Sum of charges on the left side = Sum of charges on the right side

Sum of charges on the left side = 2 *+3 = +6

Therefore 2 * x + 2= 6

2x = 6 -2 = 4

x = 4/

x = 2

Therefore the charge on iron on the right side is +2, Fe²⁺

4 0

3 years ago

A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to

melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0

3 years ago