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loris [4]
3 years ago
11

100 mL of a strong acid is completely neutralized by 100 mL of a strong base. The observed products are salt and water. Upon fur

ther investigation, the solution has a pH > 7. How is this possible?
Chemistry
1 answer:
ser-zykov [4K]3 years ago
3 0

Answer: The concentration of base is higher than the concentration of the acid.

Explanation:

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A student designed an experiment to test the effect on the magnitude of the magnetic field that is generated around the wire loo
kykrilka [37]

Answer: C ) the student’s dependent variable is the magnitude of the magnetic field that is generated

E ) the student’s independent variable is the amount of current that is being passed through the wire

5 0
3 years ago
Which of the following tool components is unlikely to present a hazard
nydimaria [60]

Gloves is the tool component which will unlikely present a hazard in this scenario.

<h3>What is Hazard?</h3>

This is referred to as potential source of harm when performing various types of activities.

Gloves will unlikely cause harm as it is used to prevent the risk of electric shock when handling electrical devices which is why it is the most appropriate choice.

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6 0
2 years ago
A client has experienced excessive losses of bicarbonate and has subsequently developed an acid–base imbalance. How will this lo
IRISSAK [1]

The kidneys will excrete increased quantities of acid.

Explanation:

The kidneys will excrete excess H+ ions in the blood (remember H+ ions are responsible for acidity) until the acid-base balance is restored in the blood. Bicarbonates, on the other hand, will be aggressively  reabsorbed by the renal tubules as the excess H+ are being excreted.

The acid base balance is mainly determined by the quantities of H⁺ and HCO₃⁻ ions in teh blood. These ions come from the dissociation of carbonic acid formed when carbon dioxide from tissues is dissolved in blood plasma.

6 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.

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7 0
2 years ago
Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.
lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

<u>E = 2.18 x 10^-18 J</u>

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

<u>E = 13.6 eV</u>

5 0
4 years ago
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