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Anestetic [448]
3 years ago
10

All single-displacement reactions can be classified as another type of reaction as well. What type of reaction is that?

Chemistry
2 answers:
user100 [1]3 years ago
8 0

combustion because one atom in wrong place will make it combust

Lyrx [107]3 years ago
5 0

Answer:

The single displacement reaction can be classified as <u><em>redox</em></u>

Explanation:

A simple substitution reaction or simple displacement reaction, called single-displacement reaction, is a reaction in which an element of a compound is substituted by another element involved in the reaction. The starting materials are always pure elements and an aqueous compound. And a new pure aqueous compound and a different pure element are generated as products. The general form of a simple substitution reaction is:

AB + C ⇒ A +BC

where C and A are pure elements; C replaces A within compound AB to form a new co, placed CB and elementary A.

A single-displacement reaction is characterized by a place of exchange of cations or anions with another to form a new product. A begins as an ion in the solution and appears elementally in the reaction products. In case of reagent C it starts in its elemental form and ends as an ion in an aqueous state forming compound BC.

On the other hand, oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The redox reaction is a single-displacement reaction.  A chemical substance goes from being an ion to a pure element. The oxidation state of a monoatomic ion in solution is equal to its charge, while the oxidation state of a pure element is equal to zero. Any element that goes from being an ion to a pure element or vice versa must be changing oxidation state.

A displaces C, which can be the H₂ molecule or another metal. If C is formed by the reduction of ions, then species A, by the conservation of mass and energy and must provide electrons: it must oxidize.

On the other hand, if A and C are metallic species, but C is in the ionic form (M+) and A in its natural state, then the displacement reaction will occur only if A is more active than C, forcing the latter to Accept the electrons to reduce to their metallic state (M).

So finally, <em><u>the single displacement reaction can be classified as redox</u></em>.

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Generally how does the first ionization energy vary as the atomic number increases going across a period
Vlad1618 [11]

Across a period I.E increases progressively from left to right

Explanation:

The trend of the first ionization energy is such that across a period I.E increases from left to right due to the decreasing  atomic radii caused by the increasing nuclear charge. This not compensated for by successive electronic shells.

  • Ionization energy is a measure of the readiness of an atom to lose an electron.
  • The lower the value, the easier it is for an atom to lose an electron.
  • Elements in group I tend to lose their electrons more readily whereas the halogens hold most tightly to them.
  • The first ionization energy is the energy needed to remove the most loosely bonded electron of an atom in the gaseous phase.

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3 0
3 years ago
Magnetite is a binary compound containing only iron and oxygen. the percent, by weight, of iron is 72.360%. what is the empirica
hram777 [196]
Answer is: <span>c. Fe</span>₃<span>O</span>₄<span>.
</span>ω(Fe) = 72,360%.
ω(O) = 100% - 72,36% = 27,64%.
For example, if we the mass of compound is 100 g:
m(Fe) = 72,36 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 72,36 g ÷ 55,85 g/mol.
n(Fe) = 1,296 mol.
n(O) = 27,64 g ÷ 16 g/mol.
n(O) = 1,727 mol.
n(Fe) : n(O) = 1,296 mol : 1,727 mol.
n(Fe) : n(O) = 1 : 1,33 or 3 : 4.
7 0
3 years ago
Read 2 more answers
Convert each quantity to the indicated units. a. 3.01g to cg. b. 6200m to km
miss Akunina [59]

a. 301 cg

b. 6.2 km

Explanation:

a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:

if        1 g is equal to 100 cg

then  3.01 g are equal to X cg

X = (3.01 × 100) / 1 = 301 cg

b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:

if         1 km is equal to 1000 m

then   Y km are equal to 6200 m

Y = (6200 × 1) / 1000 = 6.2 km

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converting units of measurement

brainly.com/question/11300981

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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.

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7 0
2 years ago
A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec
Delicious77 [7]

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

3 0
3 years ago
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