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MArishka [77]
3 years ago
6

If 1.87 g‎ of acetic acid (ch3co2h)‎ reacts with 2.31 g‎ of isopentylalcohol (c5h12o)‎ to give‎ 3.20 g of isopentylacetate (c7h1

4o2)‎, what is the percent yield of the reaction?
Chemistry
1 answer:
Orlov [11]3 years ago
8 0

The reaction of acetic acid with isopentyl alcohol is

CH3COOH + C5H12O  ---> C7H14O2  + H2O

Thus one mole of acetic acid reacts with one mole of isopentyl alcohol to give one mole isopentyl acetate

Molar mass of acetic acid = 60.05 g / mole

Moles of acetic acid used = Mass used  / Molar mass = 1.87/ 60.05 = 0.0311

Molar mass of isopentyl alcohol = 88.15 g / mole

Moles of isopentyl alcohol = Mass / Molar mass = 2.31 / 88.15 = 0.0262

Molar mass of isopentyl acetate = 130.19 g / mole

Moles of isopentyl acetate = 3.20 / 130.19 = 0.0246

As per equation 0.0262  should of isopentyl alcohol should react with 0.0262 moles of acetic acid to give 0.0262 isopentyl acetate

Thus theoretical yield = 0.0262 moles

Actual yield = 0.0246

% yield = Actual yield X 100 / theoretical yield = 0.0246 X 100 / 0.0262 = 93.89%

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Answer:

glycerol-3-phosphate, ADP, H⁺

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The reaction of converting glycerol to glycerol-3-phosphate which makes is unfavorable and is coupled with the second reaction which involves conversion of ATP to ADP which is high energetically favorable.

Reaction 1:  Glycerol + HPO₄²⁻ ⇒ Glycerol-3-phosphate + water

Reaction 2:             ATP  + H₂O ⇒ ADP + HPO₄²⁻ + H⁺

The coupled reaction of both the reactions become favorable. Thus, the overall coupled reaction is:

<u>Glycerol + ATP ⇒ Glycerol-3-phosphate + ADP + H⁺</u>

The net products are = glycerol-3-phosphate, ADP, H⁺

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Which dosage form is a semisolid preparation that contains very small solid particles that are suspended in a liquid
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A gel is a semisolid preparation that contains a gelling agent which provides stiffness to the preparation.

The gelling agent can be, for example, agarose (this gelling agent is used to prepare gels in electrophoresis).

In an agarose gel, agarose molecules are organized into three-dimensional (3D) structures similar to pores, which allow the passage of DNA fragments during electrophoresis.

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2 years ago
Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,
torisob [31]

Answer : The full balanced ionic equation will be,

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, K^+,NO_3^-

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)

The ionic equation in separated aqueous solution will be,

2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)

In this equation, K^+\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)

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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.
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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.)

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