Consider the compound whose condensed structural formula is shown below: CH3CH2COCH2CH2CH3 This compound ______ O can be reduced to form 3-hexene O can undergo only oxidation and reduction reactions O is named as 3-hexanal O can be formed by the oxidation of 3-hexanol

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Answer 1

The compound CH3CH2COCH2CH2CH3 is named as 3-hexanone.

The given condensed structural formula represents a ketone called 3-hexanone. In the IUPAC nomenclature, the position of the carbonyl group (C=O) is indicated by the lowest possible number, which in this case is 3. The carbon chain consists of six carbon atoms (hexane) with a carbonyl group attached to the third carbon atom.

The compound 3-hexanone can undergo both oxidation and reduction reactions. As a ketone, it contains a carbonyl group that can be oxidized to form a carboxylic acid or reduced to form an alcohol. Therefore, it is not limited to only oxidation or reduction reactions.

The statement that the compound can be reduced to form 3-hexene is incorrect. Reduction of 3-hexanone would yield 3-hexanol, not 3-hexene. Reduction adds hydrogen to the carbonyl group, converting it to a hydroxyl group (-OH).

The compound with the condensed structural formula CH3CH2COCH2CH2CH3 is named 3-hexanone. It is a ketone that can undergo both oxidation and reduction reactions. However, it cannot be directly reduced to form 3-hexene; instead, reduction of 3-hexanone would yield 3-hexanol. The compound can be formed by the oxidation of 3-hexanol, where the alcohol group is converted to a carbonyl group.

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Related Questions

the hybridization of the central atom in the xef4 molecule is __________.

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The hybridization of the central atom in the XeF4 (xenon tetrafluoride) molecule is sp3d2.

In XeF4, xenon (Xe) is the central atom, and it has six electron pairs around it. The electron configuration of xenon is [Kr]5s^24d^105p^6. To form bonds, xenon promotes two of its electrons from the 5s and one electron from the 5p orbitals to the empty 5d orbitals, resulting in the electron configuration [Kr]5s^24d^105p^4. The formation of four covalent bonds with fluorine requires four orbitals, so xenon hybridizes its 5s, 5p, and 5d orbitals to form six sp3d2 hybrid orbitals. These hybrid orbitals are directed towards the corners of an octahedron, with four of them participating in sigma bonds with fluorine atoms and the other two containing lone pairs. Overall, the hybridization of the central xenon atom in XeF4 is sp3d2, indicating the involvement of five atomic orbitals in the hybridization process.

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Choose reagents from the table for conversion of 1-butanol to the following substances. Use letters from the table to list reagents in the order used (first at the left). Example: ab Reagents a. NaN3 c. CrO3/H3O+ c. Dess-Martin peropdinane in CH2Cl2 d. Butylamine e. excess NH3
f. SOCl2 g. PBr3 h. Br2/NaOH, H2O i. LiAlH4 H2O j. H2/Ni, i-PrNH2 k. NaBH3CN, (CH3)2NH l. Ag2O, H2O, heat m. NaCN n. H2O, heat o. excess CH3l a) pentlylamine: b) dibutylamine:

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To convert 1-butanol to pentlylamine, the reagents used are [tex]NaN_{3}[/tex], [tex]H_{2} O[/tex], and [tex]NH_{3}[/tex] (in excess), while to convert 1-butanol to dibutyl amine, the reagents used are [tex]SOCl_{2}[/tex] and Butylamine.

To convert 1-butanol to pentlylamine, the reagents would be:

a) [tex]NaN_{3}[/tex](Sodium azide) - to perform azide substitution

b) [tex]H_{2} O[/tex]- for hydrolysis of the azide group

c) [tex]NH_{3}[/tex](Ammonia) in excess - to carry out reductive amination

Therefore, the reagents used in the conversion of 1-butanol to pentlylamine would be a) [tex]NaN_{3}[/tex], b) [tex]H_{2} O[/tex], and c) [tex]NH_{3}[/tex](in excess).

To convert 1-butanol to dibutyl amine, the reagents would be:

a) [tex]SOCl_{2}[/tex](Thionyl chloride) - to perform a nucleophilic substitution

b) Butylamine - to react with the chloride group

Therefore, the reagents used in the conversion of 1-butanol to dibutyl amine would be: a) [tex]SOCl_{2}[/tex]and b) Butylamine.

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Calculate the amount of energy needed to change 441 g of water ice at -10 degree Celsius to steam at 125 degree Celsius. The following constants may be useful:
Cm (ice)=36.57 J/(mol⋅∘C)
Cm (water)=75.40 J/(mol⋅∘C)
Cm (steam)=36.04 J/(mol⋅∘C)
ΔHfus=+6.01 kJ/mol
ΔHvap=+40.67 kJ/mol
Express your answer with the appropriate units.

Answers

Therefore, the amount of energy required to change 441 g of water ice at -10 degree Celsius to steam at 125 degree Celsius is 18.1 MJ.

The given problem is about calculating the energy needed to change 441 g of water ice at -10 degree Celsius to steam at 125 degree Celsius. The following constants may be useful:Cm (ice)=36.57 J/(mol⋅∘C)Cm (water)=75.40 J/(mol⋅∘C)Cm (steam)=36.04 J/(mol⋅∘C)ΔHfus=+6.01 kJ/molΔHvap=+40.67 kJ/molThe specific heat capacity of ice: Cm (ice) = 36.57 J/(mol °C).The ice needs to be heated from -10°C to 0°C before it can be melted. The energy required will be:ΔH = Cm (ice) * mass * ΔTΔH = 36.57 * 441 * 10 = 161617.7 JThe energy required to melt ice at 0°C is given by the latent heat of fusion: ΔHfus = 6.01 kJ/mol ΔHfus = 6010 J / molAmount of energy needed to melt 441 g of ice = (ΔHfus / Molar mass) * massAmount of energy needed to melt 441 g of ice = (6010 / 18) * 441 = 1,986,850 JThe energy required to heat the water from 0°C to 100°C will be:ΔH = Cm (water) * mass * ΔTΔH = 75.40 * 441 * 100 = 3,313,440 JThe energy required to boil the water to steam is given by the latent heat of vaporization: ΔHvap = 40.67 kJ/mol ΔHvap = 40,670 J / molAmount of energy needed to boil 441 g of water = (ΔHvap / Molar mass) * massAmount of energy needed to boil 441 g of water = (40670 / 18) * 441 = 10,270,850 JThe energy required to heat the steam from 100°C to 125°C will be:ΔH = Cm (steam) * mass * ΔTΔH = 36.04 * 441 * 25 = 399,366 JTherefore, the total amount of energy needed to change 441 g of water ice at -10°C to steam at 125°C is:ΔHtotal = ΔH1 + ΔH2 + ΔH3 + ΔH4ΔHtotal = 161617.7 + 1986850 + 3313440 + 10270850 + 399366ΔHtotal = 18,081,123.7 J or 18.1 MJ.

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calculate the hydroxide ion concentration in an aqueous solution that contains 3.50 × 10-3 m in hydronium ion.

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The hydroxide ion concentration in the given aqueous solution is 2.86 × 10-12 M.

The given aqueous solution has a hydronium ion concentration of 3.50 × 10-3 M. To calculate the hydroxide ion concentration, the following steps need to be followed:

Step 1: Write the balanced chemical equation for the dissociation of water:

H2O(l) ⇌ H+(aq) + OH-(aq)

Step 2: Write the expression for the equilibrium constant for this reaction:

Kw = [H+(aq)][OH-(aq)]

Step 3: Substitute the value of

Kw (1.0 × 10-14 M2 at 25°C) and the given hydronium ion concentration (3.50 × 10-3 M) in the expression to solve for hydroxide ion concentration:

[OH-(aq)] = Kw/[H+(aq)] = (1.0 × 10-14 M2) ÷ (3.50 × 10-3 M) = 2.86 × 10-12 M

Therefore, the hydroxide ion concentration in the given aqueous solution is 2.86 × 10-12 M.

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