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vova2212 [387]
2 years ago
14

Name the following aromatic hydrocarbon:

Chemistry
1 answer:
Tresset [83]2 years ago
8 0

3-Methylpentane is the IUPAC name for the substance.

whether in a continuous chain or a ring, the longest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature. According to a precise set of priorities, all deviations—whether they involve numerous bonds or atoms other than carbon and hydrogen—are denoted by prefixes or suffixes.

+3-Methylpentane is the IUPAC name for the substance in question. It has a lengthy chain of 5 carbon atoms, which gives it the prefix pent-, and a single bond is what gives it the postfix -ane (alkane). Given that the methyl group is present at the third carbon, it is 3-methylpentane.

Learn more about IUPAC Nomenclature here-

brainly.com/question/14379357

#SPJ9

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If 21 mL f gas is subjected to a temperature change from 10.0C to 120C and a pressure change from 1.0 atm to 15 atm, the new vol
SCORPION-xisa [38]

Answer:

V₂ = 1.41 mL

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 21 mL

Initial pressure = 1 atm

Initial temperature = 10 °C (10 +273 = 283 K)

Final temperature = 12 °C (12 +273 = 285 K)

Final volume = ?

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Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/T₁ P₂

V₂ = 1 atm × 21 mL × 285 K / 283 K × 15 atm  

V₂ = 5985 atm .mL. K / 4245 K. atm

V₂ = 1.41 mL

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