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prohojiy [21]
3 years ago
15

Which of the following explains why 1-pentyne has a slightly higher boiling point than 1-pentene? Group of answer choices The C-

C triple bond in 1-pentyne is more polar than the C-C double bond in 1-pentene. 1-Pentyne has more carbons than 1-pentene. 1-Pentyne has more carbons per hydrogen than 1-pentene. 1-Pentyne is linear while 1-pentene is trigonal planar.
Chemistry
1 answer:
nydimaria [60]3 years ago
8 0

Answer:

Correct options: (1) The C-C triple bond in 1-pentyne is more polar than the C-C double bond in 1-pentene and (2) 1-Pentyne is linear while 1-pentene is trigonal planar.

Explanation:

  • 1-pentyne has linear structure due to presence of triple bond. Therefore, 1-pentyne molecule cam come closer to each other as compared to 1-pentene (trigonal planar) and exert more van der waal intermolecular force.
  • Presence of more number of pi-electrons in triple bond makes 1-pentyne more polarizable. Hence van der waal force is more stronger in 1-pentyne as compared to 1-pentene.
  • Van der waal force is not so strong as compared to dipole-dipole force or H-bonding.
  • Therefore 1-pentyne has slightly higher boiling point than 1-pentene.

Correct options: (1) The C-C triple bond in 1-pentyne is more polar than the C-C double bond in 1-pentene and (2) 1-Pentyne is linear while 1-pentene is trigonal planar

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Answer:

Balance molecular equation:

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Balance chemical equation with physical states:

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Net ionic equation:

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The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re
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<u>Answer:</u> The temperature of the system will be 1622 K

<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

\log D=\log D_o-\frac{E_a}{2.303RT}

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D = diffusion coefficient = 2.7\times 10^{-14}m^2/s

D_o = pre-exponential constant = 1.7\times 10^{-5}m^2/s

E_a = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K

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