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spayn [35]
3 years ago
7

What is the average of 22. 10; 13. 45; 25. 10; 17. 20; 19. 10?.

SAT
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

9.595

Explanation:

The average will be the addition of the numbers and then dividing them by the total numbers that we've. This will be:

= (22.10 + 13.45 + 25.10 + 17.20 + 19.10) / 10

= 96.95 / 10

= 9.695

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For 2-methylbutane, the ∆h° of vaporization is 25. 22 kj/mol and the ∆s° of vaporization is 84. 48 j/mol・k. At 1. 00 atm and 201
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The Gibbs free energy ΔG for the vaporisation of 2-methylbutane, in KJ/mol is 8.24 KJ/mol

<h3>Data obtained from the question </h3>
  • Enthalpy change (ΔH) = 25. 22 KJ/mol
  • Change in entropy (ΔS) = 84.48 J/Kmol = 84.48 / 1000 = 8.448×10¯² KJ/Kmol
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<h3>How to determine the Gibbs free energy </h3>

The Gibbs free energy for the reaction can be obtained as follow:

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ΔG = 25.22 – 169.8048

ΔG = 8.24 KJ/mol

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