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NeTakaya
1 year ago
10

A student measured out 0.333 moles of theobromine, CyH3N402 (MM = 180.8/mol) the

Chemistry
1 answer:
Verdich [7]1 year ago
8 0

Answer

2.0 x 10²³ molecules.

Explanation

Given:

The number of moles of theobromide measured out = 0.333 moles.

MM of theobromide = 180.8 g/mol

What to find:

The number of molecules of theobromide the student measured.

To go from moles to molecules, multiply the number of moles by Avogadro's number.

The Avogadro's number = 6.02 x 10²³

1 mole of theobromide contains 6.02 x 10²³ molecules.

So, 0.333 moles of theobromide measured out will have (0.333 x 6.02 x 10²³) = 2.0 x 10²³ molecules.

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∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

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∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

  • Calculating ∆S°rxn:

∵  ∆S°rxn = ∑∆S°products - ∑∆S°reactants

<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>

<em></em>

  • Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>

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