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Anon25 [30]
2 years ago
6

Calculate the speed of light in an unknown substance whose index of refraction is 1.65. Would you expect the light to bend towar

d the normal or away from the normal when it passes from air into the substance?
Chemistry
1 answer:
Anna71 [15]2 years ago
8 0

(a) The speed of light in the unknown substance is determined 1.82 x 10⁸ m/s.

(b) The light will bend away from the normal since speed of light in air is not equal to speed of light in the substance.

<h3>What is the speed of light?</h3>

The speed of light passing from air into the substance is calculated as follows;

refractive index = speed of light in air / speed of light in the substance

speed of light in the substance = speed of light in air/refractive index

speed of light in the substance = (3 x 10⁸) / (1.65)

speed of light in the substance = 1.82 x 10⁸ m/s

Thus, the light will bend away from the normal since speed of light in air is not equal to speed of light in the substance.

Learn more about speed of light here: brainly.com/question/104425

#SPJ1

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Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl
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∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

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

  • Calculating ∆H°rxn:

∵ ∆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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