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Andreyy89
3 years ago
13

N2 molecules absorb ultraviolet light but not visible light. I2 molecules absorb both visible and ultraviolet light. Which of th

e following statements explains the observations?
More energy is required to make N2 molecules vibrate than is required to make I2 molecules vibrate.
A

More energy is required to remove an electron from an I2 molecule than is required to remove an electron from a N2 molecule.
B

Visible light does not produce transitions between electronic energy levels in the N2 molecule but does produce transitions in the I2 molecule.
C

The molecular mass of I2 is greater than the molecular mass of N2.
Chemistry
1 answer:
Umnica [9.8K]3 years ago
4 0

C. Visible light does not produce transitions between electronic energy levels in the N₂ molecule but does produce transitions in the I₂ molecule.

Explanation:

Abortion of light by molecules will produce electronic transitions from a ground level to a higher level equal to the energy of absorbed light.

In the case of nitrogen (N₂) the allowed electronic transitions are between electronic energy levels with a energy difference equal to the energy of photons of the ultraviolet light. Nitrogen will not absorb from the visible range so it is colorless.

Now iodine (I₂) have the allowed electronic transitions between electronic energy levels with a energy difference equal to the energy of photons of the visible light. As a consequence Iodine vapors have a violet color.

Learn more about:

electronic transitions

brainly.com/question/11328705

#learnwithBrainly

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Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)

The expression used will be:  

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]

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c_{p,s} = specific heat of solid benzene = 1.51J/g^oC=1.51J/g.K

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC=1.73J/g.K

\Delta H_{fusion} = enthalpy change for fusion = -9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g

Now put all the given values in the above expression, we get:

Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]

Q=-29427.312J=-29.4kJ

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

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3 years ago
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