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hammer [34]
3 years ago
8

Consider the Bohr model of the simplest atom, hydrogen. What energy photon will be released if an electron falls from the n= 2 o

rbit (excited state) to the n0 = 1 orbit (ground state)?
Chemistry
1 answer:
Nana76 [90]3 years ago
7 0
<h2>Ultraviolet Light</h2>

Explanation:

  • The energy of a photon that will be released if an electron falls from the n= 2 orbit (excited state) to the n0 = 1 orbit (ground state) is of ultraviolet light.
  • In the ultraviolet part of the spectrum, a photon having an energy of 10.2 eV has a wavelength of 1.21 x 10-7 m.
  • Hence, when an electron wants to jump or it gets excited from the first level to the second level that is from n = 1 orbit to n = 2 orbits, it must absorb a photon of ultraviolet light.
  • But,When an electron falls from n = 2 orbit to n = 1 orbit  or from n = 2 orbit(excited state) to n = 0 orbit(groubd state), it emits a photon of ultraviolet light.
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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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4. In the following, how many digits should be in the solution to have the proper number of sig figs?
yan [13]

Answer:

i guess b maybe the answer

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