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valentinak56 [21]
3 years ago
5

What did Bohr conclude about the atom after observing emission spectrum lines a. Electrons can exist in any energy state but hav

e specific states that are preferred. b. No two electrons can exist in the same quantum state within an atom. c. Possible electron energy states are quantized within an atom. d. Plum pudding tastes fantastic.
Chemistry
1 answer:
cupoosta [38]3 years ago
7 0

Answer:

Option c: Possible electron energy states are quantized within an atom.

Explanation:

The Bohr's Model of the hydrogen atom consisted of the movements of the electrons around the positively-charged nucleus in circular orbits that have a certain energy state. The energy of that orbit is given by:  

E_{n} = - \frac{Rhc}{n^{2}}

<em>Where:</em>

E(n): is the energy of an electron in a particular orbit

R: is the Rydberg constant

h: is the Plank constant

c: is the speed of light

n: is a positive integer which corresponds to the number of the orbit

The ground state energy of a electron in the hydrogen atom is equal to -13,6 eV.

Bohr's Model aims to propose that the electron is restrictedly to occupy a certain region in the atom.    

Therefore, the conclusion of Bohr after observing emission spectrum lines is that "possible electron energy states are quantized within an atom", so the correct option is c.

I hope it helps you!

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An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6
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Answer :

(a) The concentration of Pb^{2+} is, 0.0337 M

(b) The concentration of Pb^{2+} is, 6.093\times 10^{32}M

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Pb\rightarrow Pb^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

The balanced cell reaction will be,  

Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}

E^o=0.34V-(-0.13V)=0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

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

0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}

[Pb^{2+}]=0.0337M

Therefore, the concentration of Pb^{2+} is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction:  Pb^{2+}+2e^-\rightarrow Pb

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}

E^o=-0.13V-(0.34V)=-0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

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

0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}

[Pb^{2+}]=6.093\times 10^{32}M

Therefore, the concentration of Pb^{2+} is, 6.093\times 10^{32}M

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Số phân tử NaOH : Số phân tử Na2SO4 = 2:1

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Explanation:

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Answer:

293.1 mL.

Explanation:

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P₂ = 652.0 mm Hg, V₂ = ??? mL.

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3 years ago
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