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irina1246 [14]
3 years ago
8

What type of light is emitted by electrons falling to the ground state?

Chemistry
1 answer:
Solnce55 [7]3 years ago
3 0
Solar energy is the right awnser

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Which of the following is true for a chemical reaction at equilibrium?
victus00 [196]

Answer:

youre gonna have to include the answers for me to help

Explanation:

7 0
2 years ago
Dinitrogen pentoxide is used in the preparation of explosives. If 7.93 mol of
Tpy6a [65]

The volume of O₂ produced: 84.6 L

<h3>Further explanation</h3>

Given

7.93 mol of  dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

Volume of O₂

Solution

Decomposition reaction of dinitrogen pentoxide

2N₂O₅(g)→4NO₂(g)+O₂ (g)

From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

\tt V=\dfrac{nRT}{P}\\\\V=\dfrac{3.965\times 0.082\times 321}{1.23365}\\\\V=84.6~L

5 0
3 years ago
A certain orbital of the hydrogen atom has n = 4 and l = 2 what are the possible values of ml for this orbital?
Nadya [2.5K]

The ml is also called as the magnetic quantum number. The value of ml can range from –l to +l including zero. Hence all of the possible values for ml given that l = 2 are:

<span>-2, -1, 0, + 1, + 2</span>

3 0
3 years ago
2. Why can we eliminate the claim you circled?
Luda [366]

Answer: theres no image or claim

7 0
3 years ago
Read 2 more answers
When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot
viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the 'n^{th}' orbit of hydrogen atom.

Formula used :

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number  of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV

Energy of n = 2 in an hydrogen atom:

E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0
3 years ago
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