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never [62]
2 years ago
6

What happens when a electron moves from the first energy level to the second??

Chemistry
2 answers:
Kitty [74]2 years ago
6 0
The atom absorbs or emits light in discrete packets called photons, and each photon has a definite energy<span>. Only a photon with an </span>energy<span> of exactly 10.2 eV can be absorbed or emitted when the </span>electron <span>jumps between the n = 1 and n = 2 </span>energy levels.
joja [24]2 years ago
5 0
The atom absorb the light in the discrete packet in the photons
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What element does not follow the horizontal and verticals trend?
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Nitrogen, Oxygen, and Fluorine do not follow this trend. The noble gas electron configuration will be close to zero because they will not easily gain electrons.
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Substance A is mixed with water and donates 0.4% of its H+ ions. Which of the following BEST describes Substance A? (See picture
mariarad [96]

Answer:

Explanation:

honesty i just know it has to be b or d because bases don’t donate. acids donate.

7 0
1 year ago
If nitrogen and hydrogen combine in a synthesis reaction, what would the product of the reaction be? n2 + 3h2 → 3
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N2 + 3H2 ---> 2NH3
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1 year ago
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A laser emits light with a frequency of 4.69 x 10 to the 14th power s - 1 calculate the wavelength of this light.
jekas [21]

Answer:

6.40x10^-7

Explanation:

answer with work is attached.

5 0
1 year ago
4. When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II)
loris [4]

Answer:

Lead chromate form = 0.97 g

Explanation:

The Balanced chemical equation of the given reaction is as follows

              K₂CrO₄ + Pb(NO₃)₂ → PbCrO₄ + 2 KNO₃

<u>Given that</u>

        Mass of Pb(NO₃)₂ = 1 g

                      Mole=\frac{Mass}{Molar mass} = \frac{1}{331.2}= 0.003

Mole of 25 ml of 1.00 (M)  K₂CrO₄

                   Mole =   25 x 1 milimol = 25 x 10⁻³

Using mole ratio method to find<u> limiting reagent </u>

                                          \frac{Mole}{Stoichiometry}  

                      For K_{2}CrO_{4}   = \frac{25 X10^{-3} }{1}=25 X10^{-3} \\\\For Pb(NO_{3} )_{2} = \frac{0.003}{1}=0.003

Hence K₂CrO₄(potassium chromate) is a<u> limiting reagent.</u>

So lead chromate formed = mole X molar mass

                                           = 0.003 X 323.19 g

                                           = 0.97 g

∴ 0.97 g lead chromate form when a 1.00-g sample of Pb(NO₃)₂ is added to 25.0 ml of 1.00 M K₂CrO₄ solution.

3 0
1 year ago
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