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3 years ago

What do Lewis structures show?

Chemistry

2 answers:

Anon25 [30]3 years ago

6 0

Hello there!

The correct answer is option D

The valence electrons are the numbers from 1 to 8.

Best of luck!

belka [17]3 years ago

5 0

<span>Lewis structures only show the valence electrons of an atom in a molecule. Valence electrons exist in the energy level farthest away from the nucleus of an atom. Lewis structures are diagrams showing bonding of atoms with lines, and valence electrons with dots.</span>

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How many grams of chlorine gas can be produced when 50.0 grams of aluminum chloride decompose? 2AlCl3→ 2Al + 3Cl2

lianna [129]

Answer:

Explanation:

Approx.

425

⋅

Explanation:

(

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(

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→

(

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You have given a stoichiometrically balanced equation, so bravo.

The equation explicitly tells us that

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2 years ago

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emmasim [6.3K]

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3 years ago

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marusya05 [52]

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2 years ago

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3 years ago

Read 2 more answers

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

2 years ago