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3 years ago

PLSS HELP I NEED HELP RN :/

Chemistry

1 answer:

bogdanovich [222]3 years ago

5 0

Valence electrons are located on the outermost energy level and determine how reactive an atom will be

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A___________ consists of a mixture of graphite powder (a form of carbon) and clay that is baked and hardened and encased in wood

Harlamova29_29 [7]

Answer:

A_____PENCIL______ consists of a mixture of graphite powder (a form of carbon) and clay that is baked and hardened and encased in wood or paper. a. silverpoint b. pencil c. charcoal d. chalk e. paste

Explanation:

A pencil is a writing material, where the graphite is what gives the black color to the writing stroke.

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3 years ago

Please identify each element associated with the orbital notation or electron configuration. Note: One “slash” is one electron,

mylen [45]

Picture is upside

Explanation:

I can't read upsidedown

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3 years ago

Read 2 more answers

The molecular geometry of SO 3 is a. tetrahedral b. bent C. octahedral d. trigonal planar e. pyramidal

dybincka [34]

Answer:

The molecular geometry of SO3 is trigonal planar.

Explanation:

Look at the Lewis

5 0

3 years ago

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 $5.23\times 10^{-5} m$

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3 years ago

What happens when an atom of sulfur combines with two atoms of chlorine to produce SCI2?

Aloiza [94]

Answer:..A.) Each chlorine atom shares a pair of electrons with the sulfur atom

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2 years ago