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Vesnalui [34]
3 years ago
8

Rank the following atoms by number of valence electrons. Rank from most to fewest valence electrons. To rank items as equivalent

, overlap them. Rb Xe Sb I In
Chemistry
2 answers:
ExtremeBDS [4]3 years ago
7 0

Answer:

I>Sb>In>Rb>Xe

Explanation:

Valence electrons are electrons found on the outermost shell of an atom.

In the group of atoms, iodine has seven valence electrons, antimony has five, indium has three, rubidium has one and Xenon has none.

Hence the order stated in the answer

Fed [463]3 years ago
6 0

To count the number of valence electrons we look at the electronic configuration and add the electrons form the electronic shell with the highest principal quantum number.

Rb: [Kr] 5s¹ - 1 valence electron

Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons

Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons

I:    [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons

In:  [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons

Rank from most to fewest valence electrons:

Xe > I > Sb > In > Rb

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silver has an atomic mass of 107.868 amu. silver has two common isotopes. one of the isotopes has a mass of 106.906 amu and a re
Sunny_sXe [5.5K]

A. The abundance of the 2nd isotope is 48.119%

B. The mass of the 2nd isotope is 108.905 amu

Let the 1st isotope be A

Let the 2nd isotope be B

A. Determination of the abundance of the 2nd isotope

Abundance of isotope A = 51.881%.

<h3>Abundance of isotope B =? </h3>

Abundance of B = 100 – A

Abundance of B = 100 – 51.881

<h3>Abundance of B = 48.119%</h3>

B. Determination of the mass of the 2nd isotope

Atomic mass of silver = 107.868 amu.

Mass of 1st isotope (A) = 106.906 amu

Abundance of isotope A (A%) = 51.881%.

Abundance of isotope B (B%) = 48.119%

<h3>Mass of 2nd isotope (B) =? </h3>

atomic \: mass =  \frac{mass \: of \:A \times \:A\%}{100}  + \frac{mass \: of \:B \times \:B\%}{100} \\  \\ 107.868 = \frac{106.906\times \ \: 51.881}{100}  + \frac{mass \: of \:B \times \:48.119}{100} \\  \\ 107.868 = \: 55.464 + 0.48119 \times mass \: of \:B  \\  \\  collect \: like \: terms \\  \\ 0.48119 \times mass \: of \:B  = 107.868  - 55.464  \\  \\ divide \: both \: side \: by \: 0.48119 \\  \\ mass \: of \:B  = \frac{107.868  - 55.464 }{0.48119}  \\  \\ mass \: of \:B  =108.905 \: amu \\  \\

Therefore, the mass of the 2nd isotope is 108.905 amu

Learn more: brainly.com/question/7955048

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