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olga2289 [7]
3 years ago
7

Barium (Ba) has two valence electrons, and these electrons are located in the 6s subshell. Without using the periodic table, in

which group and period is barium located?
Group 2A, Period 7

Group 1A, Period 6

Group 1A, Period 7

Group 2A, Period 6
Chemistry
2 answers:
Katena32 [7]3 years ago
7 0
Barium is located in Group 2A, Period 6
artcher [175]3 years ago
5 0

Answer:

Group 2A, Period 6

Explanation:

To know where an atom is located, we need to see its electronic distribution. The major number in it will be the period (not necessary is the number of the valence, because subshell 4s is less energetic than 3d, so it will occur first!).

To know the group, we need to check the subshell more energetic, if is subshell <em>s</em>, it will be in group 1A or 2A, if it has only 1 valence electron, then its group 1A, with two valence electrons, is group 2A.

If the subshell more energetic is the subshell <em>p</em>, so it will be from 3A to 8A, we just add the number of electrons of subshell <em>p</em> with the numbers of electrons of subshell <em>s</em> of the same layer, and we'll have the number of the group.

If the subshell more energetic is the subshell <em>d</em>, it will be in group B, and the number of electrons will be the number of the group.

If the subshell more energetic is subshell <em>f</em>, then it is on group 3B, because will be in latanids or actinides.

So, Barium has the subshell 6s with 2 valence electrons, then it is in group 2A, Period 6.

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An ion with six protons, seven neutrons, and a charge of 2+ has an atomic number of ________.
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A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ions as
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Answer:

  259.497 mg,   58.84%

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BaSO₄ → Ba²⁺ + SO₄²⁻

to calculate the mole of BaSO₄

mole BaSO₄ = mass given / molar mass = 403 mg / 233.38 g/mol = 1.7268 mol

comparing the mole ratio

1.7268 mol of BaSO₄ yields 1.7268 mol of Ba²⁺

403 mg BaSO₄  yields     ( 1.7268 × 137.327 ) where 137.327 is the molar mass of Barium mol of Ba²⁺

441 mg BaSO₄  will yield   ( 1.7268 × 137.327  × 441 mg ) / 403 mg = 259 .497 mg

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