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3 years ago

The Lewis dot model of a metal atom has two dots. Which of the following is true about the atom?

Chemistry

1 answer:

nataly862011 [7]3 years ago

6 0

Answer : the statement that is true about the atom is It is present in the second group of the periodic table.

Explanation :

The Lewis dot model shows the number of valence electrons present in the atom.

Since the given metal has 2 dots, it indicates that the metal atom must have 2 electrons in its outermost shell.

Therefore option 2 : It has two electrons in the innermost energy level, is wrong because it contains 2 electrons in the outermost energy level

Option 1 : It is most likely to form covalent bonds is also wrong because a metal atom always forms ionic bond

Option 3 : It requires six other atoms to form a stable compound, is also wrong as the metal can form a stable compound by donating its 2 electrons to a non metal.

On periodic table, representative elements are arranged in groups based on their valence electrons. Therefore by knowing the outermost electrons, we can predict the group in which the element can be categorized.

Since the given metal atom contains 2 electrons, it must be present is second group.

Therefore the statement that is true about the atom is It is present in the second group of the periodic table.

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3 years ago

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2 years ago

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3 years ago

You are given a solid that is a mixture of na2so4 and k2so4.

Murljashka [212]

Here we have to calculate the amount of $SO_{4}^{2-}$ ion present in the sample.

In the sample solution 0.122g of $SO_{4}^{2-}$ ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ + $SO_{4}^{2-}$

(Na)₂SO₄=2Na⁺ + $SO_{4}^{2-}$

Thus, BaCl₂+ $SO_{4}^{2-}$ = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of $SO_{4}^{2-}$ ion is precipitated in this reaction.

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion ( $SO_{4}^{2-}$ ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present. So in 1 g of BaSO₄ $\frac{96.06}{233.3}=0.411$ g of $SO_{4}^{2-}$ ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of $SO_{4}^{2-}$ ion is present.

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3 years ago