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Illusion [34]
3 years ago
9

How would the electron configuration of nitrogen change to make a stable configuration?

Chemistry
1 answer:
AnnZ [28]3 years ago
5 0

The answers to the questions are as follows;

  1. It would gain three electrons
  2. The difference in their electronegativities.
  3. The elements have filled Valence levels
  4. potassium (K) with a 1+ charge
  5. ClO-

Question 1:

  • How would the electron configuration of nitrogen change to make a stable configuration?

Since Nitrogen has 5 Valence electrons, it needs 3 electrons to attain it's octet configuration. As such, it gains 3 electrons.

Question 2:

  • Which quantity determines how two atoms bond.

The quantity which determines how two atoms bond is The difference in their electronegativities.

Question 3:

  • Which statement best explains why the elements in Group 18 do not have electronegativity values.

This is because the elements have filled Valence levels.

Question 4:

  • Based on patterns in the periodic table, which ion has a stable valence electron configuration

The ion which has a stable Valence electron configuration is potassium (K) with a 1+ charge

Question 5;

  • Which chemical formula represents a polyatomic ion?

The chemical formula which represents a polyatomic ion is; ClO-

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If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t
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Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of CaCO_3

\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}

Given:

Mass of CaCO_3 = 0.524 g

Molar mass of CaCO_3 = 100 g/mol

\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol

Now we have to calculate the concentration of CaCO_3

\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}

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Concentration of calcium ion = Concentration of CaCO_3 = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

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