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Georgia [21]
4 years ago
9

When a very electronegative atom strips a valence electron away from its partner, __________ form?

Chemistry
2 answers:
andriy [413]4 years ago
7 0
A very electronegative atoms has a strong ability of attracting electrons towards itself. As a result, it would gain electrons more than it usually has. Therefore, it would form an anion, or a negatively charge ion. These ions readily accepts electrons from the valence of other elements.
wel4 years ago
4 0

Explanation:

When there occurs sharing of electrons between two chemically combining atoms then it tends to form a covalent bond.

As, atomic number of hydrogen is 1 and its electronic configuration is 1s^{1}.  

Hence, in order to become stable in nature it will reaction share its valence electron with another atom.  

As atomic number of chlorine is 17 and its electronic distribution is 2, 8, 7. So, in order to attain stability it will readily react a hydrogen atom who also wants to become stable.

As a result, HCl compound will be formed which is a covalent compound.

Whereas when an electron is being donated by an atom to another atom then it results in the formation of an ionic bond.

For example, sodium has one valence electron and chlorine has 7 valence electron.

Hence, chlorine being more electronegative in nature will attract the valencxe electron of sodium towards itself. Hence, an ionic compound NaCl will be formed.

Thus, we can conclude that when a very electronegative atom strips a valence electron away from its partner, an ionic compound forms.

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Answer:

\large \boxed{\text{5.50 g}}

Explanation:

Percent means " per hundred," so % m/V means the grams of something in 100 mL of  volume.

\text{m/V \%} = \dfrac{\text{mass of solute} }{\text{volume of solution}} \times 100 \, \%

Data:

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Calculations:

\text{2.20 \%} = \dfrac{\text{mass of solute} }{\text{250 mL}} \times 100 \, \%\\\\\text{Mass of solute} = \dfrac{2.20\, \% \times \text{250 mL}}{100 \%} = \text{5.50 g}\\\\\text{The mass of Co(NO$_{3}$)$_{2}$ is $\large \boxed{\textbf{5.50 g}}$}

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