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horsena [70]
2 years ago
11

Figure 2 shows what happens to the electrons in the outer shells when a sodium atom reacts with a fluorine atom.

Chemistry
1 answer:
QveST [7]2 years ago
4 0

Sodium reacts with fluorine to make a covalent bond between them.  A covalent bond is formed when there is a mutual sharing of electrons between the two adjacent atoms.

Here, Sodium is metal and fluorine is non-metal. So, the bond between them may be an ionic bond, but as the Sodium has one valence electron and fluorine is highly electronegative. Sodium wants to lose electrons and fluorine wants to gain electrons to form a chemical bond. Therefore, the sodium donates its valence electron to fluorine, forms a cation, and fluorine accepts the electron and forms an anion and they distribute the difference in the charge between them.

So, we can say that when sodium and fluorine react to form sodium fluoride with a covalent bond.

Learn more about covalent bonds here:

brainly.com/question/12732708

#SPJ10

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<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %

<u>Explanation:</u>

The given chemical formula of lead (IV) carbonate is Pb(CO_3)_2

To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:

\text{Mass percent of lead}=\frac{\text{Mass of lead}}{\text{Mass of lead (IV) carbonate}}\times 100

Mass of lead = (1 × 207.2) = 207.2 g

Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g

Putting values in above equation, we get:

\text{Mass percent of lead}=\frac{207.2g}{327.2g}\times 0100=63.32\%

Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %

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