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

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:

Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %
N₂O₃
3 moles oxgyen atoms in 1 mole .
hope this helps!
The correct answer would be 3.49 times 10^ minus 24 molecules
The answer is sodium (Na)
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The answer is A. The electrons orbit the protons, which are at the center of the atom.