<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 %
Answer:
2 moles
Explanation:
Let us first start by calculating the molecular mass of Al₂O₃.
The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.
Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.
Molecular mass of Al : 26.981539 u
Molecular mass of O: 15.999 u
Therefore, molecular mass of Al₂O₃ is:
= u
= 101.960078 u
This can be approximated to 102 u.
1mole weighs 102 u
So, 2moles will weigh 2*102 = 204 u