<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 %
Its D because
Li = 6
Be =9
C= 12
O=15
Answer:
The boiling point is the temperature at which the vapor pressure equals the pressure of gas.
The normal boiling point is the temperature at which the vapor pressure equals one atmosphere
Explanation:
One way of knowing that oxygen was the gas removed from the volume of air and not another is to know what the volume of air is made of first. When the composition of the volume of air is already identified, then next would be the process of separating these elements from each other and as to which is to be separated first. This would usually lead to knowing their masses, their boiling and freezing points, the temperatures at which they condense, and so on. This is to identify their differences to each other and use those differences to successfully separate those elements to each other.
