<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:
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Explanation:
0.000141 to kilowatt-hours. hope this helped
Because they do not have the same qualities therefore they are different
A
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1) Find the number of mols of HCl in 5.2 liters of 4.0M solution:
n = M*V(L) = 4.0 mol/L * 5.2 L = 20.8 mol
2) Find the number of mols of Mg that will react with 20.8 mol of HCl, using the coefficients of the balanced equation
[1mol Mg / 2 mol HCl] * 20.8 mol HCl = 10.4 mol Mg
3) Transform mol to mass using the atomic mass:
10.4 mol Mg * 24.3 g/mol = 252.7 g of Mg.
