Question
Silver carbonate [AgCar] has a density of 6.077 g/cm3. Calculate the mass (in fg) of carbon atoms [C] present in a AgCar crystal that has a volume of 457.6 Å3.
Answers
DrBob222
457.6 A^3 = V
457.6 A^3 x (1 x 10^-8 cm)^3 = volume in cubic centimeters.
mass = volume x density = calculate mass = approx 3E-21 grams.
mols Ag2CO3 = grams/molar mass = ?
There is 1 mol C for 1 mol Ag2CO3 so mols C = mols Ag2CO3
Then g C atoms = mols C atoms x atomic mass C atoms.
457.6 A^3 x (1 x 10^-8 cm)^3 = volume in cubic centimeters.
mass = volume x density = calculate mass = approx 3E-21 grams.
mols Ag2CO3 = grams/molar mass = ?
There is 1 mol C for 1 mol Ag2CO3 so mols C = mols Ag2CO3
Then g C atoms = mols C atoms x atomic mass C atoms.