is suppose to be V=4pir^3/3
sry. I get part A. but part B and C are confusing!
An estimation of the radius of a lead atom:
a. You are given a cube of lead that is 1.000 em on each side. The density of lead is
11.35 g/cm3. How many atoms of lead are contained in the sample.
b. Atoms are spherical; therefore, the lead atoms in this sample cannot fill all the
available space. As an approximation, assume that 60% of the space of the cube
is filled with spherical lead atoms. Calculate the volume of one lead atom from
this information. From the calculated volume (V), and the formula V = 47tr3/3, estimate the radius (r) of a lead atom.
c. Assume the lead atoms along each side of the cube touch one another. How many atoms lie along one edge of the cube?
2 answers
a. density = 11.35 g/cc. volume = 1 cm^3. grams = volume x density = ?
Remember there are 6.022E23 atoms in a mole. You have ?g/atomic mass Pb moles Pb atoms.
b. If the volume is 1 cc, then the volume of the atom must be 0.60 if it occupies 60% of the volume. Then volume of 1 Pb atom is
1/# Pb atoms in the 1 cc = xx
I have no idea what 47tr3/3 means to you (nothing to me). You probably meant Volume of a sphere = (4/3)*pi*r^3 and you can use that to calculate the radius.
Post your work if you get stuck.
Remember there are 6.022E23 atoms in a mole. You have ?g/atomic mass Pb moles Pb atoms.
b. If the volume is 1 cc, then the volume of the atom must be 0.60 if it occupies 60% of the volume. Then volume of 1 Pb atom is
1/# Pb atoms in the 1 cc = xx
I have no idea what 47tr3/3 means to you (nothing to me). You probably meant Volume of a sphere = (4/3)*pi*r^3 and you can use that to calculate the radius.
Post your work if you get stuck.
2 answers