Estimating the radius of a lead atom. (a) You are given a cube of lead that is 1.000 cm on each side. The density of lead is 11.35 g/ cm 3 . How many atoms of lead are 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 (4/3) πτ 3 for the volume of a sphere, estimate the radius ( r ) of a lead atom.
Estimating the radius of a lead atom. (a) You are given a cube of lead that is 1.000 cm on each side. The density of lead is 11.35 g/ cm 3 . How many atoms of lead are 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 (4/3) πτ 3 for the volume of a sphere, estimate the radius ( r ) of a lead atom.
(a) You are given a cube of lead that is 1.000 cm on each side. The density of lead is 11.35 g/ cm3. How many atoms of lead are 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 (4/3) πτ3 for the volume of a sphere, estimate the radius (r) of a lead atom.
(a)
Expert Solution
Interpretation Introduction
Interpretation: The number of atoms for lead in given sample of lead cube, with each side of value 1cm needed to be calculated.
Concept introduction:
Conversion formula for mass of a molecule and number moles,
Numberofmoles=MassingramsMolarmass
Equation for number of atoms is,
Number of moles×6.022×1023atoms=number of atoms
Equation for density from volume and mass is,
Density=MassVolume
Equation for finding Volume of sphere is,
Volume=(4/3)πr3
Answer to Problem 154GQ
The number of atoms of lead is 3.3×1022atoms
Explanation of Solution
The side of the lead cube is given that 1cm.
Therefore, the volume of the lead cube is,
(1cm)3=1cm3
The density of the lead cube is given as 11.35g/cm3.
Equation for mass from volume and density is,
Density×Volume=Mass
Therefore, the mass of lead cube is,
11.35g/cm3×1cm3=11.35g
Conversion formula for mass of a molecule and number moles,
Numberofmoles=MassingramsMolarmass
Therefore, the number of lead atoms is,
Numberofmoles=11.35g207.2g/mol=0.05477mol
Equation for number of atoms is,
Number of moles×6.022×1023atoms=number of atoms
Therefore, the number of lead atoms in the sample is,
0.05477×6.022×1023atoms=3.3×1022atoms
(b)
Expert Solution
Interpretation Introduction
Interpretation: The volume of one lead atom and its radius have to be calculated under given conditions.
Concept introduction:
Conversion formula for mass of a molecule and number moles,
Numberofmoles=MassingramsMolarmass
Equation for number of atoms is,
Number of moles×6.022×1023atoms=number of atoms
Equation for density from volume and mass is,
Density=MassVolume
Equation for finding Volume of sphere is,
Volume=(4/3)πr3
Answer to Problem 154GQ
The volume and radius of one lead atom is 1.8×10-23cm3 and 0.7572×10-23cm respectively.
Explanation of Solution
The volume of the lead cube is found that 1cm3.
If 60% of the cube is filled with 3.3×1022 lead atom spheres, then the volume of one lead atom is,
(a) Atoms are very small compared to objects on the macroscopic scale. The radius of a vanadium atom is 131 pm. What is this value in meters and in
centimeters?
cm
(b) The mass of a single vanadium atom is 8.46×10-23 g. Suppose enough V atoms were lined up like beads on a string to span a distance of 44.7 cm ( 18
atoms
inches). How many atoms would be required?
What mass in grams of V would be used?
Could you weigh out this amount of vanadium using a typical laboratory balance?
(c) Taking the density of vanadium metal to be 6.08 g/cm³, calculate the mass of metal needed to form a piece of V wire with the same length as the
distance in b, but with a diameter of 1.00 mm. Hint: The volume of a cylinder is T times its radius squared times its height. (V = T r² h)
How many vanadium atoms does this represent?
atoms
(a) Atoms are very small compared to objects on the macroscopic scale. The radius of a nickel atom is 125 pm. What is this value in meters
and in centimeters?
cm
-23
(b) The mass of a single nickel atom is 9.75×10 g. Suppose enough Ni atoms were lined up like beads on a string to span a distance of
31.3 cm (12 inches). How many atoms would be required?
atoms
What mass in grams of Ni would be used?
Could you weigh out this amount of nickel using a typical laboratory balance?
(c) Taking the density of nickel metal to be 8.91 g/cm, calculate the mass of metal needed to form a piece of Ni wire with the same length
as the distance in b, but with a diameter of 1.00 mm. Hint: The volume of a cylinder is n times its radius squared times its height. (V = nr
h)
How many nickel atoms does this represent?
atoms
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell