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 150GQ
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 150GQ
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,
The radius of a strontium atom is 215 pm How many strontium atoms would have to be laid side by side to span a distance of 1.65 mm?
With the advent of techniques such as scanning tunneling microscopy, it is now possible to “write” with individual atoms by manipulating and arranging atoms on an atomic surface.(A) If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? Show all work. [2](B) If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part (B) is used, what number of ruthenium atoms is needed to construct the surface? Show all work. [2]
The mass of a single palladium atom is 1.77×10-22 grams. How many palladium atoms would there be in 91.1 milligrams of palladium?
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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