We know that the relationship between angstrom and mm is [tex] 1:1x10^{-7} [/tex].
Knowing the radius of an atom of gold (1.35 angstroms), we can solve for how many atoms fit into 9.0mm:
[tex] \frac{1.35angstrom}{1} *\frac{1x10^{-7}mm}{1angstrom} =1.35x10^{-7}mm [/tex]
[tex] \frac{9.0mm}{1} [/tex] ÷ [tex] \frac{1.35x10^{-7}mm}{1} [/tex]
[tex] \frac{9.0mm}{1} *\frac{1}{1.35x10^{-7}mm}=\frac{9.0x10^{0}}{1.35x10^{-7}}= 6.7x10^{7} [/tex]
Therefore, we now know that [tex] 6.7x10^{7} [/tex] atoms of gold (Au) will line up to span 9.0mm.
The radius of an atom of gold (Au) is about 1.35 Å. Gold atoms would have to be lined up to span 9.0 mm in the amount of [tex]3.333 * 10^7[/tex]
Gold is a chemical element with the symbol [tex]Au[/tex] and atomic number 79. Its one of the higher atomic number elements that occured naturally. In its purest form, it is a bright, slightly reddish yellow, dense, soft, malleable, and ductile metal
The radius of an atom of gold (Au) is about 1.35 Å. How many gold atoms would have to be lined up to span 9.0 mm ?
Its known that the diameter of an atom is twice the radius = [tex]2*1.35 angstrom = 2.70 angstrom[/tex]
Then we convert the diameter of an atom in mm
[tex]2.70 angstrom * \frac{10^-10 m}{1 angstrom} cot \frac{10^3 mm}{1m} = 2.70 * 10^-7 mm[/tex]
Then to calculate the number of the gold atoms, we have to divide 1 mm by the diameter of an atom
[tex]\frac{9 mm}{2.7*10^-7 mm} = 3.333 * 10^7[/tex]
Grade: 9
Subject: chemistry
Chapter: atom
Keywords: span, The radius of an atom, atom, gold, mm
