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One mole of magnesium (6 × 1023 atoms) has a mass of 24 grams, as shown in the periodic table on the inside front cover of the textbook. The density of magnesium is 1.7 grams/cm3. What is the approximate diameter of a magnesium atom (length of a bond) in a solid block of the material? Make the simplifying assumption that the atoms are arranged in a "cubic" array, as shown in the figure. Remember to convert to SI units.

User Dhg
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Final answer:

To find the approximate diameter of a magnesium atom in a solid block of the material, we can use the given information. One mole of magnesium contains 6x10^23 atoms and has a mass of 24 grams. The density of magnesium is 1.7 grams/cm^3. By calculating the volume of one atom and solving for the radius, we can determine the approximate diameter of a magnesium atom in a solid block.

Step-by-step explanation:

To find the approximate diameter of a magnesium atom in a solid block of the material, we can use the given information. We know that one mole of magnesium contains 6x10^23 atoms and has a mass of 24 grams. The density of magnesium is 1.7 grams/cm^3. In a cubic array, the atoms are arranged in a face-centered cubic (FCC) structure, where each atom is touching 12 nearest neighbors. Using this information, we can calculate the approximate diameter of a magnesium atom.

First, we need to find the volume of one atom. Since the density is given in grams/cm^3, we need to convert this to grams/mm^3 to match the units of the mass. One cm^3 is equal to (10 mm)^3 = 1000 mm^3, so the density of magnesium is 1.7 grams/cm^3 = 1.7 grams/1000 mm^3 = 0.0017 grams/mm^3.

Next, we can find the volume of one atom by dividing the volume of one mole of magnesium by Avogadro's number (6x10^23). The volume of one mole of magnesium is the mass divided by the density: (24 grams) / (0.0017 grams/mm^3) = 14117.6471 mm^3.

Now, we can find the volume of one atom by dividing the volume of one mole by Avogadro's number: (14117.6471 mm^3) / (6x10^23) = 2.35294118x10^-20 mm^3.

Since the atoms are touching in an FCC structure, the volume of one atom is equal to 4/3 times pi times the radius cubed: (4/3) x pi x r^3 = 2.35294118x10^-20 mm^3. We can solve this equation for the radius, and then multiply by 2 to get the diameter.

By rearranging the equation, we have r^3 = (3/4) x (2.35294118x10^-20 mm^3) / pi. Taking the cube root of both sides, we get r = ( (3/4) x (2.35294118x10^-20 mm^3) / pi )^(1/3) = 1.603701318x10^-7 mm.

Finally, we multiply the radius by 2 to get the diameter: 2 x (1.603701318x10^-7 mm) = 3.2074026376x10^-7 mm.

User Milan Rakos
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