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University of New Mexico

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Problem 109

(a) You are given a cube of silver metal that measures 1.000 $\mathrm{cm}$ on each edge. The density of silver is 10.5 $\mathrm{g} / \mathrm{cm}^{3} .$ How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom.(c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

Answer

(a) $5.853 \times 10^{22}$ atoms

(b) $1.265 \times 10^{-23}$ cubic centimeters

(c) 14 angstroms

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## Discussion

## Video Transcript

for this question, the volume off Cuba is one centimeter cube, along with the density of 10.5. We know the formula that bull must is acquittal volume into density. Using this formula, the mass will be 10.5 lumps. The Mullah Omar's off silver is 107 so 107.87 grams off silver. Wilke will contain Abou Guardo's number off Adam. That is 6.2 into 10 days to about 23 10.5 grams off silver. Contains using the unitary method will contain 5.86 into 10 days to put 22 atoms for option B volume off the Q can be determined by same formula density Zagato mass. Per unit volume, or 74% off the volume is actually filled. So if the Cube is 100 the volume of Cuba's 100 then the volume filled with water, or William filled with any liquid will be 0.74 centimeter cube. So there are 5.86 into 10 days Super 22 atoms, so the volume off each atom can be determined by this formula, which turns out to be 1.263 into 10 dates of one negative 23 centimeter kill for option C. Let the radius off each atom are so the volume will be four by three. By our cue for the daring along, we know the volume off an atom equals 1.263 to 10 days to part 23 centimeter. Q. Therefore, our cube will be equals 9.47 to 10 days. Super negative 24 which gives us the final radius off silver at the mast 211.5. I'm strong.

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