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The radius of a silver atom is 145 pm. How many silver atoms would have to be laid side by side to span a distance of 2.31
mm?
atoms
M

Sagot :

dsdrajlin

Given that the radius of a silver atom is 145 pm, it would take [tex]7.97 \times 10^{6}[/tex] silver atoms laid side by side to span a distance of 2.31 mm.

The radius of a silver atom is 145 pm. Then, the diameter (twice the radius) is:

[tex]d = 2 \times 145 pm = 290 pm[/tex]

We will convert 290 pm to millimeters. We will use the conversion factors:

  • 1 m = 10¹² pm
  • 1 m = 10³ mm

[tex]290 pm \times \frac{1m}{10^{12}pm } \times \frac{10^{3}mm }{1m} = 2.90 \times 10^{-7} mm[/tex]

If the diameter of 1 silver atom is 2.90 × 10⁻⁷ mm, the number of silver atoms required to span a distance of 2.31 mm is:

[tex]2.31 mm \times \frac{1Ag\ atom}{2.90 \times 10^{-7}mm } = 7.97 \times 10^{6} Ag\ atom[/tex]

[tex]7.97 \times 10^{6}[/tex] silver atoms span a distance of 2.31 mm.

You can learn more about unit conversions here: brainly.com/question/19420601

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