Question

The radius of a iridium atom is 135 pm. How many iridium atoms would have to be laid side by side to span a distance of 2.33 mm? g

Answer 1

Answer:

Number of atoms, N = 17,259,259

Explanation:

Given:

Radius of Iridium = 135 pm

distance = 2.33 mm

To determine number of iridium atom that would be laid side by side to span a distance of 2.33 mm, we say let the number of the atom = N

$N =\frac{Distance}{Iridium. Radius} = \frac{2.33X10^{-3}}{135 X10^{-12}} \\\\N = 17259259.26$

Therefore, If a iridium atom has a radius of  135 pm, then 17,259,259 atoms of Iridium would be laid side by side to span a distance of 2.33 mm.

Number of atoms, N = 17,259,259

Answer 2

17,259,259 atoms of iridium are present in a distance of 2.33 mm

First, to calculate the amount of iridium atoms in a distance of 2.33 mm it is necessary to divide the two values:

                                       $N = \frac{Distance}{Radius}$

As the element radius value is in pm, it is necessary to transform this unit to suit the other:

                                         $135pm = 135\times 10^{-9} mm$

Now, we can apply the values in the expression:

                                             $N = \frac{2.33}{135\times10^{-9}}$

                                        $N = 17,259,259$ atoms

So, 17,259,259 atoms of iridium are present in a distance of 2.33 mm.

Learn more about calculation of atoms in: brainly.com/question/3320707