Question

(a) Atoms are very small compared to objects on the macroscopic scale. The radius of a platinum atom is 139 pm. What is this value in meters and in centimeters? ____m ____cm

( b) The mass of a single platinum atom is 3.24*1022 g. Suppose enough Pt atoms were lined up like beads on a string to span a distance of 48.1 cm (19 inches) How many atoms would be required? ___atoms
What mass in grams of Pt would be used? ___g
Could you weigh out this amount of platinum using a typical laboratory balance? Yes or no?

(c) Taking the density of platinum metal to be 21.4 g/cm^3, calculate the mass of metal needed to form a piece of Pt wire with the same length as the distance in b, but with a diameter of atoms 1.50 mm. Hint: The volume of a cylinder is π times its radius squared times its height (V = πr^2h) ___g

How many platinum atoms does this represent? ____atoms

Answer 1

Answer:

Kindly check explanation

Explanation:

pm = picometer

Radius of platinum atom = 139 pm

1 pm = 1 × 10^-12 m

Hence,

139 pm = 139 × 10^-12 m

1 pm = 1 × 10^-10 cm

Hence,

139 pm = 139 × 10^-10 cm

B) mass of a single platinum atom = 3.24 × 10²² g

To obtain the diameter of 1 platinum atom in cm:

139 × 10^-10 cm

Number of atoms required :

48.1 cm / 1.39 × 10^-8 cm = 34.60 × 10^8 atoms

Mass in gram of P(t) :

(3.24 × 10²²) × (34.60 × 10^8) = 112.117 × 10^(22+8)

= 112.12 × 10^30 g

C) density (d) = 21.4 g/cm^3, diameter = 1.50 mm

V = πr^2h

h = 48.1cm ; r = 1.5 / 2 = 0.75 mm = 0.075cm

Density = mass / volume

Volume = π * 0.075^2 * 48.1 = 0.8499971cm³

Mass = density × volume

Mass = 21.4g/cm³ × 0.8499971cm³

= 18.19g