A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm 3)

1476.43 cm

Explanation:

From the question given above, the following data were obtained:

Diameter = 1 mm

Number of mole of Au = 1 mole

Density of Au = 17.0 g/cm³

Length of wire =?

Next, we shall determine the mass of 1 mole of Au. This can be obtained as follow:

Number of mole of Au = 1 mole

Molar mass of Au = 197 g/mol

Mass of Au =?

Mole = mass / Molar mass

1 = mass of Au / 197

Mass of Au = 1 × 197

Mass of Au = 197 g

Next, we shall determine the volume of the Au. This can be obtained as follow:

Density of Au = 17.0 g/cm³

Mass of Au = 197 g

Volume of Au =

Density = mass /volume

17 = 197 / Volume of Au

Cross multiply

17 × volume of Au = 197

Divide both side by 17

Volume of Au = 197 / 17

Volume of Au = 11.59 cm³

Next, we shall determine the radius of the gold wire. This can be obtained as follow:

Diameter = 1 mm

Radius =?

Radius = Diameter /2

Radius = 1/2

Radius = 0.5 mm

Converting 0.5 mm to cm, we have:

10 mm = 1 cm

Therefore,

0.5 mm = 0.5 mm × 1 cm / 10 mm

0.5 mm = 0.05 cm

Thus, the radius of the gold wire is 0.05 cm

Finally, we shall determine the length of the wire. This can be obtained as follow:

Volume of Au (V) = 11.59 cm³

Radius (r) = 0.05 cm

Pi (π) = 3.14

Length (L) =?

V = πr²L

11.59 = 3.14 × 0.05² × L

11.59 = 3.14 × 0.0025 × L

11.59 = 0.00785 × L

Divide both side by 0.00785

L = 11.59 / 0.00785

L = 1476.43 cm

Thus, the length of the gold wire is 1476.43 cm

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).