Question

On your wedding day your lover gives you a gold ring of mass 3.6 g. 52 years later its mass is 3.35 g. On the average how many atoms were abrated from the ring during each second of your marriage

Answer 1

Approximately 1.53 × 10-7 grams of gold are abraded from the ring every second during a 52-year marriage.

Step-by-step explanation:

To calculate how many atoms were abraded from the gold ring during each second of your marriage, we need to determine the mass difference between the initial mass and the final mass of the ring.

Mass difference = Initial mass - Final mass = 3.6 g - 3.35 g = 0.25 g

Next, we need to convert this mass difference into grams per second. Since 52 years is equal to 1,637,760 seconds, we can divide the mass difference by the number of seconds to get the rate of abradation:

Abrasion rate = Mass difference / Number of seconds = 0.25 g / 1,637,760 s ≈ 1.53 × 10-7 g/s

Therefore, on average, approximately 1.53 × 10-7 grams of gold are abraded from the ring every second during your 52-year marriage.

Answer 2

`7.646 * 10^(23)` atoms

Step-by-step explanation:

First we find the mass of gold that was lost during the 52 years.

Mass lost = 3.6 g - 3.35 g = 0.25 g

Secondly we find the mass of each atom using gold's relative atomic mass.

mass of a gold atom = 196.96657 amu *
`1.66*10^(-27)` kg/amu

=
`3.2696 * 10^(-25)` per atom of gold

Thirdly now we can calculate the number of gold atoms lost by dividing the mass of gold lost by the mass of each gold atom.

number of gold atoms lost =
`(0.25 g)/(3.2696*10^(-25) g/atom)`

=
`7.646 * 10^(23)` atoms