For hundreds of years, alchemists searched for ways to turn various metals into gold. How would the structure of an atom of \(\frac{202}{80} \mathrm{Hg}\) (mercury) have to be changed for the atom to become an atom of \(\stackrel{197}{79}\) Au \((\) gold \() ?\)

Short Answer

Expert verified
In order for an atom of Mercury (\(\frac{202}{80} \mathrm{Hg}\)) to become an atom of Gold (\(\stackrel{197}{79}\) Au), one proton and four neutrons would need to be removed.

Step by step solution

01

Analysis of the Mercury Atom

An atom of Mercury (\(\frac{202}{80} \mathrm{Hg}\)) has 80 protons as its atomic number is 80 (found at the bottom of the fraction) and a mass number of 202 (the sum of the protons and the neutrons, found at the top of the fraction). Therefore, mercury has 122 neutrons (202-80).
02

Analysis of the Gold Atom

An atom of gold (\(\stackrel{197}{79}\) Au) has 79 protons in its nucleus (the number 79 being the atomic number) and a mass number of 197. This means gold has 118 neutrons (197-79).
03

Change in Atom Structure to Transmute Mercury to Gold

To change an atom of Mercury into an atom of Gold, it would be necessary to remove 1 proton (which will also change the atomic number from 80 to 79) and 4 neutrons (which will change the mass number from 202 to 197).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free