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You have a pure (24-karat) gold ring of mass 10.8 g. Gold has an atomic mass of 197 g/mol and an atomic number of 79. (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?

Short Answer

Expert verified
  1. The total number and charge of protons in the ring are 2.61×10244.18×105Crespectively.
  2. The number of electrons in the ring when there is no charge is2.61×1024electrons.

Step by step solution

01

(a) Determination of the total number and charge of protons in the ring.

The number atoms in 1 mole of a substance are the Avogadro number NA.

So, the number of lead atoms is,

N=nNA

Here, n is the number of moles and NA is the Avogadro Number.

Now,the atomic mass of lead is,

M = 197g/mol

So the number of atoms in a mass of m = 10.8 is,

N=6.022×1023atoms/mol10.8g197g/mol=3.300×1022atoms

Atomic number is 79, thus the number of protons is,

np=(79protons/atom)3.300×1022atoms=2.61×1024protons.

Thus,Charge of the protons is,

Q=np1.60×1019C/proton=4.18×105C

02

(b) Determination of the number of electrons in the ring when there is no charge.

The number of electrons is always equal to the number of protons. So,

ne=np=2.61×1024electrons

Thus, the number of electron son the ring is 2.61×1024electrons .

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