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To find Key relationships Examples


= atomic # of the element An atom with a
= # of electrons (in a neutral atom)

=
+30 and
= nucleon – neutrons
33 neutrons has

= mass # – neutrons

30 protons.


= A neutral atom of
= protons has
(in neutral atoms) = nuclear charge 35 electrons.
= mass # – neutrons



= atomic # charge of the ion A charged atom
with
=
(in ions) = nuclear charge – charge of the ion and

has 42 electrons


= m An atom with a
= mass # – atomic #
= mass # – electrons (in neutral atoms) has
= nucleons – protons 125 neutrons.





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An atom with a
= protons
= electrons (in neutral atoms)

= nuclear charge and
= has an atomic #

of 52. The atom is

tellurium.


= neutrons + protons An atom with
and
= (neutral atoms)
= nuclear charge + neutrons has a
= nucleons mass # of 122 amu.


An atom with a
= mass #
= protons + neutrons
65 neutrons
= neutrons + electrons
= nuclear charge + neutrons has a total of
113 nucleons.

An atom of argon
= protons
with a mass of 40
= electron (in neutral atom) amu has a nuclear
= mass # neutrons charge of


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