Question

1. fill in the missing information in the data table below. then, using the periodic table, determine whether each of the following is an atom, ion, or isotope.

element	protons	neutrons	electrons	atom (a), ion (i), or isotope (is)?	create the atom symbol
oxygen	8	8	8	a
bromine	35	45	35	a

1. determine the following information from the atom symbols:

for (_{20}^{42}ca):
protons?
electrons?
neutrons?
mass #?
atomic #?
charge?
atom, ion, or isotope?

for (_{12}^{24}mg^{2 +}):
protons?
electrons?
neutrons?
mass #?
atomic #?
charge?
atom, ion, or isotope?

Explanation:

Step1: Recall atomic - related concepts

The atomic number ($Z$) is equal to the number of protons. For a neutral atom, the number of electrons is equal to the number of protons. The mass number ($A$) is the sum of protons and neutrons ($A = p + n$). An ion has a non - zero charge, and an isotope has a different number of neutrons compared to the most common form of the element.

Step2: Analyze $_{20}^{42}Ca$

The atomic number of calcium ($Ca$) is 20, so the number of protons is 20. For a neutral atom, the number of electrons is also 20. The mass number is 42. Using $A=p + n$, we can find the number of neutrons: $n=A - p=42 - 20 = 22$. The atomic number is 20. Since there is no charge indicated, the charge is 0. It is an isotope because the mass number (42) is different from the most common mass number of calcium.

Step3: Analyze $_{12}^{24}Mg^{2+}$

The atomic number of magnesium ($Mg$) is 12, so the number of protons is 12. The ion has a charge of $2+$, which means it has lost 2 electrons. So the number of electrons is $12-2 = 10$. The mass number is 24. Using $A = p + n$, we can find the number of neutrons: $n=A - p=24 - 12 = 12$. The atomic number is 12. The charge is $2+$. It is an ion.

Answer:

For $_{20}^{42}Ca$:
Protons: 20
Electrons: 20
Neutrons: 22
Mass #: 42
Atomic #: 20
Charge: 0
Atom, ion, or isotope: Isotope

For $_{12}^{24}Mg^{2+}$:
Protons: 12
Electrons: 10
Neutrons: 12
Mass #: 24
Atomic #: 12
Charge: 2+
Atom, ion, or isotope: Ion