Give the identity of substances A and B, explaining their categories. Write the formula of: The compound formed when B and C reacted; Ions formed by A and B.

Substance A has 20 protons. The element with atomic number 20 is Calcium (Ca).
It has 20 protons and 18 electrons, meaning it has a net charge of $+2$ ($20 - 18 = 2$). Therefore, it is a calcium ion, $Ca^{2+}$.
Category: Calcium is an alkaline earth metal. Since it has lost electrons, it is a cation.

Substance B has 17 protons. The element with atomic number 17 is Chlorine (Cl).
It has 17 protons and 18 electrons, meaning it has a net charge of $-1$ ($17 - 18 = -1$). Therefore, it is a chloride ion, $Cl^{-}$.
Category: Chlorine is a halogen (a non-metal). Since it has gained an electron, it is an anion.

• Substance A: Calcium ($Ca^{2+}$). It is an alkaline earth metal and a cation.
• Substance B: Chlorine ($Cl^{-}$). It is a halogen (non-metal) and an anion.

From part a), Substance B is a chloride ion ($Cl^{-}$).
Substance C has 12 protons, which is Magnesium (Mg). It has 10 electrons, meaning it has a net charge of $+2$ ($12 - 10 = 2$). So, Substance C is a magnesium ion ($Mg^{2+}$).

To form a neutral compound, the charges must balance. One $Mg^{2+}$ ion requires two $Cl^{-}$ ions.
The formula is $\boxed{MgCl_2}$.

From part a), Substance A forms a calcium ion ($Ca^{2+}$).
From part a), Substance B forms a chloride ion ($Cl^{-}$).

The ions formed by A and B are $\boxed{Ca^{2+} \text{ and } Cl^{-}}$.

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