Answer:
a) [tex]BaF_{2}[/tex]
b) [tex]SnCl_{4}[/tex]
c) [tex]Cr_{2}O_{3}[/tex]
d) [tex]FeO[/tex]
e) [tex]Li_{3} P[/tex]
Explanation:
The charges of the ions have to be 0, so they balance.
a) [tex]Ba^{2+}[/tex] has the charge of +2 and [tex]F^{-}[/tex] has the charge of -1. So its [tex]0=(+2)+(-1)*2[/tex]. That means you need one [tex]Ba^{2+}[/tex] and two [tex]F^{-}[/tex] to balance the equation. -> [tex]BaF_{2}[/tex]
b) [tex]Sn^{4+}[/tex] has the charge of +4 and [tex]Cl^{-}[/tex] has the charge of -1. So its [tex]0=(+4)+(-1)*4[/tex]. That means you need one [tex]Sn^{4+}[/tex] and four [tex]Cl^{-}[/tex] to balance the equation. -> [tex]SnCl_{4}[/tex]
c) [tex]Cr^{3+}[/tex] has the charge of +3 and [tex]O^{2-}[/tex] has the charge of -2. So its [tex]0=(+3)*2+(-2)*3[/tex]. That means you need two [tex]Cr^{3+}[/tex] and three [tex]O^{2-}[/tex] to balance the equation. -> [tex]Cr_{2}O_{3}[/tex]
d) [tex]Fe^{2+}[/tex] has the charge of +2 and [tex]O^{2-}[/tex]has the charge of -2. So its [tex]0=(+2)+(-2)[/tex]. That means you need one [tex]Fe^{2+}[/tex] and one [tex]O^{2-}[/tex] to balance the equation. -> [tex]FeO[/tex]
e) [tex]Li^{+}[/tex] has the charge of +1 and [tex]P^{3-}[/tex]has the charge of -3. So its [tex]0=(+1)*3+(-3)[/tex]. That means you need three [tex]Li^{+}[/tex] and one [tex]P^{3-}[/tex] to balance the equation. -> [tex]Li_{3} P[/tex]
