Question

Please help if u can ty..​

Answer 1

a)
`BaF_(2)`

b)
`SnCl_(4)`

c)
`Cr_(2)O_(3)`

d)
`FeO`

e)
`Li_(3) P`

Step-by-step explanation:

The charges of the ions have to be 0, so they balance.

a)
`Ba^(2+)` has the charge of +2 and
`F^(-)` has the charge of -1. So its
`0=(+2)+(-1)*2`. That means you need one
`Ba^(2+)` and two
`F^(-)` to balance the equation. ->
`BaF_(2)`

b)
`Sn^(4+)` has the charge of +4 and
`Cl^(-)` has the charge of -1. So its
`0=(+4)+(-1)*4`. That means you need one
`Sn^(4+)` and four
`Cl^(-)` to balance the equation. ->
`SnCl_(4)`

c)
`Cr^(3+)` has the charge of +3 and
`O^(2-)` has the charge of -2. So its
`0=(+3)*2+(-2)*3`. That means you need two
`Cr^(3+)` and three
`O^(2-)` to balance the equation. ->
`Cr_(2)O_(3)`

d)
`Fe^(2+)` has the charge of +2 and
`O^(2-)`has the charge of -2. So its
`0=(+2)+(-2)`. That means you need one
`Fe^(2+)` and one
`O^(2-)` to balance the equation. ->
`FeO`

e)
`Li^(+)` has the charge of +1 and
`P^(3-)`has the charge of -3. So its
`0=(+1)*3+(-3)`. That means you need three
`Li^(+)` and one
`P^(3-)` to balance the equation. ->
`Li_(3) P`