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Sagot :
It looks like you are working with a table related to writing formulas for ternary compounds. Unfortunately, the provided table is incomplete, and there are some formatting and clarity issues. Nevertheless, I will help you understand how to write formulas for ternary compounds, which consist of three different elements including at least one polyatomic ion.
To write the chemical formula for the compounds, you need to ensure that the overall charge of the compound is zero. Here is a more structured example using parts of what is visible in the table:
### 1. Iron (II) Hydroxide
- [tex]\( \text{Iron (II) ion: Fe}^{2+} \)[/tex]
- [tex]\( \text{Hydroxide ion: OH}^{-} \)[/tex]
The iron (II) ion has a charge of +2 and the hydroxide ion has a charge of -1. You need two hydroxide ions to balance out the +2 charge of the iron (II) ion:
- Formula: [tex]\( \text{Fe(OH)}_2 \)[/tex]
### 2. Iron (III) Phosphate
- [tex]\( \text{Iron (III) ion: Fe}^{3+} \)[/tex]
- [tex]\( \text{Phosphate ion: PO}_4^{3-} \)[/tex]
The iron (III) ion has a charge of +3 and the phosphate ion has a charge of -3. Since their charges are already balanced:
- Formula: [tex]\( \text{FePO}_4 \)[/tex]
### 3. Cobalt (III) Sulfite
- [tex]\( \text{Cobalt (III) ion: Co}^{3+} \)[/tex]
- [tex]\( \text{Sulfite ion: SO}_3^{2-} \)[/tex]
The cobalt (III) ion has a charge of +3 and the sulfite ion has a charge of -2. To balance the charges, the least common multiple is 6. So we need 2 cobalt ions (each +3) and 3 sulfite ions (each -2):
- Formula: [tex]\( \text{Co}_2(\text{SO}_3)_3 \)[/tex]
### 4. Sodium Chlorate
- [tex]\( \text{Sodium ion: Na} \)[/tex]
- [tex]\( \text{Chlorate ion: ClO}_3^{-} \)[/tex]
The sodium ion has a charge of +1 and the chlorate ion has a charge of -1. Since their charges are already balanced:
- Formula: [tex]\( \text{NaClO}_3 \)[/tex]
### 5. Chromium (III) Chlorate
- [tex]\( \text{Chromium (III) ion: Cr}^{3+} \)[/tex]
- [tex]\( \text{Chlorate ion: ClO}_3^{-} \)[/tex]
The chromium (III) ion has a charge of +3 and the chlorate ion has a charge of -1. We need three chlorate ions to balance the +3 charge of the chromium ion:
- Formula: [tex]\( \text{Cr(ClO}_3\text{)}_3 \)[/tex]
For remaining entries, you can use the same approach: balance the total positive and negative charges to write the formulas. Ensure the charges on the ions add up to zero to form a neutral compound.
If you need assistance with specific ions or additional entries, feel free to ask!
To write the chemical formula for the compounds, you need to ensure that the overall charge of the compound is zero. Here is a more structured example using parts of what is visible in the table:
### 1. Iron (II) Hydroxide
- [tex]\( \text{Iron (II) ion: Fe}^{2+} \)[/tex]
- [tex]\( \text{Hydroxide ion: OH}^{-} \)[/tex]
The iron (II) ion has a charge of +2 and the hydroxide ion has a charge of -1. You need two hydroxide ions to balance out the +2 charge of the iron (II) ion:
- Formula: [tex]\( \text{Fe(OH)}_2 \)[/tex]
### 2. Iron (III) Phosphate
- [tex]\( \text{Iron (III) ion: Fe}^{3+} \)[/tex]
- [tex]\( \text{Phosphate ion: PO}_4^{3-} \)[/tex]
The iron (III) ion has a charge of +3 and the phosphate ion has a charge of -3. Since their charges are already balanced:
- Formula: [tex]\( \text{FePO}_4 \)[/tex]
### 3. Cobalt (III) Sulfite
- [tex]\( \text{Cobalt (III) ion: Co}^{3+} \)[/tex]
- [tex]\( \text{Sulfite ion: SO}_3^{2-} \)[/tex]
The cobalt (III) ion has a charge of +3 and the sulfite ion has a charge of -2. To balance the charges, the least common multiple is 6. So we need 2 cobalt ions (each +3) and 3 sulfite ions (each -2):
- Formula: [tex]\( \text{Co}_2(\text{SO}_3)_3 \)[/tex]
### 4. Sodium Chlorate
- [tex]\( \text{Sodium ion: Na} \)[/tex]
- [tex]\( \text{Chlorate ion: ClO}_3^{-} \)[/tex]
The sodium ion has a charge of +1 and the chlorate ion has a charge of -1. Since their charges are already balanced:
- Formula: [tex]\( \text{NaClO}_3 \)[/tex]
### 5. Chromium (III) Chlorate
- [tex]\( \text{Chromium (III) ion: Cr}^{3+} \)[/tex]
- [tex]\( \text{Chlorate ion: ClO}_3^{-} \)[/tex]
The chromium (III) ion has a charge of +3 and the chlorate ion has a charge of -1. We need three chlorate ions to balance the +3 charge of the chromium ion:
- Formula: [tex]\( \text{Cr(ClO}_3\text{)}_3 \)[/tex]
For remaining entries, you can use the same approach: balance the total positive and negative charges to write the formulas. Ensure the charges on the ions add up to zero to form a neutral compound.
If you need assistance with specific ions or additional entries, feel free to ask!
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