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write the formula for the following compounds: ionic compounds: do not …

Question

write the formula for the following compounds:
ionic compounds: do not forget to switch the valence charges!!
acids: go through the flow chart from the \identifying
covalent compounds, ionic, the acids!!\
ionic/covalent/acid
lithium oxide
li₂o
calcium bromide
__________
__________
nitric acid
__________
__________
phosphoric acid
__________
__________
sulfur hexafluoride
__________
__________
copper (ii) sulfate
__________
__________
ammonium nitrate
__________
__________
hydrogen fluoride
__________
__________
hydrofluoric acid
__________
__________
potassium hydroxide
__________
__________
sodium phosphate
__________
__________
dinitrogen dichloride
__________
__________

Explanation:

Step1: Analyze Calcium Bromide

Calcium (Ca) has a charge of \( +2 \) (group 2 metal), bromide (Br⁻) has a charge of \( -1 \). To balance charges, we need 2 Br⁻ for 1 Ca²⁺. So the formula is \( \text{CaBr}_2 \). Calcium bromide is ionic (metal + non - metal).

Step2: Analyze Nitric Acid

Nitric acid is an acid with the anion nitrate (\( \text{NO}_3^- \)) and \( \text{H}^+ \). The formula is \( \text{HNO}_3 \). It is an acid (and covalent in structure, but as an acid, we classify it as an acid; the bonding between N and O is covalent, but H⁺ and \( \text{NO}_3^- \) interaction is ionic - like in solution, but the compound is covalent in the molecular sense, but for the table, we note it as an acid.

Step3: Analyze Phosphoric Acid

Phosphoric acid has the anion phosphate (\( \text{PO}_4^{3 - } \)) and \( \text{H}^+ \). To balance the 3 - charge of phosphate, we need 3 \( \text{H}^+ \), so the formula is \( \text{H}_3\text{PO}_4 \). It is an acid.

Step4: Analyze Sulfur Hexafluoride

Sulfur (S) and fluorine (F). Sulfur hexafluoride is a covalent compound (non - metals). The formula is \( \text{SF}_6 \) (hexa means 6, so 6 F atoms bonded to S).

Step5: Analyze Copper (II) Sulfate

Copper (II) has a charge of \( +2 \) (\( \text{Cu}^{2+} \)), sulfate is \( \text{SO}_4^{2 - } \). The charges balance (2+ and 2 - ), so the formula is \( \text{CuSO}_4 \). It is ionic (metal ion + polyatomic ion).

Step6: Analyze Ammonium Nitrate

Ammonium is \( \text{NH}_4^+ \), nitrate is \( \text{NO}_3^- \). The charges balance (1+ and 1 - ), so the formula is \( \text{NH}_4\text{NO}_3 \). It is ionic (polyatomic ions).

Step7: Analyze Hydrogen Fluoride

Hydrogen fluoride is a covalent compound (H and F, both non - metals). The formula is \( \text{HF} \).

Step8: Analyze Hydrofluoric Acid

Hydrofluoric acid is \( \text{HF} \) in aqueous solution. It is an acid (H⁺ and F⁻; the compound \( \text{HF} \) is covalent, but as an acid, we classify it as an acid.

Step9: Analyze Potassium Hydroxide

Potassium (K⁺, group 1 metal) and hydroxide (\( \text{OH}^- \)). The formula is \( \text{KOH} \). It is ionic (metal + polyatomic ion).

Step10: Analyze Sodium Phosphate

Sodium (Na⁺, group 1 metal) and phosphate (\( \text{PO}_4^{3 - } \)). To balance the 3 - charge of phosphate, we need 3 Na⁺, so the formula is \( \text{Na}_3\text{PO}_4 \). It is ionic.

Step11: Analyze Dinitrogen Dichloride

Dinitrogen means 2 N atoms, dichloride means 2 Cl atoms. The formula is \( \text{N}_2\text{Cl}_2 \) (covalent, non - metals).

CompoundFormulaIonic/Covalent/Acid
Nitric Acid\( \text{HNO}_3 \)Acid
Phosphoric Acid\( \text{H}_3\text{PO}_4 \)Acid
Sulfur Hexafluoride\( \text{SF}_6 \)Covalent
Copper (II) Sulfate\( \text{CuSO}_4 \)Ionic
Ammonium Nitrate\( \text{NH}_4\text{NO}_3 \)Ionic
Hydrogen Fluoride\( \text{HF} \)Covalent
Hydrofluoric Acid\( \text{HF} \)Acid
Potassium Hydroxide\( \text{KOH} \)Ionic
Sodium Phosphate\( \text{Na}_3\text{PO}_4 \)Ionic
Dinitrogen Dichloride\( \text{N}_2\text{Cl}_2 \)Covalent

Answer:

The formulas and classifications for each compound are as shown in the above table.