chemistry
identifying compound names and formulas
write the name for each of the following compounds. identify each as ionic (i) or covalent (c).
__c__ 1. n₂s₃
_____ 2. nh₄cl
_____ 3. cuf
_____ 4. cuf₂
_____ 5. na₂s
_____ 6. hg(no₃)₂
_____ 7. al₂o₃
_____ 8. so₃
_____ 9. co₂
_____ 10. pbso₄
write the chemical formula for each of the following compounds. identify each as ionic (i) or covalent (c).
__i__ 11. nickel (ii) chloride
_____ 12. copper (ii) nitrate
_____ 13. sulfur dioxide
_____ 14. magnesium nitride
_____ 15. mercury (i) sulfide
_____ 16. carbon monoxide
_____ 17. iron (ii) oxide
_____ 18. diphosphorus pentoxide
_____ 19. dihydrogen monoxide
_____ 20. ammonium carbonate

Question

Accepted Answer

### Problem 1: $\ce{N_{2}S_{3}}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{N_{2}S_{3}}$ consists of non - metals (nitrogen and sulfur). Compounds formed between non - metals are covalent. So the type is covalent (C).
## Step 2: Name the Compound
For covalent compounds, we use prefixes. The prefix for 2 is "di - " and for 3 is "tri - ". So the name is dinitrogen trisulfide.
# Answer:
Type: C, Name: dinitrogen trisulfide

### Problem 2: $\ce{NH_{4}Cl}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{NH_{4}Cl}$ is composed of the ammonium ion ($\ce{NH_{4}^{+}}$) and the chloride ion ($\ce{Cl^{-}}$). Compounds with ions are ionic. So the type is ionic (I).
## Step 2: Name the Compound
It is ammonium chloride (the ammonium ion combines with the chloride ion).
# Answer:
Type: I, Name: ammonium chloride

### Problem 3: $\ce{CuF}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{CuF}$ has a metal (copper) and a non - metal (fluorine). Compounds with a metal and non - metal are ionic. So the type is ionic (I).
## Step 2: Determine Copper's Charge
Fluoride has a charge of - 1. Let the charge of copper be $x$. Since the compound is neutral, $x+( - 1)=0$, so $x = + 1$. So copper is in the + 1 oxidation state (cuprous).
## Step 3: Name the Compound
The name is copper(I) fluoride (or cuprous fluoride).
# Answer:
Type: I, Name: copper(I) fluoride (cuprous fluoride)

### Problem 4: $\ce{CuF_{2}}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{CuF_{2}}$ has a metal (copper) and a non - metal (fluorine), so it is ionic (I).
## Step 2: Determine Copper's Charge
Fluoride has a charge of - 1. Let the charge of copper be $x$. There are 2 fluoride ions, so $x+2	imes( - 1)=0$, $x = + 2$. Copper is in the + 2 oxidation state (cupric).
## Step 3: Name the Compound
The name is copper(II) fluoride (or cupric fluoride).
# Answer:
Type: I, Name: copper(II) fluoride (cupric fluoride)

### Problem 5: $\ce{Na_{2}S}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{Na_{2}S}$ has a metal (sodium) and a non - metal (sulfur), so it is ionic (I).
## Step 2: Name the Compound
Sodium has a charge of + 1, sulfur has a charge of - 2. The name is sodium sulfide (since the ratio of sodium to sulfur is 2:1 to balance the charges).
# Answer:
Type: I, Name: sodium sulfide

### Problem 6: $\ce{Hg(NO_{3})_{2}}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{Hg(NO_{3})_{2}}$ is made up of the mercury(II) ion ($\ce{Hg^{2 + }}$) and the nitrate ion ($\ce{NO_{3}^{-}}$). So it is ionic (I).
## Step 2: Name the Compound
The name is mercury(II) nitrate (mercury in + 2 oxidation state combined with nitrate ions).
# Answer:
Type: I, Name: mercury(II) nitrate

### Problem 7: $\ce{Al_{2}O_{3}}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{Al_{2}O_{3}}$ has a metal (aluminum) and a non - metal (oxygen), so it is ionic (I).
## Step 2: Name the Compound
Aluminum has a charge of + 3, oxygen has a charge of - 2. The name is aluminum oxide (the ratio 2:3 balances the charges: $2	imes( + 3)+3	imes( - 2)=0$).
# Answer:
Type: I, Name: aluminum oxide

### Problem 8: $\ce{SO_{3}}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{SO_{3}}$ consists of non - metals (sulfur and oxygen). So it is a covalent compound (C).
## Step 2: Name the Compound
Using prefixes, the prefix for 1 (sulfur, we don't use "mono - " for the first element) and for 3 is "tri - ". So the name is sulfur trioxide.
# Answer:
Type: C, Name: sulfur trioxide

### Problem 9: $\ce{CO_{2}}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{CO_{2}}$ has non - metals (carbon and oxygen), so it is covalent (C).
## Step 2: Name the Compound
Using prefixes, the prefix for 1 (carbon) is not "mono - ", and for 2 (oxygen) is "di - ". So the name is carbon dioxide.
# Answer:
Type: C, Name: carbon dioxide

### Problem 10: $\ce{PbSO_{4}}$
# Explanation:
## Step 1: Identify Bond Type
$\ce{PbSO_{4}}$ is composed of the lead ion (let's assume lead is in + 2 state here, $\ce{Pb^{2+}}$) and the sulfate ion ($\ce{SO_{4}^{2 - }}$). So it is ionic (I).
## Step 2: Name the Compound
The name is lead(II) sulfate (assuming lead is in + 2 oxidation state, as the charges of $\ce{Pb^{2+}}$ and $\ce{SO_{4}^{2 - }}$ balance).
# Answer:
Type: I, Name: lead(II) sulfate

### Problem 11: nickel (II) chloride
# Explanation:
## Step 1: Identify Bond Type
Nickel (II) chloride has a metal (nickel) and a non - metal (chlorine), so it is ionic (I).
## Step 2: Write the Formula
Nickel(II) has a charge of + 2 ($\ce{Ni^{2+}}$), chloride has a charge of - 1 ($\ce{Cl^{-}}$). To balance the charges, we need 2 chloride ions for 1 nickel ion. So the formula is $\ce{NiCl_{2}}$.
# Answer:
Type: I, Formula: $\ce{NiCl_{2}}$

### Problem 12: copper (II) nitrate
# Explanation:
## Step 1: Identify Bond Type
Copper (II) nitrate has a metal (copper) and a polyatomic ion (nitrate), so it is ionic (I).
## Step 2: Write the Formula
Copper(II) has a charge of + 2 ($\ce{Cu^{2+}}$), nitrate has a charge of - 1 ($\ce{NO_{3}^{-}}$). To balance the charges, we need 2 nitrate ions for 1 copper ion. So the formula is $\ce{Cu(NO_{3})_{2}}$.
# Answer:
Type: I, Formula: $\ce{Cu(NO_{3})_{2}}$

### Problem 13: sulfur dioxide
# Explanation:
## Step 1: Identify Bond Type
Sulfur dioxide has non - metals (sulfur and oxygen), so it is covalent (C).
## Step 2: Write the Formula
Using prefixes, "di - " for oxygen means 2 oxygen atoms. Sulfur is represented as $\ce{S}$, oxygen as $\ce{O}$. So the formula is $\ce{SO_{2}}$.
# Answer:
Type: C, Formula: $\ce{SO_{2}}$

### Problem 14: magnesium nitride
# Explanation:
## Step 1: Identify Bond Type
Magnesium nitride has a metal (magnesium) and a non - metal (nitrogen), so it is ionic (I).
## Step 2: Write the Formula
Magnesium has a charge of + 2 ($\ce{Mg^{2+}}$), nitride has a charge of - 3 ($\ce{N^{3 - }}$). To balance the charges, we need 3 magnesium ions and 2 nitride ions (since $3	imes( + 2)+2	imes( - 3)=0$). So the formula is $\ce{Mg_{3}N_{2}}$.
# Answer:
Type: I, Formula: $\ce{Mg_{3}N_{2}}$

### Problem 15: mercury (I) sulfide
# Explanation:
## Step 1: Identify Bond Type
Mercury (I) sulfide has a metal (mercury) and a non - metal (sulfur), so it is ionic (I).
## Step 2: Write the Formula
Mercury(I) exists as $\ce{Hg_{2}^{2+}}$ (a di - mercury ion with + 2 charge overall, so each $\ce{Hg}$ is + 1), sulfide has a charge of - 2 ($\ce{S^{2 - }}$). To balance the charges, we need 1 $\ce{Hg_{2}^{2+}}$ ion and 1 $\ce{S^{2 - }}$ ion. So the formula is $\ce{Hg_{2}S}$.
# Answer:
Type: I, Formula: $\ce{Hg_{2}S}$

### Problem 16: carbon monoxide
# Explanation:
## Step 1: Identify Bond Type
Carbon monoxide has non - metals (carbon and oxygen), so it is covalent (C).
## Step 2: Write the Formula
Using prefixes, "mono - " for oxygen (since there is 1 oxygen atom). Carbon is $\ce{C}$, oxygen is $\ce{O}$. So the formula is $\ce{CO}$.
# Answer:
Type: C, Formula: $\ce{CO}$

### Problem 17: iron (II) oxide
# Explanation:
## Step 1: Identify Bond Type
Iron (II) oxide has a metal (iron) and a non - metal (oxygen), so it is ionic (I).
## Step 2: Write the Formula
Iron(II) has a charge of + 2 ($\ce{Fe^{2+}}$), oxygen has a charge of - 2 ($\ce{O^{2 - }}$). The ratio of iron to oxygen is 1:1 to balance the charges. So the formula is $\ce{FeO}$.
# Answer:
Type: I, Formula: $\ce{FeO}$

### Problem 18: diphosphorus pentoxide
# Explanation:
## Step 1: Identify Bond Type
Diphosphorus pentoxide has non - metals (phosphorus and oxygen), so it is covalent (C).
## Step 2: Write the Formula
"Di - " for phosphorus means 2 phosphorus atoms, "penta - " for oxygen means 5 oxygen atoms. So the formula is $\ce{P_{2}O_{5}}$.
# Answer:
Type: C, Formula: $\ce{P_{2}O_{5}}$

### Problem 19: dihydrogen monoxide
# Explanation:
## Step 1: Identify Bond Type
Dihydrogen monoxide (water) has non - metals (hydrogen and oxygen), so it is covalent (C).
## Step 2: Write the Formula
"Dihydrogen" means 2 hydrogen atoms, "monoxide" means 1 oxygen atom. So the formula is $\ce{H_{2}O}$.
# Answer:
Type: C, Formula: $\ce{H_{2}O}$

### Problem 20: ammonium carbonate
# Explanation:
## Step 1: Identify Bond Type
Ammonium carbonate has the ammonium ion ($\ce{NH_{4}^{+}}$) and the carbonate ion ($\ce{CO_{3}^{2 - }}$), so it is ionic (I).
## Step 2: Write the Formula
Ammonium has a charge of + 1, carbonate has a charge of - 2. To balance the charges, we need 2 ammonium ions for 1 carbonate ion. So the formula is $\ce{(NH_{4})_{2}CO_{3}}$.
# Answer:
Type: I, Formula: $\ce{(NH_{4})_{2}CO_{3}}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1: $\ce{N_{2}S_{3}}$

Step 1: Identify Bond Type

Step 2: Name the Compound

Step 1: Identify Bond Type

Step 2: Name the Compound

Step 1: Identify Bond Type

Step 2: Determine Copper's Charge

Step 3: Name the Compound

Answer:

Problem 2: $\ce{NH_{4}Cl}$