unit 9: mixed ionic compounds
directions: answer each question completely. indicate how you determined the answer to
each question below.
1) write the cation, anion, and formula for each of the following mixed ionic compounds.
cation anion formula
a. barium sulfite ____ ____ _____________________
b. lead(iv) phosphate ____ ____ _____________________
c. magnesium sulfate ____ ____ _____________________
d. calcium oxide ____ ____ _____________________
e. sodium dichromate ____ ____ _____________________
f. magnesium oxide ____ ____ _____________________
g. copper(ii) nitrate ____ ____ _____________________
h. manganese(ii) oxide ____ ____ _____________________
i. silver(i) nitrate ____ ____ _____________________
j. iron(iii) oxide ____ ____ _____________________
k. sodium carbonate ____ ____ _____________________
l. copper(ii) chloride ____ ____ _____________________
m. tin(iv) oxide ____ ____ _____________________
n. barium chromate ____ ____ _____________________

Question

unit 9: mixed ionic compounds
directions: answer each question completely. indicate how you determined the answer to
each question below.
1) write the cation, anion, and formula for each of the following mixed ionic compounds.
				cation	anion	formula
	 a. barium sulfite	____	____	_____________________
	 b. lead(iv) phosphate	____	____	_____________________
	 c. magnesium sulfate	____	____	_____________________
	 d. calcium oxide	____	____	_____________________
	 e. sodium dichromate	____	____	_____________________
	 f. magnesium oxide	____	____	_____________________
	 g. copper(ii) nitrate	____	____	_____________________
	 h. manganese(ii) oxide	____	____	_____________________
	 i. silver(i) nitrate	____	____	_____________________
	 j. iron(iii) oxide	____	____	_____________________
	 k. sodium carbonate	____	____	_____________________
	 l. copper(ii) chloride	____	____	_____________________
	 m. tin(iv) oxide	____	____	_____________________
	 n. barium chromate	____	____	_____________________

Accepted Answer

### Part a: Barium Sulfite
# Explanation:
## Step 1: Identify Cation and Anion
Barium is a Group 2 metal, so its cation is $ \text{Ba}^{2+} $. Sulfite is a polyatomic ion with formula $ \text{SO}_3^{2-} $.

## Step 2: Determine Formula
To balance charges, the number of $ \text{Ba}^{2+} $ and $ \text{SO}_3^{2-} $ ions must be such that total positive charge equals total negative charge. Since both have a charge magnitude of 2, the formula is $ \text{BaSO}_3 $.

# Answer:
- Cation: $ \text{Ba}^{2+} $
- Anion: $ \text{SO}_3^{2-} $
- Formula: $ \text{BaSO}_3 $

### Part b: Lead(IV) Phosphate
# Explanation:
## Step 1: Identify Cation and Anion
Lead(IV) means the cation is $ \text{Pb}^{4+} $. Phosphate is a polyatomic ion with formula $ \text{PO}_4^{3-} $.

## Step 2: Balance Charges
To balance $ 4+ $ and $ 3- $ charges, find the least common multiple of 4 and 3, which is 12. So we need 3 $ \text{Pb}^{4+} $ ions (total positive charge $ 3 \times 4 = 12 $) and 4 $ \text{PO}_4^{3-} $ ions (total negative charge $ 4 \times 3 = 12 $). Thus, the formula is $ \text{Pb}_3(\text{PO}_4)_4 $.

# Answer:
- Cation: $ \text{Pb}^{4+} $
- Anion: $ \text{PO}_4^{3-} $
- Formula: $ \text{Pb}_3(\text{PO}_4)_4 $

### Part c: Magnesium Sulfate
# Explanation:
## Step 1: Identify Cation and Anion
Magnesium is a Group 2 metal, so its cation is $ \text{Mg}^{2+} $. Sulfate is a polyatomic ion with formula $ \text{SO}_4^{2-} $.

## Step 2: Determine Formula
The charges $ 2+ $ and $ 2- $ balance in a 1:1 ratio. So the formula is $ \text{MgSO}_4 $.

# Answer:
- Cation: $ \text{Mg}^{2+} $
- Anion: $ \text{SO}_4^{2-} $
- Formula: $ \text{MgSO}_4 $

### Part d: Calcium Oxide
# Explanation:
## Step 1: Identify Cation and Anion
Calcium is a Group 2 metal, so its cation is $ \text{Ca}^{2+} $. Oxide is a monatomic ion with formula $ \text{O}^{2-} $.

## Step 2: Determine Formula
The charges $ 2+ $ and $ 2- $ balance in a 1:1 ratio. So the formula is $ \text{CaO} $.

# Answer:
- Cation: $ \text{Ca}^{2+} $
- Anion: $ \text{O}^{2-} $
- Formula: $ \text{CaO} $

### Part e: Sodium Dichromate
# Explanation:
## Step 1: Identify Cation and Anion
Sodium is a Group 1 metal, so its cation is $ \text{Na}^+ $. Dichromate is a polyatomic ion with formula $ \text{Cr}_2\text{O}_7^{2-} $.

## Step 2: Balance Charges
To balance $ 1+ $ and $ 2- $ charges, we need 2 $ \text{Na}^+ $ ions (total positive charge $ 2 \times 1 = 2 $) and 1 $ \text{Cr}_2\text{O}_7^{2-} $ ion (total negative charge $ 2 $). Thus, the formula is $ \text{Na}_2\text{Cr}_2\text{O}_7 $.

# Answer:
- Cation: $ \text{Na}^+ $
- Anion: $ \text{Cr}_2\text{O}_7^{2-} $
- Formula: $ \text{Na}_2\text{Cr}_2\text{O}_7 $

### Part f: Magnesium Oxide
# Explanation:
## Step 1: Identify Cation and Anion
Magnesium is a Group 2 metal, so its cation is $ \text{Mg}^{2+} $. Oxide is a monatomic ion with formula $ \text{O}^{2-} $.

## Step 2: Determine Formula
The charges $ 2+ $ and $ 2- $ balance in a 1:1 ratio. So the formula is $ \text{MgO} $.

# Answer:
- Cation: $ \text{Mg}^{2+} $
- Anion: $ \text{O}^{2-} $
- Formula: $ \text{MgO} $

### Part g: Copper(II) Nitrate
# Explanation:
## Step 1: Identify Cation and Anion
Copper(II) means the cation is $ \text{Cu}^{2+} $. Nitrate is a polyatomic ion with formula $ \text{NO}_3^- $.

## Step 2: Balance Charges
To balance $ 2+ $ and $ 1- $ charges, we need 1 $ \text{Cu}^{2+} $ ion and 2 $ \text{NO}_3^- $ ions (total negative charge $ 2 \times 1 = 2 $). Thus, the formula is $ \text{Cu(NO}_3\text{)}_2 $.

# Answer:
- Cation: $ \text{Cu}^{2+} $
- Anion: $ \text{NO}_3^- $
- Formula: $ \text{Cu(NO}_3\text{)}_2 $

### Part h: Manganese(II) Oxide
# Explanation:
## Step 1: Identify Cation and Anion
Manganese(II) means the cation is $ \text{Mn}^{2+} $. Oxide is a monatomic ion with formula $ \text{O}^{2-} $.

## Step 2: Determine Formula
The charges $ 2+ $ and $ 2- $ balance in a 1:1 ratio. So the formula is $ \text{MnO} $.

# Answer:
- Cation: $ \text{Mn}^{2+} $
- Anion: $ \text{O}^{2-} $
- Formula: $ \text{MnO} $

### Part i: Silver(I) Nitrate
# Explanation:
## Step 1: Identify Cation and Anion
Silver(I) means the cation is $ \text{Ag}^+ $. Nitrate is a polyatomic ion with formula $ \text{NO}_3^- $.

## Step 2: Balance Charges
The charges $ 1+ $ and $ 1- $ balance in a 1:1 ratio. So the formula is $ \text{AgNO}_3 $.

# Answer:
- Cation: $ \text{Ag}^+ $
- Anion: $ \text{NO}_3^- $
- Formula: $ \text{AgNO}_3 $

### Part j: Iron(III) Oxide
# Explanation:
## Step 1: Identify Cation and Anion
Iron(III) means the cation is $ \text{Fe}^{3+} $. Oxide is a monatomic ion with formula $ \text{O}^{2-} $.

## Step 2: Balance Charges
To balance $ 3+ $ and $ 2- $ charges, find the least common multiple of 3 and 2, which is 6. So we need 2 $ \text{Fe}^{3+} $ ions (total positive charge $ 2 \times 3 = 6 $) and 3 $ \text{O}^{2-} $ ions (total negative charge $ 3 \times 2 = 6 $). Thus, the formula is $ \text{Fe}_2\text{O}_3 $.

# Answer:
- Cation: $ \text{Fe}^{3+} $
- Anion: $ \text{O}^{2-} $
- Formula: $ \text{Fe}_2\text{O}_3 $

### Part k: Sodium Carbonate
# Explanation:
## Step 1: Identify Cation and Anion
Sodium is a Group 1 metal, so its cation is $ \text{Na}^+ $. Carbonate is a polyatomic ion with formula $ \text{CO}_3^{2-} $.

## Step 2: Balance Charges
To balance $ 1+ $ and $ 2- $ charges, we need 2 $ \text{Na}^+ $ ions (total positive charge $ 2 \times 1 = 2 $) and 1 $ \text{CO}_3^{2-} $ ion (total negative charge $ 2 $). Thus, the formula is $ \text{Na}_2\text{CO}_3 $.

# Answer:
- Cation: $ \text{Na}^+ $
- Anion: $ \text{CO}_3^{2-} $
- Formula: $ \text{Na}_2\text{CO}_3 $

### Part l: Copper(II) Chloride
# Explanation:
## Step 1: Identify Cation and Anion
Copper(II) means the cation is $ \text{Cu}^{2+} $. Chloride is a monatomic ion with formula $ \text{Cl}^- $.

## Step 2: Balance Charges
To balance $ 2+ $ and $ 1- $ charges, we need 1 $ \text{Cu}^{2+} $ ion and 2 $ \text{Cl}^- $ ions (total negative charge $ 2 \times 1 = 2 $). Thus, the formula is $ \text{CuCl}_2 $.

# Answer:
- Cation: $ \text{Cu}^{2+} $
- Anion: $ \text{Cl}^- $
- Formula: $ \text{CuCl}_2 $

### Part m: Tin(IV) Oxide
# Explanation:
## Step 1: Identify Cation and Anion
Tin(IV) means the cation is $ \text{Sn}^{4+} $. Oxide is a monatomic ion with formula $ \text{O}^{2-} $.

## Step 2: Balance Charges
To balance $ 4+ $ and $ 2- $ charges, we need 1 $ \text{Sn}^{4+} $ ion and 2 $ \text{O}^{2-} $ ions (total negative charge $ 2 \times 2 = 4 $). Thus, the formula is $ \text{SnO}_2 $.

# Answer:
- Cation: $ \text{Sn}^{4+} $
- Anion: $ \text{O}^{2-} $
- Formula: $ \text{SnO}_2 $

### Part n: Barium Chromate
# Explanation:
## Step 1: Identify Cation and Anion
Barium is a Group 2 metal, so its cation is $ \text{Ba}^{2+} $. Chromate is a polyatomic ion with formula $ \text{CrO}_4^{2-} $.

## Step 2: Determine Formula
The charges $ 2+ $ and $ 2- $ balance in a 1:1 ratio. So the formula is $ \text{BaCrO}_4 $.

# Answer:
- Cation: $ \text{Ba}^{2+} $
- Anion: $ \text{CrO}_4^{2-} $
- Formula: $ \text{BaCrO}_4 $

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QUESTION IMAGE

Question

Explanation:

Part a: Barium Sulfite

Step 1: Identify Cation and Anion

Step 2: Determine Formula

Step 1: Identify Cation and Anion

Step 2: Balance Charges

Step 1: Identify Cation and Anion

Step 2: Determine Formula

Answer:

Part b: Lead(IV) Phosphate