write the formula for the following ionic compounds:
17. potassium bromide
18. calcium sulfate
19. copper (ii) acetate
20. zinc nitrate
21. ammonium chloride
22. barium hydroxide
23. tin (iv) oxide
write the name of the following ionic compounds: watch for roman numerals
24. sno₂
25. mg₃n₂
26. niso₄
27. cr₂o₃
28. ag₂co₃
29. cocl₃
30. ba(c₂h₃o₂)₂
calculate the molar masses for each of the formulas listed. don’t forget units!!!
31. a. na₂so₄ b. k₃po₄ c. al₂(co₃)₃
calculate percent by mass/percent composition.
32. a. na₂so₄ b. k₃po₄ c. al₂(co₃)₃

Question

write the formula for the following ionic compounds:
17. potassium bromide
18. calcium sulfate
19. copper (ii) acetate
20. zinc nitrate
21. ammonium chloride
22. barium hydroxide
23. tin (iv) oxide
write the name of the following ionic compounds: watch for roman numerals
24. sno₂
25. mg₃n₂
26. niso₄
27. cr₂o₃
28. ag₂co₃
29. cocl₃
30. ba(c₂h₃o₂)₂
calculate the molar masses for each of the formulas listed. don’t forget units!!!
31. a. na₂so₄  b. k₃po₄  c. al₂(co₃)₃
calculate percent by mass/percent composition.
32. a. na₂so₄  b. k₃po₄  c. al₂(co₃)₃

Accepted Answer

### 17. Potassium bromide
# Explanation:
## Step1: Identify ions
Potassium ion: $K^+$, Bromide ion: $Br^-$
## Step2: Balance charges
Charge of $K^+$ is +1, charge of $Br^-$ is -1. So ratio is 1:1.
# Answer: $KBr$

### 18. Calcium sulfate
# Explanation:
## Step1: Identify ions
Calcium ion: $Ca^{2+}$, Sulfate ion: $SO_4^{2-}$
## Step2: Balance charges
Charge of $Ca^{2+}$ is +2, charge of $SO_4^{2-}$ is -2. Ratio is 1:1.
# Answer: $CaSO_4$

### 19. Copper (II) acetate
# Explanation:
## Step1: Identify ions
Copper (II) ion: $Cu^{2+}$, Acetate ion: $C_2H_3O_2^-$
## Step2: Balance charges
Charge of $Cu^{2+}$ is +2, charge of $C_2H_3O_2^-$ is -1. So we need 2 acetate ions.
# Answer: $Cu(C_2H_3O_2)_2$

### 20. Zinc nitrate
# Explanation:
## Step1: Identify ions
Zinc ion: $Zn^{2+}$, Nitrate ion: $NO_3^-$
## Step2: Balance charges
Charge of $Zn^{2+}$ is +2, charge of $NO_3^-$ is -1. So we need 2 nitrate ions.
# Answer: $Zn(NO_3)_2$

### 21. Ammonium chloride
# Explanation:
## Step1: Identify ions
Ammonium ion: $NH_4^+$, Chloride ion: $Cl^-$
## Step2: Balance charges
Charge of $NH_4^+$ is +1, charge of $Cl^-$ is -1. Ratio is 1:1.
# Answer: $NH_4Cl$

### 22. Barium hydroxide
# Explanation:
## Step1: Identify ions
Barium ion: $Ba^{2+}$, Hydroxide ion: $OH^-$
## Step2: Balance charges
Charge of $Ba^{2+}$ is +2, charge of $OH^-$ is -1. So we need 2 hydroxide ions.
# Answer: $Ba(OH)_2$

### 23. Tin (IV) oxide
# Explanation:
## Step1: Identify ions
Tin (IV) ion: $Sn^{4+}$, Oxide ion: $O^{2-}$
## Step2: Balance charges
Charge of $Sn^{4+}$ is +4, charge of $O^{2-}$ is -2. So ratio of Sn to O is 1:2 (since $4 \div 2 = 2$ for O).
# Answer: $SnO_2$

### 24. $SnO_2$
# Brief Explanations:
Tin has a +4 charge (from roman numeral in formula name logic, since $O$ is -2, 2 O gives -4, so Sn is +4). So name is Tin (IV) oxide.
# Answer: Tin (IV) oxide

### 25. $Mg_3N_2$
# Brief Explanations:
Magnesium ion is $Mg^{2+}$, Nitride ion is $N^{3-}$. Name is Magnesium nitride (no roman numeral as Mg has fixed +2 charge).
# Answer: Magnesium nitride

### 26. $NiSO_4$
# Brief Explanations:
$Ni$ here has +2 charge (since $SO_4^{2-}$ is -2, so Ni is +2). $SO_4^{2-}$ is sulfate. So name is Nickel (II) sulfate.
# Answer: Nickel (II) sulfate

### 27. $Cr_2O_3$
# Brief Explanations:
O is -2, 3 O gives -6. So 2 Cr must have +6 total, so each Cr is +3. Name is Chromium (III) oxide.
# Answer: Chromium (III) oxide

### 28. $Ag_2CO_3$
# Brief Explanations:
$Ag^+$ is silver ion, $CO_3^{2-}$ is carbonate. Name is Silver carbonate (Ag has fixed +1 charge).
# Answer: Silver carbonate

### 29. $CoCl_3$
# Brief Explanations:
Cl is -1, 3 Cl gives -3. So Co is +3. Name is Cobalt (III) chloride.
# Answer: Cobalt (III) chloride

### 30. $Ba(C_2H_3O_2)_2$
# Brief Explanations:
$Ba^{2+}$ is barium ion, $C_2H_3O_2^-$ is acetate. Name is Barium acetate.
# Answer: Barium acetate

### 31a. $Na_2SO_4$
# Explanation:
## Step1: Atomic masses
Na: $22.99$ g/mol, S: $32.07$ g/mol, O: $16.00$ g/mol
## Step2: Calculate each atom's contribution
2 Na: $2 \times 22.99 = 45.98$
1 S: $32.07$
4 O: $4 \times 16.00 = 64.00$
## Step3: Sum them up
$45.98 + 32.07 + 64.00 = 142.05$ g/mol
# Answer: $142.05$ g/mol

### 31b. $K_3PO_4$
# Explanation:
## Step1: Atomic masses
K: $39.10$ g/mol, P: $30.97$ g/mol, O: $16.00$ g/mol
## Step2: Calculate each atom's contribution
3 K: $3 \times 39.10 = 117.30$
1 P: $30.97$
4 O: $4 \times 16.00 = 64.00$
## Step3: Sum them up
$117.30 + 30.97 + 64.00 = 212.27$ g/mol
# Answer: $212.27$ g/mol

### 31c. $Al_2(CO_3)_3$
# Explanation:
## Step1: Atomic masses
Al: $26.98$ g/mol, C: $12.01$ g/mol, O: $16.00$ g/mol
## Step2: Calculate each part
2 Al: $2 \times 26.98 = 53.96$
3 $CO_3$: each $CO_3$ has 1 C and 3 O. So 3 C: $3 \times 12.01 = 36.03$, 9 O: $9 \times 16.00 = 144.00$. Total for $CO_3$ part: $36.03 + 144.00 = 180.03$, times 3? Wait no, $Al_2(CO_3)_3$ has 2 Al, 3 C, 9 O.
Wait re - calculate: 2 Al: $2\times26.98 = 53.96$; 3 C: $3\times12.01 = 36.03$; 9 O: $9\times16.00 = 144.00$. Sum: $53.96+36.03 + 144.00=233.99\approx234.0$ g/mol.
# Answer: $234.0$ g/mol (or more precise $233.99$ g/mol)

### 32a. $Na_2SO_4$ (Percent composition of Na, S, O)
# Explanation:
## Step1: Molar mass of $Na_2SO_4$ is $142.05$ g/mol (from 31a)
## Step2: Percent of Na:
Molar mass of 2 Na: $2\times22.99 = 45.98$. Percent: $\frac{45.98}{142.05}\times100\%\approx32.37\%$
## Step3: Percent of S:
Molar mass of S: $32.07$. Percent: $\frac{32.07}{142.05}\times100\%\approx22.58\%$
## Step4: Percent of O:
Molar mass of 4 O: $4\times16.00 = 64.00$. Percent: $\frac{64.00}{142.05}\times100\%\approx45.05\%$
# Answer: Na: ~32.37%, S: ~22.58%, O: ~45.05%

### 32b. $K_3PO_4$ (Percent composition of K, P, O)
# Explanation:
## Step1: Molar mass of $K_3PO_4$ is $212.27$ g/mol (from 31b)
## Step2: Percent of K:
Molar mass of 3 K: $3\times39.10 = 117.30$. Percent: $\frac{117.30}{212.27}\times100\%\approx55.26\%$
## Step3: Percent of P:
Molar mass of P: $30.97$. Percent: $\frac{30.97}{212.27}\times100\%\approx14.59\%$
## Step4: Percent of O:
Molar mass of 4 O: $4\times16.00 = 64.00$. Percent: $\frac{64.00}{212.27}\times100\%\approx30.15\%$
# Answer: K: ~55.26%, P: ~14.59%, O: ~30.15%

### 32c. $Al_2(CO_3)_3$ (Percent composition of Al, C, O)
# Explanation:
## Step1: Molar mass of $Al_2(CO_3)_3$ is $233.99$ g/mol (from 31c)
## Step2: Percent of Al:
Molar mass of 2 Al: $2\times26.98 = 53.96$. Percent: $\frac{53.96}{233.99}\times100\%\approx23.06\%$
## Step3: Percent of C:
Molar mass of 3 C: $3\times12.01 = 36.03$. Percent: $\frac{36.03}{233.99}\times100\%\approx15.40\%$
## Step4: Percent of O:
Molar mass of 9 O: $9\times16.00 = 144.00$. Percent: $\frac{144.00}{233.99}\times100\%\approx61.54\%$
# Answer: Al: ~23.06%, C: ~15.40%, O: ~61.54%

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

17. Potassium bromide

Step1: Identify ions

Step2: Balance charges

Step1: Identify ions

Step2: Balance charges

Step1: Identify ions

Step2: Balance charges

Answer:

18. Calcium sulfate