for each of the following names, write the appropriate formula for it.
1) dinitrogen trioxide
2) nitrous acid
3) hydroiodic acid
4) vanadium (v) oxide
5) aluminum hydroxide
6) silver phosphate
7) potassium acetate
8) disilicon hexabromide
9) titanium (iv) nitrate
10) copper (i) phosphate
11) tetrasulfur dinitride
12) phosphoric acid
13) carbonic acid
14) manganese (ii) phosphate
15) ammonium oxide
16) carbon tetrachloride

Question

Accepted Answer

### 1) dinitrogen trioxide
# Explanation:
## Step1: Identify prefixes. "Di-" means 2, "tri-" means 3.
## Step2: Elements are nitrogen (N) and oxygen (O). Combine with counts: $N_2O_3$ (Note: Original answer had Na which is sodium, wrong. Correct is N for nitrogen.)
# Answer: $N_2O_3$

### 2) nitrous acid
# Explanation:
## Step1: Nitrous acid is an oxyacid. The nitrite ion is $NO_2^-$, and acid has $H^+$.
## Step2: Combine $H^+$ and $NO_2^-$: $HNO_2$ (charge balance: $H^+$ (1+) and $NO_2^-$ (1-), so 1:1 ratio)
# Answer: $HNO_2$

### 3) hydroiodic acid
# Explanation:
## Step1: Hydroiodic acid is a binary acid (H + nonmetal). Nonmetal is iodine (I).
## Step2: Formula for binary acid: $HI$ (H + I, 1:1 ratio as H is +1, I is -1 in acid)
# Answer: $HI$

### 4) vanadium (V) oxide
# Explanation:
## Step1: Vanadium (V) has charge +5, oxygen has charge -2.
## Step2: Use crisscross method: V^5+ and O^2- → $V_2O_5$ (5 from V to O, 2 from O to V, so subscripts 2 and 5)
# Answer: $V_2O_5$

### 5) aluminum hydroxide
# Explanation:
## Step1: Aluminum (Al) has charge +3, hydroxide ($OH^-$) has charge -1.
## Step2: Crisscross: Al^3+ and $OH^-$ → $Al(OH)_3$ (3 from Al to OH, 1 from OH to Al, so 3 OH^- per Al^3+)
# Answer: $Al(OH)_3$ (Original answer had Al only, missing $OH^-$)

### 6) silver phosphate
# Explanation:
## Step1: Silver (Ag) has charge +1 (Ag^+), phosphate ($PO_4^{3-}$) has charge -3.
## Step2: Crisscross: 3 Ag^+ to balance 1 $PO_4^{3-}$ → $Ag_3PO_4$ (Original answer had Fe which is iron, wrong. Correct is Ag for silver.)
# Answer: $Ag_3PO_4$

### 7) potassium acetate
# Explanation:
## Step1: Potassium (K) has charge +1, acetate ($C_2H_3O_2^-$) has charge -1.
## Step2: Combine 1 K^+ and 1 $C_2H_3O_2^-$ → $KC_2H_3O_2$ (Original answer had K only, missing acetate ion)
# Answer: $KC_2H_3O_2$

### 8) disilicon hexabromide
# Explanation:
## Step1: "Di-" = 2 (Si), "hexa-" = 6 (Br). Elements: silicon (Si) and bromine (Br).
## Step2: Combine with counts: $Si_2Br_6$ (Original answer had typo, correct subscripts 2 and 6)
# Answer: $Si_2Br_6$

### 9) titanium (IV) nitrate
# Explanation:
## Step1: Titanium (IV) has charge +4, nitrate ($NO_3^-$) has charge -1.
## Step2: Crisscross: 1 Ti^4+ and 4 $NO_3^-$ → $Ti(NO_3)_4$ (4 $NO_3^-$ to balance +4 charge of Ti)
# Answer: $Ti(NO_3)_4$

### 10) copper (I) phosphate
# Explanation:
## Step1: Copper (I) has charge +1 (Cu^+), phosphate ($PO_4^{3-}$) has charge -3.
## Step2: Crisscross: 3 Cu^+ to balance 1 $PO_4^{3-}$ → $Cu_3PO_4$ (Original answer had Cu only, missing phosphate ion)
# Answer: $Cu_3PO_4$

### 11) tetrasulfur dinitride
# Explanation:
## Step1: "Tetra-" = 4 (S), "di-" = 2 (N). Elements: sulfur (S) and nitrogen (N).
## Step2: Combine with counts: $S_4N_2$ (Original answer was correct here)
# Answer: $S_4N_2$

### 12) phosphoric acid
# Explanation:
## Step1: Phosphoric acid is an oxyacid. Phosphate ion is $PO_4^{3-}$, acid has $H^+$.
## Step2: Combine 3 $H^+$ (to balance -3 charge of $PO_4^{3-}$) → $H_3PO_4$ (Original answer had H3P which is wrong, correct is $H_3PO_4$)
# Answer: $H_3PO_4$

### 13) carbonic acid
# Explanation:
## Step1: Carbonic acid is $H_2CO_3$ (carbonate ion $CO_3^{2-}$, so 2 $H^+$ to balance -2 charge). Original answer was correct here.
# Answer: $H_2CO_3$

### 14) manganese (II) phosphate
# Explanation:
## Step1: Manganese (II) has charge +2 (Mn^2+), phosphate ($PO_4^{3-}$) has charge -3.
## Step2: Crisscross: 3 Mn^2+ and 2 $PO_4^{3-}$ (LCM of 2 and 3 is 6, so 3 Mn^2+ (3×2=6+) and 2 $PO_4^{3-}$ (2×3=6-)) → $Mn_3(PO_4)_2$ (Original answer had Mn only, missing phosphate ion)
# Answer: $Mn_3(PO_4)_2$

### 15) ammonium oxide
# Explanation:
## Step1: Ammonium ion is $NH_4^+$, oxide ion is $O^{2-}$.
## Step2: Crisscross: 2 $NH_4^+$ (2×1=2+) and 1 $O^{2-}$ (2-) → $(NH_4)_2O$ (Original answer had $NH_4O$ which is wrong, need 2 $NH_4^+$ to balance $O^{2-}$)
# Answer: $(NH_4)_2O$

### 16) carbon tetrachloride
# Explanation:
## Step1: "Tetra-" = 4 (Cl). Elements: carbon (C) and chlorine (Cl).
## Step2: Combine with counts: $CCl_4$ (1 C, 4 Cl)
# Answer: $CCl_4$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

1) dinitrogen trioxide

Step1: Identify prefixes. "Di-" means 2, "tri-" means 3.

Step2: Elements are nitrogen (N) and oxygen (O). Combine with counts: $N_2O_3$ (Note: Original answer had Na which is sodium, wrong. Correct is N for nitrogen.)

Step1: Nitrous acid is an oxyacid. The nitrite ion is $NO_2^-$, and acid has $H^+$.

Step2: Combine $H^+$ and $NO_2^-$: $HNO_2$ (charge balance: $H^+$ (1+) and $NO_2^-$ (1-), so 1:1 ratio)

Step1: Hydroiodic acid is a binary acid (H + nonmetal). Nonmetal is iodine (I).

Step2: Formula for binary acid: $HI$ (H + I, 1:1 ratio as H is +1, I is -1 in acid)

Answer:

2) nitrous acid