QUESTION IMAGE
Question
a. copper (ii) sulfite _________
b. iron (ii) cyanide _________
c. sulfuric acid _________
d. ferric carbonate_________
e. calcium hypochlorite_________
f. barium nitrate_________
g. sodium chlorate_________
h. magnesium sulfate _________
i. iron (iii) nitrate_________
j. lithium perchlorate _________
k. sodium hydroxide _________
To determine the chemical formulas for each compound, we use the rules of ionic compound nomenclature (for ionic compounds) and acid nomenclature (for sulfuric acid). Here are the step - by - step solutions for each:
Part a: Copper (II) sulfite
Step 1: Identify ions
Copper (II) has a charge of $Cu^{2+}$, and sulfite is $SO_{3}^{2 -}$.
Step 2: Balance charges
The charges of $Cu^{2+}$ and $SO_{3}^{2 -}$ are equal in magnitude (2+ and 2 -), so the formula is $CuSO_{3}$.
Part b: Iron (II) cyanide
Step 1: Identify ions
Iron (II) is $Fe^{2+}$, and cyanide is $CN^{-}$.
Step 2: Balance charges
To balance the 2+ charge of $Fe^{2+}$ with the 1 - charge of $CN^{-}$, we need 2 $CN^{-}$ ions. So the formula is $Fe(CN)_{2}$.
Part c: Sulfuric acid
Step 1: Recall acid nomenclature
Sulfuric acid is a strong acid. The anion is sulfate ($SO_{4}^{2 -}$), and for an acid, we combine the anion with $H^{+}$ ions.
Step 2: Balance charges
To balance the 2 - charge of $SO_{4}^{2 -}$, we need 2 $H^{+}$ ions. So the formula is $H_{2}SO_{4}$.
Part d: Ferric carbonate
Step 1: Identify ions
Ferric means iron (III), so $Fe^{3+}$, and carbonate is $CO_{3}^{2 -}$.
Step 2: Balance charges
Using the criss - cross method, the charge of Fe (3+) becomes the subscript of $CO_{3}$, and the charge of $CO_{3}$ (2 -) becomes the subscript of Fe. So we have $Fe_{2}(CO_{3})_{3}$.
Part e: Calcium hypochlorite
Step 1: Identify ions
Calcium is $Ca^{2+}$, and hypochlorite is $ClO^{-}$.
Step 2: Balance charges
To balance the 2+ charge of $Ca^{2+}$ with the 1 - charge of $ClO^{-}$, we need 2 $ClO^{-}$ ions. So the formula is $Ca(ClO)_{2}$.
Part f: Barium nitrate
Step 1: Identify ions
Barium is $Ba^{2+}$, and nitrate is $NO_{3}^{-}$.
Step 2: Balance charges
To balance the 2+ charge of $Ba^{2+}$ with the 1 - charge of $NO_{3}^{-}$, we need 2 $NO_{3}^{-}$ ions. So the formula is $Ba(NO_{3})_{2}$.
Part g: Sodium chlorate
Step 1: Identify ions
Sodium is $Na^{+}$, and chlorate is $ClO_{3}^{-}$.
Step 2: Balance charges
The charges of $Na^{+}$ (1+) and $ClO_{3}^{-}$ (1 -) are equal in magnitude, so the formula is $NaClO_{3}$.
Part h: Magnesium sulfate
Step 1: Identify ions
Magnesium is $Mg^{2+}$, and sulfate is $SO_{4}^{2 -}$.
Step 2: Balance charges
The charges of $Mg^{2+}$ and $SO_{4}^{2 -}$ are equal in magnitude (2+ and 2 -), so the formula is $MgSO_{4}$.
Part i: Iron (III) nitrate
Step 1: Identify ions
Iron (III) is $Fe^{3+}$, and nitrate is $NO_{3}^{-}$.
Step 2: Balance charges
To balance the 3+ charge of $Fe^{3+}$ with the 1 - charge of $NO_{3}^{-}$, we need 3 $NO_{3}^{-}$ ions. So the formula is $Fe(NO_{3})_{3}$.
Part j: Lithium perchlorate
Step 1: Identify ions
Lithium is $Li^{+}$, and perchlorate is $ClO_{4}^{-}$.
Step 2: Balance charges
The charges of $Li^{+}$ (1+) and $ClO_{4}^{-}$ (1 -) are equal in magnitude, so the formula is $LiClO_{4}$.
Part k: Sodium hydroxide
Step 1: Identify ions
Sodium is $Na^{+}$, and hydroxide is $OH^{-}$.
Step 2: Balance charges
The charges of $Na^{+}$ (1+) and $OH^{-}$ (1 -) are equal in magnitude, so the formula is $NaOH$.
Final Answers
a. $\boldsymbol{CuSO_{3}}$
b. $\boldsymbol{Fe(CN)_{2}}$
c. $\boldsymbol{H_{2}SO_{4}}$
d. $\boldsymbol{Fe_{2}(CO_{3})_{3}}$
e. $\boldsymbol{Ca(ClO)_{2}}$
f. $\boldsymbol{Ba(NO_{3})_{2}}$
g. $\boldsymbol{NaClO_{3}}$
h. $\boldsymbol{MgSO_{4}}$
i. $\boldsymbol{Fe(NO_{3})_{3}}$
j. $\boldsymbol{LiClO_{4}}$
k. $\boldsymbol{NaOH}$
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
To determine the chemical formulas for each compound, we use the rules of ionic compound nomenclature (for ionic compounds) and acid nomenclature (for sulfuric acid). Here are the step - by - step solutions for each:
Part a: Copper (II) sulfite
Step 1: Identify ions
Copper (II) has a charge of $Cu^{2+}$, and sulfite is $SO_{3}^{2 -}$.
Step 2: Balance charges
The charges of $Cu^{2+}$ and $SO_{3}^{2 -}$ are equal in magnitude (2+ and 2 -), so the formula is $CuSO_{3}$.
Part b: Iron (II) cyanide
Step 1: Identify ions
Iron (II) is $Fe^{2+}$, and cyanide is $CN^{-}$.
Step 2: Balance charges
To balance the 2+ charge of $Fe^{2+}$ with the 1 - charge of $CN^{-}$, we need 2 $CN^{-}$ ions. So the formula is $Fe(CN)_{2}$.
Part c: Sulfuric acid
Step 1: Recall acid nomenclature
Sulfuric acid is a strong acid. The anion is sulfate ($SO_{4}^{2 -}$), and for an acid, we combine the anion with $H^{+}$ ions.
Step 2: Balance charges
To balance the 2 - charge of $SO_{4}^{2 -}$, we need 2 $H^{+}$ ions. So the formula is $H_{2}SO_{4}$.
Part d: Ferric carbonate
Step 1: Identify ions
Ferric means iron (III), so $Fe^{3+}$, and carbonate is $CO_{3}^{2 -}$.
Step 2: Balance charges
Using the criss - cross method, the charge of Fe (3+) becomes the subscript of $CO_{3}$, and the charge of $CO_{3}$ (2 -) becomes the subscript of Fe. So we have $Fe_{2}(CO_{3})_{3}$.
Part e: Calcium hypochlorite
Step 1: Identify ions
Calcium is $Ca^{2+}$, and hypochlorite is $ClO^{-}$.
Step 2: Balance charges
To balance the 2+ charge of $Ca^{2+}$ with the 1 - charge of $ClO^{-}$, we need 2 $ClO^{-}$ ions. So the formula is $Ca(ClO)_{2}$.
Part f: Barium nitrate
Step 1: Identify ions
Barium is $Ba^{2+}$, and nitrate is $NO_{3}^{-}$.
Step 2: Balance charges
To balance the 2+ charge of $Ba^{2+}$ with the 1 - charge of $NO_{3}^{-}$, we need 2 $NO_{3}^{-}$ ions. So the formula is $Ba(NO_{3})_{2}$.
Part g: Sodium chlorate
Step 1: Identify ions
Sodium is $Na^{+}$, and chlorate is $ClO_{3}^{-}$.
Step 2: Balance charges
The charges of $Na^{+}$ (1+) and $ClO_{3}^{-}$ (1 -) are equal in magnitude, so the formula is $NaClO_{3}$.
Part h: Magnesium sulfate
Step 1: Identify ions
Magnesium is $Mg^{2+}$, and sulfate is $SO_{4}^{2 -}$.
Step 2: Balance charges
The charges of $Mg^{2+}$ and $SO_{4}^{2 -}$ are equal in magnitude (2+ and 2 -), so the formula is $MgSO_{4}$.
Part i: Iron (III) nitrate
Step 1: Identify ions
Iron (III) is $Fe^{3+}$, and nitrate is $NO_{3}^{-}$.
Step 2: Balance charges
To balance the 3+ charge of $Fe^{3+}$ with the 1 - charge of $NO_{3}^{-}$, we need 3 $NO_{3}^{-}$ ions. So the formula is $Fe(NO_{3})_{3}$.
Part j: Lithium perchlorate
Step 1: Identify ions
Lithium is $Li^{+}$, and perchlorate is $ClO_{4}^{-}$.
Step 2: Balance charges
The charges of $Li^{+}$ (1+) and $ClO_{4}^{-}$ (1 -) are equal in magnitude, so the formula is $LiClO_{4}$.
Part k: Sodium hydroxide
Step 1: Identify ions
Sodium is $Na^{+}$, and hydroxide is $OH^{-}$.
Step 2: Balance charges
The charges of $Na^{+}$ (1+) and $OH^{-}$ (1 -) are equal in magnitude, so the formula is $NaOH$.
Final Answers
a. $\boldsymbol{CuSO_{3}}$
b. $\boldsymbol{Fe(CN)_{2}}$
c. $\boldsymbol{H_{2}SO_{4}}$
d. $\boldsymbol{Fe_{2}(CO_{3})_{3}}$
e. $\boldsymbol{Ca(ClO)_{2}}$
f. $\boldsymbol{Ba(NO_{3})_{2}}$
g. $\boldsymbol{NaClO_{3}}$
h. $\boldsymbol{MgSO_{4}}$
i. $\boldsymbol{Fe(NO_{3})_{3}}$
j. $\boldsymbol{LiClO_{4}}$
k. $\boldsymbol{NaOH}$