which formula is correct for **chromium(iii) oxide**?
options: cro₂, cro, cr₂o₃, cr₃o₂
the compound is mnso₄. the anion is sulfate (so₄^(2-)). what is the complete, correct name for this compound?
why is it necessary to include a roman numeral when naming fecl₃ but not when naming nacl?
determine the charge of the gold ion in the compound au₂o₃.
options: +3, +1, +4, +2
write the chemical formula for **nickel(i) sulfide**.

Question

Accepted Answer

### First Sub - Question (Chromium(III) oxide formula)
# Explanation:
## Step1: Determine ion charges
Chromium(III) means $Cr^{3 +}$, oxide is $O^{2-}$.
## Step2: Use criss - cross method
The formula is formed by balancing charges. For $Cr^{3+}$ and $O^{2-}$, we take the absolute value of each charge and swap them. So the number of $Cr$ atoms is 2 (from $O$'s charge) and the number of $O$ atoms is 3 (from $Cr$'s charge). So the formula is $Cr_2O_3$.
# Answer: $Cr_2O_3$ (the option with $Cr_2O_3$)

### Second Sub - Question (Name of $MnSO_4$)
# Explanation:
First, find the charge of $Mn$. Sulfate ($SO_4^{2 -}$) has a charge of - 2. Let the charge of $Mn$ be $x$. In $MnSO_4$, the sum of charges is zero. So $x+( - 2)=0$, so $x = + 2$. So the name is Manganese(II) sulfate.
# Answer: Manganese(II) sulfate

### Third Sub - Question (Roman numeral in naming)
# Brief Explanations:
Iron ($Fe$) is a transition metal with variable oxidation states (can form $Fe^{2+}$ or $Fe^{3+}$ in $FeCl_2$ or $FeCl_3$). Sodium ($Na$) is an alkali metal with a fixed + 1 oxidation state. Roman numerals indicate the oxidation state of the metal, so for metals with variable states (like $Fe$) we need it, for fixed - state metals (like $Na$) we don't.
# Answer: Iron is a transition metal with variable oxidation states (can have different charges in compounds), so a Roman numeral is needed to show its charge in $FeCl_3$. Sodium is an alkali metal with a fixed + 1 charge, so no Roman numeral is needed in $NaCl$.

### Fourth Sub - Question (Charge of Gold ion in $Au_2O_3$)
# Explanation:
## Step1: Determine $O$ charge
Oxygen has a charge of - 2 in oxides.
## Step2: Calculate $Au$ charge
Let the charge of $Au$ be $x$. In $Au_2O_3$, the sum of charges is zero. So $2x+3\times(- 2)=0$. Solving for $x$: $2x = 6$, so $x = + 3$.
# Answer: + 3 (the option with + 3)

### Fifth Sub - Question (Formula for Nickel(I) sulfide)
# Explanation:
## Step1: Identify ions
Nickel(I) is $Ni^{+}$, sulfide is $S^{2-}$.
## Step2: Balance charges
To balance the charges, we need 2 $Ni^{+}$ ions (total + 2 charge) to balance the - 2 charge of $S^{2-}$. So the formula is $Ni_2S$.
# Answer: $Ni_2S$

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QUESTION IMAGE

Question

Explanation:

First Sub - Question (Chromium(III) oxide formula)

Step1: Determine ion charges

Step2: Use criss - cross method

Answer:

Second Sub - Question (Name of \(MnSO_4\))