Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Sure! To write the empirical formulas for ionic compounds formed from the given ions [tex]\(CN^-\)[/tex], [tex]\(Fe^{2+}\)[/tex], [tex]\(CrO_4^{2-}\)[/tex], and [tex]\(NH_4^+\)[/tex], we need to combine these ions to form neutral compounds. Let's determine the possible combinations step-by-step:
### 1. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CN^-\)[/tex]:
The iron ion ([tex]\(Fe^{2+}\)[/tex]) has a charge of [tex]\(2+\)[/tex], while the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. To form a neutral compound, we need two cyanide ions to balance the charge of one iron ion.
- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(2 \times CN^{-}\)[/tex]
- Formula: [tex]\(Fe(CN)_2\)[/tex]
The empirical formula is [tex]\( \text{Fe(CN)}_2 \)[/tex].
### 2. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:
The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], while the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) has a charge of [tex]\(2-\)[/tex]. To balance the charges, we need two ammonium ions for every one chromate ion.
- Charges: [tex]\(2 \times NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\((NH_4)_2CrO_4\)[/tex]
The empirical formula is [tex]\((\text{NH}_4)_2\text{CrO}_4\)[/tex].
### 3. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:
Both the iron ion ([tex]\(Fe^{2+}\)[/tex]) and the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) have charges of [tex]\(2+\)[/tex] and [tex]\(2-\)[/tex] respectively. They can combine in a 1:1 ratio to form a neutral compound.
- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\(FeCrO_4\)[/tex]
The empirical formula is [tex]\( \text{FeCrO}_4 \)[/tex].
### 4. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]:
The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], and the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. These ions can combine in a 1:1 ratio to form a neutral compound.
- Charges: [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]
- Formula: [tex]\(NH_4CN\)[/tex]
The empirical formula is [tex]\( \text{NH}_4\text{CN} \)[/tex].
In summary, the empirical formulas for the four ionic compounds that could be formed from the given ions are:
1. [tex]\( \text{Fe(CN)}_2 \)[/tex]
2. [tex]\( (\text{NH}_4)_2\text{CrO}_4 \)[/tex]
3. [tex]\( \text{FeCrO}_4 \)[/tex]
4. [tex]\( \text{NH}_4\text{CN} \)[/tex]
### 1. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CN^-\)[/tex]:
The iron ion ([tex]\(Fe^{2+}\)[/tex]) has a charge of [tex]\(2+\)[/tex], while the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. To form a neutral compound, we need two cyanide ions to balance the charge of one iron ion.
- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(2 \times CN^{-}\)[/tex]
- Formula: [tex]\(Fe(CN)_2\)[/tex]
The empirical formula is [tex]\( \text{Fe(CN)}_2 \)[/tex].
### 2. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:
The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], while the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) has a charge of [tex]\(2-\)[/tex]. To balance the charges, we need two ammonium ions for every one chromate ion.
- Charges: [tex]\(2 \times NH_4^+\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\((NH_4)_2CrO_4\)[/tex]
The empirical formula is [tex]\((\text{NH}_4)_2\text{CrO}_4\)[/tex].
### 3. Combination of [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]:
Both the iron ion ([tex]\(Fe^{2+}\)[/tex]) and the chromate ion ([tex]\(CrO_4^{2-}\)[/tex]) have charges of [tex]\(2+\)[/tex] and [tex]\(2-\)[/tex] respectively. They can combine in a 1:1 ratio to form a neutral compound.
- Charges: [tex]\(Fe^{2+}\)[/tex] and [tex]\(CrO_4^{2-}\)[/tex]
- Formula: [tex]\(FeCrO_4\)[/tex]
The empirical formula is [tex]\( \text{FeCrO}_4 \)[/tex].
### 4. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]:
The ammonium ion ([tex]\(NH_4^+\)[/tex]) has a charge of [tex]\(1+\)[/tex], and the cyanide ion ([tex]\(CN^{-}\)[/tex]) has a charge of [tex]\(1-\)[/tex]. These ions can combine in a 1:1 ratio to form a neutral compound.
- Charges: [tex]\(NH_4^+\)[/tex] and [tex]\(CN^-\)[/tex]
- Formula: [tex]\(NH_4CN\)[/tex]
The empirical formula is [tex]\( \text{NH}_4\text{CN} \)[/tex].
In summary, the empirical formulas for the four ionic compounds that could be formed from the given ions are:
1. [tex]\( \text{Fe(CN)}_2 \)[/tex]
2. [tex]\( (\text{NH}_4)_2\text{CrO}_4 \)[/tex]
3. [tex]\( \text{FeCrO}_4 \)[/tex]
4. [tex]\( \text{NH}_4\text{CN} \)[/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
