Answered

Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

[tex] CN^{-}, Fe^{2+}, CrO_4^{2-}, NH_4^{+} [/tex]

A. [tex] (Fe^{2+})(CN^{-})_2 [/tex]
B. [tex] (NH_4^{+})_2(CrO_4^{2-}) [/tex]
C. [tex] (NH_4^{+})(CN^{-}) [/tex]
D. [tex] (Fe^{2+})(CrO_4^{2-}) [/tex]

Sagot :

GinnyAnswer
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]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.