Question

roman numeral indicates the combining capacity for polyvalent elements metal polyatomic ion put the polyatomic ion in brackets magnesium dichromate potassium bicarbonate calcium sulphate silver nitrate lead ii carbonate gold iii arsenate iron iii chromate calcium nitrate lead iv carbonate symbols and combining capacities. shown as superscripts. mg+2(cro₇)-2 k+1(hco₃)-1 ca+2(so₄)-2 ag+1(no₃)-1 pb+2(co₃)-2 au+3(aso₄)-3 fe+3(cro₄)-2 ca+2(no₃)-1 pb+4(co₃)-2 symbols and combining capacities crisscrossed. now shown as subscripts. mg₂(cr₂o₇)ₙ k₁(hco₃)₁ ca₂(so₄)₁ ag₁(no₃)₁ pb₂(co₃)₁ au₃(aso₄)₁ fe₂(cro₄)₃ ca₂(no₃)₁ pb₄(co₃)₁ resulting formula reduce common multiples never show subscript 1 mg(cr₂o₇) khco₃ (crossed out) pb₂((co₃))₂ au₃(aso₄)₁ fe₂(cro₄)₃ ca₂((no₃))₂ pb₄(no₃)₄ compound name remember, the metallic ion is capitalized, metal is lowercase magnesium dichromate potass bicarbonate (crossed out) # of atoms 10 6 (crossed out) 5 10 21 17 9 18

Explanation:

Step1: Identify the problem type

This problem involves writing chemical formulas for compounds formed by metals (with Roman numerals indicating oxidation states) and polyatomic ions. It's a chemistry (Natural Science - Chemistry) problem about chemical nomenclature and formula writing.

Step2: Recall the crisscross method

The crisscross method involves taking the absolute value of the oxidation state of the metal (cation) and using it as the subscript for the polyatomic ion, and vice - versa. For example, for Magnesium (Mg²⁺) and dichromate ($\text{Cr}_2\text{O}_7^{2 - }$):
- The oxidation state of Mg is +2, and of dichromate is -2.
- Crisscrossing the absolute values: Mg gets a subscript of 2 (from dichromate's charge) and dichromate gets a subscript of 2 (from Mg's charge), but we reduce common multiples. The common multiple of 2 and 2 is 2, so we divide by 2.
- The formula becomes $\text{MgCr}_2\text{O}_7$ (since $\frac{2}{2}=1$ for Mg and $\frac{2}{2} = 1$ for $\text{Cr}_2\text{O}_7$? Wait, no. Wait, the charge of Mg is +2, charge of dichromate is -2. So the formula is $\text{Mg}_1(\text{Cr}_2\text{O}_7)_1$ or simply $\text{MgCr}_2\text{O}_7$ (because when the subscripts are 1, we don't write them).

Let's check another example, Potassium (K⁺) and bicarbonate ($\text{HCO}_3^-$):
- Charge of K is +1, charge of bicarbonate is -1.
- Crisscrossing gives K with subscript 1 and bicarbonate with subscript 1. So the formula is $\text{KHCO}_3$.

For Calcium (Ca²⁺) and sulphate ($\text{SO}_4^{2 - }$):
- Charge of Ca is +2, charge of sulphate is -2.
- Crisscrossing gives Ca with subscript 2 and sulphate with subscript 2. But we reduce by the common multiple (2), so the formula is $\text{CaSO}_4$ (since $\frac{2}{2}=1$ for both).

For Silver (Ag⁺) and nitrate ($\text{NO}_3^-$):
- Charge of Ag is +1, charge of nitrate is -1.
- Crisscrossing gives Ag with subscript 1 and nitrate with subscript 1. So the formula is $\text{AgNO}_3$.

For Lead II (Pb²⁺) and carbonate ($\text{CO}_3^{2 - }$):
- Charge of Pb is +2, charge of carbonate is -2.
- Crisscrossing gives Pb with subscript 2 and carbonate with subscript 2. Reduce by common multiple (2), formula is $\text{PbCO}_3$.

For Gold III (Au³⁺) and arsenate ($\text{AsO}_4^{3 - }$):
- Charge of Au is +3, charge of arsenate is -3.
- Crisscrossing gives Au with subscript 3 and arsenate with subscript 3. Reduce by common multiple (3), formula is $\text{AuAsO}_4$.

For Iron III (Fe³⁺) and chromate ($\text{CrO}_4^{2 - }$):
- Charge of Fe is +3, charge of chromate is -2.
- Crisscrossing: Fe gets subscript 2 (from chromate's charge -2, absolute value), chromate gets subscript 3 (from Fe's charge +3, absolute value). So the formula is $\text{Fe}_2(\text{CrO}_4)_3$.

For Calcium (Ca²⁺) and nitrate ($\text{NO}_3^-$):
- Charge of Ca is +2, charge of nitrate is -1.
- Crisscrossing: Ca gets subscript 1 (from nitrate's charge -1, absolute value), nitrate gets subscript 2 (from Ca's charge +2, absolute value). So the formula is $\text{Ca(NO}_3)_2$.

For Lead IV (Pb⁴⁺) and carbonate ($\text{CO}_3^{2 - }$):
- Charge of Pb is +4, charge of carbonate is -2.
- Crisscrossing: Pb gets subscript 2 (from carbonate's charge -2, absolute value), carbonate gets subscript 4 (from Pb's charge +4, absolute value). But we can reduce by common multiple 2. So $\text{Pb}_1(\text{CO}_3)_2$ or $\text{Pb(CO}_3)_2$ (since $\frac{4}{2}=2$ for carbonate and $\frac{2}{2}=1$ for Pb? Wait, no. Wait, the charge of Pb is +4, carbonate is -2. So to balance the charges: let the subscript of Pb be $x$ and carbonate be $y$. Then $4x+(- 2)y = 0\Rightarrow4x = 2y\Rightarrow2x=y$. The simplest ratio is $x = 1,y = 2$. So formula is $\text{Pb(CO}_3)_2$.

Answer:

The chemical formulas are written using the crisscross method to balance the charges of cations (metals) and polyatomic anions. For example:

• Magnesium dichromate: $\text{MgCr}_2\text{O}_7$
• Potassium bicarbonate: $\text{KHCO}_3$
• Calcium sulphate: $\text{CaSO}_4$
• Silver nitrate: $\text{AgNO}_3$
• Lead II carbonate: $\text{PbCO}_3$
• Gold III arsenate: $\text{AuAsO}_4$
• Iron III chromate: $\text{Fe}_2(\text{CrO}_4)_3$
• Calcium nitrate: $\text{Ca(NO}_3)_2$
• Lead IV carbonate: $\text{Pb(CO}_3)_2$

(If the question was to find a specific formula, for example, for Magnesium dichromate, the answer is $\text{MgCr}_2\text{O}_7$; for Potassium bicarbonate, $\text{KHCO}_3$ etc. Since the table has multiple entries, we can summarize the correct formula - writing process and the resulting formulas as above.)