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using the following below, you must make 20 compounds using one from co…

Question

using the following below, you must make 20 compounds using one from column i, one from column ii to make ternary ionic compounds and write their formulas. make sure you use a variety from each column. it will be collected when you are finished. please make sure that each of the twenty compounds are ternary ionic compounds. please also make sure that no more than 5 have a 1:1 ratio. ex: from column i: beryllium: be²⁺ from column ii: nitrate: no₃⁻ be(no₃)₂ beryllium nitrate column i positive ions (cations) sodium, potassium, iron (iii), beryllium, francium, copper (ii), aluminum, calcium, ammonium, gallium, lead (iv) column ii negative ions (anions) sulfide, sulfate, phosphate, chlorate, nitride, carbonate, oxide, hydroxide, acetate, nitrate,

Explanation:

Step1: Identify Ions

Column I: Potassium ($K^+$), Column II: Sulfate ($SO_4^{2-}$)

Step2: Balance Charges

2 $K^+$ (total +2) balance 1 $SO_4^{2-}$ (total -2)

Step3: Write Formula

Use criss - cross: $K_2SO_4$

Step4: Name Compound

Potassium Sulfate (cation name + anion name, sulfate for $SO_4^{2-}$)

Let's make more compounds following the same logic:

Compound 2: Sodium Carbonate

Step1: Ions

$Na^+$ (Column I), $CO_3^{2-}$ (Column II)

Step2: Balance Charges

2 $Na^+$ (total +2) balance 1 $CO_3^{2-}$ (total -2)

Step3: Formula

$Na_2CO_3$

Step4: Name

Sodium Carbonate

Compound 3: Iron (III) Hydroxide

Step1: Ions

$Fe^{3+}$ (Column I), $OH^-$ (Column II)

Step2: Balance Charges

3 $OH^-$ (total -3) balance 1 $Fe^{3+}$ (total +3)

Step3: Formula

$Fe(OH)_3$

Step4: Name

Iron (III) Hydroxide

Compound 4: Beryllium Nitrate

Step1: Ions

$Be^{2+}$ (Column I), $NO_3^-$ (Column II)

Step2: Balance Charges

2 $NO_3^-$ (total -2) balance 1 $Be^{2+}$ (total +2)

Step3: Formula

$Be(NO_3)_2$

Step4: Name

Beryllium Nitrate (as in example)

Compound 5: Francium Chlorate

Step1: Ions

$Fr^+$ (Column I), $ClO_3^-$ (Column II)

Step2: Balance Charges

1 $Fr^+$ (total +1) balance 1 $ClO_3^-$ (total -1) (1:1 ratio)

Step3: Formula

$FrClO_3$

Step4: Name

Francium Chlorate

Compound 6: Copper (II) Phosphate

Step1: Ions

$Cu^{2+}$ (Column I), $PO_4^{3-}$ (Column II)

Step2: Balance Charges

3 $Cu^{2+}$ (total +6) balance 2 $PO_4^{3-}$ (total -6)

Step3: Formula

$Cu_3(PO_4)_2$

Step4: Name

Copper (II) Phosphate

Compound 7: Aluminum Oxide

Step1: Ions

$Al^{3+}$ (Column I), $O^{2-}$ (Column II)

Step2: Balance Charges

2 $Al^{3+}$ (total +6) balance 3 $O^{2-}$ (total -6)

Step3: Formula

$Al_2O_3$

Step4: Name

Aluminum Oxide

Compound 8: Calcium Acetate

Step1: Ions

$Ca^{2+}$ (Column I), $C_2H_3O_2^-$ (Column II)

Step2: Balance Charges

2 $C_2H_3O_2^-$ (total -2) balance 1 $Ca^{2+}$ (total +2)

Step3: Formula

$Ca(C_2H_3O_2)_2$

Step4: Name

Calcium Acetate

Compound 9: Ammonium Sulfide

Step1: Ions

$NH_4^+$ (Column I), $S^{2-}$ (Column II)

Step2: Balance Charges

2 $NH_4^+$ (total +2) balance 1 $S^{2-}$ (total -2)

Step3: Formula

$(NH_4)_2S$

Step4: Name

Ammonium Sulfide

Compound 10: Gallium Nitrate

Step1: Ions

$Ga^{3+}$ (Column I), $NO_3^-$ (Column II)

Step2: Balance Charges

3 $NO_3^-$ (total -3) balance 1 $Ga^{3+}$ (total +3)

Step3: Formula

$Ga(NO_3)_3$

Step4: Name

Gallium Nitrate

Compound 11: Lead (IV) Sulfide

Step1: Ions

$Pb^{4+}$ (Column I), $S^{2-}$ (Column II)

Step2: Balance Charges

2 $S^{2-}$ (total -4) balance 1 $Pb^{4+}$ (total +4)

Step3: Formula

$PbS_2$

Step4: Name

Lead (IV) Sulfide

Compound 12: Potassium Nitride

Step1: Ions

$K^+$ (Column I), $N^{3-}$ (Column II)

Step2: Balance Charges

3 $K^+$ (total +3) balance 1 $N^{3-}$ (total -3)

Step3: Formula

$K_3N$

Step4: Name

Potassium Nitride

Compound 13: Iron (III) Sulfate

Step1: Ions

$Fe^{3+}$ (Column I), $SO_4^{2-}$ (Column II)

Step2: Balance Charges

2 $Fe^{3+}$ (total +6) balance 3 $SO_4^{2-}$ (total -6)

Step3: Formula

$Fe_2(SO_4)_3$

Step4: Name

Iron (III) Sulfate

Compound 14: Beryllium Chlorate

Step1: Ions

$Be^{2+}$ (Column I), $ClO_3^-$ (Column II)

Step2: Balance Charges

2 $ClO_3^-$ (total -2) balance 1 $Be^{2+}$ (total +2)

Step3: Formula

$Be(ClO_3)_2$

Step4: Name

Beryllium Chlorate

Compound 15: Francium Oxide

Step1: Ions

$Fr^+$ (Column I), $O^{2-}$ (Column II)

Step2: Balance Charges

2 $Fr^+$ (total +2) balanc…

Answer:

20 Ternary Ionic Compounds:

  1. $K_2SO_4$ (Potassium Sulfate)
  2. $Na_2CO_3$ (Sodium Carbonate)
  3. $Fe(OH)_3$ (Iron (III) Hydroxide)
  4. $Be(NO_3)_2$ (Beryllium Nitrate)
  5. $FrClO_3$ (Francium Chlorate)
  6. $Cu_3(PO_4)_2$ (Copper (II) Phosphate)
  7. $Al_2O_3$ (Aluminum Oxide)
  8. $Ca(C_2H_3O_2)_2$ (Calcium Acetate)
  9. $(NH_4)_2S$ (Ammonium Sulfide)
  10. $Ga(NO_3)_3$ (Gallium Nitrate)
  11. $PbS_2$ (Lead (IV) Sulfide)
  12. $K_3N$ (Potassium Nitride)
  13. $Fe_2(SO_4)_3$ (Iron (III) Sulfate)
  14. $Be(ClO_3)_2$ (Beryllium Chlorate)
  15. $Fr_2O$ (Francium Oxide)
  16. $Cu(C_2H_3O_2)_2$ (Copper (II) Acetate)
  17. $Al(OH)_3$ (Aluminum Hydroxide)
  18. $Ca_3(PO_4)_2$ (Calcium Phosphate)
  19. $NH_4NO_3$ (Ammonium Nitrate)
  20. $Ga_2S_3$ (Gallium Sulfide)