write the formulas of the compounds produced from the listed ions.
write the formulas of the compounds produced from the listed ions.
Answer
Explanation:
Step1: Combine ions
When combining ions to form compounds, the overall charge of the compound must be zero. We use the criss - cross method where the magnitude of the charge of one ion becomes the sub - script of the other ion.
Step2: For $Na^+$ and $Cl^-$
The compound is $NaCl$ since the + 1 charge of $Na^+$ and - 1 charge of $Cl^-$ balance each other.
Step3: For $Na^+$ and $CO_3^{2 - }$
Using the criss - cross method, we get $Na_2CO_3$ as we need 2 $Na^+$ ions to balance the - 2 charge of $CO_3^{2 - }$.
Step4: For $Na^+$ and $OH^-$
The compound is $NaOH$.
Step5: For $Na^+$ and $SO_4^{2 - }$
We get $Na_2SO_4$.
Step6: For $Na^+$ and $PO_4^{3 - }$
We get $Na_3PO_4$.
Step7: For $Na^+$ and $NO_3^-$
The compound is $NaNO_3$.
Step8: For $NH_4^+$ and $Cl^-$
The compound is $NH_4Cl$.
Step9: For $NH_4^+$ and $CO_3^{2 - }$
We get $(NH_4)_2CO_3$.
Step10: For $NH_4^+$ and $OH^-$
The compound is $NH_4OH$.
Step11: For $NH_4^+$ and $SO_4^{2 - }$
We get $(NH_4)_2SO_4$.
Step12: For $NH_4^+$ and $PO_4^{3 - }$
We get $(NH_4)_3PO_4$.
Step13: For $NH_4^+$ and $NO_3^-$
The compound is $NH_4NO_3$.
Step14: For $K^+$ and $Cl^-$
The compound is $KCl$.
Step15: For $K^+$ and $CO_3^{2 - }$
We get $K_2CO_3$.
Step16: For $K^+$ and $OH^-$
The compound is $KOH$.
Step17: For $K^+$ and $SO_4^{2 - }$
We get $K_2SO_4$.
Step18: For $K^+$ and $PO_4^{3 - }$
We get $K_3PO_4$.
Step19: For $K^+$ and $NO_3^-$
The compound is $KNO_3$.
Step20: For $Ca^{2 + }$ and $Cl^-$
We get $CaCl_2$.
Step21: For $Ca^{2 + }$ and $CO_3^{2 - }$
The compound is $CaCO_3$.
Step22: For $Ca^{2 + }$ and $OH^-$
We get $Ca(OH)_2$.
Step23: For $Ca^{2 + }$ and $SO_4^{2 - }$
The compound is $CaSO_4$.
Step24: For $Ca^{2 + }$ and $PO_4^{3 - }$
We get $Ca_3(PO_4)_2$.
Step25: For $Ca^{2 + }$ and $NO_3^-$
We get $Ca(NO_3)_2$.
Step26: For $Mg^{2 + }$ and $Cl^-$
We get $MgCl_2$.
Step27: For $Mg^{2 + }$ and $CO_3^{2 - }$
The compound is $MgCO_3$.
Step28: For $Mg^{2 + }$ and $OH^-$
We get $Mg(OH)_2$.
Step29: For $Mg^{2 + }$ and $SO_4^{2 - }$
The compound is $MgSO_4$.
Step30: For $Mg^{2 + }$ and $PO_4^{3 - }$
We get $Mg_3(PO_4)_2$.
Step31: For $Mg^{2 + }$ and $NO_3^-$
We get $Mg(NO_3)_2$.
Step32: For $Zn^{2 + }$ and $Cl^-$
We get $ZnCl_2$.
Step33: For $Zn^{2 + }$ and $CO_3^{2 - }$
The compound is $ZnCO_3$.
Step34: For $Zn^{2 + }$ and $OH^-$
We get $Zn(OH)_2$.
Step35: For $Zn^{2 + }$ and $SO_4^{2 - }$
The compound is $ZnSO_4$.
Step36: For $Zn^{2 + }$ and $PO_4^{3 - }$
We get $Zn_3(PO_4)_2$.
Step37: For $Zn^{2 + }$ and $NO_3^-$
We get $Zn(NO_3)_2$.
Step38: For $Fe^{3 + }$ and $Cl^-$
We get $FeCl_3$.
Step39: For $Fe^{3 + }$ and $CO_3^{2 - }$
We get $Fe_2(CO_3)_3$.
Step40: For $Fe^{3 + }$ and $OH^-$
We get $Fe(OH)_3$.
Step41: For $Fe^{3 + }$ and $SO_4^{2 - }$
We get $Fe_2(SO_4)_3$.
Step42: For $Fe^{3 + }$ and $PO_4^{3 - }$
The compound is $FePO_4$.
Step43: For $Fe^{3 + }$ and $NO_3^-$
We get $Fe(NO_3)_3$.
Step44: For $Al^{3 + }$ and $Cl^-$
We get $AlCl_3$.
Step45: For $Al^{3 + }$ and $CO_3^{2 - }$
We get $Al_2(CO_3)_3$.
Step46: For $Al^{3 + }$ and $OH^-$
We get $Al(OH)_3$.
Step47: For $Al^{3 + }$ and $SO_4^{2 - }$
We get $Al_2(SO_4)_3$.
Step48: For $Al^{3 + }$ and $PO_4^{3 - }$
The compound is $AlPO_4$.
Step49: For $Al^{3 + }$ and $NO_3^-$
We get $Al(NO_3)_3$.
Step50: For $Co^{3 + }$ and $Cl^-$
We get $CoCl_3$.
Step51: For $Co^{3 + }$ and $CO_3^{2 - }$
We get $Co_2(CO_3)_3$.
Step52: For $Co^{3 + }$ and $OH^-$
We get $Co(OH)_3$.
Step53: For $Co^{3 + }$ and $SO_4^{2 - }$
We get $Co_2(SO_4)_3$.
Step54: For $Co^{3 + }$ and $PO_4^{3 - }$
The compound is $CoPO_4$.
Step55: For $Co^{3 + }$ and $NO_3^-$
We get $Co(NO_3)_3$.
Step56: For $Fe^{2 + }$ and $Cl^-$
We get $FeCl_2$.
Step57: For $Fe^{2 + }$ and $CO_3^{2 - }$
The compound is $FeCO_3$.
Step58: For $Fe^{2 + }$ and $OH^-$
We get $Fe(OH)_2$.
Step59: For $Fe^{2 + }$ and $SO_4^{2 - }$
The compound is $FeSO_4$.
Step60: For $Fe^{2 + }$ and $PO_4^{3 - }$
We get $Fe_3(PO_4)_2$.
Step61: For $Fe^{2 + }$ and $NO_3^-$
We get $Fe(NO_3)_2$.
Step62: For $H^+$ and $Cl^-$
The compound is $HCl$.
Step63: For $H^+$ and $CO_3^{2 - }$
We get $H_2CO_3$.
Step64: For $H^+$ and $OH^-$
The compound is $H_2O$.
Step65: For $H^+$ and $SO_4^{2 - }$
We get $H_2SO_4$.
Step66: For $H^+$ and $PO_4^{3 - }$
We get $H_3PO_4$.
Step67: For $H^+$ and $NO_3^-$
The compound is $HNO_3$.
Answer:
| $Cl^-$ | $CO_3^{2 - }$ | $OH^-$ | $SO_4^{2 - }$ | $PO_4^{3 - }$ | $NO_3^-$ | |
|---|---|---|---|---|---|---|
| $Na^+$ | $NaCl$ | $Na_2CO_3$ | $NaOH$ | $Na_2SO_4$ | $Na_3PO_4$ | $NaNO_3$ |
| $NH_4^+$ | $NH_4Cl$ | $(NH_4)_2CO_3$ | $NH_4OH$ | $(NH_4)_2SO_4$ | $(NH_4)_3PO_4$ | $NH_4NO_3$ |
| $K^+$ | $KCl$ | $K_2CO_3$ | $KOH$ | $K_2SO_4$ | $K_3PO_4$ | $KNO_3$ |
| $Ca^{2 + }$ | $CaCl_2$ | $CaCO_3$ | $Ca(OH)_2$ | $CaSO_4$ | $Ca_3(PO_4)_2$ | $Ca(NO_3)_2$ |
| $Mg^{2 + }$ | $MgCl_2$ | $MgCO_3$ | $Mg(OH)_2$ | $MgSO_4$ | $Mg_3(PO_4)_2$ | $Mg(NO_3)_2$ |
| $Zn^{2 + }$ | $ZnCl_2$ | $ZnCO_3$ | $Zn(OH)_2$ | $ZnSO_4$ | $Zn_3(PO_4)_2$ | $Zn(NO_3)_2$ |
| $Fe^{3 + }$ | $FeCl_3$ | $Fe_2(CO_3)_3$ | $Fe(OH)_3$ | $Fe_2(SO_4)_3$ | $FePO_4$ | $Fe(NO_3)_3$ |
| $Al^{3 + }$ | $AlCl_3$ | $Al_2(CO_3)_3$ | $Al(OH)_3$ | $Al_2(SO_4)_3$ | $AlPO_4$ | $Al(NO_3)_3$ |
| $Co^{3 + }$ | $CoCl_3$ | $Co_2(CO_3)_3$ | $Co(OH)_3$ | $Co_2(SO_4)_3$ | $CoPO_4$ | $Co(NO_3)_3$ |
| $Fe^{2 + }$ | $FeCl_2$ | $FeCO_3$ | $Fe(OH)_2$ | $FeSO_4$ | $Fe_3(PO_4)_2$ | $Fe(NO_3)_2$ |
| $H^+$ | $HCl$ | $H_2CO_3$ | $H_2O$ | $H_2SO_4$ | $H_3PO_4$ | $HNO_3$ |