Question

Using what you have learned about the three forms of writing chemical equations, classify each of the presented equations as either a molecular equation, a complete ionic equation, or a net ionic equation. Drag the appropriate equations to their respective bins.

a)Ca2+(aq)+SO42−(aq)→CaSO4(s)b)CuSO4(aq)+2KOH(aq)→Cu(OH)2(s)+K2SO4(aq)c)Ba2+(aq)+SO42−(aq)→BaSO4(s)
d)Ag+(aq)+NO3−(aq)+I−(aq)→AgI(s)+NO3−(aq)e)Cu2+(aq)+SO42−(aq)+2K+(aq)+2OH−(aq)→Cu(OH)2(s)+2K+(aq)+SO42−(aq)f)(BaNO3)2(aq)+K2SO4(aq)→BaSO4(s)+2KNO3(aq)Molecular equationComplete ionic equation

Answer 1

Step-by-step explanation:

Net ionic equation is a chemical equation in which no spectator ions are written.

Complete ionic equation is a chemical equation in which spectator ions are written.

Molecular equation is a chemical equation in which molecuklar formula of compounds are but no ions .

a)
`Ca^(2+)(aq)+SO_4^(2-)(aq)\rightarrow CaSO_4(s)`

The given equation is net ionic equation as there are calcium ions and sulfate ions present in the chemical equation.

b)
`CuSO_4(aq)+2KOH(aq)\rightarrow Cu(OH)_2(s)+K_2SO_4(aq)`

The given equation is a molecular equation as there are no ions present but only compounds on both reactant and product sides.

c)
`Ba^(2+)(aq)+SO_4^(2-)(aq)\rightarrow BaSO_4(s)`

The given equation is net ionic equation as there are barium ions and sulfate ions present in the chemical equation.

d)
`Ag^+(aq)+NO_3^(-)(aq)+I^-(aq)\rightarrow AgI(s)+NO_3^(-)(aq)`

The given equation is complete ionic equation as there are silver ions, nitrate ions, and iodide ions present in the chemical equation.

e)
`Cu^(2+)(aq)+SO_4^(2-)(aq)+2K^+(aq)+2OH^-(aq)\rightarrow Cu(OH)_2(s)+2K^+(aq)+SO_4^(2-)(aq)`

The given equation is complete ionic equation as there are copper ions,sulfate ions, potassium ions and hydroxide ions present in the chemical equation.

f)
`Ba(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KNO_3(aq)`

The given equation is a molecular equation as there are no ions present but only compounds on both reactant and product sides.