Can someone help me on how to write equations? For example: Sodium carbonate solution with nitric acid and copper oxide with HCL. I know the balanced equations but where do I go next? How do I know which ions to keep in and which are spectators?
Can someone help me on how to write equations? For example: Sodium carbonate solution with nitric acid and copper oxide with HCL. I know the balanced equations but where do I go next? How do I know which ions to keep in and which are spectators?
Can someone help me on how to write equations? For example: Sodium carbonate solution with nitric acid and copper oxide with HCL. I know the balanced equations but where do I go next? How do I know which ions to keep in and which are spectators?
After having the balanced equation, split the ionic compounds into their ions. Then you removed the spectator ions and rewrite the ionic equation.
ok so: CuO+2HCl--->CuCl2+H2O Now split the ionic compounds into ions: Cu+2 + O2- + 2HCl---> Cu2+ + 2Cl- + H2O Is this right? How do I tell which the spectators are?
ok so: CuO+2HCl--->CuCl2+H2O Now split the ionic compounds into ions: Cu+2 + O2- + 2HCl---> Cu2+ + 2Cl- + H2O Is this right? How do I tell which the spectators are?
Looks good except for HCl. You have to split HCl into 2H+ and 2Cl-. Spectator ions appear on both sides of the equation. So here you can see that 2Cl- is a spectator ion because it appears on both sides of the equation
Looks good except for HCl. You have split HCl into 2H+ and 2Cl-. Spectator ions appear on both sides of the equation. So here you can see that 2Cl- is a spectator ion because it appears on both sides of the equation
ok so: CuO+2HCl--->CuCl2+H2O Now split the ionic compounds into ions: Cu+2 + O2- + 2HCl---> Cu2+ + 2Cl- + H2O Is this right? How do I tell which the spectators are?
They'll appear on both sides of the reaction, like Cu2+ does. Shouldn't HCl be split up into H+ and Cl-?
Yep, HCl is covalent with a permanent dipole (electrons are attracted towards the Cl more strongly because Cl is more electronagtive). But i think in this case HCl is aqueous and aqueous species always separate into their ions.
This is what i've been taught. Maybe someone has a proper explanation.
I didn't know it was ionic, I thought it was covalent.
HCl is covalent in gas form but when it's "dissolved" (I'm sure that's not the right word) in water the proton is released. That's one of the definitions of acidity anyways.
HCl is covalent in gas form but when it's "dissolved" (I'm sure that's not the right word) in water the proton is released. That's one of the definitions of acidity anyways.
Ah yes, how could I forget. It's the dissociation that makes acids acidic, am I right?
in my understanding only the dissolved ionic substances can be shown as separate ions - since copper oxide is insoluble in water it should not be shown as separate ions in the construction of your ionic equation so the e ionic equation is
in my understanding only the dissolved ionic substances can be shown as separate ions - since copper oxide is insoluble in water it should not be shown as separate ions in the construction of your ionic equation so the e ionic equation is
CuO + 2H+ --> Cu2+ + H2O
only the Cl-ions are spectator ions
Im no expert but i know this isnt right. Dont ask me why tho lol
in my understanding only the dissolved ionic substances can be shown as separate ions - since copper oxide is insoluble in water it should not be shown as separate ions in the construction of your ionic equation so the e ionic equation is
CuO + 2H+ --> Cu2+ + H2O
only the Cl-ions are spectator ions
The copper oxide is still involved in the reaction so it still separates into ions which may be why you're mistaken. I'm just going by what QuantumSuicide said as they said you're incorrect