hello i am having trouble writing these two ionic equations. could someone explain to me how to do that? 1.)K2SO4(aq) and Na2CO3(aq).Write the ionic equation.
2.)RbCl(aq) and KOH(aq).Write the ionic equation.
2. Rb and K are in the same column 1 (group 1) which is soluble; meaning that they dissolved in a solution or solvent. The answer is just "noreaction" for the Ionic and Net equation
Certainly! I can help you with that. To write ionic equations, we need to first understand the reactants and products involved.
1) K2SO4(aq) and Na2CO3(aq):
Step 1: Write the balanced molecular equation
K2SO4(aq) + Na2CO3(aq) -> ?
Step 2: Identify the soluble compounds and write their dissociation.
K2SO4(aq) -> 2K+(aq) + SO4^2-(aq)
Na2CO3(aq) -> 2Na+(aq) + CO3^2-(aq)
Step 3: Combine the dissociated ions to form possible products.
2K+(aq) + CO3^2-(aq) -> K2CO3(s)
2Na+(aq) + SO4^2-(aq) -> Na2SO4(aq)
Step 4: Write the balanced ionic equation.
2K+(aq) + CO3^2-(aq) + 2Na+(aq) + SO4^2-(aq) -> K2CO3(s) + Na2SO4(aq)
2) RbCl(aq) and KOH(aq):
Step 1: Write the balanced molecular equation
RbCl(aq) + KOH(aq) -> ?
Step 2: Identify the soluble compounds and write their dissociation.
RbCl(aq) -> Rb+(aq) + Cl-(aq)
KOH(aq) -> K+(aq) + OH-(aq)
Step 3: Combine the dissociated ions to form possible products.
Rb+(aq) + OH-(aq) -> RbOH(aq)
K+(aq) + Cl-(aq) -> KCl(aq)
Step 4: Write the balanced ionic equation.
Rb+(aq) + OH-(aq) + K+(aq) + Cl-(aq) -> RbOH(aq) + KCl(aq)
I hope this helps! Let me know if you have any further questions.
Certainly! I'd be happy to explain how to write the ionic equations for these two reactions.
1.) K2SO4(aq) + Na2CO3(aq) → ?
To write the ionic equation, we need to first write out the complete chemical equation.
The chemical equation for the reaction between K2SO4(aq) and Na2CO3(aq) is:
K2SO4(aq) + Na2CO3(aq) → 2NaKSO4(aq) + CO2(g)
Next, we need to identify the ionic compounds in the equation and separate them into their respective ions.
K2SO4(aq) can be broken down into:
2K+(aq) + SO4^2-(aq)
Na2CO3(aq) can be broken down into:
2Na+(aq) + CO3^2-(aq)
Now, we need to determine if any of the ions are spectator ions. Spectator ions are the ions that appear on both sides of the equation and do not participate in the reaction. In this case, K+ and Na+ are spectator ions.
Therefore, the ionic equation for this reaction is:
SO4^2-(aq) + CO3^2-(aq) → CO2(g)
2.) RbCl(aq) + KOH(aq) → ?
Following the same steps, let's write the complete chemical equation for the reaction between RbCl(aq) and KOH(aq):
RbCl(aq) + KOH(aq) → RbOH(aq) + KCl(aq)
Now, let's break down the ionic compounds into their respective ions:
RbCl(aq) can be broken down into:
Rb+(aq) + Cl-(aq)
KOH(aq) can be broken down into:
K+(aq) + OH-(aq)
Next, we need to identify the spectator ions. In this case, K+ is a spectator ion.
Therefore, the ionic equation for this reaction is:
Rb+(aq) + OH-(aq) → RbOH(aq)
I hope this explanation helps you understand how to write ionic equations for these reactions! Let me know if you have any further questions.
K2SO4(aq) + Na2CO3(aq) ==> 2K^+(aq) + SO4^2-(aq) + 2Na^+(aq) + CO3^2-(aq)
The second one is done the same way. If there is
1. no gas
2. no ppt
3. no slightly ionized product
then you write a simple mixing o the ions. If one of the 3 occurs, there will be a reaction and you can write a net ionic equation.
For example,
AgNO3(aq) + NaCl(aq)==> AgCl(s) + NaNO3(aq) for which the net ionic equation is
Ag^+(aq) + Cl^-(aq) ==> AgCl(s)