which choice identifies the spectator ions in the following reaction cuso4 aq bacl2 aq -> bacl2(aq) baso4(s)+cucl2(aq)

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CuSO4 + BaCl2 ==> BaSO4(s) + CuCl2.
The net ionic equation is
SO4^2-(aq) + Ba^2+(aq) ==> BaSO4(s)
The other ions are spectator ions.

find the ionic equation

To identify the spectator ions in the given reaction, you need to first write down the complete ionic equation. This equation shows all the ions present in the reaction.

The given equation is:
CuSO4(aq) + BaCl2(aq) -> BaCl2(aq) + BaSO4(s) + CuCl2(aq)

Now, let's break down the reactants and products into their respective ions:

Reactants:
CuSO4(aq) breaks down into Cu2+(aq) + SO42-(aq)
BaCl2(aq) breaks down into Ba2+(aq) + 2Cl-(aq)

Products:
BaCl2(aq) remains unchanged as it is a spectator ion.
BaSO4(s) does not break down into ions since it is a solid.
CuCl2(aq) breaks down into Cu2+(aq) + 2Cl-(aq)

From the above breakdown, we can identify the spectator ions in the reaction as Ba2+(aq) and Cl-(aq). These ions are present on both the reactant and product sides of the equation and do not participate in the overall reaction.

To identify the spectator ions in the given reaction, we first need to understand what spectator ions are. Spectator ions are the ions that do not participate in the chemical reaction and remain unchanged throughout the reaction.

In this reaction, CuSO4(aq) + BaCl2(aq) -> BaCl2(aq) + BaSO4(s) + CuCl2(aq), the spectator ions can be determined by looking at the substances that are not involved in the formation of the precipitate, BaSO4(s).

The reactants in this case are CuSO4(aq) and BaCl2(aq). The products are BaCl2(aq), BaSO4(s), and CuCl2(aq).

By comparing the reactants and products, we can see that BaCl2(aq) appears on both sides of the reaction arrow. It remains unchanged, indicating that it is a spectator ion. Similarly, CuCl2(aq) remains unchanged and is also a spectator ion.

Therefore, the spectator ions in this reaction are BaCl2(aq) and CuCl2(aq).