![]() Questions like these usually ask for the Spectator Ions, which is simply the crossed out ions we saw when we were finding the Net Ionic Equation, which is #Ca^(2+)(aq) and 2NO_3^(-)(aq)# in this case. ![]() Write the molecular equation, the complete ionic equation, and the net ionic equation for the. ![]() and net ionic equation for the reaction that occurs between aqueous solutions of cobalt(II) chloride and potassium iodate. #cancel(Ca^(2+)(aq)) + 2Cl^(-)(aq) + 2Ag^+(aq) + cancel(2NO_3^(-)(aq))rarr2AgCl(s) + cancel(Ca^(2+)(aq)) + cancel(2NO_3^(-)(aq))#Īs you can see, we are left with #2Cl^(-)(aq) + 2Ag^+(aq)rarr2AgCl(s)# Net Ionic Equation 4) silver nitrate (aq) + sodium hydroxide (aq) -> Balanced Formula Equation Complete Ionic Equation. All you have to do is cross out common ions that are on both sides to the arrow. This is simplest process out of the three. If a compound is solid or liquid, leave it as it is.Īs you can see in the answer, charges are expressed in this, and subscripts are written as coefficients to the left of the equation. To find the CIE, you have to separate each aqueous compound into its constituent ions. To find the states, use a solubility table to determine which products are soluble: Anything with nitrate is soluble (thus making #Ca(NO_3)_2# aqueous), but Cl is insoluble with Ag (thus making #AgCl# a solid). However, since Calcium's charge is 2+, while Nitrate's charge is -, you have to balance the net charge to zero, which means adding a 2 subscript for nitrate. Since Silver has a charge of +, and Chlorine's charge is -, you do not need to make any subscript changes here. Keep in mind that during this process, subscripts are ignored during the conversion. In this case, the two 'inner' elements are Silver and Chloride, while the two 'outer' elements are Calcium and Nitrate. To find the products, you simply combine the two 'inner' and two 'outer' elements together. The given compounds are reactants: Calcium Chloride #(CaCl_2)# and Silver Nitrate #(AgNO_3^-)#.
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