1.  Determine the type of reaction and predict the products: NaOH  Li + Br 2  C 2 H 4 + O 2 

2.  How to use the activity series. 1. Find the element in the compound on the table. 2. Find the Solo element 3. If the solo element is above the element in the compound then the reaction will take place.

3. 2Al (s) + 3ZnCl 2(aq)  To replace the Zinc, Aluminum must be higher on the series Cu (s) + 2NaCl (aq)  Can copper replace sodium in the compound? 3Zn (s) + 2AlCl 3(aq ) NO REACTION

4.  Above the activity series for metals, there is an activity series for Halogens.  If your solo element is a halogen, it will replace the bonded halogen as long as it is above it on the activity series.  Remember, every halogen on the series is a diatomic molecule, so when it’s by itself, there will be two of them (F 2, Br 2, …)

5. Cr (s) + Pb(NO 3 ) 2(aq)  Pt (s) + CaCl 2(aq)  Ca (s) + FeO (aq)  Cr(NO 3 ) 2(aq) + Pb (s) NO REACTION CaO (aq) + Fe (s)

6.  Determine the type of reaction and predict the products: NaOH  Li + Br 2  C 2 H 4 + O 2 

8.  Anions switch places and are each bonded to a different cation  AB + CD  AD + CB

9.  Involves determining charges, criss-crossing, and the solubility rules

10.  All double displacement reactions (in this unit) will produce a precipitate  A precipitate is a solid that’s produced during a chemical reaction in a solution  So, if a precipitate is not formed, then the reaction will not take place!!  We can determine if a precipitate is formed by looking at our solubility rules

11.  If something is soluble, then it can be dissolved by what it’s bonded to  If something is insoluble, then it cannot be dissolved

12.  If one of your products is insoluble, then its state of matter is solid and a precipitate has formed.  If one of your products is soluble, then its state of matter is aqueous and no precipitate has formed.  You must have at least one solid product for a reaction to occur.

13.  Soluble › All Nitrates, Acetates, Ammoniums, and Group 1 salts. › All Chlorides, Bromides, and Iodides, except Silver, Lead, and Mercury (I) › All Fluorides except Group 2, Lead (II), and Iron (III) › All Sulfates except Calcium, Strontium, Barium, Mercury, Lead(II), and Silver

14.  Insoluble › All Carbonates and Phosphates except Group 1 and Ammonium › All Hydroxides except Group 1, Strontium, Barium, and Ammonium › All Sulfides except Group 1, Group 2, and Ammonium › All Oxides except Group 1

15. 1. First break the reactants into their ions (find the charges!). 2. Next, swap partners for both (OI with a twist) 3. Check solubility rules to see if a solid (precipitate) has formed. 4. Write complete balanced equation with states of matter.

16.  Sodium Hydroxide + Copper (II) Sulfate  What are the Ions?  What are the reactants?  What are the potential products?  Are any potential products insoluble?  What is the complete equation

17.  Sodium phosphate + Nickel (II) chloride › NaCl and Ni 3 (PO 4 ) 2.  Lead (II) Nitrate + Potassium Iodide › PbI 2 and KNO 3  Sodium Hydroxide + Potassium Chloride › NaCl and KOH  Sodium phosphate + Lead (IV) nitrate › Pb 3 (PO 4 ) 4 and NaNO 3

18.  You already know how to do this!  This is the chemical equation with the states of matter in it.  Make sure it’s balanced!

19.  You almost know how to do this! 1. Start with the completely balanced equation. 2. Look at the solid product and make it the product of your Net Ionic equation. 3. For the reactants, put the ions that lead to the product

20.  Na 2 SO 4 + CaCl 2  2NaCl + CaSO 4 › SO 4 2- (aq) + Ca 2+ (aq)  CaSO 4 (s)  3NaOH + FeCl 3  3NaCl + Fe(OH) 3 › 3OH - (aq) + Fe 3+ (aq)  Fe(OH) 3 (s)

21.  Take the complete ionic equation and remove the spectator ions.  Spectator ions are the ions not involved in the reaction.  Ex. Na 2 SO 4 + CaCl 2  2NaCl + CaSO 4 › CaSO 4 (s) + 2NaCl (aq) › SO 4 2- (aq) + Ca 2+ (aq)  CaSO 4 (s)

22.  NaCl + AgNO 3  AgCl +NaNO 3 › Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq)  AgCl (s) + Na + (aq) + NO 3 - (aq) › Ag + (aq) + Cl - (aq)  AgCl (s)  2NaOH + CuSO 4  Cu(OH) 2 + Na 2 SO 4 › 2Na + (aq) + 2OH - (aq) + Cu 2+ (aq) + SO 4 2- (aq)  Cu(OH) 2(s) + 2Na + (aq) + SO 4 2- (aq) › Cu 2+ (aq) + 2OH - (aq)  Cu(OH) 2 (s)