2. Catalyst Pick up Papers from the front Calculate the actual yield of Potassium Sulfate if you started with 15 g of Silver (II) Sulfate and excess Potassium Phosphate, which produced Potassium Sulfate at a 75% yield. Use the following equation: AgSO 4 + K 3 PO 4  Ag 3 (PO 4 ) 2 + K 2 SO 4

3. GRADE!

4. Catalyst Pick up Papers from the front Calculate the actual yield of Potassium Sulfate if you started with 15 g of Silver (II) Sulfate and excess Potassium Phosphate, which produced Potassium Sulfate at a 75% yield. Use the following equation: AgSO 4 + K 3 PO 4  Ag 3 (PO 4 ) 2 + K 2 SO 4

5. BIG GOAL

6. Explosion! (3 MINUTES)

7. Explosion Questions 1.What is the formula for density? 2.Give an example of a chemical property. 3.Give an example of a physical property. 4.What is the law of conservation of mass? 5.What is the law of definite proportions? 6.What is the law of multiple proportions? 7.Who discovered the electron? 8.What are the three types of radiation? 9.Which particle is the biggest? 10.Which is the smallest? 11.Who is Ernest Rutherford? 12.How do you find the number of neutrons? 13.What is an isotope? 14.What is the formula for Nitrite? 15.What is the formula for Perchlorate? 16.What is the formula for Hyposulfite? 17.What is the empirical formula? 18.What is the molecular formula? 19.What is the limiting reagent? 20.What is the excess reagent? 21.What is the % yield? 22.What is % composition? 23.What does it mean to be aqueous? 24.Name 3 strong acids 25.Name 3 Strong bases

8. Solution Composition

9. ****Molarity (M): Moles of solute per Liter of solution SUPER IMPORTANT!!! Need to KNOW THIS! – all solutions are calculated based on M!

10. Step 1: Calculate the number of moles Step 2: Divide moles of solute by liters of solution. Example 1: Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.

11. Step 1: Calculate the number of moles Step 2: change the volume to L Step 3: Divide moles of solute by liters of solution. Example 2: Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.

12. Concentration of Ions

13. When we say 1 M NaCl, this means the solution was prepared using 1 mole of NaCl in 1 L of water…but it does NOT mean that the solution has 1 M of Na + and Cl -. This actually means we have 1 M Na + AND 1 M of Cl -, To convert we use MOLE RATIO, let’s try a sample problem

14. Step 1: Write out the equation (how ions are broken down) Step 2: Use a mole ratio to calculate the molarity of each ion! Example 1: Give the concentration of the following solutions: a.0.5 M Co(NO 3 ) 2 b.1 M Fe (ClO 4 ) 3

15. So… Moles of Solute = Molarity (M) x Volume of solution (L) Often times, chemists need to know the # of moles of solute in a solution:

16. Step 1: Write the dissociation reaction Step 2: Use the mole ratio to find the M of Cl - ions Step 2: Multiply the M by the Volume of solution. Example 1: Calculate the number of moles of Cl - ions in 1.75 L of 1.0x10 - 3 M Zinc (II) Chloride.

17. Example 2: Typical Blood serum is about 0.14 M NaCl. What volume of blood contains 1.0 mg NaCl?

18. So…moles before must = moles after dilution Use formula: M 1 V 1 =M 2 V 2 1: soln before, 2: soln after Dilution: adding water to a solution to achieve a lower concentration

19. Example 1: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.1 M H 2 SO 4 solution?

20. Precipitation Reactions!

21. As we’ve seen, (through last week’s lab) sometimes reaction of two liquids cause a precipitate to form! Precipitate : formation of a solid is solution through a chemical reaction

22. As Chemists, we can actually PREDICT this formation! All we need to know are the solubility rules!

23. Soluble: means dissolved in water Insoluble: cannot be dissolved in water (will be a precipitate!) Slightly Soluble: only a very small portion is dissolved so it looks insoluble ***NEED TO MEMORIZE THEM!!!!!!!!

24. Book Page:

25. Use the Rules to determine if the following are soluble or insoluble: NH 4 NO 3 Soluble Ag 2 OInsoluble Mg(C 2 H 3 O 2 ) 2 Insoluble

26. Book Page: Now we need to write reactions! NH 4 NO 3 Soluble Ag 2 OInsoluble Mg(C 2 H 3 O 2 ) 2 Insoluble

27. Writing Precipitation Reactions Step 1: Determine the formulas for the possible products using the general double displacement equation. (Remember to consider ion charges when writing your formulas!) AB + CD  AD + CB Step 2: Predict whether either of the possible products is water insoluble. *If either possible product is insoluble, a precipitation reaction takes place, and you will continue with step 3. * If neither is insoluble, write “No reaction”. Step 3: Write the physical state for each formula. 1) The insoluble product will be followed by (s). 2) Water-soluble ionic compounds will be followed by (aq). Step 4: Balance the equation. Step 5: Write the MOLECULAR Equation

28. Molecular Equation: Equation with all compounds Total Ionic Reaction: Equation written with all ions Net Ionic Reaction: Equation written with only ions involved in reaction Spectator Ion: Ions that are not directly involved in the reaction

29. Practice Sheet #1 Molecular: AgNO 3(aq) + KCl (aq)  AgCl (s) + KNO 3(aq) Total Ionic: Ag + (aq) +NO 3 ¯ (aq) + K + (aq) +Cl¯ (aq)  AgCl (s) + K + (aq) +NO 3 ¯ (aq) CROSS OUT SPECTATOR IONS to get to... Net Ionic: Ag + (aq) + Cl¯ (aq)  AgCl (s)

30. Try with: AgNO 3 (aq) + Na 2 S(aq) Step 1: Determine the formulas for the possible products using the general double displacement equation. (Remember to consider ion charges when writing your formulas!) AB + CD  AD + CB Step 2: Predict whether either of the possible products is water insoluble. *If either possible product is insoluble, a precipitation reaction takes place, and you will continue with step 3. * If neither is insoluble, write “No reaction”. Step 3: Write the physical state for each formula. 1) The insoluble product will be followed by (s). 2) Water-soluble ionic compounds will be followed by (aq). Step 4: Balance the equation. Step 5: Write the NET IONIC REACTION!