1. Entry Task: June 11 th Review Ch, 15, 19 and 16 Return your textbooks!

2. Ch. 15 & 19 Solutions & Acid- Base

4. FORGOT THE FORMULAS?? Dilutions: M 1 V 1 = M 2 V 2

5. 1. What is the percent by mass of NaHCO 3 in a solution containing 20 grams of NaHCO 3 dissolved in 600 ml H 2 O? Identify the solute___________Identify the solvent____________ NaHCO 3 Water = % NaHCO 3 20g NaHCO 3 600ml water 20g NaHCO 3 + 600ml water = 620 ml solution X 100 20g NaHCO 3 620ml solution = 3.2% NaHCO 3

6. 2. You have 1500.0 grams of bleach solution. The percent by mass of the solute sodium hypochlorite, NaOCl, is 3.62%. How many grams of NaOCl are in the solution? How many grams of solvent are in the solution? = 3.62% NaOCl Xg NaOCl 1500 g solution = 3.62 100 Xg NaHCO 3 1500 g solution = 54.3g NaOCl 1500 X 3.2 = 5430 g 100 = 1446 g Solvent 1500g solution = solvent + solute (54.3 g)

7. 3. What is the molarity of an aqueous solution containing 40.0g of glucose (C 6 H 12 O 6 ) in 1.5L of solution? 40 g of C 6 H 12 O 6 1 mole of C 6 H 12 O 6 0.222 moles of C 6 H 12 O 6 1.5 L solution = 0.148 M 180.16 g C 6 H 12 O 6

8. 4. Calculate the molarity of 1.60 L of a solution containing 1.55 g of dissolved KBr. 1.55 g of KBr 119.0 g KBr 1 mole of KBr 0.013 moles of KBr 1.60 L solution = 0.00814 M

9. 5. If I add water to 100 ml of a 0.15 M NaOH solution until the final volume is 150 ml, what will the molarity of the the diluted solution? (0.15 M)(100 ml) = (X M)(150 ml) 15 150 = 0.1 M NaOH

10. 6. How much 0.05 M HCl solution can be made by diluting 250 ml of 10 M HCl? (10 M)(250 ml) = (0.05 M)(X ml) 2500 0.05 = 50000 ml of HCl

11. _____1. proton acceptor_____12. metal and nonmetal _____2. red with pH indicator_____13. positively charged solution _____3. ionicly bonded_____14. H 2 SO 3 _____4. OH on the back of its formula_____15. covalently bonded _____5. negatively charged solution_____16. nonmetal-nonmetal _____6. does not react with metals_____17. Ca(OH) 2 _____7. reacts to carbonates_____18. does not react with carbonates _____8. H in the front of its formula_____19. proton donor _____9. hydronium ions_____20. pH of 0-6 _____10. reacts with metals_____21. hydroxide ions _____11. pH 8-14_____22. purple with pH indicator B A B B B B A A A A B B A A A A B B A A B B

12. Name the following acids and bases HF Phosphoric acid H 2 SO 3 Carbonic acid Zn(OH) 2 Barium hydroxide Hydrofluoric acid H 3 PO 4 sulfurous acid H 2 CO 3 Zinc hydroxide Ba(OH) 2

13. acidbase conjugate acidconjugate base HF + SO 3 2–  F – + HSO 3 –

14. 1. If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric acid solution (H 2 SO 4 ), what is the concentration of the H 2 SO 4 solution? KOH + H 2 SO 4  K 2 SO 4 + H 2 O X M acid 125ml acid = (2) 0.5 M base 50ml base 50 125 = 0.4M acid 22

15. Ch. 16 Energy

16. Determine the mass of a sample of silver if 705 J of heat are required to raise its temperature from 25 o C to 35 o C. The specific heat of silver is 0.235 J/g.o C. q=cm  T q = 705J c= 0.235 J/g.o C m= X g ΔT= 25-35=10 705 J 10x 0.235 300 g

17. A 7.5 g nugget of pure lead absorbs 276 J of heat. What was the final temperature of lead if the initial temperature was 25.0˚C? (The specific heat of lead is 0.129 J/g.o C) q=cm  T q = 276J c= 0.129 J/g.o C m= 7.5 g ΔT= X 276 J 7.5x 0.129 285 °C Since the temperature started at 25.0°C, the final temperature is 310°C.

18. Identify which of the following is an endothermic or exothermic process Endothermic Liquid  gas Has a positive Δ H value Solid  gas Liquid  solid Endothermic Endothermic Exothermic

19. 2C 2 H 4 O(l) + 2H 2 O(l) → 2C 2 H 6 O(l) + O 2 (g) C 2 H 6 O(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(l) ΔH = -685.5 kJ C 2 H 4 O(l) + 2.5O 2 (g) → 2CO 2 (g) + 2H 2 O(l) ΔH = -583.5 kJ 4CO 2 (g) + 6H 2 O(l) → 2C 2 H 6 O(l) + 6O 2 (g) ΔH = 1371 kJ 2C 2 H 4 O(l) + 5O 2 (g) → 4CO 2 (g) + 4H 2 O(l) ΔH = -1167 kJ 2C 2 H 4 O(l) + 2H 2 O(l) → 2C 2 H 6 O(l) + O 2 (g) ΔH= 204

20. CH 2 O(g) + H 2 (g) → CH 4 O(l) ΔH = 65 kJ N 2 H 4 (l) + H 2 (g) → 2NH 3 (g) N 2 H 4 (l) + CH 4 O(l) → CH 2 O(g) + N 2 (g) + 3H 2 (g) ΔH = -37 kJ N 2 (g) + 3H 2 (g) → 2NH 3 (g) ΔH = -46 kJ CH 4 O(l) → CH 2 O(g) + H 2 (g) ΔH = -65 kJ N 2 H 4 (l) + CH 4 O(l) → CH 2 O(g) + N 2 (g) + 3H 2 (g) ΔH = -37 kJ N 2 (g) + 3H 2 (g) → 2NH 3 (g) ΔH = -46 kJ N 2 H 4 (l) + H 2 (g) → 2NH 3 (g) ΔH= -18kJ