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June 8 th Friday

June 8 th Friday. Review over 11- 14. Break out ALL notes. June 11 th Return your books!!. Ch. 11 Moles, Empirical Formula, Molecular Formula and Hydrates. 23.6 mol Aluminum hydroxide . Al(OH) 3. Al= 26.982g X 1 = 26.982 g. O= 15.999 g X 3 = 47.997 g. H= 1.0079 g X 3 = 3.0237 g.

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June 8 th Friday

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  1. June 8th Friday Review over 11- 14 Break out ALL notes June 11th Return your books!!

  2. Ch. 11 Moles, Empirical Formula, Molecular Formula and Hydrates

  3. 23.6 mol Aluminum hydroxide Al(OH)3 Al= 26.982g X 1 = 26.982 g O= 15.999 g X 3 = 47.997 g H= 1.0079 g X 3 = 3.0237 g 1 mole = 78.0027 grams -------------------- 78.00 grams of Al(OH)3 23.6 moles of Al(OH)3 = 1841 grams of Al(OH)3 --------------------- 1 mole Al(OH)3 = 1.84 x 103 grams of Al(OH)3

  4. 3.65 x 1056 molecules of sulfur pentachloride SCl5 S= 32.065 g X 1 = 32.065 g Cl= 35.453 g X 5 = 177.265g 1 mole = 209.33 grams ------------ 3.65 x 1056molecules SCl5 ------------ 1 mole of SCl5 209.33 g of SCl5 ------------ 6.02 x 1023molecules SCl5 ---------- 1 mole of SCl5 (3.65)(209.33)/6.02 = 126.91 then 1056 – 1023 = 1033 = 1.27 x1035 grams of SCl5

  5. 45.8 g Iron III oxide Fe2O3 Fe = 55.845 X 2 = 111.69 grams O = 15.99 X 3 = 47.997 grams 1 mole = 159.687 grams 45.8 g of Fe2O3 --------------------- 1 mole Fe2O3 = 0.2868 moles of Fe2O3 159.687 grams of Fe2O3 -------------------- = 2.87 x10-1 mole of Fe2O3

  6. Empirical Formula 10.52 g Ni, 4.38g C, and 5.10g N ------- 1 mole of Ni 10.52 g of Ni = 0.179 Moles of Nickel --------- 58.693 g of Ni ------ 1 mole of C 4.38 g of C = 0.365 Moles of Carbon 12.011 g of C --------- ----- 1 mole of N 5.10 g of N = 0.364 Moles of Nitrogen 14.007 g of N ---------

  7. Empirical Formula 0.365 Moles of Carbon = 2 Moles of carbon 0.179 Moles of Nickel 0.364 Moles of Nitrogen = 2 Moles of nitrogen 0.179 Moles of Nickel 0.179 Moles of Nickel = 1 Moles of Nickel 0.179 Moles of Nickel NiC2N2

  8. Hydrates A common hydrate is forms with cobalt (II) chloride. If 11.75g of this hydrate is heated, 9.25g of anhydrous cobalt chloride remains. What is the formula and name for this hydrate? CoCl2  XH2O 11.75 g minus 9.25 g 2.5 g of H2O But how many moles is this? ------ 2.5 g of H2O 1 mole of H2O = 0.139 moles of H2O --------- 18.0 g of H2O

  9. What is the relationship to its compound? CoCl2  XH2O 9.25 g of CoCl2 ------ 9.25 g of CoCl2 1 mole of CoCl2 = 0.0712 moles of CoCl2 --------- 129.83 of CoCl2

  10. What is the relationship to its compound? 0.0712 moles of CoCl2 0.139 moles of H2O Get the ratios  divide smallest mole into others 0.139 moles of H2O = 1.95 (2) mole of H2O 0.0712 moles of CoCl2 CoCl2  2H2O Cobalt II chloride dihydrate

  11. Ch. 12 Mole-mole ratio, Mole-gram ratio, Gram-Gram (stoich), Limited-Excess, Actual Theoretical.

  12. A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction. ___NaBr + ___KCl ___KBr + ___NaCl

  13. A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction. ___NaBr + ___KCl  ___KBr + ___NaCl 25.0 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 14 gNaCl 103 g NaBr 1 mole NaBr 1 mol NaCl 20 g KCl 1 mole KCl 1 mole NaCl 58.4 g NaCl 15.6 g NaCl 1 mole KCl 74.5g KCl 1 mole NaCl

  14. Which reactant is Limited and which reactant is Excess? ___NaBr + ___KCl  ___KBr + ___NaCl Limited= NaBr 25.0 g NaBr 1 mole NaBr 58.4 g NaCl 1 mol NaCl 14 gNaCl 103 g NaBr 1 mole NaBr 1 mol NaCl Excess = KCl 20 g KCl 1 mole KCl 1 mole NaCl 58.4 g NaCl 15.6 g NaCl 1 mole KCl 74.5g KCl 1 mole NaCl

  15. A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction. ___NaBr + ___KCl  ___KBr + ___NaCl Start with the limited reactant!! Used In reaction 25.0 g NaBr 1 moleNaBr 74.5g KCl 1 mole KCl 18 g KCl 1 mole NaBr 103 g NaBr 1 mole KCl Given Used In reaction 20 grams KClminus 18 g KCl = 2 g EXCESS KCl

  16. If 12 grams of sodium chloride were produced after the reaction, what is the % yield? • Actual yield (from experiment) • Percent Yield= • X 100 • Theoretical yield (from stoich calculation) • 12 g (from experiment) • Percent Yield= • X 100 • 14 g (from stoich calculation-limit reactant) • 87%

  17. Ch. 13 Intermolecular forces, Phase changes & Energy Changes

  18. Chaos Most energy Taking in energy (endothermic) Releases energy (exothermic) Taking in energy (endothermic) Releases energy (exothermic) Taking in energy (endothermic) Order Less energy Less order More energy Releases energy (exothermic)

  19. Intramolecular Bonds

  20. Intermolecular Bonds • Intermolecular bonds are “bond” or attractions between molecules • Dispersion forces- nonpolar molecules, very weak, shifting of electrons within the molecule • H2, O2, Ne, CO2, CH4 • Dipole-dipole forces- polar molecules, stronger, unequal sharing of electrons • HCl, CH3OH, CS2 • Hydrogen bonding- polar molecules with FON, very strong. HF, H2O, NH3

  21. Critical Point Melting Freezing Liquid Solid Vaporization  Condensation Gas Triple point-- Sublimation Deposition

  22. Define an endothermic process?Provide 3 examples Endothermic process requires energy. Melting, Vaporization, Sublimation

  23. Define an exothermic process? Provide 3 examples Exothermic process releases- loses energy. Freezing, Condensation, Deposition

  24. Ch. 14 Phase changes & Energy Changes

  25. 1. Helium gas in a 2.00-L cylinder is under 1.12 atm pressure. At 36.5ºC, that same gas sample has a pressure of 2.56 atm. What was the initial temperature (˚C) of the gas in the cylinder? What Gas Law? Gay-Lussac’ Law What Formula? P1 = P2 T1 T2

  26. 1. Helium gas in a 2.00-L cylinder is under 1.12 atm pressure. At 36.5ºC, that same gas sample has a pressure of 2.56 atm. What was the initial temperature (˚C) of the gas in the cylinder? P1 = P2 T1 T2 1.12 atm = 2.56 atm 309.5 K X 346.6 = -138 ˚C = 135-273 2.56

  27. 2. Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00-L vessel at a pressure of 143 atm. What Gas Law? Ideal What Formula? PV = nRT

  28. 2. Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00-L vessel at a pressure of 143 atm. PV = nRT (143 atm)(1.00L) = (2.49)(0.0821)(X K) 143 = 699.5 - 273 0.204429 = 427 ˚C

  29. 3. Air trapped in a cylinder fitted with a piston occupies 145.7 mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston? What Gas Law? Boyles What Formula? P1V1 = P2V2

  30. 3. Air trapped in a cylinder fitted with a piston occupies 145.7 mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston? P1V1 = P2V2 (1.08 atm)(145.7 ml) = (1.43)(X ml) 157.4 = 110 ml 1.43

  31. 4. Use the reaction shown to calculate the mass of iron that must be used to obtain 0.500 L of hydrogen at STP. ___Fe(s) + ____H2O(l)  ___ Fe3O4(s)+ ___ H2(g) What Gas Law? Stoich Gas What Formula? PV = nRT

  32. 4. Use the reaction shown to calculate the mass of iron that must be used to obtain 0.500 L of hydrogen at STP. ___Fe(s) + ____H2O(l)  ___ Fe3O4(s)+ ___ H2(g) 3 4 4 PV = nRT 0.500 L H2 3 mole Fe = 0.375 L of Fe 4 mole of H2

  33. 4. Use the reaction shown to calculate the mass of iron that must be used to obtain 0.500 L of hydrogen at STP. PV = nRT (1 atm)(0.375 L) = (X)(0.0821)(273K) 0.375 = 0.0167 mol of Fe 22.4 0.0167 mol Fe 55.848 g of Fe = 0.933 g of Fe 1 mole of Fe

  34. 5. The Celsius temperature of a 3.00-L sample of gas is lowered from 80.0ºC to 30.0ºC. What will be the resulting volume of this gas? What Gas Law? Charles What Formula? V1 = V2 T1 T2

  35. 5. The Celsius temperature of a 3.00-L sample of gas is lowered from 80.0ºC to 30.0ºC. What will be the resulting volume of this gas? V1 = V2 T1 T2 3.0 L = X L 353 K 303 K 909 = 2.57 L 353

  36. 6. A weather balloon contains 14.0 L of helium at a pressure of 95.5 kPa and a temperature of 12.0°C. If this had been stored in a 1.50-L cylinder at 21.0°C, what must the pressure in the cylinder have been? What Gas Law? Combined What Formula? P1V1 = P2V2 T1 T2

  37. 6. A weather balloon contains 14.0 L of helium at a pressure of 95.5 kPa and a temperature of 12.0°C. If this had been stored in a 1.50-L cylinder at 21.0°C, what must the pressure in the cylinder have been? P1V1 = P2V2 T1 T2 = (1.50 L)( X kPa) (14.0 L)(95.5 kPa) 294 K 285 K 393078 = 919 kPa 427.5

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