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Entry Task: April 23 rd Monday

Entry Task: April 23 rd Monday. Question: P= 34 atm V= 200L Temp 293K and R = 0.0821 What is the number of moles?. Agenda:. Discuss All Gas Law ws I can… statements Review Ch. 11 – 14 (field questions). All Late assignments need to be done by P of U day.

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Entry Task: April 23 rd Monday

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  1. Entry Task: April 23rd Monday Question: P= 34 atm V= 200L Temp 293K and R = 0.0821 What is the number of moles?

  2. Agenda: • Discuss All Gas Law ws • I can… statements • Review Ch. 11 – 14 (field questions)

  3. All Late assignments need to be done by P of U day

  4. 2. 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

  5. 2. 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

  6. 4. 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

  7. 4. 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

  8. 6. 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

  9. 6. 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

  10. 8. 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

  11. 8. 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

  12. 8. 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

  13. 10. 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

  14. 10. 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

  15. 12. 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

  16. 12. 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

  17. I can… • Apply the 3 gas laws to problems involving pressure, volume and temperature of a gas. • Pressure v. Volume • Temperature v. Volume • Pressure v. Temperature

  18. Use the combined gas law equation with gas problems P1V1/T1 = P2V2/T2 Relate the amount of gas to its pressure, temperature and volume by using the ideal gas law PV = nRT I can…

  19. Use stoichiometry to find information about other substances in a gas reaction. I can…

  20. I can…from Ch. 11 • Calculate gram-mole • Calculate Empirical Formulas • Calculate Molecular formulas • Calculate Hydrates

  21. Empirical to Molecular Formula A compound was found to contain 49.98g carbon and 10.47g hydrogen. The molar mass is found to be 58.12 g/mol. Determine the molecular formula. Start with empirical formula 1 mole of C ------------- 49.98 g carbon = 4.16 Moles of Carbon --------- 12.01 of C 1 mole of H ------------- 10.47 g hydrogen = 10.39 Moles of Hydrogen --------- 1.007g of H

  22. Empirical to Molecular Formula Divide by the smallest ratio. 4.16 Moles of Carbon = 1 Moles of Carbon 4.16 Moles of Carbon CAN’T HAVE ½ a mole 10.39 Moles of Hydrogen = 2.5 Moles of hydrogen X 2 4.16 Moles of Carbon C2H5

  23. Empirical to Molecular Formula C2H5 empirical formula- get its mass ------------- 12.01 of C 2 Moles of Carbon = 24.02 g of Carbon --------- 1 mole of C ------------------- 1.007 of H 5 Moles of hydrogen = 5.04 g of Hydrogen 1 mole of H --------- Empirical mass is 29.055g

  24. Empirical mass to the fix A compound was found to contain 49.98g carbon and 10.47g hydrogen. The molar mass is found to be 58.12 g/mol. Determine the molecular formula. 58.12 g/mole 29.055 g/mole Experimental molar mass (given in problem) Mass of empirical mass 2.00 Multiply the empirical formula by 2. 2.00(C2H5) C4H10 is the molecular formula

  25. You try! Suppose you have 2.50 grams of hydrated Copper II sulfate. CuSO4  XH2O Before heating = 2.50g After heating = 1.59g How much water was driven off? 0.91 g of H2O But how many moles is this? ------ 0.91g of H2O 1 mole of H2O = 0.050 moles of H2O --------- 18.0 g of H2O

  26. What is the relationship to its compound? CuSO4  XH2O Use the mass of the anhydrous CuSO4 = 1.59 g How many moles are in this mass? ------ 1.59 g of CuSO4 1 mole of CuSO4 = 0.00996 moles of CuSO4 --------- 159.6 of CuSO4

  27. What is the relationship to its compound? = 0.00996 moles of CuSO4 = 0.050 moles of H2O Get the ratios  divide smallest mole into others 0.050 moles of H2O = 5 mole of H2O 0.00996 moles of CuSO4 CuSO4  5H2O Copper II sulfate pentahydrate

  28. I can…from Ch. 12 • Calculate gram to gram • Calculate Limited reactants • Calculate Amount of excess • Calculate % yield

  29. 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made AND how much excess. 2 K + I2 2 KI 15.0 g K 1 mole K 165.99 g KI 2 mol KI 63.68 g KI 39.098 g K 2 mole K 1 mol KI 15.0 g I2 1 mol I2 2 mol KI 165.99 g KI 19.6 g KI 1 mol I2 253.8 g I2 1 mol KI

  30. Which reactant is Limited and which reactant is Excess? 2 K + I2 2 KI Excess = K 15.0 g K 1 mole K 165.99 g KI 2 mol KI 63.68 g KI 39.098 g K 2 mole K 1 mol KI Limited= l2 15.0 g I2 1 mol I2 2 mol KI 165.99 g KI 19.6 g KI 1 mol I2 253.8 g I2 1 mol KI

  31. 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made AND how much excess. 2 K + I2 2 KI Start with the limited reactant!! Used In reaction 15.0 g I2 1 mol I2 2 mol K 39.1 g K 4.62 g K 1 mol I2 1 mol K 253.8 g I2 Given Used In reaction 10.38 g EXCESS 15.0 grams Kminus 4.62 g K =

  32. I can…from Ch. 13 • Describe the difference between intermolecular and intramolecular forces. • Distinguish the different types of IMF • Identify substance IMF by formulas

  33. Compare these two compounds • CH3CH2CH2CH3 and HF • IMF • Viscosity • Boiling point

  34. Describe the difference between intermolecular and intramolecular forces

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