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The Answer is RIGHT!

emmanuel-frank avatar
emmanuel-frank
40 diapositives · 607 Views · Nov 10, 2014

The Answer is RIGHT!. By S P and E I. PLAY BY THE RULES!. Between the 2 teams you will each have 1team leader. In the beginning of each round each team will get a chance to answer 1 question. The team that gets the answer RIGHT will compete in that round.

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  1. The Answer is RIGHT! By S P and E I

  2. PLAY BY THE RULES! • Between the 2 teams you will each have 1team leader. • In the beginning of each round each team will get a chance to answer 1 question. The team that gets the answer RIGHT will compete in that round. • In case of a tie there will be a brief question and the person who answers the fastest will win. The team who wins the most contests over all will WIN.

  3. Round Uno: Warm Up • Name situations where stoichiometry is used. (Write down what you think are the top 3 answers out of 4. The amount of points will be determined by the order you guess them in.) • 1.? • 2.? • 3.? SURVEY SAYS...

  4. Round Uno

  5. Round Dos: Stoichiometry • How many liters of CO2 is produced when 22.7g of C2H5OH is burned?(3points) • C2H5OH + 3O 2CO2 + 3H20

  6. Round Dos: Stoichiometry • Answer: • 22.7gC2H5OH(1 moleC2H5OH/46.08gC2H5OH)(2moleCO2/1moleC2H5OH)(22.4LCO2/1moleCO2)= 22.1 LCO2

  7. Round Dos: Stoichiometry • How much water is produced (in grams) when 2.5 moles of NH3 is reacted?(3points) • 4NH3+ 5O2 4NO2 + 6H2O

  8. Round Dos: Stoichiometry • Answer: • 2.5mole NH3 (6 mole H2O/4moleNH3) (18.02gH2O/1mole H2O)= 68gH2O

  9. Round Dos: Stoichiometry • In the following combustion reaction, how many moles of water are produced when 22.7g of C2H5OH is burned?(5points) • C2H5OH + 3O2 2CO2 + 3H2O

  10. Round Dos: Stoichiometry • Answer: • 22.7g C2H5OH(1moleC2H5OH/46.06gC2H5OH) (3moleH2O/1moleC2H5OH) = 1.48mole H2O

  11. Round Dos: Stoichiometry • How Many iron grams are formed when 43.7 grams of Iron Oxide react with hydrogen? 3H2 + Fe2O3 2Fe + 3H20

  12. Round Dos: Stoichiometry • Answer: • 43.7 grams Fe203(1 mole Fe203/159.7 g Fe203) (2mole Fe/ 1mole Fe203) (55.85 g Fe/ 1mole Fe)= 30.6 grams Fe

  13. Round Tres: Mole conversions • 4.25g H2SO4 to moles

  14. Round Tres: Mole conversions • Answer: 4.25g H2SO4 ( 1mole H2SO4/98.08g H2SO4) = .0433mole H2SO4

  15. Round Tres: Mole conversions • 25.0g Cl2to atoms of Chlorine

  16. Round Tres: Mole conversions • Answer: 25.0g Cl2 (6.02x10^23 molecules CO2/ 70.90g Cl2)(2atoms CL/ 1moleculeCl2)= 4.25x10^23 atoms Cl

  17. Round Tres: Mole conversions • 4.25 H2SO4 to moles

  18. Round Tres: Mole conversions • Answer: 4.25g H2SO4 (1mole H2SO4/ 98.08g H2SO4)= .0433 mole H2SO4

  19. Round Tres: Mole conversions • 60.0g of CO2 to liters

  20. Round Tres: Mole conversions • Answer: 60.0g CO2 (22.4L CO2/ 44.01g CO2)= 30.5 L CO2

  21. Round Tres: Mole conversions • 17.0 mole of KMnO4 to grams

  22. Round Tres: Mole conversions • Answer: 17.0moles KMnO4(158.05g KMnO4/ 1mole KMnO4)= 2680g KMnO4

  23. Round Tres: Mole conversions • 22.4L N2 to molecules

  24. Round Tres: Mole conversions • Answer: 22.4L N2 (6.02x10^23 molecules N2/ 22.4L N2) = 6.02x10^23 molecules N2

  25. Round Cuatro: Molecular Mass & Empirical Formulas • A compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It has a molecular weight of 318.31 g/mol. What is the molecular formula for this compound?

  26. Round Cuatro: Molecular Mass & Empirical Formulas • C-75.46g C (1 mole C/12.01g C)=6.28/1.26 mole C=5*2=10 • H- 4.43g H (1 mole H/1.01g H)=4.39/1.26 mole H=3.5*2=7 • O- 20.10g O (1 mole O/16.00g O)=1.26/1.26 mole O=1*2=2 Empirical FormulaC10H7O2- 159.17g 318.31/159.17=2 Molecular Formula C20H14O4

  27. Round Cuatro: Molecular Mass & Empirical Formulas • A combustion analysis gives the following mass %: H=9.15% C=54.53% O=36.32%. Determine the molecular formula knowing that the molecular mass=132.16.

  28. Round Cuatro: Molecular Mass & Empirical Formulas • H-9.15g H(1 mole H/1.01g H)=9.06/2.27 mole H=4 • C-54.53g C (1 mole C/12.01gC)=4.54/2.27 mole C=2 • O=36.32g O (1 mole O/16.00g O)=2.27/2.27 mole O=1 • Empirical FormulaH4C2O – 44.06 • 132.16/44.06=3 Molecular Formula-H12C6O3

  29. Round Cuatro: Molecular Mass & Empirical Formulas • This data was obtained in a lab: • Mass of copper 2.25g • Mass of sulfide .56g • Mass of Copper sulfide 2.18g • Calculate the Empirical Formula for copper sulfide.

  30. Round Cuatro: Molecular Mass & Empirical Formulas • Answer: • Cu= 2.25g Cu (1 mole/63.55g Cu)=.0354/.01747=2 • S=.56g S (1 mole S/32.00g S)=.01747/.01747=1 Cu2S

  31. Round Cuatro: Molecular Mass & Empirical Formulas • This data was obtained in a lab: • Mass hydrate (CuSO4*H2O) 8.39g • Mass H2O 3.03g • Mass anhydrate (CuSO4) 5.36g • Calculate the formula for the hydrate.

  32. Round Cuatro: Molecular Mass & Empirical Formulas • Answer: • CuSO4=5.36g CuSO4(1 mole CuSO4/159.61g CuSO4)=.03359/.03359 mole CuSO4=1 • H2O=3.03g H2O(1 mole H2O/18.02)=.1681/.03359 mole H2O=5 CuSo45H2O

  33. Round Cinco: Combined Gas Laws • 4.5 L of Carbon dioxide at 23 degrees C has a pressure of 3.2 atms. What is the pressure of the carbon dioxide at 95 degrees C at 3.4 L?

  34. Round Cinco: Combined Gas Laws • Answer: • V1=4.5 L T1=296K P1=3.2atm • V2=3.4 L T2=368K P2=x • 3.2atm(4.5L/3.4L)(368K/296K)=15.3atm • V P T P

  35. Round Cinco: Combined Gas Laws • Oxygen at 25 degrees C and 760 torr pressure occupies a volume of 21.2 L. What is the volume of oxygen gas at 133 degrees C and 830 torr?

  36. Round Cinco: Combined Gas Laws • Answer: • T1=298K P1=760torr V1=21.2L • T2=406K P2=830torr V2=x • 21.2L(406K/298K)(760torr/830torr)=26 L • T V P V

  37. Round Cinco: Combined Gas Laws • 4.3 L of methane at 5.4 kPa has a temperature of 46 degrees C. What is the temperature of methane at 5.4 L at 6.6 kPa?

  38. Round Cinco: Combined Gas Laws • Answer: • V1=4.3L P1=5.4kPa T1=319K • V2=5.4L P2=6.6kPa T2=x • 319K(5.4L/4.3L)(6.6kPa/5.4kPa)=490K • V T P T

  39. Round Cinco: Combined Gas Laws • The temperature of a gas confined in a 20.0 L metal container was raised from 25.0 degrees C to 77.5 degrees C. If the initial pressure was 800 mmHg, what was the final pressure of the gas?

  40. Round Cinco: Combined Gas Laws • Answer: • V=20.0L(stays the same) • T1= 298K P1=800mmHg • T2=350.5K P2=x • 800mmHg (350.5K/298K)=241mmHg • T P