1 / 40

The Answer is RIGHT!

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.

Download Presentation

The Answer is RIGHT!

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related