3. Figure 3.1 Schematic Diagram of a Mass Spectrometer

4. Figure 3.2 Neon Gas

16. More Stoichiometry

39. CO 2 is absorbed H 2 O is absorbed Sample is burned completely to form CO 2 and H 2 O Pure O 2 in

40. Figure 3.5 Schematic Diagram of the Combustion Device Used to Analyze Substances for Carbon and Hydrogen

45. 3.7 Balancing equations pp CH 4 + O 2  CO 2 + H 2 O Reactants Products C 1 1 O 2 3 H 4 2

46. Balancing equations pp CH 4 + O 2  CO 2 + 2 H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4

47. Balancing equations pp CH 4 + O 2  CO 2 + 2H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4 4

48. Balancing equations pp CH 4 + 2 O 2  CO 2 + 2H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4 4 4

49. Another Example H 2 + H 2 O O 2  Make a table to keep track of where you are at

50. Example pp H 2 + H 2 O O 2  Need twice as much O in the product R P H O 2 2 2 1

51. Example pp H 2 + H 2 O O 2  Changes the O R P H O 2 2 2 1 2

52. Example pp H 2 + H 2 O O 2  Also changes the H R P H O 2 2 2 1 2 2

53. Example pp H 2 + H 2 O O 2  Need twice as much H in the reactant R P H O 2 2 2 1 2 2 4

54. Example pp H 2 + H 2 O O 2  Recount R P H O 2 2 2 1 2 2 4 2

55. Example pp H 2 + H 2 O O 2  The equation is balanced, has the same number of each kind of atom on both sides R P H O 2 2 2 1 2 2 4 2 4

56. Example H 2 + H 2 O O 2  This is the answer R P H O 2 2 2 1 2 2 4 2 4 Not this

63. 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu pp 0.181 mol Fe 2 mol Fe 3 mol Cu = 0.272 mol Cu 0.272 mol Cu 1 mol Cu 63.55 g Cu = 17.3 g Cu

64. Could have done it via “spider” pp 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu = 17.3 g Cu

65. pp