Chapters 2 and 3 Review Worksheet

Chapters 2 and 3Review Worksheet

Problem 1: First find the volume: V = ℓwh V = (3.54 yd)(6.39 yd)(11.8 yd) V = 267 yd3

Problem 1 (continued): Convert the answer (267 yd3) to nm3 1 m3 267 yd3 1027 nm3 1 m3 1.3079 yd3 = 2.04 x 1029 nm3

Problem 2: Convert 9.04 x 108 in3 to Tm3 cm3 9.04 x 108 in3 1 Tm3 16.4 1042 cm3 1 in3 = 1.48 x 1048Tm3

Problem 3: Convert 947,200 m3 to mi3 947,200 1 mi3 1 ft3 1 in3 1 cm3 m3 ft3 in3 1012 1.47 x 1011 1728 16.4 cm3 m3 = 2.274 x 1022 mi3

Problem 4: D = m/V D = (5.36 kg)/(640 mL) D = 0.0084 kg/mL

Problem 5: First convert 453 hm3 to mL cm3 453 hm3 1 mL 1012 1 cm3 1 hm3 = 4.53 x 1014 mL then....................

Problem 5: (continued) Now find the volume: D = m/V so M=DV M = (0.537 g/mL)  (4.53 x 1014 mL) M = 2.43 x 1014 g

1 calorie = 4.184 J Problem 6: Problem 7: 135 cal 4.184 J = 565 J 1 cal

Problem 8: 1 cal 106 J 1.94 x 1014 MJ 1 MJ 4.184 J = 4.64 x 109 cal

Problem 9: Kelvin = Celsius + 273 Kelvin = 43  + 273 Kelvin = 230 K

Problem 10: Kelvin = Celsius + 273 293 K = Celsius + 273 293 K  273 = Celsius 20 C = Celsius

Problem 11: (i) chemical change (ii) chemical property (iii) physical change (iv) physical property

12. During electrolysis, an electric current is passed through a substance. If the substance is a compound, it may be broken down into the separate elements that form it. 14. distillation, crystallization, and chromatography 13. filtration 15. kinetic, radiant, and potential

16. solid, liquid, gas, and plasma 17. chemical, electrical, and gravitational 18. Dalton, schoolteacher 19. Democritus, philosopher 20. scanning tunneling microscope 21. Faraday, chemist

22. “elektron” is the Greek word for amber (fossilized tree sap which when rubbed with cloth would attract dust and other particles – static electricity) 23. Ben Franklin, He flew a kite with a key on its string during a thunderstorm. 24. J.J. Thomson, physicist, He discovered electrons.

25. Henri Becquerel, physicist 26. Ernest Rutherford, scientist, He discovered the nucleus (positively charged) 27. The plum-pudding model states that negative charges are distributed evenly throughout an atom’s positively-charged interior. Rutherford found that the positive charges were centrally located in a core.

28. number of protons = atomic number 29. mass number = protons + neutrons 30. Average atomic mass is a weighted average of all of the isotopes of an element. Multiply the abundance percentage by the mass number for all the isotopes and add them together.

Problem 31: (i) 194Os+4 P = 76, N = 118, E = 72 (ii) 116In+3 P = 49, N = 67, E = 46 (iii) 129Te2 P = 52, N = 77, E = 54 (iv) 227Ra P = 88, N = 139, E = 88

Problem 32: (i) 97Mo+6 (ii) 33P3 (iii) 244Pu+5

Problem 33: Average Atomic Mass = 51(.9975) + 50(.0025) Average Atomic Mass = 50.9975

Average Atomic Mass = 90(.5145)91(.1122)92(.1715)94(.1738)96(.0280) Problem 34: +______________ Average Atomic Mass = 91.3184

Problem 35: Abundance = x Ga-69 Abundance = 1  x Ga-71 69x + 71(1  x) = 69.72 69x + 71  71x = 69.72  2x =  1.28 Ga-69 = 64%Ga-71 = 36% x = 0.64

Problem 36: Abundance = x K-39 Abundance = 1  x K-41 39x + 41(1  x) = 39.0983 39x + 41  41x = 39.0983  2x =  1.9017 K-39 = 95.085%K-41 = 4.915% x = .95085

37. 217Rn  42He + 213Po 38. 138Cs  0–1e + 138Ba 39. 17F  01e + 17O 40. 7Be + 0–1e  7Li 41. 212At  42He + 208Bi 42. 117Ag  0–1e + 117Cd

43. Ba yellow-green 44. Co not visible 45. K violet 46. Cu turquoise-green 47. Na yellow 48. Sr red 49. Ca orange-red

Chapters 2 and 3 Review Worksheet