Chapters 2 and 3Review

Worksheet

Problem 1:First find the volume:

V = ℓwhV = (3.54 yd)(6.39 yd)(11.8 yd)

V = 267 yd3

Problem 1 (continued):Convert the answer (267 yd3) to nm3

1027 nm3

1 m3

1 m3

1.3079 yd3

267 yd3

= 2.04 x 1029 nm3

Problem 2:Convert 9.04 x 108 in3 to Tm3

Tm311042 cm3

9.04 x 108

1cm3

in3

16.4 in3

= 1.48 x 1048 Tm3

Problem 3:Convert 947,200 m3 to mi3

1mi3

1.47 x 1011 ft3

1ft3

1728in3

1in3

= 2.274 x 1022 mi3

cm316.41 cm3

m31012

947,200m3

Problem 4:

D = m/V

D = (5.36 kg)/(640 mL)

D = 0.0084 kg/mL

Problem 5:First convert 453 hm3 to mL

mL11 cm3

4531

cm3

hm3

1012 hm3

= 4.53 x 1014 mL

then....................

Problem 5: (continued)Now find the volume:

D = m/V so M=DV

M = (0.537 g/mL) (4.53 x 1014 mL)

M = 2.43 x 1014 g

Problem 6: 1 calorie = 4.184 J

4.184 J1 cal

135 cal = 565 J

Problem 7:

Problem 8:

cal14.184 J

1.94 x 1014

1J

MJ106 MJ

= 4.64 x 109 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.

13. filtration14. distillation, crystallization, and chromatography

15. kinetic, radiant, and potential

16. solid, liquid, gas, and plasma

17. chemical, electrical, and gravitational18. Dalton, schoolteacher19. 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 + neutrons30. 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) 129Te2 P = 52, N = 77, E = 54

(iv) 227Ra P = 88, N = 139, E = 88

Problem 32:

(i) 97Mo+6

(ii) 33P3

(iii) 244Pu+5

Problem 33:

Average Atomic Mass =

51(.9975) + 50(.0025)

Average Atomic Mass = 50.9975

Problem 34:

Average Atomic Mass =

90(.5145)91(.1122)92(.1715)94(.1738)96(.0280)

Average Atomic Mass = 91.3184

+______________

Problem 35:

Ga-69Ga-71

Abundance = x

69x + 71(1 x) = 69.72Abundance = 1 x

69x + 71 71x = 69.72 2x = 1.28x = 0.64

Ga-69 = 64%Ga-71 = 36%

Problem 36:

K-39K-41

Abundance = x

39x + 41(1 x) = 39.0983Abundance = 1 x

39x + 41 41x = 39.0983 2x = 1.9017x = .95085

K-39 = 95.085%K-41 = 4.915%

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 visible45. K violet46. Cu turquoise-green

47. Na yellow

48. Sr red 49. Ca orange-red