Chapters 2 and 3 Review Worksheet

Problem 1: First find the volume: V = ℓ  w  h V = (3.54 yd)(6.39 yd)(11.8 yd) V = 267 yd 3

Problem 1 (continued): Convert the answer (267 yd 3 ) to nm nm 3 1m3m3 1m3m yd 3 267yd 3 = 2.04 x nm 3

Problem 2: Convert 9.04 x 10  8 in 3 to Tm 3 Tm cm x 10  8 1 cm 3 in in 3 = 1.48 x 10  48 Tm 3

Problem 3: Convert 947,200  m 3 to mi 3 1mi x ft in 3 1 = x 10  22 mi 3 cm cm 3 m3m ,200 m3m3

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

Problem 5: First convert 453 hm 3 to mL mL1 1cm cm 3 hm hm 3 = 4.53 x mL then

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

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

Problem 8: cal J 1.94 x 10  14 1 J MJ 10 6 MJ = 4.64 x 10  9 cal

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

Problem 10: Kelvin =  Celsius K =  Celsius 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. filtration 14. distillation, crystallization, and chromatography 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) 194 Os +4 P = 76, N = 118, E = 72 (ii) 116 In +3 P = 49, N = 67, E = 46 (iii) 129 Te  2 P = 52, N = 77, E = 54 (iv) 227 Ra P = 88, N = 139, E = 88

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

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

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

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

Problem 36: K-39 K-41 Abundance = x 39x + 41(1  x) = Abundance = 1  x 39x + 41  41x =  2x =  x = K-39 = % K-41 = 4.915%

Rn  4 2 He Po Cs  0 –1 e Ba F  0 1 e + 17 O Be + 0 –1 e  7 Li At  4 2 He Bi Ag  0 –1 e Cd

43. Ba yellow-green 44. Co not visible 45. K violet 46. Cu turquoise-green 47. Na yellow 48. Sr red49. Ca orange-red