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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.

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Presentation on theme: "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."— Presentation transcript:

1 Chapters 2 and 3 Review Worksheet

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

3 Problem 1 (continued): Convert the answer (267 yd 3 ) to nm 3 10 27 nm 3 1m3m3 1m3m3 1.3079yd 3 267yd 3 = 2.04 x 10 29 nm 3

4 Problem 2: Convert 9.04 x 10  8 in 3 to Tm 3 Tm 3 1 10 42 cm 3 9.04 x 10  8 1 cm 3 in 3 16.4 in 3 = 1.48 x 10  48 Tm 3

5 Problem 3: Convert 947,200  m 3 to mi 3 1mi 3 1.47 x 10 11 ft 3 1 1728 in 3 1 = 2.274 x 10  22 mi 3 cm 3 16.4 1cm 3 m3m3 10 12 947,200 m3m3

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

7 Problem 5: First convert 453 hm 3 to mL mL1 1cm 3 453 1 cm 3 hm 3 10 12 hm 3 = 4.53 x 10 14 mL then....................

8 Problem 5: (continued) Now find the volume: D = m/V so M=D  V M = (0.537 g/mL)  (4.53 x 10 14 mL) M = 2.43 x 10 14 g

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

10 Problem 8: cal1 4.184J 1.94 x 10  14 1 J MJ 10 6 MJ = 4.64 x 10  9 cal

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

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

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

14 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

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

16 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.

17 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.

18 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.

19 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

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

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

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

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

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

25 37. 217 Rn  4 2 He + 213 Po 38. 138 Cs  0 –1 e + 138 Ba 39. 17 F  0 1 e + 17 O 40. 7 Be + 0 –1 e  7 Li 41. 212 At  4 2 He + 208 Bi 42. 117 Ag  0 –1 e + 117 Cd

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


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