Download Chemistry Homework Solutions: Chapter 1 and more Assignments Chemistry in PDF only on Docsity! Chapter 1 Homework: solutions 3. a) time b) density c) length d) area e) temperature f)volume g) temperature 6. The age of the artifact was determined to the nearest 100 years. Thus, extra 20 years are no significant – or in other words if we round 1920 to the nearest hundred, we get 1900 years old. 9. a) heterogeneous mixture b) homogeneous mixture c) pure substance (element) d) homogeneous mixture 18. chemical properties: dissolves in sulphuric acid, giving off hydrogen; reacts slowly with oxygen to give ZnO. physical properties: silver-gray metal; melts at 420 °C; hardness 2.5; density 7.13g/cm3 19. chemical processes: a), d), e); physical processes: b), c) 22. Take a sample (a drop), place it on a watch glass and heat. If it is a solution containing salt, the salt will be left on the watchglass after water evaporates. 24. a) 63.5 mL (can be also 6.35 cL, but cL is only rarely used these days) b) 6.5 µs c) 0.95 mm = 950 µm d) 4.23 mm3 (remember, 1mm3 is a cube with dimensions of 1mm x 1mm x 1mm) e) 0.125 mg = 125 µg f) 0.35 ng = 350 pg g) 6.54 µs 26. a) ! 9.5*10 "2 kg* 1000g 1kg = 95g b) ! 0.0023µm* 10 "6 m 1µm * 1nm 10 "9 m = 2.3nm c) ! 7.25*10 "4 s* 1ms 10 "3 s = 0.725ms 31. We need to get a thickness of gold leaf. We are given the mass and density of gold, so we can calculate volume of gold (leaf). From the area of gold leaf we can then determine it’s thickness. ! d = m V "V = m d = 200mg 19.32g /cm 3 = 200mg* 10 #3 g 1mg 19.32 g cm 3 = 0.01cm 3 ! V = A.l" l = V A = 0.01cm 3 2.4 ft *1.0 ft = 0.01cm 3 2.4 ft *1.0 ft 12in 1 ft # $ % & ' ( 2 2.54cm 1in # $ % & ' ( 2 = 5*10 )6 cm l = 5*10 )6 cm* 1m 10 )2 cm = 5*10 )8 m = 50nm 33. a) 62°F °C: (62-32)*(5/9)=17 62°F=17°C b) 216.7°C °F: 216.7*(9/5)+32=422.1 216.7°C=422.1°F c) 233°C K: 233+273.15=506 233°C=506K d) 315K °F: 315-273.15=42 315K=42°C 42*(9/5)+32=108 315K=42°C=108°F e) 2500°F K: (2500-32)*(5/9)=1400 2500°F=1400°C 1400+273.15=1600 2500°F=1400°C=1600K Note: in the last case you can clearly see the effect of carrying the calculations on with all the figures (significant and not significant) and rounding off to the appropriate number of significant figures only at the end of calculation. 36. exact: b (we can’t have a fraction of a student in the class) and e (there is exactly 1000 mm in 1m by definition, so there is exactly 109mm3 in 1m3). 38. a) 5 b) 3 c) 4 d) 4 e) 6 41. a) 21.10 b) 237.4 c) 652 d) 0.0766 42. a) -2300 b) 8.260*107 c) 3.4*104 d) 7.62*105 45. a) 76 mL b) 50 nm c) 6.88*10-4 s d) ! 0.50lb* 453.59g 1lb = 230g e) ! 1.55 kg m 3 * 1000g 1kg * 10 "3 m 3 1L =1.55 g L f) ! 5.850 gal hr * 3.7854L 1gal * 1hr 3600s = 6.151*10 "3 L s 48. a) ! 0.105in * 2.54cm 1in * 10mm 1cm = 2.67mm b) ! 0.870qt * 1L 1.0567qt * 1000mL 1L = 823mL c) ! 8.75 µm s * 10 "6 m 1µm * 1km 1000m * 3600s 1hr = 3.15*10"5 km h d) ! 4.733yd 3 * 1m 1.0936yd " # $ % & ' 3 = 3.619m 3
