Chapter 1: 23, 33, 35, 39, 41, 45, 47, 49, 53, 63, 65, 67, 69, 71, 73, 75, 77, 81, 89

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23. a. When a nail rusts, it is forming Fe3O2 (iron(III) oxide); this means that iron is combining with oxygen (chemically that looks like this: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s). Oxygen has mass (all matter has some mass), so adding oxygen means the mass is increasing.

b. Since the nail’s mass started at 23.2 g and increased to 24.1 g, the nail gained a mass of 0.9 g (via 24.1 g – 23.2 g).

33. Mass and volume are extensive because you can change the amount (mass)- you can always have more or less (same with volume) as the amount you choose is up to you.

Imagine a balloon, you can add more or remove air, which changes the volume (extensive property). Once you take the ratio (density, m/V), it now cancels out the extensiveness (more mass means more volume (again, think of a balloon), but dividing them means the same number (intensive) will be found. Ex. ^{2 𝑚}

2 𝑉 = 1^𝑚

𝑉 or Ex. ^5𝑚

5𝑉 = 1 ^𝑚

𝑉

35. One meter is about 39 inches. A yard is defined to be 36 inches. So 1 m is about the length of a yard.

39. a. c centi 1 100⁄ or 1x10^–2

b. d deci 1

⁄10 or 1x10^–1

c. G giga 1,000,000,000 or 1x10⁹ d. k kilo 1,000 or 1x10³

e. m milli 1 1,000⁄ or 1x10^–3

f. n nano 1x10^–9 or 0.000000001 (8 zeros after the leading zero) g. p pico 1x10^–12 or 0.000000000001 (11 zeros after the leading zero) h. T tera 1x10¹² or 1,000,000,000,000 (12 zeros)

41. Note: the selection for the same volume blocks are on the top right of the screen.

a. green brick: mass (m) = 18.58 g, volume (V) = 5.7 mL (drag the green brick into the water. The water starts at 25.5 mL, adding the green brick (which sinks) increases the volume up to 31.2 mL).

b. density, , = m / V = 18.58 g/5.7 mL = 3.3 g/mL d. dioptase, malachite or Paraiba tourmaline

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e. Since  = m/V; if mass increases, the density increases if the volume is constant 45. a. 704 → 7.04x10²

b. 0.03344 → 3.344x10^–2 c. 547.9 → 5.479x10² d. 22,086 → 2.2086x10⁴ e. 1,000.00 → 1.00000x10³ f. 0.0000000651 → 6.51x10^–8 g. 0.007157 → 7.157x10^–3

47. a. an hour is set to however long we say it is (a definition), exact

b. can’t have physically half a page (or any other fraction) in a book, exact c. since you measure your mass, there is some uncertainty (uncertain) d. the mass called a kg is defined, can be whatever we want, exact e. the amount you drink must be measured, uncertain

f. distance is measured (and technically the earth is made of moving plates), uncertain

49. a. 53 cm → 2 sig figs

b. 2.05x10⁸ m → 3 sig figs (ignore the x10^#) c. 86,002 J → 5 sig figs

d. 9.740x104 m/s → 4 sig figs e. 10.0613 m³ → 6 sig figs

f. 0.17 g/mL → 2 sig figs (zero is a place holder and doesn’t ‘count’).

g. 0.88400 s → 5 sig figs (the last two zeros were written down; someone bothered to write them, so they must ‘count’).

53. a. 628 x 342 = 2.15x10⁵ (multiplying and dividing, so look at least number of sig figs, both have three (3) , so answer has three sig figs too).

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b. (5.63x10²) x (7.4x10³) = 4.2x10⁶ (multiplying and dividing, so least number of sig figs is from second number; two (2) sig figs. Remember, ignore the x10^# as that does not contribute towards the number of sig figs).

c. 28.0

13.483

⁄ = 2.08 (multiplying and dividing, so least number of sig figs is from first number; three (3) sig figs).

d. 8119 x 0.000023 = 0.19 (multiplying and dividing, so least number of sig figs is from second number; two (2) sig figs. Remember, the zeros before the 23 don’t

‘count’ in sig figs as they are just placeholders).

e. 14.98 + 27,340 + 84.7593 = 27,440 (adding and subtracting, so the answer has the least number of places past the decimal. Since the second number has no places past the decimal, then answer has zero places past decimal).

f. 42.7 + 0.259 = 43.0 (adding and subtracting, so the answer has the least number of places past the decimal. Since the first number has one place past the decimal, then answer has one place past the decimal).

63. 12.0 fl. oz. x ^{1 𝑞𝑡}

32 𝑓𝑙.𝑜𝑧. x ^{1 𝐿}

1.057 𝑞𝑡 x ^{1000 𝑚𝐿}

1 𝐿 = 350 mL (3.50 x 10² mL) 65. If red blood cells diameter are 3x10^–4 in., the diameter in cm is:

3x10^–4 in. x ^{2.54 𝑐𝑚}

1.00 𝑖𝑛. = 8x10^–4 cm

67. 197 𝑙𝑏. 𝑥 453.59237 𝑔

1 𝑙𝑏. 𝑥 ^{1 𝑘𝑔}

1,000 𝑔= 𝟖𝟗. 𝟒 𝒌𝒈;

This is less than 90 kg, so the lifter is light enough to compete 69. 5 𝜇𝐿 𝑥 ^1𝑥10⁻⁶^𝐿

1 𝜇𝐿 𝑥 ^{1,000 𝑚𝐿}

1 𝐿 = 𝟓. 𝟎𝒙𝟏𝟎^−𝟑 mL

71. (Using Tables 1.3 and 1.2)

a. kg is base unit for mass… 0.13 g x ^{1 𝑘𝑔}

1,000 𝑔 = 1.3x10^–4 kg b. 232 Gg x ^1𝑥10

9 𝑔

1 𝐺𝑔 x ^{1 𝑘𝑔}

1,000 𝑔 = 2.32x10⁸ kg

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c. 5.23 pm ^{1 𝑥10}

−12 𝑚

1 𝑝𝑚 = 5.23 x10^–12 m d. 86.3 mg x ^{1 𝑥10}

−3 𝑔

1 𝑚𝑔 x ^{1 𝑘𝑔}

1,000 𝑔 = 8.63 x10^–5 kg e. 37.6 cm x ^{1 𝑥10}

−2 𝑚

1 𝑐𝑚 = 3.76 x 10^–1 m f. 54 μm x ^{1 𝑥10}⁻⁶^𝑚

1 𝜇𝑚 = 5.4 x 10^–5 m g. 1 Ts x ^{1 𝑥10}

12 𝑠

1 𝑇𝑠 = 1x10¹² s h. 27 ps x ^{1 𝑥10}

−12 𝑠

1 𝑝𝑠 = 2.7 x 10^–11 s i. 0.15 mK x ^{1 𝑥10}

−3 𝐾

1 𝑚𝐾 = 1.5x10^–4 K

73. Using the data within the chapter and the conversion table 1.6:

12.0 gal x ^{4 𝑞𝑡}

1 𝑔𝑎𝑙 x ^{0.94635 𝐿}

1 𝑞𝑡 = 45.4 L

1/2 gal. (exactly) 0.5 gal x ^{4 𝑞𝑡}

1 𝑔𝑎𝑙 x ^{0.9463 𝐿}

1 𝑞𝑡 = 1.893 L

2,240 lb. x ^{1 𝑘𝑔}

2.2046 𝑙𝑏 = 1,016.1 kg 77. a. 120 m x ^{100 𝑐𝑚}

1 𝑚 x ^{1 𝑖𝑛}

2.54 𝑐𝑚 x ^{1 𝑓𝑡}

12 𝑖𝑛 = 394 ft b. 19,565 ft x ^{12 𝑖𝑛}

1 𝑓𝑡 x ^{2.54 𝑐𝑚}

1 𝑖𝑛 x ^{1 𝑚}

100 𝑐𝑚 x ^{1 𝑘𝑚}

1,000 𝑚 = 5.9634 km c. 8.5 in x 11 in = 93.5 in²

93.5 in²x ^{(2.54 𝑐𝑚)}

2

(1 𝑖𝑛)² = 6.03 x 10² cm² d. 161 in³ x ^{(2.54 𝑐𝑚)}

3

(1 𝑖𝑛)³ x ^{1 𝑚𝐿}

(1 𝑐𝑚)³ x ^{1 𝐿}

1,000 𝑚𝐿 = 2.64 L

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e. Found part of this in question #75. Question doesn’t specify if it is a ‘long’ or a

‘short’ ton, so both are solved for:

5.6x10¹⁵ long ton x ^{2,240 𝑙𝑏}

1 𝑙𝑜𝑛𝑔 𝑡𝑜𝑛 x ^{1 𝑘𝑔}

2.2046 𝑙𝑏 = 5.7 x 10¹⁸ kg (from long ton) 5.6x10¹⁵ short ton x ^{2,000 𝑙𝑏}

1 𝑠ℎ𝑜𝑟𝑡 𝑡𝑜𝑛 x ^{1 𝑘𝑔}

2.2046 𝑙𝑏 = 5.1 x 10¹⁸ kg (from short ton) f. 32.0 lb. x ^{1 𝑘𝑔}

2.2046 𝑙𝑏 = 14.5 kg g. (using Table 1.6)

5.00 grain x ^{0.00229 𝑜𝑧}

1 𝑔𝑟𝑎𝑖𝑛 x ^{28.349 𝑔}

1 𝑜𝑧. x ^{1,000 𝑚𝑔}

1 𝑔 = 324 mg

81. Can 225 g phosphoric acid (H3PO4) fit in a 150 mL Erlenmeyer flask (density,  = 1.83 g/mL)?

1 𝑐𝑚³

1.83 𝑔 𝑥 225 𝑔 𝑥 ^{1 𝑚𝐿}

1 𝑐𝑚³ = 𝟏𝟐𝟑 𝒎𝑳; 𝒚𝒆𝒔, 𝒕𝒉𝒊𝒔 𝒘𝒊𝒍𝒍 𝒇𝒊𝒕 𝒊𝒏𝒕𝒐 𝒂 𝟏𝟓𝟎 𝒎𝑳 𝒇𝒍𝒂𝒔𝒌!

89. a. m = ? (g) of 6.00 cm³ Hg, Hg = 13.5939 g/cm³ 13.5939 𝑔 𝐻𝑔

1 𝑐𝑚³ 𝐻𝑔 𝑥 6.00 𝑐𝑚³ 𝐻𝑔 = 𝟖𝟏. 𝟔 𝒈 𝑯𝒈 b. m = ? (g) of 25.0 mL octane, octane = 0.702 g/cm³

0.702 𝑔 𝑜𝑐𝑡𝑎𝑛𝑒

1 𝑐𝑚³ 𝑜𝑐𝑡𝑎𝑛𝑒 𝑥 ^{1 𝑐𝑚}³^{𝑜𝑐𝑡𝑎𝑛𝑒}

1 𝑚𝐿 𝑜𝑐𝑡𝑎𝑛𝑒 𝑥 25.0 𝑚𝐿 𝑜𝑐𝑡𝑎𝑛𝑒 = 𝟏𝟕. 𝟔 𝒈 𝒐𝒄𝒕𝒂𝒏𝒆