Chap-1-HW-Solutions

Chap-1-HW-Solutions - Chapter One Home Work Solutions 18....

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Chapter One Home Work Solutions 18 . (a) 20 °C (above freezing) is higher than 20 °F (below freezing). (b) 100 °C (at boiling) is higher than 180 °F (below boiling) (c) 100 °F is close to body temperature, which is around 40 °C. Therefore 60 °C is higher than 100 °F. (d) – 12 °C and 20 °F are both below freezing, so let’s use the conversion equation to figure this one out: o C = 5 9 × o F 32 = 5 9 × 20 o F 32 ⎟ =− 6.7 o C – 6.7 °C is warmer than –12 °C, so 20 °F is a higher temperature than – 12 °C. 21 . V metal = V final – V initial = (37.2 mL) – (25.4 mL) = 11.8 mL d = m V = 105.5 g 11.8 mL = 8.94 g mL According to Table 1.1, this is very close to the density of copper (d = 8.93 g/mL). 22. 10.0 g lead × 1 mL lead 11.34 g lead = 0.882 mL lead The difference between the starting volume and the final volume must be the volume of the metal piece, so adding the volume of the metal piece to the starting volume gives the final volume: V final = V metal + V initial = (30.0 mL) + (0.882 mL) = 30.882 mL Rounding to one decimal place, the limit of the uncertainty, gives a final volume of 30.9 mL. 23 . V = (thickness) × (width) × (length) = (1.0 cm) × (2.0 cm) × (10.0 cm) = 20. cm 3 Using dimensional analysis, find the volume in mL: 20. cm 3 × 1mL 1cm 3 = 20. mL Find the density: d = m V = 54.0 g 20. mL = 2.7 g mL According to Table 1.1, the density that most closely matches this one is aluminum (density = 2.70 g/mL). 49
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Chap-1-HW-Solutions - Chapter One Home Work Solutions 18....

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