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Convert each temperature.a. 32 °F to °C (temperature at

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 51E Chapter 1

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 51E

Problem 51E

Convert each temperature.

a.        32 °F to °C (temperature at which water freezes)

b.        77 K to °F (temperature of liquid nitrogen)

c.        -109 °F to °C (temperature of dry ice)

d.        93.6 °F to K (body temperature)

Step-by-Step Solution:
Step 1 of 3

Solution 51E: Unit conversion (a) 5.00 days to s, Here, we have to convert day to s, by using some of the conversion factors. We know 1 day = 24 h 1 h = 60 min 1 min = 60 s Thus the conversion factors are 24 h ,60 min, 60 s 1 day 1 h 1 min Thus, 5 days = 5 days x 1 dayx 61 hinx 1 min = 432000 s 5 = 4.32 x 10 s 5 Therefore, 5 days is equal to 4.32 x 10 s (b) 0.0550 mi to m Here, we have to convert mi to m, by using some of the conversion factors. We know 1 mi = 1.6093 km 1 km =1000 m Thus the conversion factors are 1.61 mikm, 11 kmm Thus, 1.6093 km 1000 m 0.0550 mi = 0.0550 mi x 1 mi x 1 km = 88.51 m Therefore, 0.0550 mi is equal to 88.51 m (c) $1 89/gal to dollars per liter Here, we have to convert $/gal to $/L, by using some of the conversion factors. We know 1 gal = 4 qt 1 qt =1.057 L Thus the conversion factors are 1 gal, 1 qt 4 qt 1.057 L Thus, dollar 1 gal 1 qt $1 89/gal = 189 gal x 4 qtx 1.057 L = 44.7 dollar/L Therefore, $1 89/gal is equal to 44.7 dollar/L (d) 0.510 in./ms to km/hr Here, we have to convert 0.510 in./ms to km/hr, by using some of the conversion factors. We know 1 in = 2.54 cm 1000 m = 1km 1 m= 100 cm 1 ms = 10 s-3 1 h = 60 min 1 min = 60 s Thus the conversion factors are 2.54 cm , 1m , 1 km , 1 ms , 60 s 60 min 1 in 100 cm 1000 m 103 s 1 min 1 h Thus, 0.510 in./ms = 0.510 inx 2.54 cmx 1m x 1 km x 1 ms x 60 s x 60 min ms 1 in 100 cm 1000 m 103 s 1 min 1 h = 46.6 km/h Therefore, 0.510 in/ms is equal to 46.6 km/h e) 22.50 gal/min to L/s Here, we have to convert 0.510 in./ms to m/hr, by using some of the conversion factors. We know 1 gal = 4 qt 1 qt =1.057 L 1 min = 60 s Thus the conversion factors are 4qt 1gal , 11qt7 L,60 sn Thus gal 4qt 1.057 L 1 min 22.50 gal/min = 22.55 min x 1gal x 1qt x 60 s = 1.419 L/s Therefore, 22.50 gal/min is equal to 1.419 L/s or 1.42 L/s (f) 0.02500 ft to cm 3. Here, we have to convert 0.02500 ft to cm , by using some of the conversion factors. We know 1 mile = 5280 ft 1 mile= 1.6093 km 1 km= 1000 m 1 m= 100 cm Thus the conversion factors are 1 mi 1.6093 km 1000m 100 cm 5280 ft 1 mi 1 km 1 m Thus, 3 3 1 mi 1.6093 km 1000m 100 cm 3 0.02500 ft = 0.02500 ft x [ 5280 ft x 1 mi x 1 km x 1 m ] 3 3 9 6 = 0.02500 x 1/5280 x (1.6093) x 10 x 10 2 3 = 7.078 x 10 cm 3 2 3 Therefore, 0.02500 ft is equal to 7.078 x 10 cm

Step 2 of 3

Chapter 1, Problem 51E is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Convert each temperature.a. 32 °F to °C (temperature at