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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 1 - Problem 52e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 1 - Problem 52e

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# Convert each temperature.a. 212 °F to °C (temperature of

ISBN: 9780321809247 1

## Solution for problem 52E Chapter 1

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition

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Problem 52E

Convert each temperature.

a. 212 °F to °C (temperature of boiling water at sea level)

b. 22 °C to K (approximate room temperature)

c. 0.00 K to °F (coldest temperature possible, also known as absolute zero)

d. 2.735 K to °C (average temperature of the universe as measured from background black body radiation)

Step-by-Step Solution:
Step 1 of 3

Solution 52E: (a) 0.105 in. to mm, Here, we have to convert inch to mm, by using some of the conversion factors. We know 1 m =100 cm 1 inch= 2.54 cm The conversion factors are 2.1 inm 100 cm 0.105 in. = 0.105 in x 2.54 cmx 1 m x 1000 mm 1 in 100 cm 1 m = 0.2667 x 10 mm = 2.67 mm Therefore, 0.105 in. is equal to 2.67 mm (b) 0.650 qt to mL, Here, we have to convert qt to L, by using some of the conversion factors. We know 1 qt =1.057 L 1 L =1000 mL The conversion factors are 1.057 L 1000 mL 1 qt 1L 0.650 qt = 0.650 qt x 1.057xL1000 mL 1 qt 1L = 687.0 mL Therefore, 0.650 qt. is equal to 687.0 mL (c) 8.75 pm/s to km/hr, Here, we have to convert pm/s to km/hr, by using some of the conversion factors. We know 1 pm = 10 -12m 1000 m =1 km 1 min= 60 s 1 h =60 min The conversion factors are 1012 m 1km 60s 60 min 1 m , 1000 m, 1 min 1h Thus, 8.75 pm/s = 8.75 pm/s x 1012 m x 1km x 60s x 60 min -13 1 m 1000 m 1 min 1h = 315 x 10 km/h -13 Therefore, 8.75 pm/s is equal to to 315 x 10 km/h (d) 1.955 m to yd 3 3 3 Here, we have to convert m to yd , by using some of the conversion factors. We know 1 m = 1.09361 yd The conversion factor is 1.09361 yd 1 m Thus, 3 3 1.09361 yd 3 1.955 m = 1.955 m x ( 1 m ) = 2.557 yd 3 Therefore, 1.955 m is equal to to 2.557 yd 3 e) $3.99/lb to dollars per kg, Here, we have to convert dollars per lb, to dollars per kg,, by using some of the conversion factors. We know 1 lb = 453.6 g 1000 g= 1kg The conversion factor are 1lb 1000g 453.6 g 1 kg Thus,$3.99/lb = 3.99 dollarx 1lb x 1000g lb 453.6 g 1 kg = 8.796 dollar/kg Therefore, \$3.99/lb is equal to to 8.8 dollar/kg (f) 8.75 Ib/ft3 to g/ml . Here, we have to convert Ib/ft3, to /ml,, by using some of the conversion factors. We know 1 lb = 453.6 g 1000 g= 1kg 1 mile = 5280 ft 1 mile= 1.6093 km = 1.6093 x1000 m 1 km= 1000 m 1 m =100 cm 3 1 cm = 1mL The conversion factors are 453.6 g 5280ft 1mile 1 m 1 lb 1 mile 1.6093x 1000 m 100 cm Thus, 8.75 Ib/ft3 = 8.75 lb x 453.6 gx ( 5280ft ) x ( 1mile ) x ( 1 m ) ft3 1 lb 1 mile 1.6093x 1000 m 100 cm 14 15 3 = 5.84 x 10 g/4.83 x 10 cm = 1.209 x 10 g/cm 3 = 0.121 g/mL Therefore, 8.75 Ib/ft3 is equal to to 0.121 g/mL

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##### ISBN: 9780321809247

Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. The full step-by-step solution to problem: 52E from chapter: 1 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. Since the solution to 52E from 1 chapter was answered, more than 315 students have viewed the full step-by-step answer. This full solution covers the following key subjects: analysis, carry, conversions, Dimensional, dollars. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. The answer to “?Convert each temperature. a. 212 °F to °C (temperature of boiling water at sea level) b. 22 °C to K (approximate room temperature) c. 0.00 K to °F (coldest temperature possible, also known as absolute zero) d. 2.735 K to °C (average temperature of the universe as measured from background black body radiation)” is broken down into a number of easy to follow steps, and 53 words.

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