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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 1 - Problem 28e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 1 - Problem 28e

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When adding or subtracting measured quantities, what

ISBN: 9780321809247 1

Solution for problem 28E Chapter 1

Chemistry: A Molecular Approach | 3rd Edition

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Problem 28E

When adding or subtracting measured quantities, what determines the number of significant figures in the result?

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Solution 28E: (a) The temperature on a warm summer day is 87 °F We know, The relation between celsius scale and Fahrenheit scale oC= 5/9( F-32) or F=9/5( C)+32 o = 5/9(87-32) = 5/9(55) o = 30.55 C o Thus the temperature in °C is 30.55 C (b) 25 °C The relation between celsius scale and Kelvin scale is given below We know, o K= C + 273.13 = 25 + 273.13 =298.13 K Thus the temperature in Kelvin scale is 298.13 K We know The relation between celsius scale and Fahrenheit scale is given below o o C= 5/9( F-32) or F=9/5( C)+32 = 9/5(25)+32 = 45+32 = 77 F o Thus the temperature in Fahrenheit scale is 7 7 F c) Suppose that a recipe calls for an oven temperature of 400 °F. We Know The relation between celsius scale and Fahrenheit scale is given below o o o o C= 5/9( F-32) or F=9/5( C)+32 = 5/9(400-32) = 5/9x(368) = 204.4 C Thus the temperature in °C is 204.4 C o We know, The relation between celsius scale and Kelvin scale is given below o K= C + 273.13 = 204.4+273.1 =477.53 K Thus the temperature in Kelvin scale is 477.53 K (d) Liquid nitrogen boils at 77 K We know, The relation between celsius scale and Kelvin scale is given below o K= C + 273.13 Or o C = K-273.13 = 77-273.13 o = -196.13 C Thus the temperature in °C is -196.13 C o We know The relation between celsius scale and Fahrenheit scale is given below oC= 5/9( F-32) or F=9/5( C)+32 = 9/5( -196.13)+32 = -353.03+32 o = -321.03 F Thus the temperature in °F is -321.03 F o

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