Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 1 - Section 1.3 - Scientific Measurement - Checkpoint - Page 13: 1.3.1


c) -89.2$^{\circ}$C, 183.9 K

Work Step by Step

Given F = -128.6^{\circ}F Substitute this value in the equation, C = $\frac{5}{9}$(-128.6-32)=-160.6x5/9=-89.2^{\circ}C Then, we find C (temperature in $^{\circ}C$) = -89.2$^{\circ}$C To get corresponding Kelvin scale value: Use T K= t+ 273=-89.2+273=183.8K where t is the temperature in $^{\circ}$C , i.e. -89.2.
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