A can of hair spray with a pressure of 3.0 atm at 25.0°F is exposed to heat.Should the new temperature become 35.0°C, which of the following solutions computes for the new pressure in kPa?(3.0 atm)(269.26 К)(308.15 K)101.325 kPa...a. P2 == 270 kPa1 atm(3.0 atm) (298.15 К)101.325 kPab. P2 == 290 kPa(308.15 K)1 atm(3.0 atm)(308.15 К)(269.26 K)101.325 kPac. P2 == 350 kPa1 atm((3.0 atm)(308.15 K)'101.325 kPayd. P2= 310 kPa(298.15 K)1 atm

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A can of hair spray with a pressure of 3.0 atm at 25.0°F is exposed to heat. Should the new temperature become 35.0°C, which of the following solutions computes for the new pressure in kPa? (3.0 atm)(269.26 К) 101.325 kPa ... a. P2 = 270 kPa (308.15 K) 1 atm (3.0 atm)(298.15 K) 101.325 kPa b. P2 = = 290 kPa (308.15 K) 1 atm (3.0 atm)(308.15 К) (269.26 K) 101.325 kPa' с. Р. 3 = 350 kPa 1 atm ((3.0 atm)(308.15 K)' 101.325 kPa d. P2 = = 310 kPa (298.15 K) 1 atm

A can of hair spray with a pressure of 3.0 atm at 25.0°F is exposed to heat. Should the new temperature become 35.0°C, which of the following solutions computes for the new pressure in kPa? (3.0 atm)(269.26 К) 101.325 kPa ... a. P2 = 270 kPa (308.15 K) 1 atm (3.0 atm)(298.15 K) 101.325 kPa b. P2 = = 290 kPa (308.15 K) 1 atm (3.0 atm)(308.15 К) (269.26 K) 101.325 kPa' с. Р. 3 = 350 kPa 1 atm ((3.0 atm)(308.15 K)' 101.325 kPa d. P2 = = 310 kPa (298.15 K) 1 atm

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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