
Concept explainers
(a)
Interpretation:
The unit of pressure have to be converted into atmospheres.
Concept Introduction:
The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimeters of mercury) are most commonly used by chemists.
1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa
(a)

Answer to Problem 4A
105.2 kPa = 1.038 atm
Explanation of Solution
1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa
105.2 kPa = 105200 pascal= 105200101325 atm = 1.038 atm
(b)
Interpretation:
The unit of pressure have to be converted into atmospheres.
Concept Introduction:
The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimetres of mercury) are most commonly used by chemists.
1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa
(b)

Answer to Problem 4A
75.2 cm Hg = 0.989 atm
Explanation of Solution
1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa
75.2 cm Hg = 752mmHg = 752760 atm = 0.989 atm
(c)
Interpretation:
The unit of pressure have to be converted into atmospheres.
Concept Introduction:
The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimetres of mercury) are most commonly used by chemists.
1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa
(c)

Answer to Problem 4A
752mmHg = 0.989 atm
Explanation of Solution
1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa
752mmHg = 752760 atm = 0.989 atm
(d)
Interpretation:
The unit of pressure have to be converted into atmospheres.
Concept Introduction:
The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimetres of mercury) are most commonly used by chemists.
1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa
(d)

Answer to Problem 4A
767torr= 1.01 atm
Explanation of Solution
1 standard atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr = 101,325 Pa
767torr= 767760 atm = 1.01 atm
Chapter 13 Solutions
World of Chemistry, 3rd edition
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