
Concept explainers
(a)
Interpretation:
The pressure 42.7 kPa is to be converted into mmHg, atm and psi.
Concept introduction:
An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Answer to Problem 4QAP
The pressure in psi is 6.20 psi
The pressure in mm Hg is
The pressure in atm is 0.422 atm.
Explanation of Solution
The force exerted by gas over a unit surface of area is called pressure of gas. The expression of pressure is as below:
Where, P is pressure of gas, F is force exerted by gas and A is area.
The SI unit of pressure is Pascal, which is denoted as Pa. It is represented as follows;
The pressure given in kPa is 47.7 kPa. The pressure is also represented in psi, that is, pound-force square inch. The value of one Pascal is equivalent to
Therefore, the pressure in psi is 6.20 psi.
The value of one Pascal pressure is equivalent to 9.87 atm. The pressure is converted into atmosphere as follow:
Hence, the pressure in atm is 0.422 atm.
The value of one atmosphere is equivalent to 760 mm Hg. The pressure is converted into mm Hg as follow:
Hence, the pressure in mm Hg is
(b)
Interpretation:
The pressure 29.1 psi is to be converted into mmHg, atm and kPa.
Concept introduction:
An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Answer to Problem 4QAP
The pressure in kPa is 201 kPa.
The pressure in mm Hg is
The pressure in atmosphere is 1.98 atm.
Explanation of Solution
The pressure given in pound-force square inch is 29.1 psi. The value of one psi pressure is equivalent to 6.90kPa. The pressure is converted into kPa as follows:
Hence, the pressure in kPa is 201 kPa.
The value of one Pascal pressure is equivalent to 9.87 atm. The pressure is converted into atmosphere as follows:
Hence, the pressure in atmosphere is 1.98 atm.
The value of one atmosphere is equivalent to 760 mmHg. The pressure is converted into mm Hg as follows:
Hence, the pressure in mm Hg is:
(c)
Interpretation:
The pressure 0.788 atm is to be converted into mmHg, psi and kPa.
Concept introduction:
An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Answer to Problem 4QAP
The pressure in mm Hg is 599 mm Hg.
The pressure in psi is 11.6 psi.
The pressure in kPa is 79.8 kPa.
Explanation of Solution
The value of one atmosphere is equivalent to 760 mmHg. The 0.788 atmosphere pressure is converted into mm Hg as follows:
Hence, the pressure in mm Hg is 599 mm Hg.
The value of one atmosphere pressure is equivalent to 101.33 kPa. The pressure is converted into kPa as follows:
Hence, the pressure in kPa is 79.8 kPa.
The value of one Pascal is equivalent to
Hence, the pressure in psi is 11.6 psi.
(d)
Interpretation:
The pressure 1216 mm Hg is to be converted into atm, psi, and kPa.
Concept introduction:
An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Answer to Problem 4QAP
The pressure in atmosphere is 1.60 atm.
The pressure in psi is 23.5 psi.
The pressure in kPa is 162.1 kPa
Explanation of Solution
The given pressure value is 1216 mm Hg. The value of one atmosphere is equivalent to 760 mmHg. The pressure is converted into atmosphere as follows:
Therefore, the pressure in atmosphere is 1.60 atm.
The value of one atmosphere pressure is equivalent to 101.33 kPa. The pressure is converted into kPa as follows:
Hence, the pressure in kPa is 162.1 kPa.
The value of one Pascal is equivalent to
Hence, the pressure in psi is 23.5 psi.
Want to see more full solutions like this?
Chapter 5 Solutions
Chemistry: Principles and Reactions
- Problem 6-29 Identify the functional groups in the following molecules, and show the polarity of each: (a) CH3CH2C=N CH, CH, COCH (c) CH3CCH2COCH3 NH2 (e) OCH3 (b) (d) O Problem 6-30 Identify the following reactions as additions, eliminations, substitutions, or rearrangements: (a) CH3CH2Br + NaCN CH3CH2CN ( + NaBr) Acid -OH (+ H2O) catalyst (b) + (c) Heat NO2 Light + 02N-NO2 (+ HNO2) (d)arrow_forwardPredict the organic product of Y that is formed in the reaction below, and draw the skeletal ("line") structures of the missing organic product. Please include all steps & drawings & explanations.arrow_forwardPlease choose the best reagents to complete the following reactionarrow_forward
- Problem 6-17 Look at the following energy diagram: Energy Reaction progress (a) Is AG for the reaction positive or negative? Label it on the diagram. (b) How many steps are involved in the reaction? (c) How many transition states are there? Label them on the diagram. Problem 6-19 What is the difference between a transition state and an intermediate? Problem 6-21 Draw an energy diagram for a two-step reaction with Keq > 1. Label the overall AG°, transition states, and intermediate. Is AG° positive or negative? Problem 6-23 Draw an energy diagram for a reaction with Keq = 1. What is the value of AG° in this reaction?arrow_forwardProblem 6-37 Draw the different monochlorinated constitutional isomers you would obtain by the radical chlorination of the following compounds. (b) (c) Problem 6-39 Show the structure of the carbocation that would result when each of the following alkenes reacts with an acid, H+. (a) (b) (c)arrow_forwardPlease draw the major product of this reaction. Ignore inorganic byproducts and the carboxylic side productarrow_forward
- predict the product formed by the reaction of one mole each of cyclohex-2-en-1-one and lithium diethylcuprate. Assume a hydrolysis step follows the additionarrow_forwardPlease handwriting for questions 1 and 3arrow_forwardIs (CH3)3NHBr an acidic or basic salt? What happens when dissolved in aqueous solution? Doesn't it lose a Br-? Does it interact with the water? Please advise.arrow_forward
- Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningLiving By Chemistry: First Edition TextbookChemistryISBN:9781559539418Author:Angelica StacyPublisher:MAC HIGHERIntroductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningWorld of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning





