EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119227946
Author: Willard
Publisher: VST
bartleby

Concept explainers

Question
Book Icon
Chapter 12, Problem 4PE

(a)

Interpretation Introduction

Interpretation:

Conversion of 649 torr pressure into kPa has to be determined.

Concept Introduction:

Barometer is used for measurement of pressure of gas in atmospheric units (atm). Pressure can also be represented in many units such as torr, mm Hg, and pascal (Pa). These units can be related to one another as follows:

  1 atm=760 torr=760 mm Hg=101325 Pa=101.325 kPa

(a)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 649 torr in kPa units is 86.5 kPa.

Explanation of Solution

Conversion factor to convert torr to kPa is as follows:

  (101.325 kPa760 torr)

Pressure 649 torr can be converted to kPa as follows:

  Pressure(kPa)=(649 torr)(101.325 kPa760 torr)=86.5 kPa

Hence 649 torr in kPa units is 86.5 kPa.

(b)

Interpretation Introduction

Interpretation:

Conversion of 5.07 kPa pressure into atm has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 5.07 kPa in atm units is 0.0500 atm.

Explanation of Solution

Conversion factor to convert kPa to atm is as follows:

  (1 atm101.325 kPa)

Pressure 5.07 kPa can be converted to kPa as follows:

  Pressure(atm)=(5.07 kPa)(1 atm101.325 kPa)=0.0500 atm

Hence 5.07 kPa in atm units is 0.0500 atm.

(c)

Interpretation Introduction

Interpretation:

Conversion of 3.64 atm pressure into mm Hg has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 3.64 atm in mm Hg units is 2.77×103 mm Hg.

Explanation of Solution

Conversion factor to convert atm to mm Hg is as follows:

  (760 mm Hg1 atm)

Pressure 3.64 atm can be converted to mm Hg as follows:

  Pressure(mm Hg)=(3.64 atm)(760 mm Hg1 atm)=2.77×103 mm Hg

Hence 3.64 atm in mm Hg units is 2.77×103 mm Hg.

(d)

Interpretation Introduction

Interpretation:

Conversion of 803 torr pressure into atm has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 803 torr in atm units is 1.06 atm.

Explanation of Solution

Conversion factor to convert torr to atm is as follows:

  (1 atm760 torr)

Pressure 803 torr can be converted to atm as follows:

  Pressure(atm)=(803 torr)(1 atm760 torr)=1.06 atm

Hence 803 torr in atm units is 1.06 atm.

(e)

Interpretation Introduction

Interpretation:

Conversion of 1.08 atm pressure into cm Hg has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 4PE

Pressure 1.08 atm in cm Hg units is 82.1 cm Hg.

Explanation of Solution

Conversion factor to convert atm to mm Hg is as follows:

  (760 mm Hg1 atm)

Pressure 1.08 atm can be converted to cm Hg as follows:

  Pressure(cm Hg)=(1.08 atm)(760 mm Hg1 atm)(1 cm Hg10 mm Hg)=82.1 cm Hg

Hence 1.08 atm in cm Hg units is 82.1 cm Hg.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Predict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ m
Determine the structures of the missing organic molecules in the following reaction: + H₂O +H H+ Y Z ☑ ☑ Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X, Y, and Z. You may draw the structures in any arrangement that you like, so long as they aren't touching. Molecule X shows up in multiple steps, but you only have to draw its structure once. Click and drag to start drawing a structure. AP +
Please help, this is all the calculations i got!!! I will rate!!!Approx mass of KMnO in vial: 3.464 4 Moss of beaker 3×~0. z Nax200: = 29.9219 Massof weacerv after remosimgain N2C2O4. Need to fill in all the missing blanks. ง ง Approx mass of KMnO4 in vials 3.464 Mass of beaker + 3x ~0-304: 29.9219 2~0.20 Miss of beaker + 2x- 29.7239 Mass of beaker + 1x~0.2g Naz (204 29-5249 Mass of beaver after removing as qa Na₂ C₂O T1 T2 T3 Final Buiet reading Initial butet reading (int)) Hass of NaOr used for Titration -reading (mL) calculation Results: 8.5ml 17mL 27.4mL Oml Om Oml T1 T2 T3 Moles of No CO Moles of KMO used LOF KM. O used Molenty of KMNO Averagem Of KMOWL

Chapter 12 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 12.8 - Prob. 12.11PCh. 12.9 - Prob. 12.12PCh. 12.9 - Prob. 12.13PCh. 12 - Prob. 1RQCh. 12 - Prob. 2RQCh. 12 - Prob. 3RQCh. 12 - Prob. 4RQCh. 12 - Prob. 5RQCh. 12 - Prob. 6RQCh. 12 - Prob. 7RQCh. 12 - Prob. 8RQCh. 12 - Prob. 9RQCh. 12 - Prob. 10RQCh. 12 - Prob. 11RQCh. 12 - Prob. 12RQCh. 12 - Prob. 13RQCh. 12 - Prob. 14RQCh. 12 - Prob. 15RQCh. 12 - Prob. 16RQCh. 12 - Prob. 17RQCh. 12 - Prob. 18RQCh. 12 - Prob. 19RQCh. 12 - Prob. 20RQCh. 12 - Prob. 21RQCh. 12 - Prob. 22RQCh. 12 - Prob. 23RQCh. 12 - Prob. 24RQCh. 12 - Prob. 25RQCh. 12 - Prob. 26RQCh. 12 - Prob. 1PECh. 12 - Prob. 2PECh. 12 - Prob. 3PECh. 12 - Prob. 4PECh. 12 - Prob. 5PECh. 12 - Prob. 6PECh. 12 - Prob. 7PECh. 12 - Prob. 8PECh. 12 - Prob. 9PECh. 12 - Prob. 10PECh. 12 - Prob. 11PECh. 12 - Prob. 12PECh. 12 - Prob. 13PECh. 12 - Prob. 14PECh. 12 - Prob. 15PECh. 12 - Prob. 16PECh. 12 - Prob. 17PECh. 12 - Prob. 18PECh. 12 - Prob. 19PECh. 12 - Prob. 20PECh. 12 - Prob. 21PECh. 12 - Prob. 22PECh. 12 - Prob. 23PECh. 12 - Prob. 24PECh. 12 - Prob. 25PECh. 12 - Prob. 26PECh. 12 - Prob. 27PECh. 12 - Prob. 28PECh. 12 - Prob. 29PECh. 12 - Prob. 30PECh. 12 - Prob. 31PECh. 12 - Prob. 32PECh. 12 - Prob. 33PECh. 12 - Prob. 34PECh. 12 - Prob. 35PECh. 12 - Prob. 36PECh. 12 - Prob. 37PECh. 12 - Prob. 38PECh. 12 - Prob. 39PECh. 12 - Prob. 40PECh. 12 - Prob. 41PECh. 12 - Prob. 42PECh. 12 - Prob. 43PECh. 12 - Prob. 44PECh. 12 - Prob. 45PECh. 12 - Prob. 46PECh. 12 - Prob. 47PECh. 12 - Prob. 48PECh. 12 - Prob. 49PECh. 12 - Prob. 50PECh. 12 - Prob. 51PECh. 12 - Prob. 52PECh. 12 - Prob. 53PECh. 12 - Prob. 54PECh. 12 - Prob. 55AECh. 12 - Prob. 56AECh. 12 - Prob. 57AECh. 12 - Prob. 58AECh. 12 - Prob. 59AECh. 12 - Prob. 60AECh. 12 - Prob. 61AECh. 12 - Prob. 62AECh. 12 - Prob. 63AECh. 12 - Prob. 64AECh. 12 - Prob. 65AECh. 12 - Prob. 66AECh. 12 - Prob. 67AECh. 12 - Prob. 68AECh. 12 - Prob. 69AECh. 12 - Prob. 70AECh. 12 - Prob. 71AECh. 12 - Prob. 72AECh. 12 - Prob. 73AECh. 12 - Prob. 74AECh. 12 - Prob. 75AECh. 12 - Prob. 76AECh. 12 - Prob. 77AECh. 12 - Prob. 78AECh. 12 - Prob. 79AECh. 12 - Prob. 80AECh. 12 - Prob. 81AECh. 12 - Prob. 82AECh. 12 - Prob. 83AECh. 12 - Prob. 84AECh. 12 - Prob. 85AECh. 12 - Prob. 86AECh. 12 - Prob. 87AECh. 12 - Prob. 88AECh. 12 - Prob. 89AECh. 12 - Prob. 90AECh. 12 - Prob. 91AECh. 12 - Prob. 92AECh. 12 - Prob. 93AECh. 12 - Prob. 94CECh. 12 - Prob. 95CECh. 12 - Prob. 96CECh. 12 - Prob. 97CECh. 12 - Prob. 98CECh. 12 - Prob. 99CECh. 12 - Prob. 100CECh. 12 - Prob. 101CE
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
Text book image
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning