   Chapter 13, Problem 7QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# ake the indicated pressure conversions. .  45 . 2 kPa to atmospheres 755 mm Hg to atmospheres 8 0 2 torr to kilopascals 1 .0 4 atm to millimeters of mercury

Interpretation Introduction

Interpretation:

The value of pressure45.2 kPa should be converted to atmosphere.

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.

Explanation

Since, 1 kPa = 0.00987 atm

So,  45.2 kPa = 45.2 kPa × 0

Interpretation Introduction

Interpretation:

The value of pressure755 mm Hg should be converted to atmosphere.

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.

Interpretation Introduction

Interpretation:

The value of pressure 802 torr should be converted to kilopascals.

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.

Interpretation Introduction

Interpretation:

The value of pressure1.04 atm should be converted to millimeters of mercury.

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.

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