General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 1, Problem 1.80SP

(a)

Interpretation Introduction

Interpretation:

The volume in liters occupied by carbon monoxide in a room that measures 17.6long,8.80mwideand2.64mhigh has to be calculated.

Concept Introduction:

Volume can be calculated by the equation given below:

length×width×height

Volume is the ratio of mass to density and its SI unit is m3.

Volume=MassDensity

Unit conversions:

1L=1000cm31m=100cm

(a)

Expert Solution
Check Mark

Explanation of Solution

Given data:

concentration=8.00×102ppmlength=17.6mwidth=8.80mheight=2.64m

Volume can be calculated by the equation given below:

length×width×height

V=length×width×height=17.68.802.64m=409m3

Volume has to be converted into liters

409m3×(1cm1×10-2m)3×1L1000cm3=4.09×105Lair

The concentration is given by 8.00×102ppm

The volume in liters can be calculated as follows:

4.09×105Lair×8.00×102LCO1×106Lair=327LCO

(b)

Interpretation Introduction

Interpretation:

The given concentration has to be converted from mg/m3tog/L

Concept Introduction

Unit conversions:

1g=1000mg1m=100cm1L=1000cm3

(b)

Expert Solution
Check Mark

Explanation of Solution

Given data:

concentration=0.050mg/m3

The concentration has to be converted from mg/m3tog/L.

The unit conversions useful for this are given below:

1g=1000mg1m=100cm1L=1000cm3

0.050mg1m3×1×103g1mg×(1×102m1cm)3×1000cm31L=5.0×108g/L

(c)

Interpretation Introduction

Interpretation:

The given blood sugar level has to be converted from mg/dLtoμg/mL

Concept Introduction

Unit conversions:

1μg=1×103g1mL=1×102dL

(c)

Expert Solution
Check Mark

Explanation of Solution

Given data:

concentration=120mg/dL

The concentration has to be converted from mg/dLtoμg/mL.

The unit conversions useful for this are given below:

1μg=1×103g1mL=1×102dL

120mg1dL×1μg1×103mg×(1×102dL1dL)=1.20×103μg/L

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Chapter 1 Solutions

General Chemistry

Ch. 1.7 - Prob. 2PECh. 1.7 - Prob. 1RCCh. 1 - Prob. 1.1QPCh. 1 - Prob. 1.2QPCh. 1 - Prob. 1.3QPCh. 1 - Prob. 1.4QPCh. 1 - Prob. 1.5QPCh. 1 - Prob. 1.6QPCh. 1 - 1.7 Do these statements describe chemical or...Ch. 1 - 1.8 Does each of these describe a physical change...Ch. 1 - Prob. 1.9QPCh. 1 - Prob. 1.10QPCh. 1 - 1.11 Classify each of these substances as an...Ch. 1 - Prob. 1.12QPCh. 1 - Prob. 1.13QPCh. 1 - Prob. 1.14QPCh. 1 - Prob. 1.15QPCh. 1 - Prob. 1.16QPCh. 1 - Prob. 1.17QPCh. 1 - Prob. 1.18QPCh. 1 - Prob. 1.19QPCh. 1 - Prob. 1.20QPCh. 1 - Prob. 1.21QPCh. 1 - Prob. 1.22QPCh. 1 - Prob. 1.23QPCh. 1 - Prob. 1.24QPCh. 1 - Prob. 1.25QPCh. 1 - Prob. 1.26QPCh. 1 - Prob. 1.27QPCh. 1 - Prob. 1.28QPCh. 1 - Prob. 1.29QPCh. 1 - Prob. 1.30QPCh. 1 - Prob. 1.31QPCh. 1 - Prob. 1.32QPCh. 1 - 1.33 The price of gold on a certain day in 2009...Ch. 1 - Prob. 1.34QPCh. 1 - Prob. 1.35QPCh. 1 - Prob. 1.36QPCh. 1 - Prob. 1.37QPCh. 1 - Prob. 1.38QPCh. 1 - Prob. 1.39QPCh. 1 - Prob. 1.40QPCh. 1 - Prob. 1.41QPCh. 1 - Prob. 1.42QPCh. 1 - Prob. 1.43QPCh. 1 - Prob. 1.44QPCh. 1 - Prob. 1.45QPCh. 1 - Prob. 1.46QPCh. 1 - Prob. 1.47QPCh. 1 - Prob. 1.48QPCh. 1 - Prob. 1.49QPCh. 1 - Prob. 1.50QPCh. 1 - Prob. 1.51QPCh. 1 - Prob. 1.52QPCh. 1 - Prob. 1.53QPCh. 1 - Prob. 1.54QPCh. 1 - Prob. 1.55QPCh. 1 - Prob. 1.56QPCh. 1 - Prob. 1.57QPCh. 1 - Prob. 1.58QPCh. 1 - Prob. 1.59QPCh. 1 - Prob. 1.60QPCh. 1 - Prob. 1.61QPCh. 1 - Prob. 1.62QPCh. 1 - Prob. 1.63QPCh. 1 - Prob. 1.64QPCh. 1 - Prob. 1.65QPCh. 1 - 1.66 A sheet of aluminum (Al) foil has a total...Ch. 1 - Prob. 1.67QPCh. 1 - Prob. 1.68QPCh. 1 - Prob. 1.69QPCh. 1 - Prob. 1.70QPCh. 1 - Prob. 1.71QPCh. 1 - Prob. 1.73SPCh. 1 - Prob. 1.74SPCh. 1 - Prob. 1.75SPCh. 1 - Prob. 1.76SPCh. 1 - 1.77 A pycnometer is a device for measuring the...Ch. 1 - Prob. 1.78SPCh. 1 - Prob. 1.79SPCh. 1 - Prob. 1.80SPCh. 1 - Prob. 1.81SP
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