1 - 1 © chemeketa community college: chemistry for allied health ch104chemistryfor allied health...

1 - 1© Chemeketa Community College: Chemistry for Allied Health

CH104CH104

CHEMISTRY CHEMISTRY

FOR FOR

ALLIED HEALTHALLIED HEALTHCHEMEKETA COMMUNITY COLLEGE

INSTRUCTOR: Dr. Jan Cammack

http://faculty.chemeketa.edu/jcammack/


11stst Day Stuff Day Stuff

• Who are you? Are you in the right place?

–GOB

–Chemeketa elearn site

–Course Web Page

• Course Syllabus & requirements

–Who am I?

–Web CT Quizzes


CH104 Recitation & Lab: Week 1CH104 Recitation & Lab: Week 1

1. Read through

• Lab 0: Safety and

• Lab Experiment 1Experiment 1: The Scientific Method

2. Homework problems will be due each week at recitation

– Week 1 problems and online Week 1 problems and online quiz due Tuesdayquiz due Tuesday


What is ChemistryWhat is Chemistry

The Scientific MethodThe Scientific Method

How To StudyHow To Study

1 - 5© Chemeketa Community College: Chemistry for Allied Health 5

Definition of ChemistryDefinition of Chemistry

Chemistry is the study of substances in terms of

• Composition What is it made of?

• Structure How is it put together?

• Properties What characteristics does it have?

• Reactions How does it behave with other

substances?


Chemistry in Our LivesChemistry in Our Lives

Scientific Method:Thinking like a Scientist


The Scientific MethodThe Scientific MethodProcess used by scientists to explain observations.

It involves making observations writing a hypothesis doing experiments proposing a theory


Develop a theory

Experiment

Do more experiments

Question

Hypothesis

Try newtests Did it work?

Yes

Scientific methodScientific method

Observe

No


Summary of the Scientific MethodSummary of the Scientific Method


Everyday Scientific ThinkingEveryday Scientific Thinking

Observation: The sound from a CD in a CD player skips.

Hypothesis 1: The CD player is faulty.

Experiment 1: When the CD is replaced with another one, the sound from the 2nd CD is OK.

Hypothesis 2: The original CD has a defect.

Experiment 2: When I play the CD in another player, the sound still skips.

Theory: My experimental results indicate that the original CD has a defect.


Learning CheckLearning CheckThe step of scientific method indicated in the processes below is

1) observation 2) hypothesis3) experiment 4) theory

A. A blender does not work when plugged in.

B. The blender motor is broken.

C. The plug has malfunctioned.

D. The blender does not work when plugged into a different outlet.

E. The blender needs repair.


SolutionSolutionThe step of scientific method indicated in the processes below is

1) observation 2) hypothesis3) experiment 4) theory

A. A blender does not work when plugged in. 11

B. The blender motor is broken. 22

C. The plug has malfunctioned. 22

D. The blender does not work when plugged into a different outlet. 33

E. The blender needs repair. 44


Chemistry in Our LivesChemistry in Our Lives

A Study Plan for Learning Chemistry


Text Features for LearningText Features for LearningThe study features in the text include:

1.Learning Goals

• indicate what to learn

2. Concept Checks

• develop new concepts

3.Sample Problems

• provide step-by-step problem solving models

4.Guides to Problem Solving

• give directions for working out a problem


Studying for ChemistryStudying for ChemistryActivities that can help you learn chemistry

successfullyinclude the following: attend class regularly go to your instructor’s office hours form a study group with other students to discuss

problems and their solutions study and work problems every day do not wait until the night before the test to

prepare for an exam


Active LearningActive LearningActive participation will help learn the material

more quickly and with more understanding.

1.Obtain an overview from the Looking Ahead topics.

2.Form a question from the section title.

3.Read the section, and answer your question.

4.Work the Sample Problems and Study Checks.

5.Check Answers at the end of the chapter.

6.Proceed to the next section in the text and repeat above.


Learning CheckLearning CheckWhich of the following activities would be part of a successful study plan?

A. staying out late the night before an exam

B. reading the text before class

C. working problems with a study group

D. skipping the lecture 1 or 2 times a week

E. discussing a problem with the instructor


SolutionSolutionWhich of the following activities would be part of a successful study plan?

A. (No) staying out late the night before an exam

B. (Yes) reading the text before class

C. (Yes) working problems with a study groupD. (No) skipping lecture 1 or 2 times a week

E. (Yes) discussing a problem with the instructor


Chapter 1:Chapter 1:MeasurementMeasurement

Units of MeasurementUnits of Measurement

Scientific NotationScientific Notation

Significant FiguresSignificant Figures

Conversion CalculationsConversion Calculations

DensityDensity


Chapter 1 MeasurementsChapter 1 Measurements

1.1Units of Measurement


Units are importantUnits are important

Measurements in chemistry Measurements in chemistry

has little meaning, just a number

has some meaning - money

more meaning - person’s salary


Measurements in chemistryMeasurements in chemistry

English unitsEnglish units. . Still commonly used in daily life.

For Example:

Common English measures of volumeCommon English measures of volume

1 tablespoon = 3 teaspoons1 cup = 16 tablespoons1 pint = 2 cups1 quart = 2 pints1 gallon = 4 quarts1 peck = 2 gallons1 bushel = 4 pecks

Not often used in scientific workNot often used in scientific work


Metric SI Common

ConversionsLength


meter (m) meter (m)meter (m) meter (m)

1 m = 1.09 yd1 m = 1.09 yd2.54 cm = 1 in2.54 cm = 1 in

1 m = 100 cm = 1000 mm1 m = 100 cm = 1000 mm


Metric SI Common

ConversionsLength

Volume



1 m = 1.09 yd1 m = 1.09 yd2.54 cm = 1 in2.54 cm = 1 in

1 m = 100 cm = 1000 mm1 m = 100 cm = 1000 mm

liter (L) cubic meter (mliter (L) cubic meter (m33))

1 L = 1.06 qt1 L = 1.06 qt

946 mL = 1 qt946 mL = 1 qt

1 L = 1000 mL1 L = 1000 mL


Metric SI Common

ConversionsLength

Volume

Mass



1 m = 1.09 yd1 m = 1.09 yd2.54 cm = 1 in2.54 cm = 1 in

1 m = 100 cm = 1000 mm1 m = 100 cm = 1000 mm

liter (L) cubic meter (mliter (L) cubic meter (m33))

1 L = 1.06 qt1 L = 1.06 qt

946 mL = 1 qt946 mL = 1 qt

1 L = 1000 mL1 L = 1000 mL

gram (g) Kilogram (kg)gram (g) Kilogram (kg)

1 kg = 2.20 lb1 kg = 2.20 lb1 kg = 1000 g1 kg = 1000 g

454 g = 1 lb454 g = 1 lb


The amount of material in an objectmaterial in an objectMass:Mass:Mass (in g’s) of

a 1L Bowling Ball > a 1 L Balloon

Weight:Weight:

Mass Vs. WeightMass Vs. Weight

Pull of Gravity on an object.Weight of

Person on Earth > Person on Moon


Mass Vs. WeightMass Vs. Weight

How much would you weigh How much would you weigh on another planet?on another planet?

http://www.exploratorium.edu/ronh/weight/


Metric SI Common

ConversionsLength

Volume

Mass

Time

Temp


meter (m) meter (m) 1 m = 1.09 ydmeter (m) meter (m) 1 m = 1.09 yd 2.54 cm = 1 in2.54 cm = 1 in

liter (L) cubic meter (m3)

gram (g) Kilogram (kg) 1 kg = 2.20 lb

1 L = 1.06 qt

946 mL = 1 qt

Celsius (oC) Kelvin (K) oC = (oF-32)/1.8 K = oC + 273

second (s) second (s) 60 s = 1 min


For each of the following, indicate whether the unit describes

1) length, 2) mass, or 3) volume.

____ A. A bag of tomatoes is 4.5 kg.

____ B. A person is 2.0 m tall.

____ C. A medication contains 0.50 g of aspirin.

____ D. A bottle contains 1.5 L of water.

Learning CheckLearning Check


For each of the following, indicate whether the unit describes

1) length, 2) mass, or 3) volume.

____ A. A bag of tomatoes is 4.5 kg.

____ B. A person is 2.0 m tall.

____ C. A medication contains 0.50 g of aspirin.

____ D. A bottle contains 1.5 L of water.

SolutionSolution

1

2

2

3


Learning CheckLearning Check Identify the measurement that has an SI unit. A. John’s height is _____.

1) 1.5 yd 2) 6 ft 3) 2.1 m

B. The race was won in _____. 1) 19.6 s 2) 14.2 min 3) 3.5 h

C. The mass of a lemon is _____.1) 12 oz 2) 0.145 kg 3) 0.62 lb

D. The temperature is _____.1) 85 °C 2) 255 K 3) 45 °F


SolutionSolutionIdentify the measurement that has an SI unit. A. John’s height is _____.

1) 1.5 yd 2) 6 ft 3) 2.1 m

B. The race was won in _____. 1) 19.6 s 2) 14.2 min 3) 3.5 h

C. The mass of a lemon is _____.1) 12 oz 2) 0.145 kg 3) 0.62 lb

D. The temperature is _____.1) 85 °C 2) 255 K 3) 45 °F



1.2Scientific Notation


Scientific notationScientific notation

If a number is If a number is largerlarger than 1 than 1

•Move decimal point XX places leftleft to get a number between 1 and 10.

1 2 3 , 0 0 0 , 0 0 0. •The resulting number is multiplied by 10XX.

= 1.23 x 108



If a number is If a number is smallersmaller than 1 than 1•Move decimal point XX places rightright to get

a number between 1 and 10.

0. 0 0 0 0 0 0 1 2 3 = 1.23 x 10-7-7

•The resulting number is multiplied by 10-X-X.


ExamplesExamplesWrite in Scientific Notation:

25 =

8931.5 =

0.000593 =

0.0000004 =

3,210. =

2.5 x 10 2.5 x 10 11

8.9315 x 10 8.9315 x 10 33

5.93 x 10 5.93 x 10 - 4- 4

4 x 10 4 x 10 - 7- 7

3.210 x 103.210 x 1033


x 10x 10

1.44939 x 101.44939 x 10-2-2 = =


+

-1

/

x

0

2 3

4 5 6

7 8 9

.

CE

EE

log

ln

1/x

x2

cos tan

1.44939 E-2

0.01449390.0144939

On Calculator On Calculator 1.44939 (-) 21.44939 (-) 2EEEE

Means Means x 10x 10 ChangeChange

SignSign


Select the correct scientific notation for each.

A. 0.000 000 08 m

1) 8 x 108 m 2) 8 x 10-8 m 3) 0.8 x 10-7 m

B. 72 000 L

1) 7.2 x 104 L 2) 72 x 103 L 3) 7.2 x 10-4

41



SolutionSolution

Select the correct scientific notation for each.

A. 0.000 000 08 m

1) 8 x 108 m 2) 8 x 10-8 m 3) 0.8 x 10-7 m

B. 72 000 L

1) 7.2 x 104 L 2) 72 x 103 L 3) 7.2 x 10-4


Learning CheckLearning CheckWrite each as a standard number.

A. 2.0 x 10-2 s

1) 200 s 2) 0.0020 s 3) 0.020 s

B. 1.8 x 105 g

1) 180 000 g 2) 0.000 018 g 3) 18 000 g


SolutionSolutionWrite each as a standard number.

A. 2.0 x 10-2 s

1) 200 s 2) 0.0020 s 3) 0.020 s

B. 1.8 x 105 g

1) 180 000 g 2) 0.000 018 g 3) 18 000 g


1.3Measured Numbers

and Significant Figures



Measured & Exact NumbersMeasured & Exact Numbers

Exact Numbers =Exact Numbers = from counting or by definition

12 coins per package12 coins per package

12 coins 1 package

12 coins 1 package

1 package12 coins

1 package12 coins=

12 coins 1 dozen coins 12 coins

1 dozen coins 1 dozen coins

12 coins 1 dozen coins

12 coins=


Examples of Exact NumbersExamples of Exact Numbers

from numbers in a defined relationship

when objects are counted


Measured & Exact NumbersMeasured & Exact Numbers

Measured Numbers =Measured Numbers = estimated using a tool

All measurements contain some uncertaintyuncertainty.

•We make errors

•Tools have limits


AccuracyAccuracy

How close are we to the true value?

TruthTruth

PrecisionPrecision

How well do our values agree?

ConsistencyConsistency


Accuracy and precisionAccuracy and precision

Our goal!

Truth and ConsistencyTruth and Consistency

Values we can trust.


Length of object is between 6.76.7 and 6.86.8The next digit would be a guess.

Significant figuresSignificant figures

If use 6.766.76 then have error of + 0.01cm and have 3 significant figures3 significant figures.



Expresses accuracy & precision.accuracy & precision.

You can’t report values more accurate than the methods of measurement used .

6.76.766 units = 33 significant figures

Certain Digits

UncertainDigit



. l8. . . . l . . . . l9. . . . l . . . . l10. . cm

What is the length of the red line?

1) 9.0 cm

2) 9.03 cm

3) 9.04 cm


The length of the red line could be reported as

2) 9.03 cm

or 3) 9.04 cm

The estimated digit may be slightly different. Both readings are acceptable.

56

SolutionSolution

. l8. . . . l . . . . l9. . . . l . . . . l10. . cm


We can estimate the value to be 8.45 mL8.45 mL but cannot be more accurate than that.

8.45 has 3 sig figs.

Significant figuresSignificant figuresMeniscus is between

8.48.4 and 8.58.5The next digit would

be a guess.



Sig Figs don’t depend on the decimal point.Sig Figs don’t depend on the decimal point.

255 millimetersmillimeters

25.5 centimeterscentimeters

2.55 decimetersdecimeters

0.255 metersmeters

0.0255 decametersdecameters


Significant figures: Rules for zerosSignificant figures: Rules for zeros

0.00421 Leading zeroLeading zero

Captive zeros areare significant. 4012

Trailing zeros behind decimal areare significant.114.20

Captive zeroCaptive zero

Trailing zeroTrailing zero

Leading zeros are notare not significant.

33 sig figs

44 sig figs

55 sig figs


32,00032,000Are the 0’s significant?

22 sig figs =

33 sig figs =

44 sig figs =

55 sig figs =

3.2 x 103.2 x 1044

3.20 x 103.20 x 1044

3.200 x 103.200 x 1044

3.2000 x 103.2000 x 1044

32,000.32,000.



1025 km

2.00 mg

0.00570

520


ThreeThree (only trailing zero behind decimal

is significant, leading zeros are not)

FourFour (Captive zeros are significant)

ThreeThree (trailing zeros behind decimal

are significant)

TwoTwo (No decimal, zero assumed insignif)


Significant Figures in Significant Figures in Scientific NotationScientific Notation

In scientific notation: All digits, including zeros in the coefficient, are

significant.

Scientific Notation Number of Significant Figures___________

8 x 104 m

8.0 x 104 m

8.00 x 104 m

12

3


State the number of significant figures in each of the following measurements:

A. 0.030 m

B. 4.050 L

C. 0.0008 g

D. 2.80 m



State the number of significant figures in each of the following measurements:

A. 0.030 m

B. 4.050 L

C. 0.0008 g

D. 2.80 m

SolutionSolution

2

4

1

3


A. Which answer(s) contains 3 significant figures?

1) 0.4760 2) 0.00476 3) 4.76 x 103

B. All the zeros are significant in

1) 0.00307 2) 25.300 3) 2.050 x 103

C. The number of significant figures in 5.80 x 102 is

1) one 3) two 3) three



SolutionSolutionA. Which answer(s) contains 3 significant figures?

1) 0.4760 2) 0.00476 3) 4.76 x 103

B. All the zeros are significant in

1) 0.00307 2) 25.300 3) 2.050 x 103

C. The number of significant figures in 5.80 x 102 is

1) one 3) two 3) three


In which set(s) do both numbers contain the same number of significant figures?

1) 22.0 and 22.00

2) 400.0 and 4.00 x 102

3) 0.000015 and 150 000



In which set(s) do both numbers contain the same number of significant figures?

1) 22.0 and 22.00

2) 400.0 and 4.00 x 102

3) 0.000015 and 150 000

SolutionSolution

Both numbers contain two (2) significant figures.



A. Exact numbers are obtained by

1. using a measuring tool

2. counting

3. definition

B. Measured numbers are obtained by


2. counting

3. definition


SolutionSolutionA. Exact numbers are obtained by


2. counting

3. definition

B. Measured numbers are obtained by


2. counting

3. definition


Learning CheckLearning CheckClassify each of the following as exact (E) or

measured (M) numbers. Explain your answer.

A. __ Gold melts at 1064 °C.

B. __ 1 yard = 3 feet

C. __ The diameter of a red blood cell is 6 x 10-4 cm.

D. __ There are 6 hats on the shelf.

E. __ A can of soda contains 355 mL of soda.


SolutionSolutionClassify each of the following as exact (E) or

measured (M) numbers. Explain your answer.

A. __ Gold melts at 1064 °C.

B. __ 1 yard = 3 feet

C. __ The diameter of a red blood cell is 6 x 10-4 cm.

D. __ There are 6 hats on the shelf.

E. __ A can of soda contains 355 mL of soda.

M

M

M

E

E



1.4Significant Figures in

Calculations


Write with 4 Significant Figures:

2.579 0352.579 035

RoundingRounding

1st insignificant digit1st insignificant digit

34.20 22134.20 221 becomes 34.2034.20

becomes 2.5802.58055

44

>> 5 5 round up

< 5 < 5 round down.

>> 5 5 round up

< 5 < 5 round down.

Sometimes a calculated answer shows too many significant digits so we need to round.


Adding Significant ZerosAdding Significant Zeros

4

1.5

0.2

12

Sometimes a calculated answer requires moremore significant digits so we need to add zeros.

Calculated answer

Zeros added to give 3 significant figures

4.00

1.50

0.200

12.0


Learning CheckLearning CheckAdjust the following calculated answers to give

answers with three significant figures.

A. 824.75 cm

B. 0.112486 g

C. 8.2 L


SolutionSolutionAdjust the following calculated answers to give

answers with three significant figures.

A. 824.75 cm

B. 0.112486 g

C. 8.2 L

First digit dropped is greater than 4.825 cm

First digit dropped is 4.0.112 g

Significant zero is added.8.20 L


Significant figuresSignificant figuresand calculationsand calculations

An answer can’t have greater significance than the quantities used to produce it.

speed = 1.00 km 3.0 min

+

-1

/

x

0

2 3

4 5 6

7 8 9

.

CE

EE

log

ln

1/x

x2

cos tan

0.3333333333

= ??

ExampleExample How fast did you run if youwent 1.00 km in 3.0 minutes?


Simplified rules for significant figuresSimplified rules for significant figuresMultiplication & Division Problems:Multiplication & Division Problems:

• Do calculations.

•Look at sig figs for each value in calculation. (Constants don’t count.)

•Report answer with same sig figs as least significant value.

•Round off as needed.

speed = 1.00 km 3.0 min

= 0.333333333 0.333333333 kmkm minmin

= 0.33 0.33 kmkm minmin

2 sig figs

3 sig figs


Simplified rules for significant figuresSimplified rules for significant figuresAddition & Subtraction Problems:Addition & Subtraction Problems:

• Do calculations.

•Look at least significant placeplace for each value in calculation.

•Report answer to least significant place.

•Round off as needed.

1.9+ 18.65 20.55

= 20.6 = 20.6

Significant to .1

Significant to .01

Significant to .1


((1.9 + 18.651.9 + 18.65 ) ) = =2.153 2.153

Add & Sub mixed w/ Mult & Div Problems:Add & Sub mixed w/ Mult & Div Problems:

(( 20. 20.5555 )) = = 2.1532.153

•Do Addition & SubtractionAddition & Subtraction calculations 1st.

•Make note of the least significant placeleast significant place.

3 sig figs3 sig figs (after addition)

4 sig figs4 sig figs


•Do Multiplication & DivisionMultiplication & Division calculations.

9.544821189.54482118

•Round to least # sig fig.

9.549.549.549.54

((1.9 + 18.651.9 + 18.65 ) ) = =2.153 2.153

(( 20. 20.5555 )) = = 2.1532.153

3 sig figs 3 sig figs (after addition)(after addition)

Add & Sub mixed w/ Mult & Div Problems:Add & Sub mixed w/ Mult & Div Problems:

4 sig figs4 sig figs


Give an answer for the following with the correct number of significant figures:

A. 2.19 x 4.2 =

1) 9 2) 9.2 3) 9.198

B. 4.311 ÷ 0.07 =

1) 61.59 2) 62 3) 60

C. 2.54 x 0.0028 =

0.0105 x 0.060

1) 11.3 2) 11 3) 0.041



SolutionSolution

2.54 x 0.0028 0.0105 0.060 = = 11.28888889

= 11 (rounded)

Give an answer for the following with the correct number of significant figures:

A. 2.19 x 4.2 =

1) 9 2) 9.2 3) 9.198

B. 4.311 ÷ 0.07 =

1) 61.59 2) 62 3) 60

C. 2.54 x 0.0028 =

0.0105 x 0.060

1) 11.3 2) 11 3) 0.041


Learning CheckLearning Check For each calculation, round the answer to give

the correct number of decimal places.

A. 235.05 + 19.6 + 2 =

1) 257

2) 256.7

3) 256.65

B. 58.925 – 18.2 =

1) 40.725

2) 40.73

3) 40.7


SolutionSolution

235.05 +19.6 + 2 256.65 rounds to 257

58.925 –18.2 40.725 rounds to 40.7

For each calculation, round the answer to give the correct number of decimal places.

A. 235.05 + 19.6 + 2 =

1) 257

2) 256.7

3) 256.65

B. 58.925 – 18.2 =

1) 40.725

2) 40.73

3) 40.7



1.5Prefixes and Equalities


PrefixesPrefixesA prefix in front of a unit increases or decreases the size of that

unit by one or more factors of 10 indicates a numerical value

Prefix Value

1 kilometer = 1000 meters

1 kilogram = 1000 grams


Metric prefixesMetric prefixesPrefix Prefix

(Symbol)(Symbol)Factor Factor

(multiple)(multiple)

Common Common ConversionConversion

mega (M)

kilo (k)

deci (d)

centi (c)

milli (m)

micro ()

nano (n)

1,000,000 = (101,000,000 = (1066) 1Mm = 1,000,000 m) 1Mm = 1,000,000 m

1,000 = (101,000 = (1033) 1km = 1,000 m) 1km = 1,000 m

0.1 = (100.1 = (10-1-1) 1m = 10 dm ) 1m = 10 dm

0.01 = (100.01 = (10-2-2) 1m = 100 cm) 1m = 100 cm

0.001 = (100.001 = (10-3-3) 1m = 1,000 mm) 1m = 1,000 mm

0.000001 = (100.000001 = (10-6-6) 1m = 1,000,000 ) 1m = 1,000,000 mm

0.000,000,001 = (100.000,000,001 = (10-9-9) 1m = 1,000,000,000 nm) 1m = 1,000,000,000 nm


Indicate the unit that matches the description:

1. a mass that is 1000 times greater than 1 gram

1) kilogram 2) milligram 3) megagram

2. a length that is 1/100 of 1 meter

1) decimeter 2) centimeter 3) millimeter

3. a unit of time that is 1/1000 of a second

1) nanosecond 2) microsecond 3) millisecond



SolutionSolution

= 0.01 of 1 meter

= 0.001 of a sec

Indicate the unit that matches the description:

1. a mass that is 1000 times greater than 1 gram

1) kilogram 2) milligram 3) megagram

2. a length that is 1/100 of 1 meter

1) decimeter 2) centimeter 3) millimeter

3. a unit of time that is 1/1000 of a second

1) nanosecond 2) microsecond 3) millisecond


Select the unit you would use to measure

A. your height

1) millimeters 2) meters 3) kilometers

B. your mass

1) milligrams 2) grams 3) kilograms

C. the distance between two cities


D. the width of an artery




SolutionSolutionSelect the unit you would use to measure

A. your height


B. your mass

1) milligrams 2) grams 3) kilograms

C. the distance between two cities


D. the width of an artery



States the same measurement in two different unitsMetric EqualitiesMetric Equalities

Length: 1 meter is the same as 100 cm or 1000 mm.

1 m = 100 cm 1 m = 1000 mmVolume: 1 L is the same as 1000 cm3.

1 L = 10 cm X 10cm X 10 cm 1 L = 1000 mL

Mass: 1 kg = 1000 g

1 g = 1000 mg

1 mg = 0.001 g

1 mg = 1000 µg


Indicate the unit that completes each of the followingequalities:

A. 1000 m = 1) 1 mm 2) 1 km 3) 1dm

B. 0.001 g = 1) 1 mg 2) 1 kg 3) 1dg

C. 0.1 s = 1) 1 ms2) 1 cs 3) 1ds

D. 0.01 m = 1) 1 mm 2) 1 cm 3) 1dm



SolutionSolutionIndicate the unit that completes each of the followingequalities:

A. 1000 m = 1) 1 mm 2) 1 km 3) 1dm

B. 0.001 g = 1) 1 mg 2) 1 kg 3) 1dg

C. 0.1 s = 1) 1 ms2) 1 cs 3) 1ds

D. 0.01 m = 1) 1 mm 2) 1 cm 3) 1dm


Complete each of the following equalities:

A. 1 kg = 1) 10 g 2) 100 g 3) 1000 g

B. 1 mm = 1) 0.001 m 2) 0.01 m 3) 0.1 m



SolutionSolutionComplete each of the following equalities:

A. 1 kg = 1) 10 g 2) 100 g 3) 1000 g

B. 1 mm = 1) 0.001 m 2) 0.01 m 3) 0.1 m



1.6Writing Conversion Factors


See Handout Sheet of See Handout Sheet of

Common conversion factorsCommon conversion factors

&&

Handout of Handout of

Conversion ProblemsConversion Problems

Measurements in chemistry Measurements in chemistry


Some Common EqualitiesSome Common Equalities


Write equalities and conversion factors for each pair of units:

A. liters and mL

B. hours and minutes

C. meters and kilometers



Write equalities and conversion factors for each pair of units:

A. liters and mL

B. hours and minutes

C. meters and kilometers

SolutionSolution

Equality: 1 L = 1000 mL

1 L and 1000 mL 1000 mL 1 L

Equality: 1 hr = 60 min 1 hr and 60 min

60 min 1 hr

Equality: 1 km = 1000 m

1 km and 1000 m

1000 m 1 km



Write the equality and conversion factors for each of the following:

A. meters and centimeters

B. jewelry that contains 18% gold

C. one liter of gas is $ 0.95


SolutionSolution

A. meters and centimeters

1 m and 100 cm 100 cm 1 m

B. jewelry that contains 18% gold

18 g gold and 100 g jewelry

100 g jewelry 18 g gold

C. one liter of gas is $0.95

1 L and $0.95

$0.95 1 L



1.7Problem Solving


Conversion of unitsConversion of units

Example:Example: Metric Conversion Metric Conversion How many milligrams (mg) are in 5 kilograms

(kg)? Factor label methodFactor label method

• Identify your conversions factors.

1 kg1 kg = 1 = 11000 g1000 g

1000 g1000 g = 1 = 11 kg1 kg

1 g 1 g = 1 = 11000 mg1000 mg

1000 mg1000 mg = 1 = 11 g1 g


• Identify what is to the problem.

• Identify how you wantwant the answer to look.

5 kg5 kg = mg= mg

Example:Example: Metric Conversion Metric Conversion How many milligrams are in 5 kilograms?


• Multiply by conversion factors until units cancel.

• If the words work, the numbers will work.

5 kg5 kg

11

= mg= mg1000 g1000 g

1 kg1 kg

1000 mg1000 mg

1 g1 g

5,000,0005,000,000

Example:Example: Metric Conversion Metric Conversion How many milligrams are in 5 kilograms?


Example: English-Metric ConversionExample: English-Metric Conversion

You have a pen of rats each with an average weight of 0.75 lb0.75 lb. How much rubbing alcohol will it take to kill ½ of the population if theLD50 is 5000. mg/kg ?

• Identify your conversions factors.Identify your conversions factors.

1 kg Bw1 kg Bw = 1 = 15000 mg Alc5000 mg Alc

5000 mg Alc5000 mg Alc = 1 = 11 kg Bw1 kg Bw

1.0 kg Bw1.0 kg Bw = 1 = 12.2 lb Bw2.2 lb Bw

2.2 lb Bw2.2 lb Bw = 1 = 11.0 kg Bw1.0 kg Bw


0.75 lbBW0.75 lbBW

Example: English-Metric ConversionExample: English-Metric Conversion

You have a pen of rats each with an average weight of 0.75 lb0.75 lb. How much rubbing alcohol will it take to kill ½ of the population if theLD50 is 5000. mg/kg ?

1.0 kgBW

2.2 lbBW

5000. mgAlc

1 kg BW= mgAlc 1704.5451704.545

1700 mg1700 mg = 1.7 x 10= 1.7 x 1033

Identify what is uniqueunique to the problem.

Identify how you wantwant the answer to look.


Given (unique) = 1.6 days

Needed unit = ? min

Plan = days hours min

Example:Example:How many minutes are 1.6 days?

Set up problem to cancel hours (h). 1.6 days x 24 hrs x 60 min = 1 day 1 hr

2 SigFigs Exact Exact = 2 SigFigs

2300 min = 2.3 x 103


Be sure to check your unit cancellation in the setup.

The units in the conversion factors must cancel to give the correct unit for the answer.

Example: What is wrong with the following setup?1.4 day x 1 day x 1 h

24 h 60 min

Units = day2/min, which is not the unit needed

Units don’t cancel properly. Therefore, setup iswrong.

Check the Unit CancellationCheck the Unit Cancellation


==

100 mL100 mL

10 % Alcohol10 % Alcohol

PercentagesPercentages

Part Part x 100 = x 100 =WholeWhole

%%

______100100

Secret code for

10 mL Alcohol10 mL Alcohol

SolutionSolution


Percentages as Conversion FactorsPercentages as Conversion Factors

Example:Example: An athlete normally has 15 %15 % body fat. How many lbs of fat does a 7474 kg athlete have?

Secret code for

15% Body Fat =15% Body Fat = 15 lb Fat15 lb Fat 100 lb BW100 lb BW

100 lb BW100 lb BW 15 lb Fat15 lb Fat


2.2 lbBw2.2 lbBw

1.0 KbBw1.0 KbBw= = lb fatlb fat

Percentages as Conversion FactorsPercentages as Conversion Factors

• Identify what is to the problem.

• Identify how you wantwant the answer to look.

7474 KgBw KgBw

24.4224.42

24 lb fat24 lb fat

Example:Example: An athlete normally has 15 %15 % body fat. How many lbs of fat does a 7474 kg athlete have?

15 lb Fat15 lb Fat 100 lb BW100 lb BW


Learning Check:Learning Check:If the thickness of the skin fold at the waist indicates an 11% body fat, how much fat is in a person with a mass of 86 kg?


Solution:Solution:

11% body fat means 11 kg fat 100 kg

86 kg x 11 kg fat = 9.5 kg of fat

100 kg

If the thickness of the skin fold at the waist indicates an 11% body fat, how much fat is in a person with a mass of 86 kg?


1.8Density



DensityDensity

WaterWater 1.0 1.0 UrineUrine 1.01 - 1.031.01 - 1.03

AirAir 0.0013 0.0013 BoneBone 1.7 - 2.01.7 - 2.0

GoldGold 19.319.3 GasolineGasoline 0.66 - 0.690.66 - 0.69

WaterWater 1.0 1.0 UrineUrine 1.01 - 1.031.01 - 1.03

AirAir 0.0013 0.0013 BoneBone 1.7 - 2.01.7 - 2.0

GoldGold 19.319.3 GasolineGasoline 0.66 - 0.690.66 - 0.69

Density =Density = MassMass

VolumeVolume

1cccc = 1 cmcm33 = 1 mlml = 1 gg water 1cccc = 1 cmcm33 = 1 mlml = 1 gg water

g g

cmcm33

g g

mlmlor AtAt 4 4 oo C C


Example.Example.Density calculationDensity calculation

What is the density of 5.00 ml5.00 ml of serum if it has a mass of 5.230 grams5.230 grams?

d =d = mm V V

d =d = 5.230 g5.230 g 5.00 ml5.00 ml

= 1.05 = 1.05 g g mlml


Guide to Calculating DensityGuide to Calculating Density

d =d = mm V V


Osmium is a very dense metal. What is its density in g/cm3 if 50.0 g of osmium has a volume of 2.22 cm3?

1) 2.25 g/cm3

2) 22.5 g/cm3

3) 111 g/cm3



Given: mass = 50.0 g ,volume = 22.2 cm3

Need: Density

Plan: Place the mass and volume of the osmium metal in the density expression.

D = mass = 50.0 g volume 2.22 cm3

Calculator = 22.522522 g/cm3

Final answer (2 SF) = 22.5 g/cm3

SolutionSolution


Volume by DisplacementVolume by DisplacementA solid completely submerged in water displaces its own volume of water.The volume of the solid is calculated from the volume difference.

45.0 mL – 35.5 mL

= 9.5 mL = 9.5 cm3


Density Using Volume Density Using Volume DisplacementDisplacement

The density of the zinc object is calculated from its mass and volume.

mass = 68.60 g = 7.2 g/cm3

volume 9.5 cm3


What is the density (g/cm3) of 48.0 g of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added?

1) 0.17 g/cm3 2) 6.0 g/cm3 3) 380 g/cm3

33.0 mL

25.0 mL

object



Given: 48.0 g Volume of water = 25.0 mL

Volume of water + metal = 33.0 mL

Need: Density (g/cm3 )

Plan: Calculate the volume difference. Change to cm3,

and place in density expression.

33.0 mL – 25.0 mL = 8.0 mL

8.0 mL x 1 cm3 = 8.0 cm3

1 mLSet up problem:

Density = 48.0 g = 6.0 g = 6.0 g/cm3

8.0 cm3 1 cm3 (2 SF)

SolutionSolution


Sink or FloatSink or FloatIce floats in water because the density of ice is less than the density of water. Aluminum sinks in water because its density is greater than the density of water.


Which diagram correctly represents the liquid layers in the cylinder? Karo (K) syrup (1.4 g/mL), vegetable (V) oil (0.91 g/mL), water (W) (1.0 g/mL)

1 2 3

K

K

W

W

W

V

V

V

K



1)

vegetable oil (0.91 g/mL)

water (1.0 g/mL)

Karo syrup (1.4 g/mL)K

W

V

SolutionSolution


Density as a ConversionDensity as a ConversionA liquid sample with a density of 1.09 g/mL is

found to weigh 7.453 grams. What is the volume of the liquid in mLs?

A liquid sample with a density of 1.09 g/mL is found to weigh 7.453 grams. What is the volume of the liquid in mLs?

1.09 g1.09 g1 ml1 ml

1 ml1 ml1.09 g1.09 g

• Identify any conversion factors.

•How should the answer look?

7.453 g7.453 g = ml= ml

• What is unique to the problem?

1 ml1 ml1.09 g1.09 g

6.8376146.837614 = = 6.846.84 ml ml


If olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil?

1) 0.26 L 2) 0.31 L 3) 310 L



Given: D = 0.92 g/mL mass = 285 g

Need: volume in L

Plan: g mL L

Equalities: 1 mL = 0.92 g 1 L = 1000 mLSet up:

285 g x 1 mL x 1 L = 0.31 L 0.92 g 1000 mL density metric

factor factor

SolutionSolutionIf olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil?

1) 0.26 L 2) 0.31 L 3) 310 L


Specific gravitySpecific gravity

Specific Gravity =Specific Gravity = density of substance g

mldensity of reference g

ml

ReferenceReference

commonly commonly water at water at

44ooCC

•Specific Gravity is Specific Gravity is unitlessunitless..

•density = specific gravity density = specific gravity (if (if at 44ooC)C)


Specific gravitySpecific gravity

•Commonly used to test sugar in urine.

Hydrometer

•Float height will be based on Specific

Gravity.

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