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Monday September 20, 2010 Chapter 3 Math: The Central Language of Science

Ch 3 final ppt

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The Mathematics of Chemistry

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Page 1: Ch 3 final ppt

Monday September 20, 2010

Chapter 3Math: The Central Language

of Science

Page 2: Ch 3 final ppt

Measurement

Consists of 1) a number2) a unit

Ex: 1 mile10.0 grams450.00 mL

Page 3: Ch 3 final ppt

Accuracy and Percent Error

Percent error: how accurate a measurement is to the accepted value

How to calculate percent error: Accepted value – Experimental Value X 100%

Accepted value

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Sample ProblemYou measure the mass of a product in a

chemical reaction to the 3.80 g. Theoretical calculations predict that you should have obtained 3.92 g. What is the percent error?

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Precision versus Accuracy

Tuesday September 21, 2010

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SI unitsInternationally agreed upon units of measurement

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Metric SystemCommonly Used Prefixes

kilo k 1,000 103

hecto h 100 102

deka da 10 101

no prefix 1 100

deci d 0.1 10-1

centi c 0.01 10-2

milli m 0.001 10-3

micro µ 0.000001 10-6

nano n 0.000000001 10-9

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Significant FiguresMade Easy

By Miss Virginia Williams

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Significant Figures Rules:

1. All nonzero numbers are significant.2. All zeros between two nonzero digits

are significant.3. When a decimal is in the number, the

first nonzero number present and all the numbers after it are significant.

4. When the number is written in scientific notation, all the numbers to the left of the multiply sign are significant.

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Determine the number of significant digits in the following quantities:

a) 9.370 kgb) 63,000 gc) 705.06 mLd) 0.0001 se) 2300.000 m

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Significant Digits and Mathematical Operations

1) After add and sub, the answer cannot be more precise than the least precise measurementEx. 50.23m + 14.678m + 23.7m = 88.608

23.7 only goes to the tenth place, therefore the answer must be rounded to the tenth place, 88.6 m

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Significant Digits and Mathematical Operations

2) In multiplication or division, the answer cannot contain more significant digits than the measurement with the least number of significant digitsEx: (0.238 m)(0.31 m) = 0.07378 m2

The answer must be rounded off to 0.074 m2

Because 0.31 m only has 2 sig figs

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Significant Digits andMathematical Operations

3) The rule for rounding: (same as normal rounding rules)

If the digit to be dropped is less than 5, simply drop the digit.If the digit is or greater, increase the preceding digit by oneEx: 12.4489 expressed to 3 sig figs =

12.4489 expressed to 4 sig figs =12.412.45

Do Sample problems pg 51

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Monday Sept 27, 2010Scientific Notation

Allows us to write very big and very small numbers

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Scientific Notation

3.42 x 106 or

8.005 x 10-3

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Scientific Notation

3.42 x 106 means

3.42 x (10 x 10 x 10 x 10 x 10 x 10)

or 3,420,000

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Scientific Notation

8.005 x 10-3 means

or 0.008005

8.005 x (0.1 x 0.1 x 0.1)

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Scientific Notation

Let's say you wanted to convert the following number to scientific notation:

0.00000782

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Convert to Scientific Notation

1) 3,400

2) 0.000023

3) 101,000

4) 0.010

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Perform the following operations, give answer with correct number of significant digits.

a. (3.55 x 104 g) + (3.55x 103 g) = 3.91 x 104 gb. (6.22x 10-2 m) (8.4x 10-4 m) = 5.2 x 10-5 mc. (3.55x 104 cm) /(3.55x 103 cm)=1.00 x 101cmd. (6.22x 10-2 kg) - (8.4x 10-4 kg)=6.1x10-2 kg

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Dimensional Analysis –converting from one unit to another

ex: feet/sec to miles/hour

Tuesday September 28, 2010

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Dimensional AnalysisConvert 2 feet to inches

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Dimensional Analysis

• The units cancel out leaving only

• Giving us an answer of 24 in.

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Dimensional Analysis

• Convert 48 inches to feet.

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Dimensional Analysis

• Convert 5 days to hours.

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Dimensional Analysis

• How many seconds are there in 4 minutes?

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Dimensional Analysis

• Convert 27.2 cm to meters.

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Dimensional Analysis

• Convert 147 cm/s to m/s

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Dimensional Analysis

• Convert 60 miles/hour to feet/second

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Analysis of an air sample reveals that it contains 3.5 x 10-6 g/L of carbon monoxide. Express the concentration of carbon monoxide in lb/ft3. (Use 1.00 lb = 454 g; 1 in = 2.54 cm)

Page 32: Ch 3 final ppt

All matter has a property called a specific heat capacity. For silver, this specific heat capacity is 0.24 J/°C · g. How much energy (in Joules) would be required to heat 120.0 g of silver (Ag) so that its temperature changes by 32°C? Use dimensional analysis, not an equation.

Page 33: Ch 3 final ppt

Based on how you set up the problem above, what would be the equation? Fill in the rest of this expression to form your own equation (that you figured out by using dimensional analysis above.) You will use the terms “mass” “specific heat” and “temperature” and some mathematical operation signs). This answer is an equation, not a dimensional analysis setup. Energy (J) =