Water, Water Everywhere

Water, Water Everywhere

Water as Liquid - Rainwater

Water as Solid - Iceberg

Snow and Snow Flakes

Water Vapor (Steam)

Water Cycle

Water Cycle

Water Molecule

Hydrogen Bonding in Water

What Makes Soils & Rocks Have Different Colors

Where does Chemistry fit in?

• Chemistry is about the study of matters and the changes they undergo.

• Chemistry probes the fundamental units of matter in order to understand how and why they are what they are.

• Chemists always ask questions and try to find the answers.

The Central Science

• Chemistry is regarded as the central science.

• Chemistry is essential in understanding the various aspects of living and non-living things;

• It is essential in understanding natural and unnatural processes of nature.

What is Matter?

• The materials of the universe anything that has mass and occupies space

What Type of Change?

• Physical or Chemical processes;

• Physical Change:A process that alters the state of a substance, but not its fundamental composition.

• Chemical Change:A process that alters the fundamental composition of the substance and, therefore, its identity.

The Scientific Approach

1. Making Observations/collecting Data

2. Formulating Hypotheses

3. Testing the Hypotheses

4. Revising the Hypothesis

5. Summarizing Hypotheses into a Theory

6. Summarizing observations or natural behavior into a Scientific Law

Steps in the scientific Method?

1. Identify the Problems and ask Questions

2. Develop a Hypothesis based on observations

3. Test The Hypothesis (Design & Perform Experiments)

4. Collect more Data

5. Analyze Results

6. Make a Conclusion

7. Suggest further studies on the subject.

Definitions of Terms in Scientific Methods

1. Hypothesis:-a plausible or logical statement that attempts to explain the observation or data.

2. Theory :-a set of (tested) hypotheses that explain a certain behavior of nature.

3. Scientific Law :-a concise statement about a natural phenomenon or behavior.

Measurements

The Number System

• Decimal Numbers:384,400

0.08206

• Scientific Notations:3.844 x 105

8.206 x 10-2

Units of Measurements

• Units give meaning to numbers.

Without Unit With Units

384,400 ? 384,400 km (very far)

384,400 cm (not very far)

0.08206 ? 0.08206 L.atm/(K.mol)

144 ? 144 eggs

English Units

Mass: ounce (oz.), pound (lb.), ton;

Length: inches, feet, yards, miles;

Volume: pints, quarts, gallons, in3, ft3, etc.;

Area: acre, hectare, in2, ft2, yd2, mi2.

Metric Units

1. Mass: milligram (mg), gram (g), kilogram (kg),

2. Length: cm, m, km, mm, m, nm,

3. Area: cm2, m2, km2

4. Volume: mL(cm3), dL, L,, m3.

SI Units

1. Mass = kilogram (kg)

2. Length = meter (m)

3. Area = square meter (m2)

4. Volume = cubic meter (m3)

5. Temperature = Kelvin (K)

6. Energy = Joule (J)

7. Charge = Coulomb (C)

8. Time = second (s)

Prefixes in the Metric System

• Prefix Symbol 10n Decimal FormsGiga G 109 1,000,000,000Mega M 106 1,000,000kilo k 103 1,000deci d 10-1 0.1centi c 10-2 0.01milli m 10-3 0.001micro 10-6 0.000,001nano n 10-9 0.000,000,001

—————————————————————

Accuracy and Precisionin Measurements

• AccuracyThe agreement of an experimental value with the “true” or accepted value;

• PrecisionThe reproducibility of measurements of the same type;

Accuracy and Precision

Errors in Measurements

• Random errors1. values have equal chances of being high or low;2. may be minimize by taking the average of several

measurements of the same kind;

• Systematic errors1. Errors due to faulty instruments; 2. Reading is either higher or lower than the correct

value by a fixed amount;3. Weighing by differences can eliminate systematic

errors of the faulty instruments.

Significant Figures

• All non-zero digitsExample: 453.6 has 4 significant figures.

• Captive zerosExample: 1.079 has 4 significant figures.

• Trailing zeros if the decimal point is shownExample: 1080 has 3 significant figures, but 1080. or 1.080 x 103 has 4 significant figures.

• Leading zeros are not significant figuresExample: 0.02050 has 4 significant figures

How many significant figures?

(a) 0.00239(b) 0.01950(c) 1.00 x 10-3

(d) 100.40(e) 168,000(f) 0.082060(g) 144 eggs in a carton(h) Express one thousand as a value with

two significant figures.

Rounding off Calculated values

• In Multiplications and DivisionsRound off the final answer so that it has the same number of significant figures as the one with the least significant figures.

Examples:

(a) 9.546 x 3.12 = 29.8 (round off from 29.78352)

(b) 9.546/2.5 = 3.8 (round off from 3.8184)

(c) (9.546 x 3.12)/2.5 = 12 (round off from 11.913408)

Rounding off Calculated values

• In Additions and SubtractionsRound off the final answer so that it has the same number of digits after the decimal point as the data value with the least number of such digits.

Examples:

(a) 53.6 + 7.265 = 60.9 (round off from 60.865)

(b) 53.6 – 7.265 = 46.3 (round off from 46.335)

(c) 41 + 7.265 – 5.5 = 43 (round off from 42.765)

Mean, Median & Standard Deviation

• Mean = averageExample:

• Consider the following temperature values:

20.4oC, 20.6oC, 20.3oC, 20.5oC, 20.4oC, and 20.2oC;

(Is there any outlying value that we can throw away?)

• No outlying value, the mean temperature is:

(20.4 + 20.6 + 20.3 + 20.5 + 20.4 + 20.2) ÷ 6 = 20.4oC

Mean, Median & Standard Deviation

Median: 1. the middle value (for odd number samples) or

2. average of two middle values (for even number)

3. when values are arranged in ascending or descending order.

For the following temperatures:

20.4oC, 20.6oC, 20.3oC, 20.5oC, 20.4oC, and 20.2oC,

the median = 20.4oC

Mean, Median & Standard Deviation

• Standard Deviation, S =

(for n < 10, Xi = sample value; = mean value)

[Note: calculated value for std. deviation should have one significant figure only.]

For above temperatures, S = 0.1; Mean = 20.4 ± 0.1 oC

X

)1(

) - ( 2

n

XX i

Calculating Mean Value

• Consider the following masses of pennies (in grams):

2.48, 2.50, 2.52, 2.49, 2.50, 3.02, 2.49, and 2.51;

• Is there outlying value?

Yes; 3.02 does not belong in the group – can be discarded

• Outlying values should not be included when calculating the mean, median, or standard deviation.

• Average (mean) mass of pennies is,

(2.48 + 2.50 + 2.52 + 2.49 + 2.50 + 2.49 + 2.51) ÷ 7 = 2.50 g;

Calculating Standard Deviation

_________________________

-0.02 0.0004

-0.00 0.0000

0.02 0.0004

-0.01 0.0001

0.00 0.0000

-0.01 0.0001

0.01 0.0001___

Sum: 0.0011------------------------------------------

) - (

XX i2) - (

XX i

0.01 0.0135

1) - (7

0.0011

)1(

) - (

2

n

XXS i

Mean and Standard Deviation

• The correct mean value that is consistent with the precision is expressed as follows:

2.50 ± 0.01

Using Q-test to retain or reject questionable values

• Calculate Qcalc. as follows:

• Qcalc. =

• Compare Qcalc with Qtab from Table-2 at the chosen confidence level for the matching sample size;

• If Qcalc < Qtab, the questionable value is retained;

• If Qcalc > Qtab, the questionable value is can rejected.

ue)Lowest val - alue(Highest v

|neighbornearest - valueleQuestionab|

Rejection Quotient

• Rejection quotient, Qtab, at 90% confidence level• ———————————————————• Sample size Qtab ___• 4 0.76• 5 0.64• 6 0.56• 7 0.51• 8 0.47• 9 0.44• 10 0.41 ——————————

Performing Q-test on Sample Data

• Consider the following set of data values:

0.5230, 0.5325, 0.5560, 0.5250, 0.5180, and 0.5270;

• Two questionable values are: 0.5180 & 0.5560 (the lowest and highest values in the group)

• Perform Q-test at 90% confidence level on 0.5180:

• Qcalc. = 0.13 < 0.56

• (limit at 90% confidence level for sample size of 6)

• We keep 0.5180.

0.56 0.13 0.0380

0.0050

0.5180) - (0.5560

|0.5230 - 0.5180|

Performing Q-test on questionable value

• Calculate rejection quotient for 0.5560:

• Qcalc. = 0.618 > 0.56 • (limit at 90% confidence level for a sample of 6 is

0.56)• We reject 0.5560.

0.56 0.618 0.0380

0.0235

0.5180) - (0.5560

|0.5325 - 0.5560|

Calculate the mean using acceptable values

• Re-write the mean value to be consistent with the precision:

• Mean = 0.526 ± 0.005

0.52580 5

2.6290

5

0.5305) 0.5180 0.5250 0.5325 (0.5230 ,Mean

X

Calculating Standard Deviation

• 0.5230 -0.0028 7.8 x 10-6

• 0.5325 0.0067 4.5 x 10-5

• 0.5250 -0.0008 6.4 x 10-7

• 0.5180 -0.0078 6.1 x 10-5

• 0.7270 0.0012 1.4 x 10-6

= 1.16 x 10-4

2____

) - ( ) - ( XXXXX iii

0.005 0.00539 4

10x 1.16

1) - (

) - (

4-2__

n

XXS i

Mean value must be consistent with the precision

Standard deviation:1. should have one significant digit only;

2. It shows where the uncertainty appears in mean value;

3. That is, which digit on the mean contains error;

4. The mean value should be rounded off at the digit where it becomes uncertain.

Thus, the mean consistent with the precision will be:

0.526 ± 0.005

(the mean value is precise up to the third decimal place)

Problem Solving by Dimensional Analysis

• Value sought = value given x conversion factor(s)Example:

How many kilometers is 25 miles? (1 mi. = 1.609 km)

Value sought: ? km; value given = 25 miles;

conversion factor: 1 mi. = 1.609 km

? km = 25 mi. x (1.609 km/1 mi.) = 40. km

Unit Conversions

(1) Express 26 miles per gallon (mpg) to kilometers per liter (kmpL). (1 mile = 1.609 km and 1 gallon = 3.7854 L)

(Answer: 11 kmpL)

(2) If the speed of light is 3.00 x 108 m/s, what is the speed in miles per hour (mph)? (1 km = 1000 m and 1 hour = 3600 s)

(Answer: 6.71 x 108)

Temperature

• Temperature scales:1. Celsius (oC)

2. Fahrenheit (oF)

3. Kelvin (K)

Reference temperatures: freezing and boiling point of water:

Tf = 0 oC = 32 oF = 273.15 K

Tb = 100 oC = 212 oF = 373.15 K

Temperature Conversion

• Fahrenheit to Celsius:(T oF – 32 oF) x (5oC/9oF) = T oC

Example: converting 98.6oF to oC;

(98.6 oF – 32 oF) x (5oC/9oF) = 37.0 oC

Temperature Conversion

• Celsius to Fahrenheit:ToC x (9oF/5oC) + 32 oF = T oF

Example: converting 25.0oC to oF;

25.0 oC x (9oF/5oC) + 32 oF = 77.0 oF

Temperature Conversion

• Celsius to Kelvin: T oC + 273.15 = T K

• Kelvin to Celsius: T K – 273.15 = T oC

Examples: 25.0 oC to Kelvin = 25.0 + 273.15 = 298.2 K

310. K to oC = 310. – 273.15 = 27 oC

Temperature Conversion

1) What is the temperature of 65.0 oF expressed in degrees Celsius and in Kelvin?

(Answer: 18.3 oC; 291.5 K)

2) A newly invented thermometer has a T-scale that ranges from -50 T to 300 T. On this thermometer, the freezing point of water is -20 T and its boiling point is 230 T. Find a formula that would enable you to convert a T-scale temperature to degrees Celsius. What is the temperature of 92.5 T in Celsius? (Answer: 45.0 oC)

Density

• Density = Mass/Volume

(Mass = Volume x density; Volume = mass/density)

Units: g/mL or g/cm3 (for liquids or solids)

g/L (for gases)

SI unit: kg/m3

Examples: density of water = 1.00 g/mL (1.00 g/cm3);

in SI unit = 1.00 x 103 kg/m3

Determining Volumes

• Rectangular objects: V = length x width x thickness;

• Cylindrical objects: V = r2l (or r2h);

• Spherical objects: V = (4/3)r3

• Liquid displacement method: the volume of object submerged in a liquid is equal to the volume of liquid displaced by the object.

Density Determination

Example-1:

A cylindrical metal rod that is 1.00 m long and a diameter of 1.50 cm weighs 477.0 grams. What is the density of metal?

Volume = (1.50 cm)2 x 100. cm = 177 cm3

Density = 477.0 g/177 cm3 = 2.70 g/cm3

Density Determination

Example-2:

A 100-mL graduated cylinder is filled with 35.0 mL of water. When a 45.0-g sample of zinc pellets is poured into the graduate, the water level rises to 41.3 mL. Calculate the density of zinc.

Volume of zinc pellets = 41.3 mL – 35.0 mL = 6.3 mL

Density of zinc = 45.0 g/6.3 mL = 7.1 g/mL (7.1 g/cm3)

Density Calculation #1

• The mass of an empty flask is 64.25 g. When filled with water, the combined mass of flask and water is 91.75 g. However, when the flask is filled with an alcohol sample the combined mass is found to be 85.90 g. If we assume that the density of water is 1.00 g/mL, what is the density of the alcohol sample?

(Answer: 0.787 g/mL)

Density Calculation #2

A 50-mL graduated cylinder weighs 41.30 g when empty. When filled with 30.0 mL of water, the combined mass is 71.25 g. A piece of metal is dropped into the water in the graduate, which causes the water level rises to 36.9 mL. The combined mass of graduate, water and metal is 132.65 g. Calculate the densities of water and metal.

(Answer: 0.998 g/mL and 8.9 g/mL, respectively)

Classification of Matter

Matter

HomogeneousMixture

HeterogeneousMixture

PureSubstances

Elements and Compounds

Classification of Matter

• Mixture: matter with variable composition

• Homogeneous mixture:One that has a uniform appearance and composition throughout the mixture;

• Heterogeneous mixture:One that has neither uniform appearance or composition – the appearance and composition in one part of the mixture may differ from the other part;

• Substance: matter with a fixed composition

Substances

• Element:Composed of only one type of atoms – it cannot be further reduced to simpler forms.

• Compound:Composed of at least two different types of atoms combined chemically in a fixed ratio – it may be broken down into simpler forms (or reduced to the elements)

Physical Changes

Examples:

1. melting,

2. freezing,

3. evaporation,

4. condensation,

5. sublimation,

6. dissolution.

Chemical Changes

Examples:

1. combustion (burning),

2. decomposition,

3. chemical combination (synthesis),

4. fermentation,

5. corrosion,

6. oxidation and reduction,

(any chemical reactions)