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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?.

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## Water, Water Everywhere

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**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**Making Observations/collecting Data Formulating Hypotheses Testing the Hypotheses Revising the Hypothesis Summarizing Hypotheses into a Theory Summarizing observations or natural behavior into a Scientific Law**Steps in the scientific Method?**• Identify the Problems and ask Questions • Develop a Hypothesis based on observations • Test The Hypothesis (Design & Perform Experiments) • Collect more Data • Analyze Results • Make a Conclusion • Suggest further studies on the subject.**Definitions of Terms in Scientific Methods**• Hypothesis:- a plausible or logical statement that attempts to explain the observation or data. • Theory :- a set of (tested) hypotheses that explain a certain behavior of nature. • 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 UnitWith 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**• Mass: milligram (mg), gram (g), kilogram (kg), • Length: cm, m, km, mm, mm, nm, • Area: cm2, m2, km2 • Volume: mL(cm3), dL, L,, m3.**SI Units**• Mass = kilogram (kg) • Length = meter (m) • Area = square meter (m2) • Volume = cubic meter (m3) • Temperature = Kelvin (K) • Energy = Joule (J) • Charge = Coulomb (C) • Time = second (s)**Prefixes in the Metric System**• Prefix Symbol 10n Decimal Forms Giga G 109 1,000,000,000 Mega M 106 1,000,000 kilo k 103 1,000 deci d 10-1 0.1 centi c 10-2 0.01 milli m 10-3 0.001 micro m 10-6 0.000,001 nano n 10-9 0.000,000,001 —————————————————————**Accuracy and Precisionin Measurements**• Accuracy The agreement of an experimental value with the “true” or accepted value; • Precision The reproducibility of measurements of the same type;**Errors in Measurements**• Random errors • values have equal chances of being high or low; • may be minimize by taking the average of several measurements of the same kind; • Systematic errors • Errors due to faulty instruments; • Reading is either higher or lower than the correct value by a fixed amount; • Weighing by differences can eliminate systematic errors of the faulty instruments.**Significant Figures**• All non-zero digits Example: 453.6 has 4 significant figures. • Captive zeros Example: 1.079 has 4 significant figures. • Trailing zeros if the decimal point is shown Example: 1080 has 3 significant figures, but 1080. or 1.080 x 103 has 4 significant figures. • Leading zeros are not significant figures Example: 0.02050 has 4 significant figures**How many significant figures?**• 0.00239 • 0.01950 • 1.00 x 10-3 • 100.40 • 168,000 • 0.082060 • 144 eggs in a carton • Express one thousand as a value with two significant figures.**Rounding off Calculated values**• In Multiplications and Divisions Round 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 Subtractions Round 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 = average Example: • 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: • the middle value (for odd number samples) or • average of two middle values (for even number) • 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**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 ------------------------------------------**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.**Rejection Quotient**• Rejection quotient, Qtab, at 90% confidence level • ——————————————————— • Sample size Qtab ___ • 4 0.76 • 5 0.64 • 60.56 • 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.**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.**Calculate the mean using acceptable values**• Re-write the mean value to be consistent with the precision: • Mean = 0.526 ± 0.005**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 S = 1.16 x 10-4**Mean value must be consistent with the precision**Standard deviation: • should have one significant digit only; • It shows where the uncertainty appears in mean value; • That is, which digit on the mean contains error; • 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: • Celsius (oC) • Fahrenheit (oF) • 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)

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