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The Study of Matter and Its Properties in Chemistry - Study Guide | CHEM 1035, Study notes of Chemistry

Chapter One Notes Material Type: Notes; Professor: Spencer; Class: General Chemistry; Subject: Chemistry; University: Virginia Polytechnic Institute And State University; Term: Summer II 2016;

Typology: Study notes

2015/2016

Uploaded on 10/21/2016

skijack1492
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Download The Study of Matter and Its Properties in Chemistry - Study Guide | CHEM 1035 and more Study notes Chemistry in PDF only on Docsity! Chem. 1035 Chapter 1 1 of 21 Jack Dale 7/7/2016 Chapter 1 Keys to the Study of Chemistry Chemistry:  The study of matter and its properties Matter:  ANYTHING that has mass and volume MEASUREMENTS Qualitative: subjective (e.g. how red is the backpack); have no number associated with these measurements Quantitative: measurements that scientists report (have numbers associated with them) There are 3 parts to every measurement: Chem. 1035 Chapter 1 2 of 21 1. Number 2. Units 3. Uncertainty (no perfect measuring devices) Chem. 1035 Chapter 1 5 of 21 **You must know the conversion factors for the metric system but you do not have to memorize English-metric conversion factors.** Chem. 1035 Chapter 1 6 of 21 Unit Conversions Example Problem: I have 62.5 g of liquid N2. What is its mass in pounds? (62.5 g) / 1000 = 0.0625 kg * 2.205 = 0.138 lb Example Problem: What volume does 62.5 g of liquid N2 occupy? We need a conversion factor between mass and volume: Density: "Bridge between mass and volume" The density of liquid N2 is 0.808 g/mL. 62.5 g / V = 0.808 g/mL -> 62.5 / 0.808 = V -> 77.4 mL d = mass volume Chem. 1035 Chapter 1 7 of 21 Example Problem Convert the density of liquid N2 to lb/ft3 (1 mL = 1 cm3) d (metric) = 0.808 g/mL (Let’s say that there is 62.5 g (77.4 mL) of N2 -> will help in metric->English conversions) 62.5 g -> 0.138 lb 77.4 mL -> 77.4 cm3 -> 0.774 m3 * 35.31 = 27.3 ft3 0.138 lb / 27.3 ft3 = 0.00505 lb/ft3 Chem. 1035 Chapter 1 10 of 21 Intensive Properties Extensive Properties  Independent on the amount of substance present EXAMPLES  Density  Temperature  Dependent on the amount of substance present  Mass, Volume  Heat Chem. 1035 Chapter 1 11 of 21 Temperature Scales Fahrenheit ( o F): Commonly used in the U.S.; water freezes at 32oF and boils at 212oF. Celsius ( o C): Scale used in science and around the world; water freezes at 0oC and boils at 100oC. Kelvin (K): The “Absolute temperature scale”; begins at absolute zero and has only positive values; water freezes at 273 K and boils at 373 K. o oF = 1.8 C + 32 o o ( F - 32)C = 1.8 oK = C + 273 Chem. 1035 Chapter 1 12 of 21 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Celsius, °C Kelvin, K Fahrenheit, °F Boiling point of water 100 > Celsius degrees 100 > kelvins 180 > Fahrenheit degrees Freezing point of water [i Chem. 1035 Chapter 1 15 of 21 Rules For Significant Figures 1) Any non zero digit is significant. 2) Zeros between non-zero digits (captive zeros) are significant 3) Placeholding zeros on the left of the first non-zero digit are not significant; these are only used to locate the decimal point. 4) Trailing zeros following a decimal point are significant. 5) Trailing zeros in a number without a decimal point are presumed to be placeholders and are not significant. Chem. 1035 Chapter 1 16 of 21 Example Problem: Number Number Sig. Figs. Number Number Sig. Figs. 2364 4 2090 3 409 3 0.04050 4 0.0579 3 3.040 x 104 4 Exact Numbers: Certain defined quantities have an infinite number of significant figures. These include most conversion factors: 100 cm = 1 m 1 in = 2.54 cm Calculations and uncertainty: The uncertainty of the data determines the uncertainty of the results of calculations based on the data. Chem. 1035 Chapter 1 17 of 21 Rules for Significant Figures in Calculations Addition and Subtraction: The answer has the same number of decimal places as there are in the measurements with the fewest decimal places. 89.332 + 1.1 90.4 Multiplication and Division: The answer contains the same number of significant figures as in the measurement with the fewest significant figures. 134 x 25 = 3400 6.85 / 112.04 = 0.0611 Chem. 1035 Chapter 1 20 of 21 Systematic error: Either ALL higher or lower than the actual value. Part of the experimental system, often caused by faulty device or consistent error in reading. Chem. 1035 Chapter 1 21 of 21 Rules for Rounding Off Numbers 1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if three significant figures are retained and to 5.4 if two significant figures are retained. 2. If the digit removed is less than 5, the preceding number is unchanged. 0.2413 rounds to 0.241 if three significant figures are retained and to 0.24 if two significant figures are retained. 3. If the digit removed is 5, the preceding number increases by 1 if it is odd and remains unchanged if it is even. 17.75 rounds to 17.8, but 17.65 rounds to 17.6. If the 5 is followed only by zeros, rule 3 is followed; if the 5 is followed by nonzeros, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7. 4. Be sure to carry two or more additional significant figures through a multistep calculation and round off only the final answer only.
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