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GENERAL CHEMISTRY: CHAPTER 1 08/23/2016 GENERAL CHEMISTRY 1035 INTRO Chemistry: Study of matter, its properties, the changes that matter undergoes, and the energy associated with those changes Matter: Anything that has a massand occupies volume Measurements: 1. Numerical Value 2. Units 3. Estimation of error Priestley: Believed in the phlogiston theory Lavoisier: Accurate measurement, on the part of Lavoisier, lead to the demise of phlogiston theory 𝐻𝑔𝑂(𝑠) → ℎ𝑔(𝑚) +1 2𝑂2(𝑔) He demonstrated that oxygen was an element UNITS Two Systems: English System &Metric System: Typical Units: Length: Meter Mass: The quantity of matter an object containso SI Units: kilogram (kg) Volume: The amount of space any sample of matter occupies o SI Units: cubic meter (𝑚3) Time: “second” UNIT CONVERSIONS When converting from one unit to another, a conversion factor (a ration of equivalent quantities) is chosen and set up so that all units cancel except those required for the answer.
Dimensional Analysis: Conversions where units cancel Example Problem: I have 62.5g of liquid 𝑁2. What is its mass in pounds? ? 𝑙𝑏𝑠 = 62.5𝑔 ∗ 1 𝑘𝑔1000𝑔∗1 𝑙𝑏0.4536𝑘𝑔= 0.138𝑙𝑏 𝑂𝑅 62.5𝑔 ∗1𝑙𝑏453.6𝑔= 0.138𝑙𝑏 Example Problem: An ostrich can run at 45 mi/hr. What is this speed in m/min 𝑠𝑝𝑒𝑒𝑑 =𝑚𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑢𝑡𝑒=45.0𝑚𝑖1 ℎ𝑟∗ 1 ℎ𝑟60 𝑚𝑖𝑛∗1. .609 𝑘𝑚1 𝑚𝑖∗1000𝑚1 𝑘𝑚= 1207 𝑚𝑒𝑡𝑒𝑟𝑠/𝑚𝑖𝑛 Density: Ratio of mass to volume 𝑑 = 𝑚𝑎𝑠𝑠𝑣𝑜𝑙𝑢𝑚𝑒 Density is a conversion factor between mass and volume
Example Problem – Unit Conversion with Density What volume does 62.5g of liquid N2 occupy? The density of liquid N2 is 0.808g/mL 𝑉𝑜𝑙𝑢𝑚𝑒 𝑚𝐿 = 62.5𝑔 ∗1 𝑚𝐿0.808𝑔= 77.4𝑚𝐿 Example Problem – The density of liquid N2 is 0.808g/mL. Convert the density of liquid N2 to lb/ft3 (1 mL = 1cm3; 1 lb = 0.4536 kg; 1 in = 2.54 cm) 0.808 𝑔1 𝑚𝐿∗ 1 𝑘𝑔1000 𝑔∗1 𝑙𝑏0.4536 𝑘𝑔∗1 𝑚𝐿1 𝑐𝑚3∗ (2.54 𝑐𝑚1 𝑖𝑛)3∗ (12 𝑖𝑛 1 𝑓𝑡)3= 50.4 𝑙𝑏𝑓𝑡3⁄ Example Problem:Fueling stop - The dipstick indicated a volume of 7682 L of fuel in the tank. A mass of 22,300 kg of fuel is required for the trip (fuel need is calculate by mass) o Mechanics used a density of 1.77 kg/L to convert between L and kg: o Mass of fuel in tank: 7682𝐿 ∗1.77𝑘𝑔𝐿= 13,600𝑘𝑔 o Mass of fuel needed: 22,300𝑘𝑔 − 13,600𝑘𝑔 = 8,700𝑘𝑔 o But the density is 1.77lb/L NOT kg/L Density in kg/L: 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =𝑘𝑔𝐿= 1.77𝑙𝑏𝐿∗0.4536𝑘𝑔1𝑙𝑏= 0.803𝑘𝑔𝐿o Mass of fuel in tank: 7682𝐿 ∗0.803𝑘𝑔𝐿= 6,170𝑘𝑔 o Mass of fuel needed: 22,300𝑘𝑔 − 6,170𝑘𝑔 = 16,130𝑘𝑔 Example Problem – Complex Unit ConversionGold can be hammered into very thin sheets called gold leaf. A builder needs to cover a 100 ft x 82 ft ceiling with gold leaf that is five millionths of an inch thick. The density of gold is 19.32 g/cm3, and gold costs $1418 per troy ounce (1 troy ounce = 31.1034768 g). How much will it cost for the builder to purchase the necessary gold? (1 in = 2.54 cm) 100𝑓𝑡 ∗ 82𝑓𝑡 ∗ (5 ∗ 10−6)𝑖𝑛 ∗1𝑓𝑡12𝑖𝑛= 0.00342𝑓𝑡3$𝐶𝑜𝑠𝑡 = 0.00342𝑓𝑡3∗ (12𝑖𝑛1𝑓𝑡)3∗ (2.54𝑐𝑚1𝑖𝑛)3∗19.32𝑔𝑐𝑚3∗1 𝑡𝑟𝑜𝑦 𝑜𝑧31.1034768𝑔∗$1418𝑜𝑧= $85,300 PROPERTIES Intensive Properties: Independent of quantity Density, Melting Point, Freezing Point, Boiling PointExtensive Properties: Depend on mass or quantity Mass, Volume, Heat, Length
TEMPERATURE SCALES Fahrenheit (oF): Commonly used in the U.S.; water freezes at 32oF and boils at 212oF. the Fahrenheit scale has a different degree size and different zero points than the other two scales. ℉ = 1. 8 ⃘𝐶 + 32 ⃘Celsius (oC): Scale used in science and around the world; water freezes at 0oC and boils at 100oC. ⃘𝐶 = (℉−32)1.8Kelvin (K): The “Absolute temperature scale”; begins at absolute zero and has only positive values; water freezes at 273 K and boils at 373 K. The Kelvin and Celsius scales use the same size degree but their starting points differ. 𝐾 = ⃘𝐶 + 273.15 Example Problem – Temperature ConversionLiquid nitrogen has a boiling point of –196oC. What is this temperature in oF and K? ℉ =1.8℉°𝐶− (−196°𝐶) + 32°𝐶 = −321℉ POLYMERS Polymers undergo a phase change as a function of temperature called the glass transition Temperatures below the glass-transition temperature of the polymer o-ring is listed as a major factor in the Challenger explosion disaster. UNCERTANITY OF MEASUREMENTS Uncertainty due to: The precision of the measuring device, the accuracy of the observer Every measurement includes some uncertainty. The rightmost digit of any quantity is always estimated The recorded digits, certain and uncertain, are called significant figures. The greater the number of significant digits in a quantity, the greater its certainty. A measurement is only as good as the measuring device SIGNIFICANT FIGURES Any non-zero digit is significant. Zeros between non-zero digits (captive zeros) are significant Place holding zeros on the left of the first non-zero digit are not significant; these are only used to locate the decimal point. Trailing zeros following a decimal point are significant. Trailing zeros in a number without a decimal point are presumed to be placeholders and are not significant. Examples: o 100 - 1 significant figure o 100. - 3 significant figures o 100.0 - 4 significant figures o 0.04050 – 4 significant figures o 2090 – 3 significant figures o 3.040 x 104–4 significant figuresExact Numbers: Defined quantities are precise and the concept of significant figures has no meaning for these. These include most conversion factors. Counted objects are treated similarly. 100 cm = 1m 1in = 2.54cm Exact numbers do not limit the number of significant digits in a calculation
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|32 Multiplication and Division: The answer contains the same number of significant figures as in the measurement with the fewest significant figures 134 ∗25=3350 → 3400 Example Problem: If a steel Ball has a circumference of 32.5mm and weighs 4.20g, what is it’s density? 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 32.5 = 2𝜋𝑟 𝑟 = 5.172536 𝑉 = 4 3𝜋𝑟3= 579.6944 𝑚𝑚3𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =4.20𝑔579.6944 𝑚𝑚3∗ (10𝑚𝑚)31𝑐𝑚3= 7.25𝑐𝑚3 Precision: Range of measurements close in value, affected by systematic error Systematic Error: Usually come from the measuring instruments Accuracy: Close to true value, affected by random error Random Error: Caused by unknown and unpredictable changes in the experiment o Measuring instruments o Environmental conditions Rules for Rounding Off Numbers: 1. If the digit removed is more than 5, the preceding number increases by 1o Ex. 5.379 -> 5.38 -> 5.42. If the digit removed is less than 5, the preceding number is unchangedo Ex. 0.2413 -> 0.241 -> 0.243. If the digit removed is 5, the preceding number increased by 1 if it is odd and remains unchanged if it is eveno Ex. 17.75 -> 17.8 o Ex. 17.65 -> 17.6 If the 5 is followed only by zeros, rule 3 is followed; if the 5 is followed by non-zeros, rule 1 is followedo Ex. 17.6500 -> 17.6 o Ex. 17.6513 -> 17.7
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School: Virginia Polytechnic Institute and State University
Course: General Chemistry
Term: Fall 2016
Tags: Chemistry, review, study, guide, 1035, virginia, and Tech
Name: General Chemistry 1035: Test 1 Study Guide
Description: These notes cover chapters 1-3 in order to prepare for the first General Chemistry test