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BC CHM 1032 - Chemistry Chapter 1 Notes

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Chapter OneChemistry Basic: Matter and MeasurementMatter Pure Substance Mixtures Elements Compound Homogenous Heterogenous- Mixtures: Combination of 2 or more substanceso Can be separated in different substanceso Homogenous: Composition is uniformed (the same throughout) Ex: Brass (Copper and Zinc *difference is not visible)o Heterogenous: Composition is not uniformed (varies throughout, can see the difference) Ex: Water and Pennies- Pure Substances: Made up of only one type of substance - Element: Simplest type of matter because it is only made of 1 type of Atom - Atom: Smallest unit of matter o Ex: Ca, K+- Compound: 2 or more elements combined, chemically joinedo Ex: H2O- Periodic Tableo Period: Rows (Across Side)o Groups: Columns (Across Top)- Each group has a number and letter designation o A – Representative Elements (Alkali, Alkaline, No common name, Halogens, Noble Gases)o B – Transition Elements- Each group are assigned with same properties- A chemical formula identifies both the type and the number of particles for each element in a compound- SI & Metrico Kilo – 103 (k) *1000o Basic Unit – 1 / 100 (g, L, m) *1o Deci – 10-1 (d) 10o Centi – 10-2 (c) 100o Milli – 10-3 (m) 1000o Micro – 10-6 (M) / (mcg) 1,000,000- Equalitieso 1 km = 103 mo 1 m = 10-3 kmo 1g = 103 mgo 1 L = 101 dL- Equalities can be used as conversion factors to convert one unit to another using one or more of these factorso Conversion factors allow you to convert a quantity in the unit to the equivalent quantity in a larger or smaller unit{Write Equality & 2 conversion factors}A. Liters & Milliliters1L = 103 mL (Equality). 1L / 1000 mL or 1000 mL / 1L (Conversion Factors)- The use of converting units to an equivalent unit is also called Dimensional Analysiso Given Units * (desired unit / given unit)o Ex:  Step 1: Stated given and needed quantities {Given: 2.44 m Need: cm} Step 2: Write a plan to convert Step 3: State equality & conversion factors {1m = 102 cm ~~ 1m =100 cm} {1m / 100 cm or 100 cm / 1m} Step 4: Set up the problem to cancel units & calculate the answerThe standard unit for mass is the kilogramThe standard for volume is the literThe standard unit for length is the meterExponents going down is (+), going up is (-)10 dL / 1L or 1L / 10 dL{2.44 m * 100 cm / 1m = 244 cm}- In any measurement, the significant figures are the digits know with certainty plus one estimated digit.- - General form for scientific notation is C * 10no C = Coefficient (a number between 1 & 9)o n = exponent (telling the # of tens places that apply) Positive tells that the actual # is >1 Negative tells that the # is between 0 & 1o In SN, only SF (sig figs) are shown in the coefficient Ex: Identify sig fig & scientific notationA. 0.002650 m = 2.650 * 10-3 m4 sf snB. 43.026 g = 4.3026 * 101 g5 sf snC. 1044000 L = 1.044 * 106 L 4 sf sn- Multiplication & Division with measured numberso The final answer is written to have the same # sig fig as the measurement with thefewest sig figsEx: 5.0 0 cm * 3.4 0 8 cm / 2.0 cm = 8.52 cm ~~ 8.5 cm Least sig fig is 28.0 0 m / 2.0 0 m = 4. m ~~ 4.00 m Least sig figs are 3, so add zeros- Addition & Subtraction with SFAll nonzero #s are considered significant4 1 g = 2 sig figs 6.0 7 1 kg = 4 sig figs 0.03 L = 1 sig fig0.002 4 g = 2 sig figs 1 2,000 km = 2 sig figsAnswers should be given to the least number of sig digitsin the measured numberso The final answer is written o that it has the same number of decimal places as the measurement having the fewest decimal places.Ex:2.0 1 2 + 61.0 9 + 3.0 = 66.102 ~~ 66.1 Least decimal place is 1- Percent (%) = Part / whole * 100o Can convert decimal to % by multiplying by 100- Mass: A measure of the amount of material in an objecto Common unit used to measure is gram (g) Weight is determined by the pull of gravity on the object- Volume: 3- dimensional measure if the space occupied by mattero Unit typically used is milliliter (mL) {In lab}o Unit typically used is cubic centimeters (cc or cm3) {Clinic}o 1 mL = 1 cc / cm3 ; 5 mL = 1 teaspoons- Density (d): comparison of a substance’s mass to its volumeEx:1 g of water has mass of 1 mL, so d of water is 1.00 g / mLEx: What is the density (g/mL) of a 48.0 – g sample of a metal if the volume of the metal isdetermined to be 8. mL?m = 48.0 g v = 8.0 mL d =? d = m / v ~~ d = 4 8.0 g / 8.0 mL = 6.0 g/mLAnswers should be given to the least number of decimal places in the measured numbersd = m / vSolving for mass: m = dv (d*v)Solve for volume: v = m / d3 sf 2sf 2sf- Specific Gravityo The ratio of the density of a sample to the density of waterSpGr = density of sample / density of water- - The Kelvin is the SI unit of Temperatureo Tk = Tc + 273o TF = 1.8 (Tc) + 32o Tc = TF – 32 / 1.8o Tk = (TF + 459.67) * 5/9 or convert to ℉ ℃- - Solving a Temp. Problemo Step 1: State the given & needed quantitiesGiven: 21 ℃ Needed: T in ℉o Step 2: Write a temp equationTF = 1.8 (Tc) + 32o Step 3: Substitute in known values & calculate new temp.TF = 1.8 (21) + 32 => TF = 38 +32 => 70 ℉- Energy Unitless because it’s a ratioKelvinNo degree symbol in frontNo negative temp.Same size units as ℃^Normal body temp. is 98.6 ℉ or 37.0 ℃ ^Body temp. over 40 ℃ (104 ℉) is hyperthermiaCan cause convulsions, coma, or permanent brain damage^Body temp. below 35 ℃ (95 ℉) is hypothermiaPerson feels cold, has an irregular heartbeat, & has a slow breathing rate32 ℉ 0 ℃ 273 Ko Ability to do worko Stored energy is potential energy (not moving)o Energy of motion is kinetic energyo SI unit for energy is Joule (j)o A calorie is the amount of energy that increases the temp. of 1g of water by 1 ℃  1 cal = 4.184 j 1 Cal (in Nutrition) = 1000 cal- Heat is kinetic energy flowing from a warmer body to a colder one- A specific heat capacity is the amount of heat needed to raise the temp. of 1g of a substance by 1 ℃ Metals have low specific heat


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