CHEM 105N 1st Edition Lecture 6 Outline of Last Lecture I. Initial and Final UnitsII. Problem Setup and SolvingIII. Problems Outline of Current Lecture I. Percent Factor in a Problem II. DensityIII. Specific GravityIV. Practice ProblemsCurrent LectureI. Percent Factor in a Problem:- In grams of sugar, 3.1% can also be written as: 3.1g/100g of sugar- The base of a percent is always 100 and takes the unit of the partial percentageII. Density:- Is a property which relates the mass of a substance to its volume- Measured in g/L for gases- Measured in g/cm^3 OR g/ml for solids and liquids- D= Mass/volume- Mass (g)- Volume (mL, cm^3, or cc)Density of Solids through Volume Displacement:-volume is calculated from the volume of water displaced when submergedEx. To calculate the density of a rock with a mass of 4g, place it in a basin of 33mL of water. If the water rises to 45mL, what is the density of the rock?- Mass 4g- Volume 12mL (aka cm^3)- Density Mass/Volume 4/12 1/3.33 g/cm^3- To determine if an object is more dense than water, observe whether it sinks or floats.- Water has a density of 1.00 g/mLThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- If an object floats, its density is less than 1.00g/mL- If an object sinks, its density is more than 1.00g/mLIII. Specific Gravity:- Density of sample v density of water- Density of sample divided by the density of water- Uses a HydrometerIV. Practice Problems:-How many pounds of sugar are in 233 g of candy if the candy is 50% (by mass) sugar?- Candy= 233 grams- Sugar= 50%- Find in Lb-2.20lb 1kg-1,000g 1kg- (233g candy) x (50g sugar/ 100g candy) = [116.5 g sugar]- 116.5 g sugar x (1 kg sugar/ 1,000 g sugar) = [.1165 kg sugar]- .1165 kg sugar x (2.20 lb sugar/1 kg sugar) = .2563 lb sugar-If the thickness of the skin fold at the waist indicates 9% body fat, how much fat is in a person with a mass of 140 kg?- (140 kg) x (9 kg fat/100 kg) = 1260/100 = 12.6 kg
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