Step 1: Identify the units that you are given and the final units that you need. Units given:

Transcription

1 Density The Long Version PROBLEM Calculate the volume of gasoline, in liters, that you would add to your car's tank given that the density of gasoline is 0.66 g/ml and the mass of the gasoline is 1.2 kg. SOLUTION Step 1: Identify the units that you are given and the final units that you need. Units given: Density of gasoline in g/ml Units needed: Mass of gasoline in kg Liters of gasoline Step 2: Review the given and needed units. Plan how you will progress from the given to the needed unit. Plan: Start with the mass of gasoline (kg). Convert it to grams. Multiply it by the density of gasoline. Convert ml to L. Step 3: Write the equalities and conversion factors that you need g/ 1kg 0.66 g/ml OR 1 ml/0.66 g 1000 ml/ 1L Step 4: Lastly, set up your problem. You want to determine the number of liters of gasoline in 1.2 kg. x L = 1.2 kg x (1000 g/1 kg) x (1 ml/0.66 g) x (1 L / 1000 ml) = 1.8 L Using the density of gasoline, we find that there are 1.5 L in 1.2 kg. You needed to use this version (1 ml/0.66 g) of the density of gasoline conversion factor so that the units of ml and g will cancel out in the problem.

2 Density The Short Version PROBLEM You added 1.2 kg of gasoline to your tank. The density of the gasoline is 0.66 g/ml. How many liters of gasoline did you add? SOLUTION Step 1: Convert 1.2 kg to g. 1.2 kg x (1000 g/1 kg) = 1200 g Step 2: Use density to determine the volume of gasoline added in liters. Be sure your units cancel! 1200 kg x (1 ml /0.66 g) x (1 L/1000 ml) = 1.8 L

3 Energy: Heat and Temperature Each scale differs in how the units are defined. In the Celsius scale, water boils at 100 degrees and the freezing point is defined as 0 degrees. The Kelvin scale is defined by the lowest temperature possible, or absolute zero, which is equivalent to -273 degrees Celsius Atoms and Elements Atomic number = number of protons in an atom Mass number = number of protons + number of neutrons Unit Conversions between Common and Metric Units Length 1 mile(mi) = 1.61 kilometers (km) 1 yard(yd) = meter (m) Mass 1 inch(in.) = 2.54 centimeters (cm) a 1 pound (lb) = 454 grams (g) Volume 1ounce(oz) = 28.4 grams (g) 1 pound (lb) = kilogram (kg) 1 U.S. quart (qt) = liter (L) 1 U.S. pint (pt) = liter (L) 1 fluid ounce (fl oz) = 29.6 milliliters (ml) 1 gallon (gal) = 3.78 liters (L)

4 Rules 1. The first element is named first, using the elements name. 2. Second element is named as an Anion (suffix "-ide") 3. Prefixes are used to denote the number of atoms 4. "Mono" is not used to name the first element Prefix number indicated mono- 1 di- 2 tri- 3 tetra- 4 penta- 5 hexa- 6 hepta- 7 octa- 8 nona- 9 deca- 10

5 Mass Percent and Mass/Volume Percent Calculations A common mistake that is made is confusing mass percent (% m/m) with mass/volume percent (% m/v). Another common mistake is to forget to multiply by 100 to express the answer as a % value. PROBLEM Calculate the mass percent when given 30 g NaOH and 120 g H2O. The final volume of the solution is 125 ml. SOLUTION Mass percent grams of solute + grams of solvent = = 150 grams solute/solute + solvent = 30/150 = 0.20 multiply by 100 to express as a percentage: 0.20 x 100 = 20% (m/m) NaOH Mass/volume For this type of problem, you will be given the final volume of the solution, which is the total volume of the solute plus solvent. The solute occupies a certain volume. When a solution is made as a ratio of mass to volume, enough solvent is added to the solute to create a solution with the specified final volume. What does this mean in terms of this problem? You need to know the final volume of the solution. The mass of the solvent is not important. grams of solute = 30 grams ml of solution = 125 ml solute/solvent = 30 g/125 ml = 0.24 multiply by 100 to express as a percentage: 0.24 x 100 = 24% (m/v) NaOH

6 Molarity Molarity is another unit of measure. It uses the atomic weights of the atoms in the compound to calculate this value. gms = 1 mol/ mw in grams Molarity is the number of moles of a compound in one liter of solution. M = moles/ liters For example, a 1 M NaCl solution = 1 mol NaCl /1 L. A 0.3 M solution of glucose contains 0.3 mol of glucose in 1 L. Part 1 gms = 1 mol/ mw in grams To solve these types of problems, the first step is to determine the chemical's molar mass (In the examples below, the molar mass is provided.). Molarity is defined as moles of solute / liters of solution. Part 2 M = moles/ liters Make sure that your volume is expressed in liters. If it is not, you need to convert the volume.

7 Dilution of Solution In chemistry, preparation of a diluted solution from a more concentrated solution is a common occurrence. We often dilute solutions in our homes. Examples include adding bleach to a load of laundry and diluting concentrated orange juice. When a solution is diluted, the amount of solute does not change. It is the total volume of the solution that changes. Thus, the relationship between the solution concentrations and volumes can be expressed as follows: C 1V1 = C2V2. In this equation, the concentration can be expressed as a percentage (%) or in molarity (M). Example 1: What volume, in ml, of a 2.5% (m/v) NaOH solution can be prepared by diluting 50.0 ml of a 12% (m/v) NaOH solution? Beginning solution: C1 = 12%, V1 = 50 ml Ending solution: C2 = 2.5%, V2 = X (C1)(V1) = (C2)(V2) ---> (12%)(50mL) = (2.5%)(X) Solve the equation: 600 = 2.5X; thus, X(unknown volume) = 240 ml (diluted NaOH solution) Example 2: What is the percent (% m/v) of the dilute solution when 25.0 ml of 15% HCl solution is diluted to 125 ml? Beginning solution: C1 = 15%, V1 = 25 ml Ending solution: C2 = X, V2 = 125 ml (15%)(25 ml) = x(125 ml); thus, x = 3% Example 3: What is the molarity of a solution prepared with 75 ml of a 4.00 M KCL solution diluted to a volume of 0.5L? In this type of problem, you need to make sure that the units of volume match. As the definition of molarity is in liters, you want all volumes to be in liters. Beginning solution: C1 = 4.00 M, V1 = 75 ml = 0.075L Ending solution C2 = x, V2 = 0.5L (4.00 M)(0.075L) = x(0.5l); thus, x = 0.6 M KCl The following presentation discusses intermolecular forces and the states of matter.

8 Brønsted Acid-Base Theory Brønsted argued that all acid-base reactions involve the transfer of an H + ion, or proton. Water reacts with itself, for example, by transferring an H + ion from one molecule to another to form an H3O + ion and an OH - ion. According to this theory, an acid is a "proton donor" and a base is a "proton acceptor." Every Brønsted acid has a conjugate base, and vice versa.