A sample of 0.1 g of water at 100

• • metric ruler with mm scale • metric measuring cups • 6 cereal bowls or shallow pans • a small piece of raw potato to cut into six ~5 mm cubes (this square is 5 x 5 mm) • single edged razor or knife • paper towels • watch or clock • table salt, distilled or tap water • 6 beakers (250 ml or larger) or cups Methods: • Pre-mix 6 beakers of salt solutions (0%, 0.1%, 0.5%, 1%, 2.5%, 5%) in distilled water. You can use this • Prepare six small potato cubes with no skin that are all about equal in size (approximately 5 millimeters in length, width and height) and blot them dry on a paper towel. (Blot means just gently remove the surface water; no need to squeeze them!) • Mass (weigh) each to the nearest 0.01 grams, keeping them separate, and record each initial mass in Table 1. Don't wait too long before putting them into the solutions, as evaporation will occur. • Fill each bowl with one of the 6 stock solutions, keeping track of which is which! Label them. You won't be able to tell the salinity just by looking. Note which potato piece went into which bowl. • Leave one of the potato slices in each of the salt solutions for up to 24 hours so that they may gain (or lose) water by osmosis. (Keep them all in the salt water the same amount of time--leaving them overnight is likely to give the best results). • Remove the slices, blot them dry on a paper towel, carefully re-weigh them and record in the data table as final mass. Results: 1. Record your actual results in a table like ...

11. A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27ºC two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27ºC is • 330 m/s • 339 m/s • 300 m/s • 350 m/s 13. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K A, K B and K C, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then • K A K B> K C • K B> K A> K C • K B< K A< K C 17. Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δ l on applying a force F, how much force is needed to stretch the second wire by the same amount? • 9F • 6F • F • 4F

1 Essential Ideas • Introduction • 1.1 Chemistry in Context • 1.2 Phases and Classification of Matter • 1.3 Physical and Chemical Properties • 1.4 Measurements • 1.5 Measurement Uncertainty, Accuracy, and Precision • 1.6 Mathematical Treatment of Measurement Results • Key Terms • Key Equations • Summary • Exercises • 2 Atoms, Molecules, and Ions • Introduction • 2.1 Early Ideas in Atomic Theory • 2.2 Evolution of Atomic Theory • 2.3 Atomic Structure and Symbolism • 2.4 Chemical Formulas • 2.5 The Periodic Table • 2.6 Ionic and Molecular Compounds • 2.7 Chemical Nomenclature • Key Terms • Key Equations • Summary • Exercises • 6 Electronic Structure and Periodic Properties of Elements • Introduction • 6.1 Electromagnetic Energy • 6.2 The Bohr Model • 6.3 Development of Quantum Theory • 6.4 Electronic Structure of Atoms (Electron Configurations) • 6.5 Periodic Variations in Element Properties • Key Terms • Key Equations • Summary • Exercises • 7 Chemical Bonding and Molecular Geometry • Introduction • 7.1 Ionic Bonding • 7.2 Covalent Bonding • 7.3 Lewis Symbols and Structures • 7.4 Formal Charges and Resonance • 7.5 Strengths of Ionic and Covalent Bonds • 7.6 Molecular Structure and Polarity • Key Terms • Key Equations • Summary • Exercises • 9 Gases • Introduction • 9.1 Gas Pressure • 9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law • 9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions • 9.4 Effusion and Diffusion of Gases • 9.5 The Kine...

We use parts per million to express the concentrations of trace amounts, of a given More specifically, a solution's concentration in parts per millions tells you the number of parts of solute present for every #10^6 = 1,000,000# parts of solution. You can thus say that a #"1 ppm"# solution will contain exactly #"1 g"# of solute for every #10^6"g"# of solution. In your case, you know that you have #38 color(red)(cancel(color(black)("mg Pb"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = 3.8 * 10^(-2)color(white)(.)"g Pb"# in exactly #"300.0 g" = 3.000 * 10^2color(white)(.)"g solution"# This means that you can use this known composition as a conversion factor to scale up the mass of the solution to #10^6"g"# #10^6 color(red)(cancel(color(black)("g solution"))) * (3.8 * 10^(-2)color(white)(.)"g Pb")/(3.000 * 10^2color(red)(cancel(color(black)("g solution")))) = "130 g Pb"# Since this represents the mass of lead present in exactly #10^6"g"# of solution, you can say that the solution has a concentration of #color(darkgreen)(ul(color(black)("concentration"_ "ppm" = "130 ppm Pb")))# The answer is rounded to two