1. Material World

2. Particle Theory Matter is anything that has volume and mass. The particle model is a scientific model based on the idea that matter is made up of small particles. Matter is also organized into 3 phases  Solid  Liquid  Gas

3. Particle Theory

4. Atoms and Molecules An atom is the smallest particle of matter. It cannot be divided by chemical means. A collection of atoms with the same number of protons and electrons are elements A molecule is a group of two or more atoms held together by chemical bonds.

5. Atom

6. Molecule

9. Mixtures and Pure Substances A mixture consists of at least 2 different substances  This means that it has two different types of particles  They can either be homogeneous or heterogeneous. A pure substance consists of only one substance, it contains only one type of particles.  A pure substance can either be compounds or elements

10. Mixtures A heterogeneous mixture is made up of at least two substances that can be distinguished with the naked eye. A homogeneous mixture is made up of at least two substances that cannot be distinguished with the naked eye. Colloids are a specific type of homogeneous solution which can be distinguished with a microscope  Milk is an example of this where the fat globules can be seen with a microscope.

11. Heterogeneous

12. Homogeneous

13. Colloids

14. Solutions Solutions are homogeneous solution that cannot be distinguished under a microscope. There are two components Solute is dissolves in another substance. Solvent is what is dissolving a solute.

15. Properties of Pure Substances Pure substances have characteristic properties. A characteristic property is one that helps identify a pure substance or the group which the pure substance belongs. Characteristic properties can be divided into two categories  Physical  Chemical

16. Physical Properties Physical properties are characteristics which help us identify a pure substance without changing the nature of the substance in the process. There are 4 which will be discussed in class:  Melting point  Boiling point  Density  Solubility

17. Melting and Boiling Point The melting point of a substance is the temperature at which it passes from a solid to a liquid. The boiling point is the temperature where the substance passes from a liquid to a gas. These are expressed in °C Different substances have different melting/boiling points.

18. Density Density represents the mass per unit of volume. Density is calculated using the following formula ρ = m / V  ρ represents density  m represents mass  V represents volume Density is typically represented in g/mL

19. Density Example You have 25 ml of an solution It has a mass of 32 g Density of the substance is 32 g/25 mL ρ = 1.28 g/mL

20. Solubility The solubility of a substance is the maximum amount of solute that can be dissolved in a given volume of solvent. This can be expressed in g/L or in %  % m/V, % m/m, % V/V Solubility is greatly affected by the temperature of the liquid  Higher temp = Higher solubility

21. Solubility

22. Physical Properties When comparing physical properties, pressure and temperature must be taken into consideration. When comparing two different substances, they must be in the same environment. Pressure has an effect on boiling temperatures At 101.3 kPa water boils at 100°C, but at 1600m, the boiling point goes down to 94°C.

23. Chemical Properties Chemical properties help us identify a substance, but in the course of identifying a substance, the nature of the substance is changed. These are typically used as indicators to detect the presence of a particular substance. There are 6 examples which we will cover.

24. Litmus Paper The colour change of a litmus paper determines the presence of an acid or a base. Blue -> Red and Red -> Red indicate an acid Red -> Blue and Blue -> Blue indicates a base Red -> Red and Blue -> Blue indicates a neutral substance

25. Litmus Paper

26. Cobalt Chloride Cobalt chloride paper indicates the presence of water. If the paper turns pink or a different shade of blue, that indicates the presence of water.

27. Cobalt Chloride

28. Limewater Limewater indicates the presence of carbon dioxide. When exposed to carbon dioxide, it goes from clear to cloudy.

29. Limewater

30. Wood Splints When a glowing wood splint is brought to a gas and it relights, that indicates a combustible gas, such as oxygen. When a burning splint causes a reaction, that indicates the presence of an explosive gas, such as hydrogen

31. Open Flame The colour of the flame will vary depending on the presence of difference substances. A purple flame indicates potassium Green -> Barium Red -> Strontium

32. Open Flame

33. Properties of Solutions A given solution has a concentration. This can be given in one of 4 forms  g/L  A number of grams of solute per liter of solution  % m/V  A number of grams of solute per 100 mL of solution  % V/V  A number of mL of solute per 100 mL of solution  % m/m  A number of grams of solute per 100 g of solution

34. Calculating Concentration Concentration in g/L can be calculated with the following equation C = m / V where  C = Concentration  m = mass of the solute in g  V = Volume of the solution in L

35. Dilution Dilution is a laboratory technique that involves decreasing the concentration of a solution by adding solvent. If you are given a solution with a concentration of 4 g/mL and you quadruple the volume, you will end up with a solution of 1 g/mL. It is important to remember that the amount of solute does not change in a dilution, only the solvent.

36. Dilution The equation for dilution is as follows C1 · V1 = C2 · V1, where:  C1 is the concentration of the original solution  V1 is the volume of the original solution  C2 is the concentration of the new solution  V2 is the volume of the new solution

37. Dilution You have 125 mL of a solution of NaCl with a concentration of 14 g/L You add 375 mL of water to the solution. What is the new concentration? Identify which variables you know and start calculating.

38. Dilution C1 = 14 g/mL V1 = 125 mL C2 = x V2 = 125 mL + 375 mL = 500 mL 14 g/mL · 125 mL = x · 500 mL x = 14 g/mL · 125 mL / 500 mL = 3.5 g/mL