1. Resources & Energy

2. Balancing Equations Fe + Cl 2  FeCl 3  Check to see if both sides of the equation are equal. Fe – 1; Cl – 2  Fe – 1, Cl – 3  If both sides of the equation are not equal, then add coefficients to make both sides equal. *When you balance equations, NEVER change the subscripts! 2Fe + 3Cl 2 = 2FeCl 3

3. Law of Conservation of Matter  Law of Conservation of Matter states that matter cannot be created or destroyed during a chemical reaction. Ex. When a piece of paper burns, the paper under went through change, but its matter stayed the same.

4. Renewable vs. Nonrenewable  A renewable resource is energy that is created from natural sources that can be replenished by natural processes over time. Ex: Water, Oxygen, Sunlight, etc.  A nonrenewable resource is energy that cannot be replaced or reused by natural processes over time. Ex: Coal, Oil, Nuclear power, etc.

5. Physical vs. Chemical Changes  A physical change is when a substance undergoes a change that doesn’t change its identity, such as protons, neutrons or electrons.  A chemical change is when a substance undergoes a change that results in the formation of a new substance.

6. Examples of Physical Changes  Freezing water  Evaporation  Melting  Paper Crumbled  Taking the air out of a balloon  Cutting Wood  Breaking glass

7. Examples of Chemical Changes  Burning wood or paper  Rusty nail  Digesting food  Photosynthesis  Lighting a match  Blowing up C4

8. Metals  A metal is a material with properties such as:  Luster  Solid at room temperature (Except for Mercury)  High Melting Points  Density  Malleable  Ductile  Thermal & Electrical Conductors  Reacts with acid

9. Metalloids  A metalloid is a material that has properties of metals and nonmetals.  Some malleable and some conduct electricity  Manmade metalloids react with acid Ex. Boron, Silicon, Germanium, Arsenic, Antimony, Tellurium & Polonium  Some metalloids are good to use in computer chips

10. Nonmetals  A nonmetal is a material with properties such as:  Brittleness  Poor Conductors of Electricity  Not Ductile  Not Malleable  Does not react with acids

12. Exothermic vs. Endothermic Reactions  Endothermic reactions is a bond breaking step that requires energy.  Exothermic reactions is the formation of a chemical bond in which energy is released.

13. Bonds broken, energy required. ENDOTHERMIC CH 4 + 2O 2 Chemical potential energy High Low H4H4 H4H4 C H4H4 H4H4 O2O2 O2O2 O2O2 O2O2 H4H4 H4H4 C H4H4 H4H4 O2O2 O2O2 O2O2 O2O2 C + 4H + 4O 2 O2O2 O2O2 C H4H4 H4H4 H4H4 H4H4 O2O2 O2O2 CO 2 + 2H 2 O Bonds formed energy released. EXOTHERMIC *The overall reaction is exothermic because more energy was released in step two than made in step one.

14. Skelton of Previous Graph

15. Potential vs. Kinetic Energy  Potential Energy is stored energy.  Kinetic Energy is energy related to motion.

16. As Bob bikes up the hill, he provides energy to the moving parts. (Kinetic Energy) … at the top of the hill, Bob preserves his energy by staying on top of the hill. (Potential Energy)

17. Specific Heat Capacity of Water  Specific Heat Capacity is the quantity of thermal energy needed to raise the temperature of 1 gram of a material by 1 °C.  The specific heat capacity of liquid water is 4.2 J/(g  °C) or 1 Cal/(g  °C). For every gram of water and °C, 4.2 joules or 1 calorie of thermal energy is absorbed.  If a substance has a relatively high specific heat capacity, then that substance requires more energy to heat up and more reduction of energy to cool down.  If a substance has a relatively low specific heat capacity, then that substance may have quick changes in temperature.

18.  The following formula is used to calculate the total amount of thermal energy absorbed by a substance: TE = cm Δ T Thermal Energy Absorbed = Specific Heat Capacity × Change in Temperature × Mass of Substance

19. How much thermal energy was absorbed by a 20 mL sample of water that was also heated from 30 °C to 50 °C? 1. Calculate the mass of water heated. We know that the density of liquid water is 1 g/mL, so each milliliter of water has a mass of 1 g. Answer: 20 g. of water 2. Calculate the change in temperature. Find the difference of the two temperatures by subtracting the final temperature (50 °C) by the original temperature (30 °C). Answer: 20 °C increase in temperature 3. Substitute the values into the equation. TE = 4.2 J × 20 g × 20 °C 4. Solve for the amount of thermal energy absorbed by the water. Multiply. Answer: 1,680 J

20. Heat of Combustion  Heat of Combustion is the quantity of thermal energy given off when a certain amount of a substance burns.  You can calculate the total quantity of thermal energy being released when either one mole or one gram of a substance burns.  Molar Heat of Combustion is the quantity of thermal energy released from burning one mole of a substance.

21. Using your previous data to calculate the specific heat of the 20-g water sample, calculate the total heat of combustion of the paraffin wax. 4. Calculate the total mass of the burned paraffin wax. Subtract the original mass of the paraffin candle (25 g) by the final mass of the paraffin candle (23.7 g). Answer: 1.3 g of Paraffin Wax burned 5. Calculate the heat of combustion of the paraffin wax. Divide the quantity of thermal energy absorbed by the mass of paraffin burned. Answer: 1,292.31 J/g of Paraffin *To find the heat of combustion in kJ/g, divide the heat of combustion by 1000 joules. Answer: 1.29 kJ/g

22. Heat of Combustion Equations  The complete combustion of a hydrocarbon can be expressed by the following equation: Hydrocarbon + Oxygen Gas  Carbon Dioxide + Water + Thermal Energy

23. Solving Combustion Equations  Solve the equation for burning hexane: C 6 H 14 + O 2  CO 2 + H 2 O + ? kJ Thermal Energy HydrocarbonFormulaHeat of Combustion (kJ/g) Molar Heat of Combustion (kJ/mol) MethaneCH 4 55.6890 EthaneC2H6C2H6 52.01560 PropaneC3H8C3H8 50.02200 ButaneC 4 H 10 49.32859 PentaneC 5 H 12 48.83510 HexaneC 6 H 14 48.24141 HeptaneC 7 H 16 48.24817 OctaneC 8 H 18 47.85450

24. 1. Balance the equation. 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O + ? kJ Thermal Energy 3. Balance the thermal energy released. The complete combustion equation for hexane: 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O + 8282 kJ Thermal Energy 2. Substitute hexane’s molar heat of combustion into the equation. (Use the table on the previous slide!) 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O + 4141 kJ Thermal Energy

25. Any Questions?