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Regents Chemistry

Regents Chemistry. Physical Behavior of Matter. Different Phases of Matter. An element, compound or mixture can exist in the form of a solid, liquid or a gas Solid – rigid form, definite volume and shape, strong attractive forces and crystalline structure

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Regents Chemistry

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  1. Regents Chemistry Physical Behavior of Matter

  2. Different Phases of Matter • An element, compound or mixture can exist in the form of a solid, liquid or a gas • Solid – rigid form, definite volume and shape, strong attractive forces and crystalline structure • Liquid – not held together as well, can move past one another, no definite shape but definite volume • Gas – minimal attractive forces, no definite shape or volume, expand to shape of container

  3. Other Phases • Vapor – is the gaseous phase of a substance that is a liquid or a solid at normal conditions: ex: water vapor • Plasma – is a gas or vapor in which some or all of the electrons have been removed from the atoms. ex: In a planet’s core!

  4. Heating and Cooling Curves • Heating Curves: Constant rate of heating of a substance over time – endothermic process!

  5. What Can We Learn From a Heating Curve? • AB: heating of a solid, one phase present, kinetic energy increases • BC: melting of a solid (melting), two phases present, potential energy increases, kinetic energy remains constant • CD: heating of a liquid, one phase present, kinetic energy increases

  6. What Can We Learn From a Heating Curve? • DE: boiling of a liquid (Vaporization), two phases present, potential energy increases, kinetic energy remains constant • EF: heating of a gas, one phase present, kinetic energy increases ***We can tell when the kinetic energy remains constant because the temperature is not increasing!***

  7. Cooling Curves • Shows the constant rate of cooling of a gas at high temperature – an exothermic process

  8. Summary of a Cooling Curve • AB:cooling of a gas (vapor), one phase present, kinetic energy decreases • BC: condensation of the gas (vapor) to liquid, two phases present, potential energy decreases, kinetic energy remains constant • CD: cooling of a liquid, one phase present, kinetic energy decreases

  9. Summary of a Cooling Curve • DE: solidification (freezing) of a liquid, two phases present, potential energy decreases, kinetic energy remains the same • EF: cooling of a solid, one phase present, kinetic energy decreases

  10. Substances That Do Not Follow the Curves • Some substances change directly from a solid to a gas – Sublimation • Example: CO2 changes from a solid to a gas a normal atmospheric pressure • Some substances change directly from gas to a solid – Deposition

  11. Practice Problem Which portions of the graph represent times when heat is absorbed and potential energy increases while kinetic energy remains constant? worksheet

  12. Regents Chemistry • Temperature Scales

  13. Temperature Scales • Celsius ° C • Based on boiling point/freezing point of water • Kelvin K • Based on absolute zero • Fahrenheit° F • Used in U.S. and Great Britain

  14. Conversions • Key Equations Celsius to Kelvin K = °C + 273 Fahrenheit to Celsius °C = 5/9 (°F - 32) Kelvin to Celsius °C = K - 273 Celsius to Fahrenheit °F = 9/5(°C) + 32 **Add the conversions on the right to your worksheet

  15. Practice Problems • Convert 10 °C to °F °F = 9/5(°C) + 32 = 9/5 (10 °C) + 32 = 50°F • Convert 25°C to K • K = °C + 273

  16. Regents Chemistry • Measurement of Heat Energy

  17. Energy and Energy Changes • Energy is the capacity to do work. In other words, it allows us to do things! • Energy surrounds us and is involved in all of life’s daily functions. • It comes in many forms!

  18. Energy and Energy Changes • Energy can be used to change the temperature of a substance • As we heat a substance (put in heat), the vibration of molecules in a substance increases. • Example: When a solid is heated, the molecules vibrate until they break free and the substance melts.

  19. Specific Heat Capacity • The specific heat capacity of a substance is the amount of heat required to raise 1 gram of the substance by 1 degree Celsius • For water it is 4.184 J / g• C • Compared to other substances, water has a very high specific heat..what does this mean?

  20. If a substance has a high specific heat capacity… • This simply means that there will be more energy stored in every 1 gram of water. This also means it will take longer to cool down.

  21. Specific Heat Capacities • Check out the specific heat capacities of different substances!

  22. Measurement of Heat Energy • Question: Your pool containing 100,000 grams of water absorbs how much heat energy when it warms from 20 °C to 30 °C? • It is easy! We use a formula on our reference tables! q = mCT

  23. This means what?.. q = mCT • q = amount of heat absorbed or lost • m = mass in grams • C = specific heat • T = difference in temperature

  24. Back to our problem… • Question: Your pool containing 100,000 g of water absorbs how much heat energy when it warms from 20 °C to 30 °C? • q = mCT q = (100,000 g)(4.184 J / g• C) (10 °C) = q = 4,184,000 Joules!

  25. Rearranging the formula.. • You need to be able to solve for any of the variables in the equation q = mCT

  26. Making it easy.. • If we are finding the heat change during the melting or boiling phases, we can use the Heat of Fusion or the Heat of Vaporization.. • Why?? Because temperature remains constant during these periods!

  27. Heat of Fusion and Vaporization • Heat of Fusion – amount of heat energy required to melt a unit mass of a substance • For water : HOF = 334 J/g • Heat of Vaporization – amount of energy required to convert a unit mass from liquid to vapor phase • For Water: HOV = 2260 J/g

  28. Practice Problem • How many joules are required to melt 255 g of ice at 0°C? • q = m x Heat of Fusion q = 255 g x 334 J/g = 85, 170 J

  29. Measuring Heat Change • Calorie = the amount of energy(heat) required to raise the temperature of one gram of water by one Celsius degree. • 1 Calorie (cal) = 4.184 Joules (J) Metric system SI system

  30. Converting Calories to Joules • Convert 60.1 cal of energy into joules 60.1 cal X 4.184 J = 251 J 1 cal = 4.184 J 1 cal

  31. Converting Joules to Calories • Convert 50.3 J to cal 1 cal = 4.184 J 50.3 J X 1 cal = 12.0 cal 4.184 J

  32. Kilojoules and Kilocalories • The prefix kilo means 1000 • energy is often expressed in kilos because the numbers are large • We can use Dimensional Analysis to convert. 4.0 J x 1 kJ = 0.0040 kJ 1000 J

  33. Converting kilojoules to kilocalories 1 cal = 4.184 J 1000 kcal = 4184 kJ 500.0 kJ x 1000 kcal = 2092 kcal 4184 kJ

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