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“I am like that,” does not help anything. “I can be different,” does.
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“I am like that,” does not help anything. “I can be different,” does.

Dec 17, 2015

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Page 1: “I am like that,” does not help anything. “I can be different,” does.

“I am like that,” does not help anything.

“I can be different,” does.

“I am like that,” does not help anything.

“I can be different,” does.

Page 2: “I am like that,” does not help anything. “I can be different,” does.

Basick EnglishBasick English

• The bandage was wound around the wound.

• The farm was used to produce produce.

• The dump was so full that it had to refuse more refuse.

• The bandage was wound around the wound.

• The farm was used to produce produce.

• The dump was so full that it had to refuse more refuse.

Page 3: “I am like that,” does not help anything. “I can be different,” does.

• A chemistry teacher was berating the students for not learning the Periodic Table of the Elements. She said, "Why when I was your age I knew both their names and weights." One kid popped up, "Yeah, but teach, there were so few of them back then.

• A chemistry teacher was berating the students for not learning the Periodic Table of the Elements. She said, "Why when I was your age I knew both their names and weights." One kid popped up, "Yeah, but teach, there were so few of them back then.

Page 4: “I am like that,” does not help anything. “I can be different,” does.

Crush the canCrush the can

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Gas LawsGas Laws

Lesson 4Lesson 4

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ScubaScuba

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Scuba Scuba

• Invented in 1943

• By Jacques - Ives Cousteau and Emile Gagnan.

• Invented in 1943

• By Jacques - Ives Cousteau and Emile Gagnan.

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Pressure and water depthPressure and water depth

• At the surface of the water the air pressure equals 1 atm.

• For every 10m of water you add 1 atm.

• At the surface of the water the air pressure equals 1 atm.

• For every 10m of water you add 1 atm.

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ProblemProblem

• If a diver dives 40m, how much pressure is on him?

• 5 atm

• If a diver dives 40m, how much pressure is on him?

• 5 atm

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Absolute zero Absolute zero

• Substance has zero kinetic energy. All movement would stop.

• Substance has zero kinetic energy. All movement would stop.

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Diffusion Diffusion

• Process by which particles of matter fill a space because of random motion.

• (Ex: smells in the air, food coloring in liquid)

• Process by which particles of matter fill a space because of random motion.

• (Ex: smells in the air, food coloring in liquid)

Page 12: “I am like that,” does not help anything. “I can be different,” does.

Evaporation Evaporation

• Process by which particles of a liquid form a gas by escaping from the surface.

• Process by which particles of a liquid form a gas by escaping from the surface.

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3 Things that affect evaporation

3 Things that affect evaporation

• Temperature

• Surface area

• Humidity

• Temperature

• Surface area

• Humidity

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Energy Energy

• The ability to do work .

• The ability to do work .

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Heat of vaporization Heat of vaporization

• Energy absorbed when 1 kg of a liquid vaporizes at its normal boiling point.

• (The heat of vaporization of water is 2.26 x 106 J / kg)

• Energy absorbed when 1 kg of a liquid vaporizes at its normal boiling point.

• (The heat of vaporization of water is 2.26 x 106 J / kg)

Page 16: “I am like that,” does not help anything. “I can be different,” does.

Heat of fusion Heat of fusion

• Energy released as 1 kg of a substance solidifies at its freezing point.

Heat of fusion of water is 3.34 x 105 J/kg

• Energy released as 1 kg of a substance solidifies at its freezing point.

Heat of fusion of water is 3.34 x 105 J/kg

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Elastic collisions Elastic collisions

• When a gas particle rebounds without losing speed or no K.E. is lost.

• When a gas particle rebounds without losing speed or no K.E. is lost.

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Fractional distillation Fractional distillation

• When mixtures of air are separated into components and these components are separated by differences in their B.P.

• (Used in production of N2 and O2 and other gases.)

• When mixtures of air are separated into components and these components are separated by differences in their B.P.

• (Used in production of N2 and O2 and other gases.)

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Factor Label Method Factor Label Method

• Method used to convert measurements in one unit to their equivalent in a second unit .

• Method used to convert measurements in one unit to their equivalent in a second unit .

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Problem 1Problem 1

• Determine the number of moles of unknown gas if a sample occupies a volume of 148 mL at 13°C and a pressure of 107.0 kPa.

• Determine the number of moles of unknown gas if a sample occupies a volume of 148 mL at 13°C and a pressure of 107.0 kPa.

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WorkWork

• Convert ml to L• 148 ml = .148 L = V

• 13oC = 286 K = T

• P = 107.0 kPa• Plug into the formula PV=nRT

• Convert ml to L• 148 ml = .148 L = V

• 13oC = 286 K = T

• P = 107.0 kPa• Plug into the formula PV=nRT

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AnswerAnswer

• .0067 mol or .007 mol• .0067 mol or .007 mol

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QuestionQuestion

• What is it called when dry ice (solid carbon dioxide) changes to carbon dioxide gas?

• Answer: Sublimation

• What is it called when dry ice (solid carbon dioxide) changes to carbon dioxide gas?

• Answer: Sublimation

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P1V1= P2V2 P1V1= P2V2

• What is the name of the above formula?

• Boyle’s Law

• What is the name of the above formula?

• Boyle’s Law

Page 25: “I am like that,” does not help anything. “I can be different,” does.

Problem 2Problem 2

• The gas left in a used aerosol can is at a pressure of 1 atm at 27oC (room temp). If this can is thrown onto a fire, what is the internal pressure of the gas when its temperature reaches 927oC?

• The gas left in a used aerosol can is at a pressure of 1 atm at 27oC (room temp). If this can is thrown onto a fire, what is the internal pressure of the gas when its temperature reaches 927oC?

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WorkWork

• Convert temp to Kelvin • 27 + 273 = 300 K • 927 + 273 = 1200 K

• 1 atm = P2

• 300 K 1200K • 1 atm x 1200 K • 300 K• Answer = 4 atm

• Convert temp to Kelvin • 27 + 273 = 300 K • 927 + 273 = 1200 K

• 1 atm = P2

• 300 K 1200K • 1 atm x 1200 K • 300 K• Answer = 4 atm

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Problem 3Problem 3

• How many moles of helium are contained in a 5.00 L canister at 101 kPa and 30 o C?

• How many moles of helium are contained in a 5.00 L canister at 101 kPa and 30 o C?

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WorkWork

• PV = nR T

5 L x 101 kPa = n x 8.31 L x kPa x 303K mol x K

Answer = .201 mol

• PV = nR T

5 L x 101 kPa = n x 8.31 L x kPa x 303K mol x K

Answer = .201 mol