Lecture 21 - Gases Part III

Thursday, April 4, 2024

9:00 AM

There are two handouts for today: ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/modules/items/3044296"Final Stoichiometry Map and ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/modules/items/3044537"Midterm 4 Equations Class notes (1-20): https://bricejurban.github.io/CHEM101/ Assignments this week: ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28698/assignments/1014190"HW 13 Chemistry of Gases Due Sunday Review Answer Key for Midterm 3 Read Chapter 7 (I may make a short reading quiz on this) Reminders: Midterm 3 grades are entered on canvas and published on Gradescope. An answer key is outside SCNC 336. If you notice any errors in grading, please make a regrade request through Gradescope. My CIC (EDUC 107) Hours: Friday 11AM - 1PM Office Hours (SCNC 314 or Zoom): ﷟HYPERLINK "https://calendly.com/bricejurban/office-hours"By appointment Today (4/4) Gas Laws Gas Stoichiometry Looking Ahead Tuesday (4/9) & Thursday (4/11) Heat and Calorimetry Phase Diagrams Acids, Bases & Titrations Gas Laws Pressure vs Volume (Boyle's Law) 1662 P V n T Initial 1.4 atm 250 mL Final ? 400 mL Sample of gas at 30 °C in a piston has an initial pressure of 1.4 atm and is then expanded by moving the piston Untitled picture.png As we expand the gas, there is more volume for the gas to take up. So its pressure decreases. Volume vs Temperature (Charles's Law) 1780's, attributed in 1802 P V n T Initial 1.0 L -80°C = 193 K Final ? 113°C = 386 K Closed balloon at atmospheric pressure in a dry ice bath has a volume of 1.0 L is heated with steam. Untitled picture.png Untitled picture.png Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Closed balloon at atmospheric pressure in a dry ice bath has a volume of 1.0 L is heated with steam. Untitled picture.png Pressure vs Temperature (Gay-Lussac's Law) ~1805 P V n T Initial 600 torr 0 °C = 273K Final ? 100°C = 373 K Sample of neon in rigid 1.0 L container is equilibrated in an ice bath and has a pressure of 600 torr, is then placed in a boiling water bath Untitled picture.png Untitled picture.png Volume vs Amount (Avogadro's Law) ~1800 but recognized 1850 P V n T Initial 3.0 L 1.00 Final 12 L ? A balloon at constant atmospheric pressure and temperature is 3.0L in size and then blown up with more air to 12L. The initial mols of gas is 1.00 Untitled picture.png Untitled picture.png Combined Gas Law P V n T Initial 450 mmHg 600 mL 263 K Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings 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Initial 450 mmHg 600 mL 263 K Final 760 mmHg ? 273 K A weather balloon's pressure and temperature changes as it descends. How does the volume change? Ideal Gas Law PV= nRT Relates all of the variables to the gas constant R Valid under conditions where volume and intermolecular forces of particles is negligible These conditions are usually met under low pressures and high temperatures P = pressure in atm, V = volume in liters n = amount in mols T = temp in K R = gas constant = 0.0821 atm·L/(mol·K) P V n T Initial ? 4.1 L 0.20 40°C = 273 +40 = 313 K A sample of 0.20 mol of oxygen gas occupies a volume of 4.1 L at 40 °C. Untitled picture At a constant temperature, which of the gases will have the highest average kinetic energy? If at a constant temperature all of the gases will have the same energy However, the speeds will not be the same. The lightest gases (lowest molar mass) will have the fastest speeds. Untitled picture 02 All the gases will have the same average kinetic Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
However, the speeds will not be the same. The lightest gases (lowest molar mass) will have the fastest speeds. Untitled picture 02 All the gases will have the same average kinetic A gas mixture was prepared to contain 50% by mass of O₂ and Ne. When the total pressure of the mixture is 1.50 atm, what is the partial pressure of O₂? 1) Assume 100 gram sample 2) Convert 50 grams of O2 to mols 3) Convert 50 grams of N2 to mols 4) Find the mole fraction of O2 5) Multiply by the total pressure to find the partial pressure A sample of gas at a constant temperature has an initial pressure reading of 2.00 atm with a volume of 175.0 mL. After a change in pressure, the final volume reading is 550.0 mL. Calculate the final pressure. 1) Use Boyle's Law A sample of gas at a constant pressure initially has a temperature of 300.0 K with a volume of 125 mL. The volume changes to 367 mL. Calculate the final temperature. 1) Use Charles's Law The pressure of a gas sample at 30.0 °C is increased from 1.10 atm to 2.30 atm by heating at a constant volume. What is the temperature (in °C) of the gas at 2.30 atm? 1) Convert temperature to K 2) Solve the final temperature (Gay-Lussac's Law) 3) Convert back to Celsius What is the pressure in a 6.00 L tank with 20.0 grams of nitrogen gas at 385 K? 1) convert grams of N2 to mol of N2. 2) Solve the ideal gas law for Pressure Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink 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What is the temperature, in K, of 0.0450 moles of argon in a 4.00 L vessel at 0.691 atm? 1) Solve the ideal gas law for temperature Gas Stoichiometry Untitled picture.png Machine generated alternative text: P (atm), V (L), and T (K) of gas A P (atm), V (in L) of gas A at STP + by 22.4 grams of A + by molar mass of A use PV=nRT V (L), or T (K) of gas B use PV=nRT V (in L) of gas B at STP x by 22.4 Coefficient B moles of A CoetT1cent A moles of B x by molar mass of B grams ofB x by 6.022E23 + by 6.022E23 x byM or v +byMorV Stoichiometry Problems at Standard Temperature and Pressure (STP) (273 K = 0 °C and 1 atm) Untitled picture.png Machine generated alternative text: V (in L) of gas A at STP V (inL)0fgasBatSTP grams of A + by 22.4 + by molar mass of A Coefficient B moles of A Coefficent A moles of B x by 22.4 x by molar mass of B grams ofB In situations that are at atmospheric pressure and near the melting point of ice (STP), we can use a simple conversion that 1 mol of gas = 22.4 L Example: Sodium metal (0.350 g) is dropped into water and reacts completely. How much hydrogen gas in liters is produced if the reaction occurs at STP? 2 Na(s) + 2 H₂O(l) → H₂(g) + 2 NaOH(aq) Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Stoichiometry Problems that are not at STP use the Ideal Gas Law Untitled picture.png Machine generated alternative text: P (atm), V (L), and T (K) of gas A use PV=nRT P (atm), V (L), or T (K) of gas B use PV=nRT grams of A + by molar mass of A Coefficient B moles of A Coefficent A moles of B x by molar mass of B grams ofB In most situations gases are not at STP and its necessary to use PV=nRT Example: What volume, in liters, of sulfur dioxide gas at 455 K and 2.03 atm is needed to produce 100 g of sulfur trioxide 2 SO₂(g) + O₂(g) →2 SO₃(g) What is the volume in liters of hydrogen gas that would be produced by the reaction of 40.0 g of Al with excess HCl at STP according to the following reaction? 2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g) 1) Convert grams of Al to mols of Al 2) Convert mols of Al to mols of H2 Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
2) Convert mols of Al to mols of H2 3) Convert mols of H2 to volume of H2 using the fact that 1 mol = 22.4 L at STP What volume, in liters, of carbon monoxide gas at 78.5 °C and 848 torr is needed to produce 3.21 g of methanol, CH₃OH? CO(g) + 2 H₂(g) → CH₃OH(g) 1) Find molar mass of methanol: (32.042 g/mol) 2) Convert grams of methanol to mols of methanol 3) Convert mols of methanol to mols of CO 4) Use the Ideal gas Law to find volume of CO A) First convert 78.5 C to K and 848 torr to atm Heat (pick up here next week) Untitled picture.png Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings Ink Drawings
Untitled picture.png Process Before Energy Change After Endo/exothermic Melting Ice Freezing Water Terms to know The term heat refers to the energy transfer from a ________ object to a _______ object until the ___________ are the same. The units of heat (q) are __________ or _________ and the conversion is : The system is the ________________________________ and the surroundings is _________________.
The system is the ________________________________ and the surroundings is _________________. Explain the following, descibing how energy of the system changes A cup of hot coffee cools as it sits on a table The temperature of the skin decreases when sweat evaporates from the skin The Measurement of Heat Transfer (Calorimetry) Calorimetry is the measure of heat is transferred due to: _________________________________________________________ A substances resistance to temperature changes is known as the ____________________ and has the units ____________ If Beaker A contains 50.0 g of water (Cp = 4.18 J/g·°C) and Beaker B contains 50.0 g of acetone (Cp = 2.15 J/g·°C), both at the same temperature. If both beakers are heated with the same amount of energy, which liquid will have the higher final temperature?
The equation for Heat transfer is: How much energy is released when 20.0 g of water is cooled from 50°C to 30°C? (Cwater = 4.18 J/g·°C) When a 50.0 g block of copper was heated with 1046 J of energy, the temperature increased from 10.0°C to 64.3°C. What is the specific heat of copper? What is the final temperature when a 90.0 g of aluminum at 20°C releases 500 J of energy? (Cp,Al = 0.897 J/g·°C)

 

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