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Behavior of Gases Chapter 10 & 12

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**A Force Exerted by a Gas over a Given Area**

Pressure What is pressure? A Force Exerted by a Gas over a Given Area What causes pressure? Collisions of the Gas Particles with each other &the Walls of the Container That’s Pressure! Walls of the Container Gas Particle

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**Which shoes create the most pressure?**

Smaller the area of contact, larger the amount of pressure exerted by an object.

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**Units of Pressure atmospheres = atm millimeters of mercury = mmHg**

kilopascals = kPa pounds per square inch = psi 1 atm = 760mmHg = 101.3kPa = 14.7psi

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**Conversion Between Units of Pressure**

1 atm = 760mmHg = 101.3kPa = 14.7psi

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Sample Problem #1 How many kilopascals are equivalent to 880mmHg? 101.3 760 = 120 kPa

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Sample Problem #2 Calculate the number of psi that are in 2.60atm. 14.7 1

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**K = ºC + 273 Temperature ºF ºC K -459 32 212 -273 100 273 373**

ALWAYS use absolute temperature (Kelvin) when working with gases. ºF ºC K -459 32 212 -273 100 273 373 K = ºC + 273

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**Practice problems K = °C + 273 Ex) 32°C = ______ K**

K = = 305 K Try one: How much is 75°C in Kelvin? K = = 348 K

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**Behavior of Gases and the Kinetic Theory**

Kinetic refers to motion. The energy of an object has because of its motion is called kinetic energy. The Kinetic theory states that the tiny particles in all forms of matter are in constant motion.

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Watch the video segment (Kinetic Molecular Theory – Standard Deviants School Chemistry: Molecular Geometry) and fill in the missing information in your workbook.

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**Postulates of the Kinetic Molecular Theory of Gases**

A gas is composed of particles, usually molecules or atoms that are far apart from one another in comparison with their own dimensions. Particles are relatively far apart from one another and between them is empty space. Gas molecules are in constant random motion. They travel in straight paths (unless they collide with a wall of a container) and move independently of each other.

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**Postulates of the Kinetic Molecular Theory of Gases**

The molecules exert no force on each other or on the container until they collide with each other or with the walls of the container. The average kinetic energy of the molecules of a gas is proportional to the temperature. Every time a molecule collides with the wall, it exerts a force on it which we call pressure.

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**Applying this knowledge we know…**

Gases fill their containers regardless of the shape and volume of the containers. Because there is so much space between particles, gases are easily compressible. Because gases are compressible, they are used in automobile airbags and other safety devices designed to absorb the energy of an impact

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**All collisions are perfectly elastic**

All collisions are perfectly elastic. This means that during collisions kinetic energy is transferred without loss from one particle to another, and total kinetic energy remains constant. The average speed of oxygen molecules in air at 20oC is 1700 km/h. At these high speeds, the odor molecules from a hot pizza in Washington, D.C., should reach Mexico City in about 106 minutes. Why doesn’t this actually happen?

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Answer Molecules are constantly striking molecules of air and rebounding in other directions. Their path of uninterrupted travel in a straight line is short.

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**Questions: What happens when a closed container is inflated?**

Pressure increases with the additional gas particles. Pressure exerted is caused by collisions of gas particles with the walls. If the number of gas particles is changed by any factor, the pressure changes by that same factor; until the container ruptures. What happens when a closed container is inflated?

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**Double the amount of air.**

A gas inside a bicycle tire exerts a pressure of 35 pounds per square inch (psi). How much air must be pumped into the tire to produce a pressure of 70 psi? Double the amount of air. Increased by a factor of 2!

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Note!!! The relationship between the amount of gas and pressure is proportional, assuming the volume & temperature stay the same. This means more gas = higher pressure and less gas = lower pressure.

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**C. What happens to pressure when a closed container is deflated?**

*Pressure of the gas is decreased because there are fewer gas particles and less collisions. *If the number of gas particles decreases by half, the pressure decreases by half. (Note: Gas particles move from region of higher pressure to lower pressure until equilibrium is reached.)

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**Homework Complete pg. 4 in your booklet.**

Bring an empty soda can to class tomorrow.

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Bell ringer – 2/5/14 Each of these flasks contains the same number of gas molecules. In which would the pressure be lowest? Explain your answer choice. Which temperature is colder: 36°C or 278 K?

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Combined Gas Law The Combined Gas Law helps us explain what happens to gases as the pressure, temperature, and volume changes in respect to moles of a substance. Combined Gas Law P1 V1 P2V2 n1 T1 n2T2 Boyle's Law P1 V1 = P2V2 Charles's Law V V2 T T2 Avogadro's Law V1 V2 n1 n2 V1V =

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**atm mmHg torr kPa psi L mL cm3 m3**

Letter or Number Variable Name Unit Conversions P Pressure atm mmHg torr kPa psi 1 atm = 760 mmHg = 760 torr = 101.3 kPa = 14.7 psi V Volume L mL cm3 m3 1 L = 1000 mL 1 mL = 1 cm3

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***NOTE: If “n” is not given in a problem, assume it to be 1 mole.**

Letter or Number Variable Name Unit Conversions n* Moles 6.02 x 1023 molecules T Temperature K (Kelvin) K = oC + 273 1 Initial variable 2 Final variable *NOTE: If “n” is not given in a problem, assume it to be 1 mole.

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**Temperature must be in Kelvin . **

Remember: STP = 1.0 atm and 273 K or 0.0°C

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**Guided Practice Givens and Unknowns: P1 = 1.2 atm P2 = 0.90 atm**

1. A hot air balloon has a volume of 7500L at 270K and a pressure of 1.2atm. What will be the volume of the balloon if the pressure changed to 0.90atm and the temperature decreases to 230K? Givens and Unknowns: P1 = atm P2 = atm V1 = L V2 = unknown n1 = 1 mole n2 = 1 mole T1 = K T2 = K

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**(1.2 atm) (7500 L) = (0.90 atm) (V2) (1 mol) (270 K) (1 mol) (230 K)**

Substitute & Solve (Cross Multiply): (1.2 atm) (7500 L) = (0.90 atm) (V2) (1 mol) (270 K) (1 mol) (230 K) (1.2 atm)(7500 L)(1 mol)(230 K) = (0.90 atm)(V2)(1 mol)(270 K) atm*L*mol*K = (V2)(243 atm*mol*K)

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**2070000 atm*L*mol*K = (V2) 243 atm*mol*K 8518.5 L = V2 Use 2 sig figs**

L = V2 Use 2 sig figs 8500 L = V2

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**Guided Practice Givens and Unknowns: P1 = 1.0 atm P2 = unknown**

2. The volume of a gas at STP is 22.4L. At 12oC, the volume of the balloon changes to 55.0L. What is the new pressure? Givens and Unknowns: P1 = atm P2 = unknown V1 = L V2 = L n1 = 1 mole n2 = 1 mole T1 = K T2 = 12oC = K

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**(1.0 atm) (22.4 L) = (P2) (55.0 L) (1 mol) (273 K) (1 mol) (285 K)**

Substitute & Solve (Cross Multiply): (1.0 atm) (22.4 L) = (P2) (55.0 L) (1 mol) (273 K) (1 mol) (285 K) (1.0 atm)(22.4 L)(1 mol)(273 K) = (P2)(55.0L)(1 mol)(285 K) 6384 atm*L*mol*K = (P2)(15015 L*mol*K)

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**66384 atm*L*mol*K = (P2) 15015 L*mol*K 0.42517 atm = P2 Use 2 sig figs**

atm = P2 Use 2 sig figs 0.43 atm= P2

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**Complete pg. 8 in your packet.**

Homework Complete pg. 8 in your packet.

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**Soda Can Activity OBJECTIVES**

Students will demonstrate the effects of air pressure. Students will demonstrate that as a gas is heated it expands and as it cools it will contract. SAFETY Be careful of hot water.

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**What caused the can to collapse?**

As the water boils, the can becomes full of steam. When the can is inverted into the cold water bath, the temperature of the gas inside the can drops and some of the water condenses. Since the temperature drops and there are fewer gas particle collisions, the pressure inside the can decreases. Since the pressure outside the can is now much greater, this higher pressure crushes the can.

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**The Crushing Can in Real Life**

The tanker truck was steam cleaned. It was then sealed. This is the site that the workers saw the following morning.

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**Day 3: Boyle’s, Charles’, and Avogadro’s Laws**

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**The Effect of Changing Size of Container – Boyle’s Law**

WHAT IF…temperature and moles do not change and we just look at the relationship between pressure and volume. Our equation would look like this: P1 V1 = P2 V2 Boyle’s Law Boyle’s Law states that at a constant temperature, the volume of a gas is inversely proportional to the pressure exerted by that gas.

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**The Effect of Changing Size of Container – Boyle’s Law**

Think of two kids (Paul Pressure and Victor Volume) on a see-saw. If Paul goes up, Victor goes down. If Victor goes up, Paul goes down. This is an inverse relationship.

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To show animation

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Examples: If a gas is compressed from 2L to 1L, the pressure will _____________ by a factor of 2. If a gas is expanded from 1L to 3L, the pressures will _______________ by a factor of 3. Gases cool when they expand and heat when they compress. Why? increase decrease Thus, if you forget to wear your suit in space, you will_EXPLODE_!!! If the volume of a container decreases in size, the pressure of gas particles in the container increases.

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**Example Problems P1 V1 = P2 V2**

The pressure of a 3.5L balloon was determined to be 1.5atm. Assuming that the temperature remained constant, what would be the volume of the balloon if the pressure was decreased to 0.45atm? At 45oC, a certain container of gas has the volume of 580mL and a pressure of 980mmHg. What would be the new volume of the gas at 250 mmHg and 45oC?

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**The Effect of Temperature changes on Volume – Charles’s Law**

As the gas inside a balloon cools, the average KE of molecules decreases. With fewer and less collisions, the gas molecules move closer together and occupy a smaller volume than they previously did. The volume decreases, assuming no change in the amount of gas and pressure.

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Charles’ Law Charles law states: At a constant pressure, the volume of a gas is directly proportional to the temperature in Kelvin. As temperature increases, the volume increases As temperature decreases, the volume decreases.

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**Observe what happens to the balloon as liquid nitrogen is being poured on it.**

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Helpful Hint!!!

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**Guided Practice Temperature must be in Kelvin!**

The temperature of a 0.65L sample of carbon dioxide gas is 580K. If the pressure remains constant, what is the new volume of the gas if the temperature increases to 1300K? A balloon has a volume of 5.6L at a temperature of 98oC. If the volume of balloon increases to 9.5L, what will be the temperature of the gas in Celsius? Assume that the pressure remains constant.

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Avogadro’s Law Avogadro's Law (Avogadro's theory; Avogadro's hypothesis) is a principle stated in 1811 by the Italian chemist Amedeo Avogadro ( ) that "equal volumes of gases at the same temperature and pressure contain the same number of molecules regardless of their chemical nature and physical properties“. This number (Avogadro's number) is 6.02 X It is the number of molecules of any gas present in a volume of L and is the same for the lightest gas (hydrogen) as for a heavy gas such as carbon dioxide or bromine.

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Avogadro’s Law Or to put it another way, "the principle that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. Thus, the molar volume of all ideal gases, at 0° C and a pressure of 1 atm., is 22.4 liters" V = the volume of the gas n = the amount of substance of the gas

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Avogadro’s Law states that that equal volumes of all gases at the same temperature and pressure contain the same number of molecules or moles.

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**V1 = V2 V = the volume of the gas n1 n2 n = the amount of substance of the gas**

Example #1: 5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? Example #2: A cylinder with a movable piston contains .005 mol of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many moles of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.)

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**Homework Complete pg. 12-13 in your packet.**

Remember, quiz is on Monday.

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**Equation for Boyle’s Law**

State Boyle’s Law For a constant temp and mass, the pressure of a gas varies INVERSELY with volume. Graph for Boyle’s Law P V P1V1 = P2V2 Equation for Boyle’s Law Inverse Inverse or Direct? Boyle’s Law

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**Graph for Charles’s Law Equation for Charles’s Law**

State Charles’s Law For a given mass and constant pressure, the volume of a gas varies directly with temp. Graph for Charles’s Law V T Equation for Charles’s Law Direct Inverse or Direct? Charles’s Law Temperature must be in Kelvin.

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**Graph for Avogadro’s Law Equation for Avogadro’s Law**

State Avogadro’s Law For a given gas at a constant temperature and pressure, the moles varies directly with volume Graph for Avogadro’s Law n V Equation for Avogadro’s Law Direct Inverse or Direct? Avogado’s Law

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**Dalton’s Law of Partial Pressures**

Partial pressure of a gas in a mixture of gases is the pressure which that gas would exert if it were the only gas present in the container. Dalton’s Law of Partial Pressures states that that the total pressure in a gas mixture is the sum of the partial pressures of each individual gas. Ptotal = Pgas a + Pgas b + Pgas c + etc Dalton’s Law of Partial Pressures assumes each gas in the mixture is behaving like an ideal gas

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A partial pressure is the pressure a gas would have or would exert if it were alone in the container.

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**Ptotal = Pgas a + Pgas b + Pgas c + etc**

Example Problems Ptotal = Pgas a + Pgas b + Pgas c + etc Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (Poxygen) at 101.3kPa if the partial pressures of nitrogen, carbon dioxide, and other gases are 79.10kPa, 0.040kPa, and 0.94kPa, respectively. A mixture of gases contains oxygen, nitrogen, and helium. The partial pressure of oxygen is 2.1atm. The partial pressure of nitrogen in 0.21atm, and the partial pressure of helium is 7.80atm. Determine the total pressure of this mixture.

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**Complete pg. 15 in your packet.**

Homework Complete pg. 15 in your packet.

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Ideal Gas Law An ideal gas is one that follows the gas laws at ________________________________________. Such a gas would have to conform precisely to the assumptions of ________ ________. As you probably suspect, there is _________ for which this is true. An ideal gas __________ exist. Nevertheless, at many conditions of temperature and pressure, _____________ behave very much like an ideal gas. all conditions of pressure and temperature kinetic theory no gas does not real gases

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An important behavior of real gases that differs from that of a hypothetical ideal gas is that real gases can be ___________ and sometimes ____________ by cooling and by applying pressure. Ideal gasses cannot be. For example, when water vapor is cooled below 100oC at standard atmospheric pressure, it condenses to a liquid. The behavior of other real gases is similar, although lower temperature and greater pressures may be required. ****Gases behave ideally at ___________________________________. liquified solidfied high temperatures and low pressure

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**n T If we look at one side of the Combined Gas Law: P V **

and solve it for one mole at STP, you would get a “constant” (symbolized as R). (101.3 kPa)(22.4 L) = (L . kPa)/(K . mol) (1 mole)(273 K) P = Pressure V = Volume n = moles T = Temperature

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**We call this the ideal gas constant (R):**

If pressure is measured in: The ideal gas constant (R) is: kPa 8.31 (L . kPa)/(K . mol) atm (L . atm)/(K . mol) mmHg 62.4 (L . mmHg)/(K . mol) torr 62.4 (L . torr)/(K . mol)

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**SO… P = Pressure V = Volume (must be in liters) n = moles**

T = Temperature (must be in Kelvin) R = Ideal gas constant SO…

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Temp must be in Kelvin!!!!

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**Givens and Unknowns: P = V = n = R = T =**

1. Calculate the number of moles of oxygen in a 12.5 L tank containing 250 atm, measured at 22oC. Givens and Unknowns: P = V = n = R = T = Equation: PV = nRT Substitute & Solve:

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**Givens and Unknowns: P = V = n = R = T =**

2. If 4.5 g methane gas (CH4) is introduced into an evacuated 2.00 L container at 35oC, what is the pressure in the container, in atm? Givens and Unknowns: P = V = n = R = T = Equation: PV = nRT Substitute & Solve:

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**Givens and Unknowns: P = V = n = R = T = Equation: PV = nRT**

3. A balloon is filled with 0.34 moles of pure nitrogen. If the balloon is at 37 oC and is under pressure of 100 kPa, calculate the volume of the balloon. Givens and Unknowns: P = V = n = R = T = Equation: PV = nRT Substitute & Solve:

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**Complete pg. 18 in your packet.**

Homework Complete pg. 18 in your packet.

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**Example Problem N2 (g) + 3 H2 (g) 2 NH3 (g)**

Example 1: liters of nitrogen (N2) gas is reacted with excess hydrogen (H2) gas to form ammonia (NH3) gas at 304°K and a pressure of 1.02 atm. How many liters of ammonia gas is formed? Step 1: Write the balanced chemical reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) Step 2: Convert liters of N2 to moles of N2. NOTE: Since the gas is not at STP, you must convert using the Ideal Gas Law before you can do Stoichiometry!!! PV = nRT: (1.02atm) (3.00L) = (n)( L • atm) (304K) mole • K n = mol of N2

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Example Problem Step 3: Convert moles of N2 to moles of NH3 using the mole ratio of the balanced chemical equation. 0.123 mol N2 2 mol NH3 1 mol N = mol NH3 Step 4: Convert moles of NH3 to liters of NH3 using the ideal gas law. PV = nRT: (1.02atm) (V) = (0.246mol)( L • atm) (304K) mole • K V = 6.02 L of NH3

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**Example 2: Calculate the volume of hydrogen gas produced at 0**

Example 2: Calculate the volume of hydrogen gas produced at 0.0°C and 1.00 atm of pressure by reacting 12.0 g of zinc metal with excess sulfuric acid. Step 1: Write the balanced chemical reaction. Zn (s) + H2SO4 (aq) H2 (g) + ZnSO4 (aq) Step 2: Convert grams of zinc to moles of H2 using the mole ratio from the balanced equation. 12.0 g zinc 1 mole zinc mol H = mol H2 65.4 g zinc mole zinc

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**Step 3: Convert moles of H2 to liters of H2 using the ideal gas law.**

PV = nRT (1.00 atm) (V) = (0.183 mol) ( L • atm ) (273 K) mole • K V = 4.10 L of H2

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**CaCO3(s) CO2(g) + CaO(s)**

Class Practice: 3. How many grams of calcium carbonate will be needed to form 4.29 liters of carbon dioxide? The following reaction takes place at a pressure of 1 atm and a temperature of 298o K. CaCO3(s) CO2(g) + CaO(s)

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**Complete pg. 21 in your packet.**

Homework Complete pg. 21 in your packet.

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Chapter 13 Calculating Gases 1 Section 12.1 Pressure + Temperature conversions, Dalton’s + Graham’s Laws Section 13.1 The Gas Laws Section 13.2 The Ideal.

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