Module 5 Exam ReviewThis review is neither a replica of the exam nor does it contain everything in the module. It is always best practice to review all lessons and assessments. Resubmitting is a great way to review/learn and improve your grade. In this review, questions 1 – 23 are the general topics for the multiple choice type questions on the module exam; questions 24 – 36 are covered as essays on the exam. There are approximately 30 questions (about 6 are essays) on the regular exam; approximately 36 questions (about 10 are essays) on the honors exam. The number in parenthesis is the lesson. The answer key is after the review. Please join our LIVE chemistry help sessions offered each week. The calendar is on your course announcement page. Direct link: http://bit.ly/1EY3CEO

1. Identify the characteristics of a solid, liquid, gas and plasma. (5.02)

2. How does plasma form? (5.02)

3. Identify a substance that expands to fill the container it is in. (5.02)

4. How does the arrangement of the particles in a solid compare to that of a liquid and a gas? (5.02)

5. Explain how kinetic energy compares in a solid, liquid and a gas. (5.02)

6. What happens to a substance as the kinetic energy increases? (5.02)

7. What happens to a substance as the kinetic energy decreases? (5.02)

8. Using the diagram to the right, what is the approximate freezing point, melting point and boiling point? (5.02)

9. Using the diagram above, what is the state of matter at approximately 375 K? (5.02)

10. Using the diagram to the right, identify the process is occurring at each number. (5.02)

11. Using the diagram above, identify the numbers that are associated with an increase of kinetic energy. (5.02)

12. Explain the kinetic molecular theory. (5.03)

13. Explain why it is bad to overheat compressed cans of gas (i.e. spray paint cans). (5.03)

14. Explain what causes a gas to diffuse. (5.04)

15. Explain what causes a gas to effuse. (5.04)

16. Explain which gases would have the fastest velocity at a given temperature. (5.04)

17. Describe the purpose of chromatography, filtration, evaporation, and distillation. (5.05)

18. Define heterogeneous and homogeneous mixture. (5.05)

19. What is a solution? (5.05)

20. Explain the conditions that allow a substance to dissolve in water. (5.06)

21. Define molality, molarity, and percent by mass. (5.06/5.07)

22. Explain the process of dilution. (5.07)

23. Explain what colligative properties are and how they are affected. (5.08)

24. (Essay) A sample of gas occupies a volume of 225 mL at a temperature of 375 K. At what temperature will this sample of gas occupy a volume of 300. mL? Show all of the work used to solve this problem. (5.03)

25. (Essay) A sample of gas has a pressure of 1.8 atmospheres at a temperature of 275 Kelvin. What would be the pressure of the same sample of gas when the temperature is decreased to 260 Kelvin? Show all of the work used to solve this problem. (5.03)

26. (Essay) What mass, in grams, of fluorine gas (F2) is contained in a 10.3 liter tank at 25.5

degrees Celsius and 2.63 atmospheres? Show all of the work used to solve this problem. (5.04)

27. (Essay) What is the volume of 4.50 moles of oxygen gas (O2) at standard temperature

and pressure (STP)? (5.04)

28. (Essay) How many grams of oxygen gas (O2) need to be combusted to produce 9.50 L

water vapor at 273 K and 1.30 atm? Show all of the work used to solve this problem. (5.04)

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

29. (Essay) How many liters of fluorine gas, at 287 K and 1.03 atm, will react with 18.3 grams of potassium metal? Show all of the work used to solve this problem. (5.04)

2 K + F2 2 KF

30. (Essay) If 24.3 grams of NaCl are dissolved in 192 grams of water, what is the concentration of the solution in percent by mass? (5.06)

31. (Essay) What is the molarity of a 1.5-liter solution containing 98 grams of HCl? (5.06)

32. (Essay) How many moles of KCl are required to prepare 1.50 L of 2.45 M KCl? (5.06)

33. (Essay) How many milliliters of a 1.25 molar hydrobromic acid (HBr) solution would be needed to react completely with 75.5 grams of lithium metal? (5.07)

2Li + 2HBr 2LiBr + H2

34. (Essay) How many liters of 10.5 molar HCl stock solution will be needed to make 14.6 liters of a 2.5 molar HCl solution? Show the work used to solve this problem.

35. (Essay) Explain how the addition of a solute affects the vapor pressure, freezing point and boiling point of a solvent.

36. (Essay) Two solutions are boiled. The first is 2 mol of a calcium chloride (CaCl2) solution forming two ions. The second one is 2 mol of a sugar (C6H12O6) solution forming. How

will the boiling points of the two solutions compare? (5.08)

Module 5 Exam Review – ANSWER KEY1. Identify the characteristics of a solid, liquid, gas and plasma. (5.02)

Type of Matter Volume Shape GeneralSolid Definite Definite Particles packed

closely; unable to move past each other

Liquid Definite Indefinite Particles packed relatively close together; can move past each other

Gas Indefinite Indefinite Particles far apart; can move past each other

Plasma Indefinite Indefinite Electrons separated from gas; positively charged particles and negatively charged electrons; ionized gas; particles move independently of each other.

2. How does plasma form? (5.02)

Plasma forms when the temperature is high enough that the collisions cause the electrons to be removed.

3. Identify a substance that expands to fill the container it is in. (5.02)

If a substance expands to fill the container, it must be a gas. Examples include carbon dioxide gas or oxygen gas.

4. How does the arrangement of the particles in a solid compare to that of a liquid and a gas? (5.02)

The particles in a solid are packed closely together. The particles in a liquid would be

further away and finally, the particles in a gas are far away from each other.

5. Explain how kinetic energy compares in a solid, liquid and a gas. (5.02)

Kinetic energy increases from solid to liquid to gas. Of these three, gases have the highest kinetic energy, and solids have the least.

6. What happens to a substance as the kinetic energy increases? (5.02)

This is how we change the state of matter. As the kinetic energy increases, the particles begin to move faster and faster. When they move fast enough to overcome the forces of attraction holding them together, they will go into the next phase. The order of increasing kinetic energy: solid liquid gas plasma.

7. What happens to a substance as the kinetic energy decreases? (5.02)

As the kinetic energy decreases, the particles will start to slow down. When they slow down enough, they will go into the next phase with less energy.

8. Using the diagram to the right, what is the approximate freezing point, melting point and boiling point? (5.02)

Freezing point: 10°C

Melting point: 10°C

Boiling point: 25°C

9. Using the diagram above (right), what is the state of matter at approximately 375 K? (5.02)

At 375 K, the cooling curve is a horizontal line, which means the phase is changing. We are undergoing condensation and have both the gas and liquid phases present at 375K.

10.

Using the diagram to the right , identify the process is occurring at each number. (5.02)

1: freezing

2: melting

3: condensing

4: vaporization

11. Using the diagram right, identify the numbers that are associated with an decrease of kinetic energy. (5.02)

The energy decreases at # 1 and 3.

12. Explain the kinetic molecular theory. (5.03)

The kinetic molecular theory explains how a model, or ideal gas behaves. It is not exactly how gases behave, however, we can make some assumptions and at this level of studying science, we can do basic calculations. There are five points to the theory:

1. The particles of a gas are in constant, random motion

2. Gases consist of large numbers of tiny particles

3. The collisions between molecules are elastic

4. There are no forces of attraction or repulsion

5. The average kinetic energy of a gas is directly proportional to the temperature (in Kelvins)

13. Explain why it is bad to overheat compressed cans of gas (i.e. spray paint cans). (5.03)

This is Gay-Lussac’s Law (see page 6 in the “Gas Laws” section of this lesson. Since the can cannot change volume, if we heat (increase temperature), the pressure will build (increase pressure). At some point, the can will not be able to withstand the pressure and will explode.

14. Explain what causes a gas to diffuse. (5.04)

Diffusion is the concept that explains what happens when a student in the back of the room sprays perfume and eventually everyone can smell it. Diffusion is the mixing of two gases because of the random motion of the particles.

The rate of diffusion depends on the velocity. The velocity depends on the temperature and the molar mass. When comparing two gases at the same given temperature, the gas that has the smallest molar mass will diffuse faster. Less mass = higher velocity = diffuse faster. Likewise, if you have a large mass = slower velocity = diffuse slower.

15. Explain what causes a gas to effuse. (5.04)

Effusion is the concept that explains air leaking out of a balloon. When a gas is sealed in a container with a small hole, it will gradually leak out of the hole. This is because the gas particles are moving around rapidly and randomly encountering, and passing through, the hole.

The rate of effusion depends on the velocity. The velocity depends on the temperature and the molar mass. When comparing two gases at the same given temperature, the gas that has the smallest molar mass will effuse faster. Less mass = higher velocity = effuse faster. Likewise, if you have a large mass = slower velocity = effuse slower.

16. Explain which gases would have the fastest velocity at a given temperature. (5.04)

The velocity depends on the temperature and the molar mass. If two gases are at the same temperature, the gas with the greater molar mass with have the slower velocity.

17. Describe the purpose of chromatography, filtration, evaporation, and distillation. (5.05)

Separation Technique

Purpose Example

Chromatography Chromatography is a group of lab techniques that are used to separate a solution containing multiple components with different degrees of polarity.

A common type of chromatography, paper chromatography, takes place on a strip of porous, absorbent paper.

Filtration Separates an insoluble solid from a liquid

Separate pepper from water

Evaporation Separates a soluble solid from a liquid

Separate salt from water

Distillation Separates a liquid from a solution Water can be separated from salty water. Pretty similar to evaporation, except that distillation captures the evaporated liquid and condenses it in a separate container.

Side information learned/needed from this lesson: solubility. An insoluble substance is

one that does not dissolve in the liquid such as pepper in water. A soluble substance dissolves in solution such as salt or sugar in water.

18. Define heterogeneous and homogeneous mixture. (5.05)

A heterogeneous mixture is not uniform in composition. Examples include vegetable soup and granite. A homogeneous mixture is uniform in composition. An example is sugar water

19. What is a solution? (5.05-see interactive in the “Homogeneous and Heterogeneous” section)

A solution is a homogeneous mixture. Solutions can be solid, liquid, or gas.

20. Explain the factors that affect solubility. (5.06)

There are several factors that affect solubility. Intermolecular forces (the attraction between particles) is the first factor. The greater the attraction between solute and solvent, the greater it will dissolve. Polar dissolves polar and nonpolar dissolved nonpolar. Think of oil and water; oil is nonpolar and water is polar; therefore, they will not dissolve. (See “Universal Solvent” tab at top of lesson)

Dissolving a gas in a liquid: the lower the temperature and the higher the pressure, the more gas will dissolve. Think of a soda can. If it is closed, the CO2 gas stays dissolved. When we open the can, you hear a gush and see bubbles form. That is the gas coming out of solution (lower pressure = less soluble). Likewise, if the soda can sits out on the counter open, it goes ‘flat’ due to the reduced solubility of the CO2. It comes out of solution, thus making the soda taste flat.

21. Define molality, molarity, and percent by mass. (5.06/5.07)

These are all units of concentration.

Molarity (M) = moles solute Molality (m) = moles solute

Liter solution Kg solvent

% by Mass = mass of solute (g)

Total mass of solution (g)

22. Explain the process of dilution. (5.07)

Dilution is the technique of taking a little of a “stock” solution (a solution with a high concentration) and putting it into another container. Then, scientists ‘dilute’ it by adding water until they have the exact concentration they need. Dilution causes the concentration to decrease as the amount of solute (the stock) remains constant and the volume increases.

23. Explain what colligative properties are and how they are affected. (5.08)

Colligative properties are where the amount of dissolved solute affects the physical properties of a solution. The three properties are: increased boiling point, decreased vapor pressure and decreased freezing point. The more solute that is in solution, the greater the effect. Remember that ionic compounds, such as NaCl, will break apart into

its ions in solution (Na+ and Cl-). Thus if you have 1 mole of NaCl, you actually get 1

mole of Na+ and 1 mole of Cl- in solution. This equals 2 moles of ions.

24. (Essay) A sample of gas occupies a volume of 225 mL at a temperature of 375 K. At what temperature will this sample of gas occupy a volume of 300. mL? Show all of the work used to solve this problem. (5.03)

This is Charles’s law.

Make a list of what was given in the problem:

V1 = 225 mL

T1 = 375 K

T2 = ? K

V2 = 300. mL

Since we just have volume and temperature changing, we need the formula with only these two: = Now, we plug & chug: =

Cross multiply and solve for T2: (T2) =

T2 = 500. K

25. (Essay) A sample of gas has a pressure of 1.8 atmospheres at a temperature of 275 Kelvin. What would be the pressure of the same sample of gas when the temperature is decreased to 260 Kelvin? Show all of the work used to solve this problem. (5.03)

This is Gay-Lussacs’s law.

Make a list of what was given in the problem:

P1 = 1.8 atm

T1 = 275 K

P2 = ?

T2 = 260 K

Since we just have pressure and temperature changing, we need the formula with only these two: = Now, plug & chug: =

Cross multiply and solve for P2: P2 =

P2 = 1.7 atm

26. (Essay) What mass, in grams, of fluorine gas (F2) is contained in a 10.3 liter tank at 25.5

degrees Celsius and 2.63 atmospheres? Show all of the work used to solve this problem. (5.04)

Make a list of what you are given:

Mass F2 = ?

V = 10.3 L

T = 25.5 °C

P = 2.63 atm

All of these measurements are for the fluorine, so no stoichiometry is needed. Since we have values for all these, and nothing is changing, we need the ideal gas law. Recall what the ideal gas law is and what all the variables represent.

PV = nRT

Pressure (in atm)

Volume (in liters)

n is the number moles

R is the gas law constant

Temperature (in Kelvins)

Now, let’s solve by plugging & chugging!

PV = nRT

(2.63 atm) (10.3L) = n (25.5°C + 273K)

Rearrange and solve for n:

n = (2.63 atm) (10.3L)

(0.0821) (25.5°C + 273K)

n = 1.105 moles

Now that we have moles of fluorine gas, we have to convert to grams. This is molar mass (lesson 2.04, 3.09, 4.06)

1.105 mol F2 ( )= 41.9679 g = 42.0 g F2 (3 sig figs)

27. (Essay) What is the volume of 4.5 moles of oxygen gas (O2) at standard temperature and pressure (STP)? (5.04)

Make a list of what was given in the problem:

n = 4.50 moles

V = ?

STP

Since we are at STP, we use the mole ratio: 1 mol = 22.4 L (Note: you can only use this at STP)

4.50 moles X = 101 L

28. (Essay) How many grams of oxygen gas (O2) need to be combusted to produce 9.50 L

water vapor at 273 K and 1.03 atm? Show all of the work used to solve this problem. (5.04)

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Make a list of what was given:

mass = ? grams O2

V = 9.50 L H2O

T = 273 K

P = 1.30 atm

Notice that the problem asked for mass of oxygen gas, O2, but gave us information (V, T

& P) about water vapor, H2O. This means that we will need to use the ideal gas law to

solve for moles (n) of water vapor. Then, use stoichiometry to convert moles of water vapor to grams of oxygen.

n = PV n = (1.30 atm)(9.50 L) n = 0.551 moles of WATER VAPOR

RT (0.0821 )(273 K)

Now that we have moles of water vapor, H2O, we can convert to grams of O2:

0.551 moles H2O X 2 mol O2 X 32 g O2 = 17.6 g O2

2 mol H2O 1 mol O2

29. (Essay) How many liters of fluorine gas, at 287 K and 1.03 atm, will react with 18.3 grams of potassium metal? Show all of the work used to solve this problem. (5.04)

2 K + F2 2 KF

Make a list of what was given:

V = ? L F2

T = 285 K

P = 1.03 atm

mass = 18.3 g K

Notice that the problem asked for liters of F2, but gave us mass of potassium metal

(solid). We cannot use the mass of a metal in the ideal GAS law. This means that we will need to use stoichiometry first to convert grams of K to moles of F2. Then, we can

plug the moles of F2 into the ideal gas law to solve for volume.

18.3 g K X 1 mol K X 1 mol F2 = 0.234 moles F2

39.09g K 2 mol K

V = nRT V = (0.234 mol F2)(0.0821)(285K) = 5.32 L F2

P (1.03 atm)

30. (Essay) If 24.3 grams of NaCl are dissolved in 192 grams of water, what is the concentration of the solution in percent by mass? (5.06)

Make a list of what was given:

24.3 grams NaCl

192 grams H2O

This is percent by mass, so we need that formula:

% by Mass = mass of solute (g) X 100%

Total mass of solution (g)

% by Mass = mass of NaCl X 100%

mass of NaCl + mass of H2O

% by Mass = 24.3 g NaCl X 100%

24.3 g NaCl + 192 g H2O

% by Mass = 0.1123 X 100%

% by Mass = 11.2%

31. (Essay) What is the molarity of a 1.5-liter solution containing 98 grams of HCl? (5.06)

Make a list of what was given:

V = 1.5 L

Mass = 98 g HCl

The problem is asking for molarity. So, we need the formula:

Molarity (M) = moles solute

Liter solution

We need the moles of our solute. The problem gave us grams, so we will need to convert:

98 g HCl X 1 mole HCl = 2.688 mol HCl

36.45 g HCl

Now we can use the molarity formula:

Molarity (M) = moles solute = 2.688 moles HCl

Liter solution 1.5 L

Molarity = 1.8 M

32. (Essay) How many moles of KCl are required to prepare 1.50 L of 2.45 M KCl? (5.06)

Make a list of what was given:

Moles = ? moles KCl

V = 1.50 L

M = 2.45 M KCl

The problem is asking for moles and gave us volume and molarity. So, we need to rearrange the molarity the formula:

Moles solute = (Molarity)(Liter)

Moles = (2.45M)(1.50L)

Moles = 3.67 moles

33. (Essay) How many milliliters of a 1.25 molar hydrobromic acid (HBr) solution would be needed to react completely with 75.5 grams of lithium metal?

2Li + 2HBr 2LiBr + H2

Make a list of what was given:

V = ? mL HBr

M = 1.25M HBr

mass = 75.5 g Li

Well, this seems like we have some random information. But, we are given the molarity of HBr and asked for the volume. This means we can use the molarity formula: M =

But, we do not know the moles of HBr. However, we do have mass of Li, and we can

always convert from one chemical to another by using stoichiometry.

75.5 g Li X 1 mol Li X 2 mol HBr = 10.87 mol HBr

6.94 g Li 2 mol Li

Now that we have moles, we can calculate the volume using the molarity formula:

V = mol V = 10.87 mole HBr = 8.70 L HBr

M 1.25 M HBr

You may think that you are done, however the question asked for mL, so we must convert:

9.70 L HBr X 1000 mL = 970. mL

1 L

34. (Essay) How many liters of 10.5 molar HCl stock solution will be needed to make 14.6 liters of a 2.5 molar HCl solution? Show the work used to solve this problem. (5.07)

Make a list of what you are given:

V = ? HCl

M = 10.5 M HCl

V = 14.6 L HCl

M = 2.5 M HCl

Since we are given two volumes and two molarities of the same chemical, this is a simple dilution. The formula for dilutions: M1V1 = M2V2

M1V1 = M2V2

(10.5M)(V1) = (2.5M)(14.6)

V1 = 3.48 L

35. (Essay) Explain how the addition of a solute affects the vapor pressure, freezing point and boiling point of a solvent. (5.08)

When a solute is added to a solution, it disrupts the vaporization of the solvent particles, thus the vapor pressure is reduced. Since the vapor pressure is reduced, and the atmospheric pressure remains the same, it is ‘harder’ for the solvent particles to escape into the gas phase. Therefore, the boiling point increases. As for the freezing point, it will be lower. When a liquid tries to solidify (freeze), the molecules slow down and arrange themselves into a rigid crystalline structure. When we dissolve a solute, it will interfere with this process, thus preventing the crystals from forming. Since they cannot form, the solution will still drop in temperature, but it won’t freeze. So, the freezing point decreases.

36. (Essay) Two solutions are boiled. The first is 2 mol of a calcium chloride (CaCl2) solution forming two ions. The second one is 2 mol of a sugar (C6H12O6) solution forming. How

will the boiling points of the two solutions compare? (5.08)

In order to determine which substance will have the greates affect, we must look at how many particles will be in solution. Ionic compounds break apart into their respective ions in solution; covalent compounds do not.

CaCl2 is an ionic compound; it will break apart into the calcium and chloride ions. When

we look at the formula, we see that we have 1 calcium ion and 2 chloride ions. This means that for every mole of CaCl2 that dissolves, THREE moles of ions form.

Remember the subscript tells you how many. CaCl2 Ca+ + 2Cl-

For colligative properties, we need the total number of particles in solution:

2 moles CaCl2 X 3 moles ions = 6 moles of ions (particles)

1 mole CaCl2

C6H12O6 is a covalent compound. Therefore, when it is dissolved, it stays together.

So, only 1 particle is in solution. So, if we start with 2 moles of C6H12O6 we will have 2

moles of particles in solution.

When we compare them, the CaCl2 solution will have the greatest effect on the

colligative properties because it has the greater number of particles in solution.