Segment 2 Exam Review - ANSWER KEY 
Chemistry V22; ID 5657
The Element Song!
Multiple Choice:

1.	Which of the following is a correct step to convert 3.56 moles calcium to atoms of calcium? (5.01)
a.	Multiply 3.56 moles Ca by its molar mass
b.	Divide 3.56 moles Ca by its molar mass
c.	Multiply 3.56 by Avogadro’s number			
d.	Divide  3.56 by Avogadro’s number

For this one, we don’t need molar mass as we are neither  given grams nor asked for grams.The only way to convert to/from moles and atoms is using Avogadro’s number.  Since we are given moles, we need to have moles on the bottom of the fraction so the units cancel. 

3.56 moles Ca  x    6.02 x 1023 atoms Ca  =  2.14 x 1024 atoms Ca
                                     1 mol Ca

2.	How many moles are in 356 grams of gold, Au? (5.01)
a.	70117 moles Au			c.  2.56 moles Au
b.	1.81 moles Au			d. 52300 moles Au

  356 grams Au  x   1 mol Au    = 1.81 moles Au
                             196.96 g Au

3.	Calculate the number of molecules in 25.6 grams of carbon dioxide, CO2.  (5.02)
a.	3.50 X 1023 molecules CO2		c. 6.78  X 1026 molecules CO2
b.	1.87 X 10-211 molecules CO2		d. 1.54 X 1025 molecules CO2

If we have grams, we must convert it to moles first by dividing by its molar mass. Then, we multiply by Avogadro’s number

25.6 g CO2    x        1 mol CO2       x          6.02 x 1023 atoms CO2    = 3.50 X 1023 atoms CO2
	      44 g CO2		   1 mol CO2

4.	How many grams of water are in 2.45 X 1024 molecules of water? (5.02)
a.	54.5 g H2O 			c. 1.47 g H2O 
b.	2.65 X 1048 g H2O 		d. 73.3 g H2O 

2.45 X 1024 molecules H2O   x          1 mol  H2O           x    18 g H2O  = 73.25 g H2O 
                                      6.02 x 1023 molecules      1 mol H2O

5.	If we use four moles of Na2S for the following reaction, which of the following will be true? (5.04)  
Na2S  + MgCl2 → 2NaCl +  MgS
a.	Eight moles of sodium chloride will be formed
b.	Four moles of magnesium sulfide will be formed
c.	It will require four moles of magnesium chloride
d.	All are correct

We need to look at the molar ratios to determine how much we will use/make of each chemical. Once you get comfortable with ratios,  you can actually do this one in your head!  For every 1 moles of Na2S, we will use 1 mole MgCl2 and form 2 moles NaCl and 1 mol MgS.

4 moles Na2S  X   2 mol NaCl = 8 moles NaCl
    1 mole Na2S

4 moles Na2S  X  1 mole MgS  = 4 moles MgS
   1 mole Na2S

4 moles Na2S  X   1 mole MgCl2  = 4 moles MgCl2
    1 mole Na2S

6.	When 25.6 grams of sodium reacts with an excess of water, as shown in the chemical equation below, how many moles of sodium hydroxide will be produced? (5.04) 

Na(s) + H2O (l) → NaOH (aq) + H2(g)

a.	588 moles NaOH		c.  0.557 moles NaOH
b.	1.11 moles NaOH		d.  2.22 moles NaOH

We need a balanced equation, so we have to do that first:    2Na (s) + 2H2O (l) → 2NaOH (aq) + H2(g)

Now we do the stoichiometry, starting with what was given and “follow the units”.  Whatever unit/chemical we start with is the same unit/chemical that goes on the bottom of the fraction.

25.6 g Na  X  1 mol Na     X    2 mol NaOH  =  1.11 moles NaOH
                     22.98 g Na            2 mol Na 
7.	If 52.4 grams of magnesium oxide reacts with 86.5 grams of hydrochloric acid according to the following reaction, what is the maximum amount of  magnesium chloride that can be formed?  (5.06)
 MgO + HCl →  MgCl2 + H2O
a.	124 g MgCl2			c. 113 g MgCl2
b.	245 g MgCl2			d. 75 MgCl2
We need a balanced equation, so we have to do that first:  MgO + 2HCl →  MgCl2 + H2O
  
Now we do the stoichiometry to calculate the theoretical yield, but we are given exact amounts of both reactants.  So, this is a limiting reactant problem! 
 
We must do two stoichiometry problems to determine which reactant will produce the least amount of product.  That will be the limiting reactant, and that will be how much of our product we ‘theoretically’ can produce.  Since we need grams, we will calculate all the way to grams of our product:
 
52.4 g MgO X   1 mol MgO    X    1 mol MgCl2   X  95.211 g MgCl2  = 123.78 g MgCl2
		      40.305 g MgO        1 mol MgO          1 mol MgCl2   
 
86.5 g HCl  X   1 mol HCl    X    1 mol MgCl2   X  95.211 g MgCl2  = 112.95 g MgCl2
		    36.458 g HCl        2 mol HCl          1 mol MgCl2   

HCl produces a significant amount less of the product than MgO.  Therefore, the limiting reactant is HCl and the theoretical yield of MgCl2 is: 113g

8.	Refer to question #7.  When we calculate the maximum amount of the product we can make, what term do we refer to that as? (5.06)
a.	Percent Yield			c. Actual Yield
b.	Theoretical Yield		d. Yield

9.	Which of the following is an incorrect rearrangement of the percent yield formula? (5.07)
a.	Actual yield =  (percent yield)(theoretical yield)
                                              100

b.	Percent yield =     actual yield       X   100
                         theoretical yield

c.	Theoretical yield  =   actual yield    X 100
                                   percent yield

d.	Actual Yield =     percent yield      X 100
                           theoretical yield

10.	During a laboratory experiment, a 4.68 g sample of potassium chlorate was  decomposed to form potassium chloride and oxygen gas. After reaction, the percent yield of potassium chloride was calculated to be 83.5%. What was the actual yield from the reaction?  (5.07)  
2KClO3 (s)  → 2KCl (s)  +  3O2 (g)
a.	2.85 g KCl				c. 120%
b.	74%					d. 2.38 g KCl

Since we are given information about one chemical and asked to solve for another, we need a balanced equation. We are given it:  2KClO3 (s)  → 2KCl (s)  +  3O2 (g)

Make a list of what you are given and asked for.  Start with what you are given, then “follow the units”

4.68g KClO3
Actual yield = ? 

In order to solve this problem, we need the formula for percent yield.  In the formula we need the theoretical yield and the actual yield.  The limiting reactant was given to us, 4.68 KCLO3, so we need to use stoichiometry to calculate the theoretical yield.  That is how much we should be able to make.

4.68 g KClO3   X   1 mol KClO3  X   2 mol KCl    X     74.55 g  KCl = 2.85 g KCl
                          122.52 g KClO3     2 mol KClO3        1 mol KCl

If you chose 2.85g KCl as your answer, it is not correct. This is the theoretical yield.  This is how much we SHOULD HAVE been able to make.  

Now we know the theoretical yield and the percent yield. We can rearrange our formula to solve for the actual yield: 

Percent yield =     actual yield       X   100
                         theoretical yield

Actual yield =  (percent yield)(theoretical yield)
                                              100
Actual yield =  (83.5) (2.85)
                                100

Actual yield = 2.38 g KCl

11.	If the reaction of 4.75 moles of hydrogen gas with excess nitrogen gas produced an 86.5% yield of ammonia, what was the actual yield of ammonia?  (5.07) 
2H2 + N2 → 3NH3
a.	3.66 moles NH3				c. 3.17 moles NH3
b.	2.74 moles NH3				d. 1.37 moles NH3

Percent yield =     actual yield       X   100
                         theoretical yield

Actual yield =  (percent yield)(theoretical yield)
                                              100

In order to use this equation, we must calculate the theoretical yield.  This is calculated using stoichiometry to convert moles of H2 to moles of NH3. We need to have a balanced equation.  Balanced equation:      3H2 + 2N2   → 2NH3

	4.75 H2 X  2 moles NH3 = 3.167 moles NH3               This is the theoretical yield.
      		        3 mol H2
 
Actual yield =  (percent yield)(theoretical yield)
                                              100

Actual yield =  (86.5)(3.167 mol NH3)
                                              100

Actual Yield = 2.74 moles NH3

12.	Compare kinetic and potential energy; include an example of each. (6.01)  Potential energy is the ‘stored’ energy ready for use.  Kinetic energy is the energy of movement.  An example of this conversion is when a ball that is resting at the top of a hill (potential energy) starts rolling downhill (converted to kinetic energy). Be sure to describe all types of energy listed in the lesson and give examples.

13.	Define temperature. Give three units to measure temperature. (6.01)  Temperature is the measure of the average kinetic energy.  As temperature increases, so does the average kinetic energy.  This means the molecules are moving faster. We can measure temperature in ℉ (United States), ℃ (rest of the world) or K (Kelvin, SI unit)
 
14.	Identify the characteristics of a solid, liquid, gas and plasma. (6.02)

15.	Put the states of matter in order in terms of increasing kinetic energy  (6.02)
a.	Solid < liquid < gas < plasma		c. Solid < gas < liquid < plasma
b.	Gas < liquid < gas < plasma		d. Plasma < gas < liquid < solid

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.

16.	Put the states of matter in order in terms of increasing attractive forces (6.02)
a.	Solid < liquid < gas 			c. Liquid < solid < gas 
b.	gas < liquid < solid 			d. Not enough information

The more energy a substance has, the ‘easier’ it is to overcome the attractive forces.  Solids have low energy compared to gasses so it has higher attractive forces.  Gasses have high energy, thus lower attractive forces.

17.	Which of the following is true for solids compared to liquids? (6.02)
a.	Their speed of motion is higher
b.	Their attractive forces are greater
c.	Their kinetic energy is higher
d.	Their ability to bond is greater

18.	Referring to the diagram to the right, which of the following is true at 4 minutes? (6.03)
a.	All of the water is a solid
b.	Most of the solid has melted into a liquid
c.	All of the water is a liquid
d.	The temperature of the liquid is higher than the solid

19.	What happens to the particles of a gas when the gas changes into a solid? (6.03)
a.	The attractive forces between them decreases
b.	They speed up
c.	They move in constant, random motion
d.	The volume decreases

20.	Which of the following are held constant in both Charles’ and Boyle’s Laws? (6.04)
a.	Pressure and number of moles	c. Number of moles are volume
b.	Temperature and pressure		d. Number of moles 

21.	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? (6.04) Note: Be able to use all gas laws to solve for any variable.
a.	1.7 atm				c. 3.5  atm
b.	5.7 atm				d. 10.4 atm
This is Gay-Lussac's law since we have two temperatures and two pressures.  

Make a list of what was given in the problem:

P1 = 1.8 atm
T1 = 275K
P2 = ?
T2 = 260 K
Since we just have pressure and temperature changing, we need the formula with only these two:                        

P1 = P2		Like all formulas, we can rearrange: P2 = (P1)(T2)
T1    T2							            T1

 Now, plug & chug:  P2 = (1.8 atm)(260 K)
   				275 K
	 
		          P2 = 1.7 atm

22.	A sample of nitrogen gas has a volume of 0.75 L at 32 °C and 3.2 atm. What is the volume when the pressure changes to 2.7 atm and temperature is constant?
a.	Not enough information	c. 0.63 L
b.	071 L				d. 0.89 L

Make a list of what you are given:
V1 = 0.75 L
T1 = 32°C
P1 = 3.2 atm
V2 = ? L
T2 = same:  32°C
P2 = 2.7 atm

Looking at the information, temperature is held constant.  We do not need a formula with T in it.  We are changing the pressure and finding the new volume. We need a formula that relates volume to pressure.  This is Boyle’s Law

                     P1V1 = P2V2
(3.2 atm) (0.75 L) = (2.7 atm) (V2)

(3.2 atm) (0.75 L) = V2
       (2.7 atm) 

0.89 L = V2

23.	Which of the following is true for the diagram below? (7.01) Note: be able to label and describe potential energy diagrams for both endothermic and exothermic reactions.

a.	It is an endothermic reaction because the reactants have more energy than the products
b.	It is an exothermic reaction because there is an increase in enthalpy
c.	There is no activation energy needed for this reaction
d.	The enthalpy decreases so this is an exothermic reaction.

24.	Which of the following is true for activation energy? (7.01)
a.	It is the energy released or absorbed in the reaction
b.	It is the energy required to start a reaction
c.	Not all reactions require activation energy
d.	Activation energy cannot be measured

25.	Which of the following is an example of an endothermic process? (7.01)
a.	Freezing water			c.  Photosynthesis
b.	Rusting of iron		d. Combustion reactions

26.	Refer to the diagram on the right.  
a.	It is endothermic because heat is flowing in
b.	It is exothermic because heat is flowing in
c.	It can not be classified without knowing what the liquid is

27.	What is the sign for enthalpy in an exothermic reaction? (7.02)
a.	Negative because the products have more energy than the reactants
b.	Positive because the reactants have more energy than the products
c.	Negative because the reactants have more energy than the products
d.	Positive because the products have less energy than the reactants

28.	45.8 grams of copper requires 1172 J of heat to reach a final temperature of 100.5 ℃.  What was the initial temperature? Specific heat of copper is 0.385 J/g℃. (7.02)
a.	66.5 				c. 34.0 
b.	122 				d. 75.2 

mass: 45.8 g Cu
q:  1172 J
Tf: 100.5℃.  
Specific heat = 0.385 J/g℃		

		q = (mass)(specific heat)(change of temperature)
q = (mass)(specific heat)(final temp - initial temp)
q = (mass)(specific heat)(Tf  - Ti)
We need to solve for “Ti“.  Let’s start to rearrange the formula:

Tf  - Ti =                q             .      	Let’s plug & chug what we know:
			   (mass)(specific heat)

100.5 - Ti  =          1172 J        .
            (45.8 g) (0.385 J/gC)

100.5 - Ti  = 66.5

Ti  = 34.0

29.	How do we determine the change of total enthalpy? (7.02)
a.	ΔH = Hreactants - Hproducts
b.	ΔH = Activation Energy - (Hreactants - Hproducts)
c.	ΔH = Hproducts - Hreactants 
d.	ΔH = Activation Energy - (Hproducts - Hreactants)

30.	Which of the following is the best description of how a catalyst affects the rate of reaction?  (7.04)
a.	Catalysts can lower activation energy which means reactions require less activation energy to begin.
b.	Catalysts increase the temperature which allows for more reactions to occur
c.	Catalysts do not affect rate of a reaction
d.	Catalysts increase the energy of reaction which allows for more reactions to occur.

31.	20 grams of magnesium were placed in a flask with 100 mL of sulfuric acid. Which one of the following would produce 100 mL of hydrogen gas the fastest?  (7.04)
a.	Removing 5 g of magnesium
b.	Removing 50 mL of hydrogen gas
c.	Increasing the temperature by 10℃
d.	Decreasing temperature by 10℃

We know that in order to increase the rate of a reaction, we can increase temperature, increase concentration or add a catalyst.  Option a is decreasing the concentration of a reactant, which will not increase the rate of reaction.  Option b has no influence on the rate as it is a product.  Option c increases the energy, which means more collisions will occur.  Option d would actually slow the reaction a bit. 

32.	List three factors that can increase the rate of reaction. Please explain these factors using the collision theory.  (7.04)  The rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit time. The collision theory provides an explanation for how different factors can influence the rate of a chemical reaction.
●	Temperature: Increasing the temperature of a reaction increases the kinetic energy of the particles, causing them to move faster and collide more frequently. According to the collision theory, reactions occur when molecules collide with enough energy to break the chemical bonds that hold them together. Increasing the temperature increases the number of high-energy collisions, leading to a higher reaction rate.
●	Concentration: Increasing the concentration of reactants increases the number of particles in the reaction, making collisions between reactant molecules more likely. The collision theory explains that a reaction occurs when molecules collide with the correct orientation and with enough energy to break the chemical bonds. By increasing the concentration, the likelihood of successful collisions increases, leading to a higher reaction rate.
●	Catalysts: Catalysts are substances that can increase the rate of a chemical reaction without being consumed in the process. Catalysts work by providing an alternative reaction pathway with lower activation energy. The collision theory explains that reactions occur when molecules collide with enough energy to break the chemical bonds that hold them together. By lowering the activation energy, a catalyst increases the number of successful collisions, leading to a higher reaction rate.

33.	The rate of a chemical reaction may be defined as (7.04)
a.	the change in concentration of any one of the reactants or products
b.	the time it takes for the reaction to be completed
c.	the change in concentration of any one of the reactants only
d.	the change in concentration of any one of the products only

34.	Which of the following best describes whether or not all reactions achieve equilibrium? (7.05)
a.	Yes, all reactions can as long as the concentrations remain the same
b.	No, it must be a reversible reaction in order to reach equilibrium
c.	Yes, as long as there are reactants left in the vessel
d.	No because some are exothermic

35.	Which of the following is true about a system at equilibrium? (7.05)
a.	Rates of the forward and reverse reactions are complete.
b.	Forward and reverse reactions continue reacting.
c.	The reaction is complete. 
d.	Concentrations of the components are equal

36.	What property allows water to get into the pores of a paper towel and move up, against gravity? (8.01)
a.	Surface tension				c.  Polarity
b.	Capillary action				d.  Cohesion

37.	Why is water considered the Universal Solvent? (8.01)
a.	It has a high polarity			c. It is a climate regulator
b.	It has a low surface tension		d.  It has strong cohesion forces

Water is considered the universal solvent because it has the ability to dissolve a wide range of substances, both polar and nonpolar. This is due to its unique molecular structure and the polarity of its molecules.

Water is a polar molecule, meaning that it has a slightly positive end and a slightly negative end. This polarity allows water molecules to interact with other polar molecules by forming hydrogen bonds. These hydrogen bonds make it easier for polar molecules to dissolve in water, such as salt, sugar, and acids.

Water can also dissolve nonpolar substances, such as oils and fats, due to its ability to form temporary hydrogen bonds with nonpolar molecules. 

Since water is so polar and has hydrogen bonds, it is able to dissolve ionic compounds such as salts, which are held together by electrostatic forces. When a salt dissolves in water, the polar water molecules surround and separate the ions, allowing them to move freely in solution.

38.	What is the molarity of a solution prepared by dissolving 20.4 g of sodium hydroxide in enough water to make a total of 255 mL of solution? (8.02)
a.	0.0800 M				c. 0.00200 M
b.	1.67 M					d.  3.34 M

Molarity  =    moles solute
         liters solution   

In the problem, we were given the mass of NaOH.  We will need to convert to moles before we can calculate the molarity. We will use molar mass to do this (recall 5.02)

20.4 g NaOH X   1 mol NaOH  =  0.51 moles NaOH
              40.0 g NaOH   

 We need the volume to be in L:  255 mL X     1 L      =  0.255 L
					          1000 mL

Molarity  =    0.51 moles NaOH  = 1.67 M
           0.255 L solution

39.	How many moles of Na2CO3 are there in 10.0 L of 2.00 M solution?(8.02)
a.	20.0 mol					c. 10.0 mol
b.	5.00 mol					d. 2.00 mol

Molarity  =    moles solute    	Rearrange to solve for mol solute:
         liters solution   
   
Moles solute = (Molarity)(L)
Moles solute = (2.00M)(10.0L)
Moles solute = 20.0 mol

40.	What happens to a solution if we add more water to it? (8.02)
a.	It becomes more acidic			c.  It becomes more diluted
b.	It becomes more concentrated		d.  It reaches equilibrium

41.	If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution. (8.02)
a.	280 M					c.  0.28 M
b.	9.1 M					d.   0.11 M

M1V1 = M2V2

(1.6 M) (0.175 L) = (M2) (1.0 L)

x = 0.28 M

Note that 0.175 L was used rather than 175 mL. Remember to keep the volume units consistent.

42.	What could be the identity of “A”? (8.04/8.05)

a.	HCl					c. NH3
b.	NaCl					d. KOH

43.	Which of the following is the least basic (8.04/8.05)? 
a.	Acetic Acid (Vinegar); pH = 2.5	
b.	Sodium bicarbonate Solution (Baking Soda); pH = 8.5
c.	Kids mouthwash; pH = 5	
d.	Hand soap; pH = 6

44.	Which of the following best describes the picture below? (8.05)

a.	It is a strong acid because all the particles dissociated in solution
b.	It is a weak acid because all the particles dissociated in solution
c.	It is a strong acid because some of the particles dissociated in solution
d.	It is a weak acid because some of the particles dissociated in solution

The following may help for question #45:
Logarithm Review (from lesson)  	 Antilog Review (from lesson)
pH = −log⁡[ H3O+ ]		pOH = −log[ OH− ]		

⁡[ H3O+ ] = 10−pH  		[ OH− ] = 10−pOH

pH + pOH = 14                      [ H3O+ ] × [ OH− ] = 1.0 × 10−14 M

45.	What is the hydroxide concentration of a solution that is pH = 3.7? (8.05)  
a.	0.56 M				c. 1.99 X 10-4 M
b.	5.03 X 10-11 M			d. 0.482 M

We have the pH, and are looking for the hydroxide concentration, ⁡[ OH− ].   Let’s start there. None of the formulas will get me from pH to OH- directly. So, it looks like we have to do two separate calculations. 
		⁡[ H3O+ ] = 10−pH  
⁡[ H3O+ ] = 10−3.7 
⁡[ H3O+ ]  = 1.99 X 10-4 M  
(If you chose this as your answer, be careful.  Question asked for hydroxide, OH−, not hydronium, H3O+)

Now I know the hydronium concentration, I can quickly determine the hydroxide:
[ H3O+ ] × [ OH− ] = 1.0 × 10−14 M

(1.99 X 10-4 M) × [ OH− ] = 1.0 × 10−14 M    Rearrange to solve for hydroxide

1.0 X 10-14 M    =  [ OH− ]            
1.99 X 10-4 M

Now, if you took an advanced math class, you may be able to do fractions with exponents by hand :) If not, use your calculator.

5.03 X 10-11 M = [ OH− ]  

Honors Multiple Choice:

46.	(5.03H) Which of the following does NOT share the same empirical formula?
a.	H2O and H2O2			c.  C4H16 and C2H8
b.	C3H6O3 and C2H4O2		d.  CH2 and C10H20

47.	(5.03H) A compound of iron and oxygen is analyzed and found to contain  69.94%  iron and  30.06%  oxygen. Find the empirical formula of the compound.
a.	Fe4O6.				c.  Fe1O1.5
b.	FeO				d.  Fe2O3

Identify the "given"information and what the problem is asking you to "find."	
%  of  Fe = 69.94% 
%  of  O = 30.06% 

Find: Empirical formula  =Fe?O? 

Calculate: Assume a 100g sample, convert the same % values to grams.	
69.94% Fe = 69.94g Fe			30.06% O = 30.06g O

Now, convert each mass to moles: 
69.94g Fe   ×   1 mol Fe    = 1.252 mol Fe
                       55.85 g Fe
 
30.06g O  ×  1 mol O  = 1.879 mol O
                       16 g O
 
Now we need to determine the ratio of the number of atoms of each element. We can do this by dividing the number of moles of each element by the smallest number of moles.Of the two chemicals, Fe has the smallest number of moles: 1.252.  We will now divide both of the elements by the smallest number of moles: 1.252

	Fe: 1.252 mol = 1		O:  1.879 mol  = 1.5
      1.252 mol			      1.252 mol

The "non-whole number" empirical formula of the compound is  Fe1O1.5 

We want to maintain the ratio as closely as possible.  So, we cannot “just” round up.  1 : 1.5 is very different from 1:2.  Let’s try to multiply each of the moles by the smallest whole number that will convert each into a whole number.   Fe:O = 2 (1:1.5) = 2:3  Two to three, 2:3 is close to 1: 1.5

The empirical formula of the compound is  Fe2O3.

48.	(6.06H) What is the volume of 4.50 moles of oxygen gas (O2) at standard temperature and pressure (STP)? 
a.	101 L O2			c.  202 L O2	
b.	50.5 L O2			d. 150 L O2	.

STP indicates that we are at the following conditions:
Standard temperature: 273 K
Standard pressure: 1 atm

At STP, 1 mole of any gas will occupy 22.4 L of space.

4.50 moles of O2   X    22.4 L  = 100.8 L O2 
		             1 mol

49.	(6.06H) If 52.0 g of magnesium metal reacts with excess hydrochloric acid, how many liters of hydrogen gas can be produced at 285K and 0.97 atm? 
Mg (s) + 2HCl (aq)   → H2(g) + MgCl2 (aq)

a.	1254 L H2			c.  123 L H2
b.	51.6 L H2			d.  74.5 L H2

Make a list of what we are given and asked for:
Mass = 52.0 g Mg
V = ? L H2
T = 285 K
P = 0.97 atm

Notice that they gave us the mass of magnesium, which is not a gas.  We can only use the ideal gas law for gasses. So, we need to convert grams of Mg to moles of H2 before we can use the ideal gas law.

52.0 g Mg  X   1 mol Mg     X    1 mol H2    = 2.13 g H2
                     24.305 g Mg        1 mol Mg

Now we can use the ideal gas law and solve for volume:  V = nRT / P
V = 
V =  (2.13 mol H2)(0.0821 L*atm/mol*K)(285 K)
	        0.97 atm

V = 51.4 L H2

50.	(6.06H) What assumption do we make when considering all gasses ideal? 
a.	We assume there are no attractive forces (intermolecular forces) between gas molecules
b.	We assume that all gas molecules have the same volume
c.	We assume that all gas molecules move in a consistent pattern
d.	We assume there is pressure between gas molecules

51.	(7.03H) What happens to the entropy in the reaction below? 
 CO2 (g) + CaO (s)  → CaCO3 (s)
a.	Increases			c. Decreases
b.	No change			d. Not enough information

52.	(7.03H)  Hot water is poured into the same amount of cold water in a calorimeter.  Which of the following statements is correct? 
a.	Both the cold water and hot water lose heat; neither gains it.
b.	The energy change is the same for both substances.
c.	The cold water and hot water experience a net loss of energy. 
d.	Both the cold water and hot water gain heat; neither loses it.

53.	(7.03H) Which of the following formulas can be used to calculate the final temperature of a system in a calorimeter? 

a.	Tf = (q)(-Ti)			c. Tf =      q    	  +   Ti		 
        (m)(c)			             (m)(c)

b.	Tf = (q)(m)		 	d. Tf = (q)(m)(c) - Ti 
       (c)(-Ti)			             

The original equation is: 
q = (mass)(specific heat)(change of temperature)
q = (mass)(specific heat)(final temp - initial temp)
q = (m)(c)(Tf  - Ti)

We need to rearrange to solve for “Tf“. 

Tf  - Ti =       q        	
			     (m)(c)

Tf    =       q   	+  Ti
			(m)(c)

54.	(7.06H) Under what pressure and temperature conditions can we shift the equilibrium to the right for the reaction below? 
2A + B ↔  C + D     ΔH= −151kJ
a.	High pressure, low temperature		
b.	Pressure will have no effect, low temperature
c.	Low pressure, high temperature		
d.	Pressure will have no effect, high temperature

Because the forward reaction is exothermic, a lower temperature will cause the system to shift farther to the right (toward the products) in order to release more heat energy. You can think of energy as a product.  If we want more

Because the reactant side of the equation has more moles of gas than the product side, increasing the pressure will shift the system farther to the right in order to decrease the pressure of the system.

55.	(7.06H) What will happen to the equilibrium when more 2SO2 (g) is added to the following system? 
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
a.	The equilibrium shifts to the right.		c. The equilibrium shifts to the left
b.	The equilibrium does not change		d. Not enough information

Adding more reactants shifts the equilibrium in the direction of the products; therefore, the equilibrium shifts to the right.

56.	(8.06H) What is the resource that best matches the following description: Heat energy from steam surfaces from circulating groundwater, moving turbines and creating electrical energy.
a.	Geothermal energy			c. Solar energy
b.	Hydropower				d.  Wind power

57.	(8.06H)  Which resource is not a clean energy source, releases greenhouse gasses, but is good for protecting habitats? 
a.	Nuclear energy			c.  Biomass energy
b.	Geothermal energy			d.  Coal