Study Guide

Chemistry
Sample Questions

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Competency 0001
Analyze the properties of matter.

1. Archaeologists estimate the age of ancient organic artifacts by using carbon 14 dating and then corroborating with other techniques. Which of the following is the most likely explanation for the use of carbon 14 dating over the uranium–lead dating method?

  1. The carbon 14 technique does not destroy the sample.
  2. The half-life of the carbon nuclei in the samples is shorter.
  3. The organic samples do not contain detectable amounts of carbon 14.
  4. The uranium–lead approach requires the sample to be destroyed.
Enter to expand or collapse answer.Answer expanded
Correct Response: B. Uranium-lead dating is based on the half-lives of uranium 235 (704 million years) and uranium 238 (4.47 billion years). Organic artifacts degrade and disappear over time and so tend to be found when they are much younger than these half-lives. Therefore, uranium-lead dating is not well suited for dating most organic artifacts, but is better suited for dating very old rock formations. Conversely, carbon 14 has a half-life of 5,740 ± 40 years, which makes it much more ideal for dating these artifacts.

Competency 0002
Analyze chemical bonding and matter interactions and properties.

2. The alloy duralumin open parens 4 percent C U, 1 percent M G, 1 percent M N, and 94 percent A L close parens is used in the construction of aircraft bodies, racing bikes, and boats. Which of the following properties makes the alloy most suitable for these applications?

  1. hard and resistant to corrosion
  2. low density and very strong
  3. lustrous and unreactive
  4. malleable and inexpensive
Enter to expand or collapse answer.Answer expanded
Correct Response: B. To construct aircraft bodies, racing bikes, and boats, weight to strength ratios need to be as low as possible. Although all of these properties can pertain to duralumin, low density and high strength are the primary reasons for choosing this material for these applications.

Competency 0003
Apply knowledge of the structure and naming of chemical substances and periodic properties of the elements.

3. Use the table below to answer the question that follows.

Element N A M G A L S I P S C L
Atomic radius/10 to the negative 12th M 186 160 143 117 110 104 99

The primary reason for the decrease in ionic radius as the elements of a row on the periodic table are examined is:

  1. effective nuclear charge.
  2. electron-electron repulsion.
  3. number of neutrons in the nucleus.
  4. shielding by the valence electrons.
Enter to expand or collapse answer.Answer expanded
Correct Response: A. When moving from left to right across a row of the periodic table, the number of protons in the nucleus increases with each element. Hence the positive charge of the nucleus increases. This results in an increasing attractive force on the negatively charged electrons in the outer regions of the atom. This results in electrons being pulled in more tightly, causing a decrease in ionic radius.

Competency 0004
Apply knowledge of the mole and mass relationships in chemical changes.

4. Given that a cough syrup contains 0.15 g of the active ingredient, dextromethorphan  open parens H B r close parens , in 112 mL, which of the following is the gram amount of this agent contained in each 5 mL dose of the syrup?

  1. 3.4 times 10 to the negative 3
  2. 6.7 times 10 to the negative 3
  3. 1.3 times 10 to the negative 2
  4. 1.0 times 10 to the negative 1
Enter to expand or collapse answer.Answer expanded
Correct Response: B. Concentration is defined as mass over volume. Using equivalent ratios, the concentration can be represented as M sub 1 over V sub 1 = M sub 2 over V sub 2. Here, we can define m sub 1 = 0.15 g, v sub 1 = 112 mL and v2 = 5 mL. Rearrange the equivalent ratios equation for the unknown gram amount (m sub 2) gives m sub 2 = M sub 1 times V sub 2 over V sub 1 = open parens 0.15 G close parens, open parens 5 M L close parens over 112 M L or m sub 2 = 0.0067 g or m sub 2 =  6.7 times 10 to the negative 3 G .

Competency 0005
Analyze energy transfers in chemical and physical processes.

5. Use the table below to answer the question that follows.

Reaction Enthalpy Change (K J)
2 C sub 8 H sub 18 open parens L close parens  + 25 O sub 2 (G)  yields 16 C O sub 2 open parens G close parens  + 18 H sub 2 O open parens G close parens  delta H degrees equals negative 10942 K J
2C O open parens g close parens + O sub 2 open parens g close parens yields 2C O sub 2 open parens g close parens delta H degrees equals negative 566.0 K J

Octane,  C sub 8 H sub 18, is a major component of gasoline and burns in an oxygen poor environment according to the following equation.

2 C sub 8 H sub 18 (L)  + 17 O sub 2 (G)  yields 16 C O (G)  + 18 H sub 2 O (G) 

According to the data provided, which of the following values correctly predicts the approximate enthalpy change of incomplete combustion of  C sub 8 H sub 18 ?

  1. negative 5,750 k J
  2. negative 6,410 k J
  3. negative 10,400 k J
  4. negative 15,500 k J
Enter to expand or collapse answer.Answer expanded
Correct Response: B. Enthalpy of reaction is a state function, which means that the value does not depend on the pathway taken to get there. So, it is possible to calculate the enthalpy change by using the enthalpy change for the complete combustion reaction, and the enthalpy change for the reverse of the C O to C O sub 2 conversion reaction that are given in the table. Any enthalpy values used in the calculation must be adjusted by the same factors used to adjust the two reactions to get the final balanced equation, as shown here:
2 C sub 8 H sub 18 (L)  + 25 O sub 2 (G)  yields 16 C O sub 2 (G)  + 18 H sub 2 O (G)  corresponds to delta H degrees = negative 10942 K J 
8(2 yields 2 C O (G)  +  O sub 2 (G) ) corresponds to 8 times delta H degrees = 8 times 566.0 K J 
Adding these chemical equations gives:
2 C sub 8 H sub 18 (L)  + 16 C O sub 2 (G)  + 25 O sub 2 (G)  yields 16 C O sub 2 (G)  + 8 O sub 2 (G)  + 16 C O (G)  + 18 H sub 2 O (G) 
The 16 C O sub 2 's cancel on both sides, and the 8 O sub 2 's on the right side can be subtracted from the 25 O sub 2 's on the left side, giving the net equation originally given in the question. This means that the enthalpy becomes:  delta H degrees equals negative 10942 K J plus 8 times 566.0 K J equals negative 6414 K J , or approximately negative 6410 K J.

Competency 0006
Analyze the types of chemical reactions and their causes.

6. Which of the following represents the balanced equation when ethanol is completely burned in air?

  1.  C sub 2 H sub 5 O H (L)  + 3 O sub 2 (G)  yields 2 C O sub 2 (G)  + 3 H sub 2 O (G) 
  2. 2 C sub 2 H sub 5 O H (L)  +  O sub 2 (G)  yields 4 C O sub 2 (G)  + 3 H sub 2 O (G) 
  3. 2 C sub 2 H sub 5 O H (L)  + 3 O sub 2 (G)  yields 4 C O sub 2 (G)  + 6 H sub 2 O (G) 
  4. 2 C sub 2 H sub 5 O H (L)  + 7 O sub 2 (G)  yields 4 C O sub 2 (G)  + 6 H sub 2 O (G) 
Enter to expand or collapse answer.Answer expanded
Correct Response: A. Based on the principles of conservation of matter, the total mass of reactants must equal the total mass of products in a chemical reaction. In order to balance the chemical reaction, there must be an equal number of atoms for each element present on both sides of the equation. To get the total amount of products and reactants, the subscripts are used to determine how many atoms are in a single molecule. For multiple molecules, such as 3 O sub 2 , the number of molecules is multiplied by the number of atoms of each element.

Competency 0007
Apply knowledge of oxidation and reduction reactions and electrochemical cells.

7. A  Z N, C U  cell is built by a group of students, and when tested, a voltmeter read negative 1.10 volts. Which of the following actions should the students take to produce positive voltage?

  1. increasing the concentrations of the  C U to the (2 positive)  and the  Z N to the (2 positive)  ions
  2. polishing the  Z N  and the  C U  electrodes with cotton wool
  3. replacing the salt bridge electrolyte with  N A sub 2 S O sub 4 (A Q) 
  4. reversing the voltmeter leads contact with the electrodes
Enter to expand or collapse answer.Answer expanded
Correct Response: D. A  Z N, C U  cell is a chemical battery that has a copper electrode and a zinc electrode. Because of its greater electronegativity, the zinc electrode will always be the negative side of the battery. Therefore, the only way to change the sign of the voltage reading would be to reverse the voltmeter leads with respect to the electrodes.

Competency 0008
Apply knowledge of chemical equilibrium.

8. Use the equation below to answer the question that follows.

 C O (G)  + 2 H sub 2 (G)  in equilibrium with  C H sub 3 O H (g)      delta H degrees equals negative 91 K J

Methanol, an alternative biofuel for internal combustion engines, is produced according to the equation shown. Which of the following actions will most likely increase the yield of  C H sub 3 O H (g)  at equilibrium?

  1. continuous removal of C O open parens g close parens from the system
  2. controlled addition of a catalyst to the system
  3. incremental increase in the system's volume
  4. moderate cooling of the reaction system
Enter to expand or collapse answer.Answer expanded
Correct Response: D. Since delta H is less than 0, the reaction is exothermic. In an exothermic reaction, the equilibrium constant, K sub c, decreases with a decrease in temperature. This causes the reaction to shift to the right. Therefore, moderate cooling will result in an increased yield of  C H sub 3 O H (g) .

Competency 0009
Demonstrate knowledge of the kinetic molecular theory, the nature of phase changes, and the gas laws.

9. Which of the following descriptions of a region of any phase diagram is always true?

  1. A substance in the gaseous phase exists where temperature is below the triple point value and pressure is above the critical point value.
  2. A substance in the supercritical fluid state exists where temperature and pressure are above the critical point of the equilibrium phase diagram.
  3. A substance in the solid phase exists where temperature is above the critical point value and pressure is below the critical point value.
  4. A substance at the triple point value exists where temperature and pressure are above the critical point value.
Enter to expand or collapse answer.Answer expanded
Correct Response: B. Supercritical fluids exist in a region of the phase diagram where distinct liquid and gas phases do not exist. This behavior occurs above a pressure and temperature in the equilibrium phase diagram that is defined as the critical point. At and above this critical point temperature and pressure, the kinetic energy of the substance is high enough so that even higher pressures cannot condense it into a well-defined liquid phase.

Competency 0010
Demonstrate knowledge of the acid and base nature of substances.

10. A weak base is titrated against a strong acid and the equivalence point is reached at an approximate pH of 5.2. A chemist could best use which of the following indicators to monitor this titration?

  1. crystal violet ( K sub A equals 2.0 times 10 to the negative 2 )
  2. methyl orange ( K sub A equals 4.0 times 10 to the negative 4 )
  3. methyl red ( K sub A equals 1.0 times 10 to the negative 5 )
  4. phenolphthalein ( K sub A equals 4.0 times 10 to the negative 10 )
Enter to expand or collapse answer.Answer expanded
Correct Response: C. The indicator will change color when the pH of the solution is equal to the  P K sub A  of the indicator. Hence, in this application, a  P K sub A  value of 5.2 would be ideal. To calculate  K sub A  from P K sub A, use the following equation:  K sub A equals 10 open parens negative P K sub A close parens . Therefore,  K sub A equals 10 open parens negative 5.2 close parens , or  K sub A equals 1.0 times 10 open parens negative 5.2 close parens . This is approximately equal to the K sub A  value of methyl red.

Competency 0011
Analyze the properties of solutions and other aqueous mixtures.

11. To make usable solutions for a lab procedure, a teacher begins with 500 mL of a 3 M stock solution. When performing dilutions using only a 1000 mL volumetric flask, which of the following concentrations is possible?

  1. 2.5 M
  2. 2 M
  3. 1.8 M
  4. 0.5 M
Enter to expand or collapse answer.Answer expanded
Correct Response: D. In order to obtain exact concentrations, a portion of the 500 mL stock solution must be precisely drawn from the stock container and added to the 1000 mL volumetric flask. Then solvent is added to fill the volumetric flask to the final volume. With dilution, the number of moles of solute does not change with dilution, only concentration changes. Number of moles at any concentration can be calculated as concentration (c) of solute times volume (v) of solvent. Since the number of moles is unchanged, initial and final concentrations and volumes can be related by  C sub I times V sub I equals C sub F times V sub F . In this case, C sub I equals 3 M, V sub I equals 500 mL is the greatest possible initial volume. Solving for C sub F gives C sub f equals C sub I times V sub I all divided by V sub F = 3 M times 500 M L all divided by 1000 M L = 1.5 M maximum possible concentration. Therefore, no dilution can create a concentration that is greater than 1.5 M.

Competency 0012
Apply knowledge of physics concepts.

12. Two objects have charges of q sub 1 and q sub 2 respectively and are separated by a distance of 2 m so that they experience a force of magnitude F. The situation changes so that the distance between the objects is reduced to 1 m, and the charge of the first object is doubled. Which of the following expressions best represents the electric force between the two objects in the new situation?

  1. 2F
  2. 4F
  3. 8F
  4. 16F
Enter to expand or collapse answer.Answer expanded
Correct Response: C. Coulomb's law can be used to quantify the force between two charged objects using the equation F equals open parens K sub e Q sub 1 Q sub 2 close parens over open parens r squared close parens in which F is the force between the two charges, Q sub 1 and Q sub 2 are the values representing the amount of charge in each particle, r is the distance between the particles, and K sub E is Coulomb's constant. In this example, the distance is reduced by half and the charge on one of the objects is doubled, so these relative values can be used to determine the change in force by using the equation

Competency 0013
Apply knowledge of biology concepts.

13. Woodpeckers live and breed in deciduous forests and feed on nuts, berries, and insects. Which of the following density-independent factors would affect the carrying capacity of a population of woodpeckers in a forest ecosystem?

  1. introduction of blue jays, which are competitors for nuts and insects
  2. drought conditions resulting in a decrease in forest insect populations
  3. availability of decayed trees that can be used as suitable nesting sites
  4. predation by domestic cats associated with homes bordering the forest
Enter to expand or collapse answer.Answer expanded
Correct Response: B. Carrying capacity is the maximum number of individuals in a species that an environment can support, given current conditions. The carrying capacity of a given population is the result of the interaction of density-dependent and density-independent factors. Density-independent factors, such as weather and climate, affect a population in a given location regardless of that population's size. Here, part of the woodpecker's food source has been limited due to weather, which in turn decreases the carrying capacity of the area.

Competency 0014
Apply knowledge of Earth science concepts.

14. Which of the following planets has an atmosphere that mainly consists of carbon dioxide?

  1. Saturn
  2. Mercury
  3. Jupiter
  4. Venus
Enter to expand or collapse answer.Answer expanded
Correct Response: D. The Venetian atmosphere is composed of approximately 96.5 percent carbon dioxide with nitrogen present as the second largest atmospheric component and trace gases comprising less than 1 percent of the atmosphere. As a result of the presence of large amounts of this greenhouse gas, Venus is the hottest planet in the solar system.