Annual Rhode Island High School Chemistry Contest: Sample Questions
In an effort to prevent an advantage to those with more sophisticated calculators, test questions are designed to require no more than the most elementary calculations. Sample problems that require difficult calculations will be replaced with simpler versions, on the same topic, for the test.
Instructors have frequently asked for sample questions or old exams to get the flavor of future exams. This page provides some sample questions. Clicking on the "Explanation" provides the correct answer and a brief explanation to the question.
1. How many significant figures are there in the measured number 0.0000730500?A. 3
B. 4
C. 6
D. 8
E. 11
Explanation
1. How many significant figures are there in the measured number 0.0000730500?
A. 3
B. 4
C. 6
D. 8
E. 11
Significant figures (or digits) apply to measured numbers as opposed to defined numbers (e.g. the number of inches in a yard) and counted numbers (e.g. five people sat at her table). Multiplication (or division) by counted or defined numbers do not change the number of significant figures.
All non-zero digits are significant. In a number without a decimal point, all zeros between non-zero digits are significant, and no others. In a number with a decimal point, all zeros to the right of the first non-zero digit are significant and no others.
0.0000730500 is a measured number and has a decimal point and three zeros to the right of the first non-zero digit, which are, therefore, significant. It has three non-zero digits which are significant as well. Thus, there are 6 significant figures in the number.
Comment: Significant figures are a fundamental concept in general chemistry and questions about them frequently appear on the contest exam, Cf. Question 9.
2. Which of the following is not an electrolyte when dissolved in water?
A. 3
B. 4
C. 6
D. 8
E. 11
Significant figures (or digits) apply to measured numbers as opposed to defined numbers (e.g. the number of inches in a yard) and counted numbers (e.g. five people sat at her table). Multiplication (or division) by counted or defined numbers do not change the number of significant figures.
All non-zero digits are significant. In a number without a decimal point, all zeros between non-zero digits are significant, and no others. In a number with a decimal point, all zeros to the right of the first non-zero digit are significant and no others.
0.0000730500 is a measured number and has a decimal point and three zeros to the right of the first non-zero digit, which are, therefore, significant. It has three non-zero digits which are significant as well. Thus, there are 6 significant figures in the number.
Comment: Significant figures are a fundamental concept in general chemistry and questions about them frequently appear on the contest exam, Cf. Question 9.
A. Sodium chloride
B. Sugar
C. Sulfuric acid
D. Potassium hydroxide
E. Bicarbonate of soda
Explanation
2. 2. Which of the following is not an electrolyte when dissolved in water?
A. Sodium chloride
B. Sugar
C. Sulfuric acid
D. Potassium hydroxide
E. Bicarbonate of soda
Sugar is the only one of the choices that does not form ions in aqueous solution.
Comment: This problem requires the application of the definition of "electrolyte" and some general knowledge of the substances named.
3. "Amorphous" refers to:
A. Sodium chloride
B. Sugar
C. Sulfuric acid
D. Potassium hydroxide
E. Bicarbonate of soda
Sugar is the only one of the choices that does not form ions in aqueous solution.
Comment: This problem requires the application of the definition of "electrolyte" and some general knowledge of the substances named.
A. a lovers' quarrel.
B. something that acts as an acid or a base.
C. one of three types of calcination which can be performed in a vacuum.
D. a material which is combined with mercury.
E. something lacking definite crystalline structure.
Explanation
3. "Amorphous" refers to
A. a lovers\' quarrel.
B. something that acts as an acid or a base.
C. one of three types of calcination which can be performed in a vacuum.
D. a material which is combined with mercury.
E. something lacking definite crystalline structure.
The first answer is slightly plausible and humorous. The second answer would fit "amphoteric," which has a lot of the same letters as "amorphous." The third answer, while nonsense, has specificity and is the longest answer, making it attractive to those whose test-taking skills exceed their chemical skills. The fourth answer fits "amalgam," which has a few letters in common with "amorphous."
Comment: This is another general knowledge, definition question.
4. What is the pH of a 1.00 × 10-10 M aqueous solution of hydrochloric acid?
A. a lovers\' quarrel.
B. something that acts as an acid or a base.
C. one of three types of calcination which can be performed in a vacuum.
D. a material which is combined with mercury.
E. something lacking definite crystalline structure.
The first answer is slightly plausible and humorous. The second answer would fit "amphoteric," which has a lot of the same letters as "amorphous." The third answer, while nonsense, has specificity and is the longest answer, making it attractive to those whose test-taking skills exceed their chemical skills. The fourth answer fits "amalgam," which has a few letters in common with "amorphous."
Comment: This is another general knowledge, definition question.
A. -10
B. 1
C. 7
D. 10
E. 14
Explanation
4. What is the pH of a 1.00 × 10-10 M aqueous solution of hydrochloric acid?
A. -10
B. 1
C. 7
D. 10
E. 14
pH = log[H+] and [H+] = 1 × 10-7.
Comment: This sort of pH problem is frequently on our exams. It challenges those who uncritically believe whatever numbers appear on their calculators. While most students are instructed that in the case of a strong acid like HCl, the [H+] is equal to the concentration of the acid, this is an exception. pH = -log[H+]. If one plugs in [H+] = 1.00 × 10-10, the pH is 10. But this would mean that adding acid to a neutral solution has made it BASIC: a paradoxical result. At this point, the student should realize that the amount of acid added is to small to influence the pH, and the correct answer is 7, the pH of neutral water. This can also be viewed as a significant figure problem: adding 1.00 × 10-10 to 1.00 × 10-7 does not change the 1.00 × 10-7. It is not a "trick question," especially when published on the Internet.
5. All other things being equal, which of the following acids would be the best choice for making a buffer for pH = 4.26?
A. -10
B. 1
C. 7
D. 10
E. 14
pH = log[H+] and [H+] = 1 × 10-7.
Comment: This sort of pH problem is frequently on our exams. It challenges those who uncritically believe whatever numbers appear on their calculators. While most students are instructed that in the case of a strong acid like HCl, the [H+] is equal to the concentration of the acid, this is an exception. pH = -log[H+]. If one plugs in [H+] = 1.00 × 10-10, the pH is 10. But this would mean that adding acid to a neutral solution has made it BASIC: a paradoxical result. At this point, the student should realize that the amount of acid added is to small to influence the pH, and the correct answer is 7, the pH of neutral water. This can also be viewed as a significant figure problem: adding 1.00 × 10-10 to 1.00 × 10-7 does not change the 1.00 × 10-7. It is not a "trick question," especially when published on the Internet.
A. Formic acid, Molar Mass = 46.03
B. Butanoic acid, M.P. = 4.26 °C.
C. Acetoacetic acid, Ka = 2.62 × 10-4
D. Ethylphenylacetic acid, Ka = 4.27 × 10-5
E. m-chlorocinnamic acid, Ka = 5.13 × 10-5
Explanation
5. All other things being equal, which of the following acids would be the best choice for making a buffer for pH = 4.26?
A. Formic acid, Molar Mass = 46.03
B. Butanoic acid, M.P. = 4.26 °C.
C. Acetoacetic acid, Ka = 2.62 × 10-4
D. Ethylphenylacetic acid, Ka = 4.27 × 10-5
E. m-chlorocinnamic acid, Ka = 5.13 × 10-5
The buffer capacity is greatest when [acid] = [conjugate base]. According to the Henderson-Hasselbach equation,
pH = pKa - log([acid]/[conjugate base]), that would mean that
pH = pKa. The closer the concentrations of acid and conjugate base, the better the buffer. Thus, the closer the pKa to the desired pH, the better the buffer: the closer the Ka to the desired [H+], the better the buffer. A pH of 4.26 corresponds to an [H+] of 5.49 × 10-5, which is closest to m-chlorocinnamic acid, based on the information available.
Comment: This question tests, among other things, if the student knows what information is relevant: numbers close to 4.26, in one form or another, appear in most of the answers. Other questions on exams, while having all necessary information, may contain some unnecessary information as well. Life is an open-book exam: you have to know what information you need and what you don't. Problems, which contain unnecessary information, challenge those students who feel compelled to use every number in a problem. It is a weakness in tests (and homework) when questions contain only needed information. It fails to develop discrimination.
(Before the popularity of the ICE chart, the author used a “Stuff chart.”)
7.1 × 10-6 = x2/0.200
x2 = 7.1 × 10-6 × 0.200 = 1.42 ×10-6
x = 1.19 × 10-3
Comment: This is a straight-forward equilibrium problem. Other equilibrium problems may require use of either the iterative method or the quadratic formula. The "check" is that, when x has been calculated, the value of HA, 0.200 - x, is within 5% of the approximation used. Should this check fail, the above-mentioned methods would be required.
Another feature of interest is the wrong answers. Some test-takers, when they know no better, simply perform mathmatical operations until they get an answer that is in agreement with one of the multiple choices. For such students, answer A is x2, where the test-taker fails to take the square root. Similarly, answer B is the result of dividing Ka by 0.200 instead of multiplying and answer E is the square root of that quotient. Having run out of plausible mathematical operations, answer D is designed to attract those whose test-taking skills exceed their chemistry skills (chemo-savvy, if you remember the Lone Ranger). It has the 1.4 part of answer A and the same exponent as answers C and E. It will seem like the best guess.
6. What is the hydronium ion concentration of a 0.200 M solution of 3,6-dinitrophenol (Ka = 7.1×10–6)?
A. Formic acid, Molar Mass = 46.03
B. Butanoic acid, M.P. = 4.26 °C.
C. Acetoacetic acid, Ka = 2.62 × 10-4
D. Ethylphenylacetic acid, Ka = 4.27 × 10-5
E. m-chlorocinnamic acid, Ka = 5.13 × 10-5
The buffer capacity is greatest when [acid] = [conjugate base]. According to the Henderson-Hasselbach equation,
pH = pKa - log([acid]/[conjugate base]), that would mean that
pH = pKa. The closer the concentrations of acid and conjugate base, the better the buffer. Thus, the closer the pKa to the desired pH, the better the buffer: the closer the Ka to the desired [H+], the better the buffer. A pH of 4.26 corresponds to an [H+] of 5.49 × 10-5, which is closest to m-chlorocinnamic acid, based on the information available.
Comment: This question tests, among other things, if the student knows what information is relevant: numbers close to 4.26, in one form or another, appear in most of the answers. Other questions on exams, while having all necessary information, may contain some unnecessary information as well. Life is an open-book exam: you have to know what information you need and what you don't. Problems, which contain unnecessary information, challenge those students who feel compelled to use every number in a problem. It is a weakness in tests (and homework) when questions contain only needed information. It fails to develop discrimination.
(Before the popularity of the ICE chart, the author used a “Stuff chart.”)
| Stuff | []initial | Change | []final | Approximation | Check |
|---|---|---|---|---|---|
| HA | 0.200 | -x | 0.200-x | 0.200 | 0.200 - .00119 ~ 0.200 -- checks |
| H+ | 0 | x | x | ||
| A- | 0 | x | x |
7.1 × 10-6 = x2/0.200
x2 = 7.1 × 10-6 × 0.200 = 1.42 ×10-6
x = 1.19 × 10-3
Comment: This is a straight-forward equilibrium problem. Other equilibrium problems may require use of either the iterative method or the quadratic formula. The "check" is that, when x has been calculated, the value of HA, 0.200 - x, is within 5% of the approximation used. Should this check fail, the above-mentioned methods would be required.
Another feature of interest is the wrong answers. Some test-takers, when they know no better, simply perform mathmatical operations until they get an answer that is in agreement with one of the multiple choices. For such students, answer A is x2, where the test-taker fails to take the square root. Similarly, answer B is the result of dividing Ka by 0.200 instead of multiplying and answer E is the square root of that quotient. Having run out of plausible mathematical operations, answer D is designed to attract those whose test-taking skills exceed their chemistry skills (chemo-savvy, if you remember the Lone Ranger). It has the 1.4 part of answer A and the same exponent as answers C and E. It will seem like the best guess.
A. 8.4×10–4 M
B. 1.2×10–3 M
C. 2.7×10–3 M
D. 6.0×10–3 M
E. 8.4×10–3 M
Explanation
6. What is the hydronium ion concentration of a 0.200 M solution of 3,6-dinitrophenol (Ka = 7.1×10–6)?
A. 8.4×10–4 M
B. 1.2×10–3 M
C. 2.7×10–3 M
D. 6.0×10–3 M
E. 8.4×10–3 M
Let 3,6-dinitrophenol = HA. Then:
HA(aq) + H2O(L) ⇔ H3O+(aq) + A–(aq)
Ka = [H3O+][A–]/[HA] = 7.1×10–6
7.1×10–6 = x2/0.200, or x = [H3O+] = 1.2×10–3 M.
Check: [HA] = 0.200 – 0.0012 = 0.199 M ~ 0.200 (less than 5 % error).
7. The solubility of PbBr2 in pure water at 20 °C is 0.884 g/100 mL. What is Ksp for lead bromide at this temperature?
A. 8.4×10–4 M
B. 1.2×10–3 M
C. 2.7×10–3 M
D. 6.0×10–3 M
E. 8.4×10–3 M
Let 3,6-dinitrophenol = HA. Then:
HA(aq) + H2O(L) ⇔ H3O+(aq) + A–(aq)
Ka = [H3O+][A–]/[HA] = 7.1×10–6
| [HA] | [H3O+] | [A–] | |
|---|---|---|---|
| Initial | 0.200 | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 0.200 - x ~ 0.200 | x | x |
7.1×10–6 = x2/0.200, or x = [H3O+] = 1.2×10–3 M.
Check: [HA] = 0.200 – 0.0012 = 0.199 M ~ 0.200 (less than 5 % error).
A. 2.8
B. 6.9 × 10-1
C. 2.4 × 10-3
D. 5.6 × 10-5
E. 2.8 × 10-5
Explanation
7. The solubility of PbBr2 in pure water at 20 °C is 0.884 g/100 mL. What is Ksp for lead bromide at this temperature?
A. 2.8
B. 6.9 × 10-1
C. 2.4 × 10-3
D. 5.6 × 10-5
E. 2.8 × 10-5
Ksp:
PbBr2 = Pb2+ + 2 Br-
Ksp = [Pb2+][Br-]2
Ksp = (s)(2s)2 = 4s3
Solubility of Lead Bromide:
s = 0.884 g/100 mL = 8.84 g/L
Molar mass of Lead Bromide
Pb: 207.2 × 1 = 207.2
Br: 79.90 × 2 = 159.8
207.2 + 159.8 = 367.0 g/mole
Molar solubility of Lead Bromide
s = 8.84 g/L ÷ 367.0 g/mole = 0.024 moles/L
Ksp = 4s3 = 4 × (0.024)3 = 5.6 × 10-5
Comments: This is a relatively straight-forward solubility product problem, except that the solubility must be changed from grams/100 mL to moles per liter. Failing to do so leads to answers A, and B if one neglects to multiply by four. Answer C is the molar solubility. Forgetting to square the two in the [Br-] gives answer E, which has the same mantissa as answer A and exponent as answer D, making it an attractive uninformed guess. In a written rather than multiple choice exam, test-takers do not get clues if their answers are wrong. A multiple choice exam should, similarly, give as few clues as possible. Thus, the wrong answers should reflect likely errors.
8. In the following, which is the oxidizing agent?
A. 2.8
B. 6.9 × 10-1
C. 2.4 × 10-3
D. 5.6 × 10-5
E. 2.8 × 10-5
Ksp:
PbBr2 = Pb2+ + 2 Br-
Ksp = [Pb2+][Br-]2
Ksp = (s)(2s)2 = 4s3
Solubility of Lead Bromide:
s = 0.884 g/100 mL = 8.84 g/L
Molar mass of Lead Bromide
Pb: 207.2 × 1 = 207.2
Br: 79.90 × 2 = 159.8
207.2 + 159.8 = 367.0 g/mole
Molar solubility of Lead Bromide
s = 8.84 g/L ÷ 367.0 g/mole = 0.024 moles/L
Ksp = 4s3 = 4 × (0.024)3 = 5.6 × 10-5
Comments: This is a relatively straight-forward solubility product problem, except that the solubility must be changed from grams/100 mL to moles per liter. Failing to do so leads to answers A, and B if one neglects to multiply by four. Answer C is the molar solubility. Forgetting to square the two in the [Br-] gives answer E, which has the same mantissa as answer A and exponent as answer D, making it an attractive uninformed guess. In a written rather than multiple choice exam, test-takers do not get clues if their answers are wrong. A multiple choice exam should, similarly, give as few clues as possible. Thus, the wrong answers should reflect likely errors.
3 Fe2+ + CrO42– + 8 H+ → 3 Fe3+ + Cr3+ + 4 H2O
A. Fe2+
B. CrO42–
C. H+
D. Cr3+
E. H2O
Explanation
8. In the following, which is the oxidizing agent?
3 Fe2+ + CrO42– + 8 H+ → 3 Fe3+ + Cr3+ + 4 H2O
A. Fe2+
B. CrO42–
C. H+
D. Cr3+
E. H2O
An oxidizing agent causes some other species to be oxidized by gaining electrons. In this case, the chromate ion, CrO42– acts as the oxidizing agent.
Comment: this question determines if a student knows the definition of a common term used in redox chemistry.
9. Eight people standing on a scale have a total weight of 1118 pounds. To the correct number of significant figures, what is the average weight of the people on the scale?
3 Fe2+ + CrO42– + 8 H+ → 3 Fe3+ + Cr3+ + 4 H2O
A. Fe2+
B. CrO42–
C. H+
D. Cr3+
E. H2O
An oxidizing agent causes some other species to be oxidized by gaining electrons. In this case, the chromate ion, CrO42– acts as the oxidizing agent.
Comment: this question determines if a student knows the definition of a common term used in redox chemistry.
A. 100
B. 140
C. 139.8
D. 139.75
E. 1339.7
Explanation
9. Eight people standing on a scale have a total weight of 1118 pounds. To the correct number of significant figures, what is the average weight of the people on the scale?
A. 100
B. 140
C. 139.8
D. 139.75
E. 1339.7
1118/8 = 139.75, but there are only 4 sig figs in 1118. Eight is a counted number (You don't have to worry that it might be 8½ people.), and dividing by it will not affect the number of significant figures. Thus, this answer must be rounded to 4 significant figures. If the greatest insignificant digit were greater than 5, the number would be rounded up; less than 5, down. When it is exactly 5, the rounding is determined by the least significant digit. If the least significant digit is odd, the number is rounded up; if even, the number is rounded down. Since the least significant digit, 7, is odd, the number is rounded up to 139.8.
The reason for this convention is simple. Since the likelihood of the least significant digit's being odd or even is the same, the sum of a large number of figures rounded in this manner should be more accurate than the sum of the figures that were all rounded either up or down.
Comment: Answer A is for those that think that the counted 8 has only one significant figure. Answer B is for those who think it is only 8.0. Answer D is for those who believe whatever their calculator displays is significant. Answer E is for people who know to round, but not which way.
10. According to Le Chatelier's Principle, which of the following would shift the following equilibrium to the right most effectively?
A. 100
B. 140
C. 139.8
D. 139.75
E. 1339.7
1118/8 = 139.75, but there are only 4 sig figs in 1118. Eight is a counted number (You don't have to worry that it might be 8½ people.), and dividing by it will not affect the number of significant figures. Thus, this answer must be rounded to 4 significant figures. If the greatest insignificant digit were greater than 5, the number would be rounded up; less than 5, down. When it is exactly 5, the rounding is determined by the least significant digit. If the least significant digit is odd, the number is rounded up; if even, the number is rounded down. Since the least significant digit, 7, is odd, the number is rounded up to 139.8.
The reason for this convention is simple. Since the likelihood of the least significant digit's being odd or even is the same, the sum of a large number of figures rounded in this manner should be more accurate than the sum of the figures that were all rounded either up or down.
Comment: Answer A is for those that think that the counted 8 has only one significant figure. Answer B is for those who think it is only 8.0. Answer D is for those who believe whatever their calculator displays is significant. Answer E is for people who know to round, but not which way.
2 F2(g) + Si(s) = SiF4(g) . . . . . . ΔH = -1614
A. Increase the temperature.
B. Increase the volume.
C. Add SiF4.
D. Add Si.
E. Add F2.
Explanation
10. According to Le Chatelier's Principle, which of the following would shift the following equilibrium to the right most effectively?
2 F2(g) + Si(s) = SiF4(g) . . . . . . ΔH = -1614
A. Increase the temperature.
B. Increase the volume.
C. Add SiF4.
D. Add Si.
E. Add F2.
Comment: The negative enthalpy means that heat is a product. Adding product shifts the equilibrium to the left. There are more gas molecules on the left that the right, so increasing the volume would shift the equilibrium to the left. Adding a product, SiF4, would shift the equilibrium to the left. Si, while a starting material, is a solid and, as such, does not appear in the equilibrium expression. Adding Si would not affect the equilibrium.
11. The principle type of bonding in NaCl is:
2 F2(g) + Si(s) = SiF4(g) . . . . . . ΔH = -1614
A. Increase the temperature.
B. Increase the volume.
C. Add SiF4.
D. Add Si.
E. Add F2.
Comment: The negative enthalpy means that heat is a product. Adding product shifts the equilibrium to the left. There are more gas molecules on the left that the right, so increasing the volume would shift the equilibrium to the left. Adding a product, SiF4, would shift the equilibrium to the left. Si, while a starting material, is a solid and, as such, does not appear in the equilibrium expression. Adding Si would not affect the equilibrium.
A. non-polar covalent
B. polar covalent
C. van der Waals
D. ionic
E. hydrogen bonding
Explanation
11. The principle type of bonding in NaCl is:
A. non-polar covalent
B. polar covalent
C. van der Waals
D. ionic
E. hydrogen bonding
Comment: This problem calls for recognizing that NaCl will be an ionic compound based on the location of the elements in the periodic table or just plain general knowledge.
12. The hybrid orbitals whose angles to one another are 180° are:
A. non-polar covalent
B. polar covalent
C. van der Waals
D. ionic
E. hydrogen bonding
Comment: This problem calls for recognizing that NaCl will be an ionic compound based on the location of the elements in the periodic table or just plain general knowledge.
A. s
B. sp
C. sp2
D. sp3
E. p
Explanation
12. The hybrid orbitals whose angles to one another are 180° are:
A. s
B. sp
C. sp2
D. sp3
E. p
Comment: This is a general knowledge question.
13. Which of the following would you expect to have the highest boiling point?
A. s
B. sp
C. sp2
D. sp3
E. p
Comment: This is a general knowledge question.
A. CH3-CH=CH2-CH2-OH
B. CH3-CH=CH-O-CH3
C. CH2=CH-O-CH2-CH3
D. O=CH-CH2-CH2-CH3
E. CH2=CH-CH2-CH3
Explanation
13. Which of the following would you expect to have the highest boiling point?
A. CH3-CH=CH2-CH2-OH
B. CH3-CH=CH-O-CH3
C. CH2=CH-O-CH2-CH3
D. O=CH-CH2-CH2-CH3
E. CH2=CH-CH2-CH3
The alcohol in answer A is the only compound shown that can undergo hydrogen bonding, which is the strongest of the intermolecular interactions. Stronger interactions lead to more attraction between molecules and a higher boiling point.
Comment: this question examines the principles of intermolecular forces and how they can be used to predict qualitative descriptive chemistry.
14. What is the oxidation number of I in KIO4?
A. CH3-CH=CH2-CH2-OH
B. CH3-CH=CH-O-CH3
C. CH2=CH-O-CH2-CH3
D. O=CH-CH2-CH2-CH3
E. CH2=CH-CH2-CH3
The alcohol in answer A is the only compound shown that can undergo hydrogen bonding, which is the strongest of the intermolecular interactions. Stronger interactions lead to more attraction between molecules and a higher boiling point.
Comment: this question examines the principles of intermolecular forces and how they can be used to predict qualitative descriptive chemistry.
A. 3
B. 4
C. 5
D. 6
E. 7
Explanation
14. What is the oxidation number of I in KIO4?
A. 3
B. 4
C. 5
D. 6
E. 7
The sum of the oxidation numbers of the atoms in the species = the charge of the species.
K = +1, O = -2
+1 + I + (4 × -2) = 0
I - 7 = 0
I = +7
15. What is the correct equilibrium mass action expression for the following reaction?
A. 3
B. 4
C. 5
D. 6
E. 7
The sum of the oxidation numbers of the atoms in the species = the charge of the species.
K = +1, O = -2
+1 + I + (4 × -2) = 0
I - 7 = 0
I = +7
Ca(HCO3)2(s) ⇔ CaO(s) + 2 CO
A. [CaO][CO2][H2O]/[Ca(HCO3)2]
B. [Ca(HCO3)2]/[CaO][CO2][H2O]
C. [CaO][CO2]2[H2O]/[Ca(HCO3)2]
D. [Ca(HCO3)2]/CaO][CO2]2[H2O]
E. [CO2]2[H2O]
Explanation
15. What is the correct equilibrium mass action expression for the following reaction?
Ca(HCO3)2(s) ⇔ CaO(s) + 2 CO(g) + H2O(g)
A. [CaO][CO2][H2O]/[Ca(HCO3)2]
B. [Ca(HCO3)2]/[CaO][CO2][H2O]
C. [CaO][CO2]2[H2O]/[Ca(HCO3)2]
D. [Ca(HCO3)2]/CaO][CO2]2[H2O]
E. [CO2]2[H2O]
Mass action expression are found from the balanced chemical reaction: the product of the concentrations of products raised to the power of the stoichiometric coefficient divided by the product of the concentrations of reactants raised to the power of the stoichiometric coefficient. Pure solids and pure liquids do not appear in mass action expressions because their concentrations do not change as a function of reaction. Within these constraints, expression E is the correct response.
Comment: equilibrium constants and associated mass action expressions are an important component of understanding much of chemical reactivity.
16. When the following equation is balanced and the coefficient of Mo24O37 is 1, the coefficient of KMnO4 is:
Ca(HCO3)2(s) ⇔ CaO(s) + 2 CO
A. [CaO][CO2][H2O]/[Ca(HCO3)2]
B. [Ca(HCO3)2]/[CaO][CO2][H2O]
C. [CaO][CO2]2[H2O]/[Ca(HCO3)2]
D. [Ca(HCO3)2]/CaO][CO2]2[H2O]
E. [CO2]2[H2O]
Mass action expression are found from the balanced chemical reaction: the product of the concentrations of products raised to the power of the stoichiometric coefficient divided by the product of the concentrations of reactants raised to the power of the stoichiometric coefficient. Pure solids and pure liquids do not appear in mass action expressions because their concentrations do not change as a function of reaction. Within these constraints, expression E is the correct response.
Comment: equilibrium constants and associated mass action expressions are an important component of understanding much of chemical reactivity.
Mo24O37 + KMnO4 + H2SO4 → MoO3 + MnSO4 + K2SO4 + H2O
A. 11
B. 12
C. 13
D. 14
E. 15
Explanation
16. When the following equation is balanced and the coefficient of Mo24O37 is 1, the coefficient of KMnO4 is:
Mo24O37 + KMnO4 + H2SO4 → MoO3 + MnSO4 + K2SO4 + H2O
A. 11
B. 12
C. 13
D. 14
E. 15
This is an oxidation reduction reaction that can be balanced by the half-reaction method:
Oxidation half-reaction:
Mo24O37 → MO3
Balance mass (all but O and H by inspection, then O with water and H with H+):
Mo24O37 + 35 H2O → 24 MO3 + 70 H+
Balance charge (with e–):
Mo24O37 + 35 H2O → 24 MO3 + 70 H+ + 70 e–
Reduction half-reaction (ignoring the spectator ions):
MnO4– → Mn2+
Balance mass (all but O and H by inspection, then O with water and H with H+):
MnO4– + 8 H+ → Mn2+ + 4 H2O
Balance charge (with e–):
MnO4– + 8 H+ + 5 e– → Mn2+ + 4 H2O
Equalize the number of electrons in each half-reaction (multiply the reduction by 14):
Mo24O37 + 35 H2O → 24 MO3 + 70 H+ + 70 e–
14 MnO4– + 112 H+ + 70 e– → 14 Mn2+ + 56 H2O
Add the two half-reactions together, eliminating the common number of electrons:
Mo24O37 + 35 H2O + 14 MnO4– + 112 H+ → 24 MO3 + 70 H+ + 14 Mn2+ + 56 H2O
Eliminate common species and simplify:
Mo24O37 + 14 MnO4– + 42 H+ → 24 MO3 + 14 Mn2+ + 21 H2O
Add back the spectator ions:
Mo24O37 + 14 KMnO4 + 21 H2 SO4→ 24 MO3 + 14 MnSO4 + 7 K2SO4 + 21 H2O
Comment: balancing chemical reactions, either by inspection or by other means, is always found on the contest exam.
17. A chemical company sells 99.99% pure mercury at $32.80 for 50.00 g. The density of mercury is 13.546 g/mL. What is the price of mercury in dollars per mL?
Mo24O37 + KMnO4 + H2SO4 → MoO3 + MnSO4 + K2SO4 + H2O
A. 11
B. 12
C. 13
D. 14
E. 15
This is an oxidation reduction reaction that can be balanced by the half-reaction method:
Oxidation half-reaction:
Mo24O37 → MO3
Balance mass (all but O and H by inspection, then O with water and H with H+):
Mo24O37 + 35 H2O → 24 MO3 + 70 H+
Balance charge (with e–):
Mo24O37 + 35 H2O → 24 MO3 + 70 H+ + 70 e–
Reduction half-reaction (ignoring the spectator ions):
MnO4– → Mn2+
Balance mass (all but O and H by inspection, then O with water and H with H+):
MnO4– + 8 H+ → Mn2+ + 4 H2O
Balance charge (with e–):
MnO4– + 8 H+ + 5 e– → Mn2+ + 4 H2O
Equalize the number of electrons in each half-reaction (multiply the reduction by 14):
Mo24O37 + 35 H2O → 24 MO3 + 70 H+ + 70 e–
14 MnO4– + 112 H+ + 70 e– → 14 Mn2+ + 56 H2O
Add the two half-reactions together, eliminating the common number of electrons:
Mo24O37 + 35 H2O + 14 MnO4– + 112 H+ → 24 MO3 + 70 H+ + 14 Mn2+ + 56 H2O
Eliminate common species and simplify:
Mo24O37 + 14 MnO4– + 42 H+ → 24 MO3 + 14 Mn2+ + 21 H2O
Add back the spectator ions:
Mo24O37 + 14 KMnO4 + 21 H2 SO4→ 24 MO3 + 14 MnSO4 + 7 K2SO4 + 21 H2O
Comment: balancing chemical reactions, either by inspection or by other means, is always found on the contest exam.
A. $22,220/mL
B. $121.06/mL
C. $8.886/mL
D. $0.1125/mL
E. $0.004882/mL
Explanation
17. A chemical company sells 99.99% pure mercury at $32.80 for 50.00 g. The density of mercury is 13.546 g/mL. What is the price of mercury in dollars per mL?
A. $22,220/mL
B. $121.06/mL
C. $8.886/mL
D. $0.1125/mL
E. $0.004882/mL
($32.80/50.00 g)×13.546 g/mL = $8.886/mL
Comment: this is a units conversion problem, which are always found on the contest exam.
18. The density of a gas at 91.45 °C and 608 torr is 1.90 g/mL. What is the molar mass of this gas?
A. $22,220/mL
B. $121.06/mL
C. $8.886/mL
D. $0.1125/mL
E. $0.004882/mL
($32.80/50.00 g)×13.546 g/mL = $8.886/mL
Comment: this is a units conversion problem, which are always found on the contest exam.
A. 20.2
B. 37.4
C. 71.1
D. 83.8
E. 154
Explanation
18. The density of a gas at 91.45 °C and 608 torr is 1.90 g/mL. What is the molar mass of this gas?
A. 20.2
B. 37.4
C. 71.1
D. 83.8
E. 154
Pick a convenient volume of gas: 1.000 L (defined to as many significant figures as necessary to not affect any calculated answers).
Using the ideal gas law: PV = nRT or n = PV/RT
P = 608 torr×760 torr/atm = 0.800 atm
V = 1.000 L
R = 0.08206 L·atm/mole·K
T = 91.45 + 273.15 = 364.60 K
n = (0.800 atm)(1.000 L)/(0.08206 L·atm/mol·K)(364.60 K) = 0.0267 mol of the gas.
Mass of the gas = (1.000 L)×(1.90 g/L) = 1.90 g.
molar mass = mass/moles = 1.90 g/0.0267 mol = 71.1 g/mol.
Comment: this is a standard ideal gas problem.
Chemistry Contest Philosophy
A. 20.2
B. 37.4
C. 71.1
D. 83.8
E. 154
Pick a convenient volume of gas: 1.000 L (defined to as many significant figures as necessary to not affect any calculated answers).
Using the ideal gas law: PV = nRT or n = PV/RT
P = 608 torr×760 torr/atm = 0.800 atm
V = 1.000 L
R = 0.08206 L·atm/mole·K
T = 91.45 + 273.15 = 364.60 K
n = (0.800 atm)(1.000 L)/(0.08206 L·atm/mol·K)(364.60 K) = 0.0267 mol of the gas.
Mass of the gas = (1.000 L)×(1.90 g/L) = 1.90 g.
molar mass = mass/moles = 1.90 g/0.0267 mol = 71.1 g/mol.
Comment: this is a standard ideal gas problem.