AP TASK ANSWERS 9 UNITS
UNIT 00 - AP CHEMISTRY PREAMBLE
1. (a) 2.45 x 104
(b) 3.56 x 102
(c) 9.85 x 10-4
(d) 2.22 x 10-1
(e) 1.22 x 104
2. (a) 4200
1. (a) 0.0186
2. (a) 3 miles = 4828.03 m
(b) 25 lbs = 11.3398 kg
1. (a) -260
2. When determining change, one degree Celsius is the same as (equivalent to) one Kelvin.
1. (a) 4
2. (a) 809
1. (a) Y
1. (a) p = 82, e = 82, n = 128
(b) p = 16, e = 16, n =18
2. (a) 20
1. (a) Sodium chloride
(b) Strontium oxide
(c) Aluminum nitride
(d) Barium chloride
(e) Potassium oxide
(f) Copper(II) oxide
(g) Copper(I) oxide
2. (a) Mg3N2
1. (a) NH4NO3
2.(a) Sodium nitrate
(b) Potassium manganate(VII) (potassium permanganate)
(c) Calcium ethandioate (calcium oxalate)
(d) Copper(II) sulfate
(e) Copper(I) sulfate
(f) Potassium nitrite
(g) Lithium perchlorate
1. (a) N2O4
2. (a) Dinitrogen pentoxide
(b) Phosphorus trichloride
(c) Sulfur hexafluoride
(d) Water (dihydrogen monoxide)
(e) Dichlorine monoxide
UNIT 01 - ATOMS, ELEMENTS & PERIODICITY
1. (a) FeO
2. CH, CH2, CH3O, C3H6O2 – all cannot be simplified further.
3. They are not the same sample, not all can be pure.
3. Yes, BUT also consistent with an impurity that has any amount of C that is less than 92.31 since the % of C has gone down.
1. 58.5 g
2. 3.5 moles
3. 0.00234 moles
4. 0.523 moles
5. Answers here are the number of moles multiplied by Avogadro’s number (6.02 x 1023), remembering that for example, NaCl has both Na+ ions and Cl– ions and that CuBr2 contains twice as many Br– ions as Cu2+ ions, etc.
1. (a) No, different elements
(b) Yes, same element with different numbers of neutrons therefore different masses.
2. p = 17, e = 17, n = 18.
1. 35.5; Cl37 = 25%, Cl35 = 75%
2. 5 peaks. Two of the same height at 79 and 81, and three at m/z values of 158, 160 and 162, with the 160 peak twice the size of the 158 and 162 which would be identical. Think of the bromine diatomic molecule as being made up of Br atom “x”, and Br atom “y”.
Combo a. x = 79 and y = 79, when added = 158
Combo b. x = 81 and y = 79, when added = 160
Combo c. x = 79 and y = 81, when added = 160
Combo d. x = 81 and y = 81, when added = 162
160 is twice as likely as 158 and 162, so has a peak twice the height.
4.(a) That there are three isotopes of Mg
(b) (i) That it is the isotope with the greatest abundance
1. (a) [Ar] 4s1 3d10
(b) [Ar] 4s2 3d7
(c) [Ar] 4s2
(d) [He] 2s2 2p2
(e) [Ne] 3s2 3p6
(f) [Ar] 4s2 3d10 4p1
2. (a) [He] 2s2 2p6 OR [Ne]
(b) [Ne] 3s2 3p6 OR [Ar] OR [Ar] 4s0
(c) [Ne] 3s2 3p6 OR [Ar]
(d) [He] 2s2 2p6 OR [Ne]
(e) [He] 2s2 2p6 OR [Ne]
1. (a) There are 4 occupied sub-shells
(b) 1s2 2s2 2p6 3s1 in decreasing order of peak energy (i.e., the 1s2 electrons are the most difficult to remove)
2. (a) Three
3. Aluminum. It is the first element to have electrons in the 3p sub-shell. The 3p sub-shell is the fifth sub-shell to be filled
1. (a) Decreases. Greater shielding outweighs greater number of protons
(b) Increases. Same shielding, more protons
2. (a) Group 1
(b) Group 2
(c) It only has three electrons
3. Energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of singly charge ions
Number of protons and the amount of shielding (distance form the nucleus).
4. (a) RO
(b) Less mutual repulsion of electrons but same number of protons
5. Increases. Same shielding, more protons
6. Zero shielding, electrons directly adjacent to the nucleus
Isoelectronic, BUT different numbers of protons. Largest number of protons = greatest attraction = smallest species
UNIT 02 - CHEMICAL BONDING I
Na+, Cl–; NaCl
Ca2+, Cl–; CaCl2
Fe3+, Br–; FeBr3
Na+, O2-; Na2O
NaCl. The smaller radius of the chloride ion when compared to the bromide ion allows the ions in NaCl to get closer to one another than in NaBr, and therefore the Coulombic attraction is greater.
The structure to the right has the most favorable formal charges (zero on all atoms), so is the preferred structure.
UNIT 03 - CHEMICAL BONDING II & GASES
1. (a) The LDFs are approximately the same.
(b) Butan-1-ol has additional hydrogen bonding as an IMF that dramatically increases the attractions between molecules of the alcohol, thus significantly raising the boiling point when compared to pentane
2. Both molecules have LDFs, but HCl also has dipole-dipole interactions
Despite dipole-dipole interactions generally being stronger than LDFs, in instances where the molecules are large (larger surface area and more electrons), the collective interactions created by many LDFs can outweigh the stronger dipole-dipole interactions to give a greater attraction between the molecules. In this case, the larger Cl2 molecules have LDFs that are collectively stronger than the LDFs and dipole-dipole interactions in the much smaller, HCl
3. Increase in boiling points since LDFs increase with surface area and number of electrons and polarizability.
A: Ionic – poor conductor when solid because ions cannot move. High m.p., means strong, ionic bonds. Will conduct when liquid when the ions are free to move.
B: Metallic – high m.p., always a good conductor with free moving electrons.
C: Molecular – never a good conductor (no charged particles to act as charge carriers. Low m.p. due to weak IMF’s.
1. 0.864 atm
2. 1.09 atm
1. 0.272 L
2. 422 mL
1. 10.4 L
2. 288 K
1. 8.21 L
2. 37.4 L
1. 318 K
2. 247 K
1. 0.435 mols
2. 266.4 g mol-1
1. 0.0255 mols
2. 717.6 kPa
Apply P1V1 = P2V2, Boyle’s law for EACH gas, then add the new P’s (Dalton’s).
1. (a) 1363 ms-1
(b) 482 ms-1
(c) 183 ms-1
2. (a) 1419 ms-1
(b) 502 ms-1
(c) 191 ms-1
3. urms is lower at lower temperatures and lower with larger molar masses
1. 0.192 L (192 mL)
2. 0.192 L (192 mL)
3. 1.75 L
4. Add approx. 250 mL of distilled/deionized water to a 500 mL volumetric flask. Slowly and carefully add 192 mL of nitric acid using a buret. Make up to the mark with distilled/deionized water. Wear safety goggles, chemical resistant gloves and lab coat or apron at all times, and only ever add small amounts of concentrated acid to large volumes of water. Diluting concentrated acid can be a highly exothermic process, and safe dilution is essential. The use of accurately graduated glassware such as burets, pipets and volumetric flasks, ensure the accuracy of the dilution
UNIT 04 - CHEMICAL REACTIONS
1. (a) H2 + CuO → Cu + H2O
(b) 2C + O2 → 2CO
(c) Mg + H2SO4 → MgSO4 + H2
2. (a) Ca + 2H2O → H2 + Ca(OH)2
(b) 2Cu + O2 → 2CuO
(c) 4Na + O2 → 2Na2O
(d) Fe + 2HCl → FeCl2 + H2
(e) 2Fe + 3Br2 → 2FeBr3
(f) C4H8 + 6O2 → 4CO2 + 4H2O
(g) Na2CO3 + 2HI → 2NaI + CO2 + H2O
(h) CuCO3 → CuO + CO2
(i) 2Pb(NO3)2 → 2PbO + 4NO2 + O2
(j) H2SO4 + 2KOH → K2SO4 + 2H2O
(k) 2NaHCO3 → Na2CO3 + H2O + CO2
(l) 4Al + 3O2 → 2Al2O3
1. (a) Formation of steam (the reaction is sufficiently exothermic that it produces enough energy to convert the water produced to a gas)
(b) 2H2 + O2 → 2H2O
(c) Reactants and products appear in the correct ratios (i.e., the equation is balanced according to the answer given in (b). Oxygen, hydrogen and water are all consistently represented by the black and white circles (i.e., the ratio of atoms in each case are consistent with the known formulas for the substances)
1. When water boils the process can be summarized as; H2O(l) → H2O(g). As such, no new substances are formed, there is only a change of state, and only intermolecular forces are broken
2. When water is broken down into its elements, the process can be summarized by the equation, 2H2O → 2H2 + O2. As such, new substances are formed, and intra bonds (covalent) are broken
1. (a) 3.24 g
(b) 76.5 g
2. (a) 1:1
X + Br2 → XBr2
(b) Calcium (molar mass 40 gmol-1) that is consistent with being a group 2 element, and forming a 2+ ion, and reacting in the ratio in the equation in (a)
3. 34.98 g
5. 0.142 M
6. 0.981 g
7. 0.05 L or 50 mL
1. 27.7 g
2. 1.12 L
3. 83.7 gmol-1
Chlorine is limiting.
1.42 moles of product (SCl6) which is equal to 348 g
157 g of excess reactant (S) left over
1. x = 6
2. Heating until the mass does not change anymore, i.e., heating until all of the water has been driven off and the salt is completely dehydrated (anhydrous)
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)
2K+(aq) + 2OH–(aq) + 2H+(aq) + SO42-(aq) → 2K+(aq) + SO42-(aq) + 2H2O(l)
2H+(aq) + 2OH–(aq) → 2H2O(l)
1. (a) +6
2. (a) Ca + S → CaS
Ca → Ca2+ + 2e–
S + 2e– → S2-
(b) 2Al + 3Br2 → 2AlBr3
Al → Al3+ + 3e–
Br2 + 2e– → 2Br–
(c) 4Al + 3O2 → 2Al2O3
Al → Al3+ + 3e–
O2 + 4e– → 2O2-
3. (a) IO3– + 5e– + 6H+ → ½I2 + 3H2O
5I– → 2½I2 + 5e–
IO3– + 6H+ + 5I– → 3H2O + 3I2
(b) I2 + 2e– → 2I–
2S2O32- → S4O62- + 2e–
2S2O32- + I2 → S4O62- + 2I–
(c) Br2 + 2e– → 2Br–
2I– → I2 + 2e–
Br2 + 2KI → I2 + 2KBr
(d) 2MnO42- → 2MnO4– + 2e–
MnO42- + 2e– + 4H+ → MnO2 +2H2O
3MnO42- + 4H+ → 2MnO4– + MnO2 + 2H2O
(e) NOT REDOX (all oxidation states the same) so no half reactions required
2CrO42- + 2H+ → Cr2O72- + H2O
(b) It goes up, side oxygen is added.
2. (a) Decomposition.
(b) CaCO3 → CaO + CO2
(c) Mass of CaCO3 = 30.46 – 24.56 = 5.90 g
Mass of CO2 = 30.46 – 27.86 = 2.6 g
Moles of CaCO3 = 5.90/100 = 0.0590
Moles of CO2 = 2.6 /44 = 0.059
i.e., 1:1 ratio which agrees with the equation
1. Calculate the moles of potassium iodate(V) by dividing mass by molar mass, and then divide by 10 (25 mL is taken out of the 250 mL).
2. Apply ratio from the balanced equation by multiplying by 3 to get the moles of iodine.
3. Apply ratio from the balanced equation by multiplying by 2 to get the moles of thiosulfate.
4. Divide moles of thiosulfate by the volume of thiosulfate to get the concentration of this sulfate.
x = 12
1. Calculate the moles of manganate(VII) by multiplying the concentration by the volume.
2. Realize (and THIS is really the quantum leap of intelligence) that in that huge, double salt, the thing that is getting oxidized is Fe2+, so therefore the missing half reaction is, Fe2+ → Fe3+ + e–.
3. Combine the half reaction in 2. above with the manganate(VII) half reaction to get the reacting ratio (5Fe2+:1MnO4–) and therefore get the moles of Fe2+ in the 25.0 mL sample from the answer to 1., above.
4. Multiple the answer in 3. above by 10 to get moles of Fe2+ in 250 mL sample, and realize that the moles of Fe2+ also equals the moles of the WHOLE salt (since in 1 mole of the whole salt there is 1 mole of Fe2+, i.e. it’s a 1:1 ratio).
5. Calculate the molar mass of the whole salt by dividing grams (given in the Q) by moles calculated in 4. above.
6. Subtract all of the molar masses of the pieces of the salt that we know (N, H, Fe, S, O) to leave the mass of H2O.
7. The answer to 6. above should be divisible nicely by 18, i.e., should give you an integer value for x.
(a) Mg2+(aq) +2OH–(aq) → Mg(OH)2(s)
(b) Ba2+(aq) + S2-(aq) + Ni2+(aq) + SO42-(aq) → NiS(s) + BaSO4(s)
(c) No reaction
(d) Al3+(aq) +3OH–(aq) → Al(OH)3(s)
(b) 0.002 mols
(c) 0.002 mols
(d) 0.0709 g
(f) The mass of precipitate would be too large, making the moles of AgCl too large, making the moles of chloride too large, making the mass of chloride too large, making the calculated % by mass too large.
UNIT 05 - CHEMICAL KINETICS
(a) They are both catalysts since they both increase the rate of reaction, and at the end of the experiment they are regenerated, i.e., are unchanged
(b) In each case the H3O+ and the OH– are regenerated
1. The total mass of the chips was the same in experiment, since the total volume of gas produced in each experiment was the same. (The acid was in excess)
2. The small chip experiment was complete between 120 and 135 seconds. The large chip experiment was complete between 150 and 165 seconds. The reaction is complete when no more gas is produced which is illustrated on the graph when the line plateaus
3. mL/s (many other possible answer would work)
4. 2HCl(aq) + CaCO3(s) → CO2(g) + H2O(l) + CaCl2(aq)
5. When solids are present, one of the factors which affects the rate of a chemical reaction is the surface area of the solid. The reaction between the solution and the solid takes place at the surface of the solid, so a greater surface area offers more sites for collision, and therefore a faster reaction. Small chips have a larger surface area than the same mass of larger chips, so the reaction is faster with the smaller sized solid
6. Mass the chips using the balance, use the measuring cylinder to measure the volume of the acid, and then use the gas syringe to collect the carbon dioxide from the reaction occurring in a Erlenmeyer flask. See this for an example of the set up
7. Heat the acid solution to varying temperatures OR change the concentration of the acid to varying concentrations, each time keeping all other variables the same
V2O4 is the intermediate, V2O5 is the catalyst
1. Rate = k [X]2
Order w.r.t [Y] = 0
Total order = 2 + 0 = 2
k = 40000 M-1 min-1
2. Rate = k [A][B]
Total order = 1 + 1 = 2
k = 8000 M-1 s-1
2. Mechanism #2
3. X is an intermediate
1. 1st order. The graph of ln[A] versus time is a straight line, consistent with the integrated rate law for first order reactions
2. 7 minutes (approx. since ‘from graph’)
3. 1.01 x 10-1 min-1
Note: This value was calculated by taking the data across the full time range of 25 minutes. Other, very similar (but different) values can be calculated by using smaller time ranges
UNIT 06 - CHEMICAL THERMODYNAMICS I
1. (a) From propane to water
(b) 23.83 K
(c) Exothermic since energy is released during the combustion
1. (a) 7599 J
(b) 92.96 K
1. -124 kJ
2. -95 kJ
3. -1198 kJ
4. 0 (zero). Coincidentally, identical covalent bonds are broken and made so the endothermic (+ve) and exothermic (-ve) processes have the same magnitude and their sum = 0
1. Ionic bonds. Intermolecular, hydrogen bonds
2. Endothermic, bonds or forces being broken.
3. They are polar, exothermic
1. C(graphite) + 1½H2(g) + ½Br2(l) → CH3Br(l)
2. 4C(graphite) + 4H2(g) + ½O2(g) → CH3COC2H5(l)
3. Na(s) + ½N2(g) + 1½O2(g) → NaNO3(s)
4. During the formation of an element, “from its elements”, i.e., from ITSELF, there is no change. For example; Na(s) ==> Na(s) represents no change in any aspect, of which enthalpy is just one
1. C5H12(l) + 😯2(g) → 5CO2(g) + 6H2O(l)
2. CO(g) + ½O2(g) → CO2(g)
3. H2(g) + ½O2(g) → H2O(l)
4. Al(s) + ¾O2(g) → ½Al2O3(s)
1. -84 kJ
2. -2005 kJ
3. (a) -852 kJ
(b) They are all zero
(c) +852 kJ
4. (a) S(s) + O2(g) → SO2(g)
(b) -297 kJ
5. (a) -126 kJ
(b) -124 kJ
(c) Answer (a), since it is calculated from formation and combustion data that are specific to the compounds involved. In part (b), the bond enthalpies are average values and therefore not necessarily specific to the compounds involved
UNIT 07 - CHEMICAL EQUILIBRIUM
1. Kc = 4.
All experiments yield the same value for K because each reaction is carried out at the same temperature
2. x = 0.845, so moles of acid = 0.155, alcohol = 1.155, ester = 0.845, water = 0.845
3. Kc = 13.1
4. Moles of HBr = 8.62
The relatively large amount of HI suggests that the equilibrium lies heavily to the right and side (products), and therefore the value of K will be relatively large
1. (a) Kp = 3.05
(b) The expression below leads to a math equation that is beyond the scope of AP chemistry, but can be solved.
Here it is simply important to see how the expression reached. (The actual value of x = 0.58 (or 58% dissociated) is consistent with a lower total pressure according to Le Châtelier’s principle)
2. (a) Kp =13.8
(b) Kp = 21.8
(c) The expression Kp = 2.11 x 104 = leads to a math equation that is beyond the scope of AP, but can be solved. Here it is simply important to see how the expression reached. The actual value of x = 0.986 (or 98.6% dissociated) is consistent with the huge change in Kp at the higher temperature
1. 1.72 x 10-11
2. 2.36 x 10-5
3. 0.0152 M, 4.75 g/L.
4. Precipitate forms since Q > Ksp and to reach equilibrium, the concentrations of the ions in solution must be reduced. That is achieved via the formation of a ppt.
1.(a) 1.26 x 10-5
(b) 2.87 x 10-8
(c) 6.32 x 10-9
2. Addition of sulfate ions from an external source makes Q much larger than K. In order to restore equilibrium, Q must decrease in order for the value of K to be re-stored, and the reaction must shift to the side where solid barium sulfate is.
2. (a) 6.7 x 10-6
(b) 1.2 x 10-9
(c) The high concentration of hydroxide ions in a solution of pH 11, causes the equilibrium to be shifted backwards (Le Châtelier’s principle or Q v. K), and the solubility of the solid decreases. Essentially, this is a common ion problem.
UNIT 08 - ACIDS & BASES
1. Diagram B
2. Diagram C
3. Diagram A is a relatively dilute, weak acid; Diagram D is a relatively dilute, strong acid
3. 4.8 x 10-5
1. 1.82 x 10-5
3. Ethanoic acid since it has a larger Ka, meaning a greater degree of ionization
5. 3.06 x 10-5
6. 2.41 x 10-10
1. 1.78 x 10-5
3. Methylamine since it has a larger Kb, meaning a greater degree of ionization
5. 3.06 x 10-5
6. 6.95 x 10-10
2. The best buffers are generally those with an approx. 1:1 ratio of acid and conjugate base (salt) components, since this allows them to absorb any additional acid and base equally well. Since the log of 1 is equal to 0, then the pH of the buffer is equal to its pKa
3. pH = pKa + log (deprotonated/protonated) = 7.6 + log (0.67/0.33) = 7.9
UNIT 09 - CHEMICAL THERMODYNAMICS II & ELECTROCHEMISTRY
1. (a) Negative
2. (a) 400 K. Higher temperature means greater disorder (entropy)
(b) A soft metal will have flexible, metallic bonding that induces more disorder than a very rigid solid like diamond
(c) The gas at the lower pressure, since the particles have more space to move in and as such can be more disordered. A gas at a relatively high pressure, means that the particles are packed into a relatively small space with relatively little room to move around. This leads to a lower level of disorder, i.e., a lower entropy
(a) E = +3.63 V; Zn(s) + F2(g) → Zn2+(aq) + 2F–(aq)
(b) E = +0.14 V; Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g)
(c) E = +0.78 V; Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
(d) E = +0.77 V; H2(g) + 2Fe3+(aq) → 2H+(aq) + 2Fe2+(aq)
(i) (Pt) | 1/2 H2(g) | H+(aq) || Ag+(aq) | Ag(s); +0.80 V
(ii) Al(s) | Al3+(aq) || Ag+(aq) | Ag(s); -1.66 V