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Question 1

UNIT 1 Atomic Structure and Properties

(a) TOPIC 1.5 Atomic Structure and Electron Configuration

(b) TOPIC 1.2 Mass Spectra of Elements

(c) TOPIC 1.1Moles andMolar Mass

(d) TOPIC 1.1Moles andMolar Mass

(e) TOPIC 1.1Moles andMolar Mass

(f) Topic 4.2 Net Ionic Equations

1. Answer the following questions related to manganese compounds.

(a) Manganese has several common oxidation states.

(i) Write the complete electron configuration for an Mn atom in the ground state.

(ii) When manganese forms cations, electrons are lost from which subshell first? Identify both the number and letter associated with the subshell.

A student performs an experiment to produce a manganese salt of unknown composition, MnxCly(aq), and determine its empirical formula. The student places a sample of Mn(s) in a beaker containing excess HCl(aq), as represented by the following equation.

\( x\\\ MN(s) + y\\\ HCI(aq)\to Mn_{x}CI_{y}(aq) + \frac{y}{2}H_{2}(g)\)

The student heats the resulting mixture until only MnxCly(s) remains in the beaker. The data are given in the following table.

(b) Calculate the mass of Cl in the sample of MnxCly(s) remaining in the beaker.

(c) Calculate the number of moles of Cl in the sample of MnxCly(s) remaining in the beaker.

(d) The student determines that 0.0199 mol of Mn was used in the experiment. Use the data to determine the empirical formula of the MnxCly (s)

(e) The student repeats the experiment using the same amounts of Mn and HCl and notices that some of the MnxCly splatters out of the beaker as it is heated to dryness. Will the number of moles of Cl calculated for this trial be greater than, less than, or equal to the number calculated in part (c) ? Justify your answer.

(f) Another compound of manganese, MnO2, is used in alkaline batteries, represented by the following diagram. Some half-reactions are given in the table.

(i) Based on the half-reactions given in the table, write the balanced net ionic equation for the reaction that has the greatest thermodynamic favorability.

(ii) Calculate the value of E°cell for the overall reaction.

(iii) Calculate the value of ΔG° in kJ/molrxn.

(iv) A student claims that the total mass of an alkaline battery decreases as the battery operates because the anode loses mass. Do you agree with the student’s claim? Justify your answer.

▶️Answer/Explanation

(a) (i) For the correct answer:
 Accept one of the following:     

  • \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}\)
  • \([Ar] 4s^{2} 3d^{5}\)

(ii) For the correct answer, consistent with part (a)(i):

     4s

(b) For the correct calculated value:
62.673g-61.262g=1.411gCI

(c) For the correct calculated value, consistent with part (b):

\(1.411\\\ gCI*\frac{1\\\ mol\\\ CI}{35.45\\\ g CI}=0.03980\\\ mol\\\ CI\)

(d) For the correct answer, consistent with part (c):

\(\frac{0.03980\\\ mol\\\ CI}{0.0199\\\ mol\\\ Mn}=\frac{2\\\ mol\\\ CI}{1\\\ mol\\\ Mn}\to MnCI_{2}\)

(e) For the correct answer and a valid justification:

Less than. If some of the mass of aqueous \(Mn_{x}CI_{Y}\) is lost due to splattering, the final mass of the dry beaker and \(Mn_{x}CI_{Y}\)  will be decreased, which will decrease the calculated mass and number of moles of chlorine in the dry solid.

(f) (i) For the correct balanced equation:

 \(2MnO_{2}\left ( s \right )+H_{2}O\left ( l \right )+2e^{-}\to Mn_{2}O_{3}\left (s \right )+2OH^{-}{\left(aq \right)}\)

\(\frac{Zn\left ( s \right )+2OH^{-}\left ( aq \right )\to ZnO\left ( s \right )+H_{2}O\left ( l \right )+2e^{-}}{2MnO_{2}(s)Zn(s)\to Mn_{2}O_{3}+ ZnO(s)}\)

(ii) For the correct calculated value, consistent with part (f)(i):

\(E_{Cell}=0.15V-\left ( -1.28V \right )=1.43V\)

(iii) For the correct calculated value, consistent with part (f)(ii):

\(\Delta G^{\circ }=-nFE^{\circ }=\frac{2mol e\widetilde{}}{1mol_{rxn}}\times \frac{96,485C}{1mole\widetilde{}}\times \frac{1.43J}{1C}\times \frac{1kJ}{1000J}= -276kj/mol_{rxn}\)

(iv) For the correct answer and a valid justification:
Accept one of the following:
• Disagree. The battery is enclosed, so no change in the total mass will occur.
• Disagree. All reactants and products are in the solid phase, so the mass of the sealed battery will remain the same (no gases enter or exit the battery).

Question 2

Topic 6.5 Energy of Phase Changes

(a) Topic 4.2 Net Ionic Equations

(b) TOPIC 9.3 Gibbs Free Energy and Thermodynamic Favorability

(c) TOPIC 2.2 Intramolecular Force and Potential Energy

(d) TOPIC 2.5 Lewis Diagrams

(e) Topic 7.6 Properties of the Equilibrium Constant

In the gas phase, AlCl3 is a molecular substance. A reaction of gaseous AlCl3 at high temperature is represented by the following balanced equation.

Reaction 1: AlCl 3(g) → Al(g) + 3 Cl(g) ΔH°1 = ? 

(a) How many grams of Cl(g) can be formed from 1.25 mol of AlCl3(g) ?

Additional reactions that involve Al or Cl are shown in the following table.

(b) Calculate the value of ΔH°1 , in kJ/molrxn, for reaction 1 above using reactions 2, 3, and 4.

(c) A potential energy diagram for Cl2 is shown in the following graph.

(i) Based on the graph, what is the bond length, in picometers, for Cl2 ? _________
(ii) A student finds that the average Al − Cl bond length is 220 picometers and the average bond energy is 425 kJ/mol. Draw the potential energy curve for the average Al − Cl bond on the preceding graph.

(d) Three proposed Lewis diagrams for the AlCl3(g) molecule are shown.

(i) The AlCl3(g) molecule has a trigonal planar geometry. Which diagram (1, 2, or 3) can be eliminated based on geometry? Justify your choice based on VSEPR theory.

(ii) Which of the three diagrams is the best representation for the bonding in AlCl3 ? Justify your choice based on formal charges.

AlCl3 is known to dimerize reversibly in the gas phase. The dimerization equilibrium is represented by the following equation.

2 AlCl3(g) \(\rightleftharpoons\) Al2Cl6(g)

(e) Write the expression for the equilibrium constant, Kp, for this reaction.

A particle-level diagram of an equilibrium mixture of AlCl3(g) and Al2Cl6(g) at 400°C in a 25 L closed container is shown.

(f) Using the particle-level diagram, calculate the value of Kp for the reaction if the total pressure in the container is 22.1 atm.

▶️Answer/Explanation(a)  For the correct calculated value reported with the correct number of significant figures:

\(1.25\\\ mol\\\ AICL_{3}\times \frac{3\\\ mol\\\ CI}{1\\\ mol\\\ AICI_{3}}\times\frac{35.45g\\\ CI}{1\\\ mol\\\ CI}=133\\\ g\\\ CI\)

(b)  For the correct algebraic manipulation of either \(\Delta 2H^{\circ } or \Delta 4H^{\circ }\) (may be implicit):

Accept one of the following:

• Reversing reaction 2:

\(AICI_{3}\left ( g \right )\to AI\left ( s \right )+\frac{3}{2}CI_{2}\left ( G \right )\Delta H^{\circ }_{rxn}=-\left ( -583 \right )=+583kJ/mol_{rxn}\)

• Multiplying reaction 4 by \(\frac{3}{2}\)

\(\frac{3}{2}\left ( CI\left ( g\right )\to 2CI\left ( g \right ) \right )\Delta H^{\circ _{}}_{rxn}=\frac{3}{2}\left ( +243 \right )=+365kJ/mol_{rxm}\)

(c) (i) For the correct answer:

200 picometers (±10 pm)

(ii) For a curve with a minimum at an internuclear distance of 220 10 pm ± :
See sample curve below

For a curve with a minimum energy value of − ± 425 20 kJ/mol that approaches zero as the internuclear distance approaches 500pm:

(d)(i) For the correct answer and a valid justification:
Diagram 2. Al has four electron domains in Diagram 2, which would be trigonal pyramidal, not trigonal planar.

(ii) For the correct answer and a valid justification:
Diagram 1. All atoms in diagram 1 have a formal charge of zero, whereas atoms in diagrams 2 and 3 have nonzero formal charges.

(e) For the correct answer:

\(K_{p}^{}=\frac{P_{AI_{2}}CI_{2}}{\left ( P_{AICI_{3}} \right )}\)

(f) For the correct calculated value, consistent with part (e):

\(K_{P}=\frac{\chi _{AI_{2}CI_{6}\left ( P_{total} \right )}}{\chi _{AICI_{3}\left ( P_{total} \right )2}^{}}=\frac{3}{10}\)

Question 3

Topic 5.7 Introduction to Reaction Mechanisms

(a) Topic 4.2 Net Ionic Equations

(b) Topic 4.2 Net Ionic Equations

(c) Topic 4.2 Net Ionic Equations

(d) Topic 4.2 Net Ionic Equations

(e) Topic 4.2 Net Ionic Equations

(f) Topic 6.2 Energy Diagrams

(g) Topic 6.3 Heat Transfer and Thermal Equilibrium

Answer the following questions about an experiment in which CaCO3(s) is combined with HCl(aq), represented by the following balanced equation.

CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

(a) Write the balanced net ionic equation for the reaction.

A student performs an investigation to study factors that affect the rate of the reaction. In each trial the student combines 50.0 mL of HCl(aq) at 21.2° C with 1.00 g of CaCO3(s) and measures the time required for the reaction to go to completion. The data are given in the following table.

(b) The student correctly identifies that trial 5 is inconsistent with the other trials. Explain why the student’s claim is correct using the data in the table.

(c) Based on the reaction conditions and the collisions that occur between particles, explain the reason for the difference in the reaction times for trial 2 and trial 3.

(d) The student claims that the reaction is zero order with respect to HCl(aq). Do you agree or disagree with the student’s claim? Justify your answer using the student’s data.

(e) The HCl(aq) was present in excess in all trials of the experiment. Determine the molarity of the HCl(aq) in the beaker after the reaction is complete in trial 2. Assume that the volume of the mixture remains constant at 50.0 mL throughout the trial. (The molar mass of CaCO3 is 100.09 g/mol.)

CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

In order to measure the enthalpy of the reaction shown, the student repeats trial 1 by mixing 50.0 mL of HCl(aq) with 1.00 g of CaCO3(s) using a coffee cup calorimeter. The student records the temperature of the system every 20 seconds. The data are given in the following table.

(f) Is the reaction endothermic or exothermic? Justify your answer using the information in the table.

(g) Based on the experimental data, the mass of the system is 51.0 g, and the specific heat of the reaction mixture is 4.0 J / (g · ° C).

(i) Calculate the magnitude of heat transfer, q, in joules.

(ii) Calculate the enthalpy of reaction in units of kJ/molrxn. Include the algebraic sign on your answer.

▶️Answer/Explanation(a) For the correct balanced equation (state symbols not required):
Accept one of the following:

\(\bullet\\\ C_{a}CO_{3}\left ( 3 \right )+2H^{\dotplus }\left ( aq \right )\to Ca^{2}\dotplus \left ( aq \right )CO_{2}\left ( g \right )+H_{2}O\left ( l \right )\)

\(\bullet\\\ C_{a}CO_{3}\left ( 3 \right )+2H_{3}O^{\dotplus }\left ( aq \right )\to Ca^{2}\dotplus \left ( aq \right )CO_{2}\left ( g \right )+3H_{2}O\left ( l \right )\)

(b) For a correct explanation:
Accept one of the following:
• Even though the concentration of HCl is greater in trial 5 than in trial 2, the reaction time is significantly longer. Both trial 2 and 5 occur under otherwise identical conditions. The trend for trial 1 and 4 indicates that the higher concentration of HCl results in a shorter time of reaction.
• The time of reaction in trial 5, with small chunks of calcium carbonate, is longer than trial 6 with large chunks. Both trial 5 and 6 occur under otherwise identical conditions. The trend for trials 1, 2, and 3 shows that larger chunks of the     solid result in longer time of reaction.

(c)For a correct explanation of the effect of particle collisions on reaction rate:
The larger interface between the two reacting substances means there will be more collisions between the particles in a given amount of time, and thus, a higher frequency of successful collisions in which the particles react to form the products.

 (d) For the correct answer and a valid justification:
Accept one of the following:
• Disagree. If the reaction was zeroth order with respect to HCl , then changing the concentration of HCl would not affect the rate of reaction, and the time of reaction would be the same for trials in which the only difference was [HCl]. The student’s data for trials 1 and 4 (likewise for 3 and 6) show that changing [HCl] significantly alters the time of reaction.
• Disagree. The reaction appears to be first order, not zeroth order, with respect to [HCl]. Tripling [HCl] results in a reaction time that is 1/3 of that when [HCl] = 1.00 M, which means the reaction rate has also tripled, indicating a first-order process.

(e) For the correct calculated moles of HCl reacted (may be implicit):

1.00  g \(CaCO_{3}\times \frac{1mol}{100.09\\\ g}=0.00999\\\ mol\\\ CaCO_{3}\)

\(0.00999\\\ mol\\\ CaCO_{3} \times \frac{2\\\ mol\\\ HCI}{1\\\ mol\\\ CaCO_{3}}=0.0200\\\ mol\\\ HCI\\\ reacted\)

For the correct calculated [HCl] remaining, consistent with the number of moles reacted:

(f) For the correct answer and a valid justification:
Exothermic. The solution temperature increases as the reaction proceeds.

(g) (i) For the correct calculated value (sign not required):

 

\(q_{surr}=mc\Delta T=\left ( 51.0g \right )(4.0\frac{j}{g.^{\circ }C})(21.90^{\circ }C – 21.20^{\circ }C) = 140\\\ J\)

(ii) For the correct calculated value, consistent with (g)(i), and the correct sign, consistent with (f):

Question 4

Topic 8.8 Properties of Buffers

(a) Topic 8.8 Properties of Buffers

(b) Topic 8.8 Properties of Buffers

(c) Topic 8.8 Properties of Buffers

A student is asked to prepare a buffer solution made with equimolar amounts of CH3NH2 (aq) and CH3NH3Cl(s). The student uses 25.00 mL of 0.100 M CH3NH2(aq), which contains 0.00250 mol of CH3NH2, to make the buffer.

(a) Calculate the mass of CH3NH3Cl(s) that contains 0.00250 mol of CH NH3 Cl3.

The student has the following materials and equipment available.

(b) The following table contains a partial procedure for making the buffer solution. Fill in steps 1 and 4 to complete the procedure using only materials and equipment selected from the choices given. (Not all materials listed will be used. Assume that all appropriate safety measures are already in place.)

The value of Kb for CH3NH2(aq) is 4.4 ¥ 10−4, and the pH of the buffer the student prepared is 10.64.

(c) The student prepares a second buffer solution. The student uses 25.00 mL of 0.050 M CH3NH2(aq) instead of 25.00 mL of 0.100 M CH3NH2(aq), and half the mass of CH3NH3Cl(s) that was used in the first buffer. Is the pH of the second buffer greater than, less than, or equal to the pH of the first buffer? Justify your answer.

▶️Answer/Explanation

(a) For the correct calculated value:

\(0.00250\\\ mol\\\ CH_{3}NH_{3}CI \times \frac{67.52 g}{1\\\ mol} = 0.169\\\ g\)

(b) For a correct description of step 1:
          Accept one of the following:
             • Use the spatula, balance, and weighing paper to measure out exactly 0.169 g of CH3NH3CI(s).
            • Use the balance to weigh out the mass of solid in part (a).

   For a correct description of step 4:
   Rinse the buret with a small amount of 0.100 M CH3NH3(aq ), drain, and refill with 0.100 M CH3NH3(aq ).

(c) For the correct answer and a valid justification:
Equal to. The ratio of weak acid to conjugate base is still 1:1.

Question 5

Topic 8.6 Molecular Structure of Acids and Bases

(a) Topic 8.6 Molecular Structure of Acids and Bases

(b) Topic 8.6 Molecular Structure of Acids and Bases

(c) Topic 8.6 Molecular Structure of Acids and Bases

HCl is a molecular gas as a pure substance but acts as an acid in aqueous solution.

(a) A sample of HCl(g) is stored in a rigid 6.00 L container at 7.45 atm and 296 K.

(i) Calculate the number of moles of HCl(g) in the container.

(ii) The rigid 6.00 L container of HCl(g) is cooled to a temperature of 271 K. Calculate the new pressure, in atm, of the HCl(g).

(b) When HCl ionizes in aqueous solution, Cl(aq) ions are formed. In the following box, draw three water molecules with proper orientation around the Cl ion. Use to represent water molecules.

The Ka values for three acids are shown in the preceding table.

(c) The following particulate diagram represents the ionization of one of the acids in the data table. Water molecules have been omitted for clarity. Which acid (HNO2, HCl, or HClO4) is represented in the diagram? Justify your answer using the information in the table.

▶️Answer/Explanation(a) (i) For the correct calculated value:
\(n = \frac{PV}{RT} =\frac{(7.45\\\ atm)(6.00 L)}{(0.08206\frac{L.amt}{mol-K})} = 1.84\\\ mol\)
(ii) For the correct calculated value:
Accept one of the following:

(b) For a correct drawing:
The drawing should show three water molecules with a hydrogen atom (dark circle) oriented towards the Cl ion.

(c) For the correct answer and a valid justification:
HNO2 . The diagram shows most of the molecules in their un-ionized form, indicating a weak acid with a Ka value less than 1, which is consistent with HNO2.

Question 6

TOPIC 3.1 Intermolecular and Interparticle Forces

(a) TOPIC 3.1 Intermolecular and Interparticle Forces

(b) TOPIC 3.1 Intermolecular and Interparticle Forces

(c) TOPIC 3.1 Intermolecular and Interparticle Forces

Answer the following questions related to HBr(l) and HF(l).

(a) In the following table, list all of the types of intermolecular forces present in pure samples of HBr(l) and HF(l).

(b) The enthalpy of vaporization, ΔHοvap, for each liquid is provided in the following table.

(i) Based on the types and relative strengths of intermolecular forces, explain why ΔHοvap of HF(l) is greater than that of HBr(l).

(ii) Calculate the amount of thermal energy, in kJ, required to vaporize 6.85 g of HF(l).

(c) Based on the arrangement of electrons in the Br and F atoms, explain why the bond length in an HBr molecule is greater than that in an HF molecule.

 

▶️Answer/Explanation

6(a) For the correct answer:
HBr(l) : London dispersion forces, dipole-dipole attractions
HF(l) : London dispersion forces, dipole-dipole attractions, hydrogen bonding

6(b)(i) For a correct explanation:
∆HΟvap is greater for HF(l) than HBr( l) because the overall intermolecular forces in HF( )l are stronger than those in HBr(l) due to hydrogen bonding attractions present in HF(l), so more energy is required to separate the molecules in HF(l).

6(b)(ii) For the correct calculated value:

\(6.85\\\ g\\\ HF \times \frac{1\\\ mol}{20.01\\\ g}\times \frac{25.2\\\ kj}{1\\\ mol} = 8.63\\\ kJ\)

6(c) For a correct explanation:
Br has two additional occupied electron shells (n = 3 and n = 4) compared to F (n = 2). The extra electron shells increase the distance between the H and Br nuclei, giving HBr the greater bond length.

Question 7

Topic 4.8 Introduction to Acid-Base Reactions

(a) Topic 4.8 Introduction to Acid-Base Reactions

(b) Topic 4.8 Introduction to Acid-Base Reactions

(c) Topic 4.8 Introduction to Acid-Base Reactions

Strontium hydroxide dissolves in water according to the following equation. The Ksp expression for strontium hydroxide is provided.

\(Sr(OH)_{2}{s} \rightleftharpoons Sr^{2+}(aq) + 2OH^{-}(aq)\)       \(K_{sp} = [Sr^{2+}][OH^{-}]^{2}\)

(a) A student draws the particulate diagram shown to represent the ions present in an aqueous solution of Sr(OH)2. (Water molecules are intentionally omitted.) Identify the error in the student’s drawing.

(b) The student prepares a saturated solution by adding excess Sr(OH)2 (s) to distilled water and stirring until no more solid dissolves. The student then determines that [Sr2+] = 0.043 M in the solution.

(i) Calculate the value of  [OH] in the solution.

(ii) Calculate the value of Ksp for Sr(OH)2.

(c) The student prepares a second saturated solution of Sr( OH)2 in aqueous 0.10 M Sr(NO3)2 instead of water. Will the value of [OH ] in the second solution be greater than, less than, or equal to the value in the first solution? Justify your answer. (Assume constant temperature.)

▶️Answer/Explanation7(a) For a correct answer:
    Accept one of the following:
      • The student’s drawing shows an incorrect ratio of 2 Sr + and OH− ions.
      • The student’s drawing is not charge-balanced.

7(b)(i) For the correct calculated value:
        \(\frac{0.043 mol Sr^{2+}}{1L}\times \frac{2 mol OH^{-}}{1 mol Sr^{2+}} = 0.086 M OH^{-}\)

7(b)(ii) For the correct calculated value, consistent with (b)(i):
       \(K_{sp} = [Sr^{2+}][OH^{-}]^{2} = (0.043)(0.086)^{2} = 3.2 * 10^{-4}\)
7(c) For the correct answer and a valid justification:
Less than. Because the Sr(NO3 )2 aq solution already contains a common ion, Sr2+(aq )+ , the solubility of Sr(OH)2 will be decreased, resulting in a lower value of [OH ] .

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