Question
Topic – 7.4 Calculating the Equilibrium Constant
(a) – Topic- 7.4 Equilibrium Constant
(b)- Topic- 8.2 pH and pOH of Strong Acids and Base
(c)- Topic- 2.5 Lewis Diagram & 1.8 Valence Electrons
(d) (i)- Topic – 4.2 Net Ionic Equations
(ii)- Topic – 3.7 Solution and mixtures
(e)- Topic – 4.9 Oxidation-Reduction (Redox) Reactions
(f)- Topic – 5.7 Introduction to Reaction Mechanisms
(g)- Topic – 5.11 Catalysis
HCOOH(aq ) + H2O(l) ⇔ H3O+(aq) + HCOO– (aq) Ka = 1.8 × 10-4
Methanoic acid, HCOOH, ionizes according to the equation above.
(a) Write the expression for the equilibrium constant, Ka, for the reaction.
(b) Calculate the pH of a 0.25 M solution of HCOOH.
(c) In the box below, complete the Lewis electron-dot diagram for HCOOH. Show all bonding and nonbonding valence electrons.
H2NNH2(aq) + H2O(l) ⇔ H2NNH3+ (aq) + OH–(aq) Kb = 1.3 × 10-6
(d) In aqueous solution, the compound H2NNH2 reacts according to the equation above. A 50.0 mL sample of 0.25 M H2NNH2(aq) is combined with a 50.0 mL sample of 0.25 M HCOOH(aq).
(i) Write the balanced net ionic equation for the reaction that occurs when H2NNH2 is combined with HCOOH.
(ii) Is the resulting solution acidic, basic, or neutral? Justify your answer.
When a catalyst is added to a solution of HCOOH(aq), the reaction represented by the following equation occurs.
HCOOH(aq) → H2 (g) + CO2 (g)
(e) Is the reaction a redox reaction? Justify your answer.
(f) The reaction occurs in a rigid 4.3 L vessel at 25°C, and the total pressure is monitored, as shown in the graph above. The vessel originally did not contain any gas. Calculate the number of moles of CO2(g) produced in the reaction. (Assume that the amount of CO2(g) dissolved in the solution is negligible.)
(g) After the reaction has proceeded for several minutes, does the amount of catalyst increase, decrease, or remain the same? Justify your answer.
Answer/Explanation
Ans:
(a) For the correct expression:
\(K_{a}=\frac{[H_{3}O^{+}][HCOO^{-}]}{[HCOOH]}\)
(b) For the correct calculated concentration of H3O+ :
HCOOH + H2O ⇔ H3O+ + HCOO–
I 0.25 0 0
C –x +x +x
E 0.25 – x x x
Let [H3O+ ] = x, then 1.8 × 10-4 \(\frac{x^{2}}{(0.25-x)}\)
Assume x << 0.25, then 1.8 × 10-4 \(\frac{x^{2}}{(0.25)}\) ⇒ x = 0.0067 M
For the correct calculated value of pH:
pH = -log[H3O + ] = – log(0.0067) = 2.17
(c) For the correct diagram:
(d) (i) For the correct balanced equation (state symbols not required):
H2NNH2 (aq) + HCOOH(aq) → H2NNH3+ (aq) + HCOO– (aq)
(ii) For the correct answer and a valid justification:
Acidic. The Ka of H2NNH3+ is greater than the Kb of HCOO−, so the production of H3O+(aq) occurs to a greater extent than the production of OH−(aq).
(e) For the correct answer and a valid justification:
Accept one of the following:
• Yes. The oxidation number of hydrogen changes from +1 in HCOOH to zero in H2.
• Yes. The oxidation number of carbon changes from +2 in HCOOH to +4 in CO2.
(f) For the correct calculated value of the pressure of CO2 (may be implicit):
24 atm total × 1 atm CO2 / 2 atm of product = 12 atm CO2
For the correct calculated number of moles of CO2:
PV = nRT
\(n = \frac{PV}{RT}=\frac{(12 atm)(4.3 L)}{(0.08206 L atm mol^{-1}K^{-})(298 K)}=2.1 mol CO_{2}\)
(g) For the correct answer and a valid justification:
It would remain the same. In a catalyzed reaction the net amount of catalyst is constant.
Question
Topic – 1.2 Mass Spectra of Elements
(a)- Topic – 1.2 Mass Spectra of Elements
(b)- Topic – 3.1 Intermolecular and Interparticle Forces
(c)- Topic- 5.7 Introduction to Reaction Mechanisms
(d)- Topic- 9.2 Absolute Entropy and Entropy Changes
(e)- Topic- 9.2 Absolute Entropy and Entropy Changes
(f)- Topic- 9.2 Absolute Entropy and Entropy Changes
(g)- Topic- 1.6 Photoelectron Spectroscopy
(h)- Topic- 1.5 Atomic Structure
Answer the following questions about the element Si and some of its compounds.
(a) The mass spectrum of a pure sample of Si is shown below.
(i) How many protons and how many neutrons are in the nucleus of an atom of the most abundant isotope of Si ?
(ii) Write the ground-state electron configuration of Si.
Two compounds that contain Si are SiO2 and SiH4.
(b) At 161 K, SiH4 boils but SiO2 remains as a solid. Using principles of interparticle forces, explain the difference in boiling points.
At high temperatures, SiH4 decomposes to form solid silicon and hydrogen gas.
(c) Write a balanced equation for the reaction.
A table of absolute entropies of some substances is given below.
(d) Explain why the absolute molar entropy of Si(s) is less than that of H2(g).
(e) Calculate the value, in J/(mol. K), of ∆S° for the reaction.
(f) The reaction is thermodynamically favorable at all temperatures. Explain why the reaction occurs only at high temperatures.
(g) A partial photoelectron spectrum of pure Si is shown below. On the spectrum, draw the missing peak that corresponds to the electrons in the 3p sublevel.
(h) Using principles of atomic structure, explain why the first ionization energy of Ge is lower than that of Si.
(i) A single photon with a wavelength of 4.00 × 10−7 m is absorbed by the Si sample. Calculate the energy of the photon in joules.
Answer/Explanation
Ans:
(a) (i) For the correct answer:
14 protons and 14 neutrons
(ii) For the correct answer:
Accept one of the following:
• 1s2 2s2 2p6 3s2 3p2
• [Ne] 3s2 3p2
(b) For a correct explanation:
SiH4 is composed of molecules, for which the only intermolecular forces are London dispersion forces. SiO2 is a network covalent compound with covalent bonds between silicon and oxygen atoms. London dispersion forces are much weaker than covalent bonds, so SiH4 boils at a much lower temperature than SiO2.
(c) For the correct balanced equation (state symbols not required):
SiH4 ( g) → Si(s) + 2 H2(g)
(d) For a correct explanation:
The H2(g) molecules are more highly dispersed than the Si(s) atoms and, therefore, have a higher absolute molar entropy. Silicon is a solid; therefore, its atoms are in fixed positions, are less dispersed, and have a lower absolute molar entropy.
(e) For the correct calculated value:
ΔS0rxn = (18 + 2(131)) – 205 = +75 J/(molrxn • K)
(f) For a correct explanation:
High temperature is required for the reactant particles to have sufficient thermal energy to overcome the activation energy of the reaction.
(g) For the correct peak height and location:
The peak should be drawn to the right of the other peaks, and it should reach the second line above the horizontal axis.
(h) For a correct explanation:
The valence electrons of a Ge atom occupy a higher shell (n=4) than those of a Si atom (n=3), so the average distance between the nucleus and the valence electrons is greater in Ge than in Si. This greater separation results in weaker Coulombic attractions between the Ge nucleus and its valence electrons, making them less tightly bound and, therefore, easier to remove compared to those in Si.
(i) For the correct calculated value:
\(E = hv = h\left ( \frac{c}{\lambda } \right )=6.626\times 10^{-34}J\cdot s\left ( \frac{2.998\times 10^{8}ms^{-1}}{4.00\times 10^{-7}m} \right )=4.97 \times 10^{-19} J\)
Question
Topic – 4.2 Net Ionic Equations
(a)- Topic- 4.2 Net Ionic Equations
(b)- Topic- 8.1 Introduction to Acids and Bases
(c)- Topic- 3.7 Solutions and Mixtures
(d)- Topic- 3.7 Solutions and Mixtures
(e)- Topic- 3.7 Solutions and Mixtures
(f)- Topic- 3.7 Solutions and Mixtures
(g)- Topic- 3.7 Solutions and Mixtures
A student is given the task of determining the molar concentration of a CuSO4 solution using two different procedures, precipitation and spectrophotometry.
For the precipitation experiment, the student adds 20.0 mL of 0.200 M Ba(NO3)2 to 50.0 mL of the CuSO4(aq). The reaction goes to completion, and a white precipitate forms. The student filters the precipitate and dries it overnight. The data are given in the following table.
(a) Write a balanced net ionic equation for the precipitation reaction.
(b) Calculate the number of moles of precipitate formed.
(c) Calculate the molarity of the original CuSO4 solution.
For the spectrophotometry experiment, the student first makes a standard curve. The student uses a 0.1000 M solution of CuSO4(aq) to make three more solutions of known concentration (0.0500 M, 0.0300 M, and 0.0100 M) in 50.00 mL volumetric flasks.
(d) Calculate the volume of 0.1000 M CuSO4(aq) needed to make 50.00 mL of 0.0500 M CuSO4(aq).
(e) Briefly describe the procedure the student should follow to make 50.00 mL of 0.0500 M CuSO4(aq) using 0.1000 M CuSO4(aq), a 50.00 mL volumetric flask, and other standard laboratory equipment. Assume that all appropriate safety precautions will be taken.
The standard curve is given below.
(f) The absorbance of the CuSO4 solution of unknown concentration is 0.219. Determine the molarity of the solution.
(g) A second student performs the same experiment. There are a few drops of water in the cuvette before the second student adds the CuSO4(aq) solution of unknown concentration. Will this result in a CuSO4 (aq) concentration for the unknown that is greater than, less than, or equal to the concentration determined in part (f) ? Justify your answer.
Answer/Explanation
Ans:
(a) For the correct balanced equation (state symbols not required):
Ba2+(aq) + SO42-(aq) →BaSO4(s)
(b) For the correct calculated value of the mass of precipitate (may be implicit):
1.136 g – 0.764 g = 0.372 g BaSO4
For the correct calculated value of the number of moles, consistent with mass of precipitate:
\(0.372 g \times \frac{1 mol}{233.39 g}= 0.00159 mol\)
(c) For the correct calculated value, consistent with part (b):
0.00159 mol BaSO4 × \(\frac{1 mol CuSO_{4}}{1 mol BaSO_{4}}\) = 0.00159 mol CuSO4
\(\frac{0.00159 mol CuSO_{4}}{0.0500 L}\)= 0.0318 M CuSO4 (0.0319 M if decimals are carried)
(d) For the correct calculated value:
M1V1 = M2V2
\(V_{1}= \frac{(0.0500 M)(50.00 mL)}{(0.1000 M)}=25.0 mL\)
(e) For a correct technique to measure the volume of solution:
First, measure out the correct volume of 0.1000 M CuSO4 solution with a 25.0 mL volumetric pipet (graduated cylinder or buret is acceptable).
For a correct technique to dilute the solution to the final volume:
Transfer the 25.0 mL of solution to a 50.00 mL volumetric flask and dilute the solution with water up to the 50.00 mL mark.
(f) For the correct value (between 0.032 M and 0.038 M):
Accept one of the following:
• \(y = mx = \frac{0.63}{0.1000}x=6.3x\)
\(x = \frac{y}{6.3}=\frac{0.219 M}{6.3}=0.035 M\)
• Estimated value from the graph within the specified range.
(g) For the correct answer:
The concentration will be less than that determined in part (f).
For a valid justification:
The additional water will decrease the concentration of CuSO4 in the cuvette. Therefore, there will be a decrease in absorbance (according to the Beer-Lambert law). This dilution results in a lower estimated concentration of CuSO4.
Question
Topic – 1.1 Moles and Molar mass
(a)- Topic- 1.1 Moles and Molar Mass
(b)- Topic- 1.1 Molar Mass
(c)- Topic- 1.3 Elemental Compositions of Pure Substances
4 Fe(s)+ 3 O2 (g) → 2 Fe 2 O 3(s) ∆H° − 1650 kJ / molrxn
A student investigates a reaction used in hand warmers, represented above. The student mixes Fe(s) with a catalyst and sand in a small open container. The student measures the temperature of the mixture as the reaction proceeds. The data are given in the following table.
(a) The mixture (Fe(s), catalyst, and sand) has a total mass of 15.0 g and a specific heat capacity of 0.72 J/(g·°C). Calculate the amount of heat absorbed by the mixture from 0 minutes to 4 minutes.
(b) Calculate the mass of Fe(s), in grams, that reacted to generate the amount of heat calculated in part (a).
(c) In a second experiment, the student uses twice the mass of iron as that calculated in part (b) but the same mass of sand as in the first experiment. Would the maximum temperature reached in the second experiment be greater than, less than, or equal to the maximum temperature in the first experiment? Justify your answer.
Answer/Explanation
Ans:
(a) For the correct calculated value with units:
q = mc ΔT = (15.0 g)(0.72 J/(g· 0C))(39.7 0C – 22.0 0C) = 190 J
(b) For the correct calculated value of the moles of reaction, consistent with part (a) (may be implicit):
qsys = – qsurr
\(-190 J\times \frac{1 kJ}{1000 J}\times \frac{1 mol _{rxn}}{-650 kJ}=0.00012 mol_{rxn}\)
For the correct calculated value of the mass of iron:
\(0.00012mol_{rxn}\times \frac{4 mol Fe}{1 mol_{rxn}}\times \frac{55.85 g Fe}{1 mol Fe}=0.027 g Fe \) (0.026 g if decimals are carried)
(c) For the correct answer and a valid justification:
Greater than. A greater mass of iron provides a greater number of moles of reaction, which would transfer a greater quantity of thermal energy to the same mass of sand and therefore lead to a greater maximum temperature.
Question
Topic – 4.9 Oxidation-Reduction (Redox) Reactions
(a)- Topic- 4.9 Reduction Reactions
(b)- Topic- 4.7 Types of Chemical reactions
(c)- – Topic- 4.7 Types of Chemical reactions
Molten MgCl2 can be decomposed into its elements if a sufficient voltage is applied using inert electrodes. The products of the reaction are liquid Mg (at the cathode) and Cl2 gas (at the anode). A simplified representation of the cell is shown above. The reduction half-reactions related to the overall reaction in the cell are given in the table.
(a) Draw an arrow on the diagram to show the direction of electron flow through the external circuit as the cell operates.
(b) Would an applied voltage of 2.0 V be sufficient for the reaction to occur? Support your claim with a calculation as part of your answer.
(c) If the current in the cell is kept at a constant 5.00 amps, how many seconds does it take to produce 2.00 g of Mg(l) at the cathode?
Answer/Explanation
Ans:
(a) For the correct answer:
Electron flow should be indicated only in a counter-clockwise direction in the external circuit, from the Cl2 anode to the Mg cathode.
(b) For the correct answer and calculated value:
No, because 2.0 V is less than 3.73 V, which is the minimum voltage needed for electrolysis to occur.
E0cell =− 2.37 V + ( -1.36 V) = -3.73 V
(c) For the correct calculated value of moles of electrons (may be implicit):
\(2.00 g Mg\times \frac{1 mol Mg}{24.30 g Mg}\times \frac{2 mol e^{-}}{1 mol Mg}=0.165 mol e^{-}\)
For the correct calculated number of seconds:
\(0.165 mol e^{-}\times \frac{96,485 C}{1 mol e^{-}}\times \frac{1s}{5.00 C}=3180 s\)
Question
Topic – 3.3 Solids & 3.10 Solubility
(a)– Topic- 3.3 Solids
(b)- Topic- 3.10 Solubility
(c)– Topic- 3.10 Solubility
(d)– Topic- 3.10 Solubility
A student is studying the properties of CaSO4 and PbSO4. The student has samples of both compounds, which are white powders.
(a) The student tests the electrical conductivity of each solid and observes that neither solid conducts electricity. Describe the structures of the solids that account for their inability to conduct electricity.
The student places excess CaSO4(s) in a beaker containing 100 mL of water and places excess PbSO4 (s) in another beaker containing 100 mL of water. The student stirs the contents of the beakers and then measures the electrical conductivity of the solution in each beaker. The student observes that the conductivity of the solution in the beaker containing the CaSO4 (s) is higher than the conductivity of the solution in the beaker containing the PbSO4 (s).
(b) Which compound is more soluble in water, CaSO4(s) or PbSO4(s) ? Justify your answer based on the results of the conductivity test.
The left side of the diagram below shows a particulate representation of the contents of the beaker containing the CaSO4(s) from the solution conductivity experiment.
(c) Draw a particulate representation of PbSO4(s) and the ions dissolved in the solution in the beaker on the right in the diagram. Draw the particles to look like those shown to the right of the beaker. Draw an appropriate number of dissolved ions relative to the number of dissolved ions in the beaker on the left.
(d) The student attempts to increase the solubility of CaSO4(s) by adding 10.0 mL of 2 M H2SO4 (aq) to the beaker, and observes that additional precipitate forms in the beaker. Explain this observation.
Answer/Explanation
Ans:
(a) For a correct description:
Ionic solids do not have free-moving ions that are required to carry an electric current. Therefore, there is no conduction of electricity.
(b) For the correct answer and a valid justification:
CaSO4. The greater electrical conductivity of the CaSO4 solution relative to the PbSO4 solution implies a higher concentration of ions, which comes from the dissolution (dissociation) of CaSO4 to a greater extent.
(c) For a correct drawing that shows an equal number of cations and anions:
The drawing shows solid PbSO4 at the bottom of the beaker (similar to the solid shown for CaSO4) and fewer dissociated Pb2+ and SO42− ions in the solution.
(d) For a correct explanation:
The additional precipitate is CaSO4 that forms in response to the increased [SO42−] in solution. According to Le Chatelier’s principle (Q > Ksp), the introduction of SO42− as a common ion shifts the equilibrium towards the formation of more CaSO4(s).
Question
Topic –3 Properties of Substances and Mixtures
(a)– Topic- 3.4 Ideal gas law
(b)- Topic- 3.5 Kinetic Molecular Theory
(c)- Topic- 3.6 Deviation from Ideal gas law
(d)- Topic- 3.6 Deviation from Ideal gas law
A student investigates gas behavior using a rigid cylinder with a movable piston of negligible mass, as shown in the diagram above. The cylinder contains 0.325 mol of O2(g).
(a) The cylinder has a volume of 7.95 L at 25°C and 1.00 atm. Calculate the density of the O2(g), in g/L, under these conditions.
(b) Attempting to change the density of the O2(g), the student opens the valve on the side of the cylinder, pushes down on the piston to release some of the gas, and closes the valve again. The temperature of the gas remains constant at 25°C. Will this action change the density of the gas remaining in the cylinder? Justify your answer.
(c) The student tries to change the density of the O2(g) by cooling the cylinder to −55°C, which causes the volume of the gas to decrease. Using principles of kinetic molecular theory, explain why the volume of the O2(g) decreases when the temperature decreases to −55°C.
(d) The student further cools the cylinder to −180°C and observes that the measured volume of the O2(g) is substantially smaller than the volume that is calculated using the ideal gas law. Assume all equipment is functioning properly. Explain why the measured volume of the O2(g) is smaller than the calculated volume. (The boiling point of O2(l) is −183°C.)
Answer/Explanation
Ans:
(a) For the correct calculated value:
Accept one of the following:
• \(0.325 mol O_{2}\times \frac{32.00 gO_{2}}{1 mol O_{2}}=10.4 gO_{2}\)
\(D = \frac{m}{V}=\frac{10.4 g}{7.95 L}=1.31 g/L\)
• \(D = \frac{m}{V}=\frac{P(MM)}{RT}=\frac{(1.0 ati)(32.00 g/mol)}{(0.08206\frac{L\cdot atm}{mol\cdot K})(298K)}=1.31 g/L\)
(b) For the correct answer and a valid justification:
Accept one of the following:
• No, the density of the gas remains constant because P, R, and T remain constant AND the mass and volume of O2 decrease proportionately.
• A mathematical justification is shown below
\(D = \frac{m}{V}=\frac{n ~moles~ of ~O_{2}\times molar ~mass ~of ~O_{2}}{\frac{nRT}{P}}=\frac{P\times (molar ~mass ~of~ O_{2})}{RT}\)
(c) For a valid explanation:
Accept one of the following:
• As the gas cools, the average kinetic energy (speed) of the O2 molecules decreases. The molecules rebound with less energy when they collide with each other and the walls of the container. The spacing between particles decreases, causing the volume occupied by the gas to decrease.
• As the gas cools, the average kinetic energy (speed) of the O2 molecules decreases. The molecules rebound with less energy when they collide with each other and the walls of the container. The only way for the molecules to maintain a constant rate of collisions with the walls of the container (maintaining a pressure of 1.00 atm) is for the volume of the gas to decrease.
(d) For a valid explanation:
The ideal gas law assumes that gas particles do not experience interparticle attractions. As a real gas cools further, the intermolecular forces have greater effect as the average speed of the molecules decreases, resulting in inelastic collisions. To maintain a gas pressure of 1.00 atm, the volume must decrease to accommodate more collisions with less energy.