UWorld General Chemistry
When the hearing aid is operating, what half reaction occurs at the cathode? Reaction 1 (half reaction that occurs at the anode during battery discharge): LiC6 --> Li^+ + e^- + C6 Reaction 2 (balanced net reaction): LiC6 + CoO2 --> LiCoO2 + C6
A. CoO2 + Li^+ + e^- --> LiCoO2 An unknown half reaction can be found by adding the balanced net reaction to the inverse of the known half-reaction. Species in the inverse known half-reaction that do not participate in the discharge (LiC6 and C6) will cancel with the balanced net reaction, leaving only those species present in the other half reaction during the discharge. Obj: The balanced net reaction for an electrochemical cell may be found by taking the sum of the half-reactions, written to be consistent with the direction of the electron flow within the cell. Half-reactions may be scaled stoichiometrically to achieve an electron balance. Species matching in type/number on both sides of the summation are not included in net reaction.
Moving down the column, the alkaline-earth metals are observed to give increasingly vigorous reactions when forming ionic bonds with nonmetals (reactions 1-3). Based on atomic properties, this trend in activity is best explained by comparing: A. The energy required to remove an electron from each atom B. The tendency of each atom to attract electrons C. The extent to which the electron cloud of each atom can be distorted by an external charge D. The energy released when an electron is added to each atom.
A. The energy required to remove an electron from each atom Reactions 1-3 involve the formation of an ionic bond between alkaline earth metals and a nonmetal. This requires the valence electrons of the metal atom to be removed and transferred to the nonmetal atom. Of the 4 properties, only ionization energy quantifies how easily electrons are removed from an atom. Therefore, ionization energy best explains the reactivity trend. Moving down the alkaline earth metal column, ionization energy decreases. This makes removing an electron more favorable and increases reactivity. Obj: Ionization energy is the energy required to move an electron from an atom. Because of the associated ionization and electron transfer involved in forming iconic bonds, the reactivity of atoms forming ionic compounds increases as ionization energy decreases.
A phosphate buffer system based on Reaction 1 is made that contains 5 M H2PO4− (phosphorous acid or dihydrogen phosphate ion) and 0.5 M HPO4^2− (hydrogen phosphate, conjugate base of above acid). Will the pH of this system fall within the pH range for urine? A. Yes; pH = 5.8 B. Yes; pH = 6.8 - assuming pH = pKa; but in a buffer system they both differ C. No; pH = 9.1 - if ammonia buffer system is used D. No; pH = 10.1 - if pKa of the ammonia buffer system is plugged into the equation
A. Yes; pH = 5.8 Given concentration and pKa: HPO4^2- + H+ -> <-H2PO4- pKa = 6.8 0.5 M = HPO4^2- 5.0 M = H2PO4- Using the Henderson - Hasselbalch equation equation to solve for pH: pH = pKa + log A-/HA pH = 6.8 + log 0.5/5 = 6.8 + (-1) = 5.8, which is between 4.5 and 8.0 (acceptable pH range for urine) A buffer solution is made of a weak acid (HA) and its conjugate base (A-). The log of the ratio of base-to-acid determines how much pH differs from pKa. Obj: A buffer resists changes to pH. A buffer system is compromised of a weak acid (or weak base) and its conjugate. The pH of a particular buffer system can be found using the Henderson-Hasselbalch equation.
If the titration of H2PO4- in a urine sample was continued until all of the acid in the solution was neutralized, how many equivalents of NaOH would be needed to fully neutralize the solution? A. 1 B. 2 C. 3 D. 4
B. 2 The plates between 2 equivalence points has a minimal slope with little change in pH because it effectively acts as a buffer region of an acid and its conjugate base. When enough NaOH is added to consume all of a given acidic species, the pH increases rapidly and the curve shows a very steep slope. If H2PO4-is fully neutralized, 2 acidic proton must be neutralized and an equivalence point would be seen in the titration curve for each of the protons. This indicates that a total of 2 equivalents of NaOH would be required to fully neutralize the H2PO4- (one OH- from NaOH for each H+). Obj: Polyprotic acids have more than 1 proton that can dissociate in water. The titration curve of a polyprotic acid will show as many plateaus and steep climbs as the number of protons. These represent the buffer regions and equivalence points in titrations, respectively.
If radium-226 were to undergo radioactive decay by electron capture (a type of beta decay) instead of alpha emission, the resulting nucleus would be? A. 222/86 Rn (alpha decay) B. 226/87 Fr C. 226/88 Ra (gamma decay - no change in mass and atomic #'s) D. 226/89 Ac (B- decay - electron emission)
B. 226/87 Fr Radioactive beta decay can occur in 3 forms: B- decay (electron emission) in which a neutron converts into a nuclear proton and emits a electron, B+ decay (positron emission) in which a proton converts a neutron and emits a position - an electron with a + charge, and electron capture in which a proton captures an electron near the nucleus and converts into a neutron without a positron or electron emission. In all 3 forms, the mass number (protons + neutrons) remains unchanged, while the atomic number (protons) increases (B- decay) or decreases (B+ decay and electron capture). As the atomic number changes, the identity of element changes accordingly. If radium-226 underwent an electron capture, the result would yield a nucleus of the same mass number (226) but an atomic number that is 1 less (88 - 1 = 87). Therefore, francium-226 would be detected.
When comparing atoms across the same row of the period table, which of the following groups will contain the atom with the lowest second ionization energy? A. Group 1 B. Group 2 C. Group 14 D. Group 16
B. Group 2 First ionization energies tend to increase across a row and decrease down a column. The same is true for 2nd ionization energy except when the second electron being removed is NOT a valence electron. If comparing elements in a period (row) three, the nonmetals Si (Group 14) and S (Group 16) are farther to the right of the period and both have a higher second ionization energy than the metals Na and Mg. However, the second ionization energy of Na (Group 1) is much higher than that of Mg (Group 2) because Na has only 1 valence electron and removing a second electron from Na requires the loss of a core electron. The availability of a second valence electron gives Mg (Group 2) the lowest second ionization energy. The second ionization energy is the energy required to the remove the second of two electrons from an atom. The second ionization energy tends to increase across a period and decrease down a group; however, ionization involving core electrons are higher energy than those involving valence electrons.
If a student adds bromothymol blue as an indicator to determine the amount of titratable acid in 25 mL of urine according to Folin's method, what effect will the bromothymol blue have on the pH of the urine sample? A. It will increase the pH of the sample. B. It will decrease the pH of the sample C. It will neutralize the sample D. It will have no effect on the pH of the sample.
D. It will have no effect on the pH of the sample. An indicator is used in acid-base titrations to determine the endpoint of a titration. A good indicator should change color close to the equivalence point, the point at which the stoichiometric amount of titrant needed to consume all the acid or base in the solution has been added. In the case of a buffer solution, a titration is halfway to the equivalence point when a pH is achieved that is near the pKa value of the buffer's acid. Indicators DO NOT participate in the reaction Bromothymol blue is an indicator that changes from pale yellow to dark blue, from pH 6 to pH 8, but it has NO EFFECT on the pH of the solution
The chemical process that occurs during acidosis can be replicated in vitro by adding a strong acid to an ammonia buffer solution. How will the pH change in the solution if 1.0 mL of 0.5 M H2SO4 is added to a 100 mL solution containing 1 M NH3 and 1 M NH4+? A. The pH will increase slightly (<0.1 pH units) because NH3 and NH4+ are a conjugate acid/base pair B. The pH will decrease slightly (<0.1 pH units) because NH3 and NH4+ act as a buffer C. The pH will increase significantly (>1.0 pH units) because NH3 is a strong base D. The pH will decrease significantly ( >1.0 pH units) because H2SO4 is a strong acid.
B. The pH will decrease slightly (<0.1 pH units) because NH3 and NH4+ act as a buffer In the experiment described in the question, a strong acid (H2SO4) is added to an ammonia buffer system. The amount of H2SO4 added is mush less than the concentration of NH3 (weak base) and will react with the NH3 to form NH4+ (a weak acid); therefore, the pH will decrease slightly but the ammonia buffer system will resist a large change in pH. It will decrease slightly because there is a sufficient concentration of NH3 solution to use the H+ (added from H2SO4) to form NH4+ The addition of an acid will cause the pH to decrease, not increase. Obj: Buffer systems counteract H+ and OH- ions. When small amounts of a strong acid or base are added to a buffer solution, the solution pH changes only slightly. Adding a strong acid causes the pH to decrease slightly whereas adding a strong base causes the pH to increase slightly.
Which of the following is the electron configuration of the Ca^2+ ion? A. [Ne]3s^2 (electron config for Magnesium atom) B. [Ne]3s^2 3p^6 (electron config for Ca+ which doesn't form) C. [Ne]3s^2 3p^6 4s^1 D. [Ne]3s^2 3p^6 4s^2 (electron config for a Calcium atom)
B. [Ne]3s^23p^6 During ionization, the electrons in the highest energy shell/subshell are lost first. Calcium therefore loses its 2 valence electrons located in 4s. As a result, the electron configuration for the remaining electrons in the Ca+2 ion is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 which can be abbreviated as [Ne]3s^23p^6 Obj: An electron configuration accounts for all electrons held by an atom or ion listed sequentially by shell or subshell in order of increasing energy. A noble gas abbrev. may be used to represent configurations of lower-level electrons. Ionized atoms lose electrons from high energy levels first.
Equilibrium is reached in a reversible reaction when the rate of the forward reaction = the rate of the reverse reaction. The graph shows the changes in concentration during the formation of HI(g) from a mixture of H2(g) and I2(g). If the measurement is taken at equilibrium, how many moles of HI(g) are present in a 750 mL sample of the reaction mixture? A. 0.8 mol B. 1.9 mol C. 4.5 mol D. 6.0 mol
C. 4.5 mol In the graph of concentration vs. time for the reaction below H2(g) and I2(g), the constant (horizontal) regions of the curves (after 5 hours) indicate a state of equilibrium in which the concentration of the chemical species no longer change. From this region of the graph, the equilibrium concentration of HI(g) is seen to be 6.0 M. Therefore, if a 750 mL sample of the reaction were analyzed at equilibrium. 750 mL x I L/1000 mL x 6.0 mol/I L = 4.5 mol Obj: Equilibrium is achieved when 2 opposing chemical reactions occur simultaneously at the same rate such that the concentrations of the chemical species become constant. Equilibrium reaction rates are equal, but the equilibrium concentrations of chemical species may be unequal.
What are the predominant species in the phosphate buffer system and in the ammonia buffer system in an aqueous solution at a pH of 8? A. H2PO4- and NH4+ B. H2PO4- and NH3 C. HPO4^2− and NH4+ D. HPO4^2− and NH3
C. HPO4^2− and NH4+ Accordingly, consider the equilibrium: H2PO4- -><- HPO4^2− + H+ At a pH of 8.0, HPO4^2− will be the dominate species in the phosphate buffer because the pKa of H2PO4- is 6.8, which is less than 8. Likewise consider the equilibrium, NH4+ -><= NH3 + H+. In this case, the acidic NH4+ will be the dominant species in the ammonia buffer because the pKa of NH4+ is 9.1, which is greater than 8. Objective: The pKa is a measure of the acidity of a particular proton in a molecule. In a buffer system, when the pKa is equal to the pH, the amount of undissocciated acid (HA) = amount of deprotonated conjugate base (A-).
What is the mass necessary to generate 11.2 L of hydrogen gas if calcium metal reacts with water at standard temperature and pressure (STP)? A. 2.0 g - if the calculation is done using the molar mass of H2 (2.02 g/mol) instead of 18.02 g/mol for H2O B. 4.5 g - 2:1 mole ratio is flipped C. 9.0 g - 1:1 mole ratio is used D. 18.0 g Balanced Equation: Ca + 2H2O -> Ca(OH)2 + H2
D. 18.0 g Obj: At standard temperature and pressure (STP), gases in reactions can be related to the molar volume of 22.4 L/mol. Mole ratios from balanced chemical can be used to relate the moles of the gases to the moles of the reaction species. The mass of a given number of moles of a particular species can be found using its molar mass.
Which compound has the bond with the smallest dipole moment? A. H2O B. H2S C. H2Se D. H2Te
D. H2Te Electronegativity increases when moving up a column on the period table/decrease moving down a group (column) on the period table. As a result, the electronegativity difference (and dipole moment) of the hydrogen-chalcogen bond increases moving up the column. Therefore, the ranking of the dipole moments for each of the 4 bonds (largest to smallest) is as follows" H-O > H-S > H-Se > H-Te. Objective: The difference in electronegativity between 2 covalently bonded atoms is proportional to (and indicative of) the magnitude of the dipole moment along the bond between the atoms. Larger differences in electronegativity = larger dipole moments between atoms.
Which of the following alkaline-earth species are diamagnetic? I. Ca II. Ca^2+ III. Sr A. I only B. II only C. I and III only D. I, II, and III
D. I, II, and III The electron configurations of the alkaline-earth metals end with a completely fill s-block. Therefore, the electrons occupying the corresponding s orbital in the respective valence shells are paired. Accordingly, isolated atoms of Ca and Sr have no unpaired electrons are are diamagnetic (Numbers I and III). The Ca^+2 ion results from losing both electrons from the valence shell, but because all the shells below the valence shell are fully filled (no unpaired electrons , an isolated Ca^+2 ion is also diamagnetic (Number II). Obj: Electrons fill the shells and sub shells of an atom in order increasing energy according to Hund's rule and the Pauli exclusion principle. Unpaired electrons in the electron configuration of an atom or ion result in paramagnetism, but a configuration without unpaired electrons results in diamagnetism.
