8:55 PM MCAT Physical Sciences Sample Questions | ||||
#mcat practice questions #MCAT Sample Questions. Physical SciencesPassage I A series of chemical reactions was carried out to study the chemistry of lead. Reaction 1 Initially, 15.0 mL of 0.300 M Pb(NO3 )2 (aq) was mixed with 15.0 mL of 0.300 M Na2 SO4 (aq).All the Pb(NO3 )2 reacted to form Compound A, a white precipitate.Compound A was removed by filtration. Reaction 2 Next, 15.0 mL of 0.300 M KI(aq) was added to Compound A. The mixture was agitated and some of Compound A dissolved. In addition, a yellow precipitate of PbI2 (s) was formed. Reaction 3 The PbI2 (s) was separated and mixed with 15.0 mL of 0.300 M Na2 CO3 (aq). A white precipitate of PbCO3 (s) formed. All of the PbI2 (s) was converted into PbCO3 (s). Reaction 4 The PbCO3 (s) was removed by filtration and a small sample gave off a gas when treated with dilute HCl. Some sample questions on this passage are as follows:
Answer: A Explanation: Reaction 4 is shown in the following equation, which is answer choice A. PbCO3 (s) + 2 HCl(aq) = PbCl2 (aq) + CO2 (g) + H2 O(l) Answers B and D do not show a reaction involving PbCO3(s), as required by Reaction 4. Answer C shows an implausible and unbalanced equation. Thus, answer choice A is the best answer. Answer: D Explanation: Reaction 1 is shown in the following equation. Compound A, the white solid, is PbSO4 (s). Neither the reactant Pb(NO3 )2 nor the product NaNO3 can precipitate because all nitrates and sodium salts are water soluble. PbI2 cannot precipitate because iodide is not present. Thus, answer choice D is the best answer.
Answer: A Explanation: The dissolution of Pb(OH)2 (s) is represented by the following equation. Pb(OH)2 (s)=Pb2 (aq) + 2 OH-(aq) At pH 9, the concentration of OH-(aq) is greater than the concentration of OH-(aq) at pH 7. According to Le Chвtelier's principle, the additional common ion, OH-(aq), will shift the position of equilibrium to the left, and less Pb(OH)2 will dissolve. Thus, answer choice A is the best answer. Answer: B Explanation: The reactions described in the passage show that lead(II) is successively precipitated as PbSO4. PbI2. and PbCO3. This sequence shows (assuming equal anion concentrations, as must be done here) that PbCO3 is less soluble than PbI2. and PbI2 is less soluble than PbSO4. The order in which the anions precipitate Pb 2 - is: CO3 2 - then I - then SO4 2 -. When this sequence is applied to the question, answer choice B is in the correct order, and answers A, C, and D are all in the opposite order. Thus, answer choice B is the best answer.
Answer: B Explanation: The initial Na2 SO4 (aq) solution in Reaction 1 is 15 mL of 0.300 MNa2 SO4 (aq). (15.0 mL) ( 1 L ) (0.300 mol Na2 SO4 ) (2 mol Na + ) (1000 mL) (1 L Na2 SO4 (aq) ) (1 mol Na2 SO4 )
| ||||
|
| ||||
| Total comments: 0 | |