Chem Particulate Practice Key

AP Chemistry: Particulate Level Review For each requested picture (particle view, microscopic view) use circles to represent the atoms (unless otherwise indicated. Different colors should be used to represent different elements.   1) Draw microscopic views (particle views) for each of the following. Each picture should have a minimum of 5 particles.   diatomic elemental gas 5 pairs of two identical circles A mixture of a diatomic gas and a monatomic gas 5 pairs of two identical circles and 5 single circles of a different color A mixture of a gaseous element and a gaseous compound 5 pairs of two identical circles and 5 groups of three circles, two of one color and one of another An aqueous solution of an ionic compound 5 groups of two different colored circles surrounded by water molecules 2) Consider the reaction between NO(g) and O2(g) a) In the box below, show 6 NO molecules and 6 O2 molecules. 6 NO molecules and 6 O2 molecules The reaction proceeds to form NO2(g). b) Write the balanced chemical equation for the reaction: 2NO + O2 -> 2NO2 c) In the box to the right, draw the molecular representation after the reaction has gone to completion. 6 NO2 molecules d) Which substance is the limiting reactant? Justify your answer. NO is the limiting reactant. The ratio of NO to O2 is 2:1. Since there are equal numbers of NO and O2 molecules, NO will be used up first. 3) A) Draw structural formulas to represent the Bronsted-Lowry acid-base reaction between HF and H2O. HF + H2O -> H3O+ + F- B) Draw structural formulas to represent the Bronsted-Lowry acid-base reaction between HF and OH- HF + OH- -> H2O + F- Question 4: For the mixture of gases represented below, the total pressure is 1.5 atm. Calculate the partial pressure of each gas. Відкриється в новому вікні vacaero.com container with 5 blue circles and 10 red circles representing gas molecules Partial Pressure of Blue Gas: Fraction of blue gas = 5 / (5 + 10) = 1/3 Partial pressure = (1/3) * 1.5 atm = 0.5 atm Partial Pressure of Red Gas: Fraction of red gas = 10 / (5 + 10) = 2/3 Partial pressure = (2/3) * 1.5 atm = 1.0 atm Question 5: For the weak acid, HA, being titrated with a strong base (OH-), draw molecule pictures (in AP speak: particulate view or microscopic view) to show the progress of the titration. In the bottom row, compare the concentration of 1 HA and A- as modeled in the initial column. The best diagrams would show the sodium ion floating around, but they just complicate the picture. 2   1. www.gpschools.org www.gpschools.org 2. www.gpschools.org www.gpschools.org Please start with 8 acid molecules. titration process with molecule diagrams Initial[HA] >> [A-] 8 HA molecules Before ½ way point[HA] > [A-] 6 HA molecules and 2 A molecules At ½ way point[HA] = [A-] 4 HA molecules and 4 A molecules 2 HA molecules and 6 A molecules Equivalence point[HA] << [A-] 8 A molecules After equivalence point[HA] << [A-] 8 A molecules and excess OH ions Question 6: a) Draw a particle view of lead (II) nitrate solution. lead (II) nitrate solution with Pb2+ ions represented by one type of circle and NO3 ions represented by another type of circle b) Draw the particle view of the reaction mixture after the two solutions are mixed. Justify your answer. reaction mixture with PbI2 precipitate, excess Pb2+ ions, and NO3 ions Pb2+ is in excess because the ratio of Pb2+ to I- is 1:2. Since we have equal volumes of 0.10 M solutions, there are equal moles of Pb2+ and I-. However, the stoichiometry of the reaction requires 2 moles of I- for every 1 mole of Pb2+. Therefore, I- is the limiting reactant and Pb2+ is in excess. Question 7: a) Write the equation for the reaction between N2 and O2 to form NO2 gas. N2 + 2O2 -> 2NO2 b) Assume that 33% of the oxygen gas reacts to attain equilibrium. Draw the particle diagram for the equilibrium mixture. equilibrium mixture with N2, O2, and NO2 molecules c) Based on the particle diagram, is K greater or less than 1? Justify your answer. K is less than 1. We can calculate K using the equilibrium concentrations: K = [NO2]^2 / ([N2][O2]^2) From the diagram, we can see that there are more reactants (N2 and O2) than products (NO2) at equilibrium. This means that the value of K will be less than 1. Alternatively, we can reason as follows: If K > 1, the products are favored, and there would be more NO2 molecules than N2 and O2 molecules at equilibrium. If K = 1, the reactants and products are equally favored, and there would be roughly equal amounts of N2, O2, and NO2 molecules at equilibrium. Since our diagram shows more reactants than products, K must be less than 1.