CHEM 101A EXAM 1 SOLUTIONS TO VERSION 1

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1 CHEM 101A EXAM 1 SOLUTIONS TO VERSION 1 Multiple-choice questions (3 points each): Write the letter of the best answer on the line beside the question. Give only one answer for each question. B 1) If 0.1 moles of NaNO 3 and 0.2 moles of Mg(NO 3 are dissolved in enough water to make 2 liters of solution, the concentration of nitrate ions in the solution will be: A) 0.15 M B) 0.25 M C) 0.3 M D) 0.5 M E) none of these Explanation: 0.1 mol of NaNO 3 dissociates into 0.1 mol of Na + and 0.1 mol of NO mol of Mg(NO 3 dissociates into 0.2 mol of Mg 2+ and 0.4 mol of NO 3. The total number of moles of NO 3 ions is 0.5 mol, so the molarity is 0.5 mol/2 L = 0.25 M. D 2) If x grams of N 2 reacts with excess H 2 according to the equation below, what mass of NH 3 will be formed? Equation: N H 2 2 NH 3 A) 0.216x grams B) 0.5x grams C) 0.608x grams D) 1.216x grams Ex grams Explanation: x g N 2 1 mol N g N 2 2 mol NH 3 1 mol N g NH 3 1 mol NH 3 = 1.216x g NH 3 D 3) A chemist reports that a reaction had a 91.3% yield. The chemist weighed the product and found that its mass was g. Which of the following statements is true? A) The actual yield was g B) The actual yield was g C) The theoretical yield was g D) The theoretical yield was g E) More than one of these statements is true F) None of these statements is true Explanation: % yield = actual yield/theoretical yield x 100%. The actual yield here is g, so the theoretical yield is g A 4) A sample of O 2 exerts a pressure of 800 torr at 200 C. If you want to reduce the pressure to 700 torr, the temperature of the gas must be: A) 141 C B) 175 C C29 C D68 C E) none of these Explanation: Use P 1 /T 1 = P 2 /T 2, where P 1 = 800 torr, T 1 = = K, and P 2 = 700 torr. You will get T 2 = 414 K = 141 C. B 5) In the van der Waals equation, the term P + a(n/v describes: A) the pressure that the gas actually exerts on the walls of its container. B) the pressure the gas would exert if the molecules did not attract each other. C) the pressure the gas would exert if the molecules did not take up any space. D) the pressure the gas would exert if its concentration were 1 mol/l. Explanation: P is the actual pressure of the gas and a(n/v is the pressure that is lost because the molecules attract each other, so P + a(n/v is the pressure the gas would exert without the molecular attractions. A 6) If you add 120 ml of water to 40 ml of M NaCl, the concentration of NaCl in the resulting solution will be: A) M B) M C) M D) 1.20 M E) 1.80 M F) none of these Explanation: Use M 1 V 1 = M 2 V 2, where M 1 = M, V 1 = 40 ml, and V 2 = = 160 ml

2 Chem 101A Exam 1 Page 2 of 8 Problems: be sure to show your work for all calculations and to give explanations wherever you are told to do so. 1) (6 points) How many oxygen atoms are there in a g sample of Ca(NO 3? Calculating the molar mass of Ca(NO 3 : 1 Ca = 1 x = N = 2 x = O = 6 x = Total = g/mol g Ca(NO mol Ca(NO mol O O atoms 1 mol O 1 mol Ca(NO g Ca(NO 3 = mol Ca(NO 3 6 mol O 1 mol Ca(NO 3 = mol O = O atoms 2) (10 points.431 g of a compound containing vanadium and chlorine is dissolved in water. Then, excess AgNO 3 is added, and a precipitate forms. The precipitate weighs g. Using this information, determine the empirical formula of the original compound. (The symbol for vanadium is V.) The precipitate must be AgCl (it cannot be a compound containing V and NO 3, because all nitrates are water-soluble). The molar mass of AgCl is = g/mol, so g AgCl mol AgCl 1 mol AgCl g AgCl 1 mol Cl 1 mol AgCl = mol AgCl formed = mol Cl in the original compound g Cl mol Cl = g Cl in the original compound 1 mol Cl g compound g Cl = g V in the original compound g V 1 mol V g V Divide both numbers of moles by the smallest: = mol V in the original compound mol Cl mol V = 4 mol Cl / mol V mol V mol V = 1 mol V / mol V So the empirical formula is VCl 4

3 Chem 101A Exam 1 Page 3 of 8 3) (9 points) Write balanced net ionic equations for the reactions that occur when the following chemicals are mixed. Include the state of each substance (solid, liquid, gas, or aqueous). a) 0.1 M NaHCO 3 is mixed with excess 1 M HBr HCO 3 (aq) + H + (aq) CO 2 (g) + H 2 O(l) b) Solutions of Ba(OH and CrCl 3 are mixed Cr 3+ (aq) + 3 OH (aq) Cr(OH) 3 (s) c) Solid Ag 2 O is dissolved in excess 2 M HNO 3 Ag 2 O(s) + 2 H + (aq) 2 Ag + (aq) + H 2 O(l) 4) (7 points) A gaseous compound that has the empirical formula CF effuses at a rate of 0.83 ml/min. Under identical conditions, gaseous argon (Ar) effuses at a rate of 1.8 ml/min. Determine the molecular formula of the compound. Use Graham s Law to calculate the molar mass of the unknown compound. Gas 1 is Ar, so M 1 = g/mol and rate 1 = 1.8 ml/min. Gas 2 is the unknown, so rate 2 = 0.83 ml/min. M 2 is our unknown. 1.8 ml / min 0.83 ml / min = M g / mol Solving this equation gives M 2 = g/mol. If the molecular formula of the gas were CF, its molar mass would be = g/mol g / mol g / mol = So the molecular formula is 6 times the empirical formula, or C 6 F 6.

4 Chem 101A Exam 1 Page 4 of 8 5) (14 points) a) Complete the following ICE table by filling in all of the empty spaces. You may assume that the reaction goes to completion. 4 NH O 2 2 N H 2 O Initial moles x x x x Change x 0.75x + 0.5x + 1.5x Ending moles x 1.5x 2.5x Note that NH 3 is the limiting reactant here, so all of the changes are related to the amount of NH 3. For example, the amount of O 2 that is used up is: 3 mol O x mol NH 3 2 = 0.75x mol O 4 mol NH 2 3 b) If the temperature is 150 C and the volume of the container is 5.00 L, determine the total pressure in the container when the above reaction is complete. Give your answer in atmospheres, in the form Ax (so your answer will be something like x atm or 570.8x atm). Note that all substances in this reaction are gases at 150 C. The total number of moles of gases at the end of the reaction is: 0.25x + 1.5x + 2.5x = 4.25x Now we can use PV = nrt to determine the pressure. Rearranging PV = nrt to solve for P gives us P = nrt V ( )( L atm / mol K) ( K) = 4.25x mol 5.00 L = 29.5x atm 6) (5 points) A sample of xenon (Xe) and a sample of neon (Ne) have the same Boltzmann speed distribution. If the xenon is at 100 C, what is the temperature of the neon? Give your answer in Celsius degrees. The two gases have the same Boltzmann speed distribution, so they must also have the same rms speed (and the same average speed, and the same most probable speed). This allows us to find the temperature of the neon. ( )( K) J / mol K kg / mol = ( )( T Ne ) J / mol K kg / mol Solving this equation gives T Ne = K = -216 C. (Yes, the neon is very cold!)

5 Chem 101A Exam 1 Page 5 of 8 7) (10 points) Acetylene (C 2 H 2 ) is a flammable gas that is made by the following reaction: CaC 2 (s) + 2 H 2 O(l) Ca(OH (s) + C 2 H 2 (g) A chemist adds excess water to a small sample of CaC 2 and collects the C 2 H 2 over water. At an atmospheric pressure of 748 torr and a temperature of 23 C, the C 2 H 2 occupies 216 ml. The vapor pressure of water at 23 C is 21.1 torr. Calculate the mass of the C 2 H 2. We can use PV = nrt to calculate the number of moles of C 2 H 2, but we must begin by finding the actual pressure of the C 2 H torr (total) 21.1 torr (water) = torr (C 2 H 2 ) Now we use the ideal gas law. Note that the volume must be converted from milliliters to liters, the temperature must be converted from Celsius to kelvins, and we must use R = L torr/mol K (because the pressure is in torr). n = PV RT = ( torr) ( L) ( L torr / mol K) ( K) = moles of C H 2 2 Finally, convert from moles to grams mol C 2 H g 1 mol = g C 2 H 2

6 Chem 101A Exam 1 Page 6 of 8 8) (8 points) Use the basic principles of the kinetic theory to explain the following observations. Observation #1: Gases can be compressed (squeezed into a smaller volume) but liquids cannot. The kinetic theory states that the distance between gas molecules is much larger than the sizes of the molecules. Therefore, most of the volume occupied by a gas is empty space. Compressing a gas simply means reducing the amount of empty space. On the other hand, in a liquid the molecules are in contact with each other, leaving very little empty space. Compressing a liquid requires making the molecules smaller, which requires an enormous amount of pressure, so liquids cannot normally be compressed. Observation #2: hydrogen reacts very rapidly with oxygen at 500 C, but extremely slowly at 100 C. At the higher temperature, the distribution of kinetic energies shifts toward higher energies. As a result, more molecules have enough energy to react when they collide with each other. The speed of a reaction is determined by how many molecules have enough energy to react when they collide. (This is covered in the slide presentation on the website, and in experiment 6.)

7 Chem 101A Exam 1 Page 7 of 8 9) (5 points) The graph below shows the distribution of kinetic energies in N 2 (g) at 150ºC, along with the areas of two regions of the graph. Fraction per J/mol A B Area of region A = Area of region B = Kinetic energy (J/mol) In a mol sample of gaseous N 2 at 150ºC, how many moles of N 2 have kinetic energies greater than 5000 J/mol? The overall area under the curve is always 1 for the Boltzmann distribution, so the area above 5000 J/mol equals 1 the area of region A = = The fraction is defined as the number of molecules that have energies within a specific range, divided by the total number of molecules: = moles of gas with energies above 5000 J / mol mol Solving this equation tells us that moles of N 2 have energies above 5000 J/mol at 150 C.

8 Chem 101A Exam 1 Page 8 of 8 10) (8 points) Consider the apparatus below, which consists of two containers connected by a valve. The valve is opened and the gases mix (but do not react). After the gases mix, the pressure in the apparatus is 511 torr. Calculate the original pressure of the N 2. You may assume that the temperature does not change. O 2 (g) 1.00 L 287 torr N 2 (g) 1.50 L unknown pressure First, calculate the pressure that the O 2 exerts after it has expanded to fill the entire apparatus, using P 1 V 1 = P 2 V 2. (287 torr)(1.00 L) = P 2 (2.50 L) P 2 = torr The total pressure in the apparatus after the gases mix is 511 torr, so the partial pressure of the N 2 after it mixes with the O 2 is 511 torr torr = torr. Finally, use P 1 V 1 = P 2 V 2 again to calculate the initial pressure of the N 2. P 1 (1.50 L) = (396.2 torr)(2.50 L) P 1 = 660 torr

= mol NO 2 1 mol Cu Now we use the ideal gas law: atm V = mol L atm/mol K 304 K

= mol NO 2 1 mol Cu Now we use the ideal gas law: atm V = mol L atm/mol K 304 K CHEM 101A ARMSTRONG SOLUTIONS TO TOPIC C PROBLEMS 1) This problem is a straightforward application of the combined gas law. In this case, the temperature remains the same, so we can eliminate it from the