Chemistry 1 Practice Exam 4 (PDF)

Practice Exam 4 with Answers Version 1 Created @November 12, 2024 1:57 PM Class Chemistry 1 Quiz 1 Time Limit: 90 Minutes Instructions: Answer all questions. S...

Practice Exam 4 with Answers Version 1 Created @November 12, 2024 1:57 PM Class Chemistry 1 Quiz 1 Time Limit: 90 Minutes Instructions: Answer all questions. Show all your work for calculation questions. Use appropriate units where necessary. Multiple Choice Questions (3 points each) Question 1: Which of the following is an example of kinetic energy? A) A stretched spring B) Chemical bonds in gasoline C) A rolling ball D) Water at the top of a waterfall Question 2: The First Law of Thermodynamics states that: A) Energy cannot be created or destroyed, only transformed B) Entropy of the universe is always increasing C) Energy is lost in every energy transfer D) The total energy of the universe is decreasing Practice Exam 4 with Answers Version 1 1  Question 3: Which property is a state function? A) Work B) Heat C) Enthalpy D) Distance traveled Question 4: Enthalpy change (delta H) is defined as: A) The heat absorbed or released at constant volume B) The work done by a system at constant pressure C) The heat absorbed or released at constant pressure D) The change in internal energy of a system Question 5: Which of the following statements about heat capacity is true? A) It is the amount of heat required to raise the temperature of a substance by 1 degree C B) It is independent of the mass of the substance C) It is the same for all substances D) It is measured in joules per mole Short Answer Questions Question 6: (5 points) Differentiate between potential energy and kinetic energy with examples. Question 7: (5 points) Explain why enthalpy is considered a state function. Question 8: (10 points) Practice Exam 4 with Answers Version 1 2  A 50.0 g piece of metal is heated to 100.0 degrees C and then placed in 100.0 g of water at 25.0 degrees C. The final temperature of the system is 30.0 degrees C. Calculate the specific heat capacity of the metal. (Assume no heat loss to surroundings. Specific heat of water = 4.18 J/g·degrees C) Question 9: (10 points) Using Hess's Law, calculate the enthalpy change for the reaction: C(s) + (1/2) O2(g) --> CO(g) Given the following reactions and their enthalpy changes: 1. C(s) + O2(g) --> CO2(g) delta H = -393.5 kJ 2. CO(g) + (1/2) O2(g) --> CO2(g) delta H = -283.0 kJ Question 10: (5 points) Explain how the caloric content of food is related to thermochemistry. Calculations and Conceptual Questions Question 11: (5 points) Explain why gases exert pressure on the walls of their container. Question 12: (5 points) State Boyle's Law and explain its significance in terms of gas behavior. Question 13: (10 points) A sample of gas has a volume of 2.50 L at 1.00 atm and 25.0 degrees C. Calculate its volume when the pressure is increased to 2.00 atm and the temperature is increased to 50.0 degrees C. Question 14: (10 points) Calculate the number of moles of gas present in a 5.00 L container at 2.00 atm and 27.0 degrees C. Question 15: (5 points) Describe Dalton's Law of Partial Pressures. Practice Exam 4 with Answers Version 1 3  Question 16: (5 points) Explain how the Kinetic-Molecular Theory describes the motion of gas particles. Question 17: (10 points) Compare the rates of effusion of helium gas and oxygen gas. Question 18: (5 points) Explain the difference between an ideal gas and a real gas. End of Quiz 1 Answer Key for Quiz 1 Multiple Choice Questions Question 1: C) A rolling ball Explanation: Kinetic energy is the energy of motion; a rolling ball is in motion. Question 2: A) Energy cannot be created or destroyed, only transformed Explanation: This is the First Law of Thermodynamics, also known as the Law of Conservation of Energy. Question 3: C) Enthalpy Explanation: Enthalpy is a state function because it depends only on the initial and final states, not on the path taken. Question 4: C) The heat absorbed or released at constant pressure Practice Exam 4 with Answers Version 1 4  Explanation: Enthalpy change (delta H) is defined under conditions of constant pressure. Question 5: A) It is the amount of heat required to raise the temperature of a substance by 1 degree C Explanation: Heat capacity is defined per degree of temperature change. Short Answer Questions Question 6: Potential energy is stored energy due to position or composition, such as water behind a dam or a compressed spring. Kinetic energy is the energy of motion, like a moving car or a rolling ball. Question 7: Enthalpy is considered a state function because its value depends only on the current state of the system (pressure, temperature, composition) and not on the path taken to reach that state. This means that the change in enthalpy for a process is the same regardless of how the process occurs. Question 8: Calculation: 1. Heat gained by water: q_water = m_water * c_water * delta T_water q_water = 100.0 g * 4.18 J/g·degrees C * (30.0 degrees C - 25.0 degrees C) q_water = 100.0 g * 4.18 J/g·degrees C * 5.0 degrees C q_water = 2090 J 2. Heat lost by metal: q_metal = -q_water q_metal = -2090 J 3. Calculate specific heat capacity of metal: Practice Exam 4 with Answers Version 1 5  q_metal = m_metal * c_metal * delta T_metal -2090 J = 50.0 g * c_metal * (30.0 degrees C - 100.0 degrees C) -2090 J = 50.0 g * c_metal * (-70.0 degrees C) -2090 J = -3500 g·degrees C * c_metal 4. Solve for c_metal: c_metal = (-2090 J) / (-3500 g·degrees C) c_metal = 0.597 J/g·degrees C Answer: The specific heat capacity of the metal is 0.597 J/g·degrees C. Question 9: Using Hess's Law: 1. Reverse Reaction 2 to get CO(g) on the right side: CO2(g) --> CO(g) + (1/2) O2(g) delta H = +283.0 kJ 2. Add Reaction 1 and the reversed Reaction 2: Reaction 1: C(s) + O2(g) --> CO2(g) delta H = -393.5 kJ Reversed Reaction 2: CO2(g) --> CO(g) + (1/2) O2(g) delta H = +283.0 kJ Net Reaction: C(s) + (1/2) O2(g) --> CO(g) 3. Calculate net enthalpy change: delta H_net = delta H1 + delta H_reversed2 delta H_net = (-393.5 kJ) + (+283.0 kJ) delta H_net = -110.5 kJ Answer: The enthalpy change for the reaction is -110.5 kJ. Question 10: Practice Exam 4 with Answers Version 1 6  The caloric content of food represents the amount of energy released when the food is metabolized in the body. This process is similar to combustion reactions in thermochemistry, where chemical energy stored in bonds is converted into heat energy. Thermochemistry helps us understand and calculate the energy changes involved in these reactions. Calculations and Conceptual Questions Question 11: Gases exert pressure on the walls of their container because gas particles are in constant, rapid, and random motion. When these particles collide with the walls of the container, they exert force. Pressure is the result of these collisions per unit area of the container walls. Question 12: Boyle's Law states that for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure: P1 * V1 = P2 * V2 This means that if the volume decreases, the pressure increases, provided the temperature remains constant. It highlights the compressibility of gases and the relationship between pressure and volume. Question 13: Given: Initial conditions: P1 = 1.00 atm V1 = 2.50 L T1 = 25.0 degrees C = 298.15 K Final conditions: P2 = 2.00 atm V2 = ? T2 = 50.0 degrees C = 323.15 K Practice Exam 4 with Answers Version 1 7  Using the Combined Gas Law: (P1 * V1) / T1 = (P2 * V2) / T2 Solve for V2: V2 = V1 * (P1 / P2) * (T2 / T1) V2 = 2.50 L * (1.00 atm / 2.00 atm) * (323.15 K / 298.15 K) V2 = 2.50 L * 0.5 * 1.0837 V2 ≈ 1.3546 L Answer: The new volume of the gas is approximately 1.355 L. Question 14: Given: P = 2.00 atm V = 5.00 L T = 27.0 degrees C = 300.15 K R = 0.0821 L·atm/(mol·K) Using the Ideal Gas Law: PV = nRT Solve for n: n = PV / (RT) n = (2.00 atm * 5.00 L) / (0.0821 L·atm/(mol·K) * 300.15 K) n = 10.00 atm·L / (24.6465 L·atm/(mol)) n ≈ 0.4057 mol Answer: There are approximately 0.406 moles of gas in the container. Question 15: Dalton's Law of Partial Pressures states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases: Practice Exam 4 with Answers Version 1 8  P_total = P_gas1 + P_gas2 + P_gas3 +... Each gas in the mixture contributes to the total pressure independently of the others, based on its own number of moles and temperature. Question 16: The Kinetic-Molecular Theory describes gas particles as small, hard spheres in constant, random motion. It states that: Gas particles move in straight lines until they collide with each other or the container walls. The collisions are perfectly elastic, meaning there is no net loss of kinetic energy. The average kinetic energy of gas particles is directly proportional to the temperature in kelvins. This theory explains gas properties such as pressure, temperature, and volume in terms of particle motion. Question 17: Using Graham's Law of Effusion: (rate of effusion of gas1) / (rate of effusion of gas2) = sqrt(M2 / M1) Given: Molar mass of He (M1) = 4.00 g/mol Molar mass of O2 (M2) = 32.00 g/mol Calculate the ratio: (rate_He) / (rate_O2) = sqrt(32.00 g/mol / 4.00 g/mol) (rate_He) / (rate_O2) = sqrt(8) (rate_He) / (rate_O2) ≈ 2.828 Answer: Helium gas effuses approximately 2.83 times faster than oxygen gas due to its lower molar mass. Question 18: Practice Exam 4 with Answers Version 1 9  Differences between an ideal gas and a real gas: 1. Intermolecular Forces: Ideal Gas: Assumes there are no intermolecular attractions or repulsions between gas particles. Real Gas: Gas particles experience intermolecular forces, which can affect pressure and volume, especially at high pressures and low temperatures. 2. Volume of Gas Particles: Ideal Gas: Assumes gas particles have negligible volume compared to the container volume. Real Gas: Gas particles occupy a finite volume, which becomes significant under high-pressure conditions. End of Answer Key for Quiz 1 Practice Exam 4 with Answers Version 1 10