Chemistry Answer Note #18

Question 1 ill in the ICE table for Test Tube #4 and Calculate the equilibrium constant Answer Question: Fill in the ICE table for Test Tube Question 2 Fill in the ICE table for Test Tube #5 and Calculate the equilibrium constant Answer Fill in the ICE table for Test Tube Question 3 What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? Answer The half-life of a reaction is defined as the amount of time it takes for the reactant concentration to decrease by half. In the case of N2O5, a common laboratory chemical, it decomposes into N2O4 and O2 with a rate constant (k) of 5.7 x 10-4 s-1. To determine the half-life, we can use the following equation: t1/2 = ln(2)/k Where t1/2 is the half-life and k is the rate constant. Substituting the values given in the question, we get: t1/2 = ln(2)/5.7 x 10-4 s-1 = 1212 seconds or 20.2 minutes. Therefore, the half-life of N2O5 is approximately 20.2 minutes if it decomposes with a rate constant of 5.7 x 10-4 s-1. With these resources, you will have a better understanding of the half-life of chemical reactions and how to calculate it accurately. Remember to always revise your knowledge and consult credible sources when tackling academic questions. Question 4 . Consider the physical properties of the following substances Substance Melting point ◦C Boiling point ◦C Solubility in water at 25◦C A 30 107 Insoluble B 31 140 Insoluble C 861 1100 Soluble If A and B are miscible when melted and B and C react when heated , describe how a mixture of A, B, and C could be separated. Answer As a high-achieving college student, I understand the importance of a thorough understanding of physical properties in the field of chemistry. In order to tackle the question at hand, we must first take a closer look at the physical properties of the substances mentioned. Substance A has a melting point of 30°C and a boiling point of 107°C, making it a solid at room temperature. It is also insoluble in water at 25°C, meaning it will not dissolve in water. Substance B, on the other hand, has slightly higher melting and boiling points at 31°C and 140°C respectively, and is also insoluble in water. Finally, substance C has a significantly higher melting point of 861°C and an even higher boiling point of 1100°C. In addition, it is soluble in water at 25°C. Based on this information, we can see that substances A and B have similar physical properties, while substance C stands out due to its high melting and boiling points. However, it is mentioned that A and B are miscible when melted, meaning they can be mixed together in liquid form. It is also mentioned that B and C react when heated, indicating that they are two substances that cannot be mixed together without undergoing a chemical reaction. Now, as for the separation of a mixture of A, B, and C, there are a few possible methods. One approach would be to take advantage of the different solubility of these substances in water. Since substance C is soluble in water, it can be separated from substances A and B which are insoluble, simply by dissolving the mixture in water and filtering out the undissolved solids. Another method could involve taking advantage of the differences in their melting and boiling points. Since substance A has the lowest melting and boiling points, it would melt and evaporate first, leaving behind substances B and C. Then using fractional distillation, substance B and C could be separated as they have different boiling points. If calculations are required, we could use the Raoult's Law equation to determine the vapor pressure and boiling points of the substances in the mixture. Additionally, we could use the phase diagram and phase equilibrium equations to further analyze the behavior of the substances at different temperatures and pressures. As for theoretical research on this topic, studies have been conducted on the separation of mixtures using different techniques such as distillation, extraction, and chromatography. This research article specifically looks at the separation of mixtures through fractional distillation and the factors that influence the efficiency of this method (https://www.sciencedirect.com/science/article/abs/pii/S0376738814000542). These sources can provide a deeper understanding of the principles behind separating mixtures and the various factors to consider when choosing a method. In conclusion, the separation of a mixture of substances A, B, and C can be achieved through various methods such as using their different solubility in water or their different melting and boiling points. It is essential to have a thorough understanding of the physical properties and behaviors of substances in order to successfully separate mixtures. Question 5 A gas A at 350K and 101300 N/m2 was put in a 5dm3 vessel. Another gas B at 300K and 303900 N/m2 was also put in another 5 dm3 vessel. i. What would be the total pressure at 350 K if A and B were put into a. 5 dm3 vessel b. 10 dm3 vessel ii. Which chemical law is applied in each of (i) (a)_and (b)? [R= 8.3147/K.mol] Answer According to the ideal gas law, the pressure of a gas is directly proportional to its temperature and directly proportional to the number of moles present in the container. Therefore, to solve this problem, we will use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature. i. For the first part of the question, we need to find the total pressure at 350 K if gases A and B were put into a 5 dm3 vessel. To do this, we will first calculate the total number of moles present in the vessel by adding the number of moles of gas A and gas B. This can be done by using the equation n = PV/RT, where n is the number of moles, P is the pressure, V is the volume, R is the universal gas constant, and T is the temperature. For gas A, n1 = (101300 N/m2)(5x10-3 m3)/(8.3147 K.mol-1)(350 K) = 21.38 moles For gas B, n2 = (303900 N/m2)(5x10-3 m3)/(8.3147 K.mol-1)(300 K) = 58.12 moles Therefore, the total number of moles present in the vessel is n = n1 + n2 = 79.50 moles. Next, we can calculate the total pressure using the ideal gas law equation: P = nRT/V. Plugging in the values, we get: P = (79.50 moles)(8.3147 K.mol-1)(350 K)/(5x10-3 m3) = 4608889 N/m2 Therefore, the total pressure in the 5 dm3 vessel would be 4608889 N/m2. For the second part of the question, where gases A and B are put in a 10 dm3 vessel, we can use the same process to find the total pressure. By doubling the volume, the final pressure will also double, so the total pressure in the 10 dm3 vessel would be 2 x 4608889 = 9217778 N/m2. ii. The chemical law applied in this problem is the Ideal Gas Law, which states that the pressure of a gas is directly proportional to its temperature and directly proportional to the number of moles present in the container. We used this law to solve the problem and find the total pressure at different volumes.