...CHM 1101 Introductory Chemistry Dawn Fox Medeba Uzzi August, 2007 Compiled and edited by Medeba Uzzi Authors’ Note This document is an initiative by the authors in an attempt to deal with what they think may be one of the reasons contributing to the relatively high failure rate in the introductory Chemistry course (CHM 1101) at the University of Guyana. It was brought to our attention that many first year students taking CHM 1101 are unable to efficiently cope with the frenetic pace of the Semester system and even less able to deal comprehensively with the large content in CHM 1101. It is hoped that by providing this paper, students will not need to make lots of notes in lectures and so they can focus on grasping the concepts taught. The document is meant to be a guide to the topics covered in CHM 1101 and is by no means exhaustive. Students are still required to attend classes regularly and punctually and to engage meaningfully in lectures and tutorials. Further, supplemental reading of these topics in any good General Chemistry text is expected. Dawn Fox Medeba Uzzi 2 SECTION 1 – Modules A – D: section deals with the foundation for chemistry. It introduces students to matter & its classification, Atom & its structure, Periodic table and chemical rxns. Introduction to Science and Measurement What is Chemistry? – Chemistry is the study of matter and its transformations Natural sciences refer to the systematic study of the natural world (our...
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...compound that produces or reacts with 1.008 grams of hydrogen has a valence of I. It follows that the same quantity producing or reacting with 2.016 grams of hydrogen is valence II, and producing or reacting with 3.024 grams of hydrogen is valence III, etc. In this experiment, the magnesium metal will be reacted with concentrated hydrochloric acid to produce hydrogen gas according to the following chemical equation: Mg (sol) + x HCl (aq) → MgCl x (aq) + x H2 (gas) 2 The hydrogen gas produced will be collected, its volume measured, and the mass in grams calculated using the ideal gas law: PV = nRT When using this equation, the units of the gas constant (R) must be consistent with the units used for the other quantities in the equation. The Handbook of Chemistry and Physics lists the value of R as R = 0.08206 L × atm × mol –1 –1 ×K Thus, pressure (P) must be expressed in atmospheres, volume (V) in liters, and temperature (T) in Kelvin. 11A Experiment A sample of magnesium metal with an approximate mass of 0.24 grams will be reacted with excess hydrochloric acid. The hydrogen gas produced will be collected, its volume measured, and finally its mass in grams will be calculated. Chemicals Required: 6M hydrochloric acid Note: 12M HCl(aq) must be diluted. Apparatus Required: 500 mL Florence flask Glass tubing Watch glass Rubber tubing Magnesium metal, ribbon or granules 125 mL Erlenmeyer flask Metal or plastic trough 100 mL graduated...
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...Science 60 (2005) 4567 – 4580 www.elsevier.com/locate/ces The effects of particle and gas properties on the fluidization of Geldart A particles M. Ye, M.A. van der Hoef, J.A.M. Kuipers∗ Fundamentals of Chemical Reaction Engineering, Faculty of Science and Technology, University of Twente, P.O. Box 217, 7500 AE Enschede, The Netherlands Received 17 November 2004; received in revised form 8 March 2005; accepted 8 March 2005 Abstract We report on 3D computer simulations based on the soft-sphere discrete particle model (DPM) of Geldart A particles in a 3D gas-fluidized bed. The effects of particle and gas properties on the fluidization behavior of Geldart A particles are studied, with focus on the predictions of Umf and Umb , which are compared with the classical empirical correlations due to Abrahamsen and Geldart [1980. Powder Technology 26, 35–46]. It is found that the predicted minimum fluidization velocities are consistent with the correlation given by Abrahamsen and Geldart for all cases that we studied. The overshoot of the pressure drop near the minimum fluidization point is shown to be influenced by both particle–wall friction and the interparticle van der Waals forces. A qualitative agreement between the correlation and the simulation data for Umb has been found for different particle–wall friction coefficients, interparticle van der Waals forces, particle densities, particle sizes, and gas densities. For fine particles with a diameter dp < 40 m, a deviation has been found between...
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...SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and Wylen CHAPTER 3 SUBSECTION Concept-Study Guide Problems Phase diagrams General Tables Ideal Gas Compressibility Factor Review Problems PROB NO. 128-132 133-134 135-145 146-148 149, 157, 158 150-156 Correspondence table The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and the SI number refers to the corresponding SI unit problem. New 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 5th Ed. new new new new new new 61E 68E a-c 68E d-f new 70E 73E 74E new 76E SI 5 7 9 11 17 23 27 30 30 40 36 47 41 44 51 New 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 5th Ed. 77E new 79E 62E new 69E c+d 70E d 72E 64E new 81E new 71E 80E 83E 65E 66E SI 53 62 58 69 65 81 113 74 49 99 95 61 106 89 - Sonntag, Borgnakke and Wylen Concept Problems 3.128E Cabbage needs to be cooked (boiled) at 250 F. What pressure should the pressure cooker be set for? Solution: If I need liquid water at 250 F I must have a pressure that is at least the saturation pressure for this temperature. Table F.7.1: 250 F Psat = 29.823 psia. 3.129E If I have 1 ft3 of ammonia at 15 psia, 60 F how much mass is that? Ammonia Tables F.8: F.8.1 Psat = 107.64 psia at 60 F so superheated vapor. F.8.2 v = 21.5641 ft3/lbm under subheading...
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...ME 416 Computer Assisted Design of Thermal Systems Actual Compressor Calculations 1. Use guidelines for ideal gas turbomachinery to calculate ideal specific work and power. 2. Calculate volume flow rate in cubic feet per minute at the compressor inlet 3. Use the Brake Horsepower for a Generic Centrifugal Compressor graph to determine the basic brake horsepower. 4. Calculate the actual power by multiplying the basic brake horsepower by the ratio of the inlet pressure (in psig) to 14.5. This will give actual power in units of horsepower. You may wish to convert it to watts. 5. Calculate the actual specific work by dividing the actual power (in W) by the mass flow rate (in kg). 6. Calculate the actual adiabatic efficiency from ηs = w ideal w act or Wideal Wact 7. Determine the actual exit conditions by first calculating the actual exit enthalpy from h out,act = h in + w act and then using property evaluation to determine the actual exit entropy and temperature. 8. Calculate any second law parameters (such as irreversibility or reversible work) that are needed. 1 ME 416 CAD of Thermal Systems Figure 1. Brake Horsepower for a Generic Centrifugal Compressor 10000 8.0 6.0 4.0 Pressure Ratio 7.0 5.0 3.5 3.0 2.5 2.0 1.5 Basic Brake Horsepower (hp) 1000 100 1 10 100 Intake Volume Flow Rate (1000 CFM) 2 ME 416 CAD of Thermal Systems Example 1: Single Stage Compressor Air at 105 kPa and 278 K enters a single stage compressor at 5 kg/s and receives...
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...Airbags are safety features that are designed to save passengers from harm in a head-on collision. Airbags react within milliseconds of a crash, and the folded nylon bag quickly becomes inflated with nitrogen gas. The inflated airbag has the role of being a cushion for passengers and prevents them from hitting into the steering column and dashboard, which can cause painful injuries for passengers. Airbags were invented by John Hetrick in 1953 after Hetrick, his wife and young daughter got into a car accident in 1952. The family was driving around the Pennsylvania countryside when suddenly a huge rock on the road forced Hetrick to swerve into a ditch. With no restraint systems, Hetrick and his wife had to grasp their daughter back to ensure that she would not hit the dashboard and to save her life. In the 1950s, car passengers who got into accidents had very severe injuries, and those who did survive these accidents typically had horribly mutilated faces, nicknamed by doctors as “steering wheel faces”. Hetrick decided to take action to ensure the safety of other drivers that got into accidents. His idea for the airbag was based off of work he had done on torpedoes, which had an inflatable canvas cover. His idea was patented in 1953, and in this, he stated, “This invention has reference to an inflatable cushion assembly adapted to be mounted in the passenger compartment of a vehicle, and arranged to be inflated responsive to sudden slowing of the forward motion of the vehicle...
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...Boyle's Law Examples in Real Life Boyle's Law Explained In 1662, Robert Boyle discovered that when held at a constant temperature, the volume and pressure of a gas are inversely proportionate. Put simply, when the volume goes up, pressure drops, and vice versa. The mathematic equation is equally as simple: PV=K where P=Pressure, V=Volume, and K is simply a constant. This has become a basic principle in chemistry, now called "Boyle's law" and is included as a special case into the more general ideal gas law. Spray paint uses a real life application of Boyle's law to work its magic. Spray Paint While there are a couple different types of aerosol cans, one being a little more elaborate than the other, they both operate off of the same basic principle: Boyle's law. We'll examine the more elaborate of the two, since it's far more popular. We know that before you spray a can of paint you are supposed to shake it up for a while, listening as a ball bearing rattles around inside. There are two substances inside the can, one being your product (paint for example), and the other being a gas that can be pressurized so much that it retains a liquid state even when it is heated past its boiling point. This liquefied gas will be a substance that has a boiling point far below room temperature. The can is sealed, preventing this gas from boiling and turning into a gaseous state. That is, until you push down the nozzle. The moment the...
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...experiment is to measure the material in the gas phase and to see how the ideal gas experiment could apply to the physical. In this the material we used was, a sip-lock plastic bag, a large plastic beaker, thermometry, dry ice, water, and butane lighter. Introduction In this lab, the experiment focused on the relationship between the temperature and pressure of an ideal gas that was dry ice inside the closed container. Therefore, the volume remains constant because the ideal gas was in a closed container. Experiment Procedure First, all the materials were collected. After collecting all the materials was prepared, we started measuring the mass of the beaker and plastic zip-lock bag on the loading balance. After that, about 50 g of dry ice was transferred into the plastic bag and weighted the total mass and calculate the mass of carbon dioxide transferred. After that, the plastic bag was collected to inflate fully and the temperature of the gas was recorded. This was due to carbon dioxide present in the dry ice and after that we removed all the gas in the bag; mass of the bag and ice was measured again. The next lab experiment helps to determine the molecular weight of the gas. After obtaining the materials required, and the group member’s filled out the burette with water and inverted it. Then the room temperature and water temperature were recorded. A tube was then attached to butane lighter outlet and burette. The volume of the gas was inside the burette was increasing with...
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...ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. nums ; Å Rla T1 subs = Append@nums, n -> nmolsD 8V1 Ø 10, T1 Ø 298, P1 Ø 10, P2 Ø 1, R Ø 8.3144, Rla Ø 0.082057, n Ø 4.08948< Finally, this constant will convert liter-atm energy units to Joule energy units. All results are given in Joules: laToJ = 101.325 ; ü 1. Reversible, Isothermal Process In an isothermal process for an ideal gas, DU = 0 ; DH = 0 ; thus heat and work are equal and given by: P2 q = w = n R T1 LogA ÅÅÅÅÅÅÅ E J ê. subs P1 -23330.9 J 16 Notes on Gaskell Text ü 2. Reversible Adiabatic Expansion In an adiabatic expansion q = 0; and P V g is a constant. Thus the final state has 1êg g P2 V2 i P1 V1 y Å ; T2 = ÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. g -> 5 ê 3 Å V2 = j ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ z j z n Rla k P2 { P1 V 1 P2 I ÅÅÅÅÅÅÅÅÅÅÅÅÅ M P ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 ÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ Å n Rla 5ê3 3ê5 For an ideal gas cv = 3R/2; thus 3 DU = ÅÅÅÅ n R HT2 - T1 L ê. subs 2 -9147.99 or we can use 3 DU = ÅÅÅÅ HP2 V2 - P1 V1 L laToJ ê. Append@subs, g -> 5 ê 3D 2 -9148.02 For some numeric results, the final temperature and volumes were ad2 = N@8V2 , T2 < ê. Append@subs, g -> 5 ê 3DD 839.8107, 118.636< The work done is dw = -DU 9148.02 For an ideal gas c p = 5R/2; thus the enthalpy change is 5 DH = ÅÅÅÅ HP2 V2 - P1 V1 L laToJ ê. Append@subs, g -> 5 ê 3D 2 -15246.7 or Notes on Gaskell Text 17 5 DH = ÅÅÅÅ n R HT2 - T1 L ê. subs 2 -15246.7 For numerical results in the subsequent...
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...PASS MOCK EXAM – FOR PRACTICE ONLY Course: __CHEM 1001 A/T___ Facilitator: _Saumya Bansal____ Dates and locations of mock exam take-up: Wednesday 4 – 5 30 (SA 502) Wednesday 6 30 – 8 (ME 3269) , Extra session maybe on Thursday (details in class) IMPORTANT: It is most beneficial to you to write this mock midterm UNDER EXAM CONDITIONS. This means: 1. • Complete the midterm in __1.5_ hour(s). 2. • Work on your own. 3. • Keep your notes and textbook closed. 4. • Attempt every question. After the time limit, go back over your work with a different colour or on a separate piece of paper and try to do the questions you are unsure of. Record your ideas in the margins to remind yourself of what you were thinking when you take it up at PASS. The purpose of this mock exam is to give you practice answering questions in a timed setting and to help you to gauge which aspects of the course content you know well and which are in need of further development and review. Use this mock exam as a learning tool in preparing for the actual exam. Please note: • Come to the PASS session with your mock exam complete. There, you can work with other students to review your work. • Often, there is not enough time to review the entire exam in the PASS session. Decide which questions you most want to review – the facilitator may ask students to vote on which questions they want to discuss. • Facilitators do not bring copies...
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...What makes the ball curve: Soccer players can make the ball curve by applying a Force, kicking, to the ball that is not in the center of the ball itself. When the ball is struck on the side by a player the ball spins while it is moving forward. In the case of the picture below, the soccer ball was struck on the right side of the ball and is spinning counter-clockwise. What causes the ball to actually curve in the air is a difference in the pressures on either side of the soccer ball. On the left side of this soccer ball, the air is moving faster, than the right side, relative to the center of the ball. This causes a lower pressure to develop on the left side of the ball, while on the right side there is a higher pressure because the air flow is moving slower relative to the center of the ball. This difference in air pressure causes the ball to curve to the left during its flight path. This curve is known as the Magnus Effect after the physicist Gustav Magnus. The shape of the soccer ball is…round! But in mathematical terms, the soccer ball is usually in the shape of an Archimedean Solid. This solid has 32 faces, 12 are pentagons and 20 are hexagons. On the Apollo 17 mission astronauts played soccer with a 200 pound moon rock. Just imagine playing any sport with something that heavy…ouch! There are many different soccer balls in use today. Each company claims that theirs is the best one out there but in all reality soccer players only like to play with soccer balls...
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...Experiment: Avogadro’s Law Type Your Name: Date: 11/17/13 Experiment 1 ________________________________________ 1. For each gas, record the following: Propane Butane Methane a Name and formula C3H8 C4H10 CH4 b Mass of 100 mL gas (g) 0.274g 0.361g 0.100g c Molecular weight of the gas (g/mole) 44.10g/mol 58.12g/mol 16.04g/mol d Number of moles in the 100 mL sample 0.0062mol 0.0062mol 0.0062mol Average of all 3 gases: (0.0062+0.0062+0.0062) / 3 = 0.0062 2. To verify Avogadro's Law, calculate the average number of moles for the three gases along with the percent deviation for each gas, according to the formula: % deviation = |(moles of gas) - (average for all gases)| / (average for all gases) * 100% %deviation= (0.0062 -0.0062) mol / 0.0062mol *100% % deviation= 0% a Average number of moles in 100 mL for all three gases 0.0062moles b % deviation for each gas All 3 the same: 0% c Do your results confirm Avogadro's Law? Yes 4. Based on the calculated number of moles in one 1 atm of gas, how many molecules are in 1 atm of gas? (There are 6.022 x 1023 molecules/mole) Since all 3 gases have the same number of moles I will calculate 1 formula for all 3. 0.0062mol (6.022 x 1023 molecules/mol)= 0.0373364 →3.73 x 1022 molecules for each gas are in 1atm. 5. Even though the number of molecules in 1 atm of gas at constant pressure and temperature is identical, the number of atoms in the gas at STP can vary...
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...COMBINED GAS LAW SUMMARY The combined gas law is a gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law. This law states: “The ratio between the pressure-volume product divided by the temperature of a system remains constant.” This can be stated mathematically as:  Where: p is the pressure, V is the volume, T is the temperature measured in kelvins, and k is a constant (with units of energy divided by temperature). Reminder: 1atm= 760 torr = 101.3 kPa & Celsius to Kelvin= add 273 and Kelvin to Celsius= subtract 273 For comparing the same substance under two different sets of conditions, the law can be written as:  If the problem does not state which unit to give the result in, then make sure that temperature is converted into Kelvin and for the Pressure and Volume just make sure you stay constant and use the same unit on both sides of the equation. Combination of 3 Laws: Boyle's Law states that the pressure-volume product is constant:  In other words as external pressure on a gas increases the volume decreases, and vice versa. Charles's Law shows that the volume is proportional to absolute temperature:  In other words as temperature increases the volume increases, and vice versa. Gay-Lussac's Law says that the pressure is proportional to the absolute temperature:  In other words as temperature increases the pressure increases, and vice versa. Where P is the pressure, V the volume and T the absolute temperature and of an ideal gas. By combining...
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...lowest number on all * Molecular formula – gives actual whole number ratio of atoms of each element in each compound. (Molecular formula weight)/(Empirical formula weight) x compound * Formulas from analysis: * Structured formula – a formula that shows the atoms of a compound, their relative positions, and the bonds between them. * Isomers – compounds with the same molecular formula, but different properties and different arrangements of atoms * Writing chemical equations (symbols) : * + adding 2 or more chemicals together * -> Yields (Products) * (arrow forward and backward) reaction in both directions (denotes equilibrium) * (arrow up) gas evolved * (arrow down) solid precipitate forms * (s), (l), (g) solid, liquid, gas * (aq) aqueous solution * (Triangle) heating the solution * Limiting reagent – the reagent that is completely consumed in a reaction * Reagent that limits the amount of product possible * Theoretical...
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...Universal Gas Constant * What are the four measurable properties of gases? Temperature, volume, moles, pressure. * R is the universal gas constant that relate the four properties of gases in one relationship. Write down the formula equal to R. PV = R nT * What is the unit that you used for R? Atm.L\k.mol * Hydrogen gas has properties very nearly like an ideal gas. Write down the equation for the production of H2 gas Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) * In our experiment, the total pressure of the gas was calculated by adding P(hydrogen) and P(water vapor). Where did the water vapor come from? * The H2 is collected over water bath * There is also some water vapor * Pressure of H2 is different than that of H2O * Explain how did you measure the following properties of H2 gas: Temperature of H2 | thermometer | Volume of H2 | HCl in the burette we Invert burette into beaker thwn we read it from the beurette | Moles of H2 | n = m/M | * A burette which contains 0.001 moles of Mg metal and HCl solution, was inverted in water bath of 300 kelvin. The reaction produces 0.025 L of hydrogen gas with a total pressure of 1 atm. Calculate the gas constant of H2. R= PVnT = 1 × 0.0250.001 × 300 = 0.0833 atm.L\k.mol * A 0.614 mole sample of ideal gas at 12 °C occupies a volume of 4.3 L. What is the pressure of the gas? (R = 0.0821 L.atm/mol.K) 12 °C+273.15=285...
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