...A. The Mole 1.Calculate the moles of chlorine (Cl2) in 50.0 g of Cl2. 50g Cl2 (1 mole Cl270g Cl2)= 0.71 mole Cl2 0.71 moles Cl2 * 6.02 *1023 = 4.2742 * 1023 moles 2. Determine the mass of one moleculeof hydrogen sulphide gas. H= 1 *2 =2 S= 32.065* 1 = 32 Molar mass of H2S = 32g/mol 3. How many molecules of are there in 60.0 g of N2O? 60g Cl2 (1 mole Cl244g Cl2)= 1.36 moles Cl2 1.36 moles Cl2 * 6.02 *1023 = 8.1872 * 1023 moles 4. If you consumed 500g of sugar in one week, how many molecules of sugar have you consumed? 500g C12H22O11(1 mole c12H22o11342g c12H22o11)=1.46 moles c12H22o11 1.46 moles c12H22o11 (6.02 * 1023) = 8.7892 * 1023moles B. Formula / Molecular Mass 1. Magnetite Fe3O4 Fe- 56x3= 168 O- 16x4 = 64 232 g/mol 2. Sodium Sulphate Decahydrate Na2SO4 10H2O Na- 23x2= 46 S- 32x1 = 32 0- 16x16= 224 H- 1x 10 = 10 322 g/mol 3. Iron (III) Ferrocyanide, Fe[Fe(CN)6]3 Fe- 56 x 7 = 392 C- 12 x 18 = 216 N- 14 x 18 = 252 322 g/mol 4. Noncaloric sweetener saccharin , C7H5NO3S C- 12x7= 84 H- 1x5= 5 N- 14x1= 14 O- 16x3= 48 S- 32x1= 32 183 g/mol C.Percentage Composition 1. Determine the percentage composition by mass of NH4H2PO4. NH4H2PO4 N= 1 * 14 = 14 %N= 14/ 115 * 100= 12.17 % H= 6 * 1 = 6 %H= 6/115* 100= 5.22 % P= 31 * 1 = 31 %P= 31/115 * 100= 26.96% O= 4 * 16 = 64 %O= 64/115 * 100= 55.65% 115 g/mol 2. Sodium phosphate...
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...10 CHAPTER SOLUTIONS MANUAL The Mole Section 10.1 Measuring Matter page 320–324 b. 2.50 ϫ 1020 atoms Fe 2.50 ؋ 1020 atoms Fe ؋ Practice Problems ؋ 51.4 ؍ 10؊4 1 mol Fe __ 6.02 ؋ 1023 atoms Fe mol Fe pages 323–324 1. Zinc (Zn) is used to form a corrosion-inhibiting surface on galvanized steel. Determine the number of Zn atoms in 2.50 mol of Zn. 2.50 mol Zn ؋ 4201 ؋ 15.1 ؍ ___ 6.02 ؋ 1023 atoms Zn 1 mol Zn atoms of Zn 2. Calculate the number of molecules in 11.5 mol of water (H2O). 11.5 mol H O ؋ ___ 6.02 ؋ 1023 molecules H2O 2 ؋ 29.6 ؍ 1024 1 mol H2O molecules of H2O 3. Silver nitrate (AgNO3) is used to make several Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc. different silver halides used in photographic films. How many formula units of AgNO3 are there in 3.25 mol AgNO3? 3.25 mol AgNO3 ؋ 4201 ؋ 69.1 ؍ ____ 6.02 ؋ 1023 formula units AgNO3 1 mol AgNO3 formula units of AgNO3 4. Challenge Calculate the number of oxygen atoms in 5.0 mol of oxygen molecules. Oxygen is a diatomic molecule, O2. 5.00 mol O2 ؋ ___ 6.02 ؋ 1023 molecules O2 1 mol O2 2 O atoms ؋ __ 01 ؋ 20.6 ؍ 24 molecule O2 atoms O 5. How many moles contain each of the following? a. 5.75 ϫ 1024 atoms Al for each formula and convert the given number of representative particles to moles. a. 3.75 ϩ 1024 CO2 The representative particle is a molecule...
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...(CHE-111) Section no.: Semester and year: Written Assignment 3: Mass Relationships for Compounds Answer all assigned questions and problems, and show all work. 1. How many atoms are there in 5.10 moles of sulfur (S)? (6 points) 1 mole= 6.02 x 1023atoms 5.10 = (Reference: Chang 3.13) 2. How many moles of calcium (Ca) atoms are there in 77.4 g of Ca? (6 points) No of moles = mass of compound/ molar mass of compound (Reference: Chang 3.15 3. What is the mass in grams of 1.00 × 1012 lead (Pb) atoms? (8 points) Mass in grams = no of atoms x molar mass of compound / Avogadro’s number (Reference: Chang 3.19) 4. Describe how you would determine the isotopic abundance of an element from its mass spectrum. (6 points) (Reference: Chang 3.32) The mass spectrometer separates isotopes on the basis of their charge-to-mass ratio providing a spectrum with a number of peaks. Each peak represents an isotope. The mass of each isotope is determined from the magnitude of its deflection on a spectrum whereas the abundance of each isotope is given by the area of the peak. 5. Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is C9H10O. a. Calculate the percent composition by mass of C, H, and O in cinnamic alcohol. (8 points) b. How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g? (8 points) c. How many carbon atoms are in 0.469 g of cinnamic alcohol...
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... H. 88 mL 3. How many grams of propane were there in 1 atm? Choose the closest answer. I. 0.735 g J. 0.274 g K. 0.664 g L. 0.232 g 4. How many grams of butane were there in 1 atm of gas? Choose the closest answer. M. 0.774 g N. 0.361 g O. 0.664 g P. 0.214 g 5. How many grams of methane were in 1 atm of gas? Choose the closest answer. Q. 0.244 g R. 0.337 g S. 0.637 g T. 0.100 g 6. The molecular weight of propane is 44.10 g/mol. Based on the number of grams of gas measured, how many moles is this? (moles = mass / molecular weight) Choose the closest answer. U. 6.2 * 10^-3 moles V. 9.8 * 10^-2 moles W. 5.2 * 10^-3 moles X. 4.1 * 10^-3 moles 7. The molecular weight of butane is 58.14 g/mol. Based on the number of grams of gas measured, how many moles is this? (moles = mass / molecular weight) Choose the closest answer. Y. 8.1 * 10^-4 moles Z. 9.4 * 10^-2 moles [. 6.2 * 10^-3 moles \. 5.5 * 10^-1 moles 8. The molecular weight of methane is 16.04 g/mol. Based on the number of grams of gas measured, how many moles is this? (moles = mass / molecular weight) Choose the closest answer. ]. 4.1 * 10^-3 moles ^. 6.2 * 10^-3 moles _. 9.3 * 10^-4 moles `. 2.1 * 10^-3 moles 9. Which of the following statements are true? a. The number of moles in 1 atm...
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...Stoichiometry ________________________________________ Stoichiometry is simply the math behind chemistry. Given enough information, one can use stoichiometry to calculate masses, moles, and percents within a chemical equation. ________________________________________ ________________________________________ What is a Chemical Equation? In chemistry, we use symbols to represent the various chemicals. Success in chemistry depends upon developing a strong familiarity with these basic symbols. For example, the symbol "C"represents an atom of carbon, and "H" represents an atom of hydrogen. To represent a molecule of table salt, sodium chloride, we would use the notation "NaCl", where "Na" represents sodium and "Cl" represents chlorine. We call chlorine "chloride" in this case because of its connection to sodium. You should have reviewed naming schemes, or nomenclature, in earlier readings. A chemical equation is an expression of a chemical process. For example: AgNO3(aq) + NaCl(aq) ---> AgCl(s) + NaNO3(aq) In this equation, AgNO3 is mixed with NaCl. The equation shows that the reactants (AgNO3 and NaCl) react through some process (--->) to form the products (AgCl and NaNO3). Since they undergo a chemical process, they are changed fundamentally. Often chemical equations are written showing the state that each substance is in. The (s) sign means that the compound is a solid. The (l) sign means the substance is a liquid. The (aq) sign stands for aqueous in water and means...
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...Study Guide for Exame 2 CHAPTER 3: Stoichiometry * Stoichiometry – study of quantitative aspects of formulas and relations * The mole – SI unit for the amount of a substance. * The amount of matter that contains the same number of atoms as 12.0g of carbon -> 6.022 x 10^23 (Avogadro’s number) * Avogadro’s number – 6.022 x 10^23 * How to determine how many atoms of each element is in a compound: * (moles or grams)(6.022x10^23)(Number of atoms/1molecule) * Molar mass - Molar mass is the weight of one mole (or 6.022 x 1023 molecules) of any chemical compounds. * Mass % of an element in a compound: * ((Number of atoms of element)(atomic weight))/(Formula weight) * Empirical formula – Gives the lowest whole number ratio of atoms of each element in a compound (Grams)/(atomic weight) --- divide by 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...
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...Name: Date: Chemistry 1A – Final Review – Multiple Choice 1. 1.00 cm is equal to how many meters? a. 0.01 b. 100. c. 10.0 d. 0.0100 2. 1.00 cm is equal to how many inches? (1 inch = 2.54 cm) a. 0.394 b. 0.10 c. 12 d. 2.54 3. 4.50 ft is how many centimeters? a. 11.4 b. 21.3 c. 454 d. 137 4. The number 0.0048 contains how many significant digits? a. 1 b. 2 c. 3 d. 4 5. Express 0.00382 in scientific notation. a. 3.82 ( 103 b. 3.8 ( 10-3 c. 3.82 ( 10-2 d. 3.82 ( 10-3 6. 42(C is equivalent to ((F = 1.8 ( (C + 32) a. 273 K b. 5.55(F c. 108(F d. 53.3(F 7. 267(F is equivalent to a. 404 K b. 116(C c. 540 K d. 389 K 8. An object has a mass of 62 g and volume of 4.6 mL. Its density is a. 0.074 mL/g b. 285 g/mL c. 7.4 g/mL d. 13 g/mL 9. The mass of a block is 9.43 g and its density is 2.35 g/mL. The block’s volume is: a. 4.01 mL b. 0.249 mL c. 22.2 mL d. 2.49 mL 10. The mass of a piece of copper that has a volume of 9.5 mL is (dcopper = 8.92 g/mL) a. 2.6 g b. 85 g c. 0.94 g d. 1.1 g 11. An empty graduate cylinder has a mass of 54.772 g. When filled with 50.0 mL of an unknown liquid, it has a mass of 101.074 g. The density of the liquid is: a. 0.926 g/mL b. 1.08 g/mL c. 2.02 g/mL d. 1.85 g/mL 12. The conversion factor to change grams to milligrams is a. 100 mg/1 g b. 1 g/100 mg c. 1 g/1000...
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...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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...gas phase at 600 K and 1 bar with a reactor feed ratio of 3 moles H2 per 1 mole styrene. What is the equilibrium composition of the three components? C6H5CH=CH2 + H2 ( C6H5CH2CH3 ΔGf° = -19.84 kcal/mol ΔHf° = -29.5 kcal/mol R = 1.987 kcal / (mol K) ln(K298) = -ΔGf°/RT ( K298 = exp(-ΔGf°/RT) = exp(19.84 /(1.987x10-3)/298) = 3.562x1014 K600 = K298 (-ΔΗf°(1/T – 1/298)/R) = 3.562x1014 (29.5(1/600 – 1/298)/(1.987x10-3)) = 4583 νetb=1, νsty=-1, νh2=-1, ν=-1 nh20 = 3 nsty0 = 1 ysty = (1 – ε) / (4 – ε) yh2 = (3 – ε) / (4 – ε) yetb = ε / (4 – ε) K = yetb / (yh2 ysty) = [ ε / (4 – ε)] / [(1 – ε) / (4 – ε) *(3 – ε) / (4 – ε)] = [ε(4–ε)]/[(1–ε) (3–ε)] 3K -4Kε + Κε2 -4ε + ε2 = 0 (K+1)ε2 – 4(K+1)ε + 3K = 0 ( (4584)ε2 – 4(4584)ε + 3(4583) = 0 ε ’ 0.9997 ysty = 0.0001 0.162 0.160 yh2 = 0.6667 OR 0.762 OR 0.160 yetb = 0.3332 0.162 0.160 if you used incorrect R-value printed on exam if you switched DG and DH (since DH not labeled correctly) (4) In the absence of a selective catalyst, the synthesis of methanol can be accompanied by a side reaction in which coke (solid carbon) is formed (see below). If 1 mol CO and 2 mol H2 are fed to a reactor at 298 K and 0.5 bar, approximately how many moles of coke will form? CO(g) + 2H2(g) ( CH3OH(g) 2CO(g)...
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...mass of Na2SO4. must be calculated to find the moles of Na2SO4. Molar mass of Na2SO4. = 2 AW of Na + AW of S + 4 AW of O (where AW = atomic weight) = 2(22.99 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 142.05 g Na2SO4./mol Na2SO4. Find the moles of Na2SO4.: 15.0 g Na2SO4 x 1 mole Na2SO4142.05 g Na2SO4 = 0.106 mol Na2SO4 Find the volume of solution in liters. Recall from part a, the mass of solution is 415.0g. 415.0 g solution x 1 mL solution1.056 g solution x 1 L solution1000 mL solution = 0.3930 L solution Calculate the molarity: Molarity of Na2SO4 = 0.106 mol Na2SO40.3930 L solution = 0.270 M Na2SO4 1 c. The moles of Na2SO4 is already known from part b. Find the kilograms of solvent (water): 400.0 g H2O x 1 kg H2O1000 g H2O = 0.4000 kg H2O Calculate the molality of Na2SO4 : molality of Na2SO4 = 0.106 mol Na2SO40.4000 kg water = 0.265 m Na2SO4 1 d. Calculate the parts per thousand of Na2SO4 from the mass of solute and mass of solution: ppt Na2SO4 = 15.0 g Na2SO4415 g solution x 1000 = 36.1 ppt Na2SO4 1 e. Parts per million may also be expressed as mg solute/L solution. In this case, converting the molarity to ppm will be the simplest route: 0.270 mol Na2SO4L solution x 2 mol Na+1 mol Na2SO4 x 22.99 g Na+1 mol Na+ x 1000 mg Na+1 g Na+ = 1.24 x 104 ppm Na+ 1 e. For the mole fraction of sulfate, we need to have: mole fraction SO42- = mol...
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...THERMOCHEMISTRY OF NaOH AND HCl LAB Overview Teacher’s Instruction: Find the Molar Heat of Reaction for the NaOH reaction. Then, predict and calculate the change in enthalpy (ΔE) and change in heat (ΔH) when 5.00g NaOH reacts completely with HCl. Reaction Equation: NaOH(s) + HCl(aq) -> NaCl(aq) + H2O(liq) Net Ionic Equation: Na(OH)(s) + H+(aq) -> H2O(liq) + Na+(aq) The Big Question: If we combine solid NaOH and aqueous HCl, how will the temperature change? What will the change be with, specifically, 5.00g of NaOH? Scientific Background and Principle: WELL, I’ll have you know that we got our hands on a fancy-schmancy Lab Quest 2 with a temperature probe. Now this device allows us to accurately record the temperature of a given entity over a period of time; as such, by having the Lab Quest record the temperature of the system, we were able to gather the total temperature change for the reaction. In theory, the temperature should increase by 53.10o Variables * Independent Variable: Amount of NaOH * We had a theoretical value for temperature that was dependent on the amount of NaOH used; as such, we set our amount at a certain point to achieve that temperature. * Dependent Variable: Heat of Reaction (Temperature) * We measured the temperature of the reaction throughout its duration, which would have varied in intensity and duration based on the amount of NaOH we used * Controlled Variables * Light-- by enclosing the reaction in darkness...
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...Dublin Institute of Technology | Combustion | Thermodynamics Laboratory | | Name: Shiyas Basheer Student Number: D10119909 Programme/Course: DT022/3 Group: A Date of lab: 09/04/2013 Abstract Table of Contents Abstract 1 Introduction and Procedure 3 Calculations 3 Theory 3 Mass Flow rate analysis 4 Exhaust Gas 6 Results 7 Conclusion 8 Reference 9 Equivalent ratio of a gas turbine Introduction and Procedure For this experiment, a two shaft gas turbine (ET792) was used. At the core of ET792 are a so-called gas generator and a free-running power turbine. The gas generator consists of a radial compressor, a combustion chamber and a radial turbine. The compressor and turbine are mounted on a shaft. The gas turbine works as an open cyclic process, with the ambient air being drawn out and fed back in. Intake and exhaust silencers reduce the noise in operation of the power turbine. The use of propane as the combustion gas ensures clean, odorless operation. A start-up fan is used to start the gas turbine. The air is compressed and fed to the combustion chamber. Before it enters the combustion chamber the air is slowed by means of a diffuser. The air is then split into primary and secondary. The primary air mixes and burns with the fuel and the secondary air cool the exhaust gases. Relevant measuring values are recorded by sensors and indicated on the display and control panel. The energy of the exhaust gas stream is either converted into mechanical...
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...“Yet Do I Marvel,” Countee Cullen expresses an ambivalence that many of us are familiar with. It is not too difficult to sympathize with the poet’s opening lines: “I doubt that God is good, well-meaning, kind,/And Did he stoop to quibble could tell why/The little buried mole continues blind.” Here, it is made plain to us that his beliefs dictate that God is ultimately good and righteous, yet, nonetheless, life has apparently given him reasons to question his. As the title indicates, Cullen is marveling at life and human beings as God’s creations and the relationship we have with Him. Although he says he is free of doubt, skepticism appears to be present in this remark. Here perhaps the rather rhetorical question of why life is the way it is is being posed. It seems, at least up until this point, that Cullen is supposing that a truly good God would not subject His creatures to such paradoxical, even cruel, situations. Upon closer analysis of the mole and its scenario (“The little buried mole continues blind”), which may function as a metaphor for the poet himself, the possibility is not ruled out that the poet is hinting to us what seems to be an inescapably bad situation may in fact be a fixable one. A mole, like all creatures, has adapted in ways that allows it to survive, and it may be that Cullen is suggesting to us that thing are not as random and as terrible as they may seem and that there is a way out of challenging situations. The following line, though, counters this...
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...Experiment #8 Molecular Weight by freezing Point Depression Maureen Shultz November 14, 2015 I. Purpose The purpose of this experiment was to determine the molecular weight of two unknown substances by measuring the freezing point depression of an aqueous solution of the unknown substances. II. Experiment Method Equipment: Test tube Constant temperature bath Thermometer Chemicals: H2O Unknown substance #1 Unknown substance #2 To begin this experiment, I started by placing an empty test tube on the solid surface workstation and poured 10mL of H2O into the test tube. I then attached a thermometer to the test tube, documenting that the liquid is clear in color. I then placed a constant temperature bath onto the solid surface workstation. I set the temperature of the constant temperature bath to the freezing point of ethylene glycol ,-15 degrees Celsius. When the desired temperature is reached, I gently place the test tube into in the constant temperature bath. Carefully observing the test tube, I noticed an ice formation, and then documented the temperature at which the freezing point of the solution became a solid. I then carefully clear the workstation and set up for the next experiment. A second empty test tube is placed on the solid surface workstation. I then pour 2 grams of the first unknown sample substance into the empty test tube and attach a thermometer. I notice a silver substance in the bottom of the test tube. I add 10mL of...
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...measuring dry material, weighing it, and then mixing it with distilled water. This will be necessary in order to observe the reaction and calculate more. Then the water was now filtered in the paper filter, this filter would later be dried in the sunlight and weighed to measure the mass of the product. After the filtering was complete, the percent yield could be calculated, as well as the actual mass of the precipitate. Data: |Initial: CaCl2*2H20 (g) |1.0 g | |Initial: CaCl2*2H20 (moles) |.0068 | |Initial: CaCl2 (moles) |.0068 | |Initial: NaCO3 (moles) |.0012 | |Initial:...
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