Experiment 9 Aim A. To investigate the existence of hydrogen bonds between ethanol molecules. B. To measure the strength of hydrogen bond formed between ethanol molecules C. To investigate the formation of hydrogen bonds between molecules of ethyl ethanoate and trichloromethane. D. To measure the strength of hydrogen bond formed between molecules of ethyl ethanoate and trichloromethane. Procedure A. 1. 10 cm3 of ethanol was added into an insulated 50 cm3 beaker by
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SCH4U 03/04/13 Hydrogen Bonding Purpose: The purpose of this investigation is to test the concept of hydrogen bonding. Hypothesis: Given the concept of hydrogen bonding I predict that the bulky glycerol molecules limits the number of possible hydrogen bonds. If water is mixed with glycerol should make it possible for water to form many hydrogen bonds with the glycerol molecules‚ causing it to become a exothermic reaction. Because glycerol has more possibilities for hydrogen bonding I believe
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Group B: Model Answer GROUP B MODEL ANSWER Yoshida et al.‚ Flexibility of Hydrogen Bond and Lowering of Symmetry in Proton Conductor‚ Symmetry 2012‚ 4‚ 507-516. DO NOT PLAGIARISE THIS MODEL ANSWER PLAGIARISM FROM ANY SOURCE AUTOMATICALLY LEADS TO A ZERO SCORE Paragraph 1 M3H(XO4)2 compounds are used for electrolytic fuel cells (where M=K‚ Rb ‚Cs; X=S‚ Se). At different temperatures‚ the compound exhibits different degrees of symmetry. There are 3 distinct phases observed. Below 369K
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Yoshida et al.‚ Flexibility of Hydrogen Bond and Lowering of Symmetry in Proton Conductor‚ Symmetry 2012‚ 4‚ 507-516. Paragraph 1 In this paper‚ we take a look at the different phases of the Cs3H(SeO4)2 polymorphs. There are 3 different phases that is observed and each of them is influenced by the temperatures. In addition‚ in each phase‚ the polymorph exhibits different crystal structure. At the room temperature of 298K‚ it will be in phase 3 whereby it takes the structure of monoclinic-C2/m
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Study of the first element – Hydrogen. Answer the following questions:- 1. Give a reason why hydrogen can be placed in group 1[IA] and group 17[vIIA] of the periodic table. 2. What similarities does it show with group 1[IA] and group 17[VIIA]. With special reference to valency electrons and ion formaton and examples. 3. How does hydrogen occur in the free and combined state? 4. Which metals react with cold‚steam and boiling water to form their respected oxides and hydroxides ? Give examples and
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of a component causes the equilibrium to shift to the side from which the component was removed. D. Is this reaction endothermic or exothermic? How do you know? When acid is added to water the reaction is strongly exothermic. Because the hydrogen ion is so tiny‚ a large amount of charge is concentrated in a very small
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IRON (METAL) Iron is a chemical element with the symbol Fe (from Latin: ferrum) and atomic number 26. It is a metal in the first transition series. It is the most common element (by mass) forming the planet Earth as a whole‚ forming much of Earth’s outer and inner core. It is the fourth most common element in the Earth’s crust. Iron’s very common presence in rocky planets like Earth is due to its abundant production as a result of fusion in high-mass stars‚ Like other group 8 elements‚ iron
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What are Yield to Maturity (YTM) and Yield to Call (YTC)? By calculating the present and future value of bonds‚ managers can make sound decisions about their potential strengths and weaknesses as investments. Answer the following questions in this week’s Discussion 2 thread: 1. What terms (or inputs) are needed to calculate yield to maturity (YTM)? How does this compare to calculating yield to call (YTC)? To calculate the YTM you will need to use Annual Interest‚ Par value‚ Market Price
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Bonds and Their Valuation After reading this chapter‚ students should be able to: • List the four main classifications of bonds and differentiate among them. • Identify the key characteristics common to all bonds. • Calculate the value of a bond with annual or semiannual interest payments. • Explain why the market value of an outstanding fixed-rate bond will fall when interest rates rise on new bonds of equal risk‚ or vice versa. • Calculate the current yield
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perpetual bond is currently selling for RS. 95/-. The coupon rate of interest is 13.5%. The approximate discount rate is 15%. The value of the bond and the YTM is: (a) Rs. 90/- and 14.2% Value is (13.5*15%=90) and YTM is ((13.5/95)*100=14.21%) (b) Rs. 100/- and 13.5% (c) Rs. 90 and 15% (d) Rs. 90/- and 13.5% 902. In 2001‚ Meridian Ltd. has issued bonds of Rs. 10‚000/-each due in 2011 with a 14% per annum coupon rate payable at the end of each year during the life of the bond. If the required
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