CHM 3411 – Problem Set 6 Due date: Wednesday‚ March 23rd. Do all of the following problems. Show your work. 1) Consider the cyclic molecule C8H8‚ the eight carbon analogue to benzene. a) Write the secular detrerminant corresponding to the pi-bonding in C8H8. b) Using the secular determinant‚ the following energies are found for the pi-bonding molecular orbitals: 1 = + 2 2 = + 1.41 (two states) 3 = (two states) 4 = - 1.41 (two states) 5 = - 2 Give the electron
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Introduction: During this laboratory exercise will identify by the graph given which is isotonic‚ hypotonic‚ or hypertonic. The kidneys are a pair of fist-sized organs located outside the peritoneal cavity on each side of the spine. The kidney is a highly specialized organ that maintains the internal environment of the body by selectively excreting or retaining various substances according to specific body needs. The process of urine formation and adjustment of blood composition involves three
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Questions and Hypothesis: Seed germination‚ plant growth‚ and sprouting of leaves are affected by several factors: supply of nutrients‚ water‚ exposure to sunlight‚ and conditions of surroundings. Due to certain stimuli in the surroundings plant “hormones” cause plants to behave in ways that ensure the most efficient use of resources while preserving the most energy. One stimulus to consider‚ is the exposure of plants to sunlight; when plants are just sprouting and are underground‚ the sti
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Title: to get organic food Specific purpose: Organic food advocates claim that organically grown foods are safer and more nutritious than foods raised with non-organic methods such as pesticide and non-organic fertilizer use‚ or antibiotic and hormone use. Thesis statement: Many people just don’t trust these chemicals and don’t want to put them into their bodies. Since virtually all non-organically produced foods contain residues of pesticides‚ fertilizers and other chemicals‚ the only way to
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Chapter 3 1. 1 atom of X = atomic mass (amu) 1 mole f X=atomic mass in grams 1 mole of something = 6.022 x 1023 units of that substance 1 mole of a compound =66.022 x 1023 atoms 6.022 x 1023 amu =1 g 2. What is the mass of 6 atoms of Fe? Answer: 6 atoms of Fe x 55.85 amu÷atom of Fe x 1 g of Fe÷6.022 x 1023 amu = 3. How many atoms does it take to make 1 g of Gold (Au)? Answer: 197.0 g Au =1 mole of Au
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Introduction: The purpose of this lab is to see how the colour of an apple would change (browning) over time‚ when placed in different conditions. When referring to the term “browning”‚ it is meant to imply the change of colour that occurs inside the apple‚ giving an appearance that is distasteful (Di Guardo et al.‚ 2013). Specifically‚ within this lab the apple was cut into 4 pieces with 3 pieces being placed on a weigh boat‚ and a 4th piece placed in a beaker of water. These weigh boats were
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Katerina Baeza Hernandez | | | | | | | |Zn |Mg |Cu |Pb | | |(+) clear liquid but the |(+) a black spot appeared on|(-) no reaction occurred |(-) no reaction | |Pb(NO3)2
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Use this packet and your book to answer the questions throughout this packet. Organic Nomenclature - Alkanes‚ Alkenes‚ Alkynes Naming organic compounds can be a challenge to any chemist at any level. Historically‚ chemists developed names for new compounds without any systematic guidelines. In this century‚ the need for standardization was recognized. For simple molecules‚ the nomenclature system worked out by the International Union of Pure and Applied Chemists (IUPAC) works well. For complex
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Introduction In unit 7.3 the experiment tested the ability of lactase to specifically bind and interact with lactose compared to maltose. In unit 7.4 the experiment tested the role‚ if any‚ that metal ions have on the activity of lactase. My hypothesis for unit 7.3 was knowing that lactase is specific for lactose‚ lactose will separate into galactose and glucose‚ as maltose will not change (153-155). Lactase should like lactose. For unit 7.4 my hypothesis was that EDTA will remove the ions‚ and
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Limiting Reagent and Percent Yield Aim To determine the limiting reagent between the reaction of lead (II) nitrate and potassium iodide. To determine the percent yield of lead (II) iodide. Date Started: 13/4/12. Finished: 19/4/12. Data collection and processing Measurements: * Amount of distilled water: 75.0ml ± 0.5ml. * Mass of watch glass: 31.65g ± 0.01g. * Mass of watch glass + potassium iodide: 32.45g ± 0.01g. * Mass of potassium iodide: 0.8g ± 0.02g. * Mass of watch
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