Topic 5 Energetics File
... Average bond enthalpy: The average enthalpy change of breaking one mole of a bond in a gaseous atom into its constituent gaseous atoms. Born-Haber cycle: Energy cycles for the formation of ionic compounds. If there is little agreement between the theoretical and experimental values, this could indic ...
... Average bond enthalpy: The average enthalpy change of breaking one mole of a bond in a gaseous atom into its constituent gaseous atoms. Born-Haber cycle: Energy cycles for the formation of ionic compounds. If there is little agreement between the theoretical and experimental values, this could indic ...
CH 8 blackboard
... NO2 is given in kilograms, you must first convert to grams. Since the final quantity is requested in kilograms, you must convert back to kilograms at the end. The main conversion factor is the stoichiometric relationship between moles of nitrogen dioxide and moles of nitric acid. This conversion fac ...
... NO2 is given in kilograms, you must first convert to grams. Since the final quantity is requested in kilograms, you must convert back to kilograms at the end. The main conversion factor is the stoichiometric relationship between moles of nitrogen dioxide and moles of nitric acid. This conversion fac ...
2005/6 - SAASTA
... There are two types of electrolytic cells; voltaic/galvanic and electrolytic. 1. A voltaic or galvanic cell produces electricity from a chemical reaction. 2. Electrolysis uses electricity to bring about chemical change. Regardless whether a cell is a voltaic or electrolytic cell; the anode (-ve elec ...
... There are two types of electrolytic cells; voltaic/galvanic and electrolytic. 1. A voltaic or galvanic cell produces electricity from a chemical reaction. 2. Electrolysis uses electricity to bring about chemical change. Regardless whether a cell is a voltaic or electrolytic cell; the anode (-ve elec ...
Chapter 5
... have a slightly distorted square-planar geometry, in which the allyl ligand coordinates in a π manner and occupies two coordination sites. We postulate that 1 and 2 might be the π-allyl complexes formed at an early stage of the reaction. The energy required to obtain these complexes is not low, but ...
... have a slightly distorted square-planar geometry, in which the allyl ligand coordinates in a π manner and occupies two coordination sites. We postulate that 1 and 2 might be the π-allyl complexes formed at an early stage of the reaction. The energy required to obtain these complexes is not low, but ...
Lecture notes
... Heat transfer causes a change in temperature and we can measure the amount of energy (in the form of heat) required to afford a specific change in temperature. We can’t measure heat directly so instead we calculate it from the resultant change in temperature (∆T) when heat (q) is transferred between ...
... Heat transfer causes a change in temperature and we can measure the amount of energy (in the form of heat) required to afford a specific change in temperature. We can’t measure heat directly so instead we calculate it from the resultant change in temperature (∆T) when heat (q) is transferred between ...
CHAPTER-7
... b) Equilibrium constant depends on temperature. It is constant for a given reaction at constant temperature. c) The equilibrium constant for the reverse reaction is equal to inverse of equilibrium constant for the forward reaction is Krev = 1/Kc d) Equilibrium constant is applicable only when concen ...
... b) Equilibrium constant depends on temperature. It is constant for a given reaction at constant temperature. c) The equilibrium constant for the reverse reaction is equal to inverse of equilibrium constant for the forward reaction is Krev = 1/Kc d) Equilibrium constant is applicable only when concen ...
Chapter 4
... 1. Most nitrate (NO3) salts are soluble. 2. Most alkali metal (group 1A) salts and NH4+ are soluble. 3. Most Cl, Br, and I salts are soluble (except Ag+, Pb2+, Hg22+). 4. Most sulfate salts are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4). 5. Most OH are only slightly soluble (NaOH, KOH are sol ...
... 1. Most nitrate (NO3) salts are soluble. 2. Most alkali metal (group 1A) salts and NH4+ are soluble. 3. Most Cl, Br, and I salts are soluble (except Ag+, Pb2+, Hg22+). 4. Most sulfate salts are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4). 5. Most OH are only slightly soluble (NaOH, KOH are sol ...
LABORATORY MANUAL CHEMISTRY 121
... The reactant is green while the product is supposed to be burgundy red to violet, but because of problems we have had in recent years, the product is sometimes yellow or orange. To determine the rate law and rate constants for this reaction, we shall measure the half-life for each experiment. When 5 ...
... The reactant is green while the product is supposed to be burgundy red to violet, but because of problems we have had in recent years, the product is sometimes yellow or orange. To determine the rate law and rate constants for this reaction, we shall measure the half-life for each experiment. When 5 ...
Unit 11 acids and bases part 1
... This new bond is called a coordinate covalent bond since both new bonding electrons come from the same atom ...
... This new bond is called a coordinate covalent bond since both new bonding electrons come from the same atom ...
Section 3_Energetics
... Direct determination of lattice energy is very difficult because it is very difficult to get isolated sodium and chloride ions. Therefore the values are usually calculated from other experimentally determined data by applying the Hess Law. The Born-Haber Cycle is a technique of applying Hess‘s Law t ...
... Direct determination of lattice energy is very difficult because it is very difficult to get isolated sodium and chloride ions. Therefore the values are usually calculated from other experimentally determined data by applying the Hess Law. The Born-Haber Cycle is a technique of applying Hess‘s Law t ...
AP Chemistry:
... 2 MnO4-(aq) + 5 C2O42−(aq) + 16 H+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l) 63. Permanganate and oxalate ions react in an acidified solution according to the balanced equation above. How many moles of CO2(g) are produced when 20. mL of acidified 0.20 M KMnO4 solution is added to 50. mL of 0.10 M Na2 ...
... 2 MnO4-(aq) + 5 C2O42−(aq) + 16 H+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l) 63. Permanganate and oxalate ions react in an acidified solution according to the balanced equation above. How many moles of CO2(g) are produced when 20. mL of acidified 0.20 M KMnO4 solution is added to 50. mL of 0.10 M Na2 ...
(cobalamin)-dependent enzymes
... diols or amino alcohols, followed by dehydration or deamination to yield aldehydes. Second, there are the aminomutases, which catalyse the 1,2 migration of an amino group within an amino acid, and also require pyridoxal phosphate as an additional coenzyme. Finally, perhaps the most unusual type of i ...
... diols or amino alcohols, followed by dehydration or deamination to yield aldehydes. Second, there are the aminomutases, which catalyse the 1,2 migration of an amino group within an amino acid, and also require pyridoxal phosphate as an additional coenzyme. Finally, perhaps the most unusual type of i ...
Chemistry - Ysgol Bro Pedr
... compounds. It is the number of electrons that need to be added (or taken away from) an element to make it neutral. (You don’t need to learn this definition.) Eg, a calcium ion, Ca2+ needs two electrons added to it in order to make a neutral calcium atom. Its oxidation number is therefore +2 (add 2 e ...
... compounds. It is the number of electrons that need to be added (or taken away from) an element to make it neutral. (You don’t need to learn this definition.) Eg, a calcium ion, Ca2+ needs two electrons added to it in order to make a neutral calcium atom. Its oxidation number is therefore +2 (add 2 e ...
Chemistry - cloudfront.net
... 42. given a Lewis structure, be able to compute the formal charge for an atom 43. given a covalent bond between two atoms, be able to discern if the bond is polar or non-polar 44. given a molecular formula, be able to draw the Lewis structure to be able to tell if the molecule is polar or non-polar ...
... 42. given a Lewis structure, be able to compute the formal charge for an atom 43. given a covalent bond between two atoms, be able to discern if the bond is polar or non-polar 44. given a molecular formula, be able to draw the Lewis structure to be able to tell if the molecule is polar or non-polar ...
Homework Solutions Week 6
... When silver sulfate starts to precipitate, 97% of the calcium has precipitated. And when calcium from 97 to 99% precipitated, silver ion goes from 0 to 41% precipitated. 9-17 a) Why do many rivers in Box 9-1. lie on the line [HCO3-] = 2[Ca2+]? According to Box 9-1, the source of calcium in the river ...
... When silver sulfate starts to precipitate, 97% of the calcium has precipitated. And when calcium from 97 to 99% precipitated, silver ion goes from 0 to 41% precipitated. 9-17 a) Why do many rivers in Box 9-1. lie on the line [HCO3-] = 2[Ca2+]? According to Box 9-1, the source of calcium in the river ...
Oxidation-Reduction Reactions - An Introduction to Chemistry
... Oxidation-Reduction and Molecular Compounds The oxidation of nitrogen to form nitrogen monoxide is very similar to the oxidation of zinc to form zinc oxide. N2(g) + O2(g) → 2NO(g) 2Zn(s) + O2(g) → 2ZnO(s) The main difference between these reactions is that as the nitrogen monoxide forms, electrons a ...
... Oxidation-Reduction and Molecular Compounds The oxidation of nitrogen to form nitrogen monoxide is very similar to the oxidation of zinc to form zinc oxide. N2(g) + O2(g) → 2NO(g) 2Zn(s) + O2(g) → 2ZnO(s) The main difference between these reactions is that as the nitrogen monoxide forms, electrons a ...
CHAPTER 12 Study Guide
... 41. a. 11.3 mol CO, 22.5 mol H2 b. 112 g CO, 16.0 g H2 c. 11.4 g H2 42. a. 372 g F2 b. 1.32 g NH3 c. 123 g N2F4 43. The coefficients indicate the relative number of moles (or particles) of reactants and products. 44. a. 51.2 g H2O b. 5.71 × 1023 molecules NH3 c. 23.2 g Li3N 45. The amount of the lim ...
... 41. a. 11.3 mol CO, 22.5 mol H2 b. 112 g CO, 16.0 g H2 c. 11.4 g H2 42. a. 372 g F2 b. 1.32 g NH3 c. 123 g N2F4 43. The coefficients indicate the relative number of moles (or particles) of reactants and products. 44. a. 51.2 g H2O b. 5.71 × 1023 molecules NH3 c. 23.2 g Li3N 45. The amount of the lim ...
Experimental and Computational Evidence of Metal‑O2 Activation
... on the initial concentration of O2 and whether the CoAGAO was pretreated with excess H2O2. Oxygen isotope fractionation was analyzed as previously described38−47,49 using a specially constructed apparatus.64 Samples were prepared by isolating O2 before and after treatment of O2 saturated solutions c ...
... on the initial concentration of O2 and whether the CoAGAO was pretreated with excess H2O2. Oxygen isotope fractionation was analyzed as previously described38−47,49 using a specially constructed apparatus.64 Samples were prepared by isolating O2 before and after treatment of O2 saturated solutions c ...
CH 151 Companion
... accurate observation of behavior of the system of interest. Conclusions are based solely on the observed data. Only by using experimental observations to study the behavior of matter, arranging the results of such studies in an orderly fashions, correlating the observed data and testing these correl ...
... accurate observation of behavior of the system of interest. Conclusions are based solely on the observed data. Only by using experimental observations to study the behavior of matter, arranging the results of such studies in an orderly fashions, correlating the observed data and testing these correl ...
1.24 calculations and chemical reactions
... reaction. Calculate the relative molecular mass, Mr, of H2A 4.2) Sodium carbonate forms several hydrates of general formula Na2CO3.xH2O. A 2.98 g sample of one of these hydrates was dissolved in water and the solution made up to 250cm3. In a titration, a 25.0 cm3 portion of this solution required 28 ...
... reaction. Calculate the relative molecular mass, Mr, of H2A 4.2) Sodium carbonate forms several hydrates of general formula Na2CO3.xH2O. A 2.98 g sample of one of these hydrates was dissolved in water and the solution made up to 250cm3. In a titration, a 25.0 cm3 portion of this solution required 28 ...
KUT 203/2 - Chemistry Practical III (Inorganic Chemistry)
... • Understand the correlation between the colors of transition metal or metal ion with its oxidation state i.e. VO2+ (yellow), VO2+ (blue) etc. • Determine the composition of a metal complex of which the metal exists in various oxidation states by using the titration technique. • Synthesize several c ...
... • Understand the correlation between the colors of transition metal or metal ion with its oxidation state i.e. VO2+ (yellow), VO2+ (blue) etc. • Determine the composition of a metal complex of which the metal exists in various oxidation states by using the titration technique. • Synthesize several c ...
Unit #8 - consumerchem
... # of atoms of each element on the right of the "yields" arrow Many equations can be balanced by trial and error… However, the following five rules will make balancing quicker. 3) 1. Write the correct formulas: a) For all reactants to the left of the arrow. b) For all products to the right of the arr ...
... # of atoms of each element on the right of the "yields" arrow Many equations can be balanced by trial and error… However, the following five rules will make balancing quicker. 3) 1. Write the correct formulas: a) For all reactants to the left of the arrow. b) For all products to the right of the arr ...
C3 Revision Question Booklet
... Write the name of the element with the next highest atomic number after Te (tellurium) in the periodic table. ...
... Write the name of the element with the next highest atomic number after Te (tellurium) in the periodic table. ...
Lab # 18
... 3. What is a hydrate? Give an example of a hydrated substance we have used in lab. 4. Define anhydrous. How can a hydrate be changed to its anhydrous form? 5. Why would it be better to use plain popcorn instead of buttered popcorn for this experiment? 6. Write the formula for calculating percent by ...
... 3. What is a hydrate? Give an example of a hydrated substance we have used in lab. 4. Define anhydrous. How can a hydrate be changed to its anhydrous form? 5. Why would it be better to use plain popcorn instead of buttered popcorn for this experiment? 6. Write the formula for calculating percent by ...
Lewis acid catalysis
In Lewis acid catalysis of organic reactions, a metal-based Lewis acid acts as an electron pair acceptor to increase the reactivity of a substrate. Common Lewis acid catalysts are based on main group metals such as aluminum, boron, silicon, and tin, as well as many early (titanium, zirconium) and late (iron, copper, zinc) d-block metals. The metal atom forms an adduct with a lone-pair bearing electronegative atom in the substrate, such as oxygen (both sp2 or sp3), nitrogen, sulfur, and halogens. The complexation has partial charge-transfer character and makes the lone-pair donor effectively more electronegative, activating the substrate toward nucleophilic attack, heterolytic bond cleavage, or cycloaddition with 1,3-dienes and 1,3-dipoles.Many classical reactions involving carbon–carbon or carbon–heteroatom bond formation can be catalyzed by Lewis acids. Examples include the Friedel-Crafts reaction, the aldol reaction, and various pericyclic processes that proceed slowly at room temperature, such as the Diels-Alder reaction and the ene reaction. In addition to accelerating the reactions, Lewis acid catalysts are able to impose regioselectivity and stereoselectivity in many cases.Early developments in Lewis acid reagents focused on easily available compounds such as TiCl4, BF3, SnCl4, and AlCl3. The relative strengths of these (and other) Lewis acids may be estimated from NMR spectroscopy by the Childs method or the Gutmann-Beckett method. Over the years, versatile catalysts bearing ligands designed for specific applications have facilitated improvement in both reactivity and selectivity of Lewis acid-catalyzed reactions. More recently, Lewis acid catalysts with chiral ligands have become an important class of tools for asymmetric catalysis.Challenges in the development of Lewis acid catalysis include inefficient catalyst turnover (caused by catalyst affinity for the product) and the frequent requirement of two-point binding for stereoselectivity, which often necessitates the use of auxiliary groups.