Lab Manual Yr 1 organic
... Never look directly into the mouth of a flask containing a reaction mixture. Never point a test tube or reaction flask towards yourself or your neighbour. When using a separating funnel, vent frequently and remove the stopper immediately upon setting it upright for separation. Never use a thermomete ...
... Never look directly into the mouth of a flask containing a reaction mixture. Never point a test tube or reaction flask towards yourself or your neighbour. When using a separating funnel, vent frequently and remove the stopper immediately upon setting it upright for separation. Never use a thermomete ...
STUDY MATERIAL 2016-17 CHEMISTRY CLASS XII
... into cubic lattice. If the density is 2g/cm3 then find the radius of metal atom (NA = 6.022 x 1023) 4. The density of KBr is 2.75 gm cm -3 . The length of edge of the unit cell is 654 pm. Predict the type of cubic lattice.. NA=6.023 x 1023 ; at mass of K=39: Br. = 80 Hint- Calculate value of z= 4 so ...
... into cubic lattice. If the density is 2g/cm3 then find the radius of metal atom (NA = 6.022 x 1023) 4. The density of KBr is 2.75 gm cm -3 . The length of edge of the unit cell is 654 pm. Predict the type of cubic lattice.. NA=6.023 x 1023 ; at mass of K=39: Br. = 80 Hint- Calculate value of z= 4 so ...
LaBrake, Fundamentals Diagnostic Questions
... b) Most of the space of an atom is empty space. c) The nuclear model in which the positive charge is held densely in the center of the atom is no longer accepted, because it was proposed over 100 years ago. d) In a neutral atom, the number of protons in the nucleus is equal to the number of electron ...
... b) Most of the space of an atom is empty space. c) The nuclear model in which the positive charge is held densely in the center of the atom is no longer accepted, because it was proposed over 100 years ago. d) In a neutral atom, the number of protons in the nucleus is equal to the number of electron ...
Sign of enthalpy changes Exothermic vs endothermic Acid
... Problem: 0.3 mL of 1mM solution of a drug is added to a tube with a protein solution. Some of the drug reacts with the protein to form 1:1 protein-drug complex. The reaction produces a heat of –2.4 x 10–3 cal. Estimate the molar enthalpy of binding binding. ...
... Problem: 0.3 mL of 1mM solution of a drug is added to a tube with a protein solution. Some of the drug reacts with the protein to form 1:1 protein-drug complex. The reaction produces a heat of –2.4 x 10–3 cal. Estimate the molar enthalpy of binding binding. ...
- Mendeley Data
... 2.5. X-Ray Crystallography Analysis of the Vanadyl Schiff base complex By using graphite monochromated Mo-Kα radiation, the X-ray diffraction measurements were collected on a STOE IPDS-2T diffractometer. For the purpose of collecting data, the black crystal Vanadyl complex with a dimension of 0.16 ...
... 2.5. X-Ray Crystallography Analysis of the Vanadyl Schiff base complex By using graphite monochromated Mo-Kα radiation, the X-ray diffraction measurements were collected on a STOE IPDS-2T diffractometer. For the purpose of collecting data, the black crystal Vanadyl complex with a dimension of 0.16 ...
Introduction to Qualitative Analysis
... Hydrogen peroxide solutions will degrade over time due to this reaction. To slow the process and increase the shelf life of the solution, hydrogen peroxide solutions may be stored in a refrigerator. Formation of Complex Ions Many common anions and neutral molecules can donate one or more lone pairs ...
... Hydrogen peroxide solutions will degrade over time due to this reaction. To slow the process and increase the shelf life of the solution, hydrogen peroxide solutions may be stored in a refrigerator. Formation of Complex Ions Many common anions and neutral molecules can donate one or more lone pairs ...
CHAPTER 21 NONMETALLIC ELEMENTS AND THEIR COMPOUNDS
... hygroscopic). Thus, most of the white solid is NaHCO3 plus some Na2CO3. ...
... hygroscopic). Thus, most of the white solid is NaHCO3 plus some Na2CO3. ...
AP Chemistry: Chapter 13 Gaseous Equilibrium Section 1: Multiple
... If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid cat ...
... If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid cat ...
CHAPTER 19
... When you think of atoms changing oxidation states, don’t think that either atom has totally lost or totally gained any electrons. In the case of the formation of hydrogen chloride, for example, hydrogen simply has donated a share of its bonding electron to the chlorine; it has not completely transfe ...
... When you think of atoms changing oxidation states, don’t think that either atom has totally lost or totally gained any electrons. In the case of the formation of hydrogen chloride, for example, hydrogen simply has donated a share of its bonding electron to the chlorine; it has not completely transfe ...
Chemical Quantities and Aqueous Reactions
... In recent years scientists have become concerned because the amount of atmospheric carbon dioxide (CO2)—Earth’s most significant greenhouse gas in terms of its contribution to climate—is rising. More CO2 enhances the atmosphere’s ability to hold heat and may therefore lead to global warming, an incr ...
... In recent years scientists have become concerned because the amount of atmospheric carbon dioxide (CO2)—Earth’s most significant greenhouse gas in terms of its contribution to climate—is rising. More CO2 enhances the atmosphere’s ability to hold heat and may therefore lead to global warming, an incr ...
Stoichiometric Problems III: Sto c o et c ob e s
... Here is a typical problem. problem If I have the following chemical reaction: N2(g) + 3H2 (g) = 2NH3 (g) And I start with 5 moles of N2 and 5 moles of H2, which chemical (H2 or N2) is the limiting reactant? My approach to solving this kind of problem is to calculate the amount of product that would ...
... Here is a typical problem. problem If I have the following chemical reaction: N2(g) + 3H2 (g) = 2NH3 (g) And I start with 5 moles of N2 and 5 moles of H2, which chemical (H2 or N2) is the limiting reactant? My approach to solving this kind of problem is to calculate the amount of product that would ...
WHAT YOU EAT - Montana State University Extended University
... a short time), the elevated blood sugar elevates insulin and shuts off fat burning by the tissues (so the body can reduce the blood sugar spike caused by eating the carbohydrates) and usually much ...
... a short time), the elevated blood sugar elevates insulin and shuts off fat burning by the tissues (so the body can reduce the blood sugar spike caused by eating the carbohydrates) and usually much ...
Chemical Reaction Equations
... In order to make sure this happens, more of the other reactant must be present than is required An excess reagent is the reactant whose entities are present in surplus amounts, so that some remain after the reaction ends.. In our reaction: much more copper was used than needed (evidenced by the unre ...
... In order to make sure this happens, more of the other reactant must be present than is required An excess reagent is the reactant whose entities are present in surplus amounts, so that some remain after the reaction ends.. In our reaction: much more copper was used than needed (evidenced by the unre ...
Higher Chemistry Specimen Question Paper
... 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at ...
... 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at ...
Part A Completion
... © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. ...
... © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. ...
Answers to examination questions
... Q5 D The oxygen molecule (O=O) contains one double bond; the carbon dioxide molecule (O=C=O) contains two double bond and the tetrafluoroethene molecule contains one double bond. Q6 A B is trigonal planar (bond angles 120°); A, C and D are based upon a tetrahedral arrangement with four regio ...
... Q5 D The oxygen molecule (O=O) contains one double bond; the carbon dioxide molecule (O=C=O) contains two double bond and the tetrafluoroethene molecule contains one double bond. Q6 A B is trigonal planar (bond angles 120°); A, C and D are based upon a tetrahedral arrangement with four regio ...
Chapter 13
... shift to reach equilibrium, we compare the values of Q and K. There are three possible cases: 1.Q is equal to K. The system is at equilibrium; no shift will occur. 2.Q is greater than K. The system shifts to the left, consuming products and forming reactants, until equilibrium is achieved. 3.Q is le ...
... shift to reach equilibrium, we compare the values of Q and K. There are three possible cases: 1.Q is equal to K. The system is at equilibrium; no shift will occur. 2.Q is greater than K. The system shifts to the left, consuming products and forming reactants, until equilibrium is achieved. 3.Q is le ...
Ch 17 Equilibrium
... • Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant. • Using the collision model: – as the amount of NO2 builds up, there is a chance that two NO2 molecules will coll ...
... • Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant. • Using the collision model: – as the amount of NO2 builds up, there is a chance that two NO2 molecules will coll ...
Acids and Bases Unit
... This is a guessing question so be prepared for different responses. Most likely, that the light bulb will light up After testing the conductivity of the strong acid and before testing the weak acid in solution: o Did anything happen to the light bulb? Is that what we expected? If done correct ...
... This is a guessing question so be prepared for different responses. Most likely, that the light bulb will light up After testing the conductivity of the strong acid and before testing the weak acid in solution: o Did anything happen to the light bulb? Is that what we expected? If done correct ...
It`s Easy Being a Green Chemist
... way products are made. It is a science that aims to reduce or eliminate the use and/or generation of hazardous substances in the design phase of materials development. Considered a subset of sustainable chemistry, Green Chemistry is defined as: The utilization of a set of principles that reduces or ...
... way products are made. It is a science that aims to reduce or eliminate the use and/or generation of hazardous substances in the design phase of materials development. Considered a subset of sustainable chemistry, Green Chemistry is defined as: The utilization of a set of principles that reduces or ...
2014_S4_CHM_NORMAL (ALL)
... 53. Element X (atomic number 11) reacts with element Y (atomic number 16) to form an ionic compound. Each atom of X loses one electron and each atom of Y accepts two electrons to form a compound with formula X2Y. 54. Consider the following information: ...
... 53. Element X (atomic number 11) reacts with element Y (atomic number 16) to form an ionic compound. Each atom of X loses one electron and each atom of Y accepts two electrons to form a compound with formula X2Y. 54. Consider the following information: ...
Line 4: Equation
... 7. Calculate the mole ratio. In order to get a ratio, you have to divide two numbers that have the same unit. We were given moles, so we can divide line 2 by line 6. If we were given grams, we would divide the given number on line 3 by line 7. Write this number next to mole ratio. 2.5 mol/ 1 mol = 2 ...
... 7. Calculate the mole ratio. In order to get a ratio, you have to divide two numbers that have the same unit. We were given moles, so we can divide line 2 by line 6. If we were given grams, we would divide the given number on line 3 by line 7. Write this number next to mole ratio. 2.5 mol/ 1 mol = 2 ...
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.