FREQUENTLY FORGOTTEN FACTS
... shift away from the side it is on. If something is removed, then the equilibrium will shift towards that side. After the shift, whatever is being shifted towards will increase in concentration, and whatever is being shifted away from will decrease in concentration. For the equilibrium N2 (g) + 3H2 ( ...
... shift away from the side it is on. If something is removed, then the equilibrium will shift towards that side. After the shift, whatever is being shifted towards will increase in concentration, and whatever is being shifted away from will decrease in concentration. For the equilibrium N2 (g) + 3H2 ( ...
Equilibria PPT
... • A strong acid is one which is FULLY IONISED in water. It will have a high hydrogen ion concentration in this case ALL of the H+ is made so strong acid pH1 • A weak acid is one which is NOT fully ionised and is in equilibrium. It has a low hydrogen ion concentration In this case there are SOME H+ b ...
... • A strong acid is one which is FULLY IONISED in water. It will have a high hydrogen ion concentration in this case ALL of the H+ is made so strong acid pH1 • A weak acid is one which is NOT fully ionised and is in equilibrium. It has a low hydrogen ion concentration In this case there are SOME H+ b ...
Thermochemistry Questions
... (ΔH f) of solid Mg(OH)2, • given the following information: • Rxn 1: 2Mg (s) + O2 (g) → 2MgO (s) ΔH = -1203.6 kJ • Rxn 2: Mg(OH)2 (s) → MgO (s) + H2O (l) ΔH = +37.1 kJ • Rxn 3: 2H2O (l) → 2H2 (g) + O2 (g) ΔH = +571.7 kJ ...
... (ΔH f) of solid Mg(OH)2, • given the following information: • Rxn 1: 2Mg (s) + O2 (g) → 2MgO (s) ΔH = -1203.6 kJ • Rxn 2: Mg(OH)2 (s) → MgO (s) + H2O (l) ΔH = +37.1 kJ • Rxn 3: 2H2O (l) → 2H2 (g) + O2 (g) ΔH = +571.7 kJ ...
CHEMICAL REACTIONS
... e. Lewis acid (e- pr acceptor) and Lewis base (e- pr. donor) • e.g. boron trifluoride (e- deficient) and ammonia (lone pair on N) : BF3 + NH3 Æ BF3NH3 B. Decomposition/ Analysis Reactions 1. decomposition (not necessarily heated): a. hydrogen peroxide decomposes into water and oxygen. b. ammonium hy ...
... e. Lewis acid (e- pr acceptor) and Lewis base (e- pr. donor) • e.g. boron trifluoride (e- deficient) and ammonia (lone pair on N) : BF3 + NH3 Æ BF3NH3 B. Decomposition/ Analysis Reactions 1. decomposition (not necessarily heated): a. hydrogen peroxide decomposes into water and oxygen. b. ammonium hy ...
Energy Changes, Reaction Rates and Equilibrium Thermodynamics
... •The Ea is the minimum amount of energy that the reactants must possess for a reaction to occur. •Ea is called the energy barrier and the height of the barrier determines the reaction rate. •When the Ea is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow. ...
... •The Ea is the minimum amount of energy that the reactants must possess for a reaction to occur. •Ea is called the energy barrier and the height of the barrier determines the reaction rate. •When the Ea is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow. ...
Selenium dioxide catalysed oxidation of acetic acid hydrazide by
... followed by its oxidation by diprotonated bromate in a slow step. Acetic acid was found to be the oxidation product. Other kinetic data like effect of solvent polarity and ionic strength on the reaction support the proposed mechanism. Keywords. ...
... followed by its oxidation by diprotonated bromate in a slow step. Acetic acid was found to be the oxidation product. Other kinetic data like effect of solvent polarity and ionic strength on the reaction support the proposed mechanism. Keywords. ...
CHAP 1 - NCERT books
... skeletal chemical equation. Is the chemical Eq. (1.2) balanced? Let us learn about balancing a chemical equation step by step. The word-equation for Activity 1.3 may be represented as – ...
... skeletal chemical equation. Is the chemical Eq. (1.2) balanced? Let us learn about balancing a chemical equation step by step. The word-equation for Activity 1.3 may be represented as – ...
AP CHEMISTRY COURSE SYLLABUS
... with respect to the kind of textbook used, the range and depth of topics covered, the kind of laboratory work done, and the time and effort required. The aim of this course is to provide you with the conceptual framework, factual knowledge, and analytical skills necessary to deal critically with the ...
... with respect to the kind of textbook used, the range and depth of topics covered, the kind of laboratory work done, and the time and effort required. The aim of this course is to provide you with the conceptual framework, factual knowledge, and analytical skills necessary to deal critically with the ...
Journal of Physical and Chemical Reference Data
... There are between 10 and 20 million known compounds that can actually or hypothetically react with each other in an astronomical number of ways and it is therefore literally impossible to catalog all the possible heats of reaction. To get around this problem we define for each substance a standard ...
... There are between 10 and 20 million known compounds that can actually or hypothetically react with each other in an astronomical number of ways and it is therefore literally impossible to catalog all the possible heats of reaction. To get around this problem we define for each substance a standard ...
Physical Science Chapter 7 Chemical Reactions Section 7.1
... each neutral oxygen atom gains two electrons, it becomes an ion with a charge of 2–. ____________________ ___________________________________________________________________________. ...
... each neutral oxygen atom gains two electrons, it becomes an ion with a charge of 2–. ____________________ ___________________________________________________________________________. ...
Equation Writing Information
... figure out whether or not the reaction actually works, as you did in ‘Introductory’ Chemistry. You will be required to write equations in ionic form - i.e. for ionic compounds you write the anions (-) and cations (+) as separate entities. You also omit the spectator ions - those ions that do not tak ...
... figure out whether or not the reaction actually works, as you did in ‘Introductory’ Chemistry. You will be required to write equations in ionic form - i.e. for ionic compounds you write the anions (-) and cations (+) as separate entities. You also omit the spectator ions - those ions that do not tak ...
Document
... How Many Grams of N2(g) Can Be Made from 9.05 g of NH3 Reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 Are Made, What Is the Percent Yield? ...
... How Many Grams of N2(g) Can Be Made from 9.05 g of NH3 Reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 Are Made, What Is the Percent Yield? ...
Topic 5 - Chemical Reactions
... Matter can be described using both chemical and physical properties. Elements are the simplest form of matter that has a unique set of properties. A compound is a substance that contains two or more elements chemically combined in a fixed proportion. 4. Chemical formulas are used to represent compou ...
... Matter can be described using both chemical and physical properties. Elements are the simplest form of matter that has a unique set of properties. A compound is a substance that contains two or more elements chemically combined in a fixed proportion. 4. Chemical formulas are used to represent compou ...
Reactions of Metals and Their Compounds
... piece of zinc(Zn) in a small beaker or petri dish Collect 5mL of limewater(Ca(OH)2) in a test tube. Gently blow through a straw into the test tube until a change occurs. ...
... piece of zinc(Zn) in a small beaker or petri dish Collect 5mL of limewater(Ca(OH)2) in a test tube. Gently blow through a straw into the test tube until a change occurs. ...
PPT Oxidation
... • What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two halfreactions from the example: Ag+ ---> Ag Cu ---> Cu2+ • The silver is being reduced, its oxidation number g ...
... • What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two halfreactions from the example: Ag+ ---> Ag Cu ---> Cu2+ • The silver is being reduced, its oxidation number g ...
PPT Oxidation
... • What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two halfreactions from the example: Ag+ ---> Ag Cu ---> Cu2+ • The silver is being reduced, its oxidation number g ...
... • What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two halfreactions from the example: Ag+ ---> Ag Cu ---> Cu2+ • The silver is being reduced, its oxidation number g ...
Question paper - Edexcel
... 16 Two ketones, CH3COCH2CH2CH3 and CH3CH2COCH2CH3, both have Mr = 86. Which peak due to fragmentation into singly charged ions would you expect to be present in the mass spectrum of one but not the other? A 71 B 57 C 43 D (Total for Question 16 = 1 mark) TOTAL FOR SECTION A = 20 MARKS ...
... 16 Two ketones, CH3COCH2CH2CH3 and CH3CH2COCH2CH3, both have Mr = 86. Which peak due to fragmentation into singly charged ions would you expect to be present in the mass spectrum of one but not the other? A 71 B 57 C 43 D (Total for Question 16 = 1 mark) TOTAL FOR SECTION A = 20 MARKS ...
Chem 1411 Chapter 4
... Non- electrolyte. Such solutions do not conduct electricity due to the absence of ions. Water is a very effective solvent for ionic compounds because of its polar nature. The O--H bond in water is polar due to the electro negativity (electron attracting power) of ‘O’atom. The ‘O’atom of water is ric ...
... Non- electrolyte. Such solutions do not conduct electricity due to the absence of ions. Water is a very effective solvent for ionic compounds because of its polar nature. The O--H bond in water is polar due to the electro negativity (electron attracting power) of ‘O’atom. The ‘O’atom of water is ric ...
Equilibrium Constant- Keq
... Equilibrium Constant- Keq 1. In an experiment, 0.500 mol/L of hydrogen bromide gas is decomposed into hydrogen and bromine gases. a) Write the equilibrium equation and equilibrium law for this reaction. b) The equilibrium concentrations in this system are [HBr(g)] =0.240 mol/L and [H2(g)]=0.130 mol/ ...
... Equilibrium Constant- Keq 1. In an experiment, 0.500 mol/L of hydrogen bromide gas is decomposed into hydrogen and bromine gases. a) Write the equilibrium equation and equilibrium law for this reaction. b) The equilibrium concentrations in this system are [HBr(g)] =0.240 mol/L and [H2(g)]=0.130 mol/ ...
EXPERIMENT 11 (2 Weeks)!
... solutions (4-5 drops) and then look for evidence of a chemical reaction. Record any precipitate that forms and its color. If there is no reaction write N.R. Write the balanced equation for those reactions that do occur. Identify the unknown by mixing 4-5 drops of each solution with 4-5 drops of your ...
... solutions (4-5 drops) and then look for evidence of a chemical reaction. Record any precipitate that forms and its color. If there is no reaction write N.R. Write the balanced equation for those reactions that do occur. Identify the unknown by mixing 4-5 drops of each solution with 4-5 drops of your ...
7th Chemistry Unit Test Study Guide Test Date: Friday, Nov. 16
... many ATOMS of an element are in one molecule? ____________________________. Big number or little number? (circle one) In a chemical equation, which number tells us how many MOLECULES are involved in a chemical reaction? _______________________. Big number or little number? (circle one) _____________ ...
... many ATOMS of an element are in one molecule? ____________________________. Big number or little number? (circle one) In a chemical equation, which number tells us how many MOLECULES are involved in a chemical reaction? _______________________. Big number or little number? (circle one) _____________ ...
Exam 2
... Org II exam, p.2 Section A (48 points, 4 each – answer 12 of the 13 questions, marking clearly the one you choose to omit; if none is marked, the last answer will be omitted) Give the major organic product(s) of the following reactions, including stereochemistry if appropriate. ...
... Org II exam, p.2 Section A (48 points, 4 each – answer 12 of the 13 questions, marking clearly the one you choose to omit; if none is marked, the last answer will be omitted) Give the major organic product(s) of the following reactions, including stereochemistry if appropriate. ...