chemistry
... 11 The degree of polarity of a chemical bond in a molecule of a compound can be predicted by determining the difference in the (1) melting points of the elements in the compound (2) densities of the elements in the compound (3) electronegativities of the bonded atoms in a molecule of the compound (4 ...
... 11 The degree of polarity of a chemical bond in a molecule of a compound can be predicted by determining the difference in the (1) melting points of the elements in the compound (2) densities of the elements in the compound (3) electronegativities of the bonded atoms in a molecule of the compound (4 ...
1.24 calculations and chemical reactions
... 3.2. Lead(II) nitrate can be produced by the reaction between nitric acid and lead(II) oxide as shown by the equation below. PbO + 2HNO3 → Pb(NO3)2 + H2O An excess of lead(II) oxide reacted with 250cm3 of 1.25moldm–3 nitric acid. Calculate the maximum mass of lead(II) nitrate which could have been o ...
... 3.2. Lead(II) nitrate can be produced by the reaction between nitric acid and lead(II) oxide as shown by the equation below. PbO + 2HNO3 → Pb(NO3)2 + H2O An excess of lead(II) oxide reacted with 250cm3 of 1.25moldm–3 nitric acid. Calculate the maximum mass of lead(II) nitrate which could have been o ...
mark scheme - A-Level Chemistry
... M2 could be to an H+ ion and M3 an independent O − H bond break on a compound with molecular formula for H2SO4 Max any 3 of 4 marks for a correct mechanism using the wrong organic reactant or wrong organic product (if shown) or a primary carbocation. M3 must show the breaking of the O − H bond on H2 ...
... M2 could be to an H+ ion and M3 an independent O − H bond break on a compound with molecular formula for H2SO4 Max any 3 of 4 marks for a correct mechanism using the wrong organic reactant or wrong organic product (if shown) or a primary carbocation. M3 must show the breaking of the O − H bond on H2 ...
Chapter 17 lecture notes on Chemical Equilibria
... A + B ⇔ C + D Suppose we add additional reactant A or B to the system. This means we have more reactant than is necessary to establish equilibrium and the system shifts to reduce the amount of A and B. This results in an increase in the amount of C and D. The counter to this is also true. If the sys ...
... A + B ⇔ C + D Suppose we add additional reactant A or B to the system. This means we have more reactant than is necessary to establish equilibrium and the system shifts to reduce the amount of A and B. This results in an increase in the amount of C and D. The counter to this is also true. If the sys ...
Chemical Reactions
... they move about, they collide with each other and with objects in their path. Very tiny objects, such as specks of carbon in car exhaust, are buffeted by these particles, and move about erratically (Figure 4). A temperature increase indicates that these particles are moving more quickly. We can make ...
... they move about, they collide with each other and with objects in their path. Very tiny objects, such as specks of carbon in car exhaust, are buffeted by these particles, and move about erratically (Figure 4). A temperature increase indicates that these particles are moving more quickly. We can make ...
chapter 21 chemistry of the main-group elements i
... Once we return to Mg(OH)2 from MgSO4, the other substances can be made by the indicated pathways. The return reaction is: MgSO 4 (aq) + 2 NaOH(aq) Mg(OH) 2 (s) + Na 2SO 4 (aq). Then the other reactions are ...
... Once we return to Mg(OH)2 from MgSO4, the other substances can be made by the indicated pathways. The return reaction is: MgSO 4 (aq) + 2 NaOH(aq) Mg(OH) 2 (s) + Na 2SO 4 (aq). Then the other reactions are ...
Study Guide: Chemistry
... dissolving of salt in water, heating iodine Chemical Change - Bonds are broken and new bonds are formed in a compound Examples of chemical change - Rusting of iron, sodium metal reacting with water, heating magnesium in air or heating ammonium chloride, adding sulphuric acid to water, ripening of fr ...
... dissolving of salt in water, heating iodine Chemical Change - Bonds are broken and new bonds are formed in a compound Examples of chemical change - Rusting of iron, sodium metal reacting with water, heating magnesium in air or heating ammonium chloride, adding sulphuric acid to water, ripening of fr ...
CHM2210 Organic Chemistry 1
... 3. evaluating potential energy diagrams to determine the relative energy of reactants and products and to establish whether a reaction is endothermic, exothermic, endergonic or exergonic. 4. evaluating potential energy diagrams to determine the relative stability of conformers. 5. justifying the ...
... 3. evaluating potential energy diagrams to determine the relative energy of reactants and products and to establish whether a reaction is endothermic, exothermic, endergonic or exergonic. 4. evaluating potential energy diagrams to determine the relative stability of conformers. 5. justifying the ...
NATIONAL HIGH SCHOOL CHEMISTRY EXAMINATION (1995
... A. Ag+ is an oxidizing agent but Ni 2+ is a reducing agent B. Ag + is a better oxidizing agent than Ni2+ and Ag is a better reducing agent than Ni C. Ni 2+ can be reduced by silver metal D. Ag+ is a better oxidizing agent than Ni2+ and Ni is a better reducing agent than Ag E. Ni 2+ is a better oxidi ...
... A. Ag+ is an oxidizing agent but Ni 2+ is a reducing agent B. Ag + is a better oxidizing agent than Ni2+ and Ag is a better reducing agent than Ni C. Ni 2+ can be reduced by silver metal D. Ag+ is a better oxidizing agent than Ni2+ and Ni is a better reducing agent than Ag E. Ni 2+ is a better oxidi ...
Practice Test Material - Directorate of Education
... Calculate the pH of 0.10M ammonia solution. Calculate the pH after 50.0 ml of this solution is treated with 25.0 ml of 0.10M HCl. The dissociation constant of ammonia (Kb) is 1.77×10–5. Hint – In the final condition, basic buffer is formed due to the presence of NH4Cl and NH4OH in the same solution. ...
... Calculate the pH of 0.10M ammonia solution. Calculate the pH after 50.0 ml of this solution is treated with 25.0 ml of 0.10M HCl. The dissociation constant of ammonia (Kb) is 1.77×10–5. Hint – In the final condition, basic buffer is formed due to the presence of NH4Cl and NH4OH in the same solution. ...
Chemical Reactions and The Mole
... of atoms in one mole of gold atoms. If each person counted one atom per second and worked a 48-hour week, the task would take more than 8 million years. If you had a mole of pennies, you would have enough money to pay all the expenses of the United States for the next billion years. A mole of the la ...
... of atoms in one mole of gold atoms. If each person counted one atom per second and worked a 48-hour week, the task would take more than 8 million years. If you had a mole of pennies, you would have enough money to pay all the expenses of the United States for the next billion years. A mole of the la ...
Alcohols and Phenols
... Alcohols and phenols have much higher boiling points than similar alkanes and alkyl halides. A positively polarized OH hydrogen atom from one molecule is attracted to a lone pair of electrons on a negatively polarized oxygen atom of another molecule. This produces a force that holds the two molecul ...
... Alcohols and phenols have much higher boiling points than similar alkanes and alkyl halides. A positively polarized OH hydrogen atom from one molecule is attracted to a lone pair of electrons on a negatively polarized oxygen atom of another molecule. This produces a force that holds the two molecul ...
7 Organic 2
... structural isomers geometric isomers 3. Alkynes: C C 4. Aromatic hydrocarbons Reactivity of hydrocarbons • addition reactions • combustion reactions ...
... structural isomers geometric isomers 3. Alkynes: C C 4. Aromatic hydrocarbons Reactivity of hydrocarbons • addition reactions • combustion reactions ...
Page 1
... (4) different odor (5) light 26. Write chemical equations for the following, then balance the equations. a. hydrogen(g) + bromine(g) hydrogen bromide(g) H2 + Br2 2HBr ...
... (4) different odor (5) light 26. Write chemical equations for the following, then balance the equations. a. hydrogen(g) + bromine(g) hydrogen bromide(g) H2 + Br2 2HBr ...
7404_2-qp-chemistry-as-10june16-pm
... B Addition of a catalyst moves the position of EA to the right. C The area under the curve to the right of EA represents the number of molecules with enough energy to react. D The position of the peak of the curve at a higher temperature is further away from both axes. ...
... B Addition of a catalyst moves the position of EA to the right. C The area under the curve to the right of EA represents the number of molecules with enough energy to react. D The position of the peak of the curve at a higher temperature is further away from both axes. ...
Review: The preparation of Hydantoins
... concentrated hydrochloric acid and 5 g of crude product was thus obtained by recrystallization. Preparation of Hydantoin via amino actonitrile sulfate The preparation of hydantoin by Harries and Weiss has the defect of the production of glycolester hydrochloride from methyleneamino-acetonitrile and ...
... concentrated hydrochloric acid and 5 g of crude product was thus obtained by recrystallization. Preparation of Hydantoin via amino actonitrile sulfate The preparation of hydantoin by Harries and Weiss has the defect of the production of glycolester hydrochloride from methyleneamino-acetonitrile and ...
Reactions of Carboxylic Acids
... Sodium stearate has a polar end (the carboxylate end) and a nonpolar "tail“. The polar end is hydrophilic ("water-loving”). The nonpolar tail is hydrophobic ("water-hating”). In water, many stearate ions cluster together to form spherical aggregates; carboxylate ions are on the outside and nonpolar ...
... Sodium stearate has a polar end (the carboxylate end) and a nonpolar "tail“. The polar end is hydrophilic ("water-loving”). The nonpolar tail is hydrophobic ("water-hating”). In water, many stearate ions cluster together to form spherical aggregates; carboxylate ions are on the outside and nonpolar ...
HYDROCARBONS HYDROCARBONS Types of Hydrocarbons
... 5. When both directions lead to the same lowest number for one of the substituents, the direction is chosen that gives the lowest possible number to one of the remaining substituents ...
... 5. When both directions lead to the same lowest number for one of the substituents, the direction is chosen that gives the lowest possible number to one of the remaining substituents ...
Synthesis and Characterization of Coordination Compounds
... compounds is that they are formed from chemical species that have an independent existence and that this association is often readily reversible (i. e., there is an equilibrium between the solvated metal ion and the ligand). For example, NiCl2 reacts with NH3 in aqueous solution to form the compound ...
... compounds is that they are formed from chemical species that have an independent existence and that this association is often readily reversible (i. e., there is an equilibrium between the solvated metal ion and the ligand). For example, NiCl2 reacts with NH3 in aqueous solution to form the compound ...
Chemistry 11 - hrsbstaff.ednet.ns.ca
... Number toward the 2nd substituent in the direction which gives the lower number for the carbon bearing that second substituent and name any alkyl branch. ...
... Number toward the 2nd substituent in the direction which gives the lower number for the carbon bearing that second substituent and name any alkyl branch. ...
Chemical Equilibrium - Shailendra Kumar Chemistry
... 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the 2NH3(g) at equation, N2(g) + 3H2(g) constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are: (a) 4 : 5 ...
... 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the 2NH3(g) at equation, N2(g) + 3H2(g) constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are: (a) 4 : 5 ...
Principles of Drug Action I, Spring 2004
... 16. Which compound below (A-E) will undergo electrophilic aromatic substitution most readily? C. That aromatic ring with the best electron donating group will be most reactive toward electrophilic substitution. In this case only one compound has an electron donor (C), and it is a weak donor by +I. I ...
... 16. Which compound below (A-E) will undergo electrophilic aromatic substitution most readily? C. That aromatic ring with the best electron donating group will be most reactive toward electrophilic substitution. In this case only one compound has an electron donor (C), and it is a weak donor by +I. I ...
Strychnine total synthesis
Strychnine total synthesis in chemistry describes the total synthesis of the complex biomolecule strychnine. The first reported method by the group of Robert Burns Woodward in 1954 is considered a classic in this research field. At the time it formed the natural conclusion to an elaborate process of molecular structure elucidation that started with the isolation of strychnine from the beans of Strychnos ignatii by Pierre Joseph Pelletier and Joseph Bienaimé Caventou in 1818. Major contributors to the entire effort were Sir Robert Robinson with over 250 publications and Hermann Leuchs with another 125 papers in a time span of 40 years. Robinson was awarded the Nobel Prize in Chemistry in 1947 for his work on alkaloids, strychnine included. The process of chemical identification was completed with publications in 1946 by Robinson and later confirmed by Woodward in 1947. X-ray structures establishing the absolute configuration became available between 1947 and 1951 with publications from J. M. Bijvoet and J.H. Robertson .Woodward published a very brief account on the strychnine synthesis in 1954 (just 3 pages) and a lengthy one (42 pages) in 1963.Many more methods exist and reported by the research groups of Magnus, Overman, Kuehne, Rawal, Bosch, Vollhardt, Mori, Shibasaki, Li, Fukuyama Vanderwal and MacMillan. Synthetic (+)-strychnine is also known. Racemic synthesises were published by Padwa in 2007 and in 2010 by Andrade and by Reissig.In his 1963 publication Woodward quoted Sir Robert Robinson who said for its molecular size it is the most complex substance known.