![l - CMatthews](http://s1.studyres.com/store/data/011553969_1-97b112474f11ef9aa24c8396be1b80ce-300x300.png)
l - CMatthews
... 3. Calculate the Hf for the target reaction: 3 Fe2O3 (s) 2 Fe3O4 (s) + ½ O2 (g) given the data below. Explain what you needed to do to each equation to get your answer. 2 Fe (s) + 3/2 O2 (g) Fe2O3 (s) Hf = -823.41 kJ 3 Fe (s) + 2 O2 (g) Fe3O4 (s) Hf = -1120.48 kJ 4. What is the Ho in ki ...
... 3. Calculate the Hf for the target reaction: 3 Fe2O3 (s) 2 Fe3O4 (s) + ½ O2 (g) given the data below. Explain what you needed to do to each equation to get your answer. 2 Fe (s) + 3/2 O2 (g) Fe2O3 (s) Hf = -823.41 kJ 3 Fe (s) + 2 O2 (g) Fe3O4 (s) Hf = -1120.48 kJ 4. What is the Ho in ki ...
Organic Chemistry Unit Test
... 1. We did two labs involving esters. In the first, we made a series of esters. In the second, we made two polyesters. Describe at least 3 ‘real world’ uses that you could imagine for the products of either of these labs based on your observations and data table. (3 marks) ...
... 1. We did two labs involving esters. In the first, we made a series of esters. In the second, we made two polyesters. Describe at least 3 ‘real world’ uses that you could imagine for the products of either of these labs based on your observations and data table. (3 marks) ...
SAMPLE TEST PAPER KPS CHEMISTRY
... Define the term ‘osmotic pressure’. Describe, how the molecular mass of a substrate can be determined on the basis of osmotic pressure measurement? Q19. What happens when D-Glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO3 Q20. An organic compound A contains 69.77% ...
... Define the term ‘osmotic pressure’. Describe, how the molecular mass of a substrate can be determined on the basis of osmotic pressure measurement? Q19. What happens when D-Glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO3 Q20. An organic compound A contains 69.77% ...
excess
... 4. Sodium hydride, NaH, is an ionic compound. Na H a) Write the Lewis electron-dot structure for NaH. b) If NaH is placed into water (a foolish thing to do), the hydride ion is converted to hydrogen gas (H2). The resulting solution turns red litmus paper blue and has a pH OH Na>>7.H + H2O H2 + Na Wr ...
... 4. Sodium hydride, NaH, is an ionic compound. Na H a) Write the Lewis electron-dot structure for NaH. b) If NaH is placed into water (a foolish thing to do), the hydride ion is converted to hydrogen gas (H2). The resulting solution turns red litmus paper blue and has a pH OH Na>>7.H + H2O H2 + Na Wr ...
Chapter 7: Structure and Synthesis of Alkenes
... most stable chair conformation may not be the one leading to the product. this H cannot participate in the elimination since it is equatorial ...
... most stable chair conformation may not be the one leading to the product. this H cannot participate in the elimination since it is equatorial ...
a. Rank by acidity. The most acidic compound is 1, wh
... Of the following reactions that generate glycols, one reacts with periodic acid as shown above, while the second does not react with the periodic acid. Draw the structures obtained for each of these reactions. Assume proper work-up for each step. (12 points) O ...
... Of the following reactions that generate glycols, one reacts with periodic acid as shown above, while the second does not react with the periodic acid. Draw the structures obtained for each of these reactions. Assume proper work-up for each step. (12 points) O ...
Chem 3.5 Answers #7
... Aldehydes are produced by the oxidation of primary alcohols using acidified potassium dichromate solution. The aldehyde must be distilled off as it is made or it will oxidise further, up to the carboxylic acid. Ketones are made by the same oxidation reaction with secondary alcohols, but they do not ...
... Aldehydes are produced by the oxidation of primary alcohols using acidified potassium dichromate solution. The aldehyde must be distilled off as it is made or it will oxidise further, up to the carboxylic acid. Ketones are made by the same oxidation reaction with secondary alcohols, but they do not ...
Question paper - Edexcel
... be allowed to subside after each addition. 4. After the addition of iodine is complete, the mixture is heated under reflux for 30 – 60 minutes, until little or no iodine is visible. 5. The apparatus is allowed to cool and the condenser rearranged for distillation. 6. The crude 1-iodobutane is distil ...
... be allowed to subside after each addition. 4. After the addition of iodine is complete, the mixture is heated under reflux for 30 – 60 minutes, until little or no iodine is visible. 5. The apparatus is allowed to cool and the condenser rearranged for distillation. 6. The crude 1-iodobutane is distil ...
PowerPoint **
... Polar Reaction Under Basic Conditions a. Substitution and Elimination at C(sp3)-X σ bonds b. Addition of Nuclepphiles to Electrophilic π bonds c. Substitution at C(sp2)-X σ bonds d. Base-promoted Rearrangements ...
... Polar Reaction Under Basic Conditions a. Substitution and Elimination at C(sp3)-X σ bonds b. Addition of Nuclepphiles to Electrophilic π bonds c. Substitution at C(sp2)-X σ bonds d. Base-promoted Rearrangements ...
File - chemistryattweed
... Liquefying and removing the ammonia as it is produced also increase the yield of ammonia. ...
... Liquefying and removing the ammonia as it is produced also increase the yield of ammonia. ...
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
... 4. Write the Woodward Hoffmann rules for electrocyclization reaction. 5. Explain the change in the geometry of excited state molecule in a photochemical process and the variation in its physical property. 6. What is Norrish type I and II cleavage reactions? Give suitable examples. 7. What are the im ...
... 4. Write the Woodward Hoffmann rules for electrocyclization reaction. 5. Explain the change in the geometry of excited state molecule in a photochemical process and the variation in its physical property. 6. What is Norrish type I and II cleavage reactions? Give suitable examples. 7. What are the im ...
Glossary of Key Terms in Chapter Two
... quaternary ammonium salt (15.1) an amine salt with the general formula R4N+ A– (in which R– can be an alkyl or aryl group or a hydrogen atom and A– can be any anion. secondary (2˚) amine (15.1) an amine with the general formula R2-NH. tertiary (3˚) amine (15.1) an amine with the general formula R3-N ...
... quaternary ammonium salt (15.1) an amine salt with the general formula R4N+ A– (in which R– can be an alkyl or aryl group or a hydrogen atom and A– can be any anion. secondary (2˚) amine (15.1) an amine with the general formula R2-NH. tertiary (3˚) amine (15.1) an amine with the general formula R3-N ...
Exam 2 - Wake Forest University
... Do not open or begin this exam until instructed. This exam consists of 5 pages plus the cover page. Before starting the exam, check to make sure that you have all of the pages. The exam has a total of 100 points and includes 13 questions. Only legible answers written on the exam will be considered f ...
... Do not open or begin this exam until instructed. This exam consists of 5 pages plus the cover page. Before starting the exam, check to make sure that you have all of the pages. The exam has a total of 100 points and includes 13 questions. Only legible answers written on the exam will be considered f ...
Addition reactions
... Tertiary alcohols are not easily oxidised because, unlike primary and secondary alcohols, they do not have a hydrogen attached to the same carbon atom as the hydroxyl group. Of course, the opposite of oxidation is reduction and the previous two examples can also go in reverse: Example: reduction of ...
... Tertiary alcohols are not easily oxidised because, unlike primary and secondary alcohols, they do not have a hydrogen attached to the same carbon atom as the hydroxyl group. Of course, the opposite of oxidation is reduction and the previous two examples can also go in reverse: Example: reduction of ...
with answers
... Pauli Principle: No two electrons in an atom can have all four quantum numbers identical. Or an orbital can be occupied by at most two electrons, which must then have antiparallel spin. Hund’s Rule: Orbitals will be filled with electrons in order of increasing energy; orbitals of the same energy wil ...
... Pauli Principle: No two electrons in an atom can have all four quantum numbers identical. Or an orbital can be occupied by at most two electrons, which must then have antiparallel spin. Hund’s Rule: Orbitals will be filled with electrons in order of increasing energy; orbitals of the same energy wil ...
Carboxylic Acid Derivatives
... A similar procedure is used to make amides from acyl chlorides and amines (the amine must have at least one hydrogen attached to the nitrogen). ...
... A similar procedure is used to make amides from acyl chlorides and amines (the amine must have at least one hydrogen attached to the nitrogen). ...
ESTERIFICATION
... effect, putting pressure on the left side). Usually, a threefold molar excess is enough to drive the equilibrium sufficiently to the right in an esterification reaction. Either the alcohol or the acid can be used in excess. The choice can be based on cost, availability and/or ease of purification a ...
... effect, putting pressure on the left side). Usually, a threefold molar excess is enough to drive the equilibrium sufficiently to the right in an esterification reaction. Either the alcohol or the acid can be used in excess. The choice can be based on cost, availability and/or ease of purification a ...
Polymerization - WordPress.com
... a. Polymer – a large molecule consisting of repeating units. b. Repeating unit – recurring unit in a polymer c. Monomer – smallest molecule from which the polymer is made i. Naming Polymers: put poly- in front of the name of the monomer 3. There are many polymerization mechanisms. You need to know: ...
... a. Polymer – a large molecule consisting of repeating units. b. Repeating unit – recurring unit in a polymer c. Monomer – smallest molecule from which the polymer is made i. Naming Polymers: put poly- in front of the name of the monomer 3. There are many polymerization mechanisms. You need to know: ...
Polymerization
... a. Polymer – a large molecule consisting of repeating units. b. Repeating unit – recurring unit in a polymer c. Monomer – smallest molecule from which the polymer is made i. Naming Polymers: put poly- in front of the name of the monomer 3. There are many polymerization mechanisms. You need to know: ...
... a. Polymer – a large molecule consisting of repeating units. b. Repeating unit – recurring unit in a polymer c. Monomer – smallest molecule from which the polymer is made i. Naming Polymers: put poly- in front of the name of the monomer 3. There are many polymerization mechanisms. You need to know: ...
Chapter 7
... • The slow step, RDS, is the second step, the formation of the carbocation • This explains the order of reactivity with the tertiary alcohol reacting easiest, due to the tertiary carbocation being the most stable. ...
... • The slow step, RDS, is the second step, the formation of the carbocation • This explains the order of reactivity with the tertiary alcohol reacting easiest, due to the tertiary carbocation being the most stable. ...
Lab 7_Esterification
... the equilibrium to the right (to the products) by having one of the reactants in excess (in effect, putting pressure on the left side). Usually, a threefold molar excess is enough to drive the equilibrium sufficiently to the right in an esterification reaction. Either the alcohol or the acid can be ...
... the equilibrium to the right (to the products) by having one of the reactants in excess (in effect, putting pressure on the left side). Usually, a threefold molar excess is enough to drive the equilibrium sufficiently to the right in an esterification reaction. Either the alcohol or the acid can be ...
Chapter 7: Alkenes and Alkynes – Properties and Synthesis
... The cis isomer is less stable due to greater strain from crowding by the adjacent alkyl groups. ...
... The cis isomer is less stable due to greater strain from crowding by the adjacent alkyl groups. ...
E2 reactions
... Decide whether the following substrates could react by E1 or E2 (and by SN1 or SN2). Br ...
... Decide whether the following substrates could react by E1 or E2 (and by SN1 or SN2). Br ...
PHYSICAL SCIENCE PAPER 2 QUESTIONS SECTION A
... will affect product yield and the value of the equilibrium constant. Write down the respective letter and next to the letter, one of the following: increase, decrease, no effect. ...
... will affect product yield and the value of the equilibrium constant. Write down the respective letter and next to the letter, one of the following: increase, decrease, no effect. ...
File - cpprashanths Chemistry
... 2. Question No. 1-8 are very short answer questions and carry 1 mark each. 3. Question No. 9-18 are short answer questions and carry 2 marks each. 4. Question No. 19-27 are also short answer questions and carry 3 marks each. 5. Question No. 28-30 are long answer questions and carry 5 marks each. 6. ...
... 2. Question No. 1-8 are very short answer questions and carry 1 mark each. 3. Question No. 9-18 are short answer questions and carry 2 marks each. 4. Question No. 19-27 are also short answer questions and carry 3 marks each. 5. Question No. 28-30 are long answer questions and carry 5 marks each. 6. ...
Hofmann–Löffler reaction
![](https://commons.wikimedia.org/wiki/Special:FilePath/General_scheme.gif?width=300)
The Hofmann–Löffler reaction (also referred to as Hofmann–Löffler–Freytag reaction, Löffler–Freytag reaction, Löffler–Hofmann reaction, as well as Löffler's method) is an organic reaction in which a cyclic amine 2 (pyrrolidine or, in some cases, piperidine) is generated by thermal or photochemical decomposition of N-halogenated amine 1 in the presence of a strong acid (concentrated sulfuric acid or concentrated CF3CO2H). The Hofmann–Löffler–Freytag reaction proceeds via an intramolecular hydrogen atom transfer to a nitrogen-centered radical and is an example of a remote intramolecular free radical C–H functionalization.