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Study Guides for Chemistry at University of Saskatchewan (U OF S)


U OF SCHEM 115Ahiahonu PearsonFall

CHEM 115- Final Exam Guide - Comprehensive Notes for the exam ( 60 pages long!)

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Mass n2= 500. kg = 5. 00 x 105 g. Mol n2= 5. 00 x 105 g / 28 g/mol = 1. 785 x 104 mol. Mass h2= 1. 00. kg= 1. 00 x 105 g. Mol h2= 1. 00 x 105 g / 2. 01
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U OF SCHEM 250Gravel MichelFall

CHEM 250- Final Exam Guide - Comprehensive Notes for the exam ( 227 pages long!)

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The chemistry of cycloalkanes is like that of open-chain alkanes: both are nonpolar and fairly inert. Cycloalkanes are less flexible and there is much
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U OF SCHEM 115Ahiahonu PearsonFall

CHEM 115- Final Exam Guide - Comprehensive Notes for the exam ( 30 pages long!)

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Mass n2= 500. kg = 5. 00 x 105 g. Mol n2= 5. 00 x 105 g / 28 g/mol = 1. 785 x 104 mol. Mass h2= 1. 00. kg= 1. 00 x 105 g. Mol h2= 1. 00 x 105 g / 2. 01
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U OF SCHEM 250Dr.Marco SalesFall

CHEM 250- Final Exam Guide - Comprehensive Notes for the exam ( 404 pages long!)

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16. 1 monobromination of toluene gives a mixture of three bromotoluene products. Solution: o-xylene: 2. m-xylene: 3 p xylene: 1. 16. 3 when benzene is
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U OF SCHEM 250Gravel MichelFall

[CHEM 250] - Final Exam Guide - Everything you need to know! (25 pages long)

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U OF SCHEM 250Murray BremnerSpring

CHEM 250 Study Guide - Final Guide: Valence Bond Theory, Acyl Halide, Grignard Reaction

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Hybridization of nitrogen, oxygen, phosphorus and sulfur. Chapter 2 polar covalent bonds; acids and bases. Acids and bases: bronsted-lowry, acid base s
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U OF SCHEM 250Dr.Marco SalesFall

CHEM 250 Study Guide - Midterm Guide: Electrophilic Substitution, Dehydrogenation, Trifluoromethyl

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16. 1 monobromination of toluene gives a mixture of three bromotoluene products. Solution: o-xylene: 2. m-xylene: 3 p xylene: 1. 16. 3 when benzene is
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U OF SCHEM 250Dr.Marco SalesFall

CHEM 250- Midterm Exam Guide - Comprehensive Notes for the exam ( 32 pages long!)

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U OF SCHEM 115Ahiahonu PearsonFall

CHEM 115- Midterm Exam Guide - Comprehensive Notes for the exam ( 12 pages long!)

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Mass n2= 500. kg = 5. 00 x 105 g. Mol n2= 5. 00 x 105 g / 28 g/mol = 1. 785 x 104 mol. Mass h2= 1. 00. kg= 1. 00 x 105 g. Mol h2= 1. 00 x 105 g / 2. 01
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U OF SCHEM 250Majewski MarekWinter

CHEM 250 Study Guide - Midterm Guide: Covalent Bond, Stereoisomerism, Butene

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U OF SCHEM 250Gravel MichelWinter

CHEM 250 Study Guide - Midterm Guide: Enthalpy, Khan Academy, Epoxide

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Practice problems: 18, 19, 24, 26, 27, 30, 36, 38, 40, 50, 57. Chapter 2: polar covalent bonds; acids and bases. Practice problems: 22, 24, 31, 33, 36,
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U OF SCHEM 250SalesFall

Chem250-Midterm(Fall2013)-Answers.pdf

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Multiple choice answers: c, b, b, e, c, b, e, c, d, c, a, d, b, a, e, d, a, b, d, not graded. Part ii (10 marks: the visual pigment in animal cells con
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