Unit 1 -Notes 2 - e. Matter 1) Pure constant composition...

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4/8/10 e. Matter 1) Pure – constant composition 1) Mixtures – variable composition a) Homogeneous – single phase a) Heterogeneous – more than one phase
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4/8/10 Analysis of a Mixture Qualitative Analysis separation and identification of the components of the mixture Quantitative Analysis determination of the quantity of each component of the mixture
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4/8/10 Separation Techniques Crystallization separation of solids from a liquid solution Distillation separation of one component of a liquid solution from another based on boiling points Chromatography separation of one component of a liquid solution from another on a support material Solvent Extraction 16
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Distillation Separates homogeneous mixture on the basis of differences in boiling point.
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Chromatography: Separates substances on the basis of differences in solubility in a solvent.
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4/8/10 Atomic mass It is 12g/ 6.023X1023 =1.99X10-23 g The mass of one carbon atom is 1.99X10-23 g So what is atomic mass It is the mass of 6.023X1023 carbon atoms in grams, but expressed as a number 12
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4/8/10 Find the atomic mass and the identity of an element. When 1.42 g of M2SO4 were
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This note was uploaded on 04/08/2010 for the course CHM CHM140Y5 taught by Professor Dr.krish during the Spring '09 term at University of Toronto.

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Unit 1 -Notes 2 - e. Matter 1) Pure constant composition...

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