Key Question #9 - melting point than the CsF because it is...

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Lesson 9 Key Questions 41) a) The decreasing melting points from the ionic compounds from NaF to NaI relates to the ionic radius. The Na and F atom are the closest together which makes it more stable and requires more to break apart from its crystal lattice structure as opposed to the NaI which has radii are farther apart. b) This relates again to the stability of an ionic compound. Going down the list from NaF to NaI the strength of the bond decreases. This makes the NaI most soluble because it is easier to bond I - with the positive poles of the water molecules as opposed to F - which would be more difficult to bond because it has a stronger bond with the Na + atom. c) The CsF would have a higher solubility than the NaF because it is easier to bond the Cs + ion than the Na + ion with the water molecules. NaF would have a higher
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Unformatted text preview: melting point than the CsF because it is the more stable ionic compound. 42) You could do spectrography or x-ray diffraction. 43) a) It has layers of un-bonded carbon atoms within the structure which makes it more brittle. b) The arrangement of the atoms in diamond is structurally very strong. Each carbon bond has a double bond which makes it a strong bond. Since diamond has very strong bonds with the crystal lattice structure it requires more energy (heat) to make it melt. c) In layers of carbon that contain the un-bonded electrons, the free electrons are free to move throughout the layer which makes it a good conductor. 44) a) F and G d) D and E b) A and B e) F and G c) E 45) Experiment #2 is on the next pages....
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This note was uploaded on 12/13/2009 for the course ENGG eg 404 taught by Professor Weisenhaupt during the Spring '09 term at University of the Bío-Bío.

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Key Question #9 - melting point than the CsF because it is...

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