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Photon336

  • one year ago

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  1. Photon336
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    #1

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  2. Photon336
    • one year ago
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    #2 just one question

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  3. Photon336
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    #3

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  4. Photon336
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  5. Photon336
    • one year ago
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    #160 carbon tetra chloride would definitely be the lowest, b/c it has just a polar covalent bond. I think that would be a liquid. heat of fusion (s) --> (l).

  6. Rushwr
    • one year ago
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    I think I'll go with C for that question cuz I feel that H2O must be having a high heat of fusion. H2O has H bonds so in order to break that we need greater amount of heat. But I'm not sre

  7. Rushwr
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    161 I'll go with answer C

  8. Rushwr
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    196 I'll go with 1st answer

  9. Rushwr
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    208 I'll go with D

  10. Rushwr
    • one year ago
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    206 it is C

  11. Photon336
    • one year ago
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    @sweetburger

  12. anonymous
    • one year ago
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    160. You need to consider the force of the intermolecular bond between different molecules. we have: CCl4 - it's a gas at room t, so it's so it's fusion temprature is much under 0C' H20- liquid at room t, the intermolecular forces are: van der vals and h-bond. They are are weak forces, and we know that it fusion t it's' little more than 0C' NaCl and Al2O3 are bouth salt, instad we had a inic bond, and it is much stronger than van der vals or h-bond (cuz it's the attraction between partial charges, while in ionic we have entire charge). I had no idea which had higher fusing point between two salt. But wiki told me Al2O3 (2000 C') while NaCl (800 C'). So it should be the answer A.

  13. Photon336
    • one year ago
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    Like at first i thought hydrogen bonds was the strongest, but then it was heat of fusion i.e melting.

  14. anonymous
    • one year ago
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    160. i get it, Al2O3 had higher fusion t, cuz the atoms Al and O are much more eletronegative instade of Na and Cl. So following Coulombs law of the interaction between charged objects, the forse is stronger because the atoms are more charged.

  15. Photon336
    • one year ago
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    got confused by that l\[Al _{2}O _{3}\] is ionic. so you consider charges? of ions

  16. Photon336
    • one year ago
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    @Bozhena that formula only deals with the magnitude of charges right? how does that apply to the ionic compounds, don't you have to take into consideration r?

  17. Photon336
    • one year ago
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    the way i understood that was that the smaller ions have greater attraction? b/c they can get closer

  18. anonymous
    • one year ago
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    yes, i mean. We need to know which between two salt has strongest intermolecular bond (cux it's the bond you break during the fusion). So you first need to find how much charg will had each salt.|dw:1438222995551:dw| Now you can see tha each molecule could be rappresented as |dw:1438223266787:dw| U need to employ much more energy to break the interaction of force 4 than 2

  19. Photon336
    • one year ago
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    Question 206. 1 mol of every gas has 22.4L occupied of volume. \[\frac{ mol }{ 22.4L } = \frac{ 2mol }{ ? }\] 44.8 mol*L = mol(x) 44.8L = Volume

  20. Photon336
    • one year ago
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    @Bozhena I see what you did for the first part.. the aluminum has a much greater electronegativity value so it's bond is going to be stronger.

  21. Photon336
    • one year ago
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    how does columns law apply to this if we were to use that?

  22. anonymous
    • one year ago
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    196. The density is 2g/L, if you assume to have a conteiner of 1L, You can find g of gas. 2g/l * 1L= 2g You also have the mols, and you know that Molar Mass is expressed in g/mol. So 2g/1mol = 2g/mol The molecule which have MM=2g/mol is H2

  23. anonymous
    • one year ago
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    160. consider that at che atomic level r does not matter, so consider r constant (i mean it does metter but the diffrence is very very small). |dw:1438223761999:dw|

  24. Photon336
    • one year ago
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    so it will only depend on the magnitude of the charges then

  25. Photon336
    • one year ago
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    I was thinking atomic level. yeah that makes more sense now because we can only take into account the charges magnitudes (molecular level).

  26. Photon336
    • one year ago
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    161. NaCl solution is boiled, the predominate species in the gaseous phase is: I thought it was Just water..that predominates. NaCl ---> Na+ Cl- in water, i feel that the boiling point would be a little higher because of the dissolved solute. we have 2 molecule of ions dissolved. to me it made no sense that there would be gaseous NaCl b/c in an aqueous environment you would just have the ions in solution.

  27. anonymous
    • one year ago
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    208: Im not pretty sure, but i suppose that: If the velocity of gas B is twice that gas A, means that gas A do not loose his energy, so it's bumping less, so it sould be very small.The smallest gas B is He, So the answer is A.

  28. Photon336
    • one year ago
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    @Bozhena \[\frac{ r _{1} }{ r _{2} } =\frac{ \sqrt{B} }{ \sqrt{A} }\]

  29. anonymous
    • one year ago
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    161. You right, onece boiled the water will evaporate (ebullioscopic point will be how a little bit higher cuz of ions is solution.). But salt will never boil, only if the t will be hight enough, but u need to consider that onlu the melting poin on NaCl it's about 800C'.

  30. Photon336
    • one year ago
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    BTW here are the answers 160. A 161. C 206. C 208. D

  31. Photon336
    • one year ago
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    the solution manual said "that NaCl isn't a volatile solute"

  32. anonymous
    • one year ago
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    To which question i should use that formula?

  33. Photon336
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    208 but you kind of have to work backwards with the choices

  34. Photon336
    • one year ago
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    i might have written that formula wrong

  35. anonymous
    • one year ago
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    I guess that 208 it's' A, but im not sure. While the other should be right

  36. Photon336
    • one year ago
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    \[2 = \sqrt{B/A} \] 4 = B/A Kr/Ne = 80/20 = 4 \[\sqrt{4} = 2 \]

  37. Photon336
    • one year ago
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    A = molar mass of gas A B = molar mass of gas B

  38. anonymous
    • one year ago
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    yes yes, get it, all clear! :)

  39. Photon336
    • one year ago
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    r1 = velocity of gas A r2 = velocity of gas b

  40. Photon336
    • one year ago
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    No problem! take care I have to go!

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