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Photon336

  • one year ago

Another question: Going to post them in a bit.

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  1. Photon336
    • one year ago
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    Will post soon. Going to be on the rate constant, and kinetics.

  2. Photon336
    • one year ago
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    Posted here other one got cut off

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  3. Photon336
    • one year ago
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    Some more here

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  4. Photon336
    • one year ago
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    =D

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  5. Photon336
    • one year ago
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    Ok I think this is the last one for this question

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  6. Photon336
    • one year ago
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    I know this is a lot but some of them aren't that bad.

  7. Photon336
    • one year ago
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    @Rushwr @taramgrant0543664

  8. sweetburger
    • one year ago
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    Ill start with the easy one 276. Im pretty sure OH is a intermediate

  9. Photon336
    • one year ago
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    276. it's gotta be an intermediate because if it were a catalyist it would be generated in the last step again and unchanged. (Promoter and activator make no sense)

  10. sweetburger
    • one year ago
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    For 275 I'm not sure but I think that the concentration of CaCl2 would have no affect on the rate because it is a product and it isnt affecting the forward reaction. Not sure if my reasoning on this is sound though.

  11. sweetburger
    • one year ago
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    My answer also came down to that for 275 B,C, and D. would change the rate of the reaction.

  12. sweetburger
    • one year ago
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    Yea actually im certain for 275 that A would be the only answer that would not affect the reaction rate.

  13. anonymous
    • one year ago
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    Yeah, sounds right to me, I guess the concentration of \(CaCl_2\) would only affect the equilibrium constant.

  14. anonymous
    • one year ago
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    wait... I don't know for some reason I can't really decide between A and B

  15. Photon336
    • one year ago
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    289. which one is false? A. Chlorine goes from an oxidation state of 0 to -1 so it's reduced so that's true B. increasing the temperature generally does increase the rate of reaction. \[k = Ae ^{Ea/Rt}\] you increase temperature i think the quantitiy Ea/Rt goes down i believe and k goes up and reaction rate goes up as well. D yes HCL is a strong acid, and HCLO is a weak acid so that's true.

  16. sweetburger
    • one year ago
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    289 I think is C.

  17. Photon336
    • one year ago
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    289. is C. here's my justification

  18. sweetburger
    • one year ago
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    The only thing that affect the equilibrium constant is temperature from what I remember.

  19. anonymous
    • one year ago
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    Yeah it shouldn't affect the equilibrium constant, I guess the rate constant is not affected by concentrations of the products only concentration of reactants.

  20. Photon336
    • one year ago
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    Like i thought it was C for 289 b/c I notice that the number of moles of each quantity are the same. To me like an increase in pressure favors the side with fewer number of moles i think , while a decrease in pressure favors the side with more moles.. Keq

  21. Photon336
    • one year ago
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    yeah it's only affected by temperature as well

  22. Photon336
    • one year ago
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    sorry.. i was thinking in terms of what would favor the products.

  23. sweetburger
    • one year ago
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    My answer choice for 290 is B.

  24. Photon336
    • one year ago
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    Okay so 276 D. "An OH radical is formed in the rate determining step but is consumed in the next step therefore it's an intermediate" 289 C The equilibrium constant is not affected by changes in concentrations and pressures remember that equilibrium states are separate from kinetic studies.

  25. sweetburger
    • one year ago
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    Looks good.

  26. anonymous
    • one year ago
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    Yeah for 289 I don't think partial pressures come into play because we're only talking about liquids.

  27. Photon336
    • one year ago
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    290. \[k[cl _{2}]^{0.5}[H _{2}O]\]

  28. Photon336
    • one year ago
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    IF you look at the slow step the rate for formation depends on the formation of the chlorine radical and it's ability to react with water as well.

  29. Photon336
    • one year ago
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    |dw:1438546224213:dw|

  30. Photon336
    • one year ago
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    If this is what happens above.. then yo can clearly see why we also need water to make this happen and why it's also the rate depends on that too

  31. Photon336
    • one year ago
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    @Woodward I got 275 wrong why is it A

  32. sweetburger
    • one year ago
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    Wait

  33. sweetburger
    • one year ago
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    For 275 is A not the answer?

  34. Photon336
    • one year ago
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    @sweetburger sorry the answer is A.

  35. Photon336
    • one year ago
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    was asking why though?

  36. sweetburger
    • one year ago
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    I guess Ill give it a shot at explaining it. Lets say that the rate law is rate=k[CaCO3][HCl]^2. This of course might not be the rate law as it is experimentally determined but Im assuming it is. So lets say that if you changed the concentration of the product CaCl2 it would have no affect on the rate law I wrote above.

  37. Photon336
    • one year ago
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    Oh I see

  38. Photon336
    • one year ago
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    it's because it's a product and the rate doesn't depend on the products just reactants

  39. anonymous
    • one year ago
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    Yeah the partial pressure of HCl is essentially the concentration of HCl due to that whole \(P= \frac{n}{V} RT\) thing from earlier. And the concentration of a reactant is found in the rate law, not the products like \(CaCl_2\) is I guess

  40. Photon336
    • one year ago
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    \[r = k[HCl]^{2} \]

  41. Photon336
    • one year ago
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    I thought we don't include solids in the expression because their concentrations don't change much

  42. sweetburger
    • one year ago
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    Thats the equilibrium expression. What I wrote is a rate law.

  43. Photon336
    • one year ago
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    oh okay.

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