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vera_ewing

  • one year ago

chemistry question

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  1. vera_ewing
    • one year ago
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    @rvc @TrojanPoem

  2. rvc
    • one year ago
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    do you know the rate formula

  3. vera_ewing
    • one year ago
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    Yeah. I think it's either A or B not really sure...

  4. TrojanPoem
    • one year ago
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    I think this question is chemical equilibrium related. I think (A) as Water is not counted but oxygen water is counted.

  5. rvc
    • one year ago
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    yep i agree with trojan

  6. rvc
    • one year ago
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    wait

  7. rvc
    • one year ago
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    im not sure :/

  8. TrojanPoem
    • one year ago
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    Give me a second, I am gonna check.

  9. TrojanPoem
    • one year ago
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    I found this : A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration. [So H2o2 only]

  10. rvc
    • one year ago
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    oh okay

  11. vera_ewing
    • one year ago
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    So what would the answer be?

  12. rvc
    • one year ago
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    agree with trojan A

  13. vera_ewing
    • one year ago
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    What happens to the energy of the transition state when a catalyst is used? A. The energy increases. B. The energy decreases. C. The energy does not change. D. The energy varies during the catalyzed reaction.

  14. rvc
    • one year ago
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    The catalyst will decreease the energy of activation so as to increease the rate of the reaction

  15. vera_ewing
    • one year ago
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    This one?

  16. rvc
    • one year ago
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    i have already said about that

  17. sweetburger
    • one year ago
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    I think the transition state is at the top of the curve which is represented by D on the graph.

  18. TrojanPoem
    • one year ago
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    it's D.

  19. rvc
    • one year ago
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    |dw:1435413759932:dw|

  20. vera_ewing
    • one year ago
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    Wait, so it's D. A or C. D ?

  21. sweetburger
    • one year ago
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    Answer is C. D

  22. sweetburger
    • one year ago
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    confusing lol

  23. vera_ewing
    • one year ago
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    Which of the following statements is true if HI is formed in the following reaction? H2(g) + I2(g) → 2HI(g) A. Some of the collisions of the reactants have enough energy to overcome the activation energy. B. Some of the collisions between reactants have enough energy to overcome the reaction mechanism. C. There are no collisions between the reactants in the reaction. D. All of the collisions between reactants have enough energy to overcome the activation energy.

  24. sweetburger
    • one year ago
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    It isnt

  25. rvc
    • one year ago
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    for product to form collision has to occur

  26. sweetburger
    • one year ago
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    In collision theory not all of the colliosions will result in a creation of a product. Only if the collision has the proper orientation and energy will it create a product.

  27. TrojanPoem
    • one year ago
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    So B,

  28. rvc
    • one year ago
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    yep the orientation matters too but nothing is mentioned about it

  29. sweetburger
    • one year ago
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    Yes, but never in any real reaction will all collision result in a product

  30. rvc
    • one year ago
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    oops B lol

  31. sweetburger
    • one year ago
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    I cant tell the difference between A and B they look worded the same

  32. vera_ewing
    • one year ago
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    It was A.

  33. sweetburger
    • one year ago
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    That question was strange

  34. vera_ewing
    • one year ago
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  35. vera_ewing
    • one year ago
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    How about this one?

  36. sweetburger
    • one year ago
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    I think this is a zero order reaction due to the negative slope/straight line

  37. sweetburger
    • one year ago
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    Im not completely sure im a bit rusty with kinematics

  38. vera_ewing
    • one year ago
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  39. vera_ewing
    • one year ago
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    Which of these is correct?

  40. rvc
    • one year ago
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    i think it is first order reaction

  41. sweetburger
    • one year ago
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    Yea, it is I was wrong.

  42. sweetburger
    • one year ago
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    I didnt that the it was the ln[X] and not just [X] on the y axis

  43. sweetburger
    • one year ago
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    didnt notice*

  44. rvc
    • one year ago
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    yep

  45. vera_ewing
    • one year ago
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    So which one is the answer?

  46. rvc
    • one year ago
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    need to solve this

  47. rvc
    • one year ago
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    @TrojanPoem what do you think?

  48. rvc
    • one year ago
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    well guys i need to go

  49. rvc
    • one year ago
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    Sorry to go like this in between :( all the best vera

  50. TrojanPoem
    • one year ago
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    cya @rvc

  51. sweetburger
    • one year ago
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    I think i have an Idea on how to solve this

  52. sweetburger
    • one year ago
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    We first say that y=-.0952x+.25 and then we plug in 10 for x and find the cooresponding y value

  53. vera_ewing
    • one year ago
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    What did you get when you did that?

  54. sweetburger
    • one year ago
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    nope I dont think that is going to work nm

  55. abb0t
    • one year ago
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    Water is not counted in the rate constant equation.

  56. sweetburger
    • one year ago
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    What?

  57. abb0t
    • one year ago
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    Understand that a catalyst does \(\sf \color{red}{NOT}\) lower or decrease the energy of activation of a reaction, rather it finds a \(new\) pathway so that it expends less energy, and proceeds faster and consequently, lowering the requirement of the transition energy for that same reaction to proceed.

  58. abb0t
    • one year ago
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    For instance, you have: \(\sf \color{blue}{A \rightarrow D} \) without a catalyst, it might require 100 kj/mol but with the catalyst, it finds a new pathway that may only require 15 kj/mol.

  59. vera_ewing
    • one year ago
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    Ohh I got C for my answer at first, but now I think I did it wrong /:

  60. vera_ewing
    • one year ago
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    @abb0t What did you get?

  61. sweetburger
    • one year ago
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    @abb0t Could you help out with this question http://assets.openstudy.com/updates/attachments/558ea82ee4b0058b2bb701f0-vera_ewing-1435414099554-screenshot20150627at10.08.08am.png

  62. abb0t
    • one year ago
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    Well, what does the graph tell you? Is it zeroth order, first order, second order...

  63. vera_ewing
    • one year ago
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    First order right?

  64. abb0t
    • one year ago
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    Yep. If the graph is linear and has a negative slope, it's first order. Remember that.

  65. abb0t
    • one year ago
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    So, what equation should you use? Since it's linear, it will be something along the lines of: y = mx+ b which is the equation of a line.

  66. vera_ewing
    • one year ago
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    I'm not really sure. I just used the one that @sweetburger gave me...

  67. abb0t
    • one year ago
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    [A] = [A]\(_0\) - kt

  68. sweetburger
    • one year ago
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    I cant tell if the equation should be ln[A]=-kt+ln[A0] or something like [A]=[A0]e^(-kt)

  69. abb0t
    • one year ago
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    oops, forgot the ln. this is first order.

  70. sweetburger
    • one year ago
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    that subscript function is useful i need to find that

  71. abb0t
    • one year ago
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    They are both the same. Think of the natural log function, but you can use either.

  72. abb0t
    • one year ago
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    I gtg. But i've clarified a lot. I'm sure sweetburger can take it from here.

  73. sweetburger
    • one year ago
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    Oh boy

  74. vera_ewing
    • one year ago
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    Thanks @abb0t

  75. sweetburger
    • one year ago
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    Oh i think I may have figured it out

  76. vera_ewing
    • one year ago
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    Okay...

  77. sweetburger
    • one year ago
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    I literally just need to find another concentration value at a specific time and this would be so easy.

  78. vera_ewing
    • one year ago
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    Anything?

  79. sweetburger
    • one year ago
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    Nope, closest to any of the answer i have gotten is like -.018s^-1. I'm sorry, but this question is just escaping me.

  80. vera_ewing
    • one year ago
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    Okay, if I were to guess, which should I choose?

  81. sweetburger
    • one year ago
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    I dont know how much time you have to complete this, but I could try this one more way

  82. sweetburger
    • one year ago
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    No that didnt work either

  83. sweetburger
    • one year ago
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    Idk maybe try A, but I really dont know

  84. sweetburger
    • one year ago
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    OH ITS B

  85. sweetburger
    • one year ago
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    I just needed to do this i think .25/-.0952 = -2.62sec^-1

  86. vera_ewing
    • one year ago
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    It was A!!

  87. sweetburger
    • one year ago
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    Very sorry :/

  88. abb0t
    • one year ago
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    C'mon guys, that's why you have the graph! for a first-order reaction, it gives you what you have there, a straight line with the slope of the line equal to \(\sf \color{red}{-k}\) If you know that your slope is -k, plug it in, and then substitute your value into the proper equation to solve for the rate of reaction is the change in concentration of the reactants or the change in concentration of the products per unit time. Which means, you're looking for \(\sf \color{red}{\frac{M}{s}}\)

  89. sweetburger
    • one year ago
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    Im clueless @abb0t

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