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anonymous

  • one year ago

In need of Calorimetry help as soon as possible.

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  1. anonymous
    • one year ago
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    Because it's late as it is and for a full understanding, it'll just provide the document.

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  2. anonymous
    • one year ago
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    These are what need to be changed with a bit of an explanation

  3. anonymous
    • one year ago
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    Part 1 #1 Balanced reaction include phase and ions #3 q surroundings = -q system Use -q for #4 Enthalpy Change. Part 2 #1 Balanced reaction include phase and ions #3 Find moles NaOH by coverting NaOH mL to Liters then use Molarity NaOH to convert to mol NaOH The molarity of NaOH is given in the directions. #4 Divide enthalpy change from 2 by moles from 3 to calculate total enthalpy change. Percent error Recalculate percent error using corrected values in Parts 1 and 2. The set up is good; however due to prior calculation errors, the answers are incorrect. Please try this again after going back to correct the answers for Parts 1 and 2. Show the work applying Hess's Law to calculate the enthalpy change. Add #4 from part 1 to #4 of part 2. (the calculated enthalpy of each reaction) #5 Classify the reactions in the lab as endothermic or exothermic. Proceed to support this classification and attractive forces between ions and the solute or solvent.

  4. anonymous
    • one year ago
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    @welshfella @iambatman @Conqueror would you help out with this at all? c:

  5. mathmate
    • one year ago
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    Looks like you have a good understanding of the subject. Do check your calculations, and complete parts #4 and #5. If no one helps, I'll take another look when I return later in the day.

  6. anonymous
    • one year ago
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    I appreciate that, thank you mathmate c: right now my problem is that I don't actually understand it and need help getting there. This was pretty much information gathering with minor adjustments to what I do understand

  7. anonymous
    • one year ago
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    @mathmate are you available at the moment?

  8. anonymous
    • one year ago
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    @Luigi0210 would you mind stepping in?

  9. mathmate
    • one year ago
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    @alanate We'll go through your work part by part, in the same order as your work.

  10. mathmate
    • one year ago
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    To summarize, part I deals with determining the enthalpy change for the dissolution of NaOH in water, and part II deals with the enthalpy change for the neutralization of NaOH and HCl.

  11. anonymous
    • one year ago
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    Alright, thank you c:

  12. mathmate
    • one year ago
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    For Part I, here are my comments: using \(\Delta Q = mc\Delta T\) we have \(\Delta Q = 4.184\times 205.0\times (27.8-24.2)=3.087~kJ\) Note the volume is 205 (and not 200) mL. Divide 3.087 kJ by 0.06338 mol gives enthalpy change = -48.72 kJ/mol (since energy is dissipated outside of NaOH).

  13. anonymous
    • one year ago
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    Don't worry, I'm following along (the browser keeps me inactive on the page for some reason)

  14. mathmate
    • one year ago
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    So the resulting completed enthalpy equation would look like: \(NaOH_{(s)}->Na^+ + OH^-+48.7~kJ/mol\) The above equation should be part of conclusion (2).

  15. anonymous
    • one year ago
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    looks like I had that equation aside from phase symbols and the actual enthalpy c:

  16. mathmate
    • one year ago
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    For part II, we have very little numerical difference, namely due to 4.184 instead of 4.18. I get 2.517 kJ disspated heat for 0.5M*0.1L = 0.05 mol gives 2.517/0.05=50.33 kJ/mol of energy dissipated for the neutralization of aqueous NaOH and HCl. 0.05 mol is used because 100 mL of NaOH is the limiting reagent in the neutralization (assuming both reactants are exactly 0.5M).

  17. mathmate
    • one year ago
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    I will leave it to you to complete the enthalpy equation similar to part I, and which forms conclusion (3).

  18. anonymous
    • one year ago
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    Thank you for your help mathmate, I appreciate it ^-^

  19. mathmate
    • one year ago
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    Are you ok with conclusion #1 that relates to Hess's law?

  20. anonymous
    • one year ago
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    wait, no no, I'll need help with that as well, sorry, I was focused on part I and II

  21. mathmate
    • one year ago
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    Can you tell me where does Hess's law come into play in this experiment?

  22. anonymous
    • one year ago
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    hold on, I'm still calculating. As to my knowledge, Hess's law is a manifestation that enthalpy is a state function.

  23. anonymous
    • one year ago
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    Pretty much just verifying enthalpy

  24. mathmate
    • one year ago
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    exactly. You have performed two experiments, each of which is one stage of the following: \(NaOH{(s)}+HCl_{(aq)} -> NaCl_{(aq)}+H_2O_{(l)} + (?) kJ \) The amount of heat can be obtained from standard tables, which should compare with the sum of the energies of each of the two stages of experiments that you have performed.

  25. anonymous
    • one year ago
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    Sorry, I'm being distracted here. I need help with determining the moles of NaOH on question 3 of part II. I'm given mL, not grams

  26. mathmate
    • one year ago
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    mL can be converted to grams of the reactant through molarity.

  27. mathmate
    • one year ago
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    100 mL of 0.5M NaOH is 0.1L of 0.5M NaOH equals 0.1L*0.5mol/L = 0.05mol

  28. mathmate
    • one year ago
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    Note how the dimensions (units) cancel out the L to give mol.

  29. anonymous
    • one year ago
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    Ah, okay, that makes sense, sorry, it's kind of loud and distracting here. Thank you again

  30. mathmate
    • one year ago
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    no problem. If I am not around, it looks like @hwyl might be interested in helping too!

  31. anonymous
    • one year ago
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    She seems to have the same assignment c:

  32. mathmate
    • one year ago
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    But you must have different results (mass, volume, temperatures, etc.)

  33. anonymous
    • one year ago
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    that's possible but this is only hypothetical being that it's an assignment without any lab work, my masses and temperatures were a given, and not actually measured myself.

  34. mathmate
    • one year ago
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    oh! I thought they were experimental results! It would have been nice to do the experiment first hand!

  35. hwyl
    • one year ago
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    in practice they're the same thing

  36. anonymous
    • one year ago
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    I know it would definitely be, being a virtual student, we have computer visuals at least

  37. hwyl
    • one year ago
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    you use the same calculation for calorimetry but perhaps a little bit more involved because you need to measure the containers and calculate dissipation

  38. mathmate
    • one year ago
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    That's why I was surprised to see that the results in part I actually exceeded the expected results, and was looking for possible "negative leak" of energy.

  39. hwyl
    • one year ago
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    this is just to make sure that the concept is driven home the same is applied for everything you use in the process but just keep adding

  40. hwyl
    • one year ago
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    or subtracting

  41. mathmate
    • one year ago
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    Anyway, are you all set @alanate ?

  42. anonymous
    • one year ago
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    yes indeed I am c: For like the 14th time, thank you so much xD

  43. mathmate
    • one year ago
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    You're welcome! :)

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