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anonymous
 one year ago
In need of Calorimetry help as soon as possible.
anonymous
 one year ago
In need of Calorimetry help as soon as possible.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because it's late as it is and for a full understanding, it'll just provide the document.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0These are what need to be changed with a bit of an explanation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Part 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@welshfella @iambatman @Conqueror would you help out with this at all? c:

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Looks 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate are you available at the moment?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Luigi0210 would you mind stepping in?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2@alanate We'll go through your work part by part, in the same order as your work.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2To 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, thank you c:

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2For Part I, here are my comments: using \(\Delta Q = mc\Delta T\) we have \(\Delta Q = 4.184\times 205.0\times (27.824.2)=3.087~kJ\) 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).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't worry, I'm following along (the browser keeps me inactive on the page for some reason)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2So 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).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looks like I had that equation aside from phase symbols and the actual enthalpy c:

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2For 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).

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2I will leave it to you to complete the enthalpy equation similar to part I, and which forms conclusion (3).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for your help mathmate, I appreciate it ^^

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Are you ok with conclusion #1 that relates to Hess's law?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait, no no, I'll need help with that as well, sorry, I was focused on part I and II

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Can you tell me where does Hess's law come into play in this experiment?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hold on, I'm still calculating. As to my knowledge, Hess's law is a manifestation that enthalpy is a state function.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Pretty much just verifying enthalpy

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2exactly. 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, 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

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2mL can be converted to grams of the reactant through molarity.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2100 mL of 0.5M NaOH is 0.1L of 0.5M NaOH equals 0.1L*0.5mol/L = 0.05mol

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Note how the dimensions (units) cancel out the L to give mol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, okay, that makes sense, sorry, it's kind of loud and distracting here. Thank you again

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2no problem. If I am not around, it looks like @hwyl might be interested in helping too!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0She seems to have the same assignment c:

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2But you must have different results (mass, volume, temperatures, etc.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that'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.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2oh! I thought they were experimental results! It would have been nice to do the experiment first hand!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in practice they're the same thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know it would definitely be, being a virtual student, we have computer visuals at least

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you use the same calculation for calorimetry but perhaps a little bit more involved because you need to measure the containers and calculate dissipation

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2That'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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this 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

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Anyway, are you all set @alanate ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes indeed I am c: For like the 14th time, thank you so much xD
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