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anonymous

  • one year ago

I'm having trouble with Calorimetry.

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  1. anonymous
    • one year ago
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    Alright so my data was already a given: Data and Observations: Volume of H2O 205.0 mL H2O Mass of NaOH 2.535 g NaOH Initial temperature in calorimeter 24.2 °C Final temperature in calorimeter 27.8 °C

  2. anonymous
    • one year ago
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    I need help with a few of the questions below. My teacher had given me a few things I needed improvement on. Calculations: Show your work and write a short explanation with each calculation. Write out a balanced "equation" for the process you investigated in Part I, including phase symbols. NaOH(s) + H2O(l) --> NaOH(aq) Problem with this was: #1 Balanced reaction include phase and ions

  3. anonymous
    • one year ago
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    (this is just to fill in information) Calculate the number of moles of sodium hydroxide dissolved. Show your work. Moles NaOH = 2.535 grams NaOH * (1 mole/40.00 grams) = 0.06338 moles NaOH

  4. anonymous
    • one year ago
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    Calculate the amount of energy involved in this dissolving process. Show your work. q = m c (T2-T1) 200 mL H2O * 1 g/1 mL = 200 g H2O c = 4.184 J/gC T2 = 27.8 C T1 = 24.2 C Heat gained by H2O = 200 g * 4.18 J/gC * (27.8 - 24.2) = 3010 J I was told: #3 Find moles NaOH by coverting NaOH mL to Liters then use Molarity NaOH to convert to mol NaOH

  5. anonymous
    • one year ago
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    Determine the enthalpy change, per mole of sodium hydroxide dissolved. Show your work. 3010 J / 0.06338 mol NaOH = 4.75X10^4 J/mole X 1 kJ/1000 J = 47.5 kJ/mol NaOH I was told: #4 Divide enthalpy change from 2 by moles from 3 to calculate total enthalpy change.

  6. taramgrant0543664
    • one year ago
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    NaOH(s) + H2O(l) --> Na+ + OH- This formula would give you your ions and they would be aqueous

  7. taramgrant0543664
    • one year ago
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    You would have to balance it still though

  8. taramgrant0543664
    • one year ago
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    I'm not 100% sure on what the I was told #3 part is going for but I'm assuming that it's trying to say that after finding the moles of NaOH you can use that as the number of moles for water and convert it into mass from there but you have to keep in mind stoichimometric coefficients

  9. taramgrant0543664
    • one year ago
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    Don't know about that last part though

  10. anonymous
    • one year ago
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    Thank you for responding to this o.e I just kinda posted everything at first to introduce the harder questions to follow anyway. I was hoping to sort of hook someone into reading the easier parts.

  11. anonymous
    • one year ago
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    Part 2: Data and Observations: Volume of HCl solution 100.5 mL HCl Volume of NaOH solution 100.0 mL NaOH Initial temperature in calorimeter 25.2 °C Final temperature in calorimeter 28.2 °C

  12. anonymous
    • one year ago
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    Part 2 Calculations: #1 Write out a balanced equation for the reaction you investigated in Part II, including phase symbols. HCl + NaOH -> NaCl + H2O #2 Determine the enthalpy change of this reaction. The mass of the sodium hydroxide and the mass of the hydrochloric acid should be added together. q = m × c × Δt 200.5 g of solution 200.5 g * 4.18*(28.2-25.2)= 2514.27 J -2514.27/1000 = -2.514 kJ = 1.258 kj/mol #3 Determine the number of moles of NaOH. #4 Determine enthalpy per mole of NaOH. Show all of your work. I was told: #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.

  13. anonymous
    • one year ago
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    Ok I see, I can help you with understanding any of this or helping you calculate. It looks like you worked out everything previously correctly, I was quite thorough but got distracted part way through. After seeing these problems you were assigned, I guess my question is... What question do you have? Let's just start with one thing at a time because there's a lot here and it's easy to get lost in all this.

  14. anonymous
    • one year ago
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    I'm sorry that it's a lot, it's sectioned into parts and I'd just like for people to know what they're getting into with helping me. I guess what I need is a bit of proof calculating? I need the errors pointed out and sort of guided as well.

  15. anonymous
    • one year ago
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    It's fine don't apologize, I actually appreciate that you've put this much effort into trying to learn so I'm more than happy to help. So is what you've put currently here new calculations you've done, or are these the old calculations you were told were incorrectly done? I guess, can you go ahead and maybe copy one particular piece from up above and then just ask specifically about that one part what you want me to help with?

  16. anonymous
    • one year ago
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    They're the old calculations. As a whole assignment, I figured I get help with the easier small questions before I do anything (to get it this time around) to help with Part 3.

  17. anonymous
    • one year ago
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    alright so I guess I'll just use question 4 from part one. Determine the enthalpy change, per mole of sodium hydroxide dissolved. Show your work. 3010 J / 0.06338 moles NaOH = 4.75*10^4 J/mole * 1 kJ/1000 J = 47.5 what I was actually told was: "#3 (I guess she forget #3('s) for belonging to it) q surroundings = -q system Use -q for #4 Enthalpy Change." so I kinda need help with just seeing that itself. Once I've already done something, I usually just continue to see it that way.

  18. anonymous
    • one year ago
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    Ahhh ok I think I see what you mean. So enthalpy is the heat, and heat is energy. Since energy is conserved that means whatever heat you put into something had to come out of something. In this case, we heated up the water, we could tell because the temperature of the water rose. However the NaOH molecules originally had this energy in them, it was transferred out, and this was actually lost. So while what you calculated was q for the water, what you wanted was the q for the reaction itself which was -q since it lost heat in order to heat up the water. Does that answer that question for why the sign has to be negative for exothermic reactions? (reactions where heat is lost and put into the solution)

  19. anonymous
    • one year ago
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    wait, I basically missed this because of a negative sign? xD

  20. anonymous
    • one year ago
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    Yeah it sounds like it hah

  21. anonymous
    • one year ago
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    Hey, I still haven't submitted this assignment, could you help draw out these questions and help me before I have to submit it? I've been in and out of doing school work ._.

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