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anonymous

  • one year ago

Hess's law help?

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  1. anonymous
    • one year ago
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    Determine the enthalpy change for NaOH (s) + HCl (aq) → NaCl (aq) + H2O (l) using Hess's Law.

  2. anonymous
    • one year ago
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    it's in the conclusion, question 1

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  3. anonymous
    • one year ago
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    @Shalante

  4. anonymous
    • one year ago
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    @mathmate I didn't want to bother you with Hess's law, I'm sorry ;-;

  5. anonymous
    • one year ago
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    @jagr2713 wanna take a go?

  6. sweetburger
    • one year ago
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    Enthalpy of products - Enthalpy of reactants

  7. mathmate
    • one year ago
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    If A->B + H1 (part I) B+C-> D + H2 (part II) then A+C -> D + H1+H2 That's basically what Hess's law is about, applied to your situation. So use Hess's Law to find \(NaOH_{(s)} + HCl_{(aq)}~~ ->~~ NaCl_{(aq)}+H_2O_{(l)} + H\)

  8. anonymous
    • one year ago
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    so which are what? -1,006.6 kJ/moles NaOH and -48.72 kJ/moles NaOH as the products and 3.087 kJ / 0.06338 mol NaOH and 50.33 kJ / 0.05 mol NaOH as the reactants?

  9. anonymous
    • one year ago
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    nothing else would make sense in that place, that is what I'm most confused on

  10. mathmate
    • one year ago
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    The two equations required are the revised equations in conclusion (2) and conclusion (3). When you add them up, and cancel equivalent reactant and product, you will end up with the one-step equation. Do not forget to add the enthalpy changes (or kJ/mol)

  11. anonymous
    • one year ago
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    so NaOH(s)- -> Na+ + OH- + 48.7 kJ/mol + (I still haven't gotten the balanced chemical reaction and enthalpy change for Part II) = NaOH(s)+HCl(aq) −> NaCl(aq)+H2O(l)+H ?

  12. anonymous
    • one year ago
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    that's it? Without any numbers?

  13. mathmate
    • one year ago
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    Well, you will complete the balanced equation for part II, then add up the two, cancel equivalent reactant and product, and that would be the final equation. This procedure is based on Hess's law, which says that enthalpy is a state value, so no matter how you arrive at the product, the enthalpy of the product will be the same.

  14. anonymous
    • one year ago
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    how about how to get the balanced equation from part II? Like, the steps to getting it?

  15. anonymous
    • one year ago
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    For part I 1. It should be NaOH(s) +H2O(l)>>NaOH(aq)+H2O(l) |dw:1441517160519:dw|

  16. anonymous
    • one year ago
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    It is a bit messy and contains some errors ( I crossed it out and corrected it.) as I am gaming and doing this lol. (Lost focus) Supposed to be HCl not H2O, supposed to be NaCl not NaOH, supposed to be negative 1055.32kJmol not -1055.32kJ/mol. The phase forms were accidentally misplaced

  17. anonymous
    • one year ago
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    After all if you put NaOH in water, there will be dissolved NaOH and water. (NaOH would not suck all the water out). The equation is also balanced that way compared to yours.

  18. anonymous
    • one year ago
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    thank you so much for a visual

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