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anonymous
 one year ago
Hess's law help?
anonymous
 one year ago
Hess's law help?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Determine the enthalpy change for NaOH (s) + HCl (aq) → NaCl (aq) + H2O (l) using Hess's Law.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's in the conclusion, question 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate I didn't want to bother you with Hess's law, I'm sorry ;;

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jagr2713 wanna take a go?

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.0Enthalpy of products  Enthalpy of reactants

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0If 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 \(NaOH_{(s)} + HCl_{(aq)}~~ >~~ NaCl_{(aq)}+H_2O_{(l)} + H\)

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nothing else would make sense in that place, that is what I'm most confused on

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0The 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 onestep equation. Do not forget to add the enthalpy changes (or kJ/mol)

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's it? Without any numbers?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how about how to get the balanced equation from part II? Like, the steps to getting it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For part I 1. It should be NaOH(s) +H2O(l)>>NaOH(aq)+H2O(l) dw:1441517160519:dw

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

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much for a visual
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