Objective To determine of arouse ups from three exothermic fightions and look into the Hess Law Summary Based on this audition, we have education three related exothermic receptions involving atomic number 11 hydroxide. The first reception (Part A), steadfast atomic number 11 hydroxide give disjoint into body of water. The heat divulge by this answer (?H1) and this called as the heat of consequence of red-blooded NaOH. From the test we have managed to intractable that ?H1 = -41.84 KJ/mol In the plunk for reaction (Part B), an aqueous solution of NaOH is allowed to react with an aqueous solution of HCl. This is a neutralization reaction in the midst of a strong acid and strong base. Therefore the heat of reaction (?H2) is called as the heat of neutralization of HCl and NaOH solutions. The ?H2 calculated from this experiment is -6.6944KJ/mol. This is because the atomic number 1 changes when 1 jetty of H+ ions from an acid (HCl) rea cts with one mole of OH- from an alkali (NaOH) to form one mole of water molecules low the stated conditions of the experiment. In the final reaction of the experiment (Part C), strong NaOH willing react with an aqueous solution of HCl. This reaction is too the combination of the first and two reactions. The solid NaOH will dissociate into its ions as it dissolves in the acid solution which is thence alter by the acid solution.
Thus the heat of the reaction (?H3) is utter be equal to (?H1+?H2). It is called the heat of solution of solid NaOH. From our unhurriedness it known that ?H3 is -58.6 kJ/mol. When i onic solid dissolves in water, heat was libr! ated. chemical reaction 1: Dissolving solid atomic number 11 hydroxide in water. NaOH(s) ---> Na+(aq) + OH-(aq) + heat Reaction 2: Reaction of sodium hydroxide solution with dilute hydrochloric acid. Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ---> Na+(aq) + Cl-(aq) + H2O Reaction 3: Reaction of solid sodium hydroxide with dilute hydrochloric acid solution....If you want to get a blanket(a) essay, order it on our website: OrderCustomPaper.com
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