You place 40.6 ml of 0.659 m naoh in a coffee- cup calorimeter at 25.00°c and add 71.3 ml of 0.659 m hcl, also at 25.00°c. after stirring, the final temperature is 28.57°c. [assume the total volume is the sum of the individual volumes and that the final solution has the same density (1.00 g/ml) and specific heat capacity (4.184 j/gk)]. calculate the change in enthalpy (latex: \deltaδh) of the reaction in kj/mol of water formed. enter to 1 decimal place.

ΔH of reaction is 62,3 kJ/mol

Explanation:

For the reaction:

HCl + NaOH → H₂O + NaCl

The heat transfer in the reaction (q) =

q = Specific heat × mass × ΔT

The mass is 40,6 mL + 71,3 mL × (1g/mL) = 111,9 g

And ΔT is 28,57 - 25,00 = 3,57°C

q = 4,184 J/gK × 111,9 g × 3,57°C = 1671 J

The moles of NaOH are:

0,0406 L×0,659 M = 0,0268 moles

And the moles of HCl are:

0,0713 L ×0,659 M = 0,0470 moles

The moles of reaction are the moles of NaOH because are the limiting reactant. 0,0268 moles

ΔH of reaction is q/moles of reaction. Thus:

ΔH of reaction =  1671J/0,0268 moles = 62350 J/mol ≡ 62,3 kJ/mol

I hope it helps!

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

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