Question

Hello guys. I have to do these calculations but I can seemtodo it and I am stuck. So can someone please help me calculatethesevalues?

We know that the specific heat of water (same for solutionforthis expt) is 4.18 J/g C

and densities of all solutions are 1.00 g/mL

Temp of HCl: 22.2 C

Temp of NaOH: 22.0 C

so initial Temp is average of those two which is 22.1 C

final temperature of the resulting salt solution: 22.6

so;

q=mcΔT

m= 100g

c=4.18 J/g C

ΔT= 22.6 - 22.1 = 0.5 C

so q = (100)(4.18)(0.5)

q= 209 J

so q _rxn = -q _soln

so =209 J

ok now how do you calculate moles of NaOH or HCLneutralized (molesof rxn)

once we get moles of rxn we can calculate theenthalpyof neutralization of rxn

which is calculated by q_rxn/(molesofrxn)

so how do you calculate:

enthalpy change (using expt data) for OH^-+H⁺ -> H₂O _(l)

standard enthalpy for formation H2O_(l)(ΔH)

calculated enthalpy for formation OH-_(aq)(ΔH)