enthalpy change=23g*4.18J/gC*10.5C=1009J
entlalpychange/mole= -1009J/.033mol=30.8kJ/mol
which is well different from yours. It is negative, as heat was given off.
23.0mL of water
density of water is 1.00 g/mL 1g = 1ml
23g of water
Cs of water 4.18 J/(gC)
moles= 2.40g/72.0(g/mol)= .033
mass = 23g+2.4g = 25.4g
Ti= 14.5 C
Tf= 25.0 C
Calculate the enthalpy change, ΔH, for this reaction per mole of X.
I'm getting -33.4 kJ/mol
Pearson is saying that answer is wrong, but I've compared work with many other sites.
2 answers
Wow Pearson was messed up I was right the whole time. Thank you for replying, but your advice was missing some details. The mass of the unknown (like in my mole calculation) has to be included in the mass with the water to get the right amount of joules. You also have to run your calculator right to get the right answer on Pearson as they have a .1 kJ/mole tolerance.