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.
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.
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.
To calculate the enthalpy change (ΔH), you can use the equation:
ΔH = mcΔT
Where:
m is the mass of the water in grams (25.4 g in this case)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (Tf - Ti)
So first, let's calculate ΔT:
ΔT = Tf - Ti
ΔT = 25.0°C - 14.5°C
ΔT = 10.5°C
Now we can calculate ΔH:
ΔH = (25.4 g) * (4.18 J/g°C) * (10.5°C)
ΔH ≈ - 1116.09 J
Since the given answer choices are in kJ/mol, we need to convert the result to kilojoules and adjust for the number of moles.
To convert joules to kilojoules, divide the result by 1000:
ΔH = - 1116.09 J / 1000
ΔH ≈ - 1.11609 kJ
To find the value per mole, divide the ΔH by the number of moles (0.033 moles in this case):
ΔH(per mole) = - 1.11609 kJ / 0.033
ΔH(per mole) ≈ - 33.7 kJ/mol
Therefore, the correct answer should be approximately -33.7 kJ/mol. It seems like your initial calculation was close, but there might have been a slight rounding error.