Technically all three are wrong; however, #1 is ok but you didn't finish the problem. You answer of 2618.6 J is correct but that needs to be changed to kJ. That makes it 2.6186 kJ/mol and that needs to be rounded to 3 significant figures. I would round that to 2.62 kJ/mol.
#2. I think I remember doing this for you in the last day or so. I think you either didn't substitute correctly OR you made an algebra error.
#3. mass x specific heat H2O dT. You omitted the specific heat H2O from your work.
If you will show your work on #2 someone will point out the error(s).
Not sure if these are right
1) What is the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol if 5.00 moles of HCl(aq) (acid) neutralized 5.00 moles of NaOH(aq) (base) and released 13093 J of heat?
13093/5= 2618.6 then convert to kj
2) Calculate the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information:
The temperature change equals 5.06°C,
50.0 mL of 1.00 M concentration of Acid
50.0 mL of 1.00 M concentration of Base
Heat capacity of the calorimeter is 6.50 J/°C.
The specific heat of water is 4.180 J/g°C
5*5*5.06*6.50= 822.25/4.180= 196.7
3) How much heat is gained (in Joules) by the water (qwater) when a chemical reaction takes place in 100.0 mL aqueous solution and has a temperature increase of 13.26 °C? (Remember cwater = 4.180 J/°Cg)
100.0*13.26= 1326/4.180= 317.22
3 answers
I did show my work for #2
5*5*5.06*6.50= 822.25/4.180= 196.7
5*5*5.06*6.50= 822.25/4.180= 196.7
Yes, and I asked where did the 5*5 come from? I showed you how to do it in the response above that.
1 answer