Thursday

April 24, 2014

April 24, 2014

Posted by **Mary** on Sunday, June 30, 2013 at 6:02pm.

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

- Chemistry (check) -
**DrBob222**, Sunday, June 30, 2013 at 7:32pmTechnically 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).

- Chemistry (check) -
**Mary**, Sunday, June 30, 2013 at 8:06pmI did show my work for #2

5*5*5.06*6.50= 822.25/4.180= 196.7

- Chemistry (check) -
**DrBob222**, Sunday, June 30, 2013 at 9:52pmYes, and I asked where did the 5*5 come from? I showed you how to do it in the response above that.

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