Experimental procedure, part b. three student chemists measured 50.0 ml of 1.00 m naoh in separate styrofoam coffee cup calorimeters (part b). brett added 50.0 ml of 1.10 m hcl to his solution of naoh; dale added 45.5 ml of 1.10 m hcl (equal moles) to his naoh solution. lyndsay added 50.0 ml of 1.00 m hcl to her naoh solution. each student recorded the temperature change and calculated the enthalpy of neutralization. identify the student who observes a temperature change that will be different from that observed by the other two chemists. explain why and how (higher or lower) the temperature will be different.

Brett is the student that observes a different temperature change. Temperature change will be higher

Explanation:

The reaction of HCl with NaOH is:

HCl + NaOH → H₂O + NaCl + ΔH

As ΔH is in the products, the reaction produce heat. You can determine ΔH measuring change in temperature of coffee cup calorimeters. The higher moles of reaction, the higher production of heat.

The moles of NaOH that three students measured is:

0.0500L×(1.00mol / 1.00L) = 0.0500 moles

Brett measured:

0.0500L×(1.10mol/L) = 0.0550 moles of HCl

Dale measured:

0.0455L×(1.10mol/L) = 0.0500 moles of HCl

Lyndsay measured:

0.0500L×(1.00mol / 1.00L) = 0.0500 moles of HCl

Brett is the student that observes a different temperature change

That is because he's adding excess of moles of HCl that will reacts with water producing additional heat, that means he will find a higher change in temperature

I hope it helps!

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole