Experiment quiz 25

Jacob carelessly added only 40.0ml (instead of the recommended 50.0ml) of 1.1 M HCl to the 50.0ml of1.0 M NaOH. Explain the consequence of the error.

40ml HCl reacting with 50ml NaOH The mole ratio is 1:1 in reaction. Some of the NaOH would not react so (Change in H) would decrease.

The dissolution of ammonium nitrate, NH4NO3 in water is an endothermic process. Since the calorimeter is not a perfect insulator, will the enthalpy of solution, change in H for ammonium nitrate be reported as too high or too low if this heat change is ignored?

Dissolution of NH4NO3 is endothermic. Styrofoam cup is NOT a perfect insulator so some heat absorbed from surroundings. Therefore temp in cup does not drop so low as is should. (change in T) smaller than it should be, so g smaller than should be. OR heat energy lost to surroundings so temp max is lower than it should be. So (change in T) smaller than it should be.

In measuring the specific heat of a metal, Josh used the highest measured temperature for calculating the metal's specific heat rather then the extrapolated temperature. Will this decision result in a higher or lower specific heat value for the metal?

If used highest temp instead of extrapolated temp the (change in T) would be smaller for both. If divide by smaller value to get specific heat, the value would be larger than expected.

If some of the salt remains adhered to the weighing paper (and therefore is not transferred to the calorimeter), will the enthalpy of solution for the salt be reported too high or too low?

Loss of salt during transfer means less salt present in reaction so less energy.

The 200mm test tube also contained some water (besides the metal)that was subsequently added to thecalorimeter (in Part A.4). Considering a higher specific heat for water, will the temperature change in the calorimeter be higher, lower, or unaffected by this technique error?

Test tube had water at 100 C (plus metal) that was added to calorimeter. Raise the temperature of water (Tf). Also mass would not be correct. You would not account for mass due to water with the metal so q= m* c * (change in T)

The enthalpy of neutralization for all strong acid-strong base reactions should be the same within experimental error. Will that also be the case for all weak acid-strong base reactions?

The acid/Bases completely ionize for strong acid and strong base. So same amount of energy expended. Weak acid-strong base the (Change in H) would change depending on the acid.

The chemist used a thermometer that was miscalibrated by a plus 2 C over the entire thermometer scale. Will this factory error cause the reported energy of neutralization, change in H to be higher, lower, or unaffected?

You are using (change in H) so the error would carry through all temperatures and have no effect.

Heat is lost to the styrofoam calorimeter. Assuming a 6.22 C temperature change for the reaction of HCl (aq) with NaOH (aq), calculate the heat loss to the inner 2.35g Styrofoam cup. The specific heat of styrofoam is 1.34 J/g*C

q=m*C*(change in T) q=(2.35g)*(1.35J/gC)*(6.22C)=19.6 J