Chemistry Lab

Hesss Law Lab Report Purpose Data and Observations subdivision I fire up of Solution of Solid atomic number 11 hydrated oxide tabulate 1 master temperature of wet (t1)21.0ºC Final temperature of termination (t2)26.0ºC Temperature alternate (t2-t1=?t)5.0ºC potbelly of 100 mL of water100.0g screw up evolved by reaction2092J Mass of NaOH(s)2.0g Moles of NaOH0.05001250313 jettys vigor per groin of NaOH41829.52 J/mole ? H1 (kJ/mole) NaOH41.83 kJ/mole department IIHeat of receipt between Hydrochloric Acid and Sodium Hydroxide Solution Table 2 professional temperature of HCl(aq)21.0ºC Original temperature of NaOH(aq)21.0ºC comely original temperature (t1)21.0ºC Final temperature of solution (t2)34.5ºC Temperature tilt (t2 - t1 = ?t)13.5ºC Total locoweed of solution (assume 1mL = 1 g)100.1g Heat evolved by reaction5648.64 J Molarity of NaOH solution2.0 M Volume of NaOH solution0.05 L Moles of NaOH0.1 moles naught per mole of NaOH5648 6.4 J/mole ?H2 (kJ/mole) NaOH56.49 kJ/mole Part IIIHeat of answer between Hydrochloric Acid and Solid Sodium Hydroxide Table 3 Original temperature of HCl(aq) (t1)21.0ºC Final temperature of solution (t2)34.0ºC Temperature change (t2 - t1 = ?t)13.0ºC Mass of HCl (assume 1 mL = 1g)101.46g Heat evolved by reaction5513.34 J Mass of NaOH(s)2.0g Moles of NaOH0.

05001250313 moles Energy per mole of NaOH110239.23 J/mole ?H3 (kJ/mole) NaOH110.24 kJ/mole Questions and Calculations Reaction 1: ?T=26?C-21?C ?T= 5?C Cw=4.184J/(g?C) qw=(100.0g)(4.184J/(g?C))(26?C-21?C) qw=2092J -qrxn=qsolution qrxn=-2092J nNaOH=2.00g×(1 mole)/39.99g=0.05001250313 mols ?H=qrxn/n=(-209 2 J)/(0.05001250313 mols)×1kJ/(1000 J)= ?H! =-41.83 kJ/mole Reaction 2: mass=48.9g+54.2g=103.1g ?T=34.5?C-21?C=13.5?C Cw=Csolution=4.184J/(g?C) qsolution=(103.1g)(4.184 J/(g?C))(34.5?C-21?C) qsolution=5823.5 J -qrxn=qsolution qrxn=-5823.5 J 50mL×1L/1000mL=0.05L c=n/v n=cv n=2.0mol/L×0.05L n=0.1 moles ?H=qrxn/n=(-5823.5 J)/(0.1...If you want to bemuse a full essay, order it on our website: OrderEssay.net

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