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3. Water was added to remove any underacted copper(II) sulfate, iron, and iron(II) sulfate. The solution was washed to isolate the copper. 4. The water added didn't matter if the measurement w for water was exact because it wasn't included in the reaction. Wa...

11/7/24 Partner: Jeff Somvorachith 7. Standardization of a NaOH Solution Purpose: To determine the molarity of an NaOH solution by reacting a kniwn volume of each reagent, NaOH and KHP, and using the known molarity of the KHP solution to determine the mola...

7. Calculate the percent deviation between the trials. % deviation=(highest molarity - lowest molarity)/average molrity *100 8. If the % deviation is greater than 5%, repeat the titration process. Data: Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 ...

Analysis:   KHC8H4O4+NaOH->NaKC8H4O4+H2O Trial 1: 10.00mL-0.20mL=9.80mL NaOH used 2.00g/(39.10+1.008+8*12.01+4*1.008+4*10.00)*(1/1)*(9.80/1000)=0.999~1.0M Trial 2: 19.70mL-10.00mL=9.70mL NaOH used 2.00g/(39.10+1.008+8*12.01+4*1.008+4*10.00)*(1/1)*(9.70/1...

Our % deviation was 0.990%, meaning our answers between each trial was pretty close to the average. Since we got one really good titration, the % deviation means we were really close to being exact and getting a light pink color, within a few hundreths of a mi...

11/12/24 8. Vinegar Titration Lab Purpose: Determine the concentration of acetic acid in vinegar. Procedure: Using a clean funnel, fill a buret with NaOH solution, which was determined to have a molarity of 1.00M. Record the initial volume of NaOH i...

Data: Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Trial 6 Amount of vinegar 10mL 10mL 10mL 10mL 10mL 10mL Initial NaOH 1mL 9.9mL 18.0mL 26.2mL 34.4mL 36.65mL Final NaOH 9.9mL 18.0mL 26.2mL 34.4mL 42.65mL 44.6mL ...

Trial 4: 34.4mL-26.2mL=8.2mL (8.2/1000)/(10mL/1000)=0.82M Trial 5: 42.65mL-34.4mL=8.25M (8.25mL/1000)/(10mL/1000)=0.825M Trial 6: 44.6mL-36.65mL=7.95mL (7.95mL/1000)/(10mL/1000)=0.795M Average Molarity: (0.89M+0.81M+0.82M+0.82M+0.825M+0.795M)6=0.842M...

Conclusion: The purpose of the lab was to determine the concentration of acetic acid in vinegar. I achieved this by combining NaOH with vinegar until the indicator turned pink. The average concentration across the trials was determined to be 0.842M. The perce...

was left over after and what was left over was the limiting reactant. The limiting reactant turned out to be the iron, both through math and observations, as copper was all that was left after. All of the iron would have reacted to form iron(II) sulfate, leavi...

12/3/24 9. Molar Volume of a Gas Purpose: To practice calculating the molar volume of a gas by mixing magnesium metal and hydrochloric acid to form hydrogen gas, then using the mass of magnesium used and volume of hydrogen collected to calculate the volume ...

 Clamp the eudiometer into position on the ring stand. The acid flows down the tube(less dense) and reacts with the magnesium. The acid is more dilute. 8. When the magnesium has disappeared entirely and the reaction has stopped, cover the stopper with your fi...

Started slowing down as magnesium disappeared. Some HCl went out the stopper. Some bubbles stuck to the side of the tube and to the string. Magnesium eventually disappeared. Calculations: Number of moles of magnesium0.04/24.30 = 0.00165 moles of hydrog...

The formula for the Mg HCl reaction is Mg+2HCl->H2+MgCl2. Since the moles of Mg equal the moles of H2 in the equation, the moles are the same. We needed to adjust for water-vapor pressure because the water will evaporate to some degree during the experiment, m...

Partners: Jeff Somvorachith, Connor Engels 10. Additivity of Heats of Reaction: Hess's Law Purpose: To practice applying Hess's Law using a coffee cup calorimeter and confirming that the heat of one reaction should be equal to the sum of the heats for the o...

8. Rinse and dry the temperature probe, cup, and stirring rod. Dispose of the solution as directed. 9. Repeat steps 4-8, using 100.0mL of 0.5M hydrochloric acid instead of water. Use the same amount of NaOH. 10. Repeat steps 4-8, instead using 50.0mL of 1.0M...

ΔHR1= -1.875 kJ ΔHR2= -4.678 kJ ΔHR3= -2.865 kJ molR1=2.00/(22.99+16.00+1.008)= 0.05 mol molR2=2.00/(22.99+16.00+1.008)= 0.05 mol molR3=1.0*50.1000= 0.05 mol ΔH/molR1=-1.875/0.05= -37.5 kJ/mol ΔH/molR2=-4.678/0.05= -93.6 kJ/mol ΔH/molR3=-2.865/0.05...

We ended up with a 1.28% error, meaning some heat was probably lost, though it was close. This error could stem from inaccurate measuring of the NaOH or HCl, but it is more likely from losing heat to the Styrofoam cup or the atmosphere. If I were to do this la...