In this experiment, the amount of water lost in the 0.99 gram sample of hydrated salt was 0.35 grams, meaning that 35.4% of the salt’s mass was water. The unknown salt’s percent water is closest to that of Copper (II) Sulfate Pentahydrate, or CuSO4 ⋅ 5H2O. The percent error from the accepted percent water in CuSO4 ⋅ 5H2O is 1.67%, since the calculated value came out to be 0.6 less than the accepted value of 36.0%.This lab may have had some issues or sources of error, including the possibility of insufficient heating, meaning that some water may not have evaporated, that the scale was uncalibrated, or that the evaporating dish was still hot while being measured. This would have resulted in convection currents pushing up on the plate and making it seem lighter by lifting it up
Data Table: Experiment ml NaClO ml Solution B Temperature of Precipitate (degrees) 1 5 45 27.0 2 15 35 35.0 3 25 25 44.0 4 30 20 49.0 5 35 15 52.0 6 40 10 46.5 7 50 0 24.0 8 45 5 22.0 9 43 7 21.0 Graph: I eliminated the last two data points because it was making my graph weird.
Characteristic Property- Test 2- Density Materials: Triple Beam balance, distilled water, graduated cylinder, unknown 6 Procedure: first we found the mass of the empty graduated cylinder and then its mass with the now distilled unknown. After subtracting the mass of the graduated cylinder, we were able to find the volume. For every 1mL=1cm³ so there we had the volume found with the graduated cylinder. We divided the mass by the volume in order to get the density Data: We found that the density of our unknown was 0.76 g/cm3.
(0.01 moles of NaOH) x (1 mole Ca(OH)2/ 2 moles of NaOH) = 0.005 moles of Ca(OH)2 Tube 1: (0.0020 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.002 moles of Ca(OH)2 (0.002 moles of Ca(OH)2) x (74.08 grams/mole) = 0.1 grams = theoretical yield Tube 2: (0.0035 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.004 moles of Ca(OH)2 (0.004 moles of Ca(OH)2) x (74.08 grams/mole) = 0.3 grams= theoretical yield Tube 3 (0.0050 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.005 moles of Ca(OH)2 (0.005 moles of Ca(OH)2) x (74.08 grams/mole) = 0.4 grams =theoretical yield Tube
In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) /
Besides the inaccuracy in determining the freezing point and problems in the enthalpy of dissolution, the largest source of error was the temperature probe. Over the course of the multiple trials, the temperature probe would read somewhat inconsistent values. Due to this uncertainty, trials were performed on pure deionized water to see how far the probe read from zero degrees Celsius. Another source of error was in the calorimeter. Using two foam cups as a calorimeter is sufficient, although far from perfect.
To find chemical equilibrium, the following chemical equation is used in the experiment: Fe3+(aq) + SCN-(aq) FeSCN2+(aq). When iron (III) and thiocyanate react, thiocyanoiron (III) is produced. When the concentration of all ions at equilibrium are known, the equilibrium constant can be calculated by dividing the equilibrium concentration of the reactant by the equilibrium concentration of the products. In this experiment, four equilibrium systems containing different concentrations of three different ion types (Fe(NO3)3, KSCN-, and distilled water) are made and used to determine equilibrium concentrations. The equilibrium concentrations are used to calculate the concentration that all of the components of the chemical equation are at equilibrium. Using a colorimeter or spectrometer to determine the equilibrium concentration of FeSCN2+(aq) and
To graph population or disease, we needed to use exponents; in equation-form, the exponent was an X, but it could be substituted for any number, which would represent the year. You would also find the current population or number of cases and divide them by the amount the previous year (the starting number) and add that to one to find the rate, which would show you if it was growth or decay. Finally, you use the starting number as your constant or y-intercept. If you were trying to graph the decay of a population, the equation could be: y=150,000(1.5)x; if you were trying to graph decay, the equation could be: y=150,000(0.5)x. You can replace X with any number (number of years) to find the population in the future (positive number) or in the past (negative numbers).
Once the 24-hour period has passed then you will extract the shells from the vinegar beaker and record the mass of those shells and the shells that were not exposed to the vinegar. Record all data and observations on the chart labeling the vinegar shells “experimental” and the other two “control”. Then you will pour 100 ml of vinegar into a 500 ml beaker and 100 ml of salt water in the other 500 ml beaker. Set the timer to 30 minutes and drop one “control” and one “experimental” shell to each beaker at the same time. In your data table you will record the observations of what is happens to the shells while they are exposed to vinegar and the salt water every 5 minutes.
It was impossible to accurately measure the volume of liquid at any given moment, as the meniscus was moving side to side. Secondly, the distillation was ended while there was still liquid in to round bottom flask. The composition and volume of this liquid were unaccounted for in the calculated
Infer why the current that was created during this lab is called a convection current. Convection is the movement caused within a fluid when hotter, less dense water, moves upward, and colder, denser water, moves downward. I infer that the current that was created during this lab is called a convection current because the colder, denser water, moved beneath the hot water, causing the hot water to move upward. 4. How does this experiment demonstrate water density?
WHAT HAPPENED? The taller the starting ramp height was, the greater the distance the plastic container rolled up the ramp. When the height of the starting ramp was 0 cm (the control group), the average distance the container climbed up the ramp was 0 cm.
The graph shows the average volume of hydrogen that was produced from the 3 trials and the average volume of oxygen that was produced from the 3 trials across the voltage. I added the volumes of hydrogen in each trial and I divided them by 3 to get the average and I made the same thing for the volume of oxygen. The graph shows that the volume of hydrogen produced during the experiment is twice as much as the volume of oxygen. For example using the third data when I used 11 volts the average volume of hydrogen that was produced was 5.8 cm3 and the average volume of oxygen produced was 2.9 cm3