Swikriti Dasgupta HL IB Chemistry Determination of the Relationship Between Concentration and Rxn in Voltaic Cell: Research Question: How does the concentration of CuSO4 affect the voltaic cell as measured by voltmeter? Variables: Table 1: Independent Variable for Determining the Relationship Between Concentration and Rxn in Voltaic Cell: Independent Variable How it will be changed CuSO4 CuSO4 solution was diluted to 50% using distilled water. Then the diluted CuSO4 was further diluted to prepare 20%, 40%, 60%, 80%, and 100% dilution. Table 2: Dependent Variable for Determining the Relationship Between Concentration and Rxn in Voltaic Cell: Dependent Variable How it will be measured and why Voltage Using a volt meter Table 3: …show more content…
The concentrations are listed on table 4 below. Table 4: Concentration Ratios Concentration of CuSO4 (Moles) (±1.5) Ratio of CuSO4: distilled water (±1) 1.0000 20 ml of the previous concentration solution: 20 ml 0.5000 0.2500 0.1250 0.0625 Sample Calculations: Uncertainty for the concentration =Absolute uncertainty of water + absolute uncertainty of CuSO4 = 1+0.5 =1.5 Raw Data: Trials Trial 1 (±0.01) Trial 2 (±0.01) Trial 3 (±0.01) 1.000 M 1.11 1.12 1.12 0.500 M 1.10 1.10 1.11 0.250 M 1.05 1.06 1.08 0.125 M 1.02 1.03 1.01 0.625 M 0.98 0.98 0.99 Qualitative Data: Everyone in the room was using the same solutions so we ran out at the end so I did not have 40 ml to conduct my last experiment with the correct amount The wires needed to be replaced during my 3rd
Marwah Alabbad Post lab 10/21/15 Question 1: 1. Experiment 1: Number of trails NaOH concentration (M) Volume of HCl solution (mL) Initial volume of NaOH(mL) final volume of NaOH(mL) The volume of NaOH to titrate HCl (mL) Concentration of HCl (M) 1st 0.1023 25.0 10.05 36.12 26.07 0.085 2nd 0.1023 25.0 5.74 31.40 25.66 0.105 3rd 0.1023 25.0 9.84 35.52 25.68 0.105 First trail calculation: 0.02607L× (0.1023mole NaOH/1L)×(1 mol of HCL/1 mol of NaOH)×(1/0.025)= 0.085M of HCl
In the first part of the experiment, Part A, the standard solutions were prepared. As a whole, the experiment was conducted by four people, however, for Part A, the group was split in two to prepare the two different solutions. Calibrations curves were created for the standard solutions of both Red 40 and Blue 1. Each solution was treated with a serial 2-fold dilution to gain different concentrations of each solution.
Step 2: Create a 95% Confidence Interval for the ounces in the bottles. Answer: x ̅=14.87 ,s=0.5503 , n=30 , α=0.05 The level of confidence is at 95%. Use the following formula to determine the confidence interval: (x ̅-t_(α/2) (s/√n),x ̅+t_(α/2) (s/√n))
Calculation: Initial Mass(g)-Final Mass (g)=Change in Mass (g) Trial 1 74.5-62.0=12.5(g) Trial 2 272.7-271.5=1.2(g) Percent Error: 272.7-271.5 x 100 272.7 =0.440% Percent Change: 74.5-62.0 x 100 74.5 (Trial 1) =16.778% 272.7-271.5 x 100 272.7 (Trial 2) =0.440%
The quantitative solubility of the unknown compound was determined to be 29/100ml. The known solubility of sodium sulfate is 28.11g/100mL water. Using the found solubility to compare to the known solubility of sodium sulfate. This solution created in the solubility test, the conductivity of the unknown compound was tested using an Ohmmeter to measure the resistance of the solution. Resistance is the measure of a substances ability to conduct
RESTING MEMBRANE POTENTIAL When the neuron is not sending a signal at rest the membrane potential called as resting membrane potential. In this stage, permeability of K+ much greater than Na+ When a neuron is at rest, the inside of the neuron is negative relative to the outside. Although the concentrations of the different ions endeavor to balance out on both sides of the membrane, they cannot because the cell membrane sanctions only some ions to pass through channels (ion channels). At rest, potassium ions (K+) can cross through the membrane facilely. Additionally at rest, chloride ions (Cl-) and sodium ions (Na+) have a more arduous time crossing.
Hypothesis: Increasing substrate concentration will increase the initial reaction rate until it stops increasing and flattens out. Independent Variable: Substrate concentration Dependent Variable: The substrate itself, 1.0% Hydrogen Peroxide How Dependent Variable will be Measured: Hydrogen Peroxide will be used in every experiment, just with different test tubes. The amount of Hydrogen Peroxide in the mixing table is the amount that will be added to each test 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) /
Based in our experiment, we observed that an increase in the extracellular concentration of K+ increases the membrane potential of the crayfish muscle fibers thereby depolarizing these fibers. This process occurs because the ratio of K+ extracellular to intracellular was manipulated by adding KCl to the solution surrounding the muscle fibers. By increase the extracellular concentration, the K+ ions rushed inside the cell instead of their usual rushing outside. The movement of K+ ions inside the cell made the muscle fibers gain a positive voltage since there will be a change in equilibrium potential for K+ into a more positive value thereby making the membrane more positive.
The actual data is the result on our experiment vs theoretical, which is based on the calculations above. I have also learned to pay more attention to draining out all of the product completely before continuing to test the experiment, as any small drop of contaminant can veer our results into a different
Due to the unaccountability of the inconsistency in droplet size, many of the numbers may be varied because in one trial a huge droplet may count as one, but in another trial, I may have counted a small droplet as one, which causes results to possibly be
We started the experiment with part 1 of the lab using 0.20 M to find a standard absorbance curve for [FeSCN2+], we found the max absorbance which we would use would be at 432.6nm. Once we had this absorbance, we used that data to create our absorbance vs. concentration graph which was 3635x + 0.0449, with an r2. After this we did part 2 of the lab, where we got 5 different test tubes to measure the absorbance with varying amounts of 0.002 M Fe(NO3)3 and 0.002 M SCN- and H2O, and we subtracted the absorbance of test tubes 1-4 from the absorbance of test tube 1 to get the net absorbance. After this we plugged in the net absorbance from part 2 to the equation from part 1, this result gave us the concentration at equilibrium, so that we could
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
Method Eight ml of six different substrate concentrations (2000, 1500, 1250, 1000,
The chemical equation for this experiment is hydrochloric acid + sodium thiosulphate + deionised water (ranging from 25ml to 0ml in 5ml intervals) sodium chloride + deionised water (ranging from 25ml to 0ml in 5ml intervals) + sulphur dioxide + sulphur. As a scientific equation, this would be written out as, NA2S2O3 + 2HCL + H2O (ranging from 25ml to 0ml in