ipl-logo

Concentration Of Acid And Reaction Time Lab Report

1417 Words6 Pages

Concentration of acid and reaction time Aim In this experiment, I’m going to find out the relationship between concentration (mole) of acid and carbonate’s reaction. Hypothesis If the concentration of acid is stronger, then the film canister will pop (react) faster. Because the concentration of Sodium carbonate is always same in my experiment, so the factor that changes the volume of carbon dioxide is concentration of Hydrochloric acid. If the concentration gets bigger, the proportion of acid gets bigger, which means there are more particles (molecule) that will bond with other particles (molecule). For this experiment, the proportion to get a CO2 is 2HCl+Na2CO3 = CO2.., which is 2+ Na2CO3 : 1. So as the mole of hydrochloric acid is bigger, …show more content…

In beaker put 10ml of 2mole Hydrochloric acid, which is measured by beaker. In beaker from step 17, put 10 ml of distilled water, which is measured by cylinder. With the glass rod stir beaker in step 18. Repeat step 4 to 9. Repeat step 17 to 20 for 5 times. In beaker put 5ml of 2mole Hydrochloric acid, which is measured by beaker. In beaker from step 21, put 15ml of distilled water, which is measured by cylinder. With the glass rod stir beaker in step 22. Repeat stet 4 to 9. Repeat step 21 to 24 for 5 times. Diagram Results In different concentration of HCL, the time it took to react Concentration of HCl (Mol) Time it takes to react (pop) (s) Average (Time) (s) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 0.5 48.05 97.72 156.07 126.94 52.74 96.30 1.0 15.54 8.64 9.92 44.17 2.85 16.22 1.5 6.92 5.67 3.95 31.10 30.06 15.54 2.0 6.17 11.04 6.30 5.16 18.95 9.524 To calculate the average (Time) (s) (Sum of all trial in one mole)/5 = (48.05+97.72+156.07+126.94+52.74)/5 = 481.97/5 = 96.304 (4sigfig) = 96.30s I have calculated the average of the time it took to react with my hands …show more content…

When I was doing experiment, the factors that can change my result was controlled same. The factored that were controlled equally were volume of the acid, layers of paper, type of carbonate, and etc. However, the part that can be improved is the film canister. When I close the lid with paper between lid and body, the solidity that the lid is closed is different by the force that is applied to lid. Though I have closed until the surface of lid was even but as in different trials, some could be tightly closed or loosely closed. Also when I compared to other experiment, my result is similar with other results. My results are reliable. I have done 5 trials by each independent variable. By making the average of each independent variable, my result is more reliable. Not only that, the range of the independent variable is suitable. I’m trying to compare the reacting time of the in difference of Hydrochloric acid’s concentration. The range of the independent variable was 0.5M, 1.0M, 1.5M, and 2.0M. The different between the biggest concentration and smallest concentration is 1.5M, which is quite large. To know the relationship between times takes to react and concentration of acid, the difference 1.5M is appropriate to compare in different concentration. That much difference was able to notice the significant difference in results, which was 96.30 and

Show More
Related

  • Eastmamol/540 Week 4

    563 Words  | 3 Pages

    Prelab week 1 Calculations Preparation of 1.5μmol/L mixed low-level standard dilution 150μmol/L × V1=1.5μmol/L × 10ml V1=(1.5μmol/L×10ml)/(150μmol/L)=0.1ml Conversion of milliliters to microliters (0.1ml×1000)μL= 100μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=3μmol/L × 10ml V1=(3μmol/L×10ml)/(150μmol/L)=0.2ml Conversion of milliliters to microliters (0.2ml×1000)μL= 200μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=7.5μmol/L × 10ml V1=(7.5μmol/L×10ml)/(150μmol/L)=0.5ml Conversion of milliliters to microliters (0.5ml×1000)μL= 500μL Preparation of the blank samples The volumetric flask will be filled to the mark with 150μmole/L of stock solution to act as blank (reference). Additional two blanks will

    Read More

  • 6.03 Pre Lab

    993 Words  | 4 Pages

    Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029

    Read More

  • Nt1310 Unit 4 Coursework

    1433 Words  | 6 Pages

    2. Add 8cm³ of sodium carbonate to each tube using a measuring cylinder. 3. Measure out the strontium nitrate for each boiling tube and add it (boiling tube 1 contains 1cm³, test tube 2 contains 2cm³ and so on). 4.

    Read More

  • Unknown 6 Essay

    680 Words  | 3 Pages

    Unknown 6- Isopropyl Alcohol We found that unknown 6 was Isopropyl alcohol. Its chemical name is isopropanol and the chemical formula C₃H₈O but is typically called isopropyl alcohol. Isopropyl alcohol is today used as a primary ingredient in rubbing alcohol. Is smells very unpleasant and is used for disinfecting pads used by medical professionals for tasks such as sanitizing small instruments, wiping down surfaces, and cleaning a patient’s skin before an injection.

    Read More

  • Will The Bean Plants Be Affected By The Amount Of Water In Plants

    1002 Words  | 5 Pages

    For this experiment the purpose is to answer the question we asked ourselves. The question was, “Will the growth of the bean plant be affected by the amount of water used when the plants are watered?” In this experiment we will be measuring the growth of two different bean plants, the two bean plants will be watered with different amounts of water. The control group in our experiment will be watered with 15 mL of water while the experimental group will be watered with 23 mL of water. Other groups also performed this same experiment using different amounts of water for their experimental group.

    Read More

  • Mat 540 Final Paper

    801 Words  | 4 Pages

    Explorated Rinf=0.90 Calculated Rinf=0.90 10^(-4) 170 0 0 0 0 0 0 15 42.7 42.7 0.2512 -0.3 0.949711638 30 28.8 71.5 0.4206 -0.58 0 60 26.3 97.8 0.5753 -0.93 0 120 43.8 141.6 0.8329 -2.09 0 240 14.8 156.4 0.92 -3.45 0

    Read More

  • Tucson Water Research Paper

    705 Words  | 3 Pages

    Fill the well with 90ml dh20 to reach 100ml. move 10 ml of the second well to the third well. FIll the third with another 90ml dh20 to reach 100ml. Move 10 ml of the third well to the fourth well. Fill the fourth well with 90ml dh20 to reach 100ml.

    Read More

  • How Does The Amount Of Baking Soda Mixed With Vinegar Affect The Volume Of Gas Produced?

    1211 Words  | 5 Pages

    To determine the rate of reaction there are many method to be used for example, measuring the mass after the product has been added and measuring the difference in mass on the duration of a digital scale. Another method, which will be used in this experiment is using a gas syringe to measure the volume of the gas which has been produced. The cylinder inside, will be pushed out to show a quantitative presentation of the volume produced by the reaction. Hypothesis

    Read More

  • Unknown Compounds

    1027 Words  | 5 Pages

    Next, a 100 mL graduated cylinder was used to measure 60 mL of distilled water. The water was added to the compound and stirred with a glass-stirring rod until dissolved. Next, The flame test required the solution made during the solubility test. The experiment needed a metal wire that was dipped into the solution

    Read More

  • Plasma Membrane Lab Report

    712 Words  | 3 Pages

    Place sodium chloride with a concentration of 9.00 mM into left side of beaker Step 4. Place deionized water into right side of beaker Step 5. Run the experiment, once the timer hits zero record your data and flush all of the contents. Step 6. Repeat steps 3-5 with urea as your molecule.

    Read More

  • Turnip Peroxidase Lab Report

    839 Words  | 4 Pages

    TRIALS 0 SECS 20 SECS 40 SECS 60 SECS 80 SECS 100 SECS 120 SECS TRIAL 1 99.05 99.12 98.71 99.57 98.65 101.98 100.75 TRIAL 2 97.89 96.32 95.89 98.01 97.99 97.83 95.42 TRIAL 3 99.38 96.19 91.58 96.53 97.09 97.11 99.31 ENZYME CONCENTRATION: 2X ENZYME (ABSORBANCE AU) TRIALS 0 SECS 20 SECS 40 SECS 60 SECS 80 SECS 100 SECS 120 SECS TRIAL 1 0.055 0.132 0.198 0.249 0.289 0.321 0.347 TRIAL 2 0.073 0.145 0.202 0.246 0.281 0.311 0.337 TRIAL 3 0.063 0.212 0.321 0.399 0.457 0.502 0.536 ENZYME CONCENTRATION: 2X ENZYME (TRANSMITTANCE %T) TRIALS 0 SECS 20 SECS 40 SECS 60 SECS 80 SECS 100 SECS 120 SECS TRIAL 1 88.19 73.8 63.42 56.37 51.43 47.79 44.94 TRIAL 2 84.61 71.57 62.86 56.74 52.36 48.84 45.99 TRIAL 3 86.56 61.43 46.72 39.93 34.9 41.51 29.12 ENZYME CONCENTRATION:

    Read More

  • Lab: Titration Of The Unknown Acid

    628 Words  | 3 Pages

    Next, we obtained an Erlenmeyer flask for the titration. Rinse it with DI water. Next we are to dispense from the buret at least mL of the designated acid into the flask. Record the initial and final volume readings on the buret in laboratory notebook data table. Next we add 20mL of deionized water and anthocyanin indicator drop

    Read More

  • Oil Distillation Lab Report

    919 Words  | 4 Pages

    Again select the flask and select Distillation Head from the drop down menu. ➢ For the third time select the flask and choose Condenser from Distillation from the menu and for last time select the flask Distillation Take-off from the dropdown option. ➢ Select the 100 mL Graduated Cylinder from the Equipment option and put it underneath of distillation take-off.

    Read More

  • Acid Base Neutralization Lab Report

    1001 Words  | 5 Pages

    IV. Data and observations Mass of beaker (g) 174.01 Mass of beaker + NaOH pellets (g) 174.54 Mass of NaOH pellets 0.53 TRIAL 1 TRIAL 2 Mass of potassium acid phtalate (KHP) (g) 0.15 0.15 final buret reading (ml) 30.75

    Read More

  • Difference Between Acid And Base Titration

    1669 Words  | 7 Pages

    So the moles of bases and acids become same and as a result we can obtain equation which is given below to figure out unknown concentration of acid. C_acid*V_acid=C_base*V_base Finally, the obtained results from experiment compared with the theoretical values to sure from the punctuality of

    Read More

More about Concentration Of Acid And Reaction Time Lab Report

Open Document