Post-Lab Report 2

I had 500 mL solution of 0.300M HCI that I left on the bench. WhenI got back to lab the next day, the concentration of the solution was 0.5M. What volumne of my solution evaporated? 200 mL 300 mL 125 mL 500 mL 225 mL

Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 2.29   1.41   1.95   Molarity of NaOH (M) 0.100   0.100   0.100   Volume of vinegar sample (mL) 5.00   5.00   5.00   Final burette reading (mL) 50.37   49.39   49.84 Table 2. Titration data   Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 2.29   1.41   1.95   Molarity of NaOH (M) 0.100   0.100   0.100   Volume of vinegar sample (mL) 5.00   5.00   5.00   Final burette reading (mL) 50.37   49.39   49.84   Expected color at end point       Volume of NaOH used (mL) 48.08   47.98   47.89  Compute for the ff: a.   Average moles of acetic acid (mol)? b.  Average molarity of acetic acid (M)? c.  Average molarity of acetic acid (M)?

Determine the initial concentrations of the species below, after the solutions are mixed. Volume0.0350 MKMnO4(mL)   Volume0.0650 MH2C2O4(mL)   VolumeH2O(mL)   InitialConcentrationKMnO4(M)   InitialConcentrationH2C2O4(M)              (3 significant figures)    (3 significant figures) 1.00   1.00   18.00         2.00   2.00   16.00         3.00   3.00   14.00         4.00   4.00   12.00         5.00   5.00   10.00

*WRITE RESULTS AND DISCUSSIONS OF TITRATION- CONCENTRATION OF VINEGAR *WRITE CONCLUSION

I need help calculating the initial concentrations please.

what would the molarity acetic acid be?

Volume vinegar used (mL)= 5.00 mL % from labe= l5 % Molarity of NaOH = 0.0976 M Titration volume measurements   Trial 1 Trial 2 Trial 3 Inital volume NaOH (mL) 0.69   0.86   0.16   Final volume NaOH (mL) 43.47   43.36   43.03   Volume NaOH used (mL)             Average volume NaOH used (mL) Average volume NaOH used (L) Moles NaOH used (from avg volume) Moles CH3COOH Molarity CH3COOH (mol/L) Mass CH3COOH (g) % (m/v) CH3COOH in vinegar

1.10 M Acetic acid 0.900 M Sodium acetate [H30*1 (M) Solution (mL) |(mL) 1 8.00 2.00 8.71e-5M 2 7.00 3.00 5.13e-5M 6.00 4.00 3.24e-5M 4 5.00 5.00 2.17e-5 5 4.00 6.00 1.45e-5M 6. 3.00 7.00 9.31e-6M 7 2.00 8.00 1.00 9.00 00

CONCENTRATION DETERMINATION Buret contains NaOH solution *Sodium hydroxide solution: To 1 part reagent grade NaOH add 1 part distilled, carbon dioxide-free water by weight Erlenmeyer flask contains about 9ml of the acid and about 10 ml of distilled water made recently free from CO2 by boiling. DATA SHEET II Final Reading HCI Initial Reading HCI 9.4 Final Reading NaOH Initial Reading NaOH 10.65 10.55 1 ml HCI 1 mL NAOH mL NAOH mL NaOH mL HCL mL HCL Mean Values: 1 ml HCI = 1 mL NaOH = _mL NaOH mL HCL COMPUTATION:

Determination of unknown acid concentration     Trial 1 Trial 2 Trial 3 Unknown Acid number Volume of acid used (mL) 25 mL 25 mL 25 mL Initial buret reading (mL) 50 mL 50 mL 50 mL Final buret reading (mL) 14.8 mL 15 mL 14.5 mL Volume of NaOH used (mL) 50 mL 50 mL 50 mL Molarity of NaOH       Calculated molarity of HCl (M)               Show a sample calculation of HCl molarity   AVERAGE MOLARITY OF Unknown HCl Show a sample calculation of NaOH molarity

Please send me the question in 20 minutes it's very urgent plz

Please answer 1 to 5. Thank yoouuu!!!!

What is the molarity of an aqueous solution that contains 0.0720 g C2H6O2 per gram of solution . The density of the solution is 1.04 g/mL X STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 1.04 62.08 6.022 x 1023 0.0720 1.21 x 10-3 4650 1.12 1000 1.21 1 0.001 g C2H.O2 mol C2H6O2 g C2H&Oz/mol g solution/mL mL solution g C2H6O2/g solution M C2H6O2 L solution g solution

Concentration of NaOh(M)

Pre Lab Questions: (Each answer is to be written as a complete sentence) What is the reason for washing the precipitate with water in Step 9? Define precipitate. Define filtrate. In Step 2, what is the purpose for rinsing the stirring rod? read the Procedure to answer the questions  Using a balance, mass between 1.50 – 2.00 grams of sodium carbonate in a pre-massed 150mL beaker. Add 20 mL of distilled water and stir thoroughly to make sure all the crystals are dissolved. Rinse the stirring rod with a little distilled water after stirring. Using a balance, mass between 1.50 – 2.00 grams of calcium chloride dihydrate in a pre-massed 50 mL beaker. Repeat Step 2 for the solution in the 50 mL beaker. Pour the calcium chloride solution into the 150mL beaker containing the sodium carbonate solution and stir. Mass a piece of filter paper. Fold the filter paper and place it into the funnel. Wet it with a little distilled water to ensure that it is stuck to the sides of the funnel. Slowly…

SAMPLE: CANE VINEGAR % acidity in label: 4.5% %purity of KHP: 99.80% FORMULA WEIGHT of KHP: 204.22 g/mol STANDARDIZATION OF NAOH SOLUTION TRIAL 1 TRIAL 2 TRIAL 3 Weight of KHP, g Initial volume of NaOH, mL Final volume of NaOH, mL 0.1012 0.1004 0.09987 5.00 10.01 4.95 9.99 14.93 Molarity of NaOH Average Molarity of NaOH ANALYSIS OF ACETIC ACID IN A VINEGAR SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of vinegar, mL 1.00 1.00 1.00 Initial volume of NaOH, mL 14.98 22.90 30.87 Final volume of NaOH, mL 22.84 30.77 38.75 Molarity of acetic acid Average molarity of acetic acid ANALYSIS OF CARBONIC ACID IN A SODA SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of soda, mL 20.0 20.0 20.0 Initial volume of NaOH, mL 20.20 22.05 23.93 Final volume of NaOH, mL 22.03 23.89 25.8 Molarity of carbonic acid Average molarity of carbonic acid

Table 2. Titration data   Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 1.19   2.26   2.39   Molarity of NaOH (M) 0.100   0.100   0.100   Volume of vinegar sample (mL) 5.00   5.00   5.00   Final burette reading (mL) 48.55   49.43   49.99   Expected color at end point             Volume of NaOH used (mL)           Average Volume of NaOH used in liters   Average moles of NaOH used (mol)   Average moles of acetic acid (mol)   Average molarity of acetic acid (M)     Average mass of acetic acid (g)   Average mass of vinegar (g) (assume the density of vinegar is 1.00 g/mL)   Average mass % of acetic acid in vinegar   Known mass % of acetic acid in vinegar is 5.45%   Percent Error

Determination of known acid concentration     Trial 1 Trial 2 Trial 3 Volume of acid used (mL) 25 mL HCl 25 mL HCl 25 mL HCl Initial buret reading (mL) 50 mL 50 mL 50 mL Final buret reading (mL) 30 .7 mL 31.1 mL 29.3 mL Volume of NaOH used (mL) 50 mL 50 mL 50 mL Molarity of NaOH       Calculated Molarity of HCl   Show a sample calculation of HCl molarity   Average concentration of HCl from Trials 1,2, and 3   The “true” concentration of HCl is 0.10 M. The calculated average concentration of HCl is ______________M. The % error is = (???? − ?????)/????? ? 100

A 0.9182 g sample of CaBr2 is dissolved in enough water to give 500. mL of solution. What is the bromide ion concentration in this solution? 20 Multiple Cholce X 0206:02 9.19 x 10-3 M 230 x 10-2 M 272 x 10-3 M 1.84 x 10-2 M 1.25 x 10-1 M Mc Graw Hill Type here to search 近 C

Number 1 the second picture is the table

CHEMISTRY RATE WILL BE GIVEN! NO LONG EXPLANATION NEEDED.  ANSWER #3

Help me please

Identify mitigating measure/s What should be the appropriate response to minimize harm Karen is asked to mix a solution of acid and water. He then added water to the acid and the mixture started splashing.

how to find the concentration in mg/100ml ​​​​​​​btw the concentration of NBS is 0.0806 M

y/ h VG y² ki tions/Interface/acellus_engine.html?ClassID=329083093# News Share files over Blu... FL-CA - Student Inf... Aar $ All Copyright © 2003-2022 International Academy of Science. All Rights Reserved. % 8 15.0 mL of 4.0 M HCI is added to 85.0 mL of water. What is the final concentration of the acid? A .706 F6 * & F7 [?] mol HCI 8 3) Thomas Jefferso... F8 MHCI DELL prt sc apush chapter 7 ma... F10 Enter home F11 end F12 Just Short Answer q... insert y g 4 delete

Calculate the mass of acetic acid in 100ml of both samples in vinegar. given: NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate  0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume  Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3                            = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar                                           = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL                                           = 0.2627 M Concentration of NaOH = Volume of H2SO4 * Concentration of H2SO4Volume of NaOHVolume of H2SO4 * Concentration of H2SO4Volume of NaOH                                         = 10.0 mL…

A technician finds the molar mass to be 134.8 g/mol and the pK, to be 4.65 after processing a data set for a solid acid unknown. Based on the previous information and the list of acid unknowns in the Laboratory Manual, what is the best call the student can make for the identity of the solid acid unknown? picolinic acid B para-aminobenzoic acid glutamic acid monohydrate potassium hydrogen phosphate, monobasic OA. O B. O C. O D.

Part A) A 10.0 mL sample of 0.250 mol/L NiF2(aq) is mixed with 20.0 mL of 0.0900 mol/L NaOH(aq) and then diluted to a final volume of 100. mL.Calculate the concentration of Ni2+ ions in the 100 mL mixture before the reaction starts. Express your answer to three significant figures.ksp: 5.48e-16

How many milliliters of water have to be added to 122.5 mL of 0.50 M HCl to reduce the concentration to 0.10 M HCI? Vwater mL %3D eTextbook and Media Hint Save for Later Attempts: 3 of 15 used Submit Ans ост 14 tv MacBook Air 80 888 DD F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 @ 23 2$ 4 7 OC

3 Solid ammonium bromide is slowly added to 75.0 mL of a 0.283 M silver nitrate solution until the concentration of bromide ion is 0.0583 M. The percent of silver ion remaining in solution is %. Submit Answer $ 4 R F % 5 Use the References to access important values if needed for this question. Retry Entire Group 5 more group attempts remaining T G Cengage Learning | Cengage Technical Support ^ 6 MacBook Pro Y H & 7 U J * 00 8 1 ( 9 K O < ) O L P Previous Email Instructor + = Next Save and Exit

can you generate the unknown concentration using this information?