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Using the raw data, pls fill the table accordingly.
I will thumbs up if done properly.
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- The concentration of hydronium ions and hydroxide ions can vary widely in a solution, easily from 1 M to 1 x 10-14 M. This variation is too wide to fit on an analog readout. In order to make this into a smaller scale suitable for an analog readout, the log to the base 10 was used so that the read out is focused on powers of 10. Note that the log [1] = 0, and the log of 1 x 10-14 = -14. It was then decided that the numbers should be positive so the sign was changed. The way to measure concentrations of both H+ and OH- was thus defined as -log. This was then referred to as p. Thus, pH = -log[H+], pOH = -log[OH-], and pK = -log[Keq]. You need to figure out how to do this mathematical manipulation on your calculator. The acid-base reaction in pure water is H2O + H2O ⇔ H3O+1 + OH-1 and it was found that in pure water under standard conditions the [H3O+1] = [OH-1] = 1 x 10-7 M Neutral pH is defined as having a [H+] or [H3O+1] of 1 x 10-7 M. The pH of a neutral solution is thus pH = -log[1 x…(Solution BII) Prepare 25.00 mL of ~4.000x10-4 M KMnO4 solution via dilution of Solution BI (~0.00100 M KMnO4). Using the appropriately sized volumetric pipet, transfer the required amount of your solution from (BI) into a 25.00 mL volumetric flask. Fill it to the mark with distilled water, stopper it and invert several times to mix. Measure the absorbance of this solution at wavelength of 530 nm, the wavelength where this compound absorbs light most strongly. (e = 2.098 x 103 dm3/mol cm) Calculate the molarity of solution BII Volume of stock BI added via pipetteDefine Gas Chromatography and HPLC and state their differences in principle and uses. (maximum of 200 words)
- Can you please help with the table? ThanksAn analyst was measuring the performance of a HPLC column. The test solute was Toluene (5 µL). The column was a 4.6 mm i.d. column, 150 mm in length and packed with 5 µm C18 particles. The mobile phase was 40/60 water/methanol and the flow rate was 0.80 mL/min. The void time (to) was 1.86 minutes The retention time of the toluene sample was 14.50 min. The standard deviation of the peak was 7.0542 (s). 2a. The mobile phase velocity was? 0.0806 mm/s 0.1344 mm/s 1.344 mm/s 1.344 x 10³ mm/s (i) (ii) (iv) 2b. The number of theoretical plates was? (i) (ii) (iii) (iv) 2c. The height equivalent to a theoretical plate was? 1.041x10³ m 1.015x105 m 9.862x10-6 m 1.186x10¹² m (i) (ii) (iii) (iv) 4802 15210 24705 9848 2d. The reduced plate height was? (i) 2.03 (ii) 1.97 (iii) 7.55 (iv) 2.08 2e. The column was? (i) (ii) Extremely well packed Well packed Poorly packed (iii) (iv) Insufficient information to decide about the quality of the column& Spring 2023 CHEM 1411 14 General Chemistry I LAB Experiment 2 Postlab 57 minutes remaining Postlab Content Page 3 of 6 Question 2 An object with a mass of 47.2 kilograms and a volume of 47.6 liters. Find the density with units of g/cm³. Report your answer with three significant figures. Add your answer Questions Filter (5) 5 OF 5 QUESTIONS REMAINING K First Last > Continue Details & In PER Assessm 2/23/23 X ✓ • You after 76°F Time li 60 min Attem 2 atte Grading Sa
- What is compensation depth?I have tried in 4 times now and I only get one more try. I do not understand what I'm doing wrong please help me.If you have to analyze trance benzene and iron concentration in a sediment sample. You have a grinding the sample using ceramic tools in order to break down the sample to small particles. For which analytethe grinding process will cause more error? Selected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer. To b с d e Iron's analytical result will have larger positive error Benzene's analytical result will have larger positive error Iron's analytical result will have larger negative error Benzene's analytical result will have larger negative error The grinding process will not cause any error of analytical results for both analytes
- please answer the following with equations used and calculations.A student weighed out 0.150 g of protein powder and dissolved it in 100 mL of water (Solution 1). The student then diluted this solution by transferring 1 mL into a 25 mL flask and diluting with water (Solution 2). Finally, 1 mL of that solution was transferred to a test tube and combined with 4 mL Bradford reagent. The absorbance of the solution in the test tube was 0.187. Assuming that the best fit linear line of the standard curve was y = 0.04144 x + 0.01521 (μ g mL), calculate the percent protein by mass in the original protein powder.2