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- 4. A fat sample with combination of acids contain standard hydrochloric acid for blank and sample with 8mL and 5mL respectively. The normality of the standard hydrochloric acid is 0.93N and the weight of the sample is 3 grams. Calculate the saponification value.The scatter plot below shows the density of ethylene glycol solutions vs. the mass % of those solutions. Density (g/mL) 1.04+ 1.03- 1.02- 1.01 0.99- 0.98- 8 % 10 12 14 16 18 20 Mass % Q The equation for the line of best fit shown on the scatter plot is y= 0.001683 +0.9959. Use the equation of the trendline to determine the mass percent of an ethylene glycol solution with a density of 1.015 g/mL.Chemistry The amount of V was found to be 0.19% (+/-) 0.02% V (N=6). Calculate t-calc, assuming 95% confidence level.
- The standard addition method is used to analyze a sample of a river water for mercury. Solution A is made by pipetting 5.00 mL of undiluted sample in to a 10 mL volumetric flask and filling to the mark with D water. Solution B is made by pipetting 5.00 mL of undiluted sample and 3.00 mL of 15.0 ppb of Hg standard into same 10.0 mL volumetric flask and filling to the mark with DI. Solution A and B are analyzed using atomic absorption spectroscopy and give a percent transmittance values of 56 % and 33 % respectively (not blank corrected). A blank has a transmittance of 96%. Which of the following technique can be used to determine the amount of mercury in the undiluted sample of river water? O A. IR spectroscopy O B. Cold vapor atomic absorption OC. Hydride generation atomic absorption OD. Electrothermal atomic absorptionAnalysis of Percent Organic Matter of * a soil sample was conducted and the data are shown below. Calculate the Percent Organic Matter of the sample. (Write answer with 3 significant figures with no %unit) Deionized water blank Reagent blank Sample with reagent Absorbance 0.001 0.089 1.231 Standard Solution (%) Absorbance 10 20 40 80 100 0.291 0.489 0.889 1.692 2.075Standardization of Sodium Thiosulfate Solution Primary Standard used: Potassium Dichromate Formula mass of 1o standard: __________________ % Purity of 1o standard: 99.80% Trials 1 2 3 Weight of K2Cr2O7 (g)Weight of K2Cr2O7 (g) 0.0315 0.0331 0.0380 Final Volume Reading Na2S2O3 (ml) 27.50 27.50 34.90 Initial Volume Reading Na2S2O3 (ml) 1.10 0.00 1.00 Net Volume Na2S2O3 used (ml) Molarity of Na2S2O3
- Indicate which of the two filtration techniques, gravity ( G ) or vacuum ( V ), is suitable for each of the following operations. a) Hot filtration b) Removal of decolourising carbon c) Isolation of recrystallized solute from solution.Chemistry help solve part d to h. the standard deviation is 0.0043m and the mean is 0.1052m.Q/ Prepare 0.1 N and 0.1 M of 98% H,SO4 in 250 ml. M.wt of H,SO4 = 98 g/mol, and its Specific density (SP.gr.) =1.09 . tousityxio
- Q/ Prepare 0.1 N and 0.1 M of 98% H,SO, in 250 ml. M.wt of H,SO, = 98 g/mol, and its Specific density (SP.gr.) = 1.09 .- Chapte x 6 Questio x G "How m X O Chem C x E R.12 ML x A Frequen x E Advanc x * Google x E R.12 ezto.mheducation.com/ext/map/index.html? con=con&external browser%3D0&launchUrl=https%253A%252F%252Flms.mheducation.com%252 Saved apter 4 Assignment 13 3 attempts left Check my work Enter your answer in the provided box. How many grams of KOH are present in 30.0 mL of a 7.39 M KOH solution? ts g eBook Print ferencesThe standard addition method is used to analyze a sample of a river water for mercury. Solution A is made by pipetting 5.00 mL of undiluted sample in to a 10 mL volumetric flask and filling to the mark with DI water. Solution B is made by pipetting 5.00 mL of undiluted sample and 3.00 mL of 15.0 ppb of Hg standard into same 10.0 mL volumetric flask and filling to the mark with DI. Solution A and B are analyzed using atomic absorption spectroscopy and give a percent transmittance values of 56 % and 33 % respectively (not blank corrected). A blank has a transmittance of 96%. What is the corrected absorbance of both solution A and B? OA. Solution A: 0.123 Solution B: 0.463 O B. Solution A: 0.463 Solution B: 0.234 O C. Solution A: 0.123 Solution B: 0.234 O D. Solution A: 0.234 Solution B: 0.463
