Analyst Date 1% 102- 95- 90- 85- 80 75 70- 65- 60- 55- 50- 45 40 35- 30 28- 4000 Sample Name Administrator 06 3500 Administrator Thursday, March 2, 2023 1:44 PM 2862.16cm-1 2935.42cm-1 3000 2500 1703.53cm-1 cm-1 2000 Description Sample 006 Bv Administrator Date Thursday, March 02 2023 1449.28cm-1 1500 PerkinElmer Spectrum IR Version 10.7.2 Thursday, March 2, 2023 1:44 PM pmmm 749.83cm-1 1310.91cm-1 908.06cm-1 1221.50cm-1 1000 489.08cm 1118.37cm-1 500450 Quality Checks The Quality Checks do not report any warnings for the sample.

Analyst Date 1% 102- 95- 90- 85- 80 75 70- 65- 60- 55- 50- 45 40 35- 30 28- 4000 Sample Name Administrator 06 3500 Administrator Thursday, March 2, 2023 1:44 PM 2862.16cm-1 2935.42cm-1 3000 2500 1703.53cm-1 cm-1 2000 Description Sample 006 Bv Administrator Date Thursday, March 02 2023 1449.28cm-1 1500 PerkinElmer Spectrum IR Version 10.7.2 Thursday, March 2, 2023 1:44 PM pmmm 749.83cm-1 1310.91cm-1 908.06cm-1 1221.50cm-1 1000 489.08cm 1118.37cm-1 500450 Quality Checks The Quality Checks do not report any warnings for the sample.

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter2: Basic Statistical Analysis With Excel

Section: Chapter Questions

Problem 12P

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Please answer fast it’s very important and urgent I say very urgent so please answer super super fast please For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466
Initial temp: 19.06 C trial 1: 0.5 M concentration trial 2 : 1.0 M concentration 200 mL of HCl and 200 mL of NaOH are combined in an insulated container. HCl = 0.5 mol/ml x 200 x 10^-3 ml = 0.1 mol NaOH = 0.5 mol/ml x 200 x 10^-3 ml= 0.1 mol H+(aq) + OH-(aq) - - -> H2O (l) ∆H H++ = 0, ∆H OH−− = -230 kJ/mol, ∆H H2O = -286 kJ/mol ∆H reactants = ∆H H++ + ∆H OH−− = -230 + 0 = -230 kJ/mol ∆H reactants = -286 kJ/mol - ( -230 kJ/mol) = -56 kJ/mol For each trial, calculate how much energy is released during this reaction.
Please answer super fast and answer all questions and show calculations For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466
100 g of soil is leached with a strong solution of Calcium chloride such that all the exchange sites are occupied by Ca2+. The soil is subsequently leached again with a strong solution of magnesium chloride. It is determined that the resulting 100 mL leachate contains 5000 mg of Ca2+. What is the soil CEC (cmolc/kg)? The atomic wt. of Ca is 40 g/mol.please answer in word don't image upload thank you.
Sample IdentificationCodeConcentration of M%TA=2-log(%T)Q50004.00 x 10-417.90.75Q50013.20 x 10-425.00.6Q50022.40 x 10-435.70.46Q50031.60 x 10-450.20.3Q50048.000 x 10-570.80.15SampleIdentificationCode%TA=2-log(%T)AMQ0210150143.70.359518560.360.000192Q0210150244.10.355561410.360.00018Q0210150343.80.358525890.360.00017Q0210150444.10.355561410.360.00018Q0210150543.80.358525890.360.00017What was their percent error?43%Does Batch 021015 meet legal requirements?No, because it is not between 2.85 * 10(4) and 3.15 * 10(4)Well #DropsBluedye1234567891012345678910Drops 9Distilled water876543210Concne 0.26tration0.52Test Tube #0.781.041.3Solutions3Concentration (M)2.082.32.6Concentration (ppm)1:1 dilution11.82Starting Dilution21.562:1 dilution0A.Zero standard0Was your calibration curve as linear as you expected?B.Did you experience any â€œdriftâ€ of the resistance readings?C.What is the equation of your best-fit line?D.What commercial drink did you analyze?E.Assuming…
Here is a use objective for a chemical analysis to be performed at a drinking water purification plant: “Data and results collected quarterly shall be used to determine whether the concentrations of haloacetates in the treated water demonstrate compliance with the levels set by the Stage 1 Disinfection By-products Rule using Method 552.2” (a specification that sets precision, accuracy, and other requirements). Which one of the following questions best summarizes the meaning of the use objective?(i) Are haloacetate concentrations known within specified precision and accuracy?(ii) Are any haloacetates detectable in the water?(iii) Do any haloacetate concentrations exceed the regulatory limit?
Show all steps leading to the final answer po. Here’s a pdf file in accordance with the topic po: https://drive.google.com/file/d/1_FnDtXCrFKSol3RNWIG_9tNQ7IxgxD6t/view?usp=drivesdk
Qq.1. Subject :- Account
NH4+ {aq) + NO2(aq) -> N2(g) +2H2O{l} Data Initial [NH4+] Initial [NO2-] rate 1 0.0100 0.200 5.4 x10-7 2 0.0200 0.200 10.8x10-7 3 0.0400 0.200 21.5x10-7 4 0.200 0.0202 10.8x10-7 5 0.200 0.0404 21.6x10-7 6 0.200 0.0808 43.3x10-7 Find x,y,k
Was the gradual color change observed when the sodium thiosulfate (Na2S2O3) crystal was added to the aqueous solution of KI/12 in Station Cevidence of a chemical or physical change?
There's 1 drink (and you are asked to determine the glucose concentration in the drink in the units of g/100mL. (Why these units? Well, once you have the concentrations in g/100mL you will be able to compare your values with the nutritional values given on the drink bottles’ labels). The sample of the drink was diluted 1/100 (i.e. by a factor of 100). This was an essential step in the method because, without it, the machine used to analyse the glucose concentration (spectrophotometer) would have given an error as the concentration would have been too high for accurate detection. What this means for you is that the dilution factor will need to be taken into consideration in your calculations (remember the aim is to calculate the concentration in the original drink and not in the diluted drink). You measured the concentration of their diluted drink using the spectrophotometer and their results were provided to them in the units mM (millimolar). Glucose Concentration in mM of drink =…
There's 1 drink (and you are asked to determine the glucose concentration in the drink in the units of g/100mL. (Why these units? Well, once you have the concentrations in g/100mL you will be able to compare your values with the nutritional values given on the drink bottles’ labels). The sample of the drink was diluted 1/100 (i.e. by a factor of 100). This was an essential step in the method because, without it, the machine used to analyse the glucose concentration (spectrophotometer) would have given an error as the concentration would have been too high for accurate detection. What this means for you is that the dilution factor will need to be taken into consideration in your calculations (remember the aim is to calculate the concentration in the original drink and not in the diluted drink). You measured the concentration of their diluted drink using the spectrophotometer and their results were provided to them in the units mM (millimolar). Glucose Concentration in mM of drink =…