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Appl Of Ms Excel In Analytical Chemistry

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter14: Chromatography

Section: Chapter Questions

Problem 9P

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O CHEMICAL REACTIONS
Limiting reactants
Ta
Gaseous methane (CH) will react with gaseous oxygen (O,) to produce gaseous carbon dioxide (CO,) and gaseous water (H,0). Suppose 11. g of methane
is mixed with 57.0 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Round your answer to 2 significant
digits.
x10
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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.
can you answer, please? the experimental data are Reference: https://www.youtube.com/watch?v=OOXRkycKEOc&feature=youtu.be and https://www.youtube.com/watch?v=3wLJLm0QLpg&feature=youtu.be
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 =…
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
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
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…

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