कना १०मaडडापल श्ुल्कांवेर ( EcN ) vrल्वलङ्ड ज्क वतंवेe , व वटववाय P० 3cnu&4& भp8, सुdसमा पूुकलांवर (HतN ), kcN (a৭)+ Heicq৭) >HलNCY)+ kc1cqq. व हवलाशर 0न 0पाठ 8 0+ of if kcN fS न०९q+edे wwक वक ०+ मत्य , टवात्यविमर +कए ठ+ मतV oलह वे क excess वल०पकन मतV forme वे क कलड . of
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- The Ksp of Zn(OH)2 is 3.00 x 10^-174. Ksp=4.1x10-36 for Pb3(AsO4)2Calculate Eofor Pb3(AsO4)2(s) + 6e- 3Pb(s) + 2AsO43-Pb2+ +2e- Pb(s) Eo= -0.126 VWhat is the retardation factor of a pulse of dissolved benzene move through organic-rich wetland soil (organic carbon fraction of soil, foc=50%) if the effective porosity is 0.4 and the bulk density of the soil is 1.5 g/cm3. Follow these important hints: - Determine organic carbon partition coefficient (Koc) - Use this Equation: K oc = K d / f oc to calculate Kd. Select the range which includes your calculated answer. Group of answer choices 300 to 400 2.0 to 2.5 mg/l 2.0 to 2.5 2.6 to 4.8 0.5 to 1.0 2.6 to 4.8 mg/l None of these ranges include my answer.
- Calculate delt g for Pbl2(s) + Pb2+(aq) +2I-(aq) T= 25C Kp= 8.7x10-9https://m.youtube.com/watch?v=vM1SP346XBc&list=PLeJOSNLNZfHubfLdq0kOayASeUllMOGn4&index=4 I watched this the lecture video over and over and I am allowed to work with someone but I am having trouble with part B and I provided the YouTube link of the data or video attached to this labAn analyst obtained the following data for the percent compound Z in triplicates (n=3) of an insecticide preparation: 7.47, 6.98, and 7.27. Calculate the 90% confidence limit for the mean of the data assuming that only information about the precision of the method is the precision for the three data points.
- 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 =…How to convert 4.76pm to um? Showing the work. 4.76pm x___________m x_____________ um =________________um How to convert 25.0mL to cL? Showing the work. 25.0mL x______________ x ___________ = ______________cL