Calculated [H*], M Calculated [OH'], M Solution Measured pH Intermediate Final Intermediate Final value value value value 0.100 M 1.60 10 10 HCI 0.0100 M 2.47 10 2 10 HCI 0.00100 3.52 10 10 M HCI 2. 2. 2. 2. 2.
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solve missing values using data given
![Calculated [H*], M
Calculated [OH'], M
Solution
Measured pH
Intermediate
Final
Intermediate
Final
value
value
value
value
0.100
M
1.60
10
10
HCI
0.0100 M
2.47
10
2
10
HCI
0.00100
3.52
10
10
M HCI
2.
2.
2.
2.
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- 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=drivesdkBasis: https://youtu.be/Y4NMpO1xI8Uhttps://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 lab
- What 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.You can’t excess the interactive so I write down what information was being presented.how would u justify this question? my answer is correct i just do not understand the justification. this is a non graded practice worksheet
- 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 =…9. Calculate the Go in kJ/mol. 3Fe2+ (s) + 2Cr(aq) ⟶2Cr3+(aq) + 3Fe (s); Eocell=0.30 V Group of answer choices -170 +170 +87 +58 -195 10. Calculate Ecell in volts. Zn(s)|Zn2+(2.50x10-4M)||Sn2+(1.50M)|Sn(s) Eocell=+0.624 V Group of answer choices 0.736 0.635 0.512 0.848 no correct answer
- A solution containing 100 lbm KNO 3/100 lbm H 2O at 80°C is fed to a cooling crystallizer operated at 25°C. Slurry from the crystallizer (KNO3 crystals suspended in saturated solution) is fed to a filter, where the crystals are separated from the solution. Use the solubility data in Figure 6.5-1 to determine the production rate of crystals (lbm/lbm feed) and the solid-to-liquid mass ratio (lbm crystals/lbm liquid) in the slurry leaving the crystallizer.By the use of Henderson Hasselbalch equation; pH = pKa + log{[acetate ion]/[acetic acid]} 4.5 = 4.75 + log{[0.10 M]/[acetic acid]} -0.25 = log{[0.10 M]/[acetic acid]} [Acetic acid] = 0.10 M/ 10-0.25 [Acetic acid] = 0.10 M/0.56 [Acetic acid] = 0.1786 M Moles of sodium acetate dissolved in 250 mL buffer solution = 0.10 M× (250mL/1000mL) × 1L = 0.025 mol Weight (w) of sodium acetate (purity 100%) dissolved to prepare 250 mL of solution with buffer concentration of 0.10 M is calculate as follow; w100% = 0.025 mol × 82.0343 g/mol = 2.051 g Weight (w) of sodium acetate (purity 99%) is calculate as follow; w99% = 2.051 g× (100/99) = 2.072 g What was the volume of 6.12 M acetic acid HC2H3O2 needed to prepare the 250 mL acetic acid/acetate ion buffer solution required in this part? Show your calculations.If may calculations po, please show all steps leading to the final answer po. Here’s the link in accordance with the topic po: https://drive.google.com/file/d/1_FnDtXCrFKSol3RNWIG_9tNQ7IxgxD6t/view?usp=drivesdk
