Discussing the Data - Class Average vs Group 9. Use the % change my group for the unknown a. Square the spot on the best fit line that shows the sucrose concentration of the unknown. b. What is the sucrose concentration of the unknown? c. Did you get the same answer as in 3b? 10. Many times when there are differences in data it is caused by small errors made when setting up the lab, collecting data or processing the data. List 2 sources where an error could have been made.
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- Purpose: what is the concentration of the menown Solution? Hypothesis: None. Method: Use the transport Activity - dry run to get the method Independent Variable Dependent Variable (measuring) (changing) Different molarinies of sucrose Procedure: "as given" Data Table: Molarity in bag 0 (H₂0) 0.2 M Sucrose 0.4 M Sucrose 0.6 M Sucrose 0.8 M Sucrose 1.0 M Sucrose Unknown 90 38 40 initial mass (g) 20 10 10 10 10 Calculated channe final mass (g) 10.3 11.2 14.8 10.5 17.2 12.8 .2 Controlled Variables (same) Dialysis tubing, + time. (final mass-initial mass) initial mass .4 X 100 water % change my group Analysis 1. Calculate the percent change in mass for your solutions. Use the workspace in the data table for help. 2. Graph your data & the class average data. Label the independent (x axis) and dependent variable (y axis). Include a title. After plotting both sets of data one best fit STRAIGHT line. 100 Control The tube filled with water D % change class average 4 .8 46 68 30 1.0Compute for the lactic acid content of these samples given the values below. Show your solution Volume of 0.1 N NaOH in 1st Sampling: 10 mL Volume of 0.1 N NaOH in 2nd Sampling: 40 mL Volume of sample used: 10mlConvert the experimental glucose concentration from mM (millimolar) to g/100 mL. 1/100 Dilution factor, Assume 1L of Solution. Show ALL Steps 2.340 mM
- 46 50 52 60 41 46 55 Find the ff. WITH SOLUTION PLEASE MeanMedianModeSDVarianceCVQuartile 2Decile 7Percentile 25Percentile 30Table 1 Rate of pressure change (kPa/min) Sucrose solution concentration (M) 3078 0,1460 0.009095 1.0 0.9 0.8 0.7 -0.02370OMatch the common laboratory scenarios to the correct buffer required using the guide below. Molecule pKa Value(s) Acetic acid 4.8 Carbonic acid 6.1, 10.2 Citric acid 3.1, 4.8, 5.4 HEPES 7.5 Phosphoric acid 2.1, 7.1, 12.3 TRIS 8.3 Mammalian cells require a pH of about 7.4 for cell and tissue culture experiments. Elution of remaining molecules from a cation exchange column requires washing with pH 10 buffer. Dissolution of lysozyme requires pH 3 buffer to maintain activity. Elution of remaining molecules from an anion exchange column requires washing with pH 2 buffer.…
- Desired Desired Initial Volume of Stock or Volume of Concentration Volume Concentration Previous Dilution Distilled Water (C2) (V2) (C1) (V1), µL (V2- V1), µL 100 mg/dL 1.2 mL 200 mg/dL 75 mg/dL 1.2 mL 100 mg/dL* 50 mg/dL 1.2 mL 25 mg/dL 1.2 mLData for standard glucose concentrations Glucose concentration (mm) 0 0.5 1 1.5 -~~ 2 2.5 A Data for plasma samples B Patient Timepoint (hours) C D E 0 0.5 1 2 0 0.5 1 2 0 0.5 1 2 1. Draw a standard curve of absorbance at 420nm against glucose concentration in mmoles/L. We recommend you use Excel or a similar programme to insert the graph Alternatively, you can draw by hand on graph paper and insert a photo of the graph. OON 0 0.5 1 Absorbance 1 0.000 0.477 0.947 1.234 1.733 2.160 2 0 0.5 1 Absorbance Absorbance 2 0.000 0.482 0.941 1.236 1.721 2.276 Absorbance Glucose concentration Glucose concentration Abs 1 Abs 2 Average from graph (mM) in plasma (mm) 0.635 0.632 0.629 1.368 1.370 1.369 1.382 1.388 1.385 1.151 1.147 1.149 0.566 0.844 0.869 0.772 0.395 0.397 0.399 0.850 0.858 0.854 0.738 0.744 0.741 0.539 0.531 0.535 0.776 0.780 0.778 1.475 1.481 1.478 1.446 1.444 1.445 1.302 1.296 1.299 0.560 0.563 0.848 0.846 0.871 0.870 0.774 0.773 Average 0.000 0.480 0.944 1.235 1.727 2.218 0.440…Choose the FALSE statement An indicator can be used to monitor a buffer range When pH equals PKa of weak acid, the buffer capacity is the greatest Polyprotic acids have multiple stoichiometric points in their titration curve At the mid-point of titration of weak acid by strong base, the pH of the solution -does not equal the pKa of the weak acid The moles of titrant equals moles of titrate
- Please answer these question. I need these for lab. PRE-LAB ASSIGNMENT – (YOU WILL NOT BE ALLOWED TO BEGIN WORK ON LABWITHOUT THE COMPLETED CALCULATIONS) 1. COMPLETE ALL CALCULATIONS REQUIRED FOR TABLE 1 ( For all measurements, the [PNPP] concentration will be 600 μM in a 1 ml final volume.) PNPP (substrate) stock solution: 100 mM PNPP in buffer 2. IN YOUR LAB NOTEBOOK, DRAW YOUR PREDICTION OF WHAT THEFOLLOWING GRAPHS MIGHT LOOK LIKEa. pH vs. V0b. TEMP vs. V03. OUTLINE THE BEST/MOST EFFICIENT WAY TO PREPAREGiven the following information, calculate the total activity in the undiluted protein sample. Activity of 1 ml of diluted sample = 0.5 Total volume of sample = 5 ml Dilution factor = 10 25 50.5 250 2.5Table 1 - Comparison of the effect of catechol concentration on the amount of product formed. Absorbance Potato extract Absorbance 0 mins after 30mins (2nd reading) (mL) 1st reading 1 Tube # la blank 2a 3a 4a 1 1 1 dH₂O Catechol (mL) (mL) 7 5 3 1 0 2 4 6 0.00 0.060 0.033 0-05-2 Q4) Give 2 reasons for adding dH₂O to these tubes in Table 1? Time for reading: 3:21 -0.11 Absorbance: Time for reading: 3.36 Q5) Tube la serves as a control, but why is this control needed? Absorbance: 0.197 Time for reading: 3.37 Based on the data from Table 1 answer these questions: Q1) What is the name of the enzyme found in potato extract? Answer: catechol Q2) What is the substrate? Answer: THO Q3) Name of product of this enzyme catalyzed reaction? Answer: Absorbance: 0.152 Time for reading: 3:39 Absorbance: . 166 ness Catechol Benzoquinone Subtract 1st from 2nd reading -0.01 0-137 0.11.19 0.119 Q6) Notice that your 1st absorbance reading in tubes 2a-4a are quite similar but it then becomes very different…