Lab #2 - Egg Osmosis (1)

18 Students were provided with cubes of slightly alkaline gelatine of different dimensions, containing an acid-alkali indicator that is red in alkali but yellow in acid. The cubes were placed in dilute acid solution, and the time taken for the colour in the gelatine to change from red to yellow was measured. Dimensions/mm Surface area/mm² Volume/mm³ Time/minutes 10 x 10 x 10 600.0 1000.0 12.0 5x 5 x5 150.0 125.0 4.5 4 x 4 x 4 96.0 64.0 4.2 2.5 x 2.5 x 2.5 a For each block, calculate the ratio of surface area to volume (SA:V). b Plot a graph of the time taken for the colour change (vertical or y axis) against the SA:V ratio (horizontal or x axis). c Explain why the colour changes more quickly in some blocks than others. 37.5 15.6 4.0

Lab Data Verify your concentration calculation. Did you report your data to the correct number of significant figures? Stock solution Concentration iron(III) nitrate 0.40000 [Fe(NO3)a] (M) Color of iron(III) nitrate yellow Concentration potassium thiocyanate [KSCN] (M) 0.00300 Color of potassium thiocyanate colorless Volume iron(III) nitrate (mL) 10.00 Volume potassium thiocyanate (mL) 2.00 Volume DI water (mL) 8.00 Concentration iron(III) thiocyanate ion in stock solution [FESCN2"] (M) 0.000300 Color of stock solution red-orange Diluted solutions Solution 1 Solution 2 Solution 3 Volume stock solution (mL) 4.00 4.00 4.00 Volume DI water (mL) 1.00 2.00 3.00 Concentration iron(III) thiocyanate ion [FeSCN2*] (M)

Answer number 2

DATA Determine the Molar concentration of your dilutions using the formula M₁V₁ = M₂V₂. The initial molarity is 0.0001M (1x10-4M). M1 V1 1.56E-06 500uL Concentrations Tube 1 1.56E-06 Tube 2 Absorbance Tube 1 Tube 2 M2 Tube 3 Tube 3 V2 1000uL Tube 4 Tube 4 Tube 5 Tube 5 Tube 6 Tube 7 Tube 6 Tube 7 Tube 8 Tube 8 Tube 9 Tube 9 Tube 10 Tube 11 Tube 10 Tube 11 10

Time (min) 0 15 30 45 60 75 Solution A Weight (8) 4.205 4.032 3.787 3.648 3.550 3.500 Change (%) Solution B Weight (8) 3.611 -4.11% 3.606 -6.07% 3.622 -3.67% 3.662 -2.69% 3.706 1.41% 3.732 Change (%) Solution с Weight (8) 4.257 -0.138% 4.213 0.444% 4.219 4.209 Change (%) 4.210 -1.03% Solution D Weight (8) 4.728 4.513 0.142% 4.350 1.10% 0.237% 4.254 1.20% 0.024% 4.068 0.675% 4.209 -0.024% 4.140 Change (%) -4.55% -3.61% -2.21% -4.37% 1.77% Graph your results below, plotting percent weight change versus time for all 4 unknown solutions on one set of axes, labeled clearly with units. Include a title for your graph.

Assignment Score: 74.6% Resources O Hint Check Answer < Question 17 of 28 Each value represents a different aqueous solution at 25 °C. Classify each solution as acidic, basic, or neutral. Acidic Basic Neutral Answer Bank H|= 1.0 x 10-7 РОН 3 7.00 pH = 8,39 OH= 9.0 x 1o POH = 5.34 POH = 10.32 H = 5.0 x 10 [H- 3.5 x 10-D pH= 5.50 OH = 8.9 x 10

Submit 3 4. 7) 9. 10 3 of 10 Question 3 Directions: Answers must be in essay form. Outline form is not acceptable. Labeled diagrams may be used to supplement discussion, but in no case will a diagram alone suffice. It is important that you read each question completely before you begin. Water is essential to all living things. (a) Discuss THREE properties of water. В I U х? X, 5 ¢ Ω 0/10000 Word Limit (b) Explain each of the following in terms of the properties of water. You are not limited to the three properties discussed in part (a): • the role of water as a medium for the metabolic processes of cells • the ability of water to moderate temperature within living organisms and in organisms'environment • the movement of water from the roots to the leaves of plants B IU x2 X2 Ω 0/10000 W II II 5

how do we fill the chart under it

Solution Initial Mass (g) Final Mass (g) Change in Mass (g) % Sucrose Red 1.88 1.79 -0.09   Orange 1.92 0.95 -0.97   Yellow 1.95 2.40 +0.45   Green 2.00 1.35 -0.65   Blue 1.89 2.90 +1.01   Purple 1.90 2.05 +0.15   First: Calculate the change in mass for each apple block in grams (g) and write it in the chart. Include a “+” or “–” sign to signify a gain or loss in mass. Second:  According to Jasmine’s lab directions, she should have created the following sucrose solutions: 0%, 5%, 10%, 15%, 20% and 25%. Using the changes in mass, determine which color solution has which percent sucrose and fill in the answers in the chart.

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Protein Solubility 3.5 5 0.028 Absorbance 0.098 Conc.(mg/ml) 0.195 0.044 0.9% 2.786% 9.429% 11.3: % solubility 4.179% The above table indicates the concentration of protein in the diluted supernatant and the supernatant before dilution at different pH. Dilution Factor 30 22.5 pH 15 7.5 1.5 4.5 5 0.027 3 0.042 5.5 5 0.068 0.130 Protein Solubility Versus pH 6 6.5 4.5 pH Value The above figure shows a plot of protein solubility versus pH. 50 Please provide a brief discussion and explanation of the results. 7.5 7.5 50 0.032 0.053 9 8.5 50 0.054 0.100 21.429%

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Temperature (°C) before experiment Temperature (°C) after experiment Food Sample Mass (g) Mass (g) before after experiment experiment marshmallows 7. 65 g 7.41 g 19°C 22.5 °C 7.37 g 7.26 g 19°C 24°C Pumpkin seeds 7.50 g 7.31 g 19°C 24.5°C cracker Dried 7.65 g 7.61 g 19°C 21°C cranberries 7.27 g 7.24 g 19°C 20.2 °C рорсorn 7.21 g 7.19 g 19°C 21°C Rice cake 12.17 12.04 16 22.5°C almond The mass of water being used in the experiment: 100 g

estion 11 Calculate the amount (grams) of solid KOH to prepare 50 ml of 5 mM KOH solution (Molar mass: 56.1 g/mol). ABC T T T Arial 3 (12pt) T II

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Solution Absorbance mg/ml aspirin Standard solution  -  1.6 mg/mL A  0.638  0.08 mg/mL B 0.504   0.064 mg/mL C 0.376  0.048 mg/mL  D 0.259   0.032 mg/mL E 0.126  0.016 mg/mL  A = -log T  where T =  %T ÷ 100 Construct a callibration curve using the above data. Absorbance should be on the vertical axis and "mg/mL of acetylsalicylic acid" on the horizontal axis. The line should go through the origin. Using the data provided, the graph you have generated, and the procedure that was used to generate the solutions which were examined by spectroscopy, calculate the amount of acetylsalicylic acid per tablet. Commercial tablet 1 labelled as 100 mg enteric coated    Absorbance = 0.16 Commercial tablet 2 labelled as 300 mg                            Absorbance = 0.45 Student prepared tablet from practical 5                           Absorbance = 0.19   Using the data provided, the graph you have generated, and the procedure that was used…

pH 3.5 4.5 5.5 6.5 7.5 8.5 Absorbance 0.098 0.027 0.068 0.028 0.032 0.054 Concentration in diluted supernatant (mg/ml) 0.196 0.054 0.136 0.0056 0.0064 0.0108 Concentration in undiluted supernatant (mg/ml) 0.98 0.27 0.68 0.28 0.32 0.54               Formula for the amount (g) of soluble proteins in the soy flour extract: In 15 ml of soy flour extract (with 1/50 dilution), Soluble protein (g) = (C x V x F) / 1000 C = concentration (mg/ml); F = dilution factor; V = volume of solution (ml) Calculate the amount of soluble protein (in grams) in the soy flour extract

pH 3.5 4.5 5.5 6.5 7.5 8.5 Absorbance 0.098 0.027 0.068 0.028 0.032 0.054 Concentration in diluted supernatant (mg/ml) 0.196 0.054 0.136 0.0056 0.0064 0.0108 Concentration in undiluted supernatant (mg/ml) 0.98 0.27 0.68 0.28 0.32 0.54               Formula for the amount (g) of soluble proteins in the soy flour extract: In 15 ml of soy flour extract (with 1/50 dilution), Soluble protein (g) = (C x V x F) / 1000 C = concentration (mg/ml); F = dilution factor; V = volume of solution (ml) Calculate the % solubility of protein in the soy flour (Assume the soy flour contains 35% protein (w/w))

11 of 11 What volume (in µL to 1 decimal place) of stock SybrSafe solution (3.4 mg/mL) would you add to 120 mL of agarose solution to give a final concentration of 0.4 pg/mL? Answer: Submit < Previous Question Assessment Navigator Questionmark Ondemand licensed to University of Dundee DII DD 80 000 000 F10 F8 F9 F7 esc F5 F6 F3 F4 F1 F2 24 ... 6 7 * 00

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Protein Solubility 3.5 5.5 5 5 0.068 0.028 Absorbance 0.098 Conc.(mg/ml) 0.195 0.130 0.044 0.9% 2.786% 9.429% 11.3: % solubility 4.179% The above table indicates the concentration of protein in the diluted supernatant and the supernatant before dilution at different pH. Dilution Factor 30 22.5 15 PH 7.5 1.5 4.5 5 0.027 3 0.042 Protein Solubility Versus pH 6.5 6 4.5 pH Value The above figure shows a plot of protein solubility versus pH. 50 7.5 7.5 50 0.032 0.053 8.5 50 0.054 0.100 21.429% 9 Please provide a brief discussion and explanation of the results. (using isoelectric point and net charge to explain)

Calculate the concentration of protein in the diluted supernatant and the supernatant before dilution at different pH pH  3.5  4.5  5.5  6.5  7.5  8.5 Dilution Factor  5  5  5  50  50  50 Absorbance  0.098  0.027  0.068  0.028  0.032  0.054

Please explain how to calculate Relative Mobility (Rf) for gel electrophoresis (SDS-PAGE). The total distance from top of gel to bottom is is 5.4cm.

what is the calculated change over time from slope? and what condition changed the fastest?

STARTING AMOUNT + X A patient is prescribed Tigan capsules for vomiting. The drug is available in the form of 3.0 x 10² mg Tigan/capsule, and is administered in 4.0 doses/day of 1 capsule/dose. How many g Tigan does the patient receive in 2.0 days ? 3.0 x 10² 2.0 days ADD FACTOR * ( ) 2.4 x 10³ 1.2 x 10³ capsules 0.12 1 doses 2.4 0.24 1.2 10 ANSWER 0.15 1.5 x 10³ mg Tigan/capsule g Tigan 12 1.5 4.0 mg Tigan RESET 2 0.001 24 100 1000 capsule/dose doses/day.

The literature states the pediatric dose of the antibiotic clarithromycin as "7.5 mg/kg q12h." Clarithromycin is available as a suspension containing 125 mg clarithromycin/5 ml. How many millitters should be administered per dose for a child weighing 88 Ib?

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D4 Please answer the question and write out the balanced chemical equation, thank you.

1A and 1 b please

*finish the equation problem* my patients order is written in apothecaries unit dose. The package says aspirin 325 mg. I need to figure out how many tablets I need to administer In apothecaries It indicates that its 10 grains of aspirin I know 1 grain is equivalent to approximately 65 mg so all I have to do is solve the problem 65mg 1 grain set up the second fraction with the unknown quantity in the approximate position X1 grain put the fractions into a proportion 65 mg X 1 grain = 10 grains cross multiply 65mg * 10 grains = X *-1 grain Solve for X by dividing sides of the equation by 1 grain and canceling units

In a titration experiment, potassium iodate serves as the primary standard. Potassium iodate (KIO3) - with 99.4% purity, weight 0.1498 g in 100 mL solution Find the molarity of the titrant given the following data. Trial 1 2 3 Aliquot of standard 10 10 10 Net volume of titrant 33.30 33.50 33.20 Molarity X1 X2 X3