0.550-gram sample of a nondissociating enzyme was dissolved in water at 25°C to yield 50.0 mL of solution. The osmotic pressure of the solution was found to be 19.0 Torr. Calculate the molecular mass of the enzyme. I need help with the steps, make sure math is correct please! Answer: 1.08 x 104

0.550-gram sample of a nondissociating enzyme was dissolved in water at 25°C to yield 50.0 mL of solution. The osmotic pressure of the solution was found to be 19.0 Torr. Calculate the molecular mass of the enzyme. I need help with the steps, make sure math is correct please! Answer: 1.08 x 104

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter10: Fuels, Organic Chemicals, And Polymers

Section: Chapter Questions

Problem 119QRT

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A 0.550-gram sample of a nondissociating enzyme was dissolved in water at 25°C to yield 50.0 mL of solution. The osmotic pressure of the solution was found to be 19.0 Torr. Calculate the molecular mass of the enzyme. I need help with the steps Answer: 1.08 x 104
The Ksp value for calcium sulfate [CaSO4] is 2.40 X 10-5 so if a professor made 1825 mL of a CaSO4lag) solution then how many grams of sulfate (SO42) ion would be present in this solution. 0.8345 gram SO42- ion 0.8588 gram SO42- ion 0.9331 gram SO42- ion 1.0036 gram SO42- ion 0.8921 gram SO42- ion 2.. 591..docx W- LAB EXP. #3 -.docx Show All
Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Concentration iron(III) nitrate 0.00200 0.00200 0.00200 0.00200 0.00200 [Fe(NO3)3] (M) Concentration potassium thiocyanate 0.00200 0.00200 0.00200 0.00200 0.00200 [KSCN] (M) Volume Fe(NO3)3 (mL) 5.00 5.00 5.00 5.00 5.00 Volume KSCN (mL) 5.00 4.00 3.00 2.00 1.00 Volume DI water (mL) 0.00 1.00 2.00 3.00 4.00 Initial concentration [Fe3+] (M) 0.00100 0.00100 0.00100 0.00100 0.00100 Initial concentration [SCN‍] (M) 0.00100 0.000800 0.000600 0.000400 0.000200 Absorbance 0.269 0.192 0.154 0.104 0.052 Equilibrium [FeSCN2+] (M) 0.000198 0.000141 0.000113 0.0000765 0.000038 Equilibrium constant Kc Average Kc
1/T (K-¹) 2 N₂O5(g) → 4NO2(g) + O₂(g) K(S-¹) 4.87 x 10-3 1.50 x 10-³ 4.98 x 10-4 1.35 x 10-4 3.46 x 10-5 7.87 x 10-7 T (K) 338 0.0030 328 0.0030 318 0.0031 308 0.0032 298 0.0034 273 6.0037 The data in the table above provide the temperature dependence of the rate constant for the reaction shown. Complete the table with required information to 2 significant figures each. - a=-12571.42 In K, no units 5.3 - 6.5 7.6 8.9 Carefully construct a neat and well labelled graph/plot of In K (y-axis) versus 1/T (x-axis) in the appropriate quadrant (s), using EXCEL or the template below; then determine the slope and activation energy, Ea for the given reaction. Recall that the Arrhenius Equation is represented in a more useful form as In K= In A- (Ea/RT) & R= 8.31 x 10-³ kJ/mol-K. Determine by calculation, the value of Ea in kj/mole given that, slope = -E/R slope: - Ca = slope & - Ga= €ag Ea=-(-12571.42) x (8.3) x 10 Rg/mol-K) Ea=104.47 kj/md 10.3 - 14.1 Y = (-14.1)-(-5.3) = -8.8 x = (0.0037) -…
You are creating a standard curve for a protein experiment. You have to stock solution of BS a and would like to obtain dilutions. Explain how you would create a 1:1000 dilution of your protein stock solution.
An 80 kg subject consumed a snack that contained 50 g of carbohydrate. Before eating the snack the subject’s blood glucose was 5.5 mM. Calculate the concentration of glucose (mM) that would be reached in the blood after consuming the snack.Assume that all the carbohydrate in the snack was converted to glucose, rapidly absorbed and distributed throughout the extracellular water (ECW) within 30 minutes. Assume ECW is approximately 20% of body weight and the glucose is not metabolised further.
Please give me answers in 5min I will give you like sure
Calculate the molar concentration of a blood sugar level of 100 mg /dl where MW glucose = 180
A student added a total of 75 mL sodium hydroxide to titrate an amino acid solution of unknown concentration. They also calculated that they added a total of 14 mL of sodium hydroxide to arrive at pH 2.6 and a total of 32 mL of sodium hydroxide to arrvive at pH 4.8. Information about the molarity of the sodium hydroxide solution used and the initial volume of the unknown amino acid can be derived from the image. What was the initial concentration of the amino acid solution in milimolar (mM)? The calculations must be presented such that those can be recapitulated by the marker. Provide the answer with no decimal places. Show all details of your working out. State the answer in a full sentence
A technician finds the molar mass to be 134.8 g/mol and the pK, to be 4.65 after processing a data set for a solid acid unknown. Based on the previous information and the list of acid unknowns in the Laboratory Manual, what is the best call the student can make for the identity of the solid acid unknown? picolinic acid B para-aminobenzoic acid glutamic acid monohydrate potassium hydrogen phosphate, monobasic OA. O B. O C. O D.
A 1 mL sample of glycogen was calculated to contain 21 µmol (micromole) glucose. To 1 mL of this sample was added 2 mL of 2 M HCl. It was then hydrolysed by boiling the solution for 15 minutes. After boiling the hydrolysate was cooled and made up with H2O to a final volume of exactly 10 mL. The glucose was measured in this solution and found to have a concentration of 340 µg/mL (microgram/milliliter). i) Calculate the mass (mg) of glucose in the 10 mL of hydrolysate. As the 1 mL of glycogen sample was made up to a final volume of 10 mL, this mass of glucose was produced by the hydrolysis of the original 1 mL glycogen sample. ii) Calculate the amount (µmol) of glucose produced by the hydrolysis of the glycogen sample. iii) Calculate the purity of the glycogen used in the sample as % Purity = (moles of measured glucose/ moles of calculated glucose in glycogen) *100 iv) state your answer in a complete sentence. Show your working out such that the marker can easily understand it.…
Can you please provide steps for how to answer the following? This is not a graded assignment, it's just practice questions for no credit. The concentration of the stock glucosamine solution is 100 mM. A student took 0.5 ml of this stock solution and added it to a test tube containing 24.5 ml of water. The student then took 1 ml out of the diluted solution and added it to another test tube containing 9 ml of water. The concentration of glucosamine in the second test tube will be: a. 0.4 mM b. 0.8 mM c. 1.2 mM d. 1.6 mM e. 0.2 mM. If you took 10 ml out of a protein stock solation of 0.25 mg/ml and added it to a bottle containing 40 ml of buffer and 500 ul of cofactor. What will be the final concentration of the protein in the bottle? a. 55.5 ug/ml b. 45.5 ug/ml c. 52.5 ug/ml d. 49.5 ug/ml e. 65.5 ug/ml