What is the chemistry principle behind when (treatment: heavy cream) is subjected to CHILLED TEMPERATURE? What is the effect in beating time, stability (%drain) and volume foam (specific gravity) is important? see the data below Treatment (Heavy Cream) Beating Time (min) Stability (%drain) Volume of Foam (Specific Gravity) (Room Temperature) 20 No drain 0.94 Heavy Cream (chilled) 20 No drain 0.76 Heavy Cream with sugar and vanilla 20 No drain 0.88 Heavy Cream (over-whipped) 25 No drain 0.83
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What is the chemistry principle behind when (treatment: heavy cream) is subjected to CHILLED TEMPERATURE? What is the effect in beating time, stability (%drain) and volume foam (specific gravity) is important? see the data below
Treatment (Heavy Cream) |
Beating Time (min) |
Stability (%drain) |
Volume of Foam (Specific Gravity) |
(Room Temperature) |
20 |
No drain |
0.94 |
Heavy Cream (chilled) |
20 |
No drain |
0.76 |
Heavy Cream with sugar and vanilla |
20 |
No drain |
0.88 |
Heavy Cream (over-whipped) |
25 |
No drain |
0.83 |
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- The material to be heated is in a long, cylindrical tube, and is viscous enough that any flow inside the tube can be ignored, i.e. the heating is by conduction alone). Also, consider the glass of the tube to be very thin so its thermal resistance can be ignored. The tube diameter is 1 cm and the properties of the material are the same those of water, given by 0.64 W/m*K, 1000 kg/m3, and 4180 J/kg*K, for thermal conductivity, density, and specific heat, respectively. The water bath temperature is 65 °C, and the initial temperature of the material is 27 °C. Without the agitation of the water bath, the heat transfer coefficient can be considered to be 128 W/m2*K. Considering the tube to be a long cylinder, how long does it take for the coldest point to heat up to a temperature corresponding to 99% of the total possible temperature increase? If the surrounding how water is agitated using a stirrer, leading to a considerably large surface heat transfer coefficient, how long does it…The Thermal conductivity of rock salt measured at a temperature of 500 degrees C what value would be the closest ?i) Write down the equation derived from your Excel generated standard curve and describe its components; ii) provide the values of the absorbance data of the unknown sample (do NOT show the absorbance data of the glycine standards). Show all details of the working out of your calculation. Indicate all units! Provide the answer with two decimals precision iii) state the answer in a full sentence. (note: avoid mathematical symbols [=, +, -, etc] in your answer sentence).
- For the serial dilution, your stock solution must have a concentration of 3.5 mg/mL. How much diluent must be added to the 5.3 mg/mL red cell to prepare the stock solution? Show pertinent solution/s. 7. If the red cell suspension is the stock solution, what is being quantified in this test? What diluent/reactant should be used to detect your answer in number 7?The unknown metal sulfates are hygroscopic and will absorb water from air. The unknowns must thus be kept in desiccators to remove any absorbed water. Howwould your results be affected if your unknown sample was not desiccated? Would this error cause your calculation of the mass percent of sulfate in the unknown to be too high or too low? Explain.The ΔG°′ value for glucose-1-phosphate is -20.9 kJ/mol. If glucose and phosphate are both at 4.8 mM, what is the equilibrium concentration of glucose-1-phosphate?
- A glucosamine solution of unknown concentration was diluted as follows: 1 mL of the solution was diluted to 5.0 mL. A 0.3 mL aliquot of this solution was diluted with 0.7 mL of water; the absorbance of this was measured and found to be 0.440 (uncorrected, must subtract blank). What is the concentration of the original undiluted glucosamine hydrochloride solution in µmol/mL? Given: Corrected absorbance = 0.395 y= 1.165x+0.055 Standard equationDescribe the preparation of the following solutions include all mathematic calculations and equipment of the laboratory needed A.500 mm of a solution of KNO 1.50 M B. 150 g of a solution 5.0% of NaC2H3O2 C. 500 mm of a solution of NaOH 0.50 M from a solution of NaOH 6.0 MAn unknown mixture is known to contain only Ba(OH)2 (MW=171.34 g/mole) and NaOH (MW=40.0 g/mole). If the mixture is known to contain 45% by mass NaOH, and 8.0 grams of the mixture is dissolved completely in 50.0 ml of solution, answer the following. c).If 10.0 ml of a 0.2 M solution of Na2SO4 was added to the 50.0 ml solution, what would be the final concentration of Na+ in solution.
- Predict the diffusivity of human serum albumin at 293 K in water as a dilute solution and compare it with experimental data. Assume the following: The molecular weight of human serum albumin, MA= 72,300 kg/kmol The viscosity of water at 293 K is 0.897 x 10-3 Pa∙s The experimental data value is 5.93 x 10-11 m2/s.What is the change in entropy when 150 mL of cold (278 K) water is added to 150 mL of near-boiling (368 K) water? Ignore any potential heat loss to the surroundings; only consider the water. For water, Cp = 75.3 J K-1 mol-1.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))