Convert a speed of 799 mi/h to units of feet per minute. Also, show the unit analysis by dragging components into the unit-factor slots. 799 mi 1 h 5280 ft Answer Bank 1 h 60 min 1 min 60 h 12 in 60 s 3 ft I mi 1 ft Incorrect ft 799 mi/h = 527.12 min Incorrect
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- If 35,000 kg of whole milk containing 4% fat is to be separated in a 6-hour period into skim milk with 0.45% fat and cream with 45% fat, what are the mass flow rates of the two output streams from a continuous centrifuge which accomplishes this separation? (Ans; Cream=464.8335kg/h, Skim milk= 5368.4998kg/h)Downvoted for wrong solution. A river is carrying water containing 2000 mg/l Magnesium Chloride into a small lake. The lake has a naturally occurring Magnesium Chloride of 50 mg/l. If the river flow is 2500 Lmin and the lake flow rate is 1.5 m³.sec¹, what is the concentration of MgCl2 in the lake after the discharge point? Assume that the flows in the river and lake are completely mixed, that the salt is a conservative substance, and the system is at steady state.Estimate the Km and vmax from the data. [S] (M) Velocity (µM/min) 2.5 x 10-6 28 .00001 70 .00004 112 .0001 128 .002 139 .01 140 Km=.00001 vmax=140 Km=.002 vmax=112 Km=.01 vmax=140 Km=.00001 vmax=70
- A fly ash aerosol of monodisperse particles with a density of 2.0 g/mL and particle diameter of 10.0 mm has a concentration of 1000.0 mg/m3. Calculate the settling velocity in cm/s and the settling rate in g/m2-s. Please answer very soon will give rating surelyAn orange juice processing plant now produces essential oil from orange peels. one It is known that 250 kg of peel comes out of 1 ton of oranges and 2.5 g of essential oil comes out of 1 kg of peel. In a laboratory study, 250 g of bark was treated with hexane solvent and 0.548 g of essential oil was obtained in the sample cup of the rotary evaporator. Accordingly, the rotary Calculate the separation efficiency obtained in the evaporator?Ethyl alcohol (C2H5OH) may be prepared by the fermentation of glucose (C6H12O6) as indicated by the equation: yeastC6H12O6 ----> C2H5OH + CO2 74.12 mL of ethyl alcohol (specific gravity = 0.790) was collected by this fermentation pro- cess. What mass of glucose was used? SET-UP: Answer:
- 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 =…Please answer fast it’s very important and urgent I say very urgent so please answer super super fast please For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466
- Express 20,000ppb as a percentage strength A. 2% B. 0.02% C. 0.00002% D. 0.002%Hello, can someone please help me? I have already hand graphed 1 and 2 and found the v max, however I cannot find the Km value. I am confused because the plots on the linear weaver burk plot do not extend past the values that make up the line. So how am I supposed to estimate the value for the y-intercept in order to find the Km valuePlease answer super fast and answer all questions and show calculations For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466