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- The gravimetric factor used to express CoCBr6·H20 in a sample that is finally weighed as PbClBr is choose below: FW PbClBr / 6 x FW CoCBr6·H20 FW CoCBr6·H20 / FW PbClBr FW CoCBr6·H20 / 6 x FW PbClBr 6 x FW PbClBr / FW CoCBr6·H20There'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 =…
- 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)125.1 mg of streptomycin sulphate are dissolved in 10 ml of water. A GC headspace analysis is carried out in order to determine the methanol content of the drug. A peak for methanol is produced which has 73.2% of the area of a peak for a methanol standard containing 0.532 mg/100 ml of methanol in water analysed under exactly the same conditions, What is the methanol content of the streptomycin sulphate in ppm and %w/w? Answer: 311.3 ppm, 0.3113 % w/w.how??4. A fat sample with combination of acids contain standard hydrochloric acid for blank and sample with 8mL and 5mL respectively. The normality of the standard hydrochloric acid is 0.93N and the weight of the sample is 3 grams. Calculate the saponification value.
- (6.65x10^5)/(6.60x10^-6) (3.73x10^8) Please Calaculition. Only typed solution1. Calculate the experimental density of a salt solution and the percent error (same as relative error percent) using some or all the data given below. solubility of NaCl salt in water: 0.357 g/mLmass of empty graduated cylinder: 25.19g mass of graduated cylinder + salt solution: 30.47g total volume of salt solution: 4.98 mLtrue density of salt solution: 1.07 g/mLDownvoted 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.
- 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 0.466An analytical determination for manganese in an aqueous solution on gives thefollowing replicate measurements in ppm as 12.1, 12.3, 12.2, 12.2, 12.4, 12.3, 12.2, 12.4,12.2, and 12.5. Determine the following;a) The range of the datab) The meanc) Mediand) Standard deviatione) Variancef) Relative standard deviation RSD (ppt)An 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?