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- Given the following data forMass of test tube, beaker and cyclohexane = 100.17 gMass of test tube and beaker = 84.07 gFreezing point of cyclohexane = 6.59 oCMass of weighing paper + naphthalene =1.080 gMass of weighing paper = 0.928 gFreezing point solution = 5.11oCKf = 20.8oC/mDetermine the followinga. mass of cyclohexane in g (2 decimal places); _____b. mass of naphthalene in g (4 decimal places); _____c. freezing point depression (2 decimal places); _____d. molality of solution (3 significant figures); _____e. moles of naphthalene (3 significant figures); _____f. molar mass of naphthalene, experimentally (3 significant figures); _____g. % error if theoretical molar mass of naphthalene is 128.17 g/ mole, USE ABSOLUTE VALUE (3 significant figure); ____One student calibrated a 50-mL burette by using the mass of water delivered. The student used an analytical balance which was previously calibratedby STEEL (density = 7.8 g/mL). During calibration, the temperature of water was found to be 19.0oC. The density of water at this temperature is0.9984 g/mL. The results of the calibration by the student is shown in the table below:A B C D EApparent burette reading, mL 10.05 20.04 30.08 40.07 49.98Weight H2O delivered, g 10.03 20.09 30.05 40.02 49.83Note: All masses in mentioned in this section are understood to be in grams and all volumes mentioned are understood to be in milliliters. a) What is the true volume of water delivered in B?One student calibrated a 50-mL burette by using the mass of water delivered. The student used an analytical balance which was previously calibratedby STEEL (density = 7.8 g/mL). During calibration, the temperature of water was found to be 19.0oC. The density of water at this temperature is0.9984 g/mL. The results of the calibration by the student is shown in the table below:A B C D EApparent burette reading, mL 10.05 20.04 30.08 40.07 49.98Weight H2O delivered, g 10.03 20.09 30.05 40.02 49.83Note: All masses in mentioned in this section are understood to be in grams and all volumes mentioned are understood to be in milliliters. What is the correction volume of water delivered in C?A) 0.04B) 0.08C) 0.00D) -0.04E) -0.08
- One student calibrated a 50-mL burette by using the mass of water delivered. The student used an analytical balance which was previously calibratedby STEEL (density = 7.8 g/mL). During calibration, the temperature of water was found to be 19.0oC. The density of water at this temperature is0.9984 g/mL. The results of the calibration by the student is shown in the table below:A B C D EApparent burette reading, mL 10.05 20.04 30.08 40.07 49.98Weight H2O delivered, g 10.03 20.09 30.05 40.02 49.83Note: All masses in mentioned in this section are understood to be in grams and all volumes mentioned are understood to be in milliliters. What is the true volume of water delivered in D?A) 40.11B) 40.12C) 40.13D) 40.14E) 40.15Medicare does not cover the cost of this prescription medicine, which currently averages $297.78 per kilogram. The theoretical yield for the pure barium sulfate you were transferring was 500.00kg. You spilled 27.45 kg. What is the % yield of the transfer that you need to report? And also determine the consumer price of the spilled barium sulfate.Use the given equivalents, along with dimensional analysis, to convert the given unit to the unit indicated 55lb to g. 16oz = 1 lb 200lb=1T 1oz=28g 1lb= 0.45kg17 = 0.9t Please show work
- Composite decking is a manufactured substitute for woodcompounded from post-consumer plastic and wood. It isfrequently used in outdoor decks. The density of a particularcomposite decking is reported as 60.0 lb>ft3. What isthe density in kg/L? 259 kg/LUse the given equivalents, along with dimensional analysis, to convert the given unit to the unit indicated. 360 t to T 16oz = 1 lb 2000lb=1T 1oz=28g 1lb= 0.45kg17 = 0.9t Please show workOne student calibrated a 50-mL burette by using the mass of water delivered. The student used an analytical balance which was previously calibratedby STEEL (density = 7.8 g/mL). During calibration, the temperature of water was found to be 19.0oC. The density of water at this temperature is0.9984 g/mL. The results of the calibration by the student is shown in the table below: Apparent burette reading, mL A-10.05 ml, B- 20.04ml, C- 30.08ml, D- 40.07 ml, E- 49.98ml Weight H2O delivered, g A -10.03g, B- 20.09g, C- 30.05g, D- 40.02g, E- 49.83g
- One student calibrated a 50-mL burette by using the mass of water delivered. The student used an analytical balance which was previously calibrated by STEEL (density = 7.8 g/mL). During calibration, the temperature of water was found to be 19.0oC. The density of waterat this temperature is 0.9984 g/mL. The results of the calibration by the student is shown in the table below: A B C D E Apparent burette reading, mL 10.05 20.04 30.08 40.07 49.98 Weight H2O delivered, g 10.03 20.09 30.05 40.02 49.83 What is the true mass of water delivered in A?One student calibrated a 50-mL burette by using the mass of water delivered. The student used an analytical balance which was previously calibrated by STEEL (density = 7.8 g/mL). During calibration, the temperature of water was found to be 19.0oC. The density of waterat this temperature is 0.9984 g/mL. The results of the calibration by the student is shown in the table below: A B C D E Apparent burette reading, mL 10.05 20.04 30.08 40.07 49.98 Weight H2O delivered, g 10.03 20.09 30.05 40.02 49.83 What is the true volume of water delivered in A?Conversion of units using the DimensionalAnalysis(Factor Label Method) I – Give the equivalent units of measurement. 1. 1m = ____ cm = _____ mm =_____ dm =______km = _____in =____ ft =___yd=______mi2. 1kg = ____ g =______mg = _____cg = _____ dg = ____lb 3. 1h = _____s = _____ ms = ______ min = ____ yr =______ day 4. 1A = ____ MA = ____V = ______ W = ______ kW = ____mA 5. 1mol = ____ mmol = _____kmol = ______Mmol 6. 1K = ____ oC= ______ oF=_______oR 7. 1g/mL = ______g/cm3 = _____lb/in3 = _____ lb/cm3 =______kg/m3 8. 1Pa=________kPa =_______mPa = ____ GPa = _____ atm 9. 1J = ________kJ = ______ nJ = _____cal = ____k cal =_____dcal 10.1Byte = _____ GB =______ kB = _____TB=_____MB =_____