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- Round this addition of measurements to proper sig figs: 23 cm + 111.24cm + 0.0045 cmThe following data were gathered during an experiment on the Analysis of Milk : Weight of empty beaker: 26.34 g Weight of the beaker + milk sample: 38.25 g Weight of milk sample: _______________ Volume of milk: 12.0 ml Density of milk: _____________g/ml Specific gravity of milk: _____________ Theoretical value for Sp. G. of milk is 1.03 Percentage difference: ________________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 =_____
- 1. σ=[(Σd2)/(n-1)]½ n= number of trials d= measured value - mean σ=[(8.50mL-8.74mL)2+(9.72mL-8.74mL)2+(8.00mL-8.74mL)2/(3-1)]½ 2. σ=[(Σd2)/(n-1)]½ n= number of trials d= measured value - mean σ=[(10.69-11.52)2+(11.80-11.52)2+(12.07-11.52)2/(3-1)]½Specific gravity is important for diagnosing various health conditions, such as dehydration. The normal specific gravity falls between 1.002 - 1.030. Mild dehydration is indicative from specific gravity that is greater than 1.010 and the higher the value the more dehydrated one may be. A 25 year old female patient wastaken to the hospital with symptoms of extreme thirst, fatigue, and headaches. An 0.0077 liter urine sample was taken for the patient described above. The sample had a mass of 8.05 grams.Complete the table.a. 1245 kg 1.245 * 106 g 1.245 * 109 mgb. 515 km _____dm _____cmc. 122.355 s _____ms _____ksd. 3.345 kJ _____J _____mJ
- 1000 mg of granular salt is briefly stirred into a glass container of 1/4 liter volume filled with water. After briefly mixing, some salt is still seen at the bottom of the glass. The undissolved salt is filtered from the water, weighed and determined to have a mass of 200 mg. What is the dissolved phase concentration of salt remaining in the water? Provide your answer to the nearest whole number in units of mg/l. Example: 120 (no decimals, commas, or zeros after decimal)Conversion of units using the DimensionalAnalysis(Factor Label Method) I – Give the equivalent units of measurement. Show your solution 1. 1m = ____ cm = _____ mm =_____ dm =______km = _____in =____ ft =___yd =______mi 2. 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 =_____How do I write 100pm in scientific notation?
- 1.) A dense plastic-like shape has a mass of 42.985 g and has a volume of 41.96 mL. What is the density of the substance? 2.) Another plastic-like piece has a mass of 65.389 grams and displaces 53.652 grams of water (measured at 24.6 ⁰C). Is this piece likely to be composed of the same material as the sample in problem 1? Explain using the Claim-Evidence-Reasoning process.The conversion factor used to convert 55.6 kmkm to miles is _____ View Available Hint(s)for Part B 1.61 miles1 km1.61 miles1 km 1.61 km1 mile1.61 km1 mile 1 mile1.61 km1 mile1.61 km 1 km1.61 miles1 km1.61 milesWHERE u wrote substitute is that 100 x 1 so do u dot to represent multiplication