Metric Lab #2 Data, Results, Calculations                         fold value 1.         The the great(p) unwashed of the largest quiz tube was 32.08g. (32.08g) x ( honey oilmg) = 32 080mg. 1         1g         We got this cultivation by counting the tribulation tube upside down on the outstrip and recording the mass to the ne arst one-hundredth of a gram. Then, apply System Analysis, we permuteed grams to milligrams.                                 duration Measure 2.         The length of the largest test tube was 15 cm. (15cm) x (10mm) = 150mm 1 1cm We got out information employ a metric normal and rounding the results in centimeters. Then, we measured and recorded it in millimeters.                         yield Deter houruteation 3.         The ambit of the imbue paper was 128.7 g/cm3. subroutine formula¦pr2 diam of the circle was 12.8 cm. I then sh ard up that by 2 to get the radius (r). p (6.4)2= 128.7 cm2. First we measured the diameter and cleaved it into two. Then by fitting the data in the formula, we were able to come up with the area of the round filter paper.                         mass Measurement 4.         The loudness of the largest test tube was 59.7 ml. (59.7ml) x ( 1L. ) = .0597 L. 1 1000ml. We got the volume of the graduated cylinder by pickax it up to the top with body of water and pouring all of it into a measuring tube and breeding from the line where the water reaches. We utilise System Analysis to convert milliliters to liters.                         concentration Determination 5.         The slow-wittedness of ethyl group alcoholic drinkic beverage is .778 g/cm3. Density = Â!        Mass         Volume Mass = 47.84g. (Alcohol + render) ? 40.58g. (Test Tube Alone) = 6.99 Volume = 9 ml. Of ethyl alcohol. Density = 6.99g = .778g/ml3.         9 ml. First, we had to weigh the tube by itself, which was subtracted by the mass of the tube and the alcohol to image the mass of the alcohol. Then, I divided that by the make sense of alcohol to find the niggardliness of ethyl alcohol, which is less irksome than water. 6.         The density of rock # 14 was 3.104 g/cm3. Density =         Mass         Volume Mass = 22.94g (rock) Volume = 69 ml. (Water + Rock) ? 60 ml. (Water alone) = 9 ml. Density = 22.94g = 3.104 g/ml 3.         9 ml. I utilise the same process to find density as I did in #5. Except, instead of having to find the contrasting in Mass, I had to find the difference in Volume. Questions: 1.         Area of paper = Length x Width Area= 8.5in. x 11in. = 93.5 in2. 2.         How many bitonds are in a sidereal day? 1 Day = (24 hr) x (60 min) x (60 sec) = 86 400 s/day.         1 day         1 hr         1 min                         3.         Convert the undermentioned showing unit analysis: a.         35.5 km = m 35.5 km = 1000 m = 35 000 m                 1 km make: 35 000 m                 b. cxxv mm = cm                 125 mm = 1 cm = 12.
5 cm                         10mm                 Answer: 12.5 cm                 c. 8.66 kg = mg                 8.66 kg = 1000 g x 1000mg = 8 660 000 mg                         1 kg         1 g                 Answer: 8 660 000 mg                 d. 2.50 nm = m                 2.50 nm = 1 000 000 000 m = 2 cholecalciferol 000 000 m                                 1 nm                 Answer: 2 cholecalciferol 000 000 m                 e. 45.0 s = h                 45.0 s = 1 min x 1 hr = .013 hr                         60 sec 60 min                 Answer: .013 hr                 f. 9.9 mL = L                 9.9 mL = 1L . = .01 L                                 1000 mL                 Answer: .01 L Question #4 To use Unit Analysis, you engender of divide to the right the closest converting scale to the unit you are difficult to reduce or make larger. You are converting using a method that you can show your work and find where you do your mistake easily. If you want to get a across-the-board essay, identify it on our website:
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