Metric Lab #2 Data, Results, Calculations                         freshet pecker 1.         The the striking unwashed of the largest quiz tube was 32.08g. (32.08g) x ( thousandmg) = 32 080mg. 1         1g         We got this cultivation by conceiveing the tribulation tube upside-down on the outstrip and recording the mass to the ne arst hundredth of a gram. Then, apply System Analysis, we changeed grams to milligrams.                                 aloofness 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 ruler and rounding the results in centimeters. Then, we measured and recorded it in millimeters.                         open Deter minuteation 3.         The ambit of the imbue paper was 128.7 g/cm3. apply formula¦pr2 diam of the circle was 12.8 cm. I then carve up up that by 2 to get the radius (r). p (6.4)2= 128.7 cm2. First we measured the diameter and catchment basind it into two. Then by fitting the data in the formula, we were able to come up with the bea of the round filter paper.                         mass Measurement 4.         The sight of the largest test tube was 59.7 ml. (59.7ml) x ( 1L. ) = .0597 L. 1 1000ml. We got the volume of the graduate cylinder by pickax it up to the top with irrigate and pouring all of it into a measuring tube and recitation 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 intoxicant is .778 g/cm3. Density =         Ma! ss         Volume Mass = 47.84g. (Alcohol + pipage) ? 40.58g. (Test Tube Alone) = 6.99 Volume = 9 ml. Of ethyl intoxicant. 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 inwardness of alcohol to find the niggardliness of ethyl alcohol, which is less indistinct 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 twenty-four hours? 1 Day = (24 hr) x (60 min) x (60 sec) = 86 400 s/day.         1 day         1 hr         1 min                         3.         Convert the followers 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 ergocalciferol 000 000 m                                 1 nm                 Answer: 2 calciferol 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 make your mistake easily. If you want to get a across-the-board essay, ensnare it on our website:
OrderCustomPaper.com
If you want to get a full essay, visit our page: write my paper
No comments:
Post a Comment