Height/Diameter of Cylinder 9.3 cm/2.7 cmMass of water with/without aluminum 176.01 g/157.81gVolume of water with aluminum 56.1 mL/50 mL1Using your height and diameter values, calculate the volume of the 100 mL graduated cylinder up to its 50 mL mark2Assume the accepted value for the volume of a graduated cylinder up to the 50 mL mark is 50.00 mLCalculate the percentage error in your calculations from item 2Be sure to follow significant-figure rules in your calculations3Calculate the volume of the aluminumCalculate the mass of the aluminumUse these 2 values to calculate the density of the aluminumShow all your work.4Calculate the percentage the error by comparing your experimental value with the values below you found in the handbook.
For that much money I would not buy a chinese made reelYou could get an American made Ardent for around that price and from what I understand they will perform way better.
1.The area of cylinder is Pi x Radius ^2 x heightThe diameter is 2 x radius, so r 1.85 h of the total 100 mL cylinder is 9.3, but we fill it only half way (50 ml), so the height is halved to 4.65 so then we plug in: Pi x (1.85)^2 x 4.65 49.99726 cm^3 At room temp (20 C) this would also be the volume in mL: 49.997 mL(Rounded to 50.00 mL) 2percent error: 50.00-49.997 0.003 0.003/50.00 0.006% 3Vol with Al 56.1 ml (56.1 cm^3) Vol without Al 50 ml (50 cm^3) 56.1-50.0 6.1 cm^3 Mass with Al 176.01 g Mass without Al 157.81 g 176.01-157.81 18.2 g density mass/volume density g/cm^3 18.2g/6.1 cm^3 2.98 g/cm^3 4You will have to look at your handbook to find the valueThen it's: the absolute value of (experimental - actual)/experimental into a percentHope this helps(Note, the answer for question 1 is so darned close to 50.00 mL, I'm not sure if you can really call that a percentage error for question 2; I mean 0.006% is really low.
