Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Math & Sciences (https://www.askmehelpdesk.com/forumdisplay.php?f=402)
-   -   Finding diameter of cylinder given density (https://www.askmehelpdesk.com/showthread.php?t=597456)

  • Sep 15, 2011, 08:56 PM
    Rocketsurgeon
    Finding diameter of cylinder given density
    NO idea where to start on this:
    A cylindrical tube 11.6cm in length is filled with mercury (density=13.6 g/ml). The volume of a cylinder of radius are and length h is V=3.14r^2h) (that's pi, multiplied by are squared, multiplied by h) The mass of mercury needed to fill the tube is 102.5g. Calculate the inner diameter of the tube.
    While the answer would certainly be helpful, I'm looking for someone to break down to me how I can come up with it myself. We are working with dimensional analysis and I am having a really hard time seeing the relationships between conversions. Please help
  • Sep 16, 2011, 01:56 AM
    Unknown008
    Hmm, the only conversion that you need is from ml to cm^3.

    You can remember that 1000 cm^3 = 1 L

    This is a common unit used in chemistry and might help you if you are doing this :)

    Okay, so that brings it down to 1000 cm^3 = 1000 mL, or 1 cm^3 = 1 mL.

    1. Find the volume, in mL, of mercury used.
    2. Convert this to cm^3.
    3. You have to understand that once you poured mercury in the tube, the mercury also took the shape of the tube. So, the volume of mercury can also be expressed as:


    You know that the height/length of the cylinder is 11.6 cm. You know the volume from step 2. You thus have an equation with only one value to find, which is r.

    Can you try this out and post your answer? :)
  • Sep 18, 2011, 12:13 PM
    Rocketsurgeon
    Unknown-
    Firstly, a hundred thanks for your time. Just wanted to say thank you before I post another question (pre-cal based). I will be working mostly on that today but will certainly be needing to work on my chem as well either tonight or tomorrow and this will be the first problem I attack.
  • Sep 18, 2011, 12:19 PM
    Unknown008
    Sure! You're welcome and I'm glad I could help you! :)
  • Jun 12, 2012, 06:39 AM
    stephanlenzer
    Sex
  • Jun 12, 2012, 06:40 AM
    stephanlenzer
    Yes

  • All times are GMT -7. The time now is 02:54 AM.