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Watabe wa kemia msaada tafadhali..!

Discussion in 'Jukwaa la Elimu (Education Forum)' started by b5-click, Jun 18, 2012.

  1. b5-click

    b5-click JF-Expert Member

    #1
    Jun 18, 2012
    Joined: Mar 21, 2012
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    Wandugu p0leni na majukumu...
    Kwa yey0te anaye weza kunipa msaada ktk swali hili afanye hivy0 tafadhali...
    ''If an aqueous solution boil at 100.5'C, at what temperature does it freeze? (For water Kb= 0.512'C/m and Kf= 1.86'C/m)...
    Shukrani kwanza.!
     
  2. Good Guy

    Good Guy JF-Expert Member

    #2
    Jun 18, 2012
    Joined: Oct 24, 2010
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    dah long tyme but ckupata gamba la kemia advance bhure,
    ok for the boiling process
    dT=Kbxconstants
    but dt=(100.5-100)=0.5(since the liquid z water b.p=100)
    hence
    0.5=0.512xC
    hence making C subject
    c=0.5/0.512
    after dat twende kwenye 2nd step(the freezing process)
    dT=Kfxconstant
    dT=KfxC
    dT=1.86x(0.5/0.512)
    but
    dT=0-freezing point
    so (again since ts water m.p=0)
    then
    0-[1.86x(0.5/0.512)]=freezing point(u should gt -ve somethng)
    sina calculator so utajaza. Hiyo constant ni bunch of masses nmeshazisahau,ila zote zinabaki constant throughout(ts an assumption ofcourse)
     
  3. Vitendo

    Vitendo JF-Expert Member

    #3
    Jun 19, 2012
    Joined: Oct 23, 2009
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    Hivi siku hizi vitabu huko mashuleni hakuna?
     
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