give me some detail of the Na K pump?
139678
Are you lazy?\nLook here:
827357
You always get pump supplies from the pump company.They will educate you and train you on their pump.\nCongrats. This is a big move and gives you so much more control.
303247
Here is a great web site you can compare various kinds of pumps. Best Wishes..\n\nhttp://www.diabetesnet.com/diabetes_technology/insulin_pump_models.php
829590
I don't believe lows are as bad a problem with a pump as with MDI but my understanding is that you can go into DKA fairly quickly while on a pump if you should stop receiving the insulin due to a bad site or a kink in the tubing.  I can't wait to get my son on a pump. He;'s got an endo appointment at the end of the month.  You should check out the ADA's pumping forum, it's called Pump It up.  Lots of good info there.  The link is below.  Happy pumping and good A1Cs to ya!
640469
the website below should give you all the info you need...it even has pictures and everything.\n\n\nhttp://users.rcn.com/jkimball.ma.ultranet/BiologyPages/G/GasExchange.html
644652
I answered this already.\n\nhttp://answers.yahoo.com/question/index;_ylt=AmW0Em3eAdmFrnQMc4_Xilzsy6IX?qid=1006051323791\n\nWorks the same as if you were changing permeability. If you have more Na inside the cell, there is less drive for Na outside the cell to come in. Because Na only has a minor impact on the resting membrane potential, it won't change much, but it will hyperpolarize a bit (away from the Ena and towards Ek).
304632
Okay, so the first thing that happens it that the NaF completely dissociates:\nNaF ---> Na+ + F-\n\nCool.  Now, the Na+ is just floating around, so we can ignore it.  Let's deal with the F-.  The F- is in the water, so it's going to do this:\nF- + H2O <---> HF + OH-\nNote that the arrow is going in both directions; it's an equilibrium problem, so I need to set up an equilibrium equation.  \n\nBut first, because I'm dealing with OH- ions, I'll need a Kb value.  Since Kw, the product constant of water, equals Kb * Ka, and Kw= 1E-14:\n(1E-14)/(6.8E-4)= Kb = 1.47E-11\n\nNow,\nKb=[OH-][HF]/[F-]\nor\n1.47E-11 = x^2/(.00339-x)\nsolve for x (I just set it equal to zero, graph it, and find the x-intercpets on my calculator):\nx=2.23E-7 M \nSince x=[OH-], [OH-]=2.23E-7 M\n\nWe can calculate pOH by taking the negative log of [OH-]:\n-log 2.23E-7 M = 6.65\nSince pH + pOH = 14.00:\npH + 6.65 = 14.00\npH = 14.00-6.65\npH = 7.35, the answer within one one-hundredth, which is probably some minor rounding discrepancy.  Yay!
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
829252
Here you go. Good luck.\n\nhttp://www.kitco.com/
148325
http://www.accesskansas.org/kbi/ro.shtml
735223
C.  It pumps potassium in and sodium out.  It does not pump sugar.
409523
http://www.modified-motorsports.com/catalog/product_info.php?products_id=2861&osCsid=49da0f18bccd2315b3f6f3e204398da9\n\nhttp://www.aplusperformance.com/catalog/index.php?cPath=193_232&osCsid=3587d7c16d32b54684944162155401e7\n\nhttp://www.forexenterprise.com/
1151385
I am interested, but I would want to know:\na) Where is your processing plant located?\nb) Who would pay the transportation/shipping costs to Calif. ?\nc) How is it packaged for shipping ? Plastic drums?\nd) How has it been processed, liquified ?\ne) What is the salt (NaCl) content ?\n  Your question gives me very little information to satisfy my needs.
871634
stndrd enthalpy of formation for NaF (s) is -569.0 kJ/mol\n\nhttp://chemistry.about.com/library/weekly/blheatform.htm
696672
I have a drawing that shows the various types of shell and tube exchangers. I believe their functions are self evident if you see the cross sectins.\n\nSend me you email address and I'll send the drawing.
287280
Two words.......AIR PUMP
229726
Modern...with a pump.\nPrehistoric...with a bucket on a string.